shreve solution manual

Stochastic Calculus for Finance, Volume I and IIby Yan Zeng Last updated: August 20, 2007 This is a solution manual for the two-volume textbook Stochastic calculus for finance, by Steven Shreve. If you have any comments or find any typos/errors, please email me at [email protected]. The current version omits the following problems. Volume I: 1.5, 3.3, 3.4, 5.7; Volume II: 3.9, 7.1, 7.2, 7.5–7.9, 10.8, 10.9, 10.10. Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through this solution manual and communicating to me several mistakes/typos. 1 Stochastic Calculus for Finance I: The Binomial Asset Pricing Model 1. The Binomial No-Arbitrage Pricing Model 1.1. Proof. If we get the up sate, then X 1 = X 1 (H) = ∆ 0 uS 0 + (1 + r)(X 0 − ∆ 0 S 0 ); if we get the down state, then X 1 = X 1 (T) = ∆ 0 dS 0 +(1 +r)(X 0 −∆ 0 S 0 ). If X 1 has a positive probability of being strictly positive, then we must either have X 1 (H) > 0 or X 1 (T) > 0. (i) If X 1 (H) > 0, then ∆ 0 uS 0 + (1 + r)(X 0 − ∆ 0 S 0 ) > 0. Plug in X 0 = 0, we get u∆ 0 > (1 + r)∆ 0 . By condition d < 1 + r < u, we conclude ∆ 0 > 0. In this case, X 1 (T) = ∆ 0 dS 0 + (1 + r)(X 0 − ∆ 0 S 0 ) = ∆ 0 S 0 [d −(1 +r)] < 0. (ii) If X 1 (T) > 0, then we can similarly deduce ∆ 0 < 0 and hence X 1 (H) < 0. So we cannot have X 1 strictly positive with positive probability unless X 1 is strictly negative with positive probability as well, regardless the choice of the number ∆ 0 . Remark: Here the condition X 0 = 0 is not essential, as far as a property definition of arbitrage for arbitrary X 0 can be given. Indeed, for the one-period binomial model, we can define arbitrage as a trading strategy such that P(X 1 ≥ X 0 (1 +r)) = 1 and P(X 1 > X 0 (1 +r)) > 0. First, this is a generalization of the case X 0 = 0; second, it is “proper” because it is comparing the result of an arbitrary investment involving money and stock markets with that of a safe investment involving only money market. This can also be seen by regarding X 0 as borrowed from money market account. Then at time 1, we have to pay back X 0 (1 + r) to the money market account. In summary, arbitrage is a trading strategy that beats “safe” investment. Accordingly, we revise the proof of Exercise 1.1. as follows. If X 1 has a positive probability of being strictly larger than X 0 (1 + r), the either X 1 (H) > X 0 (1 + r) or X 1 (T) > X 0 (1 + r). The first case yields ∆ 0 S 0 (u−1−r) > 0, i.e. ∆ 0 > 0. So X 1 (T) = (1+r)X 0 +∆ 0 S 0 (d−1−r) < (1+r)X 0 . The second case can be similarly analyzed. Hence we cannot have X 1 strictly greater than X 0 (1 + r) with positive probability unless X 1 is strictly smaller than X 0 (1 +r) with positive probability as well. Finally, we comment that the above formulation of arbitrage is equivalent to the one in the textbook. For details, see Shreve [7], Exercise 5.7. 1.2. 1 Proof. X 1 (u) = ∆ 0 8 +Γ 0 3 − 5 4 (4∆ 0 +1.20Γ 0 ) = 3∆ 0 +1.5Γ 0 , and X 1 (d) = ∆ 0 2 − 5 4 (4∆ 0 +1.20Γ 0 ) = −3∆ 0 −1.5Γ 0 . That is, X 1 (u) = −X 1 (d). So if there is a positive probability that X 1 is positive, then there is a positive probability that X 1 is negative. Remark: Note the above relation X 1 (u) = −X 1 (d) is not a coincidence. In general, let V 1 denote the payoff of the derivative security at time 1. Suppose ¯ X 0 and ¯ ∆ 0 are chosen in such a way that V 1 can be replicated: (1 + r)( ¯ X 0 − ¯ ∆ 0 S 0 ) + ¯ ∆ 0 S 1 = V 1 . Using the notation of the problem, suppose an agent begins with 0 wealth and at time zero buys ∆ 0 shares of stock and Γ 0 options. He then puts his cash position −∆ 0 S 0 −Γ 0 ¯ X 0 in a money market account. At time one, the value of the agent’s portfolio of stock, option and money market assets is X 1 = ∆ 0 S 1 + Γ 0 V 1 −(1 +r)(∆ 0 S 0 + Γ 0 ¯ X 0 ). Plug in the expression of V 1 and sort out terms, we have X 1 = S 0 (∆ 0 + ¯ ∆ 0 Γ 0 )( S 1 S 0 −(1 +r)). Since d < (1 +r) < u, X 1 (u) and X 1 (d) have opposite signs. So if the price of the option at time zero is ¯ X 0 , then there will no arbitrage. 1.3. Proof. V 0 = 1 1+r 1+r−d u−d S 1 (H) + u−1−r u−d S 1 (T) = S0 1+r 1+r−d u−d u + u−1−r u−d d = S 0 . This is not surprising, since this is exactly the cost of replicating S 1 . Remark: This illustrates an important point. The “fair price” of a stock cannot be determined by the risk-neutral pricing, as seen below. Suppose S 1 (H) and S 1 (T) are given, we could have two current prices, S 0 and S 0 . Correspondingly, we can get u, d and u , d . Because they are determined by S 0 and S 0 , respectively, it’s not surprising that risk-neutral pricing formula always holds, in both cases. That is, S 0 = 1+r−d u−d S 1 (H) + u−1−r u−d S 1 (T) 1 +r , S 0 = 1+r−d u −d S 1 (H) + u −1−r u −d S 1 (T) 1 +r . Essentially, this is because risk-neutral pricing relies on fair price=replication cost. Stock as a replicating component cannot determine its own “fair” price via the risk-neutral pricing formula. 1.4. Proof. X n+1 (T) = ∆ n dS n + (1 +r)(X n −∆ n S n ) = ∆ n S n (d −1 −r) + (1 +r)V n = V n+1 (H) −V n+1 (T) u −d (d −1 −r) + (1 +r) ˜ pV n+1 (H) + ˜ qV n+1 (T) 1 +r = ˜ p(V n+1 (T) −V n+1 (H)) + ˜ pV n+1 (H) + ˜ qV n+1 (T) = ˜ pV n+1 (T) + ˜ qV n+1 (T) = V n+1 (T). 1.6. 2 Proof. The bank’s trader should set up a replicating portfolio whose payoff is the opposite of the option’s payoff. More precisely, we solve the equation (1 +r)(X 0 −∆ 0 S 0 ) + ∆ 0 S 1 = −(S 1 −K) + . Then X 0 = −1.20 and ∆ 0 = − 1 2 . This means the trader should sell short 0.5 share of stock, put the income 2 into a money market account, and then transfer 1.20 into a separate money market account. At time one, the portfolio consisting of a short position in stock and 0.8(1 + r) in money market account will cancel out with the option’s payoff. Therefore we end up with 1.20(1 +r) in the separate money market account. Remark: This problem illustrates why we are interested in hedging a long position. In case the stock price goes down at time one, the option will expire without any payoff. The initial money 1.20 we paid at time zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payoff at time one. Also, cf. page 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott [8], page 11-13. 1.7. Proof. The idea is the same as Problem 1.6. The bank’s trader only needs to set up the reverse of the replicating trading strategy described in Example 1.2.4. More precisely, he should short sell 0.1733 share of stock, invest the income 0.6933 into money market account, and transfer 1.376 into a separate money market account. The portfolio consisting a short position in stock and 0.6933-1.376 in money market account will replicate the opposite of the option’s payoff. After they cancel out, we end up with 1.376(1 + r) 3 in the separate money market account. 1.8. (i) Proof. v n (s, y) = 2 5 (v n+1 (2s, y + 2s) +v n+1 ( s 2 , y + s 2 )). (ii) Proof. 1.696. (iii) Proof. δ n (s, y) = v n+1 (us, y +us) −v n+1 (ds, y +ds) (u −d)s . 1.9. (i) Proof. Similar to Theorem 1.2.2, but replace r, u and d everywhere with r n , u n and d n . More precisely, set ¯ p n = 1+rn−dn un−dn and ¯ q n = 1 − ¯ p n . Then V n = ¯ p n V n+1 (H) + ¯ q n V n+1 (T) 1 +r n . (ii) Proof. ∆ n = Vn+1(H)−Vn+1(T) Sn+1(H)−Sn+1(T) = Vn+1(H)−Vn+1(T) (un−dn)Sn . (iii) 3 Proof. u n = Sn+1(H) Sn = Sn+10 Sn = 1+ 10 Sn and d n = Sn+1(T) Sn = Sn−10 Sn = 1− 10 Sn . So the risk-neutral probabilities at time n are ˜ p n = 1−dn un−dn = 1 2 and ˜ q n = 1 2 . Risk-neutral pricing implies the price of this call at time zero is 9.375. 2. Probability Theory on Coin Toss Space 2.1. (i) Proof. P(A c ) +P(A) = ¸ ω∈A c P(ω) + ¸ ω∈A P(ω) = ¸ ω∈Ω P(ω) = 1. (ii) Proof. By induction, it suffices to work on the case N = 2. When A 1 and A 2 are disjoint, P(A 1 ∪ A 2 ) = ¸ ω∈A1∪A2 P(ω) = ¸ ω∈A1 P(ω) + ¸ ω∈A2 P(ω) = P(A 1 ) + P(A 2 ). When A 1 and A 2 are arbitrary, using the result when they are disjoint, we have P(A 1 ∪ A 2 ) = P((A 1 − A 2 ) ∪ A 2 ) = P(A 1 − A 2 ) + P(A 2 ) ≤ P(A 1 ) +P(A 2 ). 2.2. (i) Proof. ¯ P(S 3 = 32) = ¯ p 3 = 1 8 , ¯ P(S 3 = 8) = 3¯ p 2 ¯ q = 3 8 , ¯ P(S 3 = 2) = 3¯ p¯ q 2 = 3 8 , and ¯ P(S 3 = 0.5) = ¯ q 3 = 1 8 . (ii) Proof. ¯ E[S 1 ] = 8 ¯ P(S 1 = 8) + 2 ¯ P(S 1 = 2) = 8¯ p + 2¯ q = 5, ¯ E[S 2 ] = 16¯ p 2 + 4 2¯ p¯ q + 1 ¯ q 2 = 6.25, and ¯ E[S 3 ] = 32 1 8 + 8 3 8 + 2 3 8 + 0.5 1 8 = 7.8125. So the average rates of growth of the stock price under ¯ P are, respectively: ¯ r 0 = 5 4 −1 = 0.25, ¯ r 1 = 6.25 5 −1 = 0.25 and ¯ r 2 = 7.8125 6.25 −1 = 0.25. (iii) Proof. P(S 3 = 32) = ( 2 3 ) 3 = 8 27 , P(S 3 = 8) = 3 ( 2 3 ) 2 1 3 = 4 9 , P(S 3 = 2) = 2 1 9 = 2 9 , and P(S 3 = 0.5) = 1 27 . Accordingly, E[S 1 ] = 6, E[S 2 ] = 9 and E[S 3 ] = 13.5. So the average rates of growth of the stock price under P are, respectively: r 0 = 6 4 −1 = 0.5, r 1 = 9 6 −1 = 0.5, and r 2 = 13.5 9 −1 = 0.5. 2.3. Proof. Apply conditional Jensen’s inequality. 2.4. (i) Proof. E n [M n+1 ] = M n +E n [X n+1 ] = M n +E[X n+1 ] = M n . (ii) Proof. E n [ Sn+1 Sn ] = E n [e σXn+1 2 e σ +e −σ ] = 2 e σ +e −σ E[e σXn+1 ] = 1. 2.5. (i) Proof. 2I n = 2 ¸ n−1 j=0 M j (M j+1 − M j ) = 2 ¸ n−1 j=0 M j M j+1 − ¸ n−1 j=1 M 2 j − ¸ n−1 j=1 M 2 j = 2 ¸ n−1 j=0 M j M j+1 + M 2 n − ¸ n−1 j=0 M 2 j+1 − ¸ n−1 j=0 M 2 j = M 2 n − ¸ n−1 j=0 (M j+1 −M j ) 2 = M 2 n − ¸ n−1 j=0 X 2 j+1 = M 2 n −n. (ii) Proof. E n [f(I n+1 )] = E n [f(I n +M n (M n+1 −M n ))] = E n [f(I n +M n X n+1 )] = 1 2 [f(I n +M n )+f(I n −M n )] = g(I n ), where g(x) = 1 2 [f(x + √ 2x +n) +f(x − √ 2x +n)], since √ 2I n +n = [M n [. 2.6. 4 Proof. E n [I n+1 −I n ] = E n [∆ n (M n+1 −M n )] = ∆ n E n [M n+1 −M n ] = 0. 2.7. Proof. We denote by X n the result of n-th coin toss, where Head is represented by X = 1 and Tail is represented by X = −1. We also suppose P(X = 1) = P(X = −1) = 1 2 . Define S 1 = X 1 and S n+1 = S n +b n (X 1 , , X n )X n+1 , where b n () is a bounded function on ¦−1, 1¦ n , to be determined later on. Clearly (S n ) n≥1 is an adapted stochastic process, and we can show it is a martingale. Indeed, E n [S n+1 − S n ] = b n (X 1 , , X n )E n [X n+1 ] = 0. For any arbitrary function f, E n [f(S n+1 )] = 1 2 [f(S n +b n (X 1 , , X n )) +f(S n −b n (X 1 , , X n ))]. Then intuitively, E n [f(S n+1 ] cannot be solely dependent upon S n when b n ’s are properly chosen. Therefore in general, (S n ) n≥1 cannot be a Markov process. Remark: If X n is regarded as the gain/loss of n-th bet in a gambling game, then S n would be the wealth at time n. b n is therefore the wager for the (n+1)-th bet and is devised according to past gambling results. 2.8. (i) Proof. Note M n = E n [M N ] and M n = E n [M N ]. (ii) Proof. In the proof of Theorem 1.2.2, we proved by induction that X n = V n where X n is defined by (1.2.14) of Chapter 1. In other words, the sequence (V n ) 0≤n≤N can be realized as the value process of a portfolio, which consists of stock and money market accounts. Since ( Xn (1+r) n ) 0≤n≤N is a martingale under ¯ P (Theorem 2.4.5), ( Vn (1+r) n ) 0≤n≤N is a martingale under ¯ P. (iii) Proof. V n (1+r) n = E n V N (1+r) N , so V 0 , V 1 1+r , , V N−1 (1+r) N−1 , V N (1+r) N is a martingale under ¯ P. (iv) Proof. Combine (ii) and (iii), then use (i). 2.9. (i) Proof. u 0 = S1(H) S0 = 2, d 0 = S1(H) S0 = 1 2 , u 1 (H) = S2(HH) S1(H) = 1.5, d 1 (H) = S2(HT) S1(H) = 1, u 1 (T) = S2(TH) S1(T) = 4 and d 1 (T) = S2(TT) S1(T) = 1. So ¯ p 0 = 1+r0−d0 u0−d0 = 1 2 , ¯ q 0 = 1 2 , ¯ p 1 (H) = 1+r1(H)−d1(H) u1(H)−d1(H) = 1 2 , ¯ q 1 (H) = 1 2 , ¯ p 1 (T) = 1+r1(T)−d1(T) u1(T)−d1(T) = 1 6 , and ¯ q 1 (T) = 5 6 . Therefore ¯ P(HH) = ¯ p 0 ¯ p 1 (H) = 1 4 , ¯ P(HT) = ¯ p 0 ¯ q 1 (H) = 1 4 , ¯ P(TH) = ¯ q 0 ¯ p 1 (T) = 1 12 and ¯ P(TT) = ¯ q 0 ¯ q 1 (T) = 5 12 . The proofs of Theorem 2.4.4, Theorem 2.4.5 and Theorem 2.4.7 still work for the random interest rate model, with proper modifications (i.e. ¯ P would be constructed according to conditional probabili- ties ¯ P(ω n+1 = H[ω 1 , , ω n ) := ¯ p n and ¯ P(ω n+1 = T[ω 1 , , ω n ) := ¯ q n . Cf. notes on page 39.). So the time-zero value of an option that pays off V 2 at time two is given by the risk-neutral pricing formula V 0 = ¯ E V2 (1+r0)(1+r1) . (ii) Proof. V 2 (HH) = 5, V 2 (HT) = 1, V 2 (TH) = 1 and V 2 (TT) = 0. So V 1 (H) = p1(H)V2(HH)+ q1(H)V2(HT) 1+r1(H) = 2.4, V 1 (T) = p1(T)V2(TH)+ q1(T)V2(TT) 1+r1(T) = 1 9 , and V 0 = p0V1(H)+ q0V1(T) 1+r0 ≈ 1. 5 (iii) Proof. ∆ 0 = V1(H)−V1(T) S1(H)−S1(T) = 2.4− 1 9 8−2 = 0.4 − 1 54 ≈ 0.3815. (iv) Proof. ∆ 1 (H) = V2(HH)−V2(HT) S2(HH)−S2(HT) = 5−1 12−8 = 1. 2.10. (i) Proof. ¯ E n [ Xn+1 (1+r) n+1 ] = ¯ E n [ ∆nYn+1Sn (1+r) n+1 + (1+r)(Xn−∆nSn) (1+r) n+1 ] = ∆nSn (1+r) n+1 ¯ E n [Y n+1 ] + Xn−∆nSn (1+r) n = ∆nSn (1+r) n+1 (u¯ p + d¯ q) + Xn−∆nSn (1+r) n = ∆nSn+Xn−∆nSn (1+r) n = Xn (1+r) n . (ii) Proof. From (2.8.2), we have ∆ n uS n + (1 +r)(X n −∆ n S n ) = X n+1 (H) ∆ n dS n + (1 +r)(X n −∆ n S n ) = X n+1 (T). So ∆ n = Xn+1(H)−Xn+1(T) uSn−dSn and X n = ¯ E n [ Xn+1 1+r ]. To make the portfolio replicate the payoff at time N, we must have X N = V N . So X n = ¯ E n [ X N (1+r) N−n ] = ¯ E n [ V N (1+r) N−n ]. Since (X n ) 0≤n≤N is the value process of the unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear equations), the no-arbitrage price of V N at time n is V n = X n = ¯ E n [ V N (1+r) N−n ]. (iii) Proof. ¯ E n [ S n+1 (1 +r) n+1 ] = 1 (1 +r) n+1 ¯ E n [(1 −A n+1 )Y n+1 S n ] = S n (1 +r) n+1 [¯ p(1 −A n+1 (H))u + ¯ q(1 −A n+1 (T))d] < S n (1 +r) n+1 [¯ pu + ¯ qd] = S n (1 +r) n . If A n+1 is a constant a, then ¯ E n [ Sn+1 (1+r) n+1 ] = Sn (1+r) n+1 (1−a)(¯ pu+¯ qd) = Sn (1+r) n (1−a). So ¯ E n [ Sn+1 (1+r) n+1 (1−a) n+1 ] = Sn (1+r) n (1−a) n . 2.11. (i) Proof. F N +P N = S N −K + (K −S N ) + = (S N −K) + = C N . (ii) Proof. C n = ¯ E n [ C N (1+r) N−n ] = ¯ E n [ F N (1+r) N−n ] + ¯ E n [ P N (1+r) N−n ] = F n +P n . (iii) Proof. F 0 = ¯ E[ F N (1+r) N ] = 1 (1+r) N ¯ E[S N −K] = S 0 − K (1+r) N . (iv) 6 Proof. At time zero, the trader has F 0 = S 0 in money market account and one share of stock. At time N, the trader has a wealth of (F 0 −S 0 )(1 +r) N +S N = −K +S N = F N . (v) Proof. By (ii), C 0 = F 0 +P 0 . Since F 0 = S 0 − (1+r) N S0 (1+r) N = 0, C 0 = P 0 . (vi) Proof. By (ii), C n = P n if and only if F n = 0. Note F n = ¯ E n [ S N −K (1+r) N−n ] = S n − (1+r) N S0 (1+r) N−n = S n −S 0 (1 +r) n . So F n is not necessarily zero and C n = P n is not necessarily true for n ≥ 1. 2.12. Proof. First, the no-arbitrage price of the chooser option at time m must be max(C, P), where C = ¯ E ¸ (S N −K) + (1 +r) N−m , and P = ¯ E ¸ (K −S N ) + (1 +r) N−m . That is, C is the no-arbitrage price of a call option at time m and P is the no-arbitrage price of a put option at time m. Both of them have maturity date N and strike price K. Suppose the market is liquid, then the chooser option is equivalent to receiving a payoff of max(C, P) at time m. Therefore, its current no-arbitrage price should be ¯ E[ max(C,P) (1+r) m ]. By the put-call parity, C = S m − K (1+r) N−m +P. So max(C, P) = P +(S m − K (1+r) N−m ) + . Therefore, the time-zero price of a chooser option is ¯ E ¸ P (1 +r) m + ¯ E ¸ (S m − K (1+r) N−m ) + (1 +r) m ¸ = ¯ E ¸ (K −S N ) + (1 +r) N + ¯ E ¸ (S m − K (1+r) N−m ) + (1 +r) m ¸ . The first term stands for the time-zero price of a put, expiring at time N and having strike price K, and the second term stands for the time-zero price of a call, expiring at time m and having strike price K (1+r) N−m . If we feel unconvinced by the above argument that the chooser option’s no-arbitrage price is ¯ E[ max(C,P) (1+r) m ], due to the economical argument involved (like “the chooser option is equivalent to receiving a payoff of max(C, P) at time m”), then we have the following mathematically rigorous argument. First, we can construct a portfolio ∆ 0 , , ∆ m−1 , whose payoff at time m is max(C, P). Fix ω, if C(ω) > P(ω), we can construct a portfolio ∆ m , , ∆ N−1 whose payoff at time N is (S N − K) + ; if C(ω) < P(ω), we can construct a portfolio ∆ m , , ∆ N−1 whose payoff at time N is (K −S N ) + . By defining (m ≤ k ≤ N −1) ∆ k (ω) = ∆ k (ω) if C(ω) > P(ω) ∆ k (ω) if C(ω) < P(ω), we get a portfolio (∆ n ) 0≤n≤N−1 whose payoff is the same as that of the chooser option. So the no-arbitrage price process of the chooser option must be equal to the value process of the replicating portfolio. In particular, V 0 = X 0 = ¯ E[ Xm (1+r) m ] = ¯ E[ max(C,P) (1+r) m ]. 2.13. (i) Proof. Note under both actual probability P and risk-neutral probability ¯ P, coin tosses ω n ’s are i.i.d.. So without loss of generality, we work on P. For any function g, E n [g(S n+1 , Y n+1 )] = E n [g( Sn+1 Sn S n , Y n + Sn+1 Sn S n )] = pg(uS n , Y n + uS n ) + qg(dS n , Y n + dS n ), which is a function of (S n , Y n ). So (S n , Y n ) 0≤n≤N is Markov under P. (ii) 7 Proof. Set v N (s, y) = f( y N+1 ). Then v N (S N , Y N ) = f( N n=0 Sn N+1 ) = V N . Suppose v n+1 is given, then V n = ¯ E n [ Vn+1 1+r ] = ¯ E n [ vn+1(Sn+1,Yn+1) 1+r ] = 1 1+r [¯ pv n+1 (uS n , Y n + uS n ) + ¯ qv n+1 (dS n , Y n + dS n )] = v n (S n , Y n ), where v n (s, y) = ¯ v n+1 (us, y +us) + ¯ v n+1 (ds, y +ds) 1 +r . 2.14. (i) Proof. For n ≤ M, (S n , Y n ) = (S n , 0). Since coin tosses ω n ’s are i.i.d. under ¯ P, (S n , Y n ) 0≤n≤M is Markov under ¯ P. More precisely, for any function h, ¯ E n [h(S n+1 )] = ¯ ph(uS n ) + ¯ h(dS n ), for n = 0, 1, , M −1. For any function g of two variables, we have ¯ E M [g(S M+1 , Y M+1 )] = ¯ E M [g(S M+1 , S M+1 )] = ¯ pg(uS M , uS M )+ ¯ qg(dS M , dS M ). And for n ≥ M+1, ¯ E n [g(S n+1 , Y n+1 )] = ¯ E n [g( Sn+1 Sn S n , Y n + Sn+1 Sn S n )] = ¯ pg(uS n , Y n +uS n )+ ¯ qg(dS n , Y n +dS n ), so (S n , Y n ) 0≤n≤N is Markov under ¯ P. (ii) Proof. Set v N (s, y) = f( y N−M ). Then v N (S N , Y N ) = f( N K=M+1 S k N−M ) = V N . Suppose v n+1 is already given. a) If n > M, then ¯ E n [v n+1 (S n+1 , Y n+1 )] = ¯ pv n+1 (uS n , Y n +uS n ) + ¯ qv n+1 (dS n , Y n +dS n ). So v n (s, y) = ¯ pv n+1 (us, y +us) + ¯ qv n+1 (ds, y +ds). b) If n = M, then ¯ E M [v M+1 (S M+1 , Y M+1 )] = ¯ pv M+1 (uS M , uS M ) + ¯ v n+1 (dS M , dS M ). So v M (s) = ¯ pv M+1 (us, us) + ¯ qv M+1 (ds, ds). c) If n < M, then ¯ E n [v n+1 (S n+1 )] = ¯ pv n+1 (uS n ) + ¯ qv n+1 (dS n ). So v n (s) = ¯ pv n+1 (us) + ¯ qv n+1 (ds). 3. State Prices 3.1. Proof. Note ¯ Z(ω) := P(ω) P(ω) = 1 Z(ω) . Apply Theorem 3.1.1 with P, ¯ P, Z replaced by ¯ P, P, ¯ Z, we get the analogous of properties (i)-(iii) of Theorem 3.1.1. 3.2. (i) Proof. ¯ P(Ω) = ¸ ω∈Ω ¯ P(ω) = ¸ ω∈Ω Z(ω)P(ω) = E[Z] = 1. (ii) Proof. ¯ E[Y ] = ¸ ω∈Ω Y (ω) ¯ P(ω) = ¸ ω∈Ω Y (ω)Z(ω)P(ω) = E[Y Z]. (iii) Proof. ˜ P(A) = ¸ ω∈A Z(ω)P(ω). Since P(A) = 0, P(ω) = 0 for any ω ∈ A. So ¯ P(A) = 0. (iv) Proof. If ¯ P(A) = ¸ ω∈A Z(ω)P(ω) = 0, by P(Z > 0) = 1, we conclude P(ω) = 0 for any ω ∈ A. So P(A) = ¸ ω∈A P(ω) = 0. (v) Proof. P(A) = 1 ⇐⇒ P(A c ) = 0 ⇐⇒ ¯ P(A c ) = 0 ⇐⇒ ¯ P(A) = 1. (vi) 8 Proof. Pick ω 0 such that P(ω 0 ) > 0, define Z(ω) = 0, if ω = ω 0 1 P(ω0) , if ω = ω 0 . Then P(Z ≥ 0) = 1 and E[Z] = 1 P(ω0) P(ω 0 ) = 1. Clearly ¯ P(Ω ` ¦ω 0 ¦) = E[Z1 Ω\{ω0} ] = ¸ ω=ω0 Z(ω)P(ω) = 0. But P(Ω ` ¦ω 0 ¦) = 1 − P(ω 0 ) > 0 if P(ω 0 ) < 1. Hence in the case 0 < P(ω 0 ) < 1, P and ¯ P are not equivalent. If P(ω 0 ) = 1, then E[Z] = 1 if and only if Z(ω 0 ) = 1. In this case ¯ P(ω 0 ) = Z(ω 0 )P(ω 0 ) = 1. And ¯ P and P have to be equivalent. In summary, if we can find ω 0 such that 0 < P(ω 0 ) < 1, then Z as constructed above would induce a probability ¯ P that is not equivalent to P. 3.5. (i) Proof. Z(HH) = 9 16 , Z(HT) = 9 8 , Z(TH) = 3 8 and Z(TT) = 15 4 . (ii) Proof. Z 1 (H) = E 1 [Z 2 ](H) = Z 2 (HH)P(ω 2 = H[ω 1 = H) + Z 2 (HT)P(ω 2 = T[ω 1 = H) = 3 4 . Z 1 (T) = E 1 [Z 2 ](T) = Z 2 (TH)P(ω 2 = H[ω 1 = T) +Z 2 (TT)P(ω 2 = T[ω 1 = T) = 3 2 . (iii) Proof. V 1 (H) = [Z 2 (HH)V 2 (HH)P(ω 2 = H[ω 1 = H) +Z 2 (HT)V 2 (HT)P(ω 2 = T[ω 1 = T)] Z 1 (H)(1 +r 1 (H)) = 2.4, V 1 (T) = [Z 2 (TH)V 2 (TH)P(ω 2 = H[ω 1 = T) +Z 2 (TT)V 2 (TT)P(ω 2 = T[ω 1 = T)] Z 1 (T)(1 +r 1 (T)) = 1 9 , and V 0 = Z 2 (HH)V 2 (HH) (1 + 1 4 )(1 + 1 4 ) P(HH) + Z 2 (HT)V 2 (HT) (1 + 1 4 )(1 + 1 4 ) P(HT) + Z 2 (TH)V 2 (TH) (1 + 1 4 )(1 + 1 2 ) P(TH) + 0 ≈ 1. 3.6. Proof. U (x) = 1 x , so I(x) = 1 x . (3.3.26) gives E[ Z (1+r) N (1+r) N λZ ] = X 0 . So λ = 1 X0 . By (3.3.25), we have X N = (1+r) N λZ = X0 Z (1 + r) N . Hence X n = ¯ E n [ X N (1+r) N−n ] = ¯ E n [ X0(1+r) n Z ] = X 0 (1 + r) n ¯ E n [ 1 Z ] = X 0 (1 +r) n 1 Zn E n [Z 1 Z ] = X0 ξn , where the second to last “=” comes from Lemma 3.2.6. 3.7. Proof. U (x) = x p−1 and so I(x) = x 1 p−1 . By (3.3.26), we have E[ Z (1+r) N ( λZ (1+r) N ) 1 p−1 ] = X 0 . Solve it for λ, we get λ = ¸ ¸ ¸ X 0 E ¸ Z p p−1 (1+r) Np p−1 ¸ p−1 = X p−1 0 (1 +r) Np (E[Z p p−1 ]) p−1 . So by (3.3.25), X N = ( λZ (1+r) N ) 1 p−1 = λ 1 p−1 Z 1 p−1 (1+r) N p−1 = X0(1+r) Np p−1 E[Z p p−1 ] Z 1 p−1 (1+r) N p−1 = (1+r) N X0Z 1 p−1 E[Z p p−1 ] . 3.8. (i) 9 Proof. d dx (U(x) −yx) = U (x) −y. So x = I(y) is an extreme point of U(x) −yx. Because d 2 dx 2 (U(x) −yx) = U (x) ≤ 0 (U is concave), x = I(y) is a maximum point. Therefore U(x) −y(x) ≤ U(I(y)) −yI(y) for every x. (ii) Proof. Following the hint of the problem, we have E[U(X N )] −E[X N λZ (1 +r) N ] ≤ E[U(I( λZ (1 +r) N ))] −E[ λZ (1 +r) N I( λZ (1 +r) N )], i.e. E[U(X N )] −λX 0 ≤ E[U(X ∗ N )] − ¯ E[ λ (1+r) N X ∗ N ] = E[U(X ∗ N )] −λX 0 . So E[U(X N )] ≤ E[U(X ∗ N )]. 3.9. (i) Proof. X n = ¯ E n [ X N (1+r) N−n ]. So if X N ≥ 0, then X n ≥ 0 for all n. (ii) Proof. a) If 0 ≤ x < γ and 0 < y ≤ 1 γ , then U(x) − yx = −yx ≤ 0 and U(I(y)) − yI(y) = U(γ) − yγ = 1 −yγ ≥ 0. So U(x) −yx ≤ U(I(y)) −yI(y). b) If 0 ≤ x < γ and y > 1 γ , then U(x) − yx = −yx ≤ 0 and U(I(y)) − yI(y) = U(0) − y 0 = 0. So U(x) −yx ≤ U(I(y)) −yI(y). c) If x ≥ γ and 0 < y ≤ 1 γ , then U(x) −yx = 1 −yx and U(I(y)) −yI(y) = U(γ) −yγ = 1 −yγ ≥ 1 −yx. So U(x) −yx ≤ U(I(y)) −yI(y). d) If x ≥ γ and y > 1 γ , then U(x) − yx = 1 − yx < 0 and U(I(y)) − yI(y) = U(0) − y 0 = 0. So U(x) −yx ≤ U(I(y)) −yI(y). (iii) Proof. Using (ii) and set x = X N , y = λZ (1+r) N , where X N is a random variable satisfying ¯ E[ X N (1+r) N ] = X 0 , we have E[U(X N )] −E[ λZ (1 +r) N X N ] ≤ E[U(X ∗ N )] −E[ λZ (1 +r) N X ∗ N ]. That is, E[U(X N )] −λX 0 ≤ E[U(X ∗ N )] −λX 0 . So E[U(X N )] ≤ E[U(X ∗ N )]. (iv) Proof. Plug p m and ξ m into (3.6.4), we have X 0 = 2 N ¸ m=1 p m ξ m I(λξ m ) = 2 N ¸ m=1 p m ξ m γ1 {λξm≤ 1 γ } . So X0 γ = ¸ 2 N m=1 p m ξ m 1 {λξm≤ 1 γ } . Suppose there is a solution λ to (3.6.4), note X0 γ > 0, we then can conclude ¦m : λξ m ≤ 1 γ ¦ = ∅. Let K = max¦m : λξ m ≤ 1 γ ¦, then λξ K ≤ 1 γ < λξ K+1 . So ξ K < ξ K+1 and X0 γ = ¸ K m=1 p m ξ m (Note, however, that K could be 2 N . In this case, ξ K+1 is interpreted as ∞. Also, note we are looking for positive solution λ > 0). Conversely, suppose there exists some K so that ξ K < ξ K+1 and ¸ K m=1 ξ m p m = X0 γ . Then we can find λ > 0, such that ξ K < 1 λγ < ξ K+1 . For such λ, we have E[ Z (1 +r) N I( λZ (1 +r) N )] = 2 N ¸ m=1 p m ξ m 1 {λξm≤ 1 γ } γ = K ¸ m=1 p m ξ m γ = X 0 . Hence (3.6.4) has a solution. 10 (v) Proof. X ∗ N (ω m ) = I(λξ m ) = γ1 {λξm≤ 1 γ } = γ, if m ≤ K 0, if m ≥ K + 1 . 4. American Derivative Securities Before proceeding to the exercise problems, we first give a brief summary of pricing American derivative securities as presented in the textbook. We shall use the notation of the book. From the buyer’s perspective: At time n, if the derivative security has not been exercised, then the buyer can choose a policy τ with τ ∈ o n . The valuation formula for cash flow (Theorem 2.4.8) gives a fair price for the derivative security exercised according to τ: V n (τ) = N ¸ k=n ¯ E n ¸ 1 {τ=k} 1 (1 +r) k−n G k = ¯ E n ¸ 1 {τ≤N} 1 (1 +r) τ−n G τ . The buyer wants to consider all the possible τ’s, so that he can find the least upper bound of security value, which will be the maximum price of the derivative security acceptable to him. This is the price given by Definition 4.4.1: V n = max τ∈Sn ¯ E n [1 {τ≤N} 1 (1+r) τ−n G τ ]. From the seller’s perspective: A price process (V n ) 0≤n≤N is acceptable to him if and only if at time n, he can construct a portfolio at cost V n so that (i) V n ≥ G n and (ii) he needs no further investing into the portfolio as time goes by. Formally, the seller can find (∆ n ) 0≤n≤N and (C n ) 0≤n≤N so that C n ≥ 0 and V n+1 = ∆ n S n+1 + (1 + r)(V n − C n − ∆ n S n ). Since ( Sn (1+r) n ) 0≤n≤N is a martingale under the risk-neutral measure ¯ P, we conclude ¯ E n ¸ V n+1 (1 +r) n+1 − V n (1 +r) n = − C n (1 +r) n ≤ 0, i.e. ( Vn (1+r) n ) 0≤n≤N is a supermartingale. This inspired us to check if the converse is also true. This is exactly the content of Theorem 4.4.4. So (V n ) 0≤n≤N is the value process of a portfolio that needs no further investing if and only if Vn (1+r) n 0≤n≤N is a supermartingale under ¯ P (note this is independent of the requirement V n ≥ G n ). In summary, a price process (V n ) 0≤n≤N is acceptable to the seller if and only if (i) V n ≥ G n ; (ii) Vn (1+r) n 0≤n≤N is a supermartingale under ¯ P. Theorem 4.4.2 shows the buyer’s upper bound is the seller’s lower bound. So it gives the price acceptable to both. Theorem 4.4.3 gives a specific algorithm for calculating the price, Theorem 4.4.4 establishes the one-to-one correspondence between super-replication and supermartingale property, and finally, Theorem 4.4.5 shows how to decide on the optimal exercise policy. 4.1. (i) Proof. V P 2 (HH) = 0, V P 2 (HT) = V P 2 (TH) = 0.8, V P 2 (TT) = 3, V P 1 (H) = 0.32, V P 1 (T) = 2, V P 0 = 9.28. (ii) Proof. V C 0 = 5. (iii) Proof. g S (s) = [4 − s[. We apply Theorem 4.4.3 and have V S 2 (HH) = 12.8, V S 2 (HT) = V S 2 (TH) = 2.4, V S 2 (TT) = 3, V S 1 (H) = 6.08, V S 1 (T) = 2.16 and V S 0 = 3.296. (iv) 11 Proof. First, we note the simple inequality max(a 1 , b 1 ) + max(a 2 , b 2 ) ≥ max(a 1 +a 2 , b 1 +b 2 ). “>” holds if and only if b 1 > a 1 , b 2 < a 2 or b 1 < a 1 , b 2 > a 2 . By induction, we can show V S n = max g S (S n ), ¯ pV S n+1 + ¯ V S n+1 1 +r ¸ ≤ max g P (S n ) +g C (S n ), ¯ pV P n+1 + ¯ V P n+1 1 +r + ¯ pV C n+1 + ¯ V C n+1 1 +r ¸ ≤ max g P (S n ), ¯ pV P n+1 + ¯ V P n+1 1 +r ¸ + max g C (S n ), ¯ pV C n+1 + ¯ V C n+1 1 +r ¸ = V P n +V C n . As to when “<” holds, suppose m = max¦n : V S n < V P n + V C n ¦. Then clearly m ≤ N −1 and it is possible that ¦n : V S n < V P n + V C n ¦ = ∅. When this set is not empty, m is characterized as m = max¦n : g P (S n ) < pV P n+1 + qV P n+1 1+r and g C (S n ) > pV C n+1 + qV C n+1 1+r or g P (S n ) > pV P n+1 + qV P n+1 1+r and g C (S n ) < pV C n+1 + qV C n+1 1+r ¦. 4.2. Proof. For this problem, we need Figure 4.2.1, Figure 4.4.1 and Figure 4.4.2. Then ∆ 1 (H) = V 2 (HH) −V 2 (HT) S 2 (HH) −S 2 (HT) = − 1 12 , ∆ 1 (T) = V 2 (TH) −V 2 (TT) S 2 (TH) −S 2 (TT) = −1, and ∆ 0 = V 1 (H) −V 1 (T) S 1 (H) −S 1 (T) ≈ −0.433. The optimal exercise time is τ = inf¦n : V n = G n ¦. So τ(HH) = ∞, τ(HT) = 2, τ(TH) = τ(TT) = 1. Therefore, the agent borrows 1.36 at time zero and buys the put. At the same time, to hedge the long position, he needs to borrow again and buy 0.433 shares of stock at time zero. At time one, if the result of coin toss is tail and the stock price goes down to 2, the value of the portfolio is X 1 (T) = (1 +r)(−1.36 −0.433S 0 ) +0.433S 1 (T) = (1 + 1 4 )(−1.36 −0.433 4) +0.433 2 = −3. The agent should exercise the put at time one and get 3 to pay off his debt. At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfolio is X 1 (H) = (1 + r)(−1.36 − 0.433S 0 ) + 0.433S 1 (H) = −0.4. The agent should borrow to buy 1 12 shares of stock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of the portfolio is X 2 (HH) = (1+r)(X 1 (H) − 1 12 S 1 (H)) + 1 12 S 2 (HH) = 0, and the agent should let the put expire. If at time two, the result of coin toss is tail and the stock price goes down to 4, the value of the portfolio is X 2 (HT) = (1 +r)(X 1 (H) − 1 12 S 1 (H)) + 1 12 S 2 (HT) = −1. The agent should exercise the put to get 1. This will pay off his debt. 4.3. Proof. We need Figure 1.2.2 for this problem, and calculate the intrinsic value process and price process of the put as follows. For the intrinsic value process, G 0 = 0, G 1 (T) = 1, G 2 (TH) = 2 3 , G 2 (TT) = 5 3 , G 3 (THT) = 1, G 3 (TTH) = 1.75, G 3 (TTT) = 2.125. All the other outcomes of G is negative. 12 For the price process, V 0 = 0.4, V 1 (T) = 1, V 1 (TH) = 2 3 , V 1 (TT) = 5 3 , V 3 (THT) = 1, V 3 (TTH) = 1.75, V 3 (TTT) = 2.125. All the other outcomes of V is zero. Therefore the time-zero price of the derivative security is 0.4 and the optimal exercise time satisfies τ(ω) = ∞ if ω 1 = H, 1 if ω 1 = T. 4.4. Proof. 1.36 is the cost of super-replicating the American derivative security. It enables us to construct a portfolio sufficient to pay off the derivative security, no matter when the derivative security is exercised. So to hedge our short position after selling the put, there is no need to charge the insider more than 1.36. 4.5. Proof. The stopping times in o 0 are (1) τ ≡ 0; (2) τ ≡ 1; (3) τ(HT) = τ(HH) = 1, τ(TH), τ(TT) ∈ ¦2, ∞¦ (4 different ones); (4) τ(HT), τ(HH) ∈ ¦2, ∞¦, τ(TH) = τ(TT) = 1 (4 different ones); (5) τ(HT), τ(HH), τ(TH), τ(TT) ∈ ¦2, ∞¦ (16 different ones). When the option is out of money, the following stopping times do not exercise (i) τ ≡ 0; (ii) τ(HT) ∈ ¦2, ∞¦, τ(HH) = ∞, τ(TH), τ(TT) ∈ ¦2, ∞¦ (8 different ones); (iii) τ(HT) ∈ ¦2, ∞¦, τ(HH) = ∞, τ(TH) = τ(TT) = 1 (2 different ones). For (i), ¯ E[1 {τ≤2} ( 4 5 ) τ G τ ] = G 0 = 1. For (ii), ¯ E[1 {τ≤2} ( 4 5 ) τ G τ ] ≤ ¯ E[1 {τ ∗ ≤2} ( 4 5 ) τ ∗ G τ ∗], where τ ∗ (HT) = 2, τ ∗ (HH) = ∞, τ ∗ (TH) = τ ∗ (TT) = 2. So ¯ E[1 {τ ∗ ≤2} ( 4 5 ) τ ∗ G τ ∗] = 1 4 [( 4 5 ) 2 1 + ( 4 5 ) 2 (1 + 4)] = 0.96. For (iii), ¯ E[1 {τ≤2} ( 4 5 ) τ G τ ] has the biggest value when τ satisfies τ(HT) = 2, τ(HH) = ∞, τ(TH) = τ(TT) = 1. This value is 1.36. 4.6. (i) Proof. The value of the put at time N, if it is not exercised at previous times, is K − S N . Hence V N−1 = max¦K−S N−1 , ¯ E N−1 [ V N 1+r ]¦ = max¦K−S N−1 , K 1+r −S N−1 ¦ = K−S N−1 . The second equality comes from the fact that discounted stock price process is a martingale under risk-neutral probability. By induction, we can show V n = K − S n (0 ≤ n ≤ N). So by Theorem 4.4.5, the optimal exercise policy is to sell the stock at time zero and the value of this derivative security is K −S 0 . Remark: We cheated a little bit by using American algorithm and Theorem 4.4.5, since they are developed for the case where τ is allowed to be ∞. But intuitively, results in this chapter should still hold for the case τ ≤ N, provided we replace “max¦G n , 0¦” with “G n ”. (ii) Proof. This is because at time N, if we have to exercise the put and K − S N < 0, we can exercise the European call to set off the negative payoff. In effect, throughout the portfolio’s lifetime, the portfolio has intrinsic values greater than that of an American put stuck at K with expiration time N. So, we must have V AP 0 ≤ V 0 +V EC 0 ≤ K −S 0 +V EC 0 . (iii) 13 Proof. Let V EP 0 denote the time-zero value of a European put with strike K and expiration time N. Then V AP 0 ≥ V EP 0 = V EC 0 − ¯ E[ S N −K (1 +r) N ] = V EC 0 −S 0 + K (1 +r) N . 4.7. Proof. V N = S N −K, V N−1 = max¦S N−1 −K, ¯ E N−1 [ V N 1+r ]¦ = max¦S N−1 −K, S N−1 − K 1+r ¦ = S N−1 − K 1+r . By induction, we can prove V n = S n − K (1+r) N−n (0 ≤ n ≤ N) and V n > G n for 0 ≤ n ≤ N − 1. So the time-zero value is S 0 − K (1+r) N and the optimal exercise time is N. 5. Random Walk 5.1. (i) Proof. E[α τ2 ] = E[α (τ2−τ1)+τ1 ] = E[α (τ2−τ1) ]E[α τ1 ] = E[α τ1 ] 2 . (ii) Proof. If we define M (m) n = M n+τm −M τm (m = 1, 2, ), then (M (m) · ) m as random functions are i.i.d. with distributions the same as that of M. So τ m+1 − τ m = inf¦n : M (m) n = 1¦ are i.i.d. with distributions the same as that of τ 1 . Therefore E[α τm ] = E[α (τm−τm−1)+(τm−1−τm−2)+···+τ1 ] = E[α τ1 ] m . (iii) Proof. Yes, since the argument of (ii) still works for asymmetric random walk. 5.2. (i) Proof. f (σ) = pe σ − qe −σ , so f (σ) > 0 if and only if σ > 1 2 (ln q − ln p). Since 1 2 (ln q − ln p) < 0, f(σ) > f(0) = 1 for all σ > 0. (ii) Proof. E n [ Sn+1 Sn ] = E n [e σXn+1 1 f(σ) ] = pe σ 1 f(σ) +qe −σ 1 f(σ) = 1. (iii) Proof. By optional stopping theorem, E[S n∧τ1 ] = E[S 0 ] = 1. Note S n∧τ1 = e σMn∧τ 1 ( 1 f(σ) ) n∧τ1 ≤ e σ·1 , by bounded convergence theorem, E[1 {τ1<∞} S τ1 ] = E[lim n→∞ S n∧τ1 ] = lim n→∞ E[S n∧τ1 ] = 1, that is, E[1 {τ1<∞} e σ ( 1 f(σ) ) τ1 ] = 1. So e −σ = E[1 {τ1<∞} ( 1 f(σ) ) τ1 ]. Let σ ↓ 0, again by bounded convergence theorem, 1 = E[1 {τ1<∞} ( 1 f(0) ) τ1 ] = P(τ 1 < ∞). (iv) Proof. Set α = 1 f(σ) = 1 pe σ +qe −σ , then as σ varies from 0 to ∞, α can take all the values in (0, 1). Write σ in terms of α, we have e σ = 1± √ 1−4pqα 2 2pα (note 4pqα 2 < 4( p+q 2 ) 2 1 2 = 1). We want to choose σ > 0, so we should take σ = ln( 1+ √ 1−4pqα 2 2pα ). Therefore E[α τ1 ] = 2pα 1+ √ 1−4pqα 2 = 1− √ 1−4pqα 2 2qα . 14 (v) Proof. ∂ ∂α E[α τ1 ] = E[ ∂ ∂α α τ1 ] = E[τ 1 α τ1−1 ], and 1 − 1 −4pqα 2 2qα = 1 2q (1 − 1 −4pqα 2 )α −1 = 1 2q [− 1 2 (1 −4pqα 2 ) − 1 2 (−4pq2α)α −1 + (1 − 1 −4pqα 2 )(−1)α 2 ]. So E[τ 1 ] = lim α↑1 ∂ ∂α E[α τ1 ] = 1 2q [− 1 2 (1 −4pq) − 1 2 (−8pq) −(1 − √ 1 −4pq)] = 1 2p−1 . 5.3. (i) Proof. Solve the equation pe σ + qe −σ = 1 and a positive solution is ln 1+ √ 1−4pq 2p = ln 1−p p = ln q −ln p. Set σ 0 = ln q −ln p, then f(σ 0 ) = 1 and f (σ) > 0 for σ > σ 0 . So f(σ) > 1 for all σ > σ 0 . (ii) Proof. As in Exercise 5.2, S n = e σMn ( 1 f(σ) ) n is a martingale, and 1 = E[S 0 ] = E[S n∧τ1 ] = E[e σMn∧τ 1 ( 1 f(σ) ) τ1∧n ]. Suppose σ > σ 0 , then by bounded convergence theorem, 1 = E[ lim n→∞ e σMn∧τ 1 ( 1 f(σ) ) n∧τ1 ] = E[1 {τ1<∞} e σ ( 1 f(σ) ) τ1 ]. Let σ ↓ σ 0 , we get P(τ 1 < ∞) = e −σ0 = p q < 1. (iii) Proof. From (ii), we can see E[1 {τ1<∞} ( 1 f(σ) ) τ1 ] = e −σ , for σ > σ 0 . Set α = 1 f(α) , then e σ = 1± √ 1−4pqα 2 2pα . We need to choose the root so that e σ > e σ0 = q p , so σ = ln( 1+ √ 1−4pqα 2 2pα ), then E[α τ1 1 {τ1<∞} ] = 1− √ 1−4pqα 2 2qα . (iv) Proof. E[τ 1 1 {τ1<∞} ] = ∂ ∂α E[α τ1 1 {τ1<∞} ][ α=1 = 1 2q [ 4pq √ 1−4pq − (1 − √ 1 −4pq)] = 1 2q [ 4pq 2q−1 − 1 + 2q − 1] = p q 1 2q−1 . 5.4. (i) Proof. E[α τ2 ] = ¸ ∞ k=1 P(τ 2 = 2k)α 2k = ¸ ∞ k=1 ( α 2 ) 2k P(τ 2 = 2k)4 k . So P(τ 2 = 2k) = (2k)! 4 k (k+1)!k! . (ii) Proof. P(τ 2 = 2) = 1 4 . For k ≥ 2, P(τ 2 = 2k) = P(τ 2 ≤ 2k) −P(τ 2 ≤ 2k −2). P(τ 2 ≤ 2k) = P(M 2k = 2) +P(M 2k ≥ 4) +P(τ 2 ≤ 2k, M 2k ≤ 0) = P(M 2k = 2) + 2P(M 2k ≥ 4) = P(M 2k = 2) +P(M 2k ≥ 4) +P(M 2k ≤ −4) = 1 −P(M 2k = −2) −P(M 2k = 0). 15 Similarly, P(τ 2 ≤ 2k −2) = 1 −P(M 2k−2 = −2) −P(M 2k−2 = 0). So P(τ 2 = 2k) = P(M 2k−2 = −2) +P(M 2k−2 = 0) −P(M 2k = −2) −P(M 2k = 0) = ( 1 2 ) 2k−2 ¸ (2k −2)! k!(k −2)! + (2k −2)! (k −1)!(k −1)! −( 1 2 ) 2k ¸ (2k)! (k + 1)!(k −1)! + (2k)! k!k! = (2k)! 4 k (k + 1)!k! ¸ 4 2k(2k −1) (k + 1)k(k −1) + 4 2k(2k −1) (k + 1)k 2 −k −(k + 1) = (2k)! 4 k (k + 1)!k! ¸ 2(k 2 −1) 2k −1 + 2(k 2 +k) 2k −1 − 4k 2 −1 2k −1 = (2k)! 4 k (k + 1)!k! . 5.5. (i) Proof. For every path that crosses level m by time n and resides at b at time n, there corresponds a reflected path that resides at time 2m−b. So P(M ∗ n ≥ m, M n = b) = P(M n = 2m−b) = ( 1 2 ) n n! (m+ n−b 2 )!( n+b 2 −m)! . (ii) Proof. P(M ∗ n ≥ m, M n = b) = n! (m+ n−b 2 )!( n+b 2 −m)! p m+ n−b 2 q n+b 2 −m . 5.6. Proof. On the infinite coin-toss space, we define M n = ¦stopping times that takes values 0, 1, , n, ∞¦ and M ∞ = ¦stopping times that takes values 0, 1, 2, ¦. Then the time-zero value V ∗ of the perpetual American put as in Section 5.4 can be defined as sup τ∈M∞ ¯ E[1 {τ<∞} (K−Sτ ) + (1+r) τ ]. For an American put with the same strike price K that expires at time n, its time-zero value V (n) is max τ∈Mn ¯ E[1 {τ<∞} (K−Sτ ) + (1+r) τ ]. Clearly (V (n) ) n≥0 is nondecreasing and V (n) ≤ V ∗ for every n. So lim n V (n) exists and lim n V (n) ≤ V ∗ . For any given τ ∈ M ∞ , we define τ (n) = ∞, if τ = ∞ τ ∧ n, if τ < ∞ , then τ (n) is also a stopping time, τ (n) ∈ M n and lim n→∞ τ (n) = τ. By bounded convergence theorem, ¯ E ¸ 1 {τ<∞} (K −S τ ) + (1 +r) τ = lim n→∞ ¯ E ¸ 1 {τ (n) <∞} (K −S τ (n) ) + (1 +r) τ (n) ≤ lim n→∞ V (n) . Take sup at the left hand side of the inequality, we get V ∗ ≤ lim n→∞ V (n) . Therefore V ∗ = lim n V (n) . Remark: In the above proof, rigorously speaking, we should use (K−S τ ) in the places of (K−S τ ) + . So this needs some justification. 5.8. (i) 16 Proof. v(S n ) = S n ≥ S n −K = g(S n ). Under risk-neutral probabilities, 1 (1+r) n v(S n ) = Sn (1+r) n is a martingale by Theorem 2.4.4. (ii) Proof. If the purchaser chooses to exercises the call at time n, then the discounted risk-neutral expectation of her payoff is ¯ E Sn−K (1+r) n = S 0 − K (1+r) n . Since lim n→∞ S 0 − K (1+r) n = S 0 , the value of the call at time zero is at least sup n S 0 − K (1+r) n = S 0 . (iii) Proof. max g(s), pv(us)+ qv(ds) 1+r ¸ = max¦s −K, pu+ qv 1+r s¦ = max¦s −K, s¦ = s = v(s), so equation (5.4.16) is satisfied. Clearly v(s) = s also satisfies the boundary condition (5.4.18). (iv) Proof. Suppose τ is an optimal exercise time, then ¯ E Sτ −K (1+r) τ 1 {τ<∞} ≥ S 0 . Then P(τ < ∞) = 0 and ¯ E K (1+r) τ 1 {τ<∞} > 0. So ¯ E Sτ −K (1+r) τ 1 {τ<∞} < ¯ E Sτ (1+r) τ 1 {τ<∞} . Since Sn (1+r) n n≥0 is a martingale under risk-neutral measure, by Fatou’s lemma, ¯ E Sτ (1+r) τ 1 {τ<∞} ≤ liminf n→∞ ¯ E Sτ∧n (1+r) τ∧n 1 {τ<∞} = liminf n→∞ ¯ E Sτ∧n (1+r) τ∧n = liminf n→∞ ¯ E[S 0 ] = S 0 . Combined, we have S 0 ≤ ¯ E Sτ −K (1+r) τ 1 {τ<∞} < S 0 . Contradiction. So there is no optimal time to exercise the perpetual American call. Simultaneously, we have shown ¯ E Sτ −K (1+r) τ 1 {τ<∞} < S 0 for any stopping time τ. Combined with (ii), we conclude S 0 is the least upper bound for all the prices acceptable to the buyer. 5.9. (i) Proof. Suppose v(s) = s p , then we have s p = 2 5 2 p s p + 2 5 s p 2 p . So 1 = 2 p+1 5 + 2 1−p 5 . Solve it for p, we get p = 1 or p = −1. (ii) Proof. Since lim s→∞ v(s) = lim s→∞ (As + B s ) = 0, we must have A = 0. (iii) Proof. f B (s) = 0 if and only if B+s 2 −4s = 0. The discriminant ∆ = (−4) 2 −4B = 4(4−B). So for B ≤ 4, the equation has roots and for B > 4, this equation does not have roots. (iv) Proof. Suppose B ≤ 4, then the equation s 2 − 4s + B = 0 has solution 2 ± √ 4 −B. By drawing graphs of 4 −s and B s , we should choose B = 4 and s B = 2 + √ 4 −B = 2. (v) Proof. To have continuous derivative, we must have −1 = − B s 2 B . Plug B = s 2 B back into s 2 B −4s B +B = 0, we get s B = 2. This gives B = 4. 6. Interest-Rate-Dependent Assets 6.2. 17 Proof. X k = S k −E k [D m (S m −K)]D −1 k − Sn Bn,m B k,m for n ≤ k ≤ m. Then E k−1 [D k X k ] = E k−1 [D k S k −E k [D m (S m −K)] − S n B n,m B k,m D k ] = D k−1 S k−1 −E k−1 [D m (S m −K)] − S n B n,m E k−1 [E k [D m ]] = D k−1 [S k−1 −E k−1 [D m (S m −K)]D −1 k−1 − S n B n,m B k−1,m ] = D k−1 X k−1 . 6.3. Proof. 1 D n ¯ E n [D m+1 R m ] = 1 D n ¯ E n [D m (1 +R m ) −1 R m ] = ¯ E n [ D m −D m+1 D n ] = B n,m −B n,m+1 . 6.4.(i) Proof. D 1 V 1 = E 1 [D 3 V 3 ] = E 1 [D 2 V 2 ] = D 2 E 1 [V 2 ]. So V 1 = D2 D1 E 1 [V 2 ] = 1 1+R1 E 1 [V 2 ]. In particular, V 1 (H) = 1 1+R1(H) V 2 (HH)P(w 2 = H[w 1 = H) = 4 21 , V 1 (T) = 0. (ii) Proof. Let X 0 = 2 21 . Suppose we buy ∆ 0 shares of the maturity two bond, then at time one, the value of our portfolio is X 1 = (1 +R 0 )(X 0 −∆B 0,2 ) + ∆ 0 B 1,2 . To replicate the value V 1 , we must have V 1 (H) = (1 +R 0 )(X 0 −∆ 0 B 0,2 ) + ∆ 0 B 1,2 (H) V 1 (T) = (1 +R 0 )(X 0 −∆ 0 B 0,2 ) + ∆ 0 B 1,2 (T). So ∆ 0 = V1(H)−V1(T) B1,2(H)−B1,2(T) = 4 3 . The hedging strategy is therefore to borrow 4 3 B 0,2 − 2 21 = 20 21 and buy 4 3 share of the maturity two bond. The reason why we do not invest in the maturity three bond is that B 1,3 (H) = B 1,3 (T)(= 4 7 ) and the portfolio will therefore have the same value at time one regardless the outcome of first coin toss. This makes impossible the replication of V 1 , since V 1 (H) = V 1 (T). (iii) Proof. Suppose we buy ∆ 1 share of the maturity three bond at time one, then to replicate V 2 at time two, we must have V 2 = (1 + R 1 )(X 1 − ∆ 1 B 1,3 ) + ∆ 1 B 2,3 . So ∆ 1 (H) = V2(HH)−V2(HT) B2,3(HH)−B2,3(HT) = − 2 3 , and ∆ 1 (T) = V2(TH)−V2(TT) B2,3(TH)−B2,3(TT) = 0. So the hedging strategy is as follows. If the outcome of first coin toss is T, then we do nothing. If the outcome of first coin toss is H, then short 2 3 shares of the maturity three bond and invest the income into the money market account. We do not invest in the maturity two bond, because at time two, the value of the bond is its face value and our portfolio will therefore have the same value regardless outcomes of coin tosses. This makes impossible the replication of V 2 . 6.5. (i) 18 Proof. Suppose 1 ≤ n ≤ m, then ¯ E m+1 n−1 [F n,m ] = ¯ E n−1 [B −1 n,m+1 (B n,m −B n,m+1 )Z n,m+1 Z −1 n−1,m+1 ] = ¯ E n−1 ¸ B n,m B n,m+1 −1 B n,m+1 D n B n−1,m+1 D n−1 = D n B n−1,m+1 D n−1 ¯ E n−1 [D −1 n ¯ E n [D m ] −D −1 n ¯ E n [D m+1 ]] = ¯ E n−1 [D m −D m+1 ] B n−1,m1 D n−1 = B n−1,m −B n−1,m+1 B n−1,m+1 = F n−1,m . 6.6. (i) Proof. The agent enters the forward contract at no cost. He is obliged to buy certain asset at time m at the strike price K = For n,m = Sn Bn,m . At time n + 1, the contract has the value ¯ E n+1 [D m (S m − K)] = S n+1 −KB n+1,m = S n+1 − SnBn+1,m Bn,m . So if the agent sells this contract at time n +1, he will receive a cash flow of S n+1 − SnBn+1,m Bn,m (ii) Proof. By (i), the cash flow generated at time n + 1 is (1 +r) m−n−1 S n+1 − S n B n+1,m B n,m = (1 +r) m−n−1 S n+1 − Sn (1+r) m−n−1 1 (1+r) m−n = (1 +r) m−n−1 S n+1 −(1 +r) m−n S n = (1 +r) m ¯ E n1 [ S m (1 +r) m ] + (1 +r) m ¯ E n [ S m (1 +r) m ] = Fut n+1,m −Fut n,m . 6.7. Proof. ψ n+1 (0) = ¯ E[D n+1 V n+1 (0)] = ¯ E[ D n 1 +r n (0) 1 {#H(ω1···ωn+1)=0} ] = ¯ E[ D n 1 +r n (0) 1 {#H(ω1···ωn)=0} 1 {ωn+1=T} ] = 1 2 ¯ E[ D n 1 +r n (0) ] = ψ n (0) 2(1 +r n (0)) . 19 For k = 1, 2, , n, ψ n+1 (k) = ¯ E ¸ D n 1 +r n (#H(ω 1 ω n )) 1 {#H(ω1···ωn+1)=k} = ¯ E ¸ D n 1 +r n (k) 1 {#H(ω1···ωn)=k} 1 {ωn+1=T} + ¯ E ¸ D n 1 +r n (k −1) 1 {#H(ω1···ωn)=k} 1 {ωn+1=H} = 1 2 ¯ E[D n V n (k)] 1 +r n (k) + 1 2 ¯ E[D n V n (k −1)] 1 +r n (k −1) = ψ n (k) 2(1 +r n (k)) + ψ n (k −1) 2(1 +r n (k −1)) . Finally, ψ n+1 (n + 1) = ¯ E[D n+1 V n+1 (n + 1)] = ¯ E ¸ D n 1 +r n (n) 1 {#H(ω1···ωn)=n} 1 {ωn+1=H} = ψ n (n) 2(1 +r n (n)) . Remark: In the above proof, we have used the independence of ω n+1 and (ω 1 , , ω n ). This is guaranteed by the assumption that ¯ p = ¯ q = 1 2 (note ξ ⊥ η if and only if E[ξ[η] = constant). In case the binomial model has stochastic up- and down-factor u n and d n , we can use the fact that ¯ P(ω n+1 = H[ω 1 , , ω n ) = p n and ¯ P(ω n+1 = T[ω 1 , , ω n ) = q n , where p n = 1+rn−dn un−dn and q n = u−1−rn un−dn (cf. solution of Exercise 2.9 and notes on page 39). Then for any X ∈ T n = σ(ω 1 , , ω n ), we have ¯ E[Xf(ω n+1 )] = ¯ E[X ¯ E[f(ω n+1 )[T n ]] = ¯ E[X(p n f(H) +q n f(T))]. 20 2 Stochastic Calculus for Finance II: Continuous-Time Models 1. General Probability Theory 1.1. (i) Proof. P(B) = P((B −A) ∪ A) = P(B −A) +P(A) ≥ P(A). (ii) Proof. P(A) ≤ P(A n ) implies P(A) ≤ lim n→∞ P(A n ) = 0. So 0 ≤ P(A) ≤ 0, which means P(A) = 0. 1.2. (i) Proof. We define a mapping φ from A to Ω as follows: φ(ω 1 ω 2 ) = ω 1 ω 3 ω 5 . Then φ is one-to-one and onto. So the cardinality of A is the same as that of Ω, which means in particular that A is uncountably infinite. (ii) Proof. Let A n = ¦ω = ω 1 ω 2 : ω 1 = ω 2 , , ω 2n−1 = ω 2n ¦. Then A n ↓ A as n → ∞. So P(A) = lim n→∞ P(A n ) = lim n→∞ [P(ω 1 = ω 2 ) P(ω 2n−1 = ω 2n )] = lim n→∞ (p 2 + (1 −p) 2 ) n . Since p 2 + (1 −p) 2 < 1 for 0 < p < 1, we have lim n→∞ (p 2 + (1 −p) 2 ) n = 0. This implies P(A) = 0. 1.3. Proof. Clearly P(∅) = 0. For any A and B, if both of them are finite, then A ∪ B is also finite. So P(A∪B) = 0 = P(A) +P(B). If at least one of them is infinite, then A∪B is also infinite. So P(A∪B) = ∞ = P(A) +P(B). Similarly, we can prove P(∪ N n=1 A n ) = ¸ N n=1 P(A n ), even if A n ’s are not disjoint. To see countable additivity property doesn’t hold for P, let A n = ¦ 1 n ¦. Then A = ∪ ∞ n=1 A n is an infinite set and therefore P(A) = ∞. However, P(A n ) = 0 for each n. So P(A) = ¸ ∞ n=1 P(A n ). 1.4. (i) Proof. By Example 1.2.5, we can construct a random variable X on the coin-toss space, which is uniformly distributed on [0, 1]. For the strictly increasing and continuous function N(x) = x −∞ 1 √ 2π e − ξ 2 2 dξ, we let Z = N −1 (X). Then P(Z ≤ a) = P(X ≤ N(a)) = N(a) for any real number a, i.e. Z is a standard normal random variable on the coin-toss space (Ω ∞ , T ∞ , P). (ii) Proof. Define X n = ¸ n i=1 Yi 2 i , where Y i (ω) = 1, if ω i = H 0, if ω i = T. Then X n (ω) → X(ω) for every ω ∈ Ω ∞ where X is defined as in (i). So Z n = N −1 (X n ) → Z = N −1 (X) for every ω. Clearly Z n depends only on the first n coin tosses and ¦Z n ¦ n≥1 is the desired sequence. 1.5. 21 Proof. First, by the information given by the problem, we have Ω ∞ 0 1 [0,X(ω)) (x)dxdP(ω) = ∞ 0 Ω 1 [0,X(ω)) (x)dP(ω)dx. The left side of this equation equals to Ω X(ω) 0 dxdP(ω) = Ω X(ω)dP(ω) = E¦X¦. The right side of the equation equals to ∞ 0 Ω 1 {x<X(ω)} dP(ω)dx = ∞ 0 P(x < X)dx = ∞ 0 (1 −F(x))dx. So E¦X¦ = ∞ 0 (1 −F(x))dx. 1.6. (i) Proof. E¦e uX ¦ = ∞ −∞ e ux 1 σ √ 2π e − (x−µ) 2 2σ 2 dx = ∞ −∞ 1 σ √ 2π e − (x−µ) 2 −2σ 2 ux 2σ 2 dx = ∞ −∞ 1 σ √ 2π e − [x−(µ+σ 2 u)] 2 −(2σ 2 uµ+σ 4 u 2 ) 2σ 2 dx = e uµ+ σ 2 u 2 2 ∞ −∞ 1 σ √ 2π e − [x−(µ+σ 2 u)] 2 2σ 2 dx = e uµ+ σ 2 u 2 2 (ii) Proof. E¦φ(X)¦ = E¦e uX ¦ = e uµ+ u 2 σ 2 2 ≥ e uµ = φ(E¦X¦). 1.7. (i) Proof. Since [f n (x)[ ≤ 1 √ 2nπ , f(x) = lim n→∞ f n (x) = 0. (ii) Proof. By the change of variable formula, ∞ −∞ f n (x)dx = ∞ −∞ 1 √ 2π e − x 2 2 dx = 1. So we must have lim n→∞ ∞ −∞ f n (x)dx = 1. (iii) Proof. This is not contradictory with the Monotone Convergence Theorem, since ¦f n ¦ n≥1 doesn’t increase to 0. 22 1.8. (i) Proof. By (1.9.1), [Y n [ = e tX −e snX t−sn = [Xe θX [ = Xe θX ≤ Xe 2tX . The last inequality is by X ≥ 0 and the fact that θ is between t and s n , and hence smaller than 2t for n sufficiently large. So by the Dominated Convergence Theorem, ϕ (t) = lim n→∞ E¦Y n ¦ = E¦lim n→∞ Y n ¦ = E¦Xe tX ¦. (ii) Proof. Since E[e tX + 1 {X≥0} ] +E[e −tX − 1 {X<0} ] = E[e tX ] < ∞ for every t ∈ R, E[e t|X| ] = E[e tX + 1 {X≥0} ] + E[e −(−t)X − 1 {X<0} ] < ∞ for every t ∈ R. Similarly, we have E[[X[e t|X| ] < ∞ for every t ∈ R. So, similar to (i), we have [Y n [ = [Xe θX [ ≤ [X[e 2t|X| for n sufficiently large, So by the Dominated Convergence Theorem, ϕ (t) = lim n→∞ E¦Y n ¦ = E¦lim n→∞ Y n ¦ = E¦Xe tX ¦. 1.9. Proof. If g(x) is of the form 1 B (x), where B is a Borel subset of R, then the desired equality is just (1.9.3). By the linearity of Lebesgue integral, the desired equality also holds for simple functions, i.e. g of the form g(x) = ¸ n i=1 1 Bi (x), where each B i is a Borel subset of R. Since any nonnegative, Borel-measurable function g is the limit of an increasing sequence of simple functions, the desired equality can be proved by the Monotone Convergence Theorem. 1.10. (i) Proof. If ¦A i ¦ ∞ i=1 is a sequence of disjoint Borel subsets of [0, 1], then by the Monotone Convergence Theorem, ¯ P(∪ ∞ i=1 A i ) equals to 1 ∪ ∞ i=1 Ai ZdP = lim n→∞ 1 ∪ n i=1 Ai ZdP = lim n→∞ 1 ∪ n i=1 Ai ZdP = lim n→∞ n ¸ i=1 Ai ZdP = ∞ ¸ i=1 ¯ P(A i ). Meanwhile, ¯ P(Ω) = 2P([ 1 2 , 1]) = 1. So ¯ P is a probability measure. (ii) Proof. If P(A) = 0, then ¯ P(A) = A ZdP = 2 A∩[ 1 2 ,1] dP = 2P(A∩ [ 1 2 , 1]) = 0. (iii) Proof. Let A = [0, 1 2 ). 1.11. Proof. ¯ E¦e uY ¦ = E¦e uY Z¦ = E¦e uX+uθ e −θX− θ 2 2 ¦ = e uθ− θ 2 2 E¦e (u−θ)X ¦ = e uθ− θ 2 2 e (u−θ) 2 2 = e u 2 2 . 1.12. Proof. First, ˆ Z = e θY − θ 2 2 = e θ(X+θ)− θ 2 2 = e θ 2 2 +θX = Z −1 . Second, for any A ∈ T, ˆ P(A) = A ˆ Zd ¯ P = (1 A ˆ Z)ZdP = 1 A dP = P(A). So P = ˆ P. In particular, X is standard normal under ˆ P, since it’s standard normal under P. 1.13. (i) 23 Proof. 1 P(X ∈ B(x, )) = 1 x+ 2 x− 2 1 √ 2π e − u 2 2 du is approximately 1 1 √ 2π e − x 2 2 = 1 √ 2π e − X 2 (¯ ω) 2 . (ii) Proof. Similar to (i). (iii) Proof. ¦X ∈ B(x, )¦ = ¦X ∈ B(y −θ, )¦ = ¦X +θ ∈ B(y, )¦ = ¦Y ∈ B(y, )¦. (iv) Proof. By (i)-(iii), P(A) P(A) is approximately √ 2π e − Y 2 (¯ ω) 2 √ 2π e − X 2 (¯ ω) 2 = e − Y 2 (¯ ω)−X 2 (¯ ω) 2 = e − (X(¯ ω)+θ) 2 −X 2 (¯ ω) 2 = e −θX(¯ ω)− θ 2 2 . 1.14. (i) Proof. ¯ P(Ω) = ¯ λ λ e −( λ−λ)X dP = ¯ λ λ ∞ 0 e −( λ−λ)x λe −λx dx = ∞ 0 ¯ λe − λx dx = 1. (ii) Proof. ¯ P(X ≤ a) = {X≤a} ¯ λ λ e −( λ−λ)X dP = a 0 ¯ λ λ e −( λ−λ)x λe −λx dx = a 0 ¯ λe − λx dx = 1 −e − λa . 1.15. (i) Proof. Clearly Z ≥ 0. Furthermore, we have E¦Z¦ = E h(g(X))g (X) f(X) = ∞ −∞ h(g(x))g (x) f(x) f(x)dx = ∞ −∞ h(g(x))dg(x) = ∞ −∞ h(u)du = 1. (ii) Proof. ¯ P(Y ≤ a) = {g(X)≤a} h(g(X))g (X) f(X) dP = g −1 (a) −∞ h(g(x))g (x) f(x) f(x)dx = g −1 (a) −∞ h(g(x))dg(x). By the change of variable formula, the last equation above equals to a −∞ h(u)du. So Y has density h under ¯ P. 24 2. Information and Conditioning 2.1. Proof. For any real number a, we have ¦X ≤ a¦ ∈ T 0 = ¦∅, Ω¦. So P(X ≤ a) is either 0 or 1. Since lim a→∞ P(X ≤ a) = 1 and lim a→∞ P(X ≤ a) = 0, we can find a number x 0 such that P(X ≤ x 0 ) = 1 and P(X ≤ x) = 0 for any x < x 0 . So P(X = x 0 ) = lim n→∞ P(x 0 − 1 n < X ≤ x 0 ) = lim n→∞ (P(X ≤ x 0 ) −P(X ≤ x 0 − 1 n )) = 1. 2.2. (i) Proof. σ(X) = ¦∅, Ω, ¦HT, TH¦, ¦TT, HH¦¦. (ii) Proof. σ(S 1 ) = ¦∅, Ω, ¦HH, HT¦, ¦TH, TT¦¦. (iii) Proof. ¯ P(¦HT, TH¦ ∩ ¦HH, HT¦) = ¯ P(¦HT¦) = 1 4 , ¯ P(¦HT, TH¦) = ¯ P(¦HT¦) + ¯ P(¦TH¦) = 1 4 + 1 4 = 1 2 , and ¯ P(¦HH, HT¦) = ¯ P(¦HH¦) + ¯ P(¦HT¦) = 1 4 + 1 4 = 1 2 . So we have ¯ P(¦HT, TH¦ ∩ ¦HH, HT¦) = ¯ P(¦HT, TH¦) ¯ P(¦HH, HT¦). Similarly, we can work on other elements of σ(X) and σ(S 1 ) and show that ¯ P(A∩ B) = ¯ P(A) ¯ P(B) for any A ∈ σ(X) and B ∈ σ(S 1 ). So σ(X) and σ(S 1 ) are independent under ¯ P. (iv) Proof. P(¦HT, TH¦ ∩ ¦HH, HT¦) = P(¦HT¦) = 2 9 , P(¦HT, TH¦) = 2 9 + 2 9 = 4 9 and P(¦HH, HT¦) = 4 9 + 2 9 = 6 9 . So P(¦HT, TH¦ ∩ ¦HH, HT¦) = P(¦HT, TH¦)P(¦HH, HT¦). Hence σ(X) and σ(S 1 ) are not independent under P. (v) Proof. Because S 1 and X are not independent under the probability measure P, knowing the value of X will affect our opinion on the distribution of S 1 . 2.3. Proof. We note (V, W) are jointly Gaussian, so to prove their independence it suffices to show they are uncorrelated. Indeed, E¦V W¦ = E¦−X 2 sin θ cos θ+XY cos 2 θ−XY sin 2 θ+Y 2 sin θ cos θ¦ = −sin θ cos θ+ 0 + 0 + sin θ cos θ = 0. 2.4. (i) 25 Proof. E¦e uX+vY ¦ = E¦e uX+vXZ ¦ = E¦e uX+vXZ [Z = 1¦P(Z = 1) +E¦e uX+vXZ [Z = −1¦P(Z = −1) = 1 2 E¦e uX+vX ¦ + 1 2 E¦e uX−vX ¦ = 1 2 [e (u+v) 2 2 +e (u−v) 2 2 ] = e u 2 +v 2 2 e uv +e −uv 2 . (ii) Proof. Let u = 0. (iii) Proof. E¦e uX ¦ = e u 2 2 and E¦e vY ¦ = e v 2 2 . So E¦e uX+vY ¦ = E¦e uX ¦E¦e vY ¦. Therefore X and Y cannot be independent. 2.5. Proof. The density f X (x) of X can be obtained by f X (x) = f X,Y (x, y)dy = {y≥−|x|} 2[x[ +y √ 2π e − (2|x|+y) 2 2 dy = {ξ≥|x|} ξ √ 2π e − ξ 2 2 dξ = 1 √ 2π e − x 2 2 . The density f Y (y) of Y can be obtained by f Y (y) = f XY (x, y)dx = 1 {|x|≥−y} 2[x[ +y √ 2π e − (2|x|+y) 2 2 dx = ∞ 0∨(−y) 2x +y √ 2π e − (2x+y) 2 2 dx + 0∧y −∞ −2x +y √ 2π e − (−2x+y) 2 2 dx = ∞ 0∨(−y) 2x +y √ 2π e − (2x+y) 2 2 dx + 0∨(−y) ∞ 2x +y √ 2π e − (2x+y) 2 2 d(−x) = 2 ∞ |y| ξ √ 2π e − ξ 2 2 d( ξ 2 ) = 1 √ 2π e − y 2 2 . So both X and Y are standard normal random variables. Since f X,Y (x, y) = f X (x)f Y (y), X and Y are not 26 independent. However, if we set F(t) = ∞ t u 2 √ 2π e − u 2 2 du, we have E¦XY ¦ = ∞ −∞ ∞ −∞ xyf X,Y (x, y)dxdy = ∞ −∞ ∞ −∞ xy1 {y≥−|x|} 2[x[ +y √ 2π e − (2|x|+y) 2 2 dxdy = ∞ −∞ xdx ∞ −|x| y 2[x[ +y √ 2π e − (2|x|+y) 2 2 dy = ∞ −∞ xdx ∞ |x| (ξ −2[x[) ξ √ 2π e − ξ 2 2 dξ = ∞ −∞ xdx( ∞ |x| ξ 2 √ 2π e − ξ 2 2 dξ −2[x[ e − x 2 2 √ 2π ) = ∞ 0 x ∞ x ξ 2 √ 2π e − ξ 2 2 dξdx + 0 −∞ x ∞ −x ξ 2 √ 2π e − ξ 2 2 dξdx = ∞ 0 xF(x)dx + 0 −∞ xF(−x)dx. So E¦XY ¦ = ∞ 0 xF(x)dx − ∞ 0 uF(u)du = 0. 2.6. (i) Proof. σ(X) = ¦∅, Ω, ¦a, b¦, ¦c, d¦¦. (ii) Proof. E¦Y [X¦ = ¸ α∈{a,b,c,d} E¦Y 1 {X=α} ¦ P(X = α) 1 {X=α} . (iii) Proof. E¦Z[X¦ = X +E¦Y [X¦ = X + ¸ α∈{a,b,c,d} E¦Y 1 {X=α} ¦ P(X = α) 1 {X=α} . (iv) Proof. E¦Z[X¦ −E¦Y [X¦ = E¦Z −Y [X¦ = E¦X[X¦ = X. 2.7. 27 Proof. Let µ = E¦Y −X¦ and ξ = E¦Y −X −µ[(¦. Note ξ is (-measurable, we have V ar(Y −X) = E¦(Y −X −µ) 2 ¦ = E¦[(Y −E¦Y [(¦) + (E¦Y [(¦ −X −µ)] 2 ¦ = V ar(Err) + 2E¦(Y −E¦Y [(¦)ξ¦ +E¦ξ 2 ¦ = V ar(Err) + 2E¦Y ξ −E¦Y [(¦ξ¦ +E¦ξ 2 ¦ = V ar(Err) +E¦ξ 2 ¦ ≥ V ar(Err). 2.8. Proof. It suffices to prove the more general case. For any σ(X)-measurable random variable ξ, E¦Y 2 ξ¦ = E¦(Y −E¦Y [X¦)ξ¦ = E¦Y ξ −E¦Y [X¦ξ¦ = E¦Y ξ¦ −E¦Y ξ¦ = 0. 2.9. (i) Proof. Consider the dice-toss space similar to the coin-toss space. Then a typical element ω in this space is an infinite sequence ω 1 ω 2 ω 3 , with ω i ∈ ¦1, 2, , 6¦ (i ∈ N). We define X(ω) = ω 1 and f(x) = 1 {odd integers} (x). Then it’s easy to see σ(X) = ¦∅, Ω, ¦ω : ω 1 = 1¦, , ¦ω : ω 1 = 6¦¦ and σ(f(X)) equals to ¦∅, Ω, ¦ω : ω 1 = 1¦ ∪ ¦ω : ω 1 = 3¦ ∪ ¦ω : ω 1 = 5¦, ¦ω : ω 1 = 2¦ ∪ ¦ω : ω 1 = 4¦ ∪ ¦ω : ω 1 = 6¦¦. So ¦∅, Ω¦ ⊂ σ(f(X)) ⊂ σ(X), and each of these containment is strict. (ii) Proof. No. σ(f(X)) ⊂ σ(X) is always true. 2.10. Proof. A g(X)dP = E¦g(X)1 B (X)¦ = ∞ −∞ g(x)1 B (x)f X (x)dx = yf X,Y (x, y) f X (x) dy1 B (x)f X (x)dx = y1 B (x)f X,Y (x, y)dxdy = E¦Y 1 B (X)¦ = E¦Y I A ¦ = A Y dP. 28 2.11. (i) Proof. We can find a sequence ¦W n ¦ n≥1 of σ(X)-measurable simple functions such that W n ↑ W. Each W n can be written in the form ¸ Kn i=1 a n i 1 A n i , where A n i ’s belong to σ(X) and are disjoint. So each A n i can be written as ¦X ∈ B n i ¦ for some Borel subset B n i of R, i.e. W n = ¸ Kn i=1 a n i 1 {X∈B n i } = ¸ Kn i=1 a n i 1 B n i (X) = g n (X), where g n (x) = ¸ Kn i=1 a n i 1 B n i (x). Define g = limsup g n , then g is a Borel function. By taking upper limits on both sides of W n = g n (X), we get W = g(X). (ii) Proof. Note E¦Y [X¦ is σ(X)-measurable. By (i), we can find a Borel function g such that E¦Y [X¦ = g(X). 3. Brownian Motion 3.1. Proof. We have T t ⊂ T u1 and W u2 −W u1 is independent of T u1 . So in particular, W u2 −W u1 is independent of T t . 3.2. Proof. E[W 2 t −W 2 s [T s ] = E[(W t −W s ) 2 + 2W t W s −2W 2 s [T s ] = t −s + 2W s E[W t −W s [T s ] = t −s. 3.3. Proof. ϕ (3) (u) = 2σ 4 ue 1 2 σ 2 u 2 +(σ 2 +σ 4 u 2 )σ 2 ue 1 2 σ 2 u 2 = e 1 2 σ 2 u 2 (3σ 4 u+σ 4 u 2 ), and ϕ (4) (u) = σ 2 ue 1 2 σ 2 u 2 (3σ 4 u+ σ 4 u 2 ) +e 1 2 σ 2 u 2 (3σ 4 + 2σ 4 u). So E[(X −µ) 4 ] = ϕ (4) (0) = 3σ 4 . 3.4. (i) Proof. Assume there exists A ∈ T, such that P(A) > 0 and for every ω ∈ A, lim n ¸ n−1 j=0 [W tj+1 −W tj [(ω) < ∞. Then for every ω ∈ A, ¸ n−1 j=0 (W tj+1 −W tj ) 2 (ω) ≤ max 0≤k≤n−1 [W t k+1 −W t k [(ω) ¸ n−1 j=0 [W tj+1 −W tj [(ω) → 0, since lim n→∞ max 0≤k≤n−1 [W t k+1 − W t k [(ω) = 0. This is a contradiction with lim n→∞ ¸ n−1 j=0 (W tj+1 − W tj ) 2 = T a.s.. (ii) Proof. Note ¸ n−1 j=0 (W tj+1 −W tj ) 3 ≤ max 0≤k≤n−1 [W t k+1 −W t k [ ¸ n−1 j=0 (W tj+1 −W tj ) 2 → 0 as n → ∞, by an argument similar to (i). 3.5. 29 Proof. E[e −rT (S T −K) + ] = e −rT ∞ 1 σ (ln K S 0 −(r− 1 2 σ 2 )T) (S 0 e (r− 1 2 σ 2 )T+σx −K) e − x 2 2T √ 2πT dx = e −rT ∞ 1 σ √ T (ln K S 0 −(r− 1 2 σ 2 )T) (S 0 e (r− 1 2 σ 2 )T+σ √ Ty −K) e − y 2 2 √ 2π dy = S 0 e − 1 2 σ 2 T ∞ 1 σ √ T (ln K S 0 −(r− 1 2 σ 2 )T) 1 √ 2π e − y 2 2 +σ √ Ty dy −Ke −rT ∞ 1 σ √ T (ln K S 0 −(r− 1 2 σ 2 )T) 1 √ 2π e − y 2 2 dy = S 0 ∞ 1 σ √ T (ln K S 0 −(r− 1 2 σ 2 )T)−σ √ T 1 √ 2π e − ξ 2 2 dξ −Ke −rT N 1 σ √ T (ln S 0 K + (r − 1 2 σ 2 )T) = Ke −rT N(d + (T, S 0 )) −Ke −rT N(d − (T, S 0 )). 3.6. (i) Proof. E[f(X t )[T t ] = E[f(W t −W s +a)[T s ][ a=Ws+µt = E[f(W t−s +a)][ a=Ws+µt = ∞ −∞ f(x +W s +µt) e − x 2 2(t−s) 2π(t −s) dx = ∞ −∞ f(y) e − (y−Ws−µs−µ(t−s)) 2 2(t−s) 2π(t −s) dy = g(X s ). So E[f(X t )[T s ] = ∞ −∞ f(y)p(t −s, X s , y)dy with p(τ, x, y) = 1 √ 2πτ e − (y−x−µτ) 2 2τ . (ii) Proof. E[f(S t )[T s ] = E[f(S 0 e σXt )[T s ] with µ = v σ . So by (i), E[f(S t )[T s ] = ∞ −∞ f(S 0 e σy ) 1 2π(t −s) e − (y−Xs−µ(t−s)) 2 2(t−s) dy S0e σy =z = ∞ 0 f(z) 1 2π(t −s) e − ( 1 σ ln z S 0 − 1 σ ln Ss S 0 −µ(t−s)) 2 2 dz σz = ∞ 0 f(z) e − (ln z Ss −v(t−s)) 2 2σ 2 (t−s) σz 2π(t −s) dz = ∞ 0 f(z)p(t −s, S s , z)dz = g(S s ). 3.7. (i) 30 Proof. E Zt Zs [T s = E[exp¦σ(W t −W s ) +σµ(t −s) −(σµ + σ 2 2 )(t −s)¦] = 1. (ii) Proof. By optional stopping theorem, E[Z t∧τm ] = E[Z 0 ] = 1, that is, E[exp¦σX t∧τm −(σµ + σ 2 2 )t ∧ τ m ¦] = 1. (iii) Proof. If µ ≥ 0 and σ > 0, Z t∧τm ≤ e σm . By bounded convergence theorem, E[1 {τm<∞} Z τm ] = E[ lim t→∞ Z t∧τm ] = lim t→∞ E[Z t∧τm ] = 1, since on the event ¦τ m = ∞¦, Z t∧τm ≤ e σm− 1 2 σ 2 t → 0 as t → ∞. Therefore, E[e σm−(σµ+ σ 2 2 )τm ] = 1. Let σ ↓ 0, by bounded convergence theorem, we have P(τ m < ∞) = 1. Let σµ + σ 2 2 = α, we get E[e −ατm ] = e −σm = e mµ−m √ 2α+µ 2 . (iv) Proof. We note for α > 0, E[τ m e −ατm ] < ∞ since xe −αx is bounded on [0, ∞). So by an argument similar to Exercise 1.8, E[e −ατm ] is differentiable and ∂ ∂α E[e −ατm ] = −E[τ m e −ατm ] = e mµ−m √ 2α+µ 2 −m 2α +µ 2 . Let α ↓ 0, by monotone increasing theorem, E[τ m ] = m µ < ∞ for µ > 0. (v) Proof. By σ > −2µ > 0, we get σµ + σ 2 2 > 0. Then Z t∧τm ≤ e σm and on the event ¦τ m = ∞¦, Z t∧τm ≤ e σm−( σ 2 2 +σµ)t → 0 as t → ∞. Therefore, E[e σm−(σµ+ σ 2 2 )τm 1 {τm<∞} ] = E[ lim t→∞ Z t∧τm ] = lim t→∞ E[Z t∧τm ] = 1. Let σ ↓ −2µ, then we get P(τ m < ∞) = e 2µm = e −2|µ|m < 1. Set α = σµ + σ 2 2 . So we get E[e −ατm ] = E[e −ατm 1 {τm<∞} ] = e −σm = e mµ−m √ 2α+µ 2 . 3.8. (i) Proof. ϕ n (u) = ¯ E[e u 1 √ n Mnt,n ] = ( ¯ E[e u √ n X1,n ]) nt = (e u √ n ¯ p n +e − u √ n ¯ q n ) nt = ¸ e u √ n r n + 1 −e − σ √ n e σ √ n −e − σ √ n +e − u √ n − r n −1 +e σ √ n e σ √ n −e − σ √ n ¸ nt . (ii) 31 Proof. ϕ 1 x 2 (u) = ¸ e ux rx 2 + 1 −e −σx e σx −e −σx −e −ux rx 2 + 1 −e σx e σx −e −σx t x 2 . So, ln ϕ 1 x 2 (u) = t x 2 ln ¸ (rx 2 + 1)(e ux −e −ux ) +e (σ−u)x −e −(σ−u)x e σx −e −σx = t x 2 ln ¸ (rx 2 + 1) sinh ux + sinh(σ −u)x sinh σx = t x 2 ln ¸ (rx 2 + 1) sinh ux + sinh σxcosh ux −cosh σxsinh ux sinh σx = t x 2 ln ¸ cosh ux + (rx 2 + 1 −cosh σx) sinh ux sinh σx . (iii) Proof. cosh ux + (rx 2 + 1 −cosh σx) sinh ux sinh σx = 1 + u 2 x 2 2 +O(x 4 ) + (rx 2 + 1 −1 − σ 2 x 2 2 +O(x 4 ))(ux +O(x 3 )) σx +O(x 3 ) = 1 + u 2 x 2 2 + (r − σ 2 2 )ux 3 +O(x 5 ) σx +O(x 3 ) +O(x 4 ) = 1 + u 2 x 2 2 + (r − σ 2 2 )ux 3 (1 +O(x 2 )) σx(1 +O(x 2 )) +O(x 4 ) = 1 + u 2 x 2 2 + rux 2 σ − 1 2 σux 2 +O(x 4 ). (iv) Proof. ln ϕ 1 x 2 = t x 2 ln(1 + u 2 x 2 2 + ru σ x 2 − σux 2 2 +O(x 4 )) = t x 2 ( u 2 x 2 2 + ru σ x 2 − σux 2 2 +O(x 4 )). So lim x↓0 ln ϕ 1 x 2 (u) = t( u 2 2 + ru σ − σu 2 ), and ¯ E[e u 1 √ n Mnt,n ] = ϕ n (u) → 1 2 tu 2 + t( r σ − σ 2 )u. By the one-to-one correspondence between distribution and moment generating function, ( 1 √ n M nt,n ) n converges to a Gaussian random variable with mean t( r σ − σ 2 ) and variance t. Hence ( σ √ n M nt,n ) n converges to a Gaussian random variable with mean t(r − σ 2 2 ) and variance σ 2 t. 4. Stochastic Calculus 4.1. 32 Proof. Fix t and for any s < t, we assume s ∈ [t m , t m+1 ) for some m. Case 1. m = k. Then I(t)−I(s) = ∆ t k (M t −M t k )−∆ t k (M s −M t k ) = ∆ t k (M t −M s ). So E[I(t)−I(s)[T t ] = ∆ t k E[M t −M s [T s ] = 0. Case 2. m < k. Then t m ≤ s < t m+1 ≤ t k ≤ t < t k+1 . So I(t) −I(s) = k−1 ¸ j=m ∆ tj (M tj+1 −M tj ) + ∆ t k (M s −M t k ) −∆ tm (M s −M tm ) = k−1 ¸ j=m+1 ∆ tj (M tj+1 −M tj ) + ∆ t k (M t −M t k ) + ∆ tm (M tm+1 −M s ). Hence E[I(t) −I(s)[T s ] = k−1 ¸ j=m+1 E[∆ tj E[M tj+1 −M tj [T tj ][T s ] +E[∆ t k E[M t −M t k [T t k ][T s ] + ∆ tm E[M tm+1 −M s [T s ] = 0. Combined, we conclude I(t) is a martingale. 4.2. (i) Proof. We follow the simplification in the hint and consider I(t k ) −I(t l ) with t l < t k . Then I(t k ) −I(t l ) = ¸ k−1 j=l ∆ tj (W tj+1 −W tj ). Since ∆ t is a non-random process and W tj+1 −W tj ⊥ T tj ⊃ T t l for j ≥ l, we must have I(t k ) −I(t l ) ⊥ T t l . (ii) Proof. We use the notation in (i) and it is clear I(t k ) − I(t l ) is normal since it is a linear combination of independent normal random variables. Furthermore, E[I(t k ) − I(t l )] = ¸ k−1 j=l ∆ tj E[W tj+1 − W tj ] = 0 and V ar(I(t k ) −I(t l )) = ¸ k−1 j=l ∆ 2 tj V ar(W tj+1 −W tj ) = ¸ k−1 j=l ∆ 2 tj (t j+1 −t j ) = t k t l ∆ 2 u du. (iii) Proof. E[I(t) −I(s)[T s ] = E[I(t) −I(s)] = 0, for s < t. (iv) Proof. For s < t, E[I 2 (t) − t 0 ∆ 2 u du −(I 2 (s) − s 0 ∆ 2 u du)[T s ] = E[I 2 (t) −I 2 (s) − t s ∆ 2 u du[T s ] = E[(I(t) −I(s)) 2 + 2I(t)I(s) −2I 2 (s)[T s ] − t s ∆ 2 u du = E[(I(t) −I(s)) 2 ] + 2I(s)E[I(t) −I(s)[T s ] − t s ∆ 2 u du = t s ∆ 2 u du + 0 − t s ∆ 2 u du = 0. 33 4.3. Proof. I(t) −I(s) = ∆ 0 (W t1 −W 0 ) + ∆ t1 (W t2 −W t1 ) −∆ 0 (W t1 −W 0 ) = ∆ t1 (W t2 −W t1 ) = W s (W t −W s ). (i) I(t) −I(s) is not independent of T s , since W s ∈ T s . (ii) E[(I(t) − I(s)) 4 ] = E[W 4 s ]E[(W t − W s ) 4 ] = 3s 3(t − s) = 9s(t − s) and 3E[(I(t) − I(s)) 2 ] = 3E[W 2 s ]E[(W t −W s ) 2 ] = 3s(t −s). Since E[(I(t) −I(s)) 4 ] = 3E[(I(t) −I(s)) 2 ], I(t) −I(s) is not normally distributed. (iii) E[I(t) −I(s)[T s ] = W s E[W t −W s [T s ] = 0. (iv) E[I 2 (t) − t 0 ∆ 2 u du −(I 2 (s) − s 0 ∆ 2 u du)[T s ] = E[(I(t) −I(s)) 2 + 2I(t)I(s) −2I 2 (s) − t s W 2 u du[T s ] = E[W 2 s (W t −W s ) 2 + 2W s (W t −W s ) −W 2 s (t −s)[T s ] = W 2 s E[(W t −W s ) 2 ] + 2W s E[W t −W s [T s ] −W 2 s (t −s) = W 2 s (t −s) −W 2 s (t −s) = 0. 4.4. Proof. (Cf. Øksendal [3], Exercise 3.9.) We first note that ¸ j Bt j +t j+1 2 (B tj+1 −B tj ) = ¸ j Bt j +t j+1 2 (B tj+1 −Bt j +t j+1 2 ) +B tj (Bt j +t j+1 2 −B tj ) + ¸ j (Bt j +t j+1 2 −B tj ) 2 . The first term converges in L 2 (P) to T 0 B t dB t . For the second term, we note E ¸ ¸ j Bt j +t j+1 2 −B tj 2 − t 2 ¸ 2 ¸ ¸ ¸ = E ¸ ¸ j Bt j +t j+1 2 −B tj 2 − ¸ j t j+1 −t j 2 ¸ 2 ¸ ¸ ¸ = ¸ j, k E ¸ Bt j +t j+1 2 −B tj 2 − t j+1 −t j 2 Bt k +t k+1 2 −B t k 2 − t k+1 −t k 2 = ¸ j E ¸ B 2 t j+1 −t j 2 − t j+1 −t j 2 2 ¸ = ¸ j 2 t j+1 −t j 2 2 ≤ T 2 max 1≤j≤n [t j+1 −t j [ → 0, 34 since E[(B 2 t −t) 2 ] = E[B 4 t −2tB 2 t +t 2 ] = 3E[B 2 t ] 2 −2t 2 +t 2 = 2t 2 . So ¸ j Bt j +t j+1 2 (B tj+1 −B tj ) → T 0 B t dB t + T 2 = 1 2 B 2 T in L 2 (P). 4.5. (i) Proof. d ln S t = dS t S t − 1 2 d'S` t S 2 t = 2S t dS t −d'S` t 2S 2 t = 2S t (α t S t dt +σ t S t dW t ) −σ 2 t S 2 t dt 2S 2 t = σ t dW t + (α t − 1 2 σ 2 t )dt. (ii) Proof. ln S t = ln S 0 + t 0 σ s dW s + t 0 (α s − 1 2 σ 2 s )ds. So S t = S 0 exp¦ t 0 σ s dW s + t 0 (α s − 1 2 σ 2 s )ds¦. 4.6. Proof. Without loss of generality, we assume p = 1. Since (x p ) = px p−1 , (x p ) = p(p −1)x p−2 , we have d(S p t ) = pS p−1 t dS t + 1 2 p(p −1)S p−2 t d'S` t = pS p−1 t (αS t dt +σS t dW t ) + 1 2 p(p −1)S p−2 t σ 2 S 2 t dt = S p t [pαdt +pσdW t + 1 2 p(p −1)σ 2 dt] = S p t p[σdW t + (α + p −1 2 σ 2 )dt]. 4.7. (i) Proof. dW 4 t = 4W 3 t dW t + 1 2 4 3W 2 t d'W` t = 4W 3 t dW t + 6W 2 t dt. So W 4 T = 4 T 0 W 3 t dW t + 6 T 0 W 2 t dt. (ii) Proof. E[W 4 T ] = 6 T 0 tdt = 3T 2 . (iii) Proof. dW 6 t = 6W 5 t dW t + 1 2 6 5W 4 t dt. So W 6 T = 6 T 0 W 5 t dW t +15 T 0 W 4 t dt. Hence E[W 6 T ] = 15 T 0 3t 2 dt = 15T 3 . 4.8. 35 Proof. d(e βt R t ) = βe βt R t dt +e βt dR t = e βt(αdt+σdWt) . Hence e βt R t = R 0 + t 0 e βs (αds +σdW s ) = R 0 + α β (e βt −1) +σ t 0 e βs dW s , and R t = R 0 e −βt + α β (1 −e −βt ) +σ t 0 e −β(t−s) dW s . 4.9. (i) Proof. Ke −r(T−t) N (d − ) = Ke −r(T−t) e − d 2 − 2 √ 2π = Ke −r(T−t) e − (d + −σ √ T−t) 2 2 √ 2π = Ke −r(T−t) e σ √ T−td+ e − σ 2 (T−t) 2 N (d + ) = Ke −r(T−t) x K e (r+ σ 2 2 )(T−t) e − σ 2 (T−t) 2 N (d + ) = xN (d + ). (ii) Proof. c x = N(d + ) +xN (d + ) ∂ ∂x d + (T −t, x) −Ke −r(T−t) N (d − ) ∂ ∂x d − (T −t, x) = N(d + ) +xN (d + ) ∂ ∂x d + (T −t, x) −xN (d + ) ∂ ∂x d + (T −t, x) = N(d + ). (iii) Proof. c t = xN (d + ) ∂ ∂x d + (T −t, x) −rKe −r(T−t) N(d − ) −Ke −r(T−t) N (d − ) ∂ ∂t d − (T −t, x) = xN (d + ) ∂ ∂t d + (T −t, x) −rKe −r(T−t) N(d − ) −xN (d + ) ¸ ∂ ∂t d + (T −t, x) + σ 2 √ T −t = −rKe −r(T−t) N(d − ) − σx 2 √ T −t N (d + ). (iv) 36 Proof. c t +rxc x + 1 2 σ 2 x 2 c xx = −rKe −r(T−t) N(d − ) − σx 2 √ T −t N (d + ) +rxN(d + ) + 1 2 σ 2 x 2 N (d + ) ∂ ∂x d + (T −t, x) = rc − σx 2 √ T −t N (d + ) + 1 2 σ 2 x 2 N (d + ) 1 σ √ T −tx = rc. (v) Proof. For x > K, d + (T − t, x) > 0 and lim t↑T d + (T − t, x) = lim τ↓0 d + (τ, x) = ∞. lim t↑T d − (T − t, x) = lim τ↓0 d − (τ, x) = lim τ↓0 1 σ √ τ ln x K + 1 σ (r + 1 2 σ 2 ) √ τ −σ √ τ = ∞. Similarly, lim t↑T d ± = −∞ for x ∈ (0, K). Also it’s clear that lim t↑T d ± = 0 for x = K. So lim t↑T c(t, x) = xN(lim t↑T d + ) −KN(lim t↑T d − ) = x −K, if x > K 0, if x ≤ K = (x −K) + . (vi) Proof. It is easy to see lim x↓0 d ± = −∞. So for t ∈ [0, T], lim x↓0 c(t, x) = lim x↓0 xN(lim x↓ d + (T − t, x)) − Ke −r(T−t) N(lim x↓0 d − (T −t, x)) = 0. (vii) Proof. For t ∈ [0, T], it is clear lim x→∞ d ± = ∞. Note lim x→∞ x(N(d + ) −1) = lim x→∞ N (d + ) ∂ ∂x d + −x −2 = lim x→∞ N (d + ) 1 σ √ T−t −x −1 . By the expression of d + , we get x = K exp¦σ √ T −td + −(T −t)(r + 1 2 σ 2 )¦. So we have lim x→∞ x(N(d + ) −1) = lim x→∞ N (d + ) −x σ √ T −t = lim d+→∞ e − d 2 + 2 √ 2π −Ke σ √ T−td+−(T−t)(r+ 1 2 σ 2 ) σ √ T −t = 0. Therefore lim x→∞ [c(t, x) −(x −e −r(T−t) K)] = lim x→∞ [xN(d + ) −Ke −r(T−t)N(d−) −x +Ke −r(T−t) ] = lim x→∞ [x(N(d + ) −1) +Ke −r(T−t) (1 −N(d − ))] = lim x→∞ x(N(d + ) −1) +Ke −r(T−t) (1 −N( lim x→∞ d − )) = 0. 4.10. (i) 37 Proof. We show (4.10.16) + (4.10.9) ⇐⇒ (4.10.16) + (4.10.15), i.e. assuming X has the representation X t = ∆ t S t + Γ t M t , “continuous-time self-financing condition” has two equivalent formulations, (4.10.9) or (4.10.15). Indeed, dX t = ∆ t dS t +Γ t dM t +(S t d∆ t +dS t d∆ t +M t dΓ t +dM t dΓ t ). So dX t = ∆ t dS t +Γ t dM t ⇐⇒ S t d∆ t +dS t d∆ t +M t dΓ t +dM t dΓ t = 0, i.e. (4.10.9) ⇐⇒ (4.10.15). (ii) Proof. First, we clarify the problems by stating explicitly the given conditions and the result to be proved. We assume we have a portfolio X t = ∆ t S t + Γ t M t . We let c(t, S t ) denote the price of call option at time t and set ∆ t = c x (t, S t ). Finally, we assume the portfolio is self-financing. The problem is to show rN t dt = ¸ c t (t, S t ) + 1 2 σ 2 S 2 t c xx (t, S t ) dt, where N t = c(t, S t ) −∆ t S t . Indeed, by the self-financing property and ∆ t = c x (t, S t ), we have c(t, S t ) = X t (by the calculations in Subsection 4.5.1-4.5.3). This uniquely determines Γ t as Γ t = X t −∆ t S t M t = c(t, S t ) −c x (t, S t )S t M t = N t M t . Moreover, dN t = ¸ c t (t, S t )dt +c x (t, S t )dS t + 1 2 c xx (t, S t )d'S t ` t −d(∆ t S t ) = ¸ c t (t, S t ) + 1 2 c xx (t, S t )σ 2 S 2 t dt + [c x (t, S t )dS t −d(X t −Γ t M t )] = ¸ c t (t, S t ) + 1 2 c xx (t, S t )σ 2 S 2 t dt +M t dΓ t +dM t dΓ t + [c x (t, S t )dS t + Γ t dM t −dX t ]. By self-financing property, c x (t, S t )dt + Γ t dM t = ∆ t dS t + Γ t dM t = dX t , so ¸ c t (t, S t ) + 1 2 c xx (t, S t )σ 2 S 2 t dt = dN t −M t dΓ t −dM t dΓ t = Γ t dM t = Γ t rM t dt = rN t dt. 4.11. Proof. First, we note c(t, x) solves the Black-Scholes-Merton PDE with volatility σ 1 : ∂ ∂t +rx ∂ ∂x + 1 2 x 2 σ 2 1 ∂ 2 ∂x 2 −r c(t, x) = 0. So c t (t, S t ) +rS t c x (t, S t ) + 1 2 σ 2 1 S 2 t c xx (t, S t ) −rc(t, S t ) = 0, and dc(t, S t ) = c t (t, S t )dt +c x (t, S t )(αS t dt +σ 2 S t dW t ) + 1 2 c xx (t, S t )σ 2 2 S 2 t dt = ¸ c t (t, S t ) +αc x (t, S t )S t + 1 2 σ 2 2 S 2 t c xx (t, S t ) dt +σ 2 S t c x (t, S t )dW t = ¸ rc(t, S t ) + (α −r)c x (t, S t )S t + 1 2 S 2 t (σ 2 2 −σ 2 1 )c xx (t, S t ) dt +σ 2 S t c x (t, S t )dW t . 38 Therefore dX t = ¸ rc(t, S t ) + (α −r)c x (t, S t )S t + 1 2 S 2 t (σ 2 2 −σ 2 1 )σ xx (t, S t ) +rX t −rc(t, S t ) +rS t c x (t, S t ) − 1 2 (σ 2 2 −σ 2 1 )S 2 t c xx (t, S t ) −c x (t, S t )αS t dt + [σ 2 S t c x (t, S t ) −c x (t, S t )σ 2 S t ]dW t = rX t dt. This implies X t = X 0 e rt . By X 0 , we conclude X t = 0 for all t ∈ [0, T]. 4.12. (i) Proof. By (4.5.29), c(t, x) − p(t, x) = x − e −r(T−t) K. So p x (t, x) = c x (t, x) − 1 = N(d + (T − t, x)) − 1, p xx (t, x) = c xx (t, x) = 1 σx √ T−t N (d + (T −t, x)) and p t (t, x) = c t (t, x) +re −r(T−t) K = −rKe −r(T−t) N(d − (T −t, x)) − σx 2 √ T −t N (d + (T −t, x)) +rKe −r(T−t) = rKe −r(T−t) N(−d − (T −t, x)) − σx 2 √ T −t N (d + (T −t, x)). (ii) Proof. For an agent hedging a short position in the put, since ∆ t = p x (t, x) < 0, he should short the underlying stock and put p(t, S t ) −p x (t, S t )S t (> 0) cash in the money market account. (iii) Proof. By the put-call parity, it suffices to show f(t, x) = x −Ke −r(T−t) satisfies the Black-Scholes-Merton partial differential equation. Indeed, ∂ ∂t + 1 2 σ 2 x 2 ∂ 2 ∂x 2 +rx ∂ ∂x −r f(t, x) = −rKe −r(T−t) + 1 2 σ 2 x 2 0 +rx 1 −r(x −Ke −r(T−t) ) = 0. Remark: The Black-Scholes-Merton PDE has many solutions. Proper boundary conditions are the key to uniqueness. For more details, see Wilmott [8]. 4.13. Proof. We suppose (W 1 , W 2 ) is a pair of local martingale defined by SDE dW 1 (t) = dB 1 (t) dW 2 (t) = α(t)dB 1 (t) +β(t)dB 2 (t). (1) We want to find α(t) and β(t) such that (dW 2 (t)) 2 = [α 2 (t) +β 2 (t) + 2ρ(t)α(t)β(t)]dt = dt dW 1 (t)dW 2 (t) = [α(t) +β(t)ρ(t)]dt = 0. (2) Solve the equation for α(t) and β(t), we have β(t) = 1 √ 1−ρ 2 (t) and α(t) = − ρ(t) √ 1−ρ 2 (t) . So W 1 (t) = B 1 (t) W 2 (t) = t 0 −ρ(s) √ 1−ρ 2 (s) dB 1 (s) + t 0 1 √ 1−ρ 2 (s) dB 2 (s) (3) 39 is a pair of independent BM’s. Equivalently, we have B 1 (t) = W 1 (t) B 2 (t) = t 0 ρ(s)dW 1 (s) + t 0 1 −ρ 2 (s)dW 2 (s). (4) 4.14. (i) Proof. Clearly Z j ∈ T tj+1 . Moreover E[Z j [T tj ] = f (W tj )E[(W tj+1 −W tj ) 2 −(t j+1 −t j )[T tj ] = f (W tj )(E[W 2 tj+1−tj ] −(t j+1 −t j )) = 0, since W tj+1 −W tj is independent of T tj and W t ∼ N(0, t). Finally, we have E[Z 2 j [T tj ] = [f (W tj )] 2 E[(W tj+1 −W tj ) 4 −2(t j+1 −t j )(W tj+1 −W tj ) 2 + (t j+1 −t j ) 2 [T tj ] = [f (W tj )] 2 (E[W 4 tj+1−tj ] −2(t j+1 −t j )E[W 2 tj+1−tj ] + (t j+1 −t j ) 2 ) = [f (W tj )] 2 [3(t j+1 −t j ) 2 −2(t j+1 −t j ) 2 + (t j+1 −t j ) 2 ] = 2[f (W tj )] 2 (t j+1 −t j ) 2 , where we used the independence of Browian motion increment and the fact that E[X 4 ] = 3E[X 2 ] 2 if X is Gaussian with mean 0. (ii) Proof. E[ ¸ n−1 j=0 Z j ] = E[ ¸ n−1 j=0 E[Z j [T tj ]] = 0 by part (i). (iii) Proof. V ar[ n−1 ¸ j=0 Z j ] = E[( n−1 ¸ j=0 Z j ) 2 ] = E[ n−1 ¸ j=0 Z 2 j + 2 ¸ 0≤i<j≤n−1 Z i Z j ] = n−1 ¸ j=0 E[E[Z 2 j [T tj ]] + 2 ¸ 0≤i<j≤n−1 E[Z i E[Z j [T tj ]] = n−1 ¸ j=0 E[2[f (W tj )] 2 (t j+1 −t j ) 2 ] = n−1 ¸ j=0 2E[(f (W tj )) 2 ](t j+1 −t j ) 2 ≤ 2 max 0≤j≤n−1 [t j+1 −t j [ n−1 ¸ j=0 E[(f (W tj )) 2 ](t j+1 −t j ) → 0, since ¸ n−1 j=0 E[(f (W tj )) 2 ](t j+1 −t j ) → T 0 E[(f (W t )) 2 ]dt < ∞. 4.15. (i) 40 Proof. B i is a local martingale with (dB i (t)) 2 = ¸ d ¸ j=1 σ ij (t) σ i (t) dW j (t) ¸ 2 = d ¸ j=1 σ 2 ij (t) σ 2 i (t) dt = dt. So B i is a Brownian motion. (ii) Proof. dB i (t)dB k (t) = d ¸ j=1 σ ij (t) σ i (t) dW j (t) ¸ ¸ ¸ d ¸ l=1 σ kl (t) σ k (t) dW l (t) ¸ = ¸ 1≤j, l≤d σ ij (t)σ kl (t) σ i (t)σ k (t) dW j (t)dW l (t) = d ¸ j=1 σ ij (t)σ kj (t) σ i (t)σ k (t) dt = ρ ik (t)dt. 4.16. Proof. To find the m independent Brownion motion W 1 (t), , W m (t), we need to find A(t) = (a ij (t)) so that (dB 1 (t), , dB m (t)) tr = A(t)(dW 1 (t), , dW m (t)) tr , or equivalently (dW 1 (t), , dW m (t)) tr = A(t) −1 (dB 1 (t), , dB m (t)) tr , and (dW 1 (t), , dW m (t)) tr (dW 1 (t), , dW m (t)) = A(t) −1 (dB 1 (t), , dB m (t)) tr (dB 1 (t), , dB m (t))(A(t) −1 ) tr dt = I m×m dt, where I m×m is the mm unit matrix. By the condition dB i (t)dB k (t) = ρ ik (t)dt, we get (dB 1 (t), , dB m (t)) tr (dB 1 (t), , dB m (t)) = C(t). So A(t) −1 C(t)(A(t) −1 ) tr = I m×m , which gives C(t) = A(t)A(t) tr . This motivates us to define A as the square root of C. Reverse the above analysis, we obtain a formal proof. 4.17. Proof. We will try to solve all the sub-problems in a single, long solution. We start with the general X i : X i (t) = X i (0) + t 0 θ i (u)du + t 0 σ i (u)dB i (u), i = 1, 2. 41 The goal is to show lim ↓0 C() V 1 ()V 2 () = ρ(t 0 ). First, for i = 1, 2, we have M i () = E[X i (t 0 +) −X i (t 0 )[T t0 ] = E ¸ t0+ t0 Θ i (u)du + t0+ t0 σ i (u)dB i (u)[T t0 = Θ i (t 0 ) +E ¸ t0+ t0 (Θ i (u) −Θ i (t 0 ))du[T t0 . By Conditional Jensen’s Inequality, E ¸ t0+ t0 (Θ i (u) −Θ i (t 0 ))du[T t0 ≤ E ¸ t0+ t0 [Θ i (u) −Θ i (t 0 )[du[T t0 Since 1 t0+ t0 [Θ i (u) − Θ i (t 0 )[du ≤ 2M and lim →0 1 t0+ t0 [Θ i (u) − Θ i (t 0 )[du = 0 by the continuity of Θ i , the Dominated Convergence Theorem under Conditional Expectation implies lim →0 1 E ¸ t0+ t0 [Θ i (u) −Θ i (t 0 )[du[T t0 = E ¸ lim →0 1 t0+ t0 [Θ i (u) −Θ i (t 0 )[du[T t0 = 0. So M i () = Θ i (t 0 ) +o(). This proves (iii). To calculate the variance and covariance, we note Y i (t) = t 0 σ i (u)dB i (u) is a martingale and by Itˆo’s formula Y i (t)Y j (t) − t 0 σ i (u)σ j (u)du is a martingale (i = 1, 2). So E[(X i (t 0 +) −X i (t 0 ))(X j (t 0 +) −X j (t 0 ))[T t0 ] = E ¸ Y i (t 0 +) −Y i (t 0 ) + t0+ t0 Θ i (u)du Y j (t 0 +) −Y j (t 0 ) + t0+ t0 Θ j (u)du [T t0 = E [(Y i (t 0 +) −Y i (t 0 )) (Y j (t 0 +) −Y j (t 0 )) [T t0 ] +E ¸ t0+ t0 Θ i (u)du t0+ t0 Θ j (u)du[T t0 +E ¸ (Y i (t 0 +) −Y i (t 0 )) t0+ t0 Θ j (u)du[T t0 +E ¸ (Y j (t 0 +) −Y j (t 0 )) t0+ t0 Θ i (u)du[T t0 = I +II +III +IV. I = E[Y i (t 0 +)Y j (t 0 +) −Y i (t 0 )Y j (t 0 )[T t0 ] = E ¸ t0+ t0 σ i (u)σ j (u)ρ ij (t)dt[T t0 . By an argument similar to that involved in the proof of part (iii), we conclude I = σ i (t 0 )σ j (t 0 )ρ ij (t 0 ) +o() and II = E ¸ t0+ t0 (Θ i (u) −Θ i (t 0 ))du t0+ t0 Θ j (u)du[T t0 + Θ i (t 0 )E ¸ t0+ t0 Θ j (u)du[T t0 = o() + (M i () −o())M j () = M i ()M j () +o(). 42 By Cauchy’s inequality under conditional expectation (note E[XY [T] defines an inner product on L 2 (Ω)), III ≤ E ¸ [Y i (t 0 +) −Y i (t 0 )[ t0+ t0 [Θ j (u)[du[T t0 ≤ M E[(Y i (t 0 +) −Y i (t 0 )) 2 [T t0 ] ≤ M E[Y i (t 0 +) 2 −Y i (t 0 ) 2 [T t0 ] ≤ M E[ t0+ t0 Θ i (u) 2 du[T t0 ] ≤ M M √ = o() Similarly, IV = o(). In summary, we have E[(X i (t 0 +) −X i (t 0 ))(X j (t 0 +) −X j (t 0 ))[T t0 ] = M i ()M j () +σ i (t 0 )σ j (t 0 )ρ ij (t 0 ) +o() +o(). This proves part (iv) and (v). Finally, lim ↓0 C() V 1 ()V 2 () = lim ↓0 ρ(t 0 )σ 1 (t 0 )σ 2 (t 0 ) +o() (σ 2 1 (t 0 ) +o())(σ 2 2 (t 0 ) +o()) = ρ(t 0 ). This proves part (vi). Part (i) and (ii) are consequences of general cases. 4.18. (i) Proof. d(e rt ζ t ) = (de −θWt− 1 2 θ 2 t ) = −e −θWt− 1 2 θ 2 t θdW t = −θ(e rt ζ t )dW t , where for the second “=”, we used the fact that e −θWt− 1 2 θ 2 t solves dX t = −θX t dW t . Since d(e rt ζ t ) = re rt ζ t dt +e rt dζ t , we get dζ t = −θζ t dW t −rζ t dt. (ii) Proof. d(ζ t X t ) = ζ t dX t +X t dζ t +dX t dζ t = ζ t (rX t dt + ∆ t (α −r)S t dt + ∆ t σS t dW t ) +X t (−θζ t dW t −rζ t dt) +(rX t dt + ∆ t (α −r)S t dt + ∆ t σS t dW t )(−θζ t dW t −rζ t dt) = ζ t (∆ t (α −r)S t dt + ∆ t σS t dW t ) −θX t ζ t dW t −θ∆ t σS t ζ t dt = ζ t ∆ t σS t dW t −θX t ζ t dW t . So ζ t X t is a martingale. (iii) Proof. By part (ii), X 0 = ζ 0 X 0 = E[ζ T X t ] = E[ζ T V T ]. (This can be seen as a version of risk-neutral pricing, only that the pricing is carried out under the actual probability measure.) 4.19. (i) Proof. B t is a local martingale with [B] t = t 0 sign(W s ) 2 ds = t. So by L´evy’s theorem, B t is a Brownian motion. 43 (ii) Proof. d(B t W t ) = B t dW t + sign(W t )W t dW t + sign(W t )dt. Integrate both sides of the resulting equation and the expectation, we get E[B t W t ] = t 0 E[sign(W s )]ds = t 0 E[1 {Ws≥0} −1 {Ws<0} ]ds = 1 2 t − 1 2 t = 0. (iii) Proof. By Itˆ o’s formula, dW 2 t = 2W t dW t +dt. (iv) Proof. By Itˆ o’s formula, d(B t W 2 t ) = B t dW 2 t +W 2 t dB t +dB t dW 2 t = B t (2W t dW t +dt) +W 2 t sign(W t )dW t +sign(W t )dW t (2W t dW t +dt) = 2B t W t dW t +B t dt +sign(W t )W 2 t dW t + 2sign(W t )W t dt. So E[B t W 2 t ] = E[ t 0 B s ds] + 2E[ t 0 sign(W s )W s ds] = t 0 E[B s ]ds + 2 t 0 E[sign(W s )W s ]ds = 2 t 0 (E[W s 1 {Ws≥0} ] −E[W s 1 {Ws<0} ])ds = 4 t 0 ∞ 0 x e − x 2 2s √ 2πs dxds = 4 t 0 s 2π ds = 0 = E[B t ] E[W 2 t ]. Since E[B t W 2 t ] = E[B t ] E[W 2 t ], B t and W t are not independent. 4.20. (i) Proof. f(x) = x −K, if x ≥ K 0, if x < K. So f (x) = 1, if x > K undefined, if x = K 0, if x < K and f (x) = 0, if x = K undefined, if x = K. (ii) Proof. E[f(W T )] = ∞ K (x − K) e − x 2 2T √ 2πT dx = T 2π e − K 2 2T − KΦ(− K √ T ) where Φ is the distribution function of standard normal random variable. If we suppose T 0 f (W t )dt = 0, the expectation of RHS of (4.10.42) is equal to 0. So (4.10.42) cannot hold. (iii) 44 Proof. This is trivial to check. (iv) Proof. If x = K, lim n→∞ f n (x) = 1 8n = 0; if x > K, for n large enough, x ≥ K + 1 2n , so lim n→∞ f n (x) = lim n→∞ (x −K) = x −K; if x < K, for n large enough, x ≤ K − 1 2n , so lim n→∞ f n (x) = lim n→∞ 0 = 0. In summary, lim n→∞ f n (x) = (x −K) + . Similarly, we can show lim n→∞ f n (x) = 0, if x < K 1 2 , if x = K 1, if x > K. (5) (v) Proof. Fix ω, so that W t (ω) < K for any t ∈ [0, T]. Since W t (ω) can obtain its maximum on [0, T], there exists n 0 , so that for any n ≥ n 0 , max 0≤t≤T W t (ω) < K − 1 2n . So L K (T)(ω) = lim n→∞ n T 0 1 (K− 1 2n ,K+ 1 2n ) (W t (ω))dt = 0. (vi) Proof. Take expectation on both sides of the formula (4.10.45), we have E[L K (T)] = E[(W T −K) + ] > 0. So we cannot have L K (T) = 0 a.s.. 4.21. (i) Proof. There are two problems. First, the transaction cost could be big due to active trading; second, the purchases and sales cannot be made at exactly the same price K. For more details, see Hull [2]. (ii) Proof. No. The RHS of (4.10.26) is a martingale, so its expectation is 0. But E[(S T − K) + ] > 0. So X T = (S T −K) + . 5. Risk-Neutral Pricing 5.1. (i) Proof. df(X t ) = f (X t )dt + 1 2 f (X t )d'X` t = f(X t )(dX t + 1 2 d'X` t ) = f(X t ) ¸ σ t dW t + (α t −R t − 1 2 σ 2 t )dt + 1 2 σ 2 t dt = f(X t )(α t −R t )dt +f(X t )σ t dW t . This is formula (5.2.20). 45 (ii) Proof. d(D t S t ) = S t dD t + D t dS t + dD t dS t = −S t R t D t dt + D t α t S t dt + D t σ t S t dW t = D t S t (α t − R t )dt + D t S t σ t dW t . This is formula (5.2.20). 5.2. Proof. By Lemma 5.2.2., ¯ E[D T V T [T t ] = E D T V T Z T Zt [T t . Therefore (5.2.30) is equivalent to D t V t Z t = E[D T V T Z T [T t ]. 5.3. (i) Proof. c x (0, x) = d dx ¯ E[e −rT (xe σ W T +(r− 1 2 σ 2 )T −K) + ] = ¯ E ¸ e −rT d dx h(xe σ W T +(r− 1 2 σ 2 )T ) = ¯ E e −rT e σ W T +(r− 1 2 σ 2 )T 1 {xe σ W T +(r− 1 2 σ 2 )T >K} = e − 1 2 σ 2 T ¯ E e σ W T 1 { W T > 1 σ (ln K x −(r− 1 2 σ 2 )T)} = e − 1 2 σ 2 T ¯ E ¸ e σ √ T W T √ T 1 { W T √ T −σ √ T> 1 σ √ T (ln K x −(r− 1 2 σ 2 )T)−σ √ T} = e − 1 2 σ 2 T ∞ −∞ 1 √ 2π e − z 2 2 e σ √ Tz 1 {z−σ √ T>−d+(T,x)} dz = ∞ −∞ 1 √ 2π e − (z−σ √ T) 2 2 1 {z−σ √ T>−d+(T,x)} dz = N(d + (T, x)). (ii) Proof. If we set ´ Z T = e σ W T − 1 2 σ 2 T and ´ Z t = ¯ E[ ´ Z T [T t ], then ´ Z is a ¯ P-martingale, ´ Z t > 0 and E[ ´ Z T ] = ¯ E[e σ W T − 1 2 σ 2 T ] = 1. So if we define ´ P by d ´ P = Z T d ¯ P on T T , then ´ P is a probability measure equivalent to ¯ P, and c x (0, x) = ¯ E[ ´ Z T 1 {S T >K} ] = ´ P(S T > K). Moreover, by Girsanov’s Theorem, ´ W t = W t + t 0 (−σ)du = W t − σt is a ´ P-Brownian motion (set Θ = −σ in Theorem 5.4.1.) (iii) Proof. S T = xe σ W T +(r− 1 2 σ 2 )T = xe σ W T +(r+ 1 2 σ 2 )T . So ´ P(S T > K) = ´ P(xe σ W T +(r+ 1 2 σ 2 )T > K) = ´ P ´ W T √ T > −d + (T, x) = N(d + (T, x)). 46 5.4. First, a few typos. In the SDE for S, “σ(t)d W(t)” → “σ(t)S(t)d W(t)”. In the first equation for c(0, S(0)), E → ¯ E. In the second equation for c(0, S(0)), the variable for BSM should be BSM ¸ T, S(0); K, 1 T T 0 r(t)dt, 1 T T 0 σ 2 (t)dt ¸ . (i) Proof. d ln S t = dSt St − 1 2S 2 t d'S` t = r t dt +σ t d W t − 1 2 σ 2 t dt. So S T = S 0 exp¦ T 0 (r t − 1 2 σ 2 t )dt + T 0 σ t d W t ¦. Let X = T 0 (r t − 1 2 σ 2 t )dt + T 0 σ t d W t . The first term in the expression of X is a number and the second term is a Gaussian random variable N(0, T 0 σ 2 t dt), since both r and σ ar deterministic. Therefore, S T = S 0 e X , with X ∼ N( T 0 (r t − 1 2 σ 2 t )dt, T 0 σ 2 t dt),. (ii) Proof. For the standard BSM model with constant volatility Σ and interest rate R, under the risk-neutral measure, we have S T = S 0 e Y , where Y = (R− 1 2 Σ 2 )T +Σ W T ∼ N((R− 1 2 Σ 2 )T, Σ 2 T), and ¯ E[(S 0 e Y −K) + ] = e RT BSM(T, S 0 ; K, R, Σ). Note R = 1 T (E[Y ] + 1 2 V ar(Y )) and Σ = 1 T V ar(Y ), we can get ¯ E[(S 0 e Y −K) + ] = e E[Y ]+ 1 2 V ar(Y ) BSM T, S 0 ; K, 1 T E[Y ] + 1 2 V ar(Y ) , 1 T V ar(Y ) . So for the model in this problem, c(0, S 0 ) = e − T 0 rtdt ¯ E[(S 0 e X −K) + ] = e − T 0 rtdt e E[X]+ 1 2 V ar(X) BSM T, S 0 ; K, 1 T E[X] + 1 2 V ar(X) , 1 T V ar(X) = BSM ¸ T, S 0 ; K, 1 T T 0 r t dt, 1 T T 0 σ 2 t dt ¸ . 5.5. (i) Proof. Let f(x) = 1 x , then f (x) = − 1 x 2 and f (x) = 2 x 3 . Note dZ t = −Z t Θ t dW t , so d 1 Z t = f (Z t )dZ t + 1 2 f (Z t )dZ t dZ t = − 1 Z 2 t (−Z t )Θ t dW t + 1 2 2 Z 3 t Z 2 t Θ 2 t dt = Θ t Z t dW t + Θ 2 t Z t dt. (ii) Proof. By Lemma 5.2.2., for s, t ≥ 0 with s < t, M s = ¯ E[ M t [T s ] = E Zt Mt Zs [T s . That is, E[Z t M t [T s ] = Z s M s . So M = Z M is a P-martingale. (iii) 47 Proof. d M t = d M t 1 Z t = 1 Z t dM t +M t d 1 Z t +dM t d 1 Z t = Γ t Z t dW t + M t Θ t Z t dW t + M t Θ 2 t Z t dt + Γ t Θ t Z t dt. (iv) Proof. In part (iii), we have d M t = Γ t Z t dW t + M t Θ t Z t dW t + M t Θ 2 t Z t dt + Γ t Θ t Z t dt = Γ t Z t (dW t + Θ t dt) + M t Θ t Z t (dW t + Θ t dt). Let ¯ Γ t = Γt+MtΘt Zt , then d M t = ¯ Γ t d W t . This proves Corollary 5.3.2. 5.6. Proof. By Theorem 4.6.5, it suffices to show W i (t) is an T t -martingale under ¯ P and [ W i , W j ](t) = tδ ij (i, j = 1, 2). Indeed, for i = 1, 2, W i (t) is an T t -martingale under ¯ P if and only if W i (t)Z t is an T t -martingale under P, since ¯ E[ W i (t)[T s ] = E ¸ W i (t)Z t Z s [T s ¸ . By Itˆo’s product formula, we have d( W i (t)Z t ) = W i (t)dZ t +Z t d W i (t) +dZ t d W i (t) = W i (t)(−Z t )Θ(t) dW t +Z t (dW i (t) + Θ i (t)dt) + (−Z t Θ t dW t )(dW i (t) + Θ i (t)dt) = W i (t)(−Z t ) d ¸ j=1 Θ j (t)dW j (t) +Z t (dW i (t) + Θ i (t)dt) −Z t Θ i (t)dt = W i (t)(−Z t ) d ¸ j=1 Θ j (t)dW j (t) +Z t dW i (t) This shows W i (t)Z t is an T t -martingale under P. So W i (t) is an T t -martingale under ¯ P. Moreover, [ W i , W j ](t) = ¸ W i + · 0 Θ i (s)ds, W j + · 0 Θ j (s)ds (t) = [W i , W j ](t) = tδ ij . Combined, this proves the two-dimensional Girsanov’s Theorem. 5.7. (i) Proof. Let a be any strictly positive number. We define X 2 (t) = (a +X 1 (t))D(t) −1 . Then P X 2 (T) ≥ X 2 (0) D(T) = P(a +X 1 (T) ≥ a) = P(X 1 (T) ≥ 0) = 1, and P X 2 (T) > X2(0) D(T) = P(X 1 (T) > 0) > 0, since a is arbitrary, we have proved the claim of this problem. Remark: The intuition is that we invest the positive starting fund a into the money market account, and construct portfolio X 1 from zero cost. Their sum should be able to beat the return of money market account. (ii) 48 Proof. We define X 1 (t) = X 2 (t)D(t) −X 2 (0). Then X 1 (0) = 0, P(X 1 (T) ≥ 0) = P X 2 (T) ≥ X 2 (0) D(T) = 1, P(X 1 (T) > 0) = P X 2 (T) > X 2 (0) D(T) > 0. 5.8. The basic idea is that for any positive ¯ P-martingale M, dM t = M t 1 Mt dM t . By Martingale Repre- sentation Theorem, dM t = ¯ Γ t d W t for some adapted process ¯ Γ t . So dM t = M t ( Γt Mt )d W t , i.e. any positive martingale must be the exponential of an integral w.r.t. Brownian motion. Taking into account discounting factor and apply Itˆo’s product rule, we can show every strictly positive asset is a generalized geometric Brownian motion. (i) Proof. V t D t = ¯ E[e − T 0 Rudu V T [T t ] = ¯ E[D T V T [T t ]. So (D t V t ) t≥0 is a ¯ P-martingale. By Martingale Represen- tation Theorem, there exists an adapted process ¯ Γ t , 0 ≤ t ≤ T, such that D t V t = t 0 ¯ Γ s d W s , or equivalently, V t = D −1 t t 0 ¯ Γ s d W s . Differentiate both sides of the equation, we get dV t = R t D −1 t t 0 ¯ Γ s d W s dt +D −1 t ¯ Γ t d W t , i.e. dV t = R t V t dt + Γt Dt dW t . (ii) Proof. We prove the following more general lemma. Lemma 1. Let X be an almost surely positive random variable (i.e. X > 0 a.s.) defined on the probability space (Ω, (, P). Let T be a sub σ-algebra of (, then Y = E[X[T] > 0 a.s. Proof. By the property of conditional expectation Y t ≥ 0 a.s. Let A = ¦Y = 0¦, we shall show P(A) = 0. In- deed, note A ∈ T, 0 = E[Y I A ] = E[E[X[T]I A ] = E[XI A ] = E[X1 A∩{X≥1} ] + ¸ ∞ n=1 E[X1 A∩{ 1 n >X≥ 1 n+1 } ] ≥ P(A∩¦X ≥ 1¦)+ ¸ ∞ n=1 1 n+1 P(A∩¦ 1 n > X ≥ 1 n+1 ¦). So P(A∩¦X ≥ 1¦) = 0 and P(A∩¦ 1 n > X ≥ 1 n+1 ¦) = 0, ∀n ≥ 1. This in turn implies P(A) = P(A∩¦X > 0¦) = P(A∩¦X ≥ 1¦) + ¸ ∞ n=1 P(A∩¦ 1 n > X ≥ 1 n+1 ¦) = 0. By the above lemma, it is clear that for each t ∈ [0, T], V t = ¯ E[e − T t Rudu V T [T t ] > 0 a.s.. Moreover, by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3.4)), we have the following stronger result: for a.s. ω, V t (ω) > 0 for any t ∈ [0, T]. (iii) Proof. By (ii), V > 0 a.s., so dV t = V t 1 Vt dV t = V t 1 Vt R t V t dt + Γt Dt d W t = V t R t dt + V t Γt VtDt d W t = R t V t dt + σ t V t d W t , where σ t = Γt VtDt . This shows V follows a generalized geometric Brownian motion. 5.9. Proof. c(0, T, x, K) = xN(d + ) − Ke −rT N(d − ) with d ± = 1 σ √ T (ln x K + (r ± 1 2 σ 2 )T). Let f(y) = 1 √ 2π e − y 2 2 , then f (y) = −yf(y), c K (0, T, x, K) = xf(d + ) ∂d + ∂y −e −rT N(d − ) −Ke −rT f(d − ) ∂d − ∂y = xf(d + ) −1 σ √ TK −e −rT N(d − ) +e −rT f(d − ) 1 σ √ T , 49 and c KK (0, T, x, K) = xf(d + ) 1 σ √ TK 2 − x σ √ TK f(d + )(−d + ) ∂d + ∂y −e −rT f(d − ) ∂d − ∂y + e −rT σ √ T (−d − )f(d − ) d − ∂y = x σ √ TK 2 f(d + ) + xd + σ √ TK f(d + ) −1 Kσ √ T −e −rT f(d − ) −1 Kσ √ T − e −rT d − σ √ T f(d − ) −1 Kσ √ T = f(d + ) x K 2 σ √ T [1 − d + σ √ T ] + e −rT f(d − ) Kσ √ T [1 + d − σ √ T ] = e −rT Kσ 2 T f(d − )d + − x K 2 σ 2 T f(d + )d − . 5.10. (i) Proof. At time t 0 , the value of the chooser option is V (t 0 ) = max¦C(t 0 ), P(t 0 )¦ = max¦C(t 0 ), C(t 0 ) − F(t 0 )¦ = C(t 0 ) + max¦0, −F(t 0 )¦ = C(t 0 ) + (e −r(T−t0) K −S(t 0 )) + . (ii) Proof. By the risk-neutral pricing formula, V (0) = ¯ E[e −rt0 V (t 0 )] = ¯ E[e −rt0 C(t 0 )+(e −rT K−e −rt0 S(t 0 ) + ] = C(0) + ¯ E[e −rt0 (e −r(T−t0) K − S(t 0 )) + ]. The first term is the value of a call expiring at time T with strike price K and the second term is the value of a put expiring at time t 0 with strike price e −r(T−t0) K. 5.11. Proof. We first make an analysis which leads to the hint, then we give a formal proof. (Analysis) If we want to construct a portfolio X that exactly replicates the cash flow, we must find a solution to the backward SDE dX t = ∆ t dS t +R t (X t −∆ t S t )dt −C t dt X T = 0. Multiply D t on both sides of the first equation and apply Itˆo’s product rule, we get d(D t X t ) = ∆ t d(D t S t ) − C t D t dt. Integrate from 0 to T, we have D T X T − D 0 X 0 = T 0 ∆ t d(D t S t ) − T 0 C t D t dt. By the terminal condition, we get X 0 = D −1 0 ( T 0 C t D t dt − T 0 ∆ t d(D t S t )). X 0 is the theoretical, no-arbitrage price of the cash flow, provided we can find a trading strategy ∆ that solves the BSDE. Note the SDE for S gives d(D t S t ) = (D t S t )σ t (θ t dt + dW t ), where θ t = αt−Rt σt . Take the proper change of measure so that W t = t 0 θ s ds +W t is a Brownian motion under the new measure ¯ P, we get T 0 C t D t dt = D 0 X 0 + T 0 ∆ t d(D t S t ) = D 0 X 0 + T 0 ∆ t (D t S t )σ t d W t . This says the random variable T 0 C t D t dt has a stochastic integral representation D 0 X 0 + T 0 ∆ t D t S t σ t d W t . This inspires us to consider the martingale generated by T 0 C t D t dt, so that we can apply Martingale Rep- resentation Theorem and get a formula for ∆ by comparison of the integrands. 50 (Formal proof) Let M T = T 0 C t D t dt, and M t = ¯ E[M T [T t ]. Then by Martingale Representation Theo- rem, we can find an adapted process ¯ Γ t , so that M t = M 0 + t 0 ¯ Γ t d W t . If we set ∆ t = Γt DtStσt , we can check X t = D −1 t (D 0 X 0 + t 0 ∆ u d(D u S u ) − t 0 C u D u du), with X 0 = M 0 = ¯ E[ T 0 C t D t dt] solves the SDE dX t = ∆ t dS t +R t (X t −∆ t S t )dt −C t dt X T = 0. Indeed, it is easy to see that X satisfies the first equation. To check the terminal condition, we note X T D T = D 0 X 0 + T 0 ∆ t D t S t σ t d W t − T 0 C t D t dt = M 0 + T 0 ¯ Γ t d W t −M T = 0. So X T = 0. Thus, we have found a trading strategy ∆, so that the corresponding portfolio X replicates the cash flow and has zero terminal value. So X 0 = ¯ E[ T 0 C t D t dt] is the no-arbitrage price of the cash flow at time zero. Remark: As shown in the analysis, d(D t X t ) = ∆ t d(D t S t ) − C t D t dt. Integrate from t to T, we get 0 − D t X t = T t ∆ u d(D u S u ) − T t C u D u du. Take conditional expectation w.r.t. T t on both sides, we get −D t X t = − ¯ E[ T t C u D u du[T t ]. So X t = D −1 t ¯ E[ T t C u D u du[T t ]. This is the no-arbitrage price of the cash flow at time t, and we have justified formula (5.6.10) in the textbook. 5.12. (i) Proof. d ¯ B i (t) = dB i (t) +γ i (t)dt = ¸ d j=1 σij(t) σi(t) dW j (t) + ¸ d j=1 σij(t) σi(t) Θ j (t)dt = ¸ d j=1 σij(t) σi(t) d W j (t). So B i is a martingale. Since d ¯ B i (t)d ¯ B i (t) = ¸ d j=1 σij(t) 2 σi(t) 2 dt = dt, by L´evy’s Theorem, ¯ B i is a Brownian motion under ¯ P. (ii) Proof. dS i (t) = R(t)S i (t)dt +σ i (t)S i (t)d ¯ B i (t) + (α i (t) −R(t))S i (t)dt −σ i (t)S i (t)γ i (t)dt = R(t)S i (t)dt +σ i (t)S i (t)d ¯ B i (t) + d ¸ j=1 σ ij (t)Θ j (t)S i (t)dt −S i (t) d ¸ j=1 σ ij (t)Θ j (t)dt = R(t)S i (t)dt +σ i (t)S i (t)d ¯ B i (t). (iii) Proof. d ¯ B i (t)d ¯ B k (t) = (dB i (t) +γ i (t)dt)(dB j (t) +γ j (t)dt) = dB i (t)dB j (t) = ρ ik (t)dt. (iv) Proof. By Itˆ o’s product rule and martingale property, E[B i (t)B k (t)] = E[ t 0 B i (s)dB k (s)] +E[ t 0 B k (s)dB i (s)] +E[ t 0 dB i (s)dB k (s)] = E[ t 0 ρ ik (s)ds] = t 0 ρ ik (s)ds. Similarly, by part (iii), we can show ¯ E[ ¯ B i (t) ¯ B k (t)] = t 0 ρ ik (s)ds. (v) 51 Proof. By Itˆ o’s product formula, E[B 1 (t)B 2 (t)] = E[ t 0 sign(W 1 (u))du] = t 0 [P(W 1 (u) ≥ 0) −P(W 1 (u) < 0)]du = 0. Meanwhile, ¯ E[ ¯ B 1 (t) ¯ B 2 (t)] = ¯ E[ t 0 sign(W 1 (u))du = t 0 [ ¯ P(W 1 (u) ≥ 0) − ¯ P(W 1 (u) < 0)]du = t 0 [ ¯ P( W 1 (u) ≥ u) − ¯ P( W 1 (u) < u)]du = t 0 2 1 2 − ¯ P( W 1 (u) < u) du < 0, for any t > 0. So E[B 1 (t)B 2 (t)] = ¯ E[ ¯ B 1 (t) ¯ B 2 (t)] for all t > 0. 5.13. (i) Proof. ¯ E[W 1 (t)] = ¯ E[ W 1 (t)] = 0 and ¯ E[W 2 (t)] = ¯ E[ W 2 (t) − t 0 W 1 (u)du] = 0, for all t ∈ [0, T]. (ii) Proof. ¯ Cov[W 1 (T), W 2 (T)] = ¯ E[W 1 (T)W 2 (T)] = ¯ E ¸ T 0 W 1 (t)dW 2 (t) + T 0 W 2 (t)dW 1 (t) ¸ = ¯ E ¸ T 0 W 1 (t)(d W 2 (t) − W 1 (t)dt) ¸ + ¯ E ¸ T 0 W 2 (t)d W 1 (t) ¸ = − ¯ E ¸ T 0 W 1 (t) 2 dt ¸ = − T 0 tdt = − 1 2 T 2 . 5.14. Equation (5.9.6) can be transformed into d(e −rt X t ) = ∆ t [d(e −rt S t ) −ae −rt dt] = ∆ t e −rt [dS t −rS t dt − adt]. So, to make the discounted portfolio value e −rt X t a martingale, we are motivated to change the measure in such a way that S t −r t 0 S u du−at is a martingale under the new measure. To do this, we note the SDE for S is dS t = α t S t dt+σS t dW t . Hence dS t −rS t dt−adt = [(α t −r)S t −a]dt+σS t dW t = σS t (αt−r)St−a σSt dt +dW t . Set θ t = (αt−r)St−a σSt and W t = t 0 θ s ds +W t , we can find an equivalent probability measure ¯ P, under which S satisfies the SDE dS t = rS t dt +σS t d W t +adt and W t is a BM. This is the rational for formula (5.9.7). This is a good place to pause and think about the meaning of “martingale measure.” What is to be a martingale? The new measure ¯ P should be such that the discounted value process of the replicating 52 portfolio is a martingale, not the discounted price process of the underlying. First, we want D t X t to be a martingale under ¯ P because we suppose that X is able to replicate the derivative payoff at terminal time, X T = V T . In order to avoid arbitrage, we must have X t = V t for any t ∈ [0, T]. The difficulty is how to calculate X t and the magic is brought by the martingale measure in the following line of reasoning: V t = X t = D −1 t ¯ E[D T X T [T t ] = D −1 t ¯ E[D T V T [T t ]. You can think of martingale measure as a calculational convenience. That is all about martingale measure! Risk neutral is a just perception, referring to the actual effect of constructing a hedging portfolio! Second, we note when the portfolio is self-financing, the discounted price process of the underlying is a martingale under ¯ P, as in the classical Black-Scholes-Merton model without dividends or cost of carry. This is not a coincidence. Indeed, we have in this case the relation d(D t X t ) = ∆ t d(D t S t ). So D t X t being a martingale under ¯ P is more or less equivalent to D t S t being a martingale under ¯ P. However, when the underlying pays dividends, or there is cost of carry, d(D t X t ) = ∆ t d(D t S t ) no longer holds, as shown in formula (5.9.6). The portfolio is no longer self-financing, but self-financing with consumption. What we still want to retain is the martingale property of D t X t , not that of D t S t . This is how we choose martingale measure in the above paragraph. Let V T be a payoff at time T, then for the martingale M t = ¯ E[e −rT V T [T t ], by Martingale Representation Theorem, we can find an adapted process ¯ Γ t , so that M t = M 0 + t 0 ¯ Γ s d W s . If we let ∆ t = Γte rt σSt , then the value of the corresponding portfolio X satisfies d(e −rt X t ) = ¯ Γ t d W t . So by setting X 0 = M 0 = ¯ E[e −rT V T ], we must have e −rt X t = M t , for all t ∈ [0, T]. In particular, X T = V T . Thus the portfolio perfectly hedges V T . This justifies the risk-neutral pricing of European-type contingent claims in the model where cost of carry exists. Also note the risk-neutral measure is different from the one in case of no cost of carry. Another perspective for perfect replication is the following. We need to solve the backward SDE dX t = ∆ t dS t −a∆ t dt +r(X t −∆ t S t )dt X T = V T for two unknowns, X and ∆. To do so, we find a probability measure ¯ P, under which e −rt X t is a martingale, then e −rt X t = ¯ E[e −rT V T [T t ] := M t . Martingale Representation Theorem gives M t = M 0 + t 0 ¯ Γ u d W u for some adapted process ¯ Γ. This would give us a theoretical representation of ∆ by comparison of integrands, hence a perfect replication of V T . (i) Proof. As indicated in the above analysis, if we have (5.9.7) under ¯ P, then d(e −rt X t ) = ∆ t [d(e −rt S t ) − ae −rt dt] = ∆ t e −rt σS t d W t . So (e −rt X t ) t≥0 , where X is given by (5.9.6), is a ¯ P-martingale. (ii) Proof. By Itˆo’s formula, dY t = Y t [σd W t + (r − 1 2 σ 2 )dt] + 1 2 Y t σ 2 dt = Y t (σd W t + rdt). So d(e −rt Y t ) = σe −rt Y t d W t and e −rt Y t is a ¯ P-martingale. Moreover, if S t = S 0 Y t +Y t t 0 a Ys ds, then dS t = S 0 dY t + t 0 a Y s dsdY t +adt = S 0 + t 0 a Y s ds Y t (σd W t +rdt) +adt = S t (σd W t +rdt) +adt. This shows S satisfies (5.9.7). Remark: To obtain this formula for S, we first set U t = e −rt S t to remove the rS t dt term. The SDE for U is dU t = σU t d W t + ae −rt dt. Just like solving linear ODE, to remove U in the d W t term, we consider V t = U t e −σ Wt . Itˆo’s product formula yields dV t = e −σ Wt dU t +U t e −σ Wt (−σ)d W t + 1 2 σ 2 dt +dU t e −σ Wt (−σ)d W t + 1 2 σ 2 dt = e −σ Wt ae −rt dt − 1 2 σ 2 V t dt. 53 Note V appears only in the dt term, so multiply the integration factor e 1 2 σ 2 t on both sides of the equation, we get d(e 1 2 σ 2 t V t ) = ae −rt−σ Wt+ 1 2 σ 2 t dt. Set Y t = e σ Wt+(r− 1 2 σ 2 )t , we have d(S t /Y t ) = adt/Y t . So S t = Y t (S 0 + t 0 ads Ys ). (iii) Proof. ¯ E[S T [T t ] = S 0 ¯ E[Y T [T t ] + ¯ E ¸ Y T t 0 a Y s ds +Y T T t a Y s ds[T t ¸ = S 0 ¯ E[Y T [T t ] + t 0 a Y s ds ¯ E[Y T [T t ] +a T t ¯ E ¸ Y T Y s [T t ds = S 0 Y t ¯ E[Y T−t ] + t 0 a Y s dsY t ¯ E[Y T−t ] +a T t ¯ E[Y T−s ]ds = S 0 + t 0 a Y s ds Y t e r(T−t) +a T t e r(T−s) ds = S 0 + t 0 ads Y s Y t e r(T−t) − a r (1 −e r(T−t) ). In particular, ¯ E[S T ] = S 0 e rT − a r (1 −e rT ). (iv) Proof. d ¯ E[S T [T t ] = ae r(T−t) dt + S 0 + t 0 ads Y s (e r(T−t) dY t −rY t e r(T−t) dt) + a r e r(T−t) (−r)dt = S 0 + t 0 ads Y s e r(T−t) σY t d W t . So ¯ E[S T [T t ] is a ¯ P-martingale. As we have argued at the beginning of the solution, risk-neutral pricing is valid even in the presence of cost of carry. So by an argument similar to that of '5.6.2, the process ¯ E[S T [T t ] is the futures price process for the commodity. (v) Proof. We solve the equation ¯ E[e −r(T−t) (S T − K)[T t ] = 0 for K, and get K = ¯ E[S T [T t ]. So For S (t, T) = Fut S (t, T). (vi) Proof. We follow the hint. First, we solve the SDE dX t = dS t −adt +r(X t −S t )dt X 0 = 0. By our analysis in part (i), d(e −rt X t ) = d(e −rt S t ) − ae −rt dt. Integrate from 0 to t on both sides, we get X t = S t − S 0 e rt + a r (1 − e rt ) = S t − S 0 e rt − a r (e rt − 1). In particular, X T = S T − S 0 e rT − a r (e rT − 1). Meanwhile, For S (t, T) = Fut s (t, T) = ¯ E[S T [T t ] = S 0 + t 0 ads Ys Y t e r(T−t) − a r (1−e r(T−t) ). So For S (0, T) = S 0 e rT − a r (1 −e rT ) and hence X T = S T −For S (0, T). After the agent delivers the commodity, whose value is S T , and receives the forward price For S (0, T), the portfolio has exactly zero value. 54 6. Connections with Partial Differential Equations 6.1. (i) Proof. Z t = 1 is obvious. Note the form of Z is similar to that of a geometric Brownian motion. So by Itˆo’s formula, it is easy to obtain dZ u = b u Z u du +σ u Z u dW u , u ≥ t. (ii) Proof. If X u = Y u Z u (u ≥ t), then X t = Y t Z t = x 1 = x and dX u = Y u dZ u +Z u dY u +dY u Z u = Y u (b u Z u du +σ u Z u dW u ) +Z u a u −σ u γ u Z u du + γ u Z u dW u +σ u Z u γ u Z u du = [Y u b u Z u + (a u −σ u γ u ) +σ u γ u ]du + (σ u Z u Y u +γ u )dW u = (b u X u +a u )du + (σ u X u +γ u )dW u . Remark: To see how to find the above solution, we manipulate the equation (6.2.4) as follows. First, to remove the term b u X u du, we multiply on both sides of (6.2.4) the integrating factor e − u t bvdv . Then d(X u e − u t bvdv ) = e − u t bvdv (a u du + (γ u +σ u X u )dW u ). Let ¯ X u = e − u t bvdv X u , ¯ a u = e − u t bvdv a u and ¯ γ u = e − u t bvdv γ u , then ¯ X satisfies the SDE d ¯ X u = ¯ a u du + (¯ γ u +σ u ¯ X u )dW u = (¯ a u du + ¯ γ u dW u ) +σ u ¯ X u dW u . To deal with the term σ u ¯ X u dW u , we consider ˆ X u = ¯ X u e − u t σvdWv . Then d ˆ X u = e − u t σvdWv [(¯ a u du + ¯ γ u dW u ) +σ u ¯ X u dW u ] + ¯ X u e − u t σvdWv (−σ u )dW u + 1 2 e − u t σvdWv σ 2 u du +(¯ γ u +σ u ¯ X u )(−σ u )e − u t σvdWv du = ˆ a u du + ˆ γ u dW u +σ u ˆ X u dW u −σ u ˆ X u dW u + 1 2 ˆ X u σ 2 u du −σ u (ˆ γ u +σ u ˆ X u )du = (ˆ a u −σ u ˆ γ u − 1 2 ˆ X u σ 2 u )du + ˆ γ u dW u , where ˆ a u = ¯ a u e − u t σvdWv and ˆ γ u = ¯ γ u e − u t σvdWv . Finally, use the integrating factor e u t 1 2 σ 2 v dv , we have d ˆ X u e 1 2 u t σ 2 v dv = e 1 2 u t σ 2 v dv (d ˆ X u + ˆ X u 1 2 σ 2 u du) = e 1 2 u t σ 2 v dv [(ˆ a u −σ u ˆ γ u )du + ˆ γ u dW u ]. Write everything back into the original X, a and γ, we get d X u e − u t bvdv− u t σvdWv+ 1 2 u t σ 2 v dv = e 1 2 u t σ 2 v dv− u t σvdWv− u t bvdv [(a u −σ u γ u )du +γ u dW u ], i.e. d X u Z u = 1 Z u [(a u −σ u γ u )du +γ u dW u ] = dY u . This inspired us to try X u = Y u Z u . 6.2. (i) 55 Proof. The portfolio is self-financing, so for any t ≤ T 1 , we have dX t = ∆ 1 (t)df(t, R t , T 1 ) + ∆ 2 (t)df(t, R t , T 2 ) +R t (X t −∆ 1 (t)f(t, R t , T 1 ) −∆ 2 (t)f(t, R t , T 2 ))dt, and d(D t X t ) = −R t D t X t dt +D t dX t = D t [∆ 1 (t)df(t, R t , T 1 ) + ∆ 2 (t)df(t, R t , T 2 ) −R t (∆ 1 (t)f(t, R t , T 1 ) + ∆ 2 (t)f(t, R t , T 2 ))dt] = D t [∆ 1 (t) f t (t, R t , T 1 )dt +f r (t, R t , T 1 )dR t + 1 2 f rr (t, R t , T 1 )γ 2 (t, R t )dt +∆ 2 (t) f t (t, R t , T 2 )dt +f r (t, R t , T 2 )dR t + 1 2 f rr (t, R t , T 2 )γ 2 (t, R t )dt −R t (∆ 1 (t)f(t, R t , T 1 ) + ∆ 2 (t)f(t, R t , T 2 ))dt] = ∆ 1 (t)D t [−R t f(t, R t , T 1 ) +f t (t, R t , T 1 ) +α(t, R t )f r (t, R t , T 1 ) + 1 2 γ 2 (t, R t )f rr (t, R t , T 1 )]dt +∆ 2 (t)D t [−R t f(t, R t , T 2 ) +f t (t, R t , T 2 ) +α(t, R t )f r (t, R t , T 2 ) + 1 2 γ 2 (t, R t )f rr (t, R t , T 2 )]dt +D t γ(t, R t )[D t γ(t, R t )[∆ 1 (t)f r (t, R t , T 1 ) + ∆ 2 (t)f r (t, R t , T 2 )]]dW t = ∆ 1 (t)D t [α(t, R t ) −β(t, R t , T 1 )]f r (t, R t , T 1 )dt + ∆ 2 (t)D t [α(t, R t ) −β(t, R t , T 2 )]f r (t, R t , T 2 )dt +D t γ(t, R t )[∆ 1 (t)f r (t, R t , T 1 ) + ∆ 2 (t)f r (t, R t , T 2 )]dW t . (ii) Proof. Let ∆ 1 (t) = S t f r (t, R t , T 2 ) and ∆ 2 (t) = −S t f r (t, R t , T 1 ), then d(D t X t ) = D t S t [β(t, R t , T 2 ) −β(t, R t , T 1 )]f r (t, R t , T 1 )f r (t, R t , T 2 )dt = D t [[β(t, R t , T 1 ) −β(t, R t , T 2 )]f r (t, R t , T 1 )f r (t, R t , T 2 )[dt. Integrate from 0 to T on both sides of the above equation, we get D T X T −D 0 X 0 = T 0 D t [[β(t, R t , T 1 ) −β(t, R t , T 2 )]f r (t, R t , T 1 )f r (t, R t , T 2 )[dt. If β(t, R t , T 1 ) = β(t, R t , T 2 ) for some t ∈ [0, T], under the assumption that f r (t, r, T) = 0 for all values of r and 0 ≤ t ≤ T, D T X T − D 0 X 0 > 0. To avoid arbitrage (see, for example, Exercise 5.7), we must have for a.s. ω, β(t, R t , T 1 ) = β(t, R t , T 2 ), ∀t ∈ [0, T]. This implies β(t, r, T) does not depend on T. (iii) Proof. In (6.9.4), let ∆ 1 (t) = ∆(t), T 1 = T and ∆ 2 (t) = 0, we get d(D t X t ) = ∆(t)D t ¸ −R t f(t, R t , T) +f t (t, R t , T) +α(t, R t )f r (t, R t , T) + 1 2 γ 2 (t, R t )f rr (t, R t , T) dt +D t γ(t, R t )∆(t)f r (t, R t , T)dW t . This is formula (6.9.5). If f r (t, r, T) = 0, then d(D t X t ) = ∆(t)D t −R t f(t, R t , T) +f t (t, R t , T) + 1 2 γ 2 (t, R t )f rr (t, R t , T) dt. We choose ∆(t) = sign ¸ −R t f(t, R t , T) +f t (t, R t , T) + 1 2 γ 2 (t, R t )f rr (t, R t , T) ¸ . To avoid arbitrage in this case, we must have f t (t, R t , T) + 1 2 γ 2 (t, R t )f rr (t, R t , T) = R t f(t, R t , T), or equivalently, for any r in the range of R t , f t (t, r, T) + 1 2 γ 2 (t, r)f rr (t, r, T) = rf(t, r, T). 56 6.3. Proof. We note d ds e − s 0 bvdv C(s, T) = e − s 0 bvdv [C(s, T)(−b s ) +b s C(s, T) −1] = −e − s 0 bvdv . So integrate on both sides of the equation from t to T, we obtain e − T 0 bvdv C(T, T) −e − t 0 bvdv C(t, T) = − T t e − s 0 bvdv ds. Since C(T, T) = 0, we have C(t, T) = e t 0 bvdv T t e − s 0 bvdv ds = T t e t s bvdv ds. Finally, by A (s, T) = −a(s)C(s, T) + 1 2 σ 2 (s)C 2 (s, T), we get A(T, T) −A(t, T) = − T t a(s)C(s, T)ds + 1 2 T t σ 2 (s)C 2 (s, T)ds. Since A(T, T) = 0, we have A(t, T) = T t (a(s)C(s, T) − 1 2 σ 2 (s)C 2 (s, T))ds. 6.4. (i) Proof. By the definition of ϕ, we have ϕ (t) = e 1 2 σ 2 T t C(u,T)du 1 2 σ 2 (−1)C(t, T) = − 1 2 ϕ(t)σ 2 C(t, T). So C(t, T) = − 2ϕ (t) φ(t)σ 2 . Differentiate both sides of the equation ϕ (t) = − 1 2 ϕ(t)σ 2 C(t, T), we get ϕ (t) = − 1 2 σ 2 [ϕ (t)C(t, T) +ϕ(t)C (t, T)] = − 1 2 σ 2 [− 1 2 ϕ(t)σ 2 C 2 (t, T) +ϕ(t)C (t, T)] = 1 4 σ 4 ϕ(t)C 2 (t, T) − 1 2 σ 2 ϕ(t)C (t, T). So C (t, T) = 1 4 σ 4 ϕ(t)C 2 (t, T) −ϕ (t) / 1 2 ϕ(t)σ 2 = 1 2 σ 2 C 2 (t, T) − 2ϕ (t) σ 2 ϕ(t) . (ii) Proof. Plug formulas (6.9.8) and (6.9.9) into (6.5.14), we get − 2ϕ (t) σ 2 ϕ(t) + 1 2 σ 2 C 2 (t, T) = b(−1) 2ϕ (t) σ 2 ϕ(t) + 1 2 σ 2 C 2 (t, T) −1, i.e. ϕ (t) −bϕ (t) − 1 2 σ 2 ϕ(t) = 0. (iii) Proof. The characteristic equation of ϕ (t) − bϕ (t) − 1 2 σ 2 ϕ(t) = 0 is λ 2 − bλ − 1 2 σ 2 = 0, which gives two roots 1 2 (b ± √ b 2 + 2σ 2 ) = 1 2 b ±γ with γ = 1 2 √ b 2 + 2σ 2 . Therefore by standard theory of ordinary differential equations, a general solution of ϕ is ϕ(t) = e 1 2 bt (a 1 e γt + a 2 e −γt ) for some constants a 1 and a 2 . It is then easy to see that we can choose appropriate constants c 1 and c 2 so that ϕ(t) = c 1 1 2 b +γ e −( 1 2 b+γ)(T−t) − c 2 1 2 b −γ e −( 1 2 b−γ)(T−t) . 57 (iv) Proof. From part (iii), it is easy to see ϕ (t) = c 1 e −( 1 2 b+γ)(T−t) −c 2 e −( 1 2 b−γ)(T−t) . In particular, 0 = C(T, T) = − 2ϕ (T) σ 2 ϕ(T) = − 2(c 1 −c 2 ) σ 2 ϕ(T) . So c 1 = c 2 . (v) Proof. We first recall the definitions and properties of sinh and cosh: sinh z = e z −e −z 2 , cosh z = e z +e −z 2 , (sinh z) = cosh z, and (cosh z) = sinh z. Therefore ϕ(t) = c 1 e − 1 2 b(T−t) ¸ e −γ(T−t) 1 2 b +γ − e γ(T−t) 1 2 b −γ = c 1 e − 1 2 b(T−t) ¸ 1 2 b −γ 1 4 b 2 −γ 2 e −γ(T−t) − 1 2 b +γ 1 4 b 2 −γ 2 e γ(T−t) = 2c 1 σ 2 e − 1 2 b(T−t) ¸ −( 1 2 b −γ)e −γ(T−t) + ( 1 2 b +γ)e γ(T−t) = 2c 1 σ 2 e − 1 2 b(T−t) [b sinh(γ(T −t)) + 2γ cosh(γ(T −t))]. and ϕ (t) = 1 2 b 2c 1 σ 2 e − 1 2 b(T−t) [b sinh(γ(T −t)) + 2γ cosh(γ(T −t))] + 2c 1 σ 2 e − 1 2 b(T−t) [−γb cosh(γ(T −t)) −2γ 2 sinh(γ(T −t))] = 2c 1 e − 1 2 b(T−t) ¸ b 2 2σ 2 sinh(γ(T −t)) + bγ σ 2 cosh(γ(T −t)) − bγ σ 2 cosh(γ(T −t)) − 2γ 2 σ 2 sinh(γ(T −t)) = 2c 1 e − 1 2 b(T−t) b 2 −4γ 2 2σ 2 sinh(γ(T −t)) = −2c 1 e − 1 2 b(T−t) sinh(γ(T −t)). This implies C(t, T) = − 2ϕ (t) σ 2 ϕ(t) = sinh(γ(T −t)) γ cosh(γ(T −t)) + 1 2 b sinh(γ(T −t)) . (vi) Proof. By (6.5.15) and (6.9.8), A (t, T) = 2aϕ (t) σ 2 ϕ(t) . Hence A(T, T) −A(t, T) = T t 2aϕ (s) σ 2 ϕ(s) ds = 2a σ 2 ln ϕ(T) ϕ(t) , and A(t, T) = − 2a σ 2 ln ϕ(T) ϕ(t) = − 2a σ 2 ln ¸ γe 1 2 b(T−t) γ cosh(γ(T −t)) + 1 2 b sinh(γ(T −t)) ¸ . 58 6.5. (i) Proof. Since g(t, X 1 (t), X 2 (t)) = E[h(X 1 (T), X 2 (T))[T t ] and e −rt f(t, X 1 (t), X 2 (t)) = E[e −rT h(X 1 (T), X 2 (T))[T t ], iterated conditioning argument shows g(t, X 1 (t), X 2 (t)) and e −rt f(t, X 1 (t), X 2 (t)) ar both martingales. (ii) and (iii) Proof. We note dg(t, X 1 (t), X 2 (t)) = g t dt +g x1 dX 1 (t) +g x2 dX 2 (t) + 1 2 g x1x2 dX 1 (t)dX 1 (t) + 1 2 g x2x2 dX 2 (t)dX 2 (t) +g x1x2 dX 1 (t)dX 2 (t) = ¸ g t +g x1 β 1 +g x2 β 2 + 1 2 g x1x1 (γ 2 11 +γ 2 12 + 2ργ 11 γ 12 ) +g x1x2 (γ 11 γ 21 +ργ 11 γ 22 +ργ 12 γ 21 +γ 12 γ 22 ) + 1 2 g x2x2 (γ 2 21 +γ 2 22 + 2ργ 21 γ 22 ) dt + martingale part. So we must have g t +g x1 β 1 +g x2 β 2 + 1 2 g x1x1 (γ 2 11 +γ 2 12 + 2ργ 11 γ 12 ) +g x1x2 (γ 11 γ 21 +ργ 11 γ 22 +ργ 12 γ 21 +γ 12 γ 22 ) + 1 2 g x2x2 (γ 2 21 +γ 2 22 + 2ργ 21 γ 22 ) = 0. Taking ρ = 0 will give part (ii) as a special case. The PDE for f can be similarly obtained. 6.6. (i) Proof. Multiply e 1 2 bt on both sides of (6.9.15), we get d(e 1 2 bt X j (t)) = e 1 2 bt X j (t) 1 2 bdt + (− b 2 X j (t)dt + 1 2 σdW j (t) = e 1 2 bt 1 2 σdW j (t). So e 1 2 bt X j (t) − X j (0) = 1 2 σ t 0 e 1 2 bu dW j (u) and X j (t) = e − 1 2 bt X j (0) + 1 2 σ t 0 e 1 2 bu dW j (u) . By Theorem 4.4.9, X j (t) is normally distributed with mean X j (0)e − 1 2 bt and variance e −bt 4 σ 2 t 0 e bu du = σ 2 4b (1 −e −bt ). (ii) Proof. Suppose R(t) = ¸ d j=1 X 2 j (t), then dR(t) = d ¸ j=1 (2X j (t)dX j (t) +dX j (t)dX j (t)) = d ¸ j=1 2X j (t)dX j (t) + 1 4 σ 2 dt = d ¸ j=1 −bX 2 j (t)dt +σX j (t)dW j (t) + 1 4 σ 2 dt = d 4 σ 2 −bR(t) dt +σ R(t) d ¸ j=1 X j (t) R(t) dW j (t). Let B(t) = ¸ d j=1 t 0 Xj(s) √ R(s) dW j (s), then B is a local martingale with dB(t)dB(t) = ¸ d j=1 X 2 j (t) R(t) dt = dt. So by L´evy’s Theorem, B is a Brownian motion. Therefore dR(t) = (a − bR(t))dt + σ R(t)dB(t) (a := d 4 σ 2 ) and R is a CIR interest rate process. 59 (iii) Proof. By (6.9.16), X j (t) is dependent on W j only and is normally distributed with mean e − 1 2 bt X j (0) and variance σ 2 4b [1 −e −bt ]. So X 1 (t), , X d (t) are i.i.d. normal with the same mean µ(t) and variance v(t). (iv) Proof. E e uX 2 j (t) = ∞ −∞ e ux 2 e − (x−µ(t)) 2 2v(t) dx 2πv(t) = ∞ −∞ e − (1−2uv(t))x 2 −2µ(t)x+µ 2 (t) 2v(t) 2πv(t) dx = ∞ −∞ 1 2πv(t) e − ( x− µ(t) 1−2uv(t) ) 2 + µ 2 (t) 1−2uv(t) − µ 2 (t) (1−2uv(t)) 2 2v(t)/(1−2uv(t)) dx = ∞ −∞ 1 −2uv(t) 2πv(t) e − ( x− µ(t) 1−2uv(t) ) 2 2v(t)/(1−2uv(t)) dx e − µ 2 (t)(1−2uv(t))−µ 2 (t) 2v(t)(1−2uv(t)) 1 −2uv(t) = e − uµ 2 (t) 1−2uv(t) 1 −2uv(t) . (v) Proof. By R(t) = ¸ d j=1 X 2 j (t) and the fact X 1 (t), , X d (t) are i.i.d., E[e uR(t) ] = (E[e uX 2 1 (t) ]) d = (1 −2uv(t)) − d 2 e udµ 2 (t) 1−2uv(t) = (1 −2uv(t)) − 2a σ 2 e − e −bt uR(0) 1−2uv(t) . 6.7. (i) Proof. e −rt c(t, S t , V t ) = ¯ E[e −rT (S T −K) + [T t ] is a martingale by iterated conditioning argument. Since d(e −rt c(t, S t , V t )) = e −rt ¸ c(t, S t , V t )(−r) +c t (t, S t , V t ) +c s (t, S t , V t )rS t +c v (t, S t , V t )(a −bV t ) + 1 2 c ss (t, S t , V t )V t S 2 t + 1 2 c vv (t, S t , V t )σ 2 V t +c sv (t, S t , V t )σV t S t ρ dt + martingale part, we conclude rc = c t +rsc s +c v (a −bv) + 1 2 c ss vs 2 + 1 2 c vv σ 2 v +c sv σsvρ. This is equation (6.9.26). (ii) 60 Proof. Suppose c(t, s, v) = sf(t, log s, v) −e −r(T−t) Kg(t, log s, v), then c t = sf t (t, log s, v) −re −r(T−t) Kg(t, log s, v) −e −r(T−t) Kg t (t, log s, v), c s = f(t, log s, v) +sf s (t, log s, v) 1 s −e −r(T−t) Kg s (t, log s, v) 1 s , c v = sf v (t, log s, v) −e −r(T−t) Kg v (t, log s, v), c ss = f s (t, log s, v) 1 s +f ss (t, log s, v) 1 s −e −r(T−t) Kg ss (t, log s, v) 1 s 2 +e −r(T−t) Kg s (t, log s, v) 1 s 2 , c sv = f v (t, log s, v) +f sv (t, log s, v) −e −r(T−t) K s g sv (t, log s, v), c vv = sf vv (t, log s, v) −e −r(T−t) Kg vv (t, log s, v). So c t +rsc s + (a −bv)c v + 1 2 s 2 vc ss +ρσsvc sv + 1 2 σ 2 vc vv = sf t −re −r(T−t) Kg −e −r(T−t) Kg t +rsf +rsf s −rKe −r(T−t) g s + (a −bv)(sf v −e −r(T−t) Kg v ) + 1 2 s 2 v ¸ − 1 s f s + 1 s f ss −e −r(T−t) K s 2 g ss +e −r(T−t) K g s s 2 +ρσsv f v +f sv −e −r(T−t) K s g sv + 1 2 σ 2 v(sf vv −e −r(T−t) Kg vv ) = s ¸ f t + (r + 1 2 v)f s + (a −bv +ρσv)f v + 1 2 vf ss +ρσvf sv + 1 2 σ 2 vf vv −Ke −r(T−t) ¸ g t + (r − 1 2 v)g s +(a −bv)g v + 1 2 vg ss +ρσvg sv + 1 2 σ 2 vg vv +rsf −re −r(T−t) Kg = rc. That is, c satisfies the PDE (6.9.26). (iii) Proof. First, by Markov property, f(t, X t , V t ) = E[1 {X T ≥log K} [T t ]. So f(T, X t , V t ) = 1 {X T ≥log K} , which implies f(T, x, v) = 1 {x≥log K} for all x ∈ R, v ≥ 0. Second, f(t, X t , V t ) is a martingale, so by differentiating f and setting the dt term as zero, we have the PDE (6.9.32) for f. Indeed, df(t, X t , V t ) = ¸ f t (t, X t , V t ) +f x (t, X t , V t )(r + 1 2 V t ) +f v (t, X t , V t )(a −bv t +ρσV t ) + 1 2 f xx (t, X t , V t )V t + 1 2 f vv (t, X t , V t )σ 2 V t +f xv (t, X t , V t )σV t ρ dt + martingale part. So we must have f t + (r + 1 2 v)f x + (a −bv +ρσv)f v + 1 2 f xx v + 1 2 f vv σ 2 v +σvρf xv = 0. This is (6.9.32). (iv) Proof. Similar to (iii). (v) Proof. c(T, s, v) = sf(T, log s, v) − e −r(T−t) Kg(T, log s, v) = s1 {log s≥log K} − K1 {log s≥log K} = 1 {s≥K} (s − K) = (s −K) + . 6.8. 61 Proof. We follow the hint. Suppose h is smooth and compactly supported, then it is legitimate to exchange integration and differentiation: g t (t, x) = ∂ ∂t ∞ 0 h(y)p(t, T, x, y)dy = ∞ 0 h(y)p t (t, T, x, y)dy, g x (t, x) = ∞ 0 h(y)p x (t, T, x, y)dy, g xx (t, x) = ∞ 0 h(y)p xx (t, T, x, y)dy. So (6.9.45) implies ∞ 0 h(y) p t (t, T, x, y) +β(t, x)p x (t, T, x, y) + 1 2 γ 2 (t, x)p xx (t, T, x, y) dy = 0. By the ar- bitrariness of h and assuming β, p t , p x , v, p xx are all continuous, we have p t (t, T, x, y) +β(t, x)p x (t, T, x, y) + 1 2 γ 2 (t, x)p xx (t, T, x, y) = 0. This is (6.9.43). 6.9. Proof. We first note dh b (X u ) = h b (X u )dX u + 1 2 h b (X u )dX u dX u = h b (X u )β(u, X u ) + 1 2 γ 2 (u, X u )h b (X u ) du+ h b (X u )γ(u, X u )dW u . Integrate on both sides of the equation, we have h b (X T ) −h b (X t ) = T t ¸ h b (X u )β(u, X u ) + 1 2 γ 2 (u, X u )h b (X u ) du + martingale part. Take expectation on both sides, we get E t,x [h b (X T ) −h b (X t )] = ∞ −∞ h b (y)p(t, T, x, y)dy −h(x) = T t E t,x [h b (X u )β(u, X u ) + 1 2 γ 2 (u, X u )h b (X u )]du = T t ∞ −∞ ¸ h b (y)β(u, y) + 1 2 γ 2 (u, y)h b (y) p(t, u, x, y)dydu. Since h b vanishes outside (0, b), the integration range can be changed from (−∞, ∞) to (0, b), which gives (6.9.48). By integration-by-parts formula, we have b 0 β(u, y)p(t, u, x, y)h b (y)dy = h b (y)β(u, y)p(t, u, x, y)[ b 0 − b 0 h b (y) ∂ ∂y (β(u, y)p(t, u, x, y))dy = − b 0 h b (y) ∂ ∂y (β(u, y)p(t, u, x, y))dy, and b 0 γ 2 (u, y)p(t, u, x, y)h b (y)dy = − b 0 ∂ ∂y (γ 2 (u, y)p(t, u, x, y))h b (y)dy = b 0 ∂ 2 ∂y (γ 2 (u, y)p(t, u, x, y))h b (y)dy. Plug these formulas into (6.9.48), we get (6.9.49). Differentiate w.r.t. T on both sides of (6.9.49), we have b 0 h b (y) ∂ ∂T p(t, T, x, y)dy = − b 0 ∂ ∂y [β(T, y)p(t, T, x, y)]h b (y)dy + 1 2 b 0 ∂ 2 ∂y 2 [γ 2 (T, y)p(t, T, x, y)]h b (y)dy, 62 that is, b 0 h b (y) ¸ ∂ ∂T p(t, T, x, y) + ∂ ∂y (β(T, y)p(t, T, x, y)) − 1 2 ∂ 2 ∂y 2 (γ 2 (T, y)p(t, T, x, y)) dy = 0. This is (6.9.50). By (6.9.50) and the arbitrariness of h b , we conclude for any y ∈ (0, ∞), ∂ ∂T p(t, T, x, y) + ∂ ∂y (β(T, y)p(t, T, x, y)) − 1 2 ∂ 2 ∂y 2 (γ 2 (T, y)p(t, T, x, y)) = 0. 6.10. Proof. Under the assumption that lim y→∞ (y −K)ry¯ p(0, T, x, y) = 0, we have − ∞ K (y−K) ∂ ∂y (ry¯ p(0, T, x, y))dy = −(y−K)ry¯ p(0, T, x, y)[ ∞ K + ∞ K ry¯ p(0, T, x, y)dy = ∞ K ry¯ p(0, T, x, y)dy. If we further assume (6.9.57) and (6.9.58), then use integration-by-parts formula twice, we have 1 2 ∞ K (y −K) ∂ 2 ∂y 2 (σ 2 (T, y)y 2 ¯ p(0, T, x, y))dy = 1 2 ¸ (y −K) ∂ ∂y (σ 2 (T, y)y 2 ¯ p(0, T, x, y))[ ∞ K − ∞ K ∂ ∂y (σ 2 (T, y)y 2 ¯ p(0, T, x, y))dy = − 1 2 (σ 2 (T, y)y 2 ¯ p(0, T, x, y)[ ∞ K ) = 1 2 σ 2 (T, K)K 2 ¯ p(0, T, x, K). Therefore, c T (0, T, x, K) = −rc(0, T, x, K) +e −rT ∞ K (y −K)¯ p T (0, T, x, y)dy = −re −rT ∞ K (y −K)¯ p(0, T, x, y)dy +e −rT ∞ K (y −K)¯ p T (0, T, x, y)dy = −re −rT ∞ K (y −K)¯ p(0, T, x, y)dy −e −rT ∞ K (y −K) ∂ ∂y (ry¯ p(t, T, x, y))dy +e −rT ∞ K (y −K) 1 2 ∂ 2 ∂y 2 (σ 2 (T, y)y 2 ¯ p(t, T, x, y))dy = −re −rT ∞ K (y −K)¯ p(0, T, x, y)dy +e −rT ∞ K ry¯ p(0, T, x, y)dy +e −rT 1 2 σ 2 (T, K)K 2 ¯ p(0, T, x, K) = re −rT K ∞ K ¯ p(0, T, x, y)dy + 1 2 e −rT σ 2 (T, K)K 2 ¯ p(0, T, x, K) = −rKc K (0, T, x, K) + 1 2 σ 2 (T, K)K 2 c KK (0, T, x, K). 7. Exotic Options 7.1. (i) 63 Proof. Since δ ± (τ, s) = 1 σ √ τ [log s + (r ± 1 2 σ 2 )τ] = log s σ τ − 1 2 + r± 1 2 σ 2 σ √ τ, ∂ ∂t δ ± (τ, s) = log s σ (− 1 2 )τ − 3 2 ∂τ ∂t + r ± 1 2 σ 2 σ 1 2 τ − 1 2 ∂τ ∂t = − 1 2τ ¸ log s σ 1 √ τ (−1) − r ± 1 2 σ 2 σ √ τ(−1) = − 1 2τ 1 σ √ τ ¸ −log ss + (r ± 1 2 σ 2 )τ) = − 1 2τ δ ± (τ, 1 s ). (ii) Proof. ∂ ∂x δ ± (τ, x c ) = ∂ ∂x 1 σ √ τ ¸ log x c + (r ± 1 2 σ 2 )τ = 1 xσ √ τ , ∂ ∂x δ ± (τ, c x ) = ∂ ∂x 1 σ √ τ ¸ log c x + (r ± 1 2 σ 2 )τ = − 1 xσ √ τ . (iii) Proof. N (δ ± (τ, s)) = 1 √ 2π e − δ ± (τ,s) 2 = 1 √ 2π e − (log s+rτ) 2 ±σ 2 τ(log s+rτ)+ 1 4 σ 4 τ 2 2σ 2 τ . Therefore N (δ + (τ, s)) N (δ − (τ, s)) = e − 2σ 2 τ(log s+rτ) 2σ 2 τ = e −rτ s and e −rτ N (δ − (τ, s)) = sN (δ + (τ, s)). (iv) Proof. N (δ ± (τ, s)) N (δ ± (τ, s −1 )) = e − [ (log s+rτ) 2 −(log 1 s +rτ) 2 ] ±σ 2 τ(log s−log 1 s ) 2σ 2 τ = e − 4rτ log s±2σ 2 τ log s 2σ 2 τ = e −( 2r σ 2 ±1) log s = s −( 2r σ 2 ±1) . So N (δ ± (τ, s −1 )) = s ( 2r σ 2 ±1) N (δ ± (τ, s)). (v) Proof. δ + (τ, s) −δ − (τ, s) = 1 σ √ τ log s + (r + 1 2 σ 2 )τ − 1 σ √ τ log s + (r − 1 2 σ 2 )τ = 1 σ √ τ σ 2 τ = σ √ τ. (vi) Proof. δ ± (τ, s) −δ ± (τ, s −1 ) = 1 σ √ τ log s + (r ± 1 2 σ 2 )τ − 1 σ √ τ log s −1 + (r ± 1 2 σ 2 )τ = 2 log s σ √ τ . (vii) Proof. N (y) = 1 √ 2π e − y 2 2 , so N (y) = 1 √ 2π e − y 2 2 (− y 2 2 ) = −yN (y). 64 To be continued ... 7.3. Proof. We note S T = S 0 e σ W T = S t e σ( W T − Wt) , ´ W T − ´ W t = ( W T − W t ) + α(T − t) is independent of T t , sup t≤u≤T ( ´ W u − ´ W t ) is independent of T t , and Y T = S 0 e σ M T = S 0 e σ sup t≤u≤T Wu 1 { Mt≤sup t≤u≤T Wt} +S 0 e σ Mt 1 { Mt>sup t≤u≤T Wu} = S t e σ sup t≤u≤T ( Wu− Wt) 1 { Y t S t ≤e σ sup t≤u≤T ( Wu− W t ) } +Y t 1 { Y t S t ≤e σ sup t≤u≤T ( Wu− W t ) } . So E[f(S T , Y T )[T t ] = E[f(x S T−t S0 , x Y T−t S0 1 { y x ≤ Y T−t S 0 } + y1 { y x ≤ Y T−t S 0 } )], where x = S t , y = Y t . Therefore E[f(S T , Y T )[T t ] is a Borel function of (S t , Y t ). 7.4. Proof. By Cauchy’s inequality and the monotonicity of Y , we have [ m ¸ j=1 (Y tj −Y tj−1 )(S tj −S tj−1 )[ ≤ m ¸ j=1 [Y tj −Y tj−1 [[S tj −S tj−1 [ ≤ m ¸ j=1 (Y tj −Y tj−1 ) 2 m ¸ j=1 (S tj −S tj−1 ) 2 ≤ max 1≤j≤m [Y tj −Y tj−1 [(Y T −Y 0 ) m ¸ j=1 (S tj −S tj−1 ) 2 . If we increase the number of partition points to infinity and let the length of the longest subinterval max 1≤j≤m [t j −t j−1 [ approach zero, then ¸ m j=1 (S tj −S tj−1 ) 2 → [S] T −[S] 0 < ∞ and max 1≤j≤m [Y tj − Y tj−1 [ → 0 a.s. by the continuity of Y . This implies ¸ m j=1 (Y tj −Y tj−1 )(S tj −S tj−1 ) → 0. 8. American Derivative Securities 8.1. Proof. v L (L+) = (K −L)(− 2r σ 2 )( x L ) − 2r σ 2 −1 1 L x=L = − 2r σ 2 L (K − L). So v L (L+) = v L (L−) if and only if − 2r σ 2 L (K −L) = −1. Solve for L, we get L = 2rK 2r+σ 2 . 8.2. Proof. By the calculation in Section 8.3.3, we can see v 2 (x) ≥ (K 2 −x) + ≥ (K 1 −x) + , rv 2 (x) −rxv 2 (x) − 1 2 σ 2 x 2 v 2 (x) ≥ 0 for all x ≥ 0, and for 0 ≤ x < L 1∗ < L 2∗ , rv 2 (x) −rxv 2 (x) − 1 2 σ 2 x 2 v 2 (x) = rK 2 > rK 1 > 0. So the linear complementarity conditions for v 2 imply v 2 (x) = (K 2 −x) + = K 2 −x > K 1 −x = (K 1 −x) + on [0, L 1∗ ]. Hence v 2 (x) does not satisfy the third linear complementarity condition for v 1 : for each x ≥ 0, equality holds in either (8.8.1) or (8.8.2) or both. 8.3. (i) 65 Proof. Suppose x takes its values in a domain bounded away from 0. By the general theory of linear differential equations, if we can find two linearly independent solutions v 1 (x), v 2 (x) of (8.8.4), then any solution of (8.8.4) can be represented in the form of C 1 v 1 +C 2 v 2 where C 1 and C 2 are constants. So it suffices to find two linearly independent special solutions of (8.8.4). Assume v(x) = x p for some constant p to be determined, (8.8.4) yields x p (r−pr− 1 2 σ 2 p(p−1)) = 0. Solve the quadratic equation 0 = r−pr− 1 2 σ 2 p(p−1) = (− 1 2 σ 2 p −r)(p −1), we get p = 1 or − 2r σ 2 . So a general solution of (8.8.4) has the form C 1 x +C 2 x − 2r σ 2 . (ii) Proof. Assume there is an interval [x 1 , x 2 ] where 0 < x 1 < x 2 < ∞, such that v(x) ≡ 0 satisfies (8.3.19) with equality on [x 1 , x 2 ] and satisfies (8.3.18) with equality for x at and immediately to the left of x 1 and for x at and immediately to the right of x 2 , then we can find some C 1 and C 2 , so that v(x) = C 1 x+C 2 x − 2r σ 2 on [x 1 , x 2 ]. If for some x 0 ∈ [x 1 , x 2 ], v(x 0 ) = v (x 0 ) = 0, by the uniqueness of the solution of (8.8.4), we would conclude v ≡ 0. This is a contradiction. So such an x 0 cannot exist. This implies 0 < x 1 < x 2 < K (if K ≤ x 2 , v(x 2 ) = (K −x 2 ) + = 0 and v (x 2 )=the right derivative of (K −x) + at x 2 , which is 0). 1 Thus we have four equations for C 1 and C 2 : C 1 x 1 +C 2 x − 2r σ 2 1 = K −x 1 C 1 x 2 +C 2 x − 2r σ 2 2 = K −x 2 C 1 − 2r σ 2 C 2 x − 2r σ 2 −1 1 = −1 C 1 − 2r σ 2 C 2 x − 2r σ 2 −1 2 = −1. Since x 1 = x 2 , the last two equations imply C 2 = 0. Plug C 2 = 0 into the first two equations, we have C 1 = K−x1 x1 = K−x2 x2 ; plug C 2 = 0 into the last two equations, we have C 1 = −1. Combined, we would have x 1 = x 2 . Contradiction. Therefore our initial assumption is incorrect, and the only solution v that satisfies the specified conditions in the problem is the zero solution. (iii) Proof. If in a right neighborhood of 0, v satisfies (8.3.19) with equality, then part (i) implies v(x) = C 1 x + C 2 x − 2r σ 2 for some constants C 1 and C 2 . Then v(0) = lim x↓0 v(x) = 0 < (K − 0) + , i.e. (8.3.18) will be violated. So we must have rv − rxv − 1 2 σ 2 x 2 v > 0 in a right neighborhood of 0. According to (8.3.20), v(x) = (K−x) + near o. So v(0) = K. We have thus concluded simultaneously that v cannot satisfy (8.3.19) with equality near 0 and v(0) = K, starting from first principles (8.3.18)-(8.3.20). (iv) Proof. This is already shown in our solution of part (iii): near 0, v cannot satisfy (8.3.19) with equality. (v) Proof. If v satisfy (K−x) + with equality for all x ≥ 0, then v cannot have a continuous derivative as stated in the problem. This is a contradiction. (vi) 1 Note we have interpreted the condition “v(x) satisfies (8.3.18) with equality for x at and immediately to the right of x 2 ” as “v(x 2 ) = (K − x 2 ) + and v (x 2 ) =the right derivative of (K − x) + at x 2 .” This is weaker than “v(x) = (K − x) in a right neighborhood of x 2 .” 66 Proof. By the result of part (i), we can start with v(x) = (K −x) + on [0, x 1 ] and v(x) = C 1 x +C 2 x − 2r σ 2 on [x 1 , ∞). By the assumption of the problem, both v and v are continuous. Since (K−x) + is not differentiable at K, we must have x 1 ≤ K.This gives us the equations K −x 1 = (K −x 1 ) + = C 1 x 1 +C 2 x − 2r σ 2 1 −1 = C 1 − 2r σ 2 C 2 x − 2r σ 2 −1 1 . Because v is assumed to be bounded, we must have C 1 = 0 and the above equations only have two unknowns: C 2 and x 1 . Solve them for C 2 and x 1 , we are done. 8.4. (i) Proof. This is already shown in part (i) of Exercise 8.3. (ii) Proof. We solve for A, B the equations AL − 2r σ 2 +BL = K −L − 2r σ 2 AL − 2r σ 2 −1 +B = −1, and we obtain A = σ 2 KL 2r σ 2 σ 2 +2r , B = 2rK L(σ 2 +2r) −1. (iii) Proof. By (8.8.5), B > 0. So for x ≥ K, f(x) ≥ BK > 0 = (K −x) + . If L ≤ x < K, f(x) −(K −x) + = σ 2 KL 2r σ 2 σ 2 + 2r x − 2r σ 2 + 2rKx L(σ 2 + 2r) −K = x − 2r σ 2 KL 2r σ 2 σ 2 + 2r( x L ) 2r σ 2 +1 −(σ 2 + 2r)( x L ) 2r σ 2 (σ 2 + 2r)L . Let g(θ) = σ 2 + 2rθ 2r σ 2 +1 − (σ 2 + 2r)θ 2r σ 2 with θ ≥ 1. Then g(1) = 0 and g (θ) = 2r( 2r σ 2 + 1)θ 2r σ 2 − (σ 2 + 2r) 2r σ 2 θ 2r σ 2 −1 = 2r σ 2 (σ 2 + 2r)θ 2r σ 2 −1 (θ − 1) ≥ 0. So g(θ) ≥ 0 for any θ ≥ 1. This shows f(x) ≥ (K − x) + for L ≤ x < K. Combined, we get f(x) ≥ (K −x) + for all x ≥ L. (iv) Proof. Since lim x→∞ v(x) = lim x→∞ f(x) = ∞ and lim x→∞ v L∗ (x) = lim x→∞ (K − L ∗ )( x L∗ ) − 2r σ 2 = 0, v(x) and v L∗ (x) are different. By part (iii), v(x) ≥ (K − x) + . So v satisfies (8.3.18). For x ≥ L, rv − rxv − 1 2 σ 2 x 2 v = rf −rxf − 1 2 σ 2 x 2 f = 0. For 0 ≤ x ≤ L, rv −rxv − 1 2 σ 2 x 2 v = r(K−x) +rx = rK. Combined, rv −rxv − 1 2 σ 2 x 2 v ≥ 0 for x ≥ 0. So v satisfies (8.3.19). Along the way, we also showed v satisfies (8.3.20). In summary, v satisfies the linear complementarity condition (8.3.18)-(8.3.20), but v is not the function v L∗ given by (8.3.13). (v) Proof. By part (ii), B = 0 if and only if 2rK L(σ 2 +2r) − 1 = 0, i.e. L = 2rK 2r+σ 2 . In this case, v(x) = Ax − 2r σ 2 = σ 2 K σ 2 +2r ( x L ) − 2r σ 2 = (K −L)( x L ) − 2r σ 2 = v L∗ (x), on the interval [L, ∞). 8.5. The difficulty of the dividend-paying case is that from Lemma 8.3.4, we can only obtain ¯ E[e −(r−a)τ L ], not ¯ E[e −rτ L ]. So we have to start from Theorem 8.3.2. (i) 67 Proof. By (8.8.9), S t = S 0 e σ Wt+(r−a− 1 2 σ 2 )t . Assume S 0 = x, then S t = L if and only if − W t − 1 σ (r − a − 1 2 σ 2 )t = 1 σ log x L . By Theorem 8.3.2, ¯ E[e −rτ L ] = e − 1 σ log x L 1 σ (r−a− 1 2 σ 2 )+ 1 σ 2 (r−a− 1 2 σ 2 ) 2 +2r . If we set γ = 1 σ 2 (r − a − 1 2 σ 2 ) + 1 σ 1 σ 2 (r −a − 1 σ 2 ) 2 + 2r, we can write ¯ E[e −rτ L ] as e −γ log x L = ( x L ) −γ . So the risk-neutral expected discounted pay off of this strategy is v L (x) = K −x, 0 ≤ x ≤ L (K −L)( x L ) −γ , x > L. (ii) Proof. ∂ ∂L v L (x) = −( x L ) −γ (1 − γ(K−L) L ). Set ∂ ∂L v L (x) = 0 and solve for L ∗ , we have L ∗ = γK γ+1 . (iii) Proof. By Itˆ o’s formula, we have d e −rt v L∗ (S t ) = e −rt ¸ −rv L∗ (S t ) +v L∗ (S t )(r −a)S t + 1 2 v L∗ (S t )σ 2 S 2 t dt +e −rt v L∗ (S t )σS t d W t . If x > L ∗ , −rv L∗ (x) +v L∗ (x)(r −a)x + 1 2 v L∗ (x)σ 2 x 2 = −r(K −L ∗ ) x L ∗ −γ + (r −a)x(K −L ∗ )(−γ) x −γ−1 L −γ ∗ + 1 2 σ 2 x 2 (−γ)(−γ −1)(K −L ∗ ) x −γ−2 L −γ ∗ = (K −L ∗ ) x L ∗ −γ ¸ −r −(r −a)γ + 1 2 σ 2 γ(γ + 1) . By the definition of γ, if we define u = r −a − 1 2 σ 2 , we have r + (r −a)γ − 1 2 σ 2 γ(γ + 1) = r − 1 2 σ 2 γ 2 +γ(r −a − 1 2 σ 2 ) = r − 1 2 σ 2 u σ 2 + 1 σ u 2 σ 2 + 2r 2 + u σ 2 + 1 σ u 2 σ 2 + 2r u = r − 1 2 σ 2 u 2 σ 4 + 2u σ 3 u 2 σ 2 + 2r + 1 σ 2 u 2 σ 2 + 2r + u 2 σ 2 + u σ u 2 σ 2 + 2r = r − u 2 2σ 2 − u σ u 2 σ 2 + 2r − 1 2 u 2 σ 2 + 2r + u 2 σ 2 + u σ u 2 σ 2 + 2r = 0. If x < L ∗ , −rv L∗ (x) +v L∗ (x)(r −a)x + 1 2 v L∗ (x)σ 2 x 2 = −r(K −x) +(−1)(r −a)x = −rK +ax. Combined, we get d e −rt v L∗ (S t ) = −e −rt 1 {St<L∗} (rK −aS t )dt +e −rt v L∗ (S t )σS t d W t . 68 Following the reasoning in the proof of Theorem 8.3.5, we only need to show 1 {x<L∗} (rK−ax) ≥ 0 to finish the solution. This is further equivalent to proving rK − aL ∗ ≥ 0. Plug L ∗ = γK γ+1 into the expression and note γ ≥ 1 σ 1 σ 2 (r −a − 1 2 σ 2 ) 2 + 1 σ 2 (r −a − 1 2 σ 2 ) ≥ 0, the inequality is further reduced to r(γ +1) −aγ ≥ 0. We prove this inequality as follows. Assume for some K, r, a and σ (K and σ are assumed to be strictly positive, r and a are assumed to be non-negative), rK − aL ∗ < 0, then necessarily r < a, since L ∗ = γK γ+1 ≤ K. As shown before, this means r(γ + 1) − aγ < 0. Define θ = r−a σ , then θ < 0 and γ = 1 σ 2 (r − a − 1 2 σ 2 ) + 1 σ 1 σ 2 (r −a − 1 2 σ 2 ) 2 + 2r = 1 σ (θ − 1 2 σ) + 1 σ (θ − 1 2 σ) 2 + 2r. We have r(γ + 1) −aγ < 0 ⇐⇒ (r −a)γ +r < 0 ⇐⇒ (r −a) ¸ 1 σ (θ − 1 2 σ) + 1 σ (θ − 1 2 σ) 2 + 2r ¸ +r < 0 ⇐⇒ θ(θ − 1 2 σ) +θ (θ − 1 2 σ) 2 + 2r +r < 0 ⇐⇒ θ (θ − 1 2 σ) 2 + 2r < −r −θ(θ − 1 2 σ)(< 0) ⇐⇒ θ 2 [(θ − 1 2 σ) 2 + 2r] > r 2 +θ 2 (θ − 1 2 σ) 2 + 2θr(θ − 1 2 σ 2 ) ⇐⇒ 0 > r 2 −θrσ 2 ⇐⇒ 0 > r −θσ 2 . Since θσ 2 < 0, we have obtained a contradiction. So our initial assumption is incorrect, and rK −aL ∗ ≥ 0 must be true. (iv) Proof. The proof is similar to that of Corollary 8.3.6. Note the only properties used in the proof of Corollary 8.3.6 are that e −rt v L∗ (S t ) is a supermartingale, e −rt∧τ L∗ v L∗ (S t ∧τ L∗ ) is a martingale, and v L∗ (x) ≥ (K−x) + . Part (iii) already proved the supermartingale-martingale property, so it suffices to show v L∗ (x) ≥ (K −x) + in our problem. Indeed, by γ ≥ 0, L ∗ = γK γ+1 < K. For x ≥ K > L ∗ , v L∗ (x) > 0 = (K−x) + ; for 0 ≤ x < L ∗ , v L∗ (x) = K −x = (K −x) + ; finally, for L ∗ ≤ x ≤ K, d dx (v L∗ (x) −(K −x)) = −γ(K −L ∗ ) x −γ−1 L −γ ∗ + 1 ≥ −γ(K −L ∗ ) L −γ−1 ∗ L −γ ∗ + 1 = −γ(K − γK γ + 1 ) 1 γK γ+1 + 1 = 0. and (v L∗ (x) − (K − x))[ x=L∗ = 0. So for L ∗ ≤ x ≤ K, v L∗ (x) − (K − x) + ≥ 0. Combined, we have v L∗ (x) ≥ (K −x) + ≥ 0 for all x ≥ 0. 8.6. Proof. By Lemma 8.5.1, X t = e −rt (S t −K) + is a submartingale. For any τ ∈ Γ 0,T , Theorem 8.8.1 implies ¯ E[e −rT (S T −K) + ] ≥ ¯ E[e −rτ∧T (S τ∧T −K) + ] ≥ E[e −rτ (S τ −K) + 1 {τ<∞} ] = E[e −rτ (S τ −K) + ], where we take the convention that e −rτ (S τ −K) + = 0 when τ = ∞. Since τ is arbitrarily chosen, ¯ E[e −rT (S T − K) + ] ≥ max τ∈Γ 0,T ¯ E[e −rτ (S τ −K) + ]. The other direction “≤” is trivial since T ∈ Γ 0,T . 8.7. 69 Proof. Suppose λ ∈ [0, 1] and 0 ≤ x 1 ≤ x 2 , we have f((1 − λ)x 1 + λx 2 ) ≤ (1 − λ)f(x 1 ) + λf(x 2 ) ≤ (1 −λ)h(x 1 ) +λh(x 2 ). Similarly, g((1 −λ)x 1 +λx 2 ) ≤ (1 −λ)h(x 1 ) +λh(x 2 ). So h((1 −λ)x 1 +λx 2 ) = max¦f((1 −λ)x 1 +λx 2 ), g((1 −λ)x 1 +λx 2 )¦ ≤ (1 −λ)h(x 1 ) +λh(x 2 ). That is, h is also convex. 9. Change of Num´eraire To provide an intuition for change of num´eraire, we give a summary of results for change of num´eraire in discrete case. This summary is based on Shiryaev [5]. Consider a model of financial market ( ¯ B, ¯ B, S) as in [1] Definition 2.1.1 or [5] page 383. Here ¯ B and ¯ B are both one-dimensional while S could be a vector price process. Suppose ¯ B and ¯ B are both strictly positive, then both of them can be chosen as num´eaire. Several results hold under this model. First, no-arbitrage and completeness properties of market are independent of the choice of num´eraire (see, for example, Shiryaev [5] page 413 Remark and page 481). Second, if the market is arbitrage-free, then corresponding to ¯ B (resp. ¯ B), there is an equivalent probability ¯ P (resp. ¯ P), such that ¯ B B , S B (resp. B ¯ B , S ¯ B ) is a martingale under ¯ P (resp. ¯ P). Third, if the market is both arbitrage-free and complete, we have the relation d ¯ P = ¯ B T ¯ B T 1 E ¯ B0 B0 d ¯ P. Finally, if f T is a European contingent claim with maturity N and the market is both arbitrage-free and complete, then ¯ B t ¯ E ¸ f T ¯ B T [T t = ¯ B t ¯ E ¸ f T ¯ B T [T t . That is, the price of f T is independent of the choice of num´eraire. The above theoretical results can be applied to market involving foreign money market account. We con- sider the following market: a domestic money market account M (M 0 = 1), a foreign money market account M f (M f 0 = 1), a (vector) asset price process S called stock. Suppose the domestic vs. foreign currency exchange rate is Q. Note Q is not a traded asset. Denominated by domestic currency, the traded assets are (M, M f Q, S), where M f Q can be seen as the price process of one unit foreign currency. Domestic risk- neutral measure ¯ P is such that M f Q M , S M is a ¯ P-martingale. Denominated by foreign currency, the traded assets are M f , M Q , S Q . Foreign risk-neutral measure ¯ P f is such that M QM f , S QM f is a ¯ P f -martingale. This is a change of num´eraire in the market denominated by domestic currency, from M to M f Q. If we assume the market is arbitrage-free and complete, the foreign risk-neutral measure is d ¯ P f = Q T M f T M T E Q0M f 0 M0 d ¯ P = Q T D T M f T Q 0 d ¯ P on T T . Under the above set-up, for a European contingent claim f T , denominated in domestic currency, its payoff in foreign currency is f T /Q T . Therefore its foreign price is ¯ E f D f T f T D f t Q T [T t . Convert this price into domestic currency, we have Q t ¯ E f D f T f T D f t Q T [T t . Use the relation between ¯ P f and ¯ P on T T and the Bayes formula, we get Q t ¯ E f ¸ D f T f T D f t Q T [T t ¸ = ¯ E ¸ D T f T D t [T t . The RHS is exactly the price of f T in domestic market if we apply risk-neutral pricing. 9.1. (i) 70 Proof. For any 0 ≤ t ≤ T, by Lemma 5.5.2, E (M2) ¸ M 1 (T) M 2 (T) T t = E ¸ M 2 (T) M 2 (t) M 1 (T) M 2 (T) T t = E[M 1 (T)[T t ] M 2 (t) = M 1 (t) M 2 (t) . So M1(t) M2(t) is a martingale under P M2 . (ii) Proof. Let M 1 (t) = D t S t and M 2 (t) = D t N t /N 0 . Then ¯ P (N) as defined in (9.2.6) is P (M2) as defined in Remark 9.2.5. Hence M1(t) M2(t) = St Nt N 0 is a martingale under ¯ P (N) , which implies S (N) t = St Nt is a martingale under ¯ P (N) . 9.2. (i) Proof. Since N −1 t = N −1 0 e −ν Wt−(r− 1 2 ν 2 )t , we have d(N −1 t ) = N −1 0 e −ν Wt−(r− 1 2 ν 2 )t [−νd W t −(r − 1 2 ν 2 )dt + 1 2 ν 2 dt] = N −1 t (−νd ´ W t −rdt). (ii) Proof. d ´ M t = M t d 1 N t + 1 N t dM t +d 1 N t dM t = ´ M t (−νd ´ W t −rdt) +r ´ M t dt = −ν ´ M t d ´ W t . Remark: This can also be obtained directly from Theorem 9.2.2. (iii) Proof. d ´ X t = d X t N t = X t d 1 N t + 1 N t dX t +d 1 N t dX t = (∆ t S t + Γ t M t )d 1 N t + 1 N t (∆ t dS t + Γ t dM t ) +d 1 N t (∆ t dS t + Γ t dM t ) = ∆ t ¸ S t d 1 N t + 1 N t dS t +d 1 N t dS t + Γ t ¸ M t d 1 N t + 1 N t dM t +d 1 N t dM t = ∆ t d ´ S t + Γ t d ´ M t . 9.3. To avoid singular cases, we need to assume −1 < ρ < 1. (i) Proof. N t = N 0 e ν W3(t)+(r− 1 2 ν 2 )t . So dN −1 t = d(N −1 0 e −ν W3(t)−(r− 1 2 ν 2 )t ) = N −1 0 e −ν W3(t)−(r− 1 2 ν 2 )t ¸ −νd W 3 (t) −(r − 1 2 ν 2 )dt + 1 2 ν 2 dt = N −1 t [−νd W 3 (t) −(r −ν 2 )dt], 71 and dS (N) t = N −1 t dS t +S t dN −1 t +dS t dN −1 t = N −1 t (rS t dt +σS t d W 1 (t)) +S t N −1 t [−νd W 3 (t) −(r −ν 2 )dt] = S (N) t (rdt +σd W 1 (t)) +S (N) t [−νd W 3 (t) −(r −ν 2 )dt] −σS (N) t ρdt = S (N) t (ν 2 −σρ)dt +S (N) t (σd W 1 (t) −νd W 3 (t)). Define γ = σ 2 −2ρσν +ν 2 and W 4 (t) = σ γ W 1 (t) − ν γ W 3 (t), then W 4 is a martingale with quadratic variation [ W 4 ] t = σ 2 γ 2 t −2 σν γ 2 ρt + ν 2 r 2 t = t. By L´evy’s Theorem, W 4 is a BM and therefore, S (N) t has volatility γ = σ 2 −2ρσν +ν 2 . (ii) Proof. This problem is the same as Exercise 4.13, we define W 2 (t) = −ρ √ 1−ρ 2 W 1 (t) + 1 √ 1−ρ 2 W 3 (t), then W 2 is a martingale, with (d W 2 (t)) 2 = − ρ 1 −ρ 2 d W 1 (t) + 1 1 −ρ 2 d W 3 (t) 2 = ρ 2 1 −ρ 2 + 1 1 −ρ 2 − 2ρ 2 1 −ρ 2 dt = dt, and d W 2 (t)d W 1 (t) = − ρ √ 1−ρ 2 dt + ρ √ 1−ρ 2 dt = 0. So W 2 is a BM independent of W 1 , and dN t = rN t dt + νN t d W 3 (t) = rN t dt +νN t [ρd W 1 (t) + 1 −ρ 2 d W 2 (t)]. (iii) Proof. Under ¯ P, ( W 1 , W 2 ) is a two-dimensional BM, and dS t = rS t dt +σS t d W 1 (t) = rS t dt +S t (σ, 0) d W 1 (t) d W 2 (t) dN t = rN t dt +νN t d W 3 (t) = rN t dt +N t (νρ, ν 1 −ρ 2 ) d W 1 (t) d W 2 (t) . So under ¯ P, the volatility vector for S is (σ, 0), and the volatility vector for N is (νρ, ν 1 −ρ 2 ). By Theorem 9.2.2, under the measure ¯ P (N) , the volatility vector for S (N) is (v 1 , v 2 ) = (σ −νρ, −ν 1 −ρ 2 . In particular, the volatility of S (N) is v 2 1 +v 2 2 = (σ −νρ) 2 + (−ν 1 −ρ 2 ) 2 = σ 2 −2νρσ +ν 2 , consistent with the result of part (i). 9.4. Proof. From (9.3.15), we have M f t Q t = M f 0 Q 0 e t 0 σ2(s)d W3(s)+ t 0 (Rs− 1 2 σ 2 2 (s))ds . So D f t Q t = D f 0 Q −1 0 e − t 0 σ2(s)d W3(s)− t 0 (Rs− 1 2 σ 2 2 (s))ds 72 and d D f t Q t = D f t Q t [−σ 2 (t)d W 3 (t) −(R t − 1 2 σ 2 2 (t))dt + 1 2 σ 2 2 (t)dt] = D f t Q t [−σ 2 (t)d W 3 (t) −(R t −σ 2 2 (t))dt]. To get (9.3.22), we note d M t D f t Q t = M t d D f t Q t + D f t Q t dM t +dM t d D f t Q t = M t D f t Q t [−σ 2 (t)d W 3 (t) −(R t −σ 2 2 (t))dt] + R t M t D f t Q t dt = − M t D f t Q t (σ 2 (t)d W 3 (t) −σ 2 2 (t)dt) = − M t D f t Q t σ 2 (t)d W f 3 (t). To get (9.3.23), we note d D f t S t Q t = D f t Q t dS t +S t d D f t Q t +dS t d D f t Q t = D f t Q t S t (R t dt +σ 1 (t)d W 1 (t)) + S t D f t Q t [−σ 2 (t)d W 3 (t) −(R t −σ 2 2 (t))dt] +S t σ 1 (t)d W 1 (t) D f t Q t (−σ 2 (t))d W 3 (t) = D f t S t Q t [σ 1 (t)d W 1 (t) −σ 2 (t)d W 3 (t) +σ 2 2 (t)dt −σ 1 (t)σ 2 (t)ρ t dt] = D f t S t Q t [σ 1 (t)d W f 1 (t) −σ 2 d W f 3 (t)]. 9.5. Proof. We combine the solutions of all the sub-problems into a single solution as follows. The payoff of a quanto call is ( S T Q T −K) + units of domestic currency at time T. By risk-neutral pricing formula, its price at time t is ¯ E[e −r(T−t) ( S T Q T − K) + [T t ]. So we need to find the SDE for St Qt under risk-neutral measure ¯ P. By formula (9.3.14) and (9.3.16), we have S t = S 0 e σ1 W1(t)+(r− 1 2 σ 2 1 )t and Q t = Q 0 e σ2 W3(t)+(r−r f − 1 2 σ 2 2 )t = Q 0 e σ2ρ W1(t)+σ2 √ 1−ρ 2 W2(t)+(r−r f − 1 2 σ 2 2 )t . So St Qt = S0 Q0 e (σ1−σ2ρ) W1(t)−σ2 √ 1−ρ 2 W2(t)+(r f + 1 2 σ 2 2 − 1 2 σ 2 1 )t . Define σ 4 = (σ 1 −σ 2 ρ) 2 +σ 2 2 (1 −ρ 2 ) = σ 2 1 −2ρσ 1 σ 2 +σ 2 2 and W 4 (t) = σ 1 −σ 2 ρ σ 4 W 1 (t) − σ 2 1 −ρ 2 σ 4 W 2 (t). Then W 4 is a martingale with [ W 4 ] t = (σ1−σ2ρ) 2 σ 2 4 t + σ2(1−ρ 2 ) σ 2 4 t +t. So W 4 is a Brownian motion under ¯ P. So if we set a = r −r f +ρσ 1 σ 2 −σ 2 2 , we have S t Q t = S 0 Q 0 e σ4 W4(t)+(r−a− 1 2 σ 2 4 )t and d S t Q t = S t Q t [σ 4 d W 4 (t) + (r −a)dt]. 73 Therefore, under ¯ P, St Qt behaves like dividend-paying stock and the price of the quanto call option is like the price of a call option on a dividend-paying stock. Thus formula (5.5.12) gives us the desired price formula for quanto call option. 9.6. (i) Proof. d + (t) −d − (t) = 1 σ √ T−t σ 2 (T −t) = σ √ T −t. So d − (t) = d + (t) −σ √ T −t. (ii) Proof. d + (t)+d − (t) = 2 σ √ T−t log For S (t,T) K . So d 2 + (t)−d 2 − (t) = (d + (t)+d − (t))(d + (t)−d − (t)) = 2 log For S (t,T) K . (iii) Proof. For S (t, T)e −d 2 + (t)/2 −Ke −d 2 − (t) = e −d 2 + (t)/2 [For S (t, T) −Ke d 2 + (t)/2−d 2 − (t)/2 ] = e −d 2 + (t)/2 [For S (t, T) −Ke log For S (t,T) K ] = 0. (iv) Proof. dd + (t) = 1 2 √ 1σ (T −t) 3 [log For S (t, T) K + 1 2 σ 2 (T −t)]dt + 1 σ √ T −t ¸ dFor S (t, T) For S (t, T) − (dFor S (t, T)) 2 2For S (t, T) 2 − 1 2 σdt = 1 2σ (T −t) 3 log For S (t, T) K dt + σ 4 √ T −t dt + 1 σ √ T −t (σd W T (t) − 1 2 σ 2 dt − 1 2 σ 2 dt) = 1 2σ(T −t) 3/2 log For S (t, T) K dt − 3σ 4 √ T −t dt + d W T (t) √ T −t . (v) Proof. dd − (t) = dd + (t) −d(σ √ T −t) = dd + (t) + σdt 2 √ T−t . (vi) Proof. By (iv) and (v), (dd − (t)) 2 = (dd + (t)) 2 = dt T−t . (vii) Proof. dN(d + (t)) = N (d + (t))dd + (t)+ 1 2 N (d + (t))(dd + (t)) 2 = 1 √ 2π e − d 2 + (t) 2 dd + (t)+ 1 2 1 √ 2π e − d 2 + (t) 2 (−d + (t)) dt T−t . (viii) 74 Proof. dN(d − (t)) = N (d − (t))dd − (t) + 1 2 N (d − (t))(dd − (t)) 2 = 1 √ 2π e − d 2 − (t) 2 dd + (t) + σdt 2 √ T −t + 1 2 e − d 2 − (t) 2 √ 2π (−d − (t)) dt T −t = 1 √ 2π e −d 2 − (t)/2 dd + (t) + σe −d 2 − (t)/2 2 2π(T −t) dt + e − d 2 − (t)(σ √ T−t−d + (t)) 2 2(T −t) √ 2π dt = 1 √ 2π e −d 2 − (t)/2 dd + (t) + σe −d 2 − (t)/2 2π(T −t) dt − d + (t)e − d 2 − (t) 2 2(T −t) √ 2π dt. (ix) Proof. dFor S (t, T)dN(d + (t)) = σFor S (t, T)d W T (t) e −d 2 + (t)/2 √ 2π 1 √ T −t d ´ W T (t) = σFor S (t, T)e −d 2 + (t)/2 2π(T −t) dt. (x) Proof. For S (t, T)dN(d + (t)) +dFor S (t, T)dN(d + (t)) −KdN(d − (t)) = For S (t, T) ¸ 1 √ 2π e −d 2 + (t)/2 dd + (t) − d + (t) 2(T −t) √ 2π e −d 2 + (t)/2 dt + σFor S (t, T)e −d 2 + (t)/2 2π(T −t) dt −K ¸ e −d 2 − (t)/2 √ 2π dd + (t) + σ 2π(T −t) e −d 2 − (t)/2 dt − d + (t) 2(T −t) √ 2π e −d 2 − (t)/2 dt ¸ = ¸ For S (t, T)d + (t) 2(T −t) √ 2π e −d 2 + (t)/2 + σFor S (t, T)e −d 2 + (t)/2 2π(T −t) − Kσe −d 2 − (t)/2 2π(T −t) − Kd + (t) 2(T −t) √ 2π e −d 2 − (t)/2 ¸ dt + 1 √ 2π For S (t, T)e −d 2 + (t)/2 −Ke −d 2 − (t)/2 dd + (t) = 0. The last “=” comes from (iii), which implies e −d 2 − (t)/2 = For S (t,T) K e −d 2 + (t)/2 . 10. Term-Structure Models 10.1. (i) Proof. Using the notation I 1 (t), I 2 (t), I 3 (t) and I 4 (t) introduced in the problem, we can write Y 1 (t) and Y 2 (t) as Y 1 (t) = e −λ1t Y 1 (0) +e −λ1t I 1 (t) and Y 2 (t) = λ21 λ1−λ2 (e −λ1t −e −λ2t )Y 1 (0) +e −λ2t Y 2 (0) + λ21 λ1−λ2 e −λ1t I 1 (t) −e −λ2t I 2 (t) −e −λ2t I 3 (t), if λ 1 = λ 2 ; −λ 21 te −λ1t Y 1 (0) +e −λ1t Y 2 (0) −λ 21 te −λ1t I 1 (t) −e −λ1t I 4 (t) +e −λ1t I 3 (t), if λ 1 = λ 2 . 75 Since all the I k (t)’s (k = 1, , 4) are normally distributed with zero mean, we can conclude ¯ E[Y 1 (t)] = e −λ1t Y 1 (0) and ¯ E[Y 2 (t)] = λ21 λ1−λ2 (e −λ1t −e −λ2t )Y 1 (0) +e −λ2t Y 2 (0), if λ 1 = λ 2 ; −λ 21 te −λ1t Y 1 (0) +e −λ1t Y 2 (0), if λ 1 = λ 2 . (ii) Proof. The calculation relies on the following fact: if X t and Y t are both martingales, then X t Y t −[X, Y ] t is also a martingale. In particular, ¯ E[X t Y t ] = ¯ E¦[X, Y ] t ¦. Thus ¯ E[I 2 1 (t)] = t 0 e 2λ1u du = e 2λ1t −1 2λ 1 , ¯ E[I 1 (t)I 2 (t)] = t 0 e (λ1+λ2)u du = e (λ1+λ2)t −1 λ 1 +λ 2 , ¯ E[I 1 (t)I 3 (t)] = 0, ¯ E[I 1 (t)I 4 (t)] = t 0 ue 2λ1u du = 1 2λ 1 ¸ te 2λ1t − e 2λ1t −1 2λ 1 and ¯ E[I 2 4 (t)] = t 0 u 2 e 2λ1u du = t 2 e 2λ1t 2λ 1 − te 2λ1t 2λ 2 1 + e 2λ1t −1 4λ 3 1 . (iii) Proof. Following the hint, we have ¯ E[I 1 (s)I 2 (t)] = ¯ E[J 1 (t)I 2 (t)] = t 0 e (λ1+λ2)u 1 {u≤s} du = e (λ1+λ2)s −1 λ 1 +λ 2 . 10.2. (i) Proof. Assume B(t, T) = E[e − T t Rsds [T t ] = f(t, Y 1 (t), Y 2 (t)). Then d(D t B(t, T)) = D t [−R t f(t, Y 1 (t), Y 2 (t))dt+ df(t, Y 1 (t), Y 2 (t))]. By Itˆo’s formula, df(t, Y 1 (t), Y 2 (t)) = [f t (t, Y 1 (t), Y 2 (t)) +f y1 (t, Y 1 (t), Y 2 (t))(µ −λ 1 Y 1 (t)) +f y2 (t, Y 1 (t), Y 2 (t))(−λ 2 )Y 2 (t)] +f y1y2 (t, Y 1 (t), Y 2 (t))σ 21 Y 1 (t) + 1 2 f y1y1 (t, Y 1 (t), Y 2 (t))Y 1 (t) + 1 2 f y2y2 (t, Y 1 (t), Y 2 (t))(σ 2 21 Y 1 (t) +α +βY 1 (t))]dt + martingale part. Since D t B(t, T) is a martingale, we must have ¸ −(δ 0 +δ 1 y 1 +δ 2 y 2 ) + ∂ ∂t + (µ −λ 1 y 1 ) ∂ ∂y 1 −λ 2 y 2 ∂ ∂y 2 + 1 2 2σ 21 y 1 ∂ 2 ∂y 1 ∂y 2 +y 1 ∂ 2 ∂y 2 1 + (σ 2 21 y 1 +α +βy 1 ) ∂ 2 ∂y 2 2 f = 0. (ii) 76 Proof. If we suppose f(t, y 1 , y 2 ) = e −y1C1(T−t)y2C2(T−t)−A(T−t) , then ∂ ∂t f = [y 1 C 1 (T − t) + y 2 C 2 (T − t) + A (T − t)]f, ∂ ∂y1 f = −C 1 (T − t)f, ∂f ∂y2 = −C 2 (T − t)f, ∂ 2 f ∂y1∂y2 = C 1 (T − t)C 2 (T − t)f, ∂ 2 f ∂y 2 1 = C 2 1 (T − t)f, and ∂ 2 f ∂y 2 2 = C 2 2 (T −t)f. So the PDE in part (i) becomes −(δ 0 +δ 1 y 1 +δ 2 y 2 )+y 1 C 1 +y 2 C 2 +A −(µ−λ 1 y 1 )C 1 +λ 2 y 2 C 2 + 1 2 2σ 21 y 1 C 1 C 2 +y 1 C 2 1 + (σ 2 21 y 1 +α +βy 1 )C 2 2 = 0. Sorting out the LHS according to the independent variables y 1 and y 2 , we get −δ 1 +C 1 +λ 1 C 1 +σ 21 C 1 C 2 + 1 2 C 2 1 + 1 2 (σ 2 21 +β)C 2 2 = 0 −δ 2 +C 2 +λ 2 C 2 = 0 −δ 0 +A −µC 1 + 1 2 αC 2 2 = 0. In other words, we can obtain the ODEs for C 1 , C 2 and A as follows C 1 = −λ 1 C 1 −σ 21 C 1 C 2 − 1 2 C 2 1 − 1 2 (σ 2 21 +β)C 2 2 +δ 1 different from (10.7.4), check! C 2 = −λ 2 C 2 +δ 2 A = µC 1 − 1 2 αC 2 2 +δ 0 . 10.3. (i) Proof. d(D t B(t, T)) = D t [−R t f(t, T, Y 1 (t), Y 2 (t))dt +df(t, T, Y 1 (t), Y 2 (t))] and df(t, T, Y 1 (t), Y 2 (t)) = [f t (t, T, Y 1 (t), Y 2 (t)) +f y1 (t, T, Y 1 (t), Y 2 (t))(−λ 1 Y 1 (t)) +f y2 (t, T, Y 1 (t), Y 2 (t))(−λ 21 Y 1 (t) −λ 2 Y 2 (t)) + 1 2 f y1y1 (t, T, Y 1 (t), Y 2 (t)) + 1 2 f y2y2 (t, T, Y 1 (t), Y 2 (t))]dt + martingale part. Since D t B(t, T) is a martingale under risk-neutral measure, we have the following PDE: ¸ −(δ 0 (t) +δ 1 y 1 +δ 2 y 2 ) + ∂ ∂t −λ 1 y 1 ∂ ∂y 1 −(λ 21 y 1 +λ 2 y 2 ) ∂ ∂y 2 + 1 2 ∂ 2 ∂y 2 1 + 1 2 ∂ ∂y 2 2 f(t, T, y 1 , y 2 ) = 0. Suppose f(t, T, y 1 , y 2 ) = e −y1C1(t,T)−y2C2(t,T)−A(t,T) , then f t (t, T, y 1 , y 2 ) = −y 1 d dt C 1 (t, T) −y 2 d dt C 2 (t, T) − d dt A(t, T) f(t, T, y 1 , y 2 ), f y1 (t, T, y 1 , y 2 ) = −C 1 (t, T)f(t, T, y 1 , y 2 ), f y2 (t, T, y 1 , y 2 ) = −C 2 (t, T)f(t, T, y 1 , y 2 ), f y1y2 (t, T, y 1 , y 2 ) = C 1 (t, T)C 2 (t, T)f(t, T, y 1 , y 2 ), f y1y1 (t, T, y 1 , y 2 ) = C 2 1 (t, T)f(t, T, y 1 , y 2 ), f y2y2 (t, T, y 1 , y 2 ) = C 2 2 (t, T)f(t, T, y 1 , y 2 ). So the PDE becomes −(δ 0 (t) +δ 1 y 1 +δ 2 y 2 ) + −y 1 d dt C 1 (t, T) −y 2 d dt C 2 (t, T) − d dt A(t, T) +λ 1 y 1 C 1 (t, T) +(λ 21 y 1 +λ 2 y 2 )C 2 (t, T) + 1 2 C 2 1 (t, T) + 1 2 C 2 2 (t, T) = 0. 77 Sorting out the terms according to independent variables y 1 and y 2 , we get −δ 0 (t) − d dt A(t, T) + 1 2 C 2 1 (t, T) + 1 2 C 2 2 (t, T) = 0 −δ 1 − d dt C 1 (t, T) +λ 1 C 1 (t, T) +λ 21 C 2 (t, T) = 0 −δ 2 − d dt C 2 (t, T) +λ 2 C 2 (t, T) = 0. That is d dt C 1 (t, T) = λ 1 C 1 (t, T) +λ 21 C 2 (t, T) −δ 1 d dt C 2 (t, T) = λ 2 C 2 (t, T) −δ 2 d dt A(t, T) = 1 2 C 2 1 (t, T) + 1 2 C 2 2 (t, T) −δ 0 (t). (ii) Proof. For C 2 , we note d dt [e −λ2t C 2 (t, T)] = −e −λ2t δ 2 from the ODE in (i). Integrate from t to T, we have 0 −e −λ2t C 2 (t, T) = −δ 2 T t e −λ2s ds = δ2 λ2 (e −λ2T −e −λ2t ). So C 2 (t, T) = δ2 λ2 (1 −e −λ2(T−t) ). For C 1 , we note d dt (e −λ1t C 1 (t, T)) = (λ 21 C 2 (t, T) −δ 1 )e −λ1t = λ 21 δ 2 λ 2 (e −λ1t −e −λ2T+(λ2−λ1)t ) −δ 1 e −λ1t . Integrate from t to T, we get −e −λ1t C 1 (t, T) = − λ21δ2 λ2λ1 (e −λ1T −e −λ1t ) − λ21δ2 λ2 e −λ2T e (λ 2 −λ 1 )T −e (λ 2 −λ 1 )t λ2−λ1 + δ1 λ1 (e −λ1T −e −λ1T ) if λ 1 = λ 2 − λ21δ2 λ2λ1 (e −λ1T −e −λ1t ) − λ21δ2 λ2 e −λ2T (T −t) + δ1 λ1 (e −λ1T −e −λ1T ) if λ 1 = λ 2 . So C 1 (t, T) = λ21δ2 λ2λ1 (e −λ1(T−t) −1) + λ21δ2 λ2 e −λ 1 (T−t) −e −λ 2 (T−t) λ2−λ1 − δ1 λ1 (e −λ1(T−t) −1) if λ 1 = λ 2 λ21δ2 λ2λ1 (e −λ1(T−t) −1) + λ21δ2 λ2 e −λ2T+λ1t (T −t) − δ1 λ1 (e −λ1(T−t) −1) if λ 1 = λ 2 . . (iii) Proof. From the ODE d dt A(t, T) = 1 2 (C 2 1 (t, T) +C 2 2 (t, T)) −δ 0 (t), we get A(t, T) = T t ¸ δ 0 (s) − 1 2 (C 2 1 (s, T) +C 2 2 (s, T)) ds. (iv) Proof. We want to find δ 0 so that f(0, T, Y 1 (0), Y 2 (0)) = e −Y1(0)C1(0,T)−Y2(0)C2(0,T)−A(0,T) = B(0, T) for all T > 0. Take logarithm on both sides and plug in the expression of A(t, T), we get log B(0, T) = −Y 1 (0)C 1 (0, T) −Y 2 (0)C 2 (0, T) + T 0 ¸ 1 2 (C 2 1 (s, T) +C 2 2 (s, T)) −δ 0 (s) ds. Taking derivative w.r.t. T, we have ∂ ∂T log B(0, T) = −Y 1 (0) ∂ ∂T C 1 (0, T) −Y 2 (0) ∂ ∂T C 2 (0, T) + 1 2 C 2 1 (T, T) + 1 2 C 2 2 (T, T) −δ 0 (T). 78 So δ 0 (T) = −Y 1 (0) ∂ ∂T C 1 (0, T) −Y 2 (0) ∂ ∂T C 2 (0, T) − ∂ ∂T log B(0, T) = −Y 1 (0) δ 1 e −λ1T − λ21δ2 λ2 e −λ2T −Y 2 (0)δ 2 e −λ2T − ∂ ∂T log B(0, T) if λ 1 = λ 2 −Y 1 (0) δ 1 e −λ1T −λ 21 δ 2 e −λ2T T −Y 2 (0)δ 2 e −λ2T − ∂ ∂T log B(0, T) if λ 1 = λ 2 . 10.4. (i) Proof. d ´ X t = dX t +Ke −Kt t 0 e Ku Θ(u)dudt −Θ(t)dt = −KX t dt + Σd ´ B t +Ke −Kt t 0 e Ku Θ(u)dudt = −K ´ X t dt + Σd ´ B t . (ii) Proof. W t = CΣ ¯ B t = 1 σ1 0 − ρ σ1 √ 1−ρ 2 1 σ2 √ 1−ρ 2 σ 1 0 0 σ 2 ¯ B t = 1 0 − ρ √ 1−ρ 2 1 √ 1−ρ 2 ¯ B t . So W is a martingale with ' W 1 ` t = ' ¯ B 1 ` t = t, ' W 2 ` t = '− ρ √ 1−ρ 2 ¯ B 1 + 1 √ 1−ρ 2 ¯ B 2 ` t = ρ 2 t 1−ρ 2 + t 1−ρ 2 −2 ρ 1−ρ 2 ρt = ρ 2 +1−2ρ 2 1−ρ 2 t = t, and ' W 1 , W 2 ` t = ' ¯ B 1 , − ρ √ 1−ρ 2 ¯ B 1 + 1 √ 1−ρ 2 ¯ B 2 ` t = − ρt √ 1−ρ 2 + ρt √ 1−ρ 2 = 0. Therefore W is a two-dimensional BM. Moreover, dY t = Cd ´ X t = −CK ¯ X t dt+CΣd ¯ B t = −CKC −1 Y t dt+d W t = −ΛY t dt+d W t , where Λ = CKC −1 = 1 σ1 0 − ρ σ1 √ 1−ρ 2 1 σ2 √ 1−ρ 2 λ 1 0 −1 λ 2 1 [C[ ¸ 1 σ2 √ 1−ρ 2 0 ρ σ1 √ 1−ρ 2 1 σ1 ¸ = λ1 σ1 0 − ρλ1 σ1 √ 1−ρ 2 − 1 σ2 √ 1−ρ 2 λ2 σ2 √ 1−ρ 2 σ 1 0 ρσ 2 σ 2 1 −ρ 2 = λ 1 0 ρσ2(λ2−λ1)−σ1 σ2 √ 1−ρ 2 λ 2 . (iii) Proof. X t = ´ X t +e −Kt t 0 e Ku Θ(u)du = C −1 Y t +e −Kt t 0 e Ku Θ(u)du = σ 1 0 ρσ 2 σ 2 1 −ρ 2 Y 1 (t) Y 2 (t) +e −Kt t 0 e Ku Θ(u)du = σ 1 Y 1 (t) ρσ 2 Y 1 (t) +σ 2 1 −ρ 2 Y 2 (t) +e −Kt t 0 e Ku Θ(u)du. 79 So R t = X 2 (t) = ρσ 2 Y 1 (t)+σ 2 1 −ρ 2 Y 2 (t)+δ 0 (t), where δ 0 (t) is the second coordinate of e −Kt t 0 e Ku Θ(u)du and can be derived explicitly by Lemma 10.2.3. Then δ 1 = ρσ 2 and δ 2 = σ 2 1 −ρ 2 . 10.5. Proof. We note C(t, T) and A(t, T) are dependent only on T −t. So C(t, t +¯ τ) and A(t, t +¯ τ) aare constants when ¯ τ is fixed. So d dt L t = − B(t, t + ¯ τ)[−C(t, t + ¯ τ)R (t) −A(t, t + ¯ τ)] ¯ τB(t, t + ¯ τ) = 1 ¯ τ [C(t, t + ¯ τ)R (t) +A(t, t + ¯ τ)] = 1 ¯ τ [C(0, ¯ τ)R (t) +A(0, ¯ τ)]. Hence L(t 2 )−L(t 1 ) = 1 ¯ τ C(0, ¯ τ)[R(t 2 )−R(t 1 )]+ 1 ¯ τ A(0, ¯ τ)(t 2 −t 1 ). Since L(t 2 )−L(t 1 ) is a linear transformation, it is easy to verify that their correlation is 1. 10.6. (i) Proof. If δ 2 = 0, then dR t = δ 1 dY 1 (t) = δ 1 (−λ 1 Y 1 (t)dt +d W 1 (t)) = δ 1 ( δ0 δ1 − Rt δ1 )λ 1 dt +d W 1 (t) = (δ 0 λ 1 − λ 1 R t )dt +δ 1 d W 1 (t). So a = δ 0 λ 1 and b = λ 1 . (ii) Proof. dR t = δ 1 dY 1 (t) +δ 2 dY 2 (t) = −δ 1 λ 1 Y 1 (t)dt +λ 1 d W 1 (t) −δ 2 λ 21 Y 1 (t)dt −δ 2 λ 2 Y 2 (t)dt +δ 2 d W 2 (t) = −Y 1 (t)(δ 1 λ 1 +δ 2 λ 21 )dt −δ 2 λ 2 Y 2 (t)dt +δ 1 d W 1 (t) +δ 2 d W 2 (t) = −Y 1 (t)λ 2 δ 1 dt −δ 2 λ 2 Y 2 (t)dt +δ 1 d W 1 (t) +δ 2 d W 2 (t) = −λ 2 (Y 1 (t)δ 1 +Y 2 (t)δ 2 )dt +δ 1 d W 1 (t) +δ 2 d W 2 (t) = −λ 2 (R t −δ 0 )dt + δ 2 1 +δ 2 2 ¸ δ 1 δ 2 1 +δ 2 2 d W 1 (t) + δ 2 δ 2 1 +δ 2 2 d W 2 (t) ¸ . So a = λ 2 δ 0 , b = λ 2 , σ = δ 2 1 +δ 2 2 and ¯ B t = δ1 √ δ 2 1 +δ 2 2 W 1 (t) + δ2 √ δ 2 1 +δ 2 2 W 2 (t). 10.7. (i) Proof. We use the canonical form of the model as in formulas (10.2.4)-(10.2.6). By (10.2.20), dB(t, T) = df(t, Y 1 (t), Y 2 (t)) = de −Y1(t)C1(T−t)−Y2(t)C2(T−t)−A(T−t) = dt term +B(t, T)[−C 1 (T −t)d W 1 (t) −C 2 (T −t)d W 2 (t)] = dt term +B(t, T)(−C 1 (T −t), −C 2 (T −t)) d W 1 (t) d W 2 (t) . So the volatility vector of B(t, T) under ¯ P is (−C 1 (T − t), −C 2 (T − t)). By (9.2.5), W T j (t) = t 0 C j (T − u)du + W j (t) (j = 1, 2) form a two-dimensional ¯ P T −BM. (ii) 80 Proof. Under the T-forward measure, the numeraire is B(t, T). By risk-neutral pricing, at time zero the risk-neutral price V 0 of the option satisfies V 0 B(0, T) = ¯ E T ¸ 1 B(T, T) (e −C1( ¯ T−T)Y1(T)−C2( ¯ T−T)Y2(T)−A( ¯ T−T) −K) + . Note B(T, T) = 1, we get (10.7.19). (iii) Proof. We can rewrite (10.2.4) and (10.2.5) as dY 1 (t) = −λ 1 Y 1 (t)dt +d W T 1 (t) −C 1 (T −t)dt dY 2 (t) = −λ 21 Y 1 (t)dt −λ 2 Y 2 (t)dt +d W T 2 (t) −C 2 (T −t)dt. Then Y 1 (t) = Y 1 (0)e −λ1t + t 0 e λ1(s−t) d W T 1 (s) − t 0 C 1 (T −s)e λ1(s−t) ds Y 2 (t) = Y 0 e −λ2t −λ 21 t 0 Y 1 (s)e λ2(s−t) ds + t 0 e λ2(s−t) d W 2 (s) − t 0 C 2 (T −s)e λ2(s−t) ds. So (Y 1 , Y 2 ) is jointly Gaussian and X is therefore Gaussian. (iv) Proof. First, we recall the Black-Scholes formula for call options: if dS t = µS t dt +σS t d W t , then ¯ E[e −µT (S 0 e σW T +(µ− 1 2 σ 2 )T −K) + ] = S 0 N(d + ) −Ke −µT N(d − ) with d ± = 1 σ √ T (log S0 K +(µ± 1 2 σ 2 )T). Let T = 1, S 0 = 1 and ξ = σW 1 +(µ− 1 2 σ 2 ), then ξ d = N(µ− 1 2 σ 2 , σ 2 ) and ¯ E[(e ξ −K) + ] = e µ N(d + ) −KN(d − ), where d ± = 1 σ (−log K + (µ ± 1 2 σ 2 )) (different from the problem. Check!). Since under ¯ P T , X d = N(µ − 1 2 σ 2 , σ 2 ), we have B(0, T) ¯ E T [(e X −K) + ] = B(0, T)(e µ N(d + ) −KN(d − )). 10.11. Proof. On each payment date T j , the payoff of this swap contract is δ(K −L(T j−1 , T j−1 )). Its no-arbitrage price at time 0 is δ(KB(0, T j ) −B(0, T j )L(0, T j−1 )) by Theorem 10.4. So the value of the swap is n+1 ¸ j=1 δ[KB(0, T j ) −B(0, T j )L(0, T j−1 )] = δK n+1 ¸ j=1 B(0, T j ) −δ n+1 ¸ j=1 B(0, T j )L(0, T j−1 ). 10.12. 81 Proof. Since L(T, T) = 1−B(T,T+δ) δB(T,T+δ) ∈ T T , we have ¯ E[D(T +δ)L(T, T)] = ¯ E[ ¯ E[D(T +δ)L(T, T)[T T ]] = ¯ E ¸ 1 −B(T, T +δ) δB(T, T +δ) ¯ E[D(T +δ)[T T ] = ¯ E ¸ 1 −B(T, T +δ) δB(T, T +δ) D(T)B(T, T +δ) = ¯ E ¸ D(T) −D(T)B(T, T +δ) δ = B(0, T) −B(0, T +δ) δ = B(0, T +δ)L(0, T). 11. Introduction to Jump Processes 11.1. (i) Proof. First, M 2 t = N 2 t −2λtN t + λ 2 t 2 . So E[M 2 t ] < ∞. f(x) = x 2 is a convex function. So by conditional Jensen’s inequality, E[f(M t )[T s ] ≥ f(E[M t [T s ]) = f(M s ), ∀s ≤ t. So M 2 t is a submartingale. (ii) Proof. We note M t has independent and stationary increment. So ∀s ≤ t, E[M 2 t − M 2 s [T s ] = E[(M t − M s ) 2 [T s ] + E[(M t − M s ) 2M s [T s ] = E[M 2 t−s ] + 2M s E[M t−s ] = V ar(N t−s ) + 0 = λ(t − s). That is, E[M 2 t −λt[T s ] = M 2 s −λs. 11.2. Proof. P(N s+t = k[N s = k) = P(N s+t − N s = 0[N s = k) = P(N t = 0) = e −λt = 1 − λt + O(t 2 ). Similarly, we have P(N s+t = k + 1[N s = k) = P(N t = 1) = (λt) 1 1! e −λt = λt(1 − λt + O(t 2 )) = λt + O(t 2 ), and P(N s+t ≥ k + 2[N 2 = k) = P(N t ≥ 2) = ¸ ∞ k=2 (λt) k k! e −λt = O(t 2 ). 11.3. Proof. For any t ≤ u, we have E ¸ S u S t T t = E[(σ + 1) Nt−Nu e −λσ(t−u) [T t ] = e −λσ(t−u) E[(σ + 1) Nt−u ] = e −λσ(t−u) E[e Nt−u log(σ+1) ] = e −λσ(t−u) e λ(t−u)(e log(σ+1) −1) (by (11.3.4)) = e −λσ(t−u) e λσ(t−u) = 1. So S t = E[S u [T t ] and S is a martingale. 11.4. 82 Proof. The problem is ambiguous in that the relation between N 1 and N 2 is not clearly stated. According to page 524, paragraph 2, we would guess the condition should be that N 1 and N 2 are independent. Suppose N 1 and N 2 are independent. Define M 1 (t) = N 1 (t) − λ 1 t and M 2 (t) = N 2 (t) − λ 2 t. Then by independence E[M 1 (t)M 2 (t)] = E[M 1 (t)]E[M 2 (t)] = 0. Meanwhile, by Itˆo’s product formula, M 1 (t)M 2 (t) = t 0 M 1 (s−)dM 2 (s) + t 0 M 2 (s−)dM 1 (s) +[M 1 , M 2 ] t . Both t 0 M 1 (s−)dM 2 (s) and t 0 M 2 (s−)dM 1 (s) are mar- tingales. So taking expectation on both sides, we get 0 = 0 + E¦[M 1 , M 2 ] t ¦ = E[ ¸ 0<s≤t ∆N 1 (s)∆N 2 (s)]. Since ¸ 0<s≤t ∆N 1 (s)∆N 2 (s) ≥ 0 a.s., we conclude ¸ 0<s≤t ∆N 1 (s)∆N 2 (s) = 0 a.s. By letting t = 1, 2, , we can find a set Ω 0 of probability 1, so that ∀ω ∈ Ω 0 , ¸ 0<s≤t ∆N 1 (s)∆N 2 (s) = 0 for all t > 0. Therefore N 1 and N 2 can have no simultaneous jump. 11.5. Proof. We shall prove the whole path of N 1 is independent of the whole path of N 2 , following the scheme suggested by page 489, paragraph 1. Fix s ≥ 0, we consider X t = u 1 (N 1 (t)−N 1 (s))+u 2 (N 2 (t)−N 2 (s))−λ 1 (e u1 −1)(t−s)−λ 2 (e u2 −1)(t−s), t > s. Then by Itˆo’s formula for jump process, we have e Xt −e Xs = t s e Xu dX c u + 1 2 t s e Xu dX c u dX c u + ¸ s<u≤t (e Xu −e Xu− ) = t s e Xu [−λ 1 (e u1 −1) −λ 2 (e u2 −1)]du + ¸ 0<u≤t (e Xu −e Xu− ). Since ∆X t = u 1 ∆N 1 (t)+u 2 ∆N 2 (t) and N 1 , N 2 have no simultaneous jump, e Xu −e Xu− = e Xu− (e ∆Xu −1) = e Xu− [(e u1 −1)∆N 1 (u) + (e u2 −1)∆N 2 (u)]. So e Xt −1 = t s e Xu− [−λ 1 (e u1 −1) −λ 2 (e u2 −1)]du + ¸ s<u≤t e Xu− [(e u1 −1)∆N 1 (u) + (e u2 −1)∆N 2 (u)] = t s e Xu− [(e u1 −1)d(N 1 (u) −λ 1 u) −(e u2 −1)d(N 2 (u) −λ 2 u)]. This shows (e Xt ) t≥s is a martingale w.r.t. (T t ) t≥s . So E[e Xt ] ≡ 1, i.e. E[e u1(N1(t)−N1(s))+u2(N2(t)−N2(s)) ] = e λ1(e u 1 −1)(t−s) e λ2(e u 2 −1)(t−s) = E[e u1(N1(t)−N1(s)) ]E[e u2(N2(t)−N2(s)) ]. This shows N 1 (t) −N 1 (s) is independent of N 2 (t) −N 2 (s). Now, suppose we have 0 ≤ t 1 < t 2 < t 3 < < t n , then the vector (N 1 (t 1 ), , N 1 (t n )) is independent of (N 2 (t 1 ), , N 2 (t n )) if and only if (N 1 (t 1 ), N 1 (t 2 ) − N 1 (t 1 ), , N 1 (t n ) − N 1 (t n−1 )) is independent of (N 2 (t 1 ), N 2 (t 2 ) −N 2 (t 1 ), , N 2 (t n ) −N 2 (t n−1 )). Let t 0 = 0, then E[e n i=1 ui(N1(ti)−N1(ti−1))+ n j=1 vj(N2(tj)−N2(tj−1)) ] = E[e n−1 i=1 ui(N1(ti)−N1(ti−1))+ n−1 j=1 vj(N2(tj)−N2(tj−1)) E[e un(N1(tn)−N1(tn−1))+vn(N2(tn)−N2(tn−1)) [T tn−1 ]] = E[e n−1 i=1 ui(N1(ti)−N1(ti−1))+ n−1 j=1 vj(N2(tj)−N2(tj−1)) ]E[e un(N1(tn)−N1(tn−1))+vn(N2(tn)−N2(tn−1)) ] = E[e n−1 i=1 ui(N1(ti)−N1(ti−1))+ n−1 j=1 vj(N2(tj)−N2(tj−1)) ]E[e un(N1(tn)−N1(tn−1)) ]E[e vn(N2(tn)−N2(tn−1)) ], where the second equality comes from the independence of N i (t n ) −N i (t n−1 ) (i = 1, 2) relative to T tn−1 and the third equality comes from the result obtained in the above paragraph. Working by induction, we have E[e n i=1 ui(N1(ti)−N1(ti−1))+ n j=1 vj(N2(tj)−N2(tj−1)) ] = n ¸ i=1 E[e ui(N1(ti)−N1(ti−1)) ] n ¸ j=1 E[e vj(N2(tj)−N2(tj−1)) ] = E[e n i=1 ui(N1(ti)−N1(ti−1)) ]E[e n j=1 vj(N2(tj)−N2(tj−1)) ]. 83 This shows the whole path of N 1 is independent of the whole path of N 2 . 11.6. Proof. Let X t = u 1 W t − 1 2 u 2 1 t +u 2 Q t −λt(ϕ(u 2 ) −1) where ϕ is the moment generating function of the jump size Y . Itˆo’s formula for jump process yields e Xt −1 = t 0 e Xs (u 1 dW s − 1 2 u 2 1 ds −λ(ϕ(u 2 ) −1)ds) + 1 2 t 0 e Xs u 2 1 ds + ¸ 0<s≤t (e Xs −e Xs− ). Note ∆X t = u 2 ∆Q t = u 2 Y Nt ∆N t , where N t is the Poisson process associated with Q t . So e Xt − e Xt− = e Xt− (e ∆Xt − 1) = e Xt− (e u2Y N t − 1)∆N t . Consider the compound Poisson process H t = ¸ Nt i=1 (e u2Yi − 1), then H t −λE[e u2Y N t −1]t = H t −λ(ϕ(u 2 ) −1)t is a martingale, e Xt −e Xt− = e Xt− ∆H t and e Xt −1 = t 0 e Xs (u 1 dW s − 1 2 u 2 1 ds −λ(ϕ(u 2 ) −1)ds) + 1 2 t 0 e Xs u 2 1 ds + t 0 e Xs− dH s = t 0 e Xs u 1 dW s + t 0 e Xs− d(H s −λ(ϕ(u 2 ) −1)s). This shows e Xt is a martingale and E[e Xt ] ≡ 1. So E[e u1Wt+u2Qt ] = e 1 2 u1t e λt(ϕ(u2)−1)t = E[e u1Wt ]E[e u2Qt ]. This shows W t and Q t are independent. 11.7. Proof. E[h(Q T )[T t ] = E[h(Q T −Q t +Q t )[T t ] = E[h(Q T−t +x)][ x=Qt = g(t, Q t ), where g(t, x) = E[h(Q T−t + x)]. References [1] F. Delbaen and W. Schachermayer. The mathematics of arbitrage. Springer, 2006. [2] J. Hull. Options, futures, and other derivatives. Fourth Edition. Prentice-Hall International Inc., New Jersey, 2000. [3] B. Øksendal. Stochastic differential equations: An introduction with applications. Sixth edition. Springer- Verlag, Berlin, 2003. [4] D. Revuz and M. Yor. Continous martingales and Brownian motion. Third edition. Springer-Verlag, Berline, 1998. [5] A. N. Shiryaev. Essentials of stochastic finance: facts, models, theory. World Scientific, Singapore, 1999. [6] S. Shreve. Stochastic calculus for finance I. The binomial asset pricing model. Springer-Verlag, New York, 2004. [7] S. Shreve. Stochastic calculus for finance II. Continuous-time models. Springer-Verlag, New York, 2004. [8] P. Wilmott. The mathematics of financial derivatives: A student introduction. Cambridge University Press, 1995. 84 5 Proof. X1 (u) = ∆0 × 8 + Γ0 × 3 − 5 (4∆0 + 1.20Γ0 ) = 3∆0 + 1.5Γ0 , and X1 (d) = ∆0 × 2 − 4 (4∆0 + 1.20Γ0 ) = 4 −3∆0 − 1.5Γ0 . That is, X1 (u) = −X1 (d). So if there is a positive probability that X1 is positive, then there is a positive probability that X1 is negative. Remark: Note the above relation X1 (u) = −X1 (d) is not a coincidence. In general, let V1 denote the ¯ ¯ payoff of the derivative security at time 1. Suppose X0 and ∆0 are chosen in such a way that V1 can be ¯ 0 − ∆0 S0 ) + ∆0 S1 = V1 . Using the notation of the problem, suppose an agent begins ¯ ¯ replicated: (1 + r)(X with 0 wealth and at time zero buys ∆0 shares of stock and Γ0 options. He then puts his cash position ¯ −∆0 S0 − Γ0 X0 in a money market account. At time one, the value of the agent’s portfolio of stock, option and money market assets is ¯ X1 = ∆0 S1 + Γ0 V1 − (1 + r)(∆0 S0 + Γ0 X0 ). Plug in the expression of V1 and sort out terms, we have ¯ X1 = S0 (∆0 + ∆0 Γ0 )( S1 − (1 + r)). S0 ¯ Since d < (1 + r) < u, X1 (u) and X1 (d) have opposite signs. So if the price of the option at time zero is X0 , then there will no arbitrage. 1.3. S0 1 Proof. V0 = 1+r 1+r−d S1 (H) + u−1−r S1 (T ) = 1+r 1+r−d u + u−1−r d = S0 . This is not surprising, since u−d u−d u−d u−d this is exactly the cost of replicating S1 . Remark: This illustrates an important point. The “fair price” of a stock cannot be determined by the risk-neutral pricing, as seen below. Suppose S1 (H) and S1 (T ) are given, we could have two current prices, S0 and S0 . Correspondingly, we can get u, d and u , d . Because they are determined by S0 and S0 , respectively, it’s not surprising that risk-neutral pricing formula always holds, in both cases. That is, 1+r−d u−d S1 (H) S0 = + u−1−r u−d S1 (T ) 1+r , S0 = 1+r−d u −d S1 (H) + u −1−r u −d S1 (T ) 1+r . Essentially, this is because risk-neutral pricing relies on fair price=replication cost. Stock as a replicating component cannot determine its own “fair” price via the risk-neutral pricing formula. 1.4. Proof. Xn+1 (T ) = = ∆n dSn + (1 + r)(Xn − ∆n Sn ) ∆n Sn (d − 1 − r) + (1 + r)Vn pVn+1 (H) + q Vn+1 (T ) ˜ ˜ Vn+1 (H) − Vn+1 (T ) (d − 1 − r) + (1 + r) = u−d 1+r = p(Vn+1 (T ) − Vn+1 (H)) + pVn+1 (H) + q Vn+1 (T ) ˜ ˜ ˜ = pVn+1 (T ) + q Vn+1 (T ) ˜ ˜ = Vn+1 (T ). 1.6. 2 Proof. The bank’s trader should set up a replicating portfolio whose payoff is the opposite of the option’s payoff. More precisely, we solve the equation (1 + r)(X0 − ∆0 S0 ) + ∆0 S1 = −(S1 − K)+ . 1 Then X0 = −1.20 and ∆0 = − 2 . This means the trader should sell short 0.5 share of stock, put the income 2 into a money market account, and then transfer 1.20 into a separate money market account. At time one, the portfolio consisting of a short position in stock and 0.8(1 + r) in money market account will cancel out with the option’s payoff. Therefore we end up with 1.20(1 + r) in the separate money market account. Remark: This problem illustrates why we are interested in hedging a long position. In case the stock price goes down at time one, the option will expire without any payoff. The initial money 1.20 we paid at time zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payoff at time one. Also, cf. page 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott [8], page 11-13. 1.7. Proof. The idea is the same as Problem 1.6. The bank’s trader only needs to set up the reverse of the replicating trading strategy described in Example 1.2.4. More precisely, he should short sell 0.1733 share of stock, invest the income 0.6933 into money market account, and transfer 1.376 into a separate money market account. The portfolio consisting a short position in stock and 0.6933-1.376 in money market account will replicate the opposite of the option’s payoff. After they cancel out, we end up with 1.376(1 + r)3 in the separate money market account. 1.8. (i) 2 s s Proof. vn (s, y) = 5 (vn+1 (2s, y + 2s) + vn+1 ( 2 , y + 2 )). (ii) Proof. 1.696. (iii) Proof. δn (s, y) = vn+1 (us, y + us) − vn+1 (ds, y + ds) . (u − d)s 1.9. (i) Proof. Similar to Theorem 1.2.2, but replace r, u and d everywhere with rn , un and dn . More precisely, set pn = 1+rn −dn and qn = 1 − pn . Then un −dn Vn = pn Vn+1 (H) + qn Vn+1 (T ) . 1 + rn (ii) Proof. ∆n = (iii) 3 Vn+1 (H)−Vn+1 (T ) Sn+1 (H)−Sn+1 (T ) = Vn+1 (H)−Vn+1 (T ) . (un −dn )Sn 10 10 Proof. un = Sn+1 (H) = Sn +10 = 1+ Sn and dn = Sn+1 (T ) = Sn −10 = 1− Sn . So the risk-neutral probabilities Sn Sn Sn Sn at time n are pn = u1−dnn = 1 and qn = 1 . Risk-neutral pricing implies the price of this call at time zero is ˜ ˜ 2 2 n −d 9.375. 2. Probability Theory on Coin Toss Space 2.1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By induction, it suffices to work on the case N = 2. When A1 and A2 are disjoint, P (A1 ∪ A2 ) = ω∈A1 ∪A2 P (ω) = ω∈A1 P (ω) + ω∈A2 P (ω) = P (A1 ) + P (A2 ). When A1 and A2 are arbitrary, using the result when they are disjoint, we have P (A1 ∪ A2 ) = P ((A1 − A2 ) ∪ A2 ) = P (A1 − A2 ) + P (A2 ) ≤ P (A1 ) + P (A2 ). 2.2. (i) 1 3 1 Proof. P (S3 = 32) = p3 = 8 , P (S3 = 8) = 3p2 q = 3 , P (S3 = 2) = 3pq 2 = 8 , and P (S3 = 0.5) = q 3 = 8 . 8 ω∈Ac P (ω) + ω∈A P (ω) = ω∈Ω P (ω) = 1. (ii) Proof. E[S1 ] = 8P (S1 = 8) + 2P (S1 = 2) = 8p + 2q = 5, E[S2 ] = 16p2 + 4 · 2pq + 1 · q 2 = 6.25, and 3 1 E[S3 ] = 32 · 1 + 8 · 8 + 2 · 3 + 0.5 · 8 = 7.8125. So the average rates of growth of the stock price under P 8 8 5 are, respectively: r0 = 4 − 1 = 0.25, r1 = 6.25 − 1 = 0.25 and r2 = 7.8125 − 1 = 0.25. 5 6.25 (iii) 8 1 Proof. P (S3 = 32) = ( 2 )3 = 27 , P (S3 = 8) = 3 · ( 2 )2 · 1 = 4 , P (S3 = 2) = 2 · 1 = 2 , and P (S3 = 0.5) = 27 . 3 3 3 9 9 9 Accordingly, E[S1 ] = 6, E[S2 ] = 9 and E[S3 ] = 13.5. So the average rates of growth of the stock price 9 6 under P are, respectively: r0 = 4 − 1 = 0.5, r1 = 6 − 1 = 0.5, and r2 = 13.5 − 1 = 0.5. 9 2.3. Proof. Apply conditional Jensen’s inequality. 2.4. (i) Proof. En [Mn+1 ] = Mn + En [Xn+1 ] = Mn + E[Xn+1 ] = Mn . (ii) 2 n+1 Proof. En [ SSn ] = En [eσXn+1 eσ +e−σ ] = 2 σXn+1 ] eσ +e−σ E[e = 1. 2.5. (i) 2 2 Proof. 2In = 2 j=0 Mj (Mj+1 − Mj ) = 2 j=0 Mj Mj+1 − j=1 Mj − j=1 Mj = 2 j=0 Mj Mj+1 + n−1 n−1 n−1 n−1 2 2 2 2 2 2 2 2 Mn − j=0 Mj+1 − j=0 Mj = Mn − j=0 (Mj+1 − Mj ) = Mn − j=0 Xj+1 = Mn − n. n−1 n−1 n−1 n−1 n−1 (ii) Proof. En [f (In+1 )] = En [f (In + Mn (Mn+1 − Mn ))] = En [f (In + Mn Xn+1 )] = 1 [f (In + Mn ) + f (In − Mn )] = 2 √ √ √ g(In ), where g(x) = 1 [f (x + 2x + n) + f (x − 2x + n)], since 2In + n = |Mn |. 2 2.6. 4 5 and Theorem 2. Therefore in general.4. Note Mn = En [MN ] and Mn = En [MN ]. Cf. then Sn would be the wealth at time n. V1 1+r . Clearly (Sn )n≥1 is an adapted stochastic process. 1+r1 (T )−d1 (T ) u1 (T )−d1 (T ) 1 12 1 = 6 . VN (1+r)N −1 (1+r)N is a martingale under P . ( (1+r)n )0≤n≤N is a martingale under P . q1 (H) = 2 .2.14) of Chapter 1. where Head is represented by X = 1 and Tail is 1 represented by X = −1.4. 4 5 q0 q1 (T ) = 12 . VN −1 . S 1 1 0 −d So p0 = 1+r−d0 0 = 2 . ωn ) := pn and P (ωn+1 = T |ω1 . with proper modifications (i. We denote by Xn the result of n-th coin toss.7. · · · . 2. ωn ) := qn . En [f (Sn+1 )] = 1 [f (Sn + bn (X1 . · · · . In the proof of Theorem 1. · · · . V2 (T H) = 1 and V2 (T T ) = 0. notes on page 39. Xn )En [Xn+1 ] = 0.e. Xn which consists of stock and money market accounts. 2. Then 2 intuitively. (iv) Proof. 5 .8. (ii) Proof. bn is therefore the wager for the (n+1)-th bet and is devised according to past gambling results. (iii) Proof.Proof. Define S1 = X1 and Sn+1 = n Sn +bn (X1 . So the time-zero value of an option that pays off V2 at time two is given by the risk-neutral pricing formula V0 = E (1+r0V2 1 ) . Xn )) + f (Sn − bn (X1 . Therefore P (HH) = p0 p1 (H) = 1 .4. ···. to be determined later on. Vn (1+r)n = En VN (1+r)N .4.7 still work for the random interest rate model. Xn )Xn+1 . u0 = u1 (H) = = S2 (HH) S1 (H) = 1. For any arbitrary function f .2. and V0 = p0 V1 (H)+q0 V1 (T ) 1+r0 ≈ 1. En [f (Sn+1 ] cannot be solely dependent upon Sn when bn ’s are properly chosen. Indeed. Xn ))]. and we can show it is a martingale. )(1+r (ii) Proof. · · · .4.5. V1 (T ) = p1 (T )V2 (T H)+q1 (T )V2 (T T ) 1+r1 (T ) p1 (H)V2 (HH)+q1 (H)V2 (HT ) 1+r1 (H) = = 1 9. the sequence (Vn )0≤n≤N can be realized as the value process of a portfolio. and P (HT ) = p0 q1 (H) = P (T H) = q0 p1 (T ) = and P (T T ) = The proofs of Theorem 2. where bn (·) is a bounded function on {−1. p1 (H) u0 5 q1 (T ) = 6 . d0 = S1S0 = 2 . · · · . we proved by induction that Xn = Vn where Xn is defined by (1. (Sn )n≥1 cannot be a Markov process. p1 (T ) = 2 1 4. V2 (HH) = 5. Since ( (1+r)n )0≤n≤N is a martingale under P (Theorem Vn 2.9. so V0 . S0 (T and d1 (T ) = S21 (TT)) = 1. In other words. V2 (HT ) = 1. Theorem 2. P would be constructed according to conditional probabilities P (ωn+1 = H|ω1 . · · · . u1 (T ) = S2 (T H) S1 (T ) =4 1+r1 (H)−d1 (H) u1 (H)−d1 (H) 1 = 1 . (i) Proof. 2. So V1 (H) = 2. (i) (H) S1 (H) 1 = 2. Proof. d1 (H) = S2 (HT ) S1 (H) = 1. En [Sn+1 − Sn ] = bn (X1 .5). then use (i). Combine (ii) and (iii). Remark: If Xn is regarded as the gain/loss of n-th bet in a gambling game.2.4. q0 = 2 . 1} . Proof. En [In+1 − In ] = En [∆n (Mn+1 − Mn )] = ∆n En [Mn+1 − Mn ] = 0.). We also suppose P (X = 1) = P (X = −1) = 2 . To make the portfolio replicate the payoff at time N . (iii) Proof. ∆1 (H) = 2. Since (Xn )0≤n≤N is the value process of the N unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear VN equations). (i) Proof. = ∆n Sn (1+r)n+1 En [Yn+1 ] + Xn −∆n Sn (1+r)n = ∆n Sn (1+r)n+1 (up + dq) + Xn −∆n Sn (1+r)n = ∆n Sn +Xn −∆n Sn (1+r)n = (ii) Proof. (iv) 6 . Cn = En [ (1+r)N −n ] = En [ (1+r)N −n ] + En [ (1+r)N −n ] = Fn + Pn . then En [ (1+r)n+1 ] = Sn (1+r)n (1−a)n .4− 9 8−2 = 0.11. (ii) CN FN PN Proof. From (2. En [ Sn+1 ] (1 + r)n+1 = = < = 1 En [(1 − An+1 )Yn+1 Sn ] (1 + r)n+1 Sn [p(1 − An+1 (H))u + q(1 − An+1 (T ))d] (1 + r)n+1 Sn [pu + qd] (1 + r)n+1 Sn . (i) Xn+1 Proof. = Sn (1+r)n (1−a). (iii) FN Proof. ∆0 = (iv) Proof. we have ∆n uSn + (1 + r)(Xn − ∆n Sn ) = Xn+1 (H) ∆n dSn + (1 + r)(Xn − ∆n Sn ) = Xn+1 (T ). F0 = E[ (1+r)N ] = 1 (1+r)N E[SN − K] = S0 − K (1+r)N .10. FN + PN = SN − K + (K − SN )+ = (SN − K)+ = CN . (1 + r)n Sn (1+r)n+1 (1−a)(pu+qd) Sn+1 If An+1 is a constant a.8.3815. the no-arbitrage price of VN at time n is Vn = Xn = En [ (1+r)N −n ].4 − 1 54 ≈ 0. V2 (HH)−V2 (HT ) S2 (HH)−S2 (HT ) V1 (H)−V1 (T ) S1 (H)−S1 (T ) = 1 2. So Xn = En [ (1+r)N −n ] = En [ (1+r)N −n ]. we 1+r VN X must have XN = VN . = 5−1 12−8 = 1.(iii) Proof. En [ (1+r)n+1 ] = En [ ∆n Yn+1 Sn + (1+r)n+1 (1+r)(Xn −∆n Sn ) ] (1+r)n+1 Xn (1+r)n .2). Sn+1 So En [ (1+r)n+1 (1−a)n+1 ] = 2. So ∆n = Xn+1 (H)−Xn+1 (T ) uSn −dSn and Xn = En [ Xn+1 ]. and P = E . ∆N −1 whose payoff at time N is (SN − K)+ . we get a portfolio (∆n )0≤n≤N −1 whose payoff is the same as that of the chooser option. Suppose the market is liquid. Yn ).P ) ]. At time zero. · · · .P ) ]. Proof. Cn = Pn if and only if Fn = 0. which is a function of (Sn . Yn + uSn ) + qg(dSn . C0 = F0 + P0 . P ) at time m. By (ii). By defining (m ≤ k ≤ N − 1) ∆k (ω) = ∆k (ω) ∆k (ω) if C(ω) > P (ω) if C(ω) < P (ω). Therefore. (v) Proof. So the no-arbitrage price process of the chooser option must be equal to the value process of the replicating portfolio. 2. First. we can construct a portfolio ∆0 .d. (ii) 7 Sn+1 Sn Sn )] . First. if C(ω) > P (ω). So n+1 without loss of generality. So max(C. C0 = P0 . Both of them have maturity date N and strike price K. whose payoff at time m is max(C. then the chooser option is equivalent to receiving a payoff of max(C. P ). · · · . Yn + = pg(uSn . (1 + r)N −m (1 + r)N −m That is. At time N . If we feel unconvinced by the above argument that the chooser option’s no-arbitrage price is E[ max(C. expiring at time N and having strike price K. (1+r)m K K By the put-call parity. Note under both actual probability P and risk-neutral probability P . (1+r)m 2. V0 = X0 = E[ (1+r)m ] = E[ max(C. P ). Since F0 = S0 − (vi) SN −K Proof. For any function g. (1+r)N S0 (1+r)N −n (1+r)N S0 (1+r)N = 0.12. ∆m−1 . and the K second term stands for the time-zero price of a call. So (Sn . the no-arbitrage price of the chooser option at time m must be max(C.Proof.. coin tosses ωn ’s are i. Yn+1 )] = En [g( SSn Sn . (i) Proof. Note Fn = En [ (1+r)N −n ] = Sn − So Fn is not necessarily zero and Cn = Pn is not necessarily true for n ≥ 1.i. where C=E (SN − K)+ (K − SN )+ . we can construct a portfolio ∆m . the trader has a wealth of (F0 − S0 )(1 + r)N + SN = −K + SN = FN . Fix ω. the trader has F0 = S0 in money market account and one share of stock. if C(ω) < P (ω). P ) = P + (Sm − (1+r)N −m )+ . the time-zero price of a chooser option is E K (Sm − (1+r)N −m )+ P +E (1 + r)m (1 + r)m =E K (Sm − (1+r)N −m )+ (K − SN )+ . then we have the following mathematically rigorous argument. En [g(Sn+1 . By (ii). · · · . P ) at time m”). C = Sm − (1+r)N −m + P . its current no-arbitrage price should be E[ max(C. +E (1 + r)N (1 + r)m The first term stands for the time-zero price of a put. Yn )0≤n≤N is Markov under P . we work on P .13.P ) ]. expiring at time m and having strike price (1+r)N −m . = Sn − S0 (1 + r)n . C is the no-arbitrage price of a call option at time m and P is the no-arbitrage price of a put option at time m. Therefore. ∆N −1 whose payoff at time N is (K − SN )+ . we can construct a portfolio ∆m . Yn + dSn ). (1+r)m due to the economical argument involved (like “the chooser option is equivalent to receiving a payoff of max(C. In Xm particular. y) = f ( Ny ). y) = vn+1 (us. P . Yn + dSn ). then = 1 1+r [pvn+1 (uSn . by P (Z > 0) = 1. Z(ω)P (ω). Set vN (s. P . Then vN (SN . YM +1 )] = pvM +1 (uSM . uSM )+ n+1 n+1 qg(dSM .Proof. YN ) = f ( N K=M +1 Sk N −M ) = VN . (vi) ω∈A ω∈Ω ω∈Ω P (ω) = ω∈Ω Z(ω)P (ω) = E[Z] = 1. y) = pvn+1 (us. (v) Proof. State Prices 3. If P (A) = ω∈A Z(ω)P (ω) = 0. us) + qvM +1 (ds. En [h(Sn+1 )] = ph(uSn ) + h(dSn ). Suppose vn+1 is already given. For any function g of two variables. for any function h. And for n ≥ M +1. Z. Proof. (ii) y Proof. Set vN (s.d. Suppose vn+1 is given.1 with P . Yn ). Y (ω)P (ω) = ω∈Ω Y (ω)Z(ω)P (ω) = E[Y Z]. we have EM [g(SM +1 . 0). E[Y ] = (iii) ˜ Proof. a) If n > M . So vM (s) = pvM +1 (us.1. we conclude P (ω) = 0 for any ω ∈ A. (i) Proof. SM +1 )] = pg(uSM .1. P (ω) = 0 for any ω ∈ A. then En [vn+1 (Sn+1 )] = pvn+1 (uSn ) + qvn+1 (dSn ). 3. b) If n = M . Yn ) = (Sn . M − 1. So P (A) = 0. for n = 0. More precisely. (Sn . y) = f ( N −M ).i. Since coin tosses ωn ’s are i. P (A) = (iv) Proof. y + ds). Yn + uSn ) + qvn+1 (dSn .2. Yn + dSn ). (Sn . Note Z(ω) := P (ω) P (ω) = 1 Z(ω) . So P (A) = ω∈A P (ω) = 0. 8 . y + us) + qvn+1 (ds. Yn+1 )] = pvn+1 (uSn . y + us) + vn+1 (ds. Apply Theorem 3. Yn + dSn )] = vn (Sn . P (A) = 1 ⇐⇒ P (Ac ) = 0 ⇐⇒ P (Ac ) = 0 ⇐⇒ P (A) = 1. we get the analogous of properties (i)-(iii) of Theorem 3. YN ) = f ( +1 Vn = where En [ Vn+1 ] 1+r = n+1 En [ vn+1 (S1+r . En [g(Sn+1 . Yn )0≤n≤N is Markov under P . c) If n < M . Yn+1 )] = En [g( SSn Sn . so (Sn . Yn + SSn Sn )] = pg(uSn .1. 1. uSM ) + vn+1 (dSM . dSM ). So vn (s) = pvn+1 (us) + qvn+1 (ds). then En [vn+1 (Sn+1 . (i) Proof. ds). Since P (A) = 0. For n ≤ M .1. Yn )0≤n≤M is Markov under P . under P . Z replaced by P . y + ds) . 1+r 2. 3. · · · . vn (s. dSM ).Yn+1 ) ] N n=0 Sn N +1 ) = VN . YM +1 )] = EM [g(SM +1 . Yn +uSn )+ qg(dSn .14. Yn + uSn ) + qvn+1 (dSn . P (Ω) = (ii) Proof. So vn (s. Then vN (SN . then EM [vM +1 (SM +1 . then E[Z] = 1 if and only if Z(ω0 ) = 1. (i) 9 . then Z as constructed above would induce a probability P that is not equivalent to P . ω=ω0 Clearly P (Ω \ {ω0 }) = E[Z1Ω\{ω0 } ] = Z(ω)P (ω) = 0. P and P are not equivalent. · P (ω0 ) = 1.3. 9 16 . Z(T H) = 8 3 8 and Z(T T ) = 15 4 . Z (3.Proof. By (3.3. But P (Ω \ {ω0 }) = 1 − P (ω0 ) > 0 if P (ω0 ) < 1.8. In this case P (ω0 ) = Z(ω0 )P (ω0 ) = 1.25). 1 P (ω0 ) . V1 (H) = [Z2 (HH)V2 (HH)P (ω2 = H|ω1 = H) + Z2 (HT )V2 (HT )P (ω2 = T |ω1 = T )] = 2. Pick ω0 such that P (ω0 ) > 0.5. Z λZ Proof.6.3.26). If P (ω0 ) = 1. Hence in the case 0 < P (ω0 ) < 1. we get  p−1 1 1  λ=  X0 p E 1 Z p−1 Np    = p−1 X0 (1 + r)N p (E[Z p−1 ])p−1 1 p . Solve it for λ. Z(HT ) = 9 . (1+r)N λZ so I(x) = = 1 Z] 1 x. ] ] 3. (iii) Proof.26) gives E[ (1+r)N 1 X0 (1 + r)n Zn En [Z · X0 N Z (1 + r) . Proof. where ξ Hence Xn = (1+r)N λZ X En [ (1+r)N −n ] N ] = X0 . we have E[ (1+r)N ( (1+r)N ) p−1 ] = X0 .25).4. (i) Proof. 1 1 1 1 P (HT ) + 1 (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 1 ) 2 3.6. Z1 (T )(1 + r1 (T )) 9 3 4. (1+r) p−1 λZ So by (3. if we can find ω0 such that 0 < P (ω0 ) < 1. In summary.7. Z1 (H)(1 + r1 (H)) [Z2 (T H)V2 (T H)P (ω2 = H|ω1 = T ) + Z2 (T T )V2 (T T )P (ω2 = T |ω1 = T )] 1 = . U (x) = have XN = 1 x.2. And P and P have to be equivalent. By (3. Z(HH) = (ii) Proof. U (x) = xp−1 and so I(x) = x p−1 . 3. So λ = = En [ X0 (1+r) Z n 1 X0 . Z1 (T ) = V1 (T ) = and V0 = Z2 (HH)V2 (HH) Z2 (HT )V2 (HT ) Z2 (T H)V2 (T H) P (HH) + P (T H) + 0 ≈ 1. Z1 (H) = E1 [Z2 ](H) = Z2 (HH)P (ω2 = H|ω1 = H) + Z2 (HT )P (ω2 = T |ω1 = H) = 3 E1 [Z2 ](T ) = Z2 (T H)P (ω2 = H|ω1 = T ) + Z2 (T T )P (ω2 = T |ω1 = T ) = 2 . if ω = ω0 Then P (Z ≥ 0) = 1 and E[Z] = if ω = ω0 . define Z(ω) = 1 P (ω0 ) 0. 0 = Xn . we 1 ] = X0 (1 + r)n En [ Z ] = the second to last “=” comes from Lemma 3.3. 3. XN = ( (1+r)N ) p−1 = 1 1 Np λ p−1 Z p−1 N (1+r) p−1 = X0 (1+r) p−1 E[Z p p−1 Z p−1 N (1+r) p−1 = (1+r)N X0 Z p−1 E[Z p p−1 1 . Plug pm and ξm into (3.6. we have 2N 2N X0 = m=1 pm ξm I(λξm ) = m=1 1 pm ξm γ1{λξm ≤ γ } . 1 d) If x ≥ γ and y > γ . (iii) XN λZ Proof. So U (x) − yx ≤ U (I(y)) − yI(y). 3. For such λ. Because dx2 (U (x) − yx) = U (x) ≤ 0 (U is concave). E[U (XN )] − λX0 ≤ E[U (XN )] − E[ (1+r)N XN ] = E[U (XN )] − λX0 . where XN is a random variable satisfying E[ (1+r)N ] = X0 . N (ii) 1 Proof. however.4).6. y = (1+r)N . 1 c) If x ≥ γ and 0 < y ≤ γ . we then can conclude 1 1 1 λξm ≤ γ } = ∅. then Xn ≥ 0 for all n. then U (x) − yx = 1 − yx and U (I(y)) − yI(y) = U (γ) − yγ = 1 − yγ ≥ 1 − yx.e. 1 b) If 0 ≤ x < γ and y > γ . we have λZ λZ ∗ E[U (XN )] − E[ XN ] ≤ E[U (XN )] − E[ X ∗ ].4) has a solution. dx (U (x) − yx) = U (x) − y. suppose there exists some K so that ξK < ξK+1 and K X0 1 m=1 ξm pm = γ . In this case. So U (x) − yx ≤ U (I(y)) − yI(y). So E[U (XN )] ≤ E[U (XN )]. a) If 0 ≤ x < γ and 0 < y ≤ γ . (1 + r)N (1 + r)N N ∗ ∗ That is.6. 2 (ii) Proof. Using (ii) and set x = XN . So E[U (XN )] ≤ E[U (XN )]. note γ > 0. note = . So U (x) − yx ≤ U (I(y)) − yI(y). So if XN ≥ 0. Also. Conversely. N N N (1 + r) (1 + r) (1 + r) (1 + r)N λ ∗ ∗ ∗ ∗ i. So X0 γ X0 γ {m : = we are looking for positive solution λ > 0). Therefore U (x) − y(x) ≤ U (I(y)) − yI(y) for every x. Suppose there is a solution λ to (3. then λξK ≤ γ < λξK+1 . 10 2N K 2N X0 1 m=1 pm ξm 1{λξm ≤ γ } . So x = I(y) is an extreme point of U (x) − yx. we have E[U (XN )] − E[XN λZ λZ λZ λZ ] ≤ E[U (I( ))] − E[ I( )]. then U (x) − yx = −yx ≤ 0 and U (I(y)) − yI(y) = U (0) − y · 0 = 0. So U (x) − yx ≤ U (I(y)) − yI(y). (i) X Proof.9. then U (x) − yx = −yx ≤ 0 and U (I(y)) − yI(y) = U (γ) − yγ = 1 − yγ ≥ 0. (iv) Proof. x = I(y) is a maximum point. Let K = max{m : λξm ≤ γ }. Xn = En [ (1+r)N −n ]. then U (x) − yx = 1 − yx < 0 and U (I(y)) − yI(y) = U (0) − y · 0 = 0. such that ξK < λγ < ξK+1 . Following the hint of the problem. So ξK < ξK+1 and K N m=1 pm ξm (Note. Then we can find λ > 0. ξK+1 is interpreted as ∞.d d Proof.4). E[U (XN )] − λX0 ≤ E[U (XN )] − λX0 . N (1 + r) (1 + r)N m=1 m=1 Hence (3. we have Z λZ 1 E[ I( )] = pm ξm 1{λξm ≤ γ } γ = pm ξm γ = X0 . that K could be 2 . We apply Theorem 4.4. This is exactly the content of Theorem 4. This inspired us to check if the converse is also true. Theorem 4. if the derivative security has not been exercised. From the seller’s perspective: A price process (Vn )0≤n≤N is acceptable to him if and only if at time n. − n+1 n (1 + r) (1 + r) (1 + r)n Vn i. we first give a brief summary of pricing American derivative securities as presented in the textbook.16 and V0S = 3.8. This is the price given by 1 Definition 4.1. Since ( (1+r)n )0≤n≤N is a martingale under the risk-neutral measure P . (ii) Proof. V1P (T ) = 2. V0C = 5. Theorem 4. (1 + r)k−n (1 + r)τ −n The buyer wants to consider all the possible τ ’s. 0≤n≤N Theorem 4.4. a price process (Vn )0≤n≤N is acceptable to the seller if and only if (i) Vn ≥ Gn . ( (1+r)n )0≤n≤N is a supermartingale. then the buyer can choose a policy τ with τ ∈ Sn . In summary.e.4.3 and have V2S (HH) = 12. V2P (HH) = 0. So it gives the price acceptable to both.08. So (Vn )0≤n≤N is the value process of a portfolio that needs no further investing if and only if Vn (1+r)n Vn (1+r)n is a supermartingale under P (note this is independent of the requirement 0≤n≤N Vn ≥ Gn ).4. so that he can find the least upper bound of security value. if m ≥ K + 1 4. Formally.4.4.4.1: Vn = maxτ ∈Sn En [1{τ ≤N } (1+r)τ −n Gτ ]. Theorem 4.5 shows how to decide on the optimal exercise policy.28. V1P (H) = 0.4 establishes the one-to-one correspondence between super-replication and supermartingale property. XN (ω m ) = I(λξm ) = γ1{λξm ≤ γ } = γ. V0P = 9. V2P (HT ) = V2P (T H) = 0. American Derivative Securities Before proceeding to the exercise problems. we conclude En Cn Vn+1 Vn =− ≤ 0.4. V1S (H) = 6. We shall use the notation of the book. 0. From the buyer’s perspective: At time n.2 shows the buyer’s upper bound is the seller’s lower bound.(v) ∗ 1 Proof. 4. (ii) is a supermartingale under P . which will be the maximum price of the derivative security acceptable to him. and finally. (iv) 11 . (iii) Proof.8.32. The valuation formula for cash flow (Theorem 2. (i) Proof. if m ≤ K . he can construct a portfolio at cost Vn so that (i) Vn ≥ Gn and (ii) he needs no further investing into the portfolio as time goes by.4.4. gS (s) = |4 − s|. V2P (T T ) = 3. V1S (T ) = 2.296. the seller can find (∆n )0≤n≤N and (Cn )0≤n≤N so that Cn ≥ 0 and Sn Vn+1 = ∆n Sn+1 + (1 + r)(Vn − Cn − ∆n Sn ). V2S (HT ) = V2S (T H) = 2.8) gives a fair price for the derivative security exercised according to τ : N Vn (τ ) = k=n En 1{τ =k} 1 1 Gk = En 1{τ ≤N } Gτ . V2S (T T ) = 3.3 gives a specific algorithm for calculating the price. 433S0 ) + 0. we need Figure 4. b2 ) ≥ max(a1 + a2 . The agent should exercise the put at time one and get 3 to pay off his debt. G2 (T H) = 3 . At time two.2 for this problem.433S0 ) + 0. b1 + b2 ). to hedge the long position. Then ∆1 (H) = and ∆0 = V2 (HH) − V2 (HT ) 1 V2 (T H) − V2 (T T ) = − . the result of coin toss is tail and the stock price goes down to 4. All the other outcomes of G is negative. When this set is not empty. We need Figure 1.36 at time zero and buys the put. “>” holds if and only if b1 > a1 .4. P P pVn+1 + Vn+1 1+r + max gC (Sn ). 4.433 × 2 = −3.1. At time one. G0 = 0. ≤ max gP (Sn ).2. the value of the portfolio 1 is X1 (H) = (1 + r)(−1. b1 ) + max(a2 . Figure 4. Proof. S1 (H) − S1 (T ) The optimal exercise time is τ = inf{n : Vn = Gn }. we can show S Vn = max gS (Sn ). if the result of coin toss is head and the stock price goes up to 8. and the agent should let the put expire. he needs to borrow again and buy 0. At the same time. 12 .1 and Figure 4. 3 G3 (T T H) = 1.36 − 0. τ (HT ) = 2. if the result of coin toss is tail and the stock price goes down to 2. 2 For the intrinsic value process. the agent borrows 1. Proof. This will pay off his debt.125. τ (T H) = τ (T T ) = 1.433 × 4) + 0.75.Proof.433S1 (H) = −0.4. if the result of coin toss is head and the stock price goes up to 16.36 − 0. the value of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) − 12 S1 (H)) + 12 S2 (HH) = 0. 1+r 4. The agent should exercise the put to get 1. By induction. suppose m = max{n : Vn < Vn + Vn }. ∆1 (T ) = = −1.36 − 0. the value of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) − 12 S1 (H)) + 12 S2 (HT ) = −1. If at time two. At time one. S P C As to when “<” holds.433.2. S2 (HH) − S2 (HT ) 12 S2 (T H) − S2 (T T ) V1 (H) − V1 (T ) ≈ −0. First. G3 (T HT ) = 1.2. we note the simple inequality max(a1 .433 shares of stock at time zero. b2 > a2 . G1 (T ) = 1. and calculate the intrinsic value process and price process of the put as follows.3. P C = Vn + Vn . So τ (HH) = ∞. the value of the portfolio 1 is X1 (T ) = (1 + r)(−1. Then clearly m ≤ N − 1 and it is possible S P C that {n : Vn < Vn + Vn } = ∅.4. S S pVn+1 + Vn+1 1+r C P P pV C + Vn+1 pVn+1 + Vn+1 + n+1 1+r 1+r C C pVn+1 + Vn+1 1+r ≤ max gP (Sn ) + gC (Sn ). The agent should borrow to buy 12 shares of stock. b2 < a2 or b1 < a1 .2.433S1 (T ) = (1 + 4 )(−1. Therefore. For this problem. G2 (T T ) = 5 . m is characterized as m = max{n : gP (Sn ) < P P pVn+1 +qVn+1 1+r and gC (Sn ) > C C pVn+1 +qVn+1 1+r or gP (Sn ) > P P pVn+1 +qVn+1 1+r and gC (Sn ) < C C pVn+1 +qVn+1 }. G3 (T T T ) = 2. τ (T H). τ (HH) = ∞. τ ∗ (T H) = τ ∗ (T T ) = 2.5.5. where τ ∗ (HT ) = 5 5 1 4 4 ∗ 2. (3) τ (HT ) = τ (HH) = 1. 5 This value is 1. For 5 (iii). since they are developed for the case where τ is allowed to be ∞. So to hedge our short position after selling the put. V3 (T T H) = 1. ∞} (4 different ones). the following stopping times do not exercise (i) τ ≡ 0. provided we replace “max{Gn . τ ∗ (HH) = ∞.6. ∞}. we must have V0AP ≤ V0 + V0EC ≤ K − S0 + V0EC . E[1{τ ≤2} ( 4 )τ Gτ ] = G0 = 1.4. For (ii). (ii) τ (HT ) ∈ {2. So by Theorem 4. τ (HH) = ∞. E[1{τ ≤2} ( 5 )τ Gτ ] ≤ E[1{τ ∗ ≤2} ( 4 )τ Gτ ∗ ]. All the other outcomes of V is zero. But intuitively. τ (T H) = τ (T T ) = 1 (2 different ones). 4. (i) Proof. V3 (T HT ) = 1.125. Hence VN −1 = VN K max{K − SN −1 . V1 (T H) = 3 . τ (T H) = τ (T T ) = 1 (4 different ones). (iii) 13 . The stopping times in S0 are (1) τ ≡ 0. Remark: We cheated a little bit by using American algorithm and Theorem 4. ∞}. ∞}. Therefore the time-zero price of the derivative security is 0. (iii) τ (HT ) ∈ {2. This is because at time N .96. is K − SN . (4) τ (HT ). τ (HH). τ (HH) ∈ {2. τ (T T ) ∈ {2. τ (T T ) ∈ {2. (ii) Proof. 0}” with “Gn ”. there is no need to charge the insider more than 1. τ (T H). So. 1. 4. the optimal exercise policy is to sell the stock at time zero and the value of this derivative security is K − S0 . 4. τ (T T ) ∈ {2. EN −1 [ 1+r ]} = max{K − SN −1 . ∞} (8 different ones). Proof.5.36 is the cost of super-replicating the American derivative security. When the option is out of money. So E[1{τ ∗ ≤2} ( 5 )τ Gτ ∗ ] = 4 [( 4 )2 · 1 + ( 5 )2 (1 + 4)] = 0. 1 if ω1 = T . E[1{τ ≤2} ( 4 )τ Gτ ] has the biggest value when τ satisfies τ (HT ) = 2. (5) τ (HT ).75. τ (T H) = τ (T T ) = 1. V1 (T ) = 1. results in this chapter should still hold for the case τ ≤ N . τ (HH) = ∞. In effect. if we have to exercise the put and K − SN < 0.36. The second equality comes from the fact that discounted stock price process is a martingale under risk-neutral probability. we can show Vn = K − Sn (0 ≤ n ≤ N ). V1 (T T ) = 3 .2 5 For the price process. ∞} (16 different ones). It enables us to construct a portfolio sufficient to pay off the derivative security. The value of the put at time N . throughout the portfolio’s lifetime.36. V3 (T T T ) = 2. we can exercise the European call to set off the negative payoff. V0 = 0. if it is not exercised at previous times.4. 1+r − SN −1 } = K − SN −1 . Proof.4 and the optimal exercise time satisfies τ (ω) = ∞ if ω1 = H.4.4. ∗ 4 For (i). no matter when the derivative security is exercised. the portfolio has intrinsic values greater than that of an American put stuck at K with expiration time N . By induction. (2) τ ≡ 1. τ (T H). 7. En [ SSn ] = En [eσXn+1 f (σ) ] = peσ f (σ) + qe−σ f (σ) = 1. (iii) 1 Proof. SN −1 − 1+r } = SN −1 − 1+r . with (m) distributions the same as that of M . 1 1 = E[1{τ1 <∞} ( f (0) )τ1 ] = P (τ1 < ∞). then as σ varies from 0 to ∞. E[Sn∧τ1 ] = E[S0 ] = 1. so f (σ) > 0 if and only if σ > f (σ) > f (0) = 1 for all σ > 0. with distributions the same as that of τ1 . 1 2 (ln q − ln p). (ii) Proof. again by bounded convergence theorem. By optional stopping theorem. by bounded convergence theorem. EN −1 [ 1+r ]} = max{SN −1 − K. (iv) 1 peσ +qe−σ .Proof. VN K K Proof. K By induction. (i) Proof. · · · ). Yes. 1). If we define Mn = Mn+τm − Mτm (m = 1. VN −1 = max{SN −1 − K. (i) Proof.1. VN = SN − K. So e−σ = E[1{τ1 <∞} ( f (σ) )τ1 ]. (m) (m) (iii) Proof. Let V0EP denote the time-zero value of a European put with strike K and expiration time N . Random Walk 5. 5. we have eσ = 2pα 2 √ √ 1+ 1−4pqα2 1−4pqα2 τ1 √2pα 2 = 1− 2qα should take σ = ln( ).i. √ 1± 1−4pqα2 (note 4pqα2 < 4( p+q )2 · 12 = 1). Since 1 2 (ln q − ln p) < 0. Therefore E[ατm ] = E[α(τm −τm−1 )+(τm−1 −τm−2 )+···+τ1 ] = E[ατ1 ]m . so we 14 . 2pα 1+ 1−4pqα Proof. 2. Note Sn∧τ1 = eσMn∧τ1 ( f (σ) )n∧τ1 ≤ eσ·1 . Then V0AP ≥ V0EP = V0EC − E[ K SN − K ] = V0EC − S0 + . E[1{τ1 <∞} Sτ1 ] = E[limn→∞ Sn∧τ1 ] = limn→∞ E[Sn∧τ1 ] = 1. (ii) 1 1 1 n+1 Proof.i. Let σ ↓ 0. E[ατ2 ] = E[α(τ2 −τ1 )+τ1 ] = E[α(τ2 −τ1 ) ]E[ατ1 ] = E[ατ1 ]2 . 1 1 E[1{τ1 <∞} eσ ( f (σ) )τ1 ] = 1. we can prove Vn = Sn − (1+r)N −n (0 ≤ n ≤ N ) and Vn > Gn for 0 ≤ n ≤ N − 1. So τm+1 − τm = inf{n : Mn = 1} are i. since the argument of (ii) still works for asymmetric random walk. Set α = 1 f (σ) = Write σ 0.2.d. α can take all the values in (0. We want to choose σ > in terms of α. 5. f (σ) = peσ − qe−σ . (1 + r)N (1 + r)N 4. Therefore E[α ] = . then (M· )m as random functions are i.d. that is. So the K time-zero value is S0 − (1+r)N and the optimal exercise time is N . As in Exercise 5.3. √ (ii) 1 1 Proof. P (τ2 = 2) = 1 .4. So f (σ) > 1 for all σ > σ0 . Set p σ0 = ln q − ln p. we can see E[1{τ1 <∞} ( f (σ) )τ1 ] = e−σ . Sn = eσMn ( f (σ) )n is a martingale. for σ > σ0 . we get P (τ1 < ∞) = e−σ0 = (iii) p q < 1. and 1 = E[S0 ] = E[Sn∧τ1 ] = E[eσMn∧τ1 ( f (σ) )τ1 ∧n ]. 1 2p−1 . 5. ∂ τ1 ∂α E[α ] − 4pq)− 2 (−8pq) − (1 − 1 √ 1 − 4pq)] = 1−4pq Proof. 4 P (τ2 ≤ 2k) = = = = P (M2k = 2) + P (M2k ≥ 4) + P (τ2 ≤ 2k. 1 = E[ lim eσMn∧τ1 ( n→∞ 1 n∧τ1 1 τ1 ) ] = E[1{τ1 <∞} eσ ( ) ]. For k ≥ 2. Set α = f (α) . E[τ1 1{τ1 <∞} ] = p 1 q 2q−1 .2. So P (τ2 = 2k) = (2k)! . From (ii). and 1− 1 − 4pqα2 2qα = = So E[τ1 ] = limα↑1 5. f (σ) f (σ) Let σ ↓ σ0 . then eσ = . M2k ≤ 0) P (M2k = 2) + 2P (M2k ≥ 4) P (M2k = 2) + P (M2k ≥ 4) + P (M2k ≤ −4) 1 − P (M2k = −2) − P (M2k = 0). P (τ2 = 2k) = P (τ2 ≤ 2k) − P (τ2 ≤ 2k − 2). so σ = ln( 2pα 2qα (iv) Proof. then f (σ0 ) = 1 and f (σ) > 0 for σ > σ0 . E[ατ2 ] = (ii) Proof. Solve the equation peσ + qe−σ = 1 and a positive solution is ln 1+ 2p = ln 1−p = ln q − ln p. then E[ατ1 1{τ1 <∞} ] = .(v) Proof. ∞ k=1 ∂ τ1 ∂α E[α 1{τ1 <∞} ]|α=1 = 1 √ 4pq 2q [ 1−4pq − (1 − √ 1 − 4pq)] = 1 4pq 2q [ 2q−1 − 1 + 2q − 1] = P (τ2 = 2k)α2k = ∞ α 2k k=1 ( 2 ) P (τ2 = 2k)4k . (i) 1 (1 − 1 − 4pqα2 )α−1 2q 1 1 1 [− (1 − 4pqα2 )− 2 (−4pq2α)α−1 + (1 − 2q 2 = 1 1 2q [− 2 (1 1 − 4pqα2 )(−1)α2 ]. need to choose the root so that eσ > eσ0 = p . ∂ τ1 ∂α E[α ] ∂ = E[ ∂α ατ1 ] = E[τ1 ατ1 −1 ]. √ 1± 1−4pqα2 1 1 Proof. (i) Proof. We 2pα √ √ 1− 1−4pqα2 1+ 1−4pqα2 q ). 4k (k+1)!k! 15 . Suppose σ > σ0 . then by bounded convergence theorem. τ (n) ∈ Mn τ ∧ n. For an American put with (1+r) ) the same strike price K that expires at time n. rigorously speaking. For every path that crosses level m by time n and resides at b at time n.Similarly. On the infinite coin-toss space. Remark: In the above proof. if τ < ∞ and limn→∞ τ (n) = τ . (1+r) + Clearly (V (n) )n≥0 is nondecreasing and V (n) ≤ V ∗ for every n.6. · · · }. n. By bounded convergence theorem. ∗ P (Mn ≥ m. then τ (n) is also a stopping time. · · · . ∞} and M∞ = {stopping times that takes values 0. = lim E 1{τ (n) <∞} τ n→∞ n→∞ (1 + r) (1 + r)τ (n) Take sup at the left hand side of the inequality. So P (τ2 = 2k) = = = = = P (M2k−2 = −2) + P (M2k−2 = 0) − P (M2k = −2) − P (M2k = 0) 1 (2k − 2)! (2k − 2)! 1 (2k)! (2k)! ( )2k−2 + − ( )2k + 2 k!(k − 2)! (k − 1)!(k − 1)! 2 (k + 1)!(k − 1)! k!k! (2k)! 4 4 (k + 1)k(k − 1) + (k + 1)k 2 − k − (k + 1) 4k (k + 1)!k! 2k(2k − 1) 2k(2k − 1) (2k)! 2(k 2 − 1) 2(k 2 + k) 4k 2 − 1 + − 4k (k + 1)!k! 2k − 1 2k − 1 2k − 1 (2k)! . we define Mn = {stopping times that takes values 0. Therefore V ∗ = limn V (n) . (i) Proof. 5. P (τ2 ≤ 2k − 2) = 1 − P (M2k−2 = −2) − P (M2k−2 = 0). Mn = b) = n! (m + n−b n+b 2 )!( 2 − m)! pm+ n−b 2 q n+b 2 −m . 1.8. (ii) Proof.5. Mn = b) = P (Mn = 2m − b) = ( )n 2 (m + n! n−b n+b 2 )!( 2 − m)! . there corresponds a reflected path that resides at time 2m − b. So this needs some justification.4 can be defined as supτ ∈M∞ E[1{τ <∞} (K−Sτ τ ]. its time-zero value V (n) is maxτ ∈Mn E[1{τ <∞} (K−Sτ τ ]. we should use (K − Sτ ) in the places of (K − Sτ )+ . ∞. if τ = ∞ For any given τ ∈ M∞ . 2. we get V ∗ ≤ limn→∞ V (n) . So limn V (n) exists and limn V (n) ≤ V ∗ . Proof. 5. 4k (k + 1)!k! 5. (i) 16 . 1. Then the time-zero value V ∗ of the perpetual )+ American put as in Section 5. we define τ (n) = . E 1{τ <∞} (K − Sτ (n) )+ (K − Sτ )+ ≤ lim V (n) . So 1 ∗ P (Mn ≥ m. Since lims→∞ v(s) = lims→∞ (As + (iii) Proof. (ii) Proof. v(Sn ) = Sn ≥ Sn −K = g(Sn ). the equation has roots and for B > 4. pu+qv s} = max{s − K.4. To have continuous derivative. = max{s − K. we must have −1 = − s2 . Plug B = s2 back into s2 − 4sB + B = 0. we conclude S0 is the least upper bound for all the prices acceptable to the buyer. by Fatou’s lemma. Solve it for p. Suppose τ is an optimal exercise time. we have Sτ −K shown E (1+r)τ 1{τ <∞} < S0 for any stopping time τ . Suppose v(s) = sp . we have S0 ≤ E (1+r)τ 1{τ <∞} < S0 .4. Interest-Rate-Dependent Assets 6. By drawing graphs of 6. so equation (5. we must have A = 0. pv(us)+qv(ds) 1+r (iv) Proof.4. by Theorem 2.16) is 1+r satisfied. Clearly v(s) = s also satisfies the boundary condition (5. So E Sτ −K (1+r)τ 1{τ <∞} < E Sτ (1+r)τ Sτ (1+r)τ 1{τ <∞} .Proof. we get p = 1 = 0. If the purchaser chooses to exercises the call at time n. then the equation s2 − 4s + √ = 0 has solution 2 ± B 4 − s and B . the value of the call at time zero is at least supn S0 − (iii) Proof.9. fB (s) = 0 if and only if B + s2 − 4s = 0. Since limn→∞ S0 − (1+r)n = S0 . Simultaneously. Contradiction. (iv) Proof. This gives B = 4. Since is a martingale n≥0 Sτ ∧n (1+r)τ ∧n 1{τ <∞} under risk-neutral measure. 1{τ <∞} ≥ S0 . (ii) 1 (1+r)n v(Sn ) = Sn (1+r)n is a martingale Proof. then we have sp = 2 2p sp + 5 or p = −1. Then P (τ < ∞) = 0 and Sn (1+r)n 1{τ <∞} > 0. Under risk-neutral probabilities. √ 4 − B. 17 . this equation does not have roots. Combined.4. s (v) B Proof. B s) 2 sp 5 2p . Suppose B ≤ 4. B B B we get sB = 2. 5. The discriminant ∆ = (−4)2 − 4B = 4(4 − B).2. max g(s). then the discounted risk-neutral expectation Sn −K K K of her payoff is E (1+r)n = S0 − (1+r)n . we should choose B = 4 and sB = 2 + 4 − B = 2. (i) Proof. s} = s = v(s). So 1 = 2p+1 5 + 21−p 5 . So there is no optimal time to exercise the perpetual American call. So for B ≤ 4. E 1{τ <∞} ≤ lim inf n→∞ E = Sτ −K Sτ ∧n lim inf n→∞ E (1+r)τ ∧n = lim inf n→∞ E[S0 ] = S0 . then E E K (1+r)τ Sτ −K (1+r)τ K (1+r)n = S0 . Combined with (ii).18). m − Bn. then short 3 shares of the maturity three bond and invest the income into the money market account.m+1 . we must have D2 D1 E1 [V2 ] = 1 1+R1 E1 [V2 ]. If the outcome of first coin toss is H. then at time one. Dn Dn Dn 6. In particular.2 ) + ∆0 B1. 6. (iii) Proof. D1 V1 = E1 [D3 V3 ] = E1 [D2 V2 ] = D2 E1 [V2 ].3.3 ) + ∆1 B2.3 (T H)−V2 (T(T)T ) = 0.3 (H) = B1. 1 (ii) 2 Proof.2 ) + ∆0 B1. the value of the bond is its face value and our portfolio will therefore have the same value regardless outcomes of coin tosses. the value of our portfolio is X1 = (1 + R0 )(X0 − ∆B0.4. we must have V2 = (1 + R1 )(X1 − ∆1 B1.m Sn −1 = Dk−1 [Sk−1 − Ek−1 [Dm (Sm − K)]Dk−1 − Bk−1.2 (H)−B1 (T(T ) = 4 .(i) Proof. then to replicate V2 at time V2 (HH)−V two. To replicate the value V1 . We do not invest in the maturity two bond. V1 (H) = (1 + R0 )(X0 − ∆0 B0. Suppose we buy ∆0 shares of the maturity two bond. This makes impossible the replication of V1 .2 (H) V1 (T ) = (1 + R0 )(X0 − ∆0 B0.3 (T )(= 4 ) and the portfolio will therefore have the same value at time one regardless the 7 outcome of first coin toss.3 2 T .−1 Proof.m ] Bn.2 ) + ∆0 B1.2 . V1 (T ) = 0.m Sn = Dk−1 Sk−1 − Ek−1 [Dm (Sm − K)] − Ek−1 [Ek [Dm ]] Bn. Let X0 = 21 . So V1 = 4 V1 (H) = 1+R1 (H) V2 (HH)P (w2 = H|w1 = H) = 21 . 1 1 Dm − Dm+1 En [Dm+1 Rm ] = En [Dm (1 + Rm )−1 Rm ] = En [ ] = Bn.m = Dk−1 Xk−1 . So ∆1 (H) = B2.3 (HT V2 T ∆1 (T ) = B2. If the outcome of first coin toss is (T H)−B2. Then Ek−1 [Dk Xk ] Sn Bk. then we do nothing. (i) 18 .2 (T ). Proof.m for n ≤ k ≤ m. So the hedging strategy is as follows.2 − 21 = 20 and buy 4 3 3 21 3 1. Xk = Sk − Ek [Dm (Sm − K)]Dk − Sn Bn.5.3 (HH)−B2 (HT ) ) = − 2 .3 . because at time two. and 3 2.2 share of the maturity two bond. Suppose we buy ∆1 share of the maturity three bond at time one.m Bk. = Ek−1 [Dk Sk − Ek [Dm (Sm − K)] − 6. V1 (H)−V ) 2 So ∆0 = B1. The reason why we do not invest in the maturity three bond is that B1. This makes impossible the replication of V2 . since V1 (H) = V1 (T ).m Dk ] Bn. The hedging strategy is therefore to borrow 4 B0. m .m+1 ] = En−1 = = Bn.m+1 = Fn−1. (i) Proof.m Bn.7.m Sn Bn+1. By (i).m+1 Dn−1 En−1 [Dm − Dm+1 ] Bn−1. The agent enters the forward contract at no cost. 2(1 + rn (0)) 19 . 6.m+1 Dn Bn−1.m+1 = Bn−1. Suppose 1 ≤ n ≤ m. At time n + 1. Bn.m − Bn−1. Proof.m+1 )Zn. 6.m+1 Zn−1. the cash flow generated at time n + 1 is (1 + r)m−n−1 Sn+1 − = = (1 + r) m−n−1 Sn Bn+1.m = BSn .m+1 Bn.m Bn. then m+1 En−1 [Fn.m = Sn+1 − flow of Sn+1 − (ii) Proof.6.m+1 (Bn.m Sn (1+r)m−n−1 1 (1+r)m−n + r)m−n Sn Sn+1 − (1 + r)m−n−1 Sn+1 − (1 Sm Sm ] + (1 + r)m En [ ] = (1 + r)m En1 [ m (1 + r) (1 + r)m = F utn+1.m −1 Bn. he will receive a cash Sn Bn+1.m − F utn.m ] −1 −1 = En−1 [Bn.m − Bn. ψn+1 (0) = E[Dn+1 Vn+1 (0)] Dn = E[ 1{#H(ω1 ···ωn+1 )=0} ] 1 + rn (0) Dn = E[ 1{#H(ω1 ···ωn )=0} 1{ωn+1 =T } ] 1 + rn (0) 1 Dn E[ ] = 2 1 + rn (0) ψn (0) = .m+1 Dn−1 Dn −1 −1 En−1 [Dn En [Dm ] − Dn En [Dm+1 ]] Bn−1. the contract has the value En+1 [Dm (Sm − K)] = n. He is obliged to buy certain asset at time m at the strike price K = F orn.m .m So if the agent sells this contract at time n + 1.m1 Dn−1 Bn−1.m .m Sn+1 − KBn+1.Proof. we have E[Xf (ωn+1 )] = E[X E[f (ωn+1 )|Fn ]] = E[X(pn f (H) + qn f (T ))]. ψn+1 (n + 1) = E[Dn+1 Vn+1 (n + 1)] = E Dn ψn (n) 1{#H(ω1 ···ωn )=n} 1{ωn+1 =H} = . · · · . · · · . we can use the fact that P (ωn+1 = H|ω1 . ωn ) = pn and P (ωn+1 = T |ω1 . ωn ). 2(1 + rn (k)) 2(1 + rn (k − 1)) Remark: In the above proof. Then for any X ∈ Fn = σ(ω1 . ψn+1 (k) = E Dn 1{#H(ω1 ···ωn+1 )=k} 1 + rn (#H(ω1 · · · ωn )) Dn Dn = E 1{#H(ω1 ···ωn )=k} 1{ωn+1 =T } + E 1{#H(ω1 ···ωn )=k} 1{ωn+1 =H} 1 + rn (k) 1 + rn (k − 1) = = Finally. · · · . · · · .9 and un −dn un −dn notes on page 39). 2. In case the binomial model has stochastic up. · · · . we have used the independence of ωn+1 and (ω1 . where pn = 1+rn −dn and qn = u−1−rn (cf.and down-factor un and dn . 1 + rn (n) 2(1 + rn (n)) 1 E[Dn Vn (k)] 1 E[Dn Vn (k − 1)] + 2 1 + rn (k) 2 1 + rn (k − 1) ψn (k) ψn (k − 1) + .For k = 1. 20 . This is guaranteed 1 by the assumption that p = q = 2 (note ξ ⊥ η if and only if E[ξ|η] = constant). solution of Exercise 2. ωn ) = qn . ωn ). n. Proof. (ii) Proof. then A ∪ B is also infinite. Then φ is one-to-one and onto.2. So 0 ≤ P (A) ≤ 0. which is uniformly distributed on [0. 21 . So P (A ∪ B) = 0 = P (A) + P (B). So the cardinality of A is the same as that of Ω. which means P (A) = 0. 1. So P (A) = lim P (An ) = lim [P (ω1 = ω2 ) · · · P (ω2n−1 = ω2n )] = lim (p2 + (1 − p)2 )n . x ξ2 where Yi (ω) = 1. we have limn→∞ (p2 + (1 − p)2 )n = 0. Similarly. (i) Proof. However. if both of them are finite.4. So P (A ∪ B) = N ∞ = P (A) + P (B). we can construct a random variable X on the coin-toss space. Then P (Z ≤ a) = P (X ≤ N (a)) = N (a) for any real number a. we can prove P (∪N An ) = n=1 P (An ). (ii) Proof. Then A = ∪∞ An is an infinite n=1 ∞ set and therefore P (A) = ∞.5. General Probability Theory 1. 1. then A ∪ B is also finite. By Example 1. let An = { n }. P ). We define a mapping φ from A to Ω as follows: φ(ω1 ω2 · · · ) = ω1 ω3 ω5 · · · .5. P (B) = P ((B − A) ∪ A) = P (B − A) + P (A) ≥ P (A). (i) Proof. n=1 1 To see countable additivity property doesn’t hold for P . i.2. even if An ’s are not disjoint. So P (A) = n=1 P (An ). If at least one of them is infinite. Let An = {ω = ω1 ω2 · · · : ω1 = ω2 . which means in particular that A is uncountably infinite. Then An ↓ A as n → ∞. Clearly Zn depends only on the first n coin tosses and {Zn }n≥1 is the desired sequence. if ωi = H 0. (i) Proof. ω2n−1 = ω2n }. For the strictly increasing and continuous function N (x) = −∞ √1 e− 2 dξ.3. 1. Define Xn = n Yi i=1 2i .1. Then Xn (ω) → X(ω) for every ω ∈ Ω∞ where X is defined as in (i). F∞ . 1. if ωi = T . we let 2π Z = N −1 (X). (ii) Proof.2 Stochastic Calculus for Finance II: Continuous-Time Models 1. 1]. n→∞ n→∞ n→∞ Since p2 + (1 − p)2 < 1 for 0 < p < 1. P (An ) = 0 for each n. Z is a standard normal random variable on the coin-toss space (Ω∞ .e. · · · . So Zn = N −1 (Xn ) → Z = N −1 (X) for every ω. For any A and B. Clearly P (∅) = 0. This implies P (A) = 0. P (A) ≤ P (An ) implies P (A) ≤ limn→∞ P (An ) = 0. By the change of variable formula.6. we have ∞ ∞ 1[0. ∞ (1 0 − F (x))dx. First. f (x) = limn→∞ fn (x) = 0. E{φ(X)} = E{euX } = euµ+ 1.X(ω)) (x)dP (ω)dx. fn (x)dx = ∞ x2 ∞ √1 e− 2 −∞ 2π dx = 1. Since |fn (x)| ≤ (ii) Proof. since {fn }n≥1 doesn’t increase to 0.Proof. by the information given by the problem. 22 . ∞ −∞ √1 . (i) Proof.X(ω)) (x)dxdP (ω) = Ω 0 0 Ω 1[0. (i) Proof. The left side of this equation equals to X(ω) dxdP (ω) = Ω 0 Ω X(ω)dP (ω) = E{X}. So we must have n→∞ lim fn (x)dx = 1. The right side of the equation equals to ∞ ∞ ∞ 1{x<X(ω)} dP (ω)dx = 0 Ω 0 P (x < X)dx = 0 (1 − F (x))dx. E{euX } = = = = = (x−µ)2 1 eux √ e− 2σ2 dx σ 2π −∞ ∞ (x−µ)2 −2σ 2 ux 1 2σ 2 √ e− dx −∞ σ 2π ∞ [x−(µ+σ 2 u)]2 −(2σ 2 uµ+σ 4 u2 ) 1 2σ 2 √ e− dx −∞ σ 2π ∞ [x−(µ+σ 2 u)]2 σ 2 u2 1 2σ 2 √ e− euµ+ 2 dx −∞ σ 2π ∞ euµ+ σ 2 u2 2 (ii) Proof. −∞ (iii) Proof. So E{X} = 1. This is not contradictory with the Monotone Convergence Theorem. 2nπ u2 σ 2 2 ≥ euµ = φ(E{X}).7. for any A ∈ F. Proof. Since any nonnegative. and hence smaller than 2t for n sufficiently large. (i) Proof. where each Bi is a Borel subset of R. the desired equality also holds for simple functions.9.12.1] dP = 2P (A ∩ [ 1 . 1. By the linearity of Lebesgue integral.1). Second. ϕ (t) = limn→∞ E{Yn } = E{limn→∞ Yn } = E{XetX }. If {Ai }∞ is a sequence of disjoint Borel subsets of [0.e. (ii) Proof. So P = P normal under P . |Yn | = e t−sn fact that θ is between t and sn . The last inequality is by X ≥ 0 and the Proof.10.9. then P (A) = (iii) Proof.1. 1]. then the desired equality is just (1. ϕ (t) = limn→∞ E{Yn } = E{limn→∞ Yn } = E{XetX }. we have E[|X|et|X| ] < ∞ for every t ∈ R. First. 2 1. Let A = [0. In particular. g of the n form g(x) = i=1 1Bi (x). P (Ω) = 2P ([ 2 .11.13. 1. By (1. the desired equality can be proved by the Monotone Convergence Theorem. E[et|X| ] = E[etX 1{X≥0} ] + − E[e−(−t)X 1{X<0} ] < ∞ for every t ∈ R.3). 2 2 2 1. P (A) = A ZdP = ˆ ˆ . similar to (i). So. θ θ θ ˆ ˆ ˆ Proof.8. 2 } = euθ− θ2 2 E{e(u−θ)X } = euθ− θ2 2 e (u−θ)2 2 =e u2 2 . So by the Dominated Convergence Theorem. (i) −e = |XeθX | = XeθX ≤ Xe2tX . 1]) = 1. Since E[etX 1{X≥0} ] + E[e−tX 1{X<0} ] = E[etX ] < ∞ for every t ∈ R. If P (A) = 0.9. So P is a probability measure. 1 ). (i) 23 . 1 Meanwhile. X is standard normal under P . then by the Monotone Convergence Theorem. Similarly. E{euY } = E{euY Z} = E{euX+uθ e−θX− θ2 2 A ZdP = 2 1 A∩[ 2 . Z = eθY − 2 = eθ(X+θ)− 2 = e 2 +θX = Z −1 . 1]) = 0. tX sn X (ii) Proof. Proof. 1. Borel-measurable function g is the limit of an increasing sequence of simple functions. since it’s standard ˆ (1A Z)ZdP = 1A dP = P (A). i=1 P (∪∞ Ai ) equals to i=1 n ∞ + − + 1∪∞ Ai ZdP = i=1 n→∞ lim 1∪n Ai ZdP = lim i=1 n→∞ 1∪n Ai ZdP = lim i=1 n→∞ ZdP = i=1 Ai i=1 P (Ai ). where B is a Borel subset of R. i. If g(x) is of the form 1B (x). So by the Dominated Convergence Theorem. we have |Yn | = |XeθX | ≤ |X|e2t|X| for n sufficiently large. Furthermore. P (Y ≤ a) = {g(X)≤a} h(g(X))g (X) dP = f (X) g −1 (a) −∞ h(g(x))g (x) f (x)dx = f (x) a −∞ g −1 (a) h(g(x))dg(x). )} = {Y ∈ B(y. (ii) 1 P (X ∈ B(x. )} = {X + θ ∈ B(y. {X ∈ B(x. )) = 1 u2 x+ 2 1 √ e− 2 x− 2 2π du is approximately x 1 √1 e− 2 2π 2 · = X √1 e− 2π 2 (ω) ¯ 2 . P (A) P (A) √ √ is approximately e− Y 2 (ω) ¯ 2 Y 2 (ω)−X 2 (ω) ¯ ¯ 2 (X(ω)+θ)2 −X 2 (ω) ¯ ¯ 2 θ2 2 2π e 2π X 2 (ω) ¯ − 2 = e− = e− ω = e−θX(¯ )− .Proof. )}. (i) Proof. Similar to (i). h(u)du. Proof. (ii) Proof. (iii) Proof. we have E{Z} = E h(g(X))g (X) f (X) ∞ = −∞ h(g(x))g (x) f (x)dx = f (x) ∞ ∞ h(g(x))dg(x) = −∞ −∞ h(u)du = 1. Clearly Z ≥ 0. −∞ By the change of variable formula. (i) Proof. 0 1. (iv) Proof. 1. (ii) Proof. P (Ω) = λ −(λ−λ)X λ e dP = λ λ ∞ ∞ e−(λ−λ)x λe−λx dx = 0 0 λe−λx dx = 1. So Y has density h under 24 .14. the last equation above equals to P.15. P (X ≤ a) = {X≤a} λ −(λ−λ)X e dP = λ a 0 λ −(λ−λ)x −λx e λe dx = λ a λe−λx dx = 1 − e−λa . )} = {X ∈ B(y − θ. By (i)-(iii). So we have 4 4 1 4 + 1 4 1 = 2. E{V W } = E{−X 2 sin θ cos θ +XY cos2 θ −XY sin2 θ +Y 2 sin θ cos θ} = − sin θ cos θ + 0 + 0 + sin θ cos θ = 0. Hence σ(X) and σ(S1 ) are not independent under P . 2. σ(S1 ) = {∅. So σ(X) and σ(S1 ) are independent under P . (i) Proof. so to prove their independence it suffices to show they are uncorrelated. Indeed. (iv) 2 Proof.3. we can work on other elements of σ(X) and σ(S1 ) and show that P (A ∩ B) = P (A)P (B) for any A ∈ σ(X) and B ∈ σ(S1 ). 2. T H} ∩ {HH. we have {X ≤ a} ∈ F0 = {∅. Proof. So P ({HT. For any real number a. We note (V. (i) 25 . P ({HT. HT }) = P ({HT }) = 4 . So P (X ≤ a) is either 0 or 1. σ(X) = {∅. {T T. HT }) = P ({HT. Ω. T H} ∩ {HH. Information and Conditioning 2. HT }) = 9 9 9 4 2 6 9 + 9 = 9 . T H})P ({HH. T H}. we can find a number x0 such that P (X ≤ x0 ) = 1 and P (X ≤ x) = 0 for any x < x0 . n→∞ n n 2. {T H. T H} ∩ {HH. T H})P ({HH. HT }). P ({HT. W ) are jointly Gaussian. {HH. (ii) Proof. HT }. T H}) = P ({HT }) + P ({T H}) = 1 and P ({HH. Ω}. Ω. P ({HT. Since lima→∞ P (X ≤ a) = 1 and lima→∞ P (X ≤ a) = 0. {HT.4. P ({HT. T T }}. Similarly. Because S1 and X are not independent under the probability measure P .2. HT }). (v) Proof. T H}) = 2 + 2 = 4 and P ({HH. knowing the value of X will affect our opinion on the distribution of S1 .2. P ({HT. HT }) = P ({HT.1. HT }) = P ({HT }) = 9 . So P (X = x0 ) = lim P (x0 − n→∞ 1 1 < X ≤ x0 ) = lim (P (X ≤ x0 ) − P (X ≤ x0 − )) = 1. T H} ∩ {HH. (iii) 1 Proof. HT }) = P ({HH}) + P ({HT }) = 1 + 1 = 2 . HH}}. Proof. X and Y are not 26 . 2. 2π 2π The density fY (y) of Y can be obtained by fY (y) = = = = = = fXY (x.5. So E{euX+vY } = E{euX }E{evY }. The density fX (x) of X can be obtained by fX (x) = fX. y)dy = {y≥−|x|} u2 2 and E{evY } = e v2 2 . (iii) Proof. E{euX+vY } = E{euX+vXZ } = E{euX+vXZ |Z = 1}P (Z = 1) + E{euX+vXZ |Z = −1}P (Z = −1) 1 1 = E{euX+vX } + E{euX−vX } 2 2 (u−v)2 1 (u+v)2 = [e 2 + e 2 ] 2 uv u2 +v 2 e + e−uv = e 2 . y) = fX (x)fY (y). Let u = 0.Proof.Y (x.Y (x. Therefore X and Y cannot 2|x| + y − (2|x|+y)2 2 √ dy = e 2π {ξ≥|x|} ξ2 x2 ξ 1 √ e− 2 dξ = √ e− 2 . 2π ∞ 0∨(−y) ∞ 2x + y − (2x+y)2 2 √ e d(−x) 2π So both X and Y are standard normal random variables. 2 (ii) Proof. Since fX. y)dx 2|x| + y − (2|x|+y)2 2 e 1{|x|≥−y} √ dx 2π ∞ 0∧y 2x + y − (2x+y)2 −2x + y − (−2x+y)2 2 2 √ √ e dx + e dx 2π 2π 0∨(−y) −∞ 2x + y − (2x+y)2 2 √ e dx + 2π 0∨(−y) ∞ ξ2 ξ ξ √ e− 2 d( ) 2 2 2π |y| 1 − y2 √ e 2. E{euX } = e be independent. Proof. d} E{Y 1{X=α} } 1{X=α} . However. P (X = α) (iii) Proof. P (X = α) (iv) Proof.d} E{Y 1{X=α} } 1{X=α} . E{Z|X} − E{Y |X} = E{Z − Y |X} = E{X|X} = X.c. 2. 27 .6. {a. if we set F (t) = ∞ ∞ u2 − u2 √ e 2 t 2π ∞ du.7. E{Y |X} = α∈{a. σ(X) = {∅.b. y)dxdy 2|x| + y − (2|x|+y)2 2 xy1{y≥−|x|} √ dxdy e 2π −∞ −∞ ∞ ∞ 2|x| + y − (2|x|+y)2 2 dy xdx y √ e 2π −∞ −|x| ∞ ∞ ξ2 ξ (ξ − 2|x|) √ e− 2 dξ xdx 2π |x| −∞ ∞ ∞ = = = = xdx( −∞ ∞ |x| ∞ ξ2 ξ2 e− 2 √ e− 2 dξ − 2|x| √ ) 2π 2π x2 = 0 ∞ x x ξ2 ξ2 √ e− 2 dξdx + 2π 0 ∞ x −∞ −x ξ2 ξ2 √ e− 2 dξdx 2π 0 = 0 xF (x)dx + −∞ xF (−x)dx. d}}. (i) ∞ 0 xF (x)dx − ∞ 0 uF (u)du = 0. {c.independent. So E{XY } = 2. b}.b.c. Ω. E{Z|X} = X + E{Y |X} = X + α∈{a. Proof.Y (x. (ii) Proof. we have E{XY } = −∞ ∞ −∞ ∞ xyfX. {ω : ω1 = 6}} and σ(f (X)) equals to {∅.10. g(X)dP A ∞ = E{g(X)1B (X)} = −∞ g(x)1B (x)fX (x)dx yfX. Consider the dice-toss space similar to the coin-toss space. (i) Proof. 28 .Y (x. {ω : ω1 = 1}. No.Proof. We define X(ω) = ω1 and f (x) = 1{odd integers} (x). 2. we have V ar(Y − X) = = = = E{(Y − X − µ)2 } E{[(Y − E{Y |G}) + (E{Y |G} − X − µ)]2 } V ar(Err) + 2E{(Y − E{Y |G})ξ} + E{ξ 2 } V ar(Err) + 2E{Y ξ − E{Y |G}ξ} + E{ξ 2 } = V ar(Err) + E{ξ 2 } ≥ V ar(Err). with ωi ∈ {1. Then a typical element ω in this space is an infinite sequence ω1 ω2 ω3 · · · .8. y)dxdy = = = E{Y 1B (X)} = E{Y IA } = A Y dP. {ω : ω1 = 1} ∪ {ω : ω1 = 3} ∪ {ω : ω1 = 5}. Proof. {ω : ω1 = 2} ∪ {ω : ω1 = 4} ∪ {ω : ω1 = 6}}. 6} (i ∈ N). and each of these containment is strict. So {∅. Note ξ is G-measurable. It suffices to prove the more general case. Ω. For any σ(X)-measurable random variable ξ. Proof. · · · .Y (x. Let µ = E{Y − X} and ξ = E{Y − X − µ|G}. y) dy1B (x)fX (x)dx fX (x) y1B (x)fX.9. 2. E{Y2 ξ} = E{(Y − E{Y |X})ξ} = E{Y ξ − E{Y |X}ξ} = E{Y ξ} − E{Y ξ} = 0. Ω. Then it’s easy to see σ(X) = {∅. 2. σ(f (X)) ⊂ σ(X) is always true. (ii) Proof. 2. · · · . Ω} ⊂ σ(f (X)) ⊂ σ(X). We can find a sequence {Wn }n≥1 of σ(X)-measurable simple functions such that Wn ↑ W .2.4. Note E{Y |X} is σ(X)-measurable. This is a contradiction with limn→∞ j=0 (Wtj+1 − Wtj )2 = T a. So E[(X − µ)4 ] = ϕ(4) (0) = 3σ 4 . since limn→∞ max0≤k≤n−1 |Wtk+1 − Wtk |(ω) = 0. and ϕ(4) (u) = σ 2 ue 2 σ 1 2 2 σ 4 u2 ) + e 2 σ u (3σ 4 + 2σ 4 u). E[Wt2 − Ws |Fs ] = E[(Wt − Ws )2 + 2Wt Ws − 2Ws |Fs ] = t − s + 2Ws E[Wt − Ws |Fs ] = t − s. Proof. Wn = i=1 an 1{X∈Bi } = i=1 an 1Bi (X) = gn (X). By taking upper limits on i both sides of Wn = gn (X). So in particular. we can find a Borel function g such that E{Y |X} = g(X). Then for every ω ∈ A. Wu2 − Wu1 is independent of Ft . We have Ft ⊂ Fu1 and Wu2 − Wu1 is independent of Fu1 . j=0 (Wtj+1 −Wtj )2 (ω) ≤ max0≤k≤n−1 |Wtk+1 −Wtk |(ω) j=0 |Wtj+1 −Wtj |(ω) → n−1 0.1. Proof.3. i. then g is a Borel function. So each An can be i i i i Kn Kn n n n n written as {X ∈ Bi } for some Borel subset Bi of R. Define g = lim sup gn .2. Note j=0 (Wtj+1 − Wtj )3 ≤ max0≤k≤n−1 |Wtk+1 − Wtk | argument similar to (i).s. 3. Each Wn Kn can be written in the form i=1 an 1An . (ii) Proof. (ii) Proof. limn j=0 |Wtj+1 − Wtj |(ω) < n−1 n−1 ∞. n−1 n−1 j=0 (Wtj+1 n−1 1 2 2 1 2 2 1 2 2 1 2 u2 (3σ 4 u+ − Wtj )2 → 0 as n → ∞. ϕ(3) (u) = 2σ 4 ue 2 σ u +(σ 2 +σ 4 u2 )σ 2 ue 2 σ u = e 2 σ u (3σ 4 u+σ 4 u2 ). 3. 2 2 Proof. i i Kn n where gn (x) = i=1 an 1Bi (x). 3. Brownian Motion 3. (i) Proof. such that P (A) > 0 and for every ω ∈ A.5. by an 29 . Assume there exists A ∈ F. we get W = g(X). 3. 3.e. (i) Proof. where An ’s belong to σ(X) and are disjoint..11. By (i). 3. (i) Proof. y)dy with p(τ. Ss . x. = 3. So E[f (Xt )|Fs ] = (ii) Proof. ∞ −∞ f (y)p(t − s.7. z)dz g(Ss ). y) = √ 1 e− 2πτ (y−x−µτ )2 2τ . S0 )) − Ke −rT N (d− (T.6. Xs . 1 2π(t − s) 1 2π(t − s) e − E[f (St )|Fs ] = −∞ S0 eσy =z ∞ f (S0 eσy ) f (z) 0 ∞ e− (y−Xs −µ(t−s))2 2(t−s) dy dz σz = S ( 1 ln z − 1 ln s −µ(t−s))2 σ S0 σ S0 2 = 0 ∞ f (z) e − (ln z −v(t−s))2 Ss 2σ 2 (t−s) σz 2π(t − s) dz = 0 f (z)p(t − s. S0 )). So by (i).Proof. E[f (St )|Fs ] = E[f (S0 eσXt )|Fs ] with µ = ∞ v σ. E[f (Xt )|Ft ] = E[f (Wt − Ws + a)|Fs ]|a=Ws +µt = E[f (Wt−s + a)]|a=Ws +µt ∞ = −∞ ∞ f (x + Ws + µt) e− e− 2(t−s) 2π(t − s) dy x2 dx (y−Ws −µs−µ(t−s))2 2(t−s) = −∞ f (y) 2π(t − s) = g(Xs ). (i) 30 . E[e−rT (ST − K)+ ] ∞ = e −rT 1 K σ (ln S0 (S0 e −(r− 1 σ 2 )T ) 2 ∞ (r− 1 σ 2 )T +σx 2 e− 2T − K) √ dx 2πT e− 2 − K) √ dy 2π ∞ 1 √ K S0 y2 x2 = = = = e −rT σ 1 √ T (S0 e (ln K S0 √ (r− 1 σ 2 )T +σ T y 2 −(r− 1 σ 2 )T ) 2 S0 e− 2 σ ∞ 1 2 ∞ T σ 1 √ T (ln K S0 −(r− 1 σ 2 )T ) 2 √ √ y2 1 √ e− 2 +σ T y dy − Ke−rT 2π σ T (ln −(r− 1 σ 2 )T ) 2 y2 1 √ e− 2 dy 2π S0 σ 1 √ T (ln K S0 −(r− 1 σ 2 )T )−σ 2 T ξ2 1 √ e− 2 dξ − Ke−rT N 2π S 1 1 √ (ln 0 + (r − σ 2 )T ) K 2 σ T Ke −rT N (d+ (T. Therefore. E[τm ] = (v) Proof. E[eσm−(σµ+ 2 2 σ ↓ 0. E[exp{σXt∧τm − (σµ + 1.n ] = (E[e σ − √n u √ X1. Then Zt∧τm ≤ eσm and on the event {τm = ∞}. Zt∧τm ≤ +σµ)t → 0 as t → ∞. So by an argument similar to Exercise 1. If µ ≥ 0 and σ > 0. ϕn (u) = = E[e e 1 u √n Mnt. Zt∧τm ≤ eσm . E[τm e−ατm ] < ∞ since xe−αx is bounded on [0. (ii) 31 . E (ii) Zt Zs |Fs = E[exp{σ(Wt − Ws ) + σµ(t − s) − (σµ + σ2 2 )(t − s)}] = 1. E[Zt∧τm ] = E[Z0 ] = 1. By bounded convergence theorem. By σ > −2µ > 0. t→∞ t→∞ 2 Let σ ↓ −2µ. (iii) Proof. E[1{τm <∞} Zτm ] = E[ lim Zt∧τm ] = lim E[Zt∧τm ] = 1. by bounded convergence theorem. we get σµ + e 2 σm−( σ 2 −m 2α + µ2 . we have P (τm < ∞) = 1. we get 2 √ 2 E[e−ατm ] = e−σm = emµ−m 2α+µ . m µ < ∞ for µ > 0. Zt∧τm ≤ eσm− 2 σ t → 0 as t → ∞. Let (iv) Proof.n n ])nt = (e u − √n u √ n pn + e −e u − √n σ √ n qn )nt nt u √ n r n +1−e σ √ n σ − √n e −e +e r −n − 1 + e e σ √ n σ − √n .8. Let σµ + σ = α. Proof. σ2 2 > 0. t→∞ 2 σ2 2 )t ∧ τm }] = t→∞ since on the event {τm = ∞}. 3. E[eσm−(σµ+ σ2 2 )τm 1{τm <∞} ] = E[ lim Zt∧τm ] = lim E[Zt∧τm ] = 1. 1 σ2 )τm ] = 1. by monotone increasing theorem. that is. then we get P (τm < ∞) = e2µm = e−2|µ|m < 1.8. (i) Proof. We note for α > 0. So we get 2 √ 2 E[e−ατm ] = E[e−ατm 1{τm <∞} ] = e−σm = emµ−m 2α+µ . E[e−ατm ] is differentiable and √ ∂ 2 E[e−ατm ] = −E[τm e−ατm ] = emµ−m 2α+µ ∂α Let α ↓ 0. By optional stopping theorem.Proof. ∞). Set α = σµ + σ . Therefore. ln ϕ 1 x2 1 x2 (u) = e ux rx2 + 1 − e−σx eσx − e−σx −e −ux rx2 + 1 − eσx eσx − e−σx t x2 .Proof. By the one-to-one 2 2 variable with mean t(r − 4.n )n converges to a Gaussian r σ σ random variable with mean t( σ − 2 ) and variance t. σ2 2 ) and variance σ 2 t. Stochastic Calculus 4.1. cosh ux + = = (rx2 + 1 − cosh σx) sinh ux sinh σx 2 2 (rx2 + 1 − 1 − σ 2x + O(x4 ))(ux + O(x3 )) u2 x2 1+ + O(x4 ) + 2 σx + O(x3 ) 2 (r − σ )ux3 + O(x5 ) u2 x2 2 1+ + + O(x4 ) 2 σx + O(x3 ) 2 (r − σ )ux3 (1 + O(x2 )) u2 x2 2 = 1+ + + O(x4 ) 2 σx(1 + O(x2 )) u2 x2 rux2 1 = 1+ + − σux2 + O(x4 ).n r ] = ϕn (u) → 1 tu2 + t( σ − σ )u. sinh σx ln (iii) Proof. 32 . and E[e 1 u √n Mnt. ϕ So. x2 2 σ 2 x 2 σ 2 2 So limx↓0 ln ϕ 1 correspondence between distribution and moment generating function. ln ϕ 1 x2 = t u2 x2 ru 2 σux2 t u2 x2 ru 2 σux2 ln(1 + + x − + O(x4 )) = 2 ( + x − + O(x4 )). 2 σ 2 (iv) Proof. (u) = = = = t x2 t x2 t x2 t x2 (rx2 + 1)(eux − e−ux ) + e(σ−u)x − e−(σ−u)x eσx − e−σx (rx2 + 1) sinh ux + sinh(σ − u)x ln sinh σx 2 (rx + 1) sinh ux + sinh σx cosh ux − cosh σx sinh ux ln sinh σx 2 (rx + 1 − cosh σx) sinh ux ln cosh ux + . ( √n Mnt.n )n converges to a Gaussian random 1 x2 (u) = t( u + 2 ru σ − σu 2 ). Hence ( √n Mnt. Proof. m = k. So E[I(t)−I(s)|Ft ] = ∆tk E[Mt − Ms |Fs ] = 0. (i) Proof. Fix t and for any s < t. (iv) Proof. Then I(t)−I(s) = ∆tk (Mt −Mtk )−∆tk (Ms −Mtk ) = ∆tk (Mt −Ms ). for s < t. 4. Then I(tk ) − I(tl ) = k−1 j=l ∆tj (Wtj+1 − Wtj ). E[I(tk ) − I(tl )] = j=l ∆tj E[Wtj+1 − Wtj ] = 0 and V ar(I(tk ) − I(tl )) = (iii) Proof. Furthermore. u E[I 2 (t) − 0 ∆2 du − (I 2 (s) − u 0 t ∆2 du)|Fs ] u = E[I 2 (t) − I 2 (s) − s ∆2 du|Fs ] u t = E[(I(t) − I(s))2 + 2I(t)I(s) − 2I 2 (s)|Fs ] − s ∆2 du u t = E[(I(t) − I(s))2 ] + 2I(s)E[I(t) − I(s)|Fs ] − s t t ∆2 du u = s ∆2 du + 0 − u s ∆2 du u = 0. Case 1. Hence E[I(t) − I(s)|Fs ] k−1 = j=m+1 E[∆tj E[Mtj+1 − Mtj |Ftj ]|Fs ] + E[∆tk E[Mt − Mtk |Ftk ]|Fs ] + ∆tm E[Mtm+1 − Ms |Fs ] = 0. Case 2. We use the notation in (i) and it is clear I(tk ) − I(tl ) is normal since it is a linear combination of k−1 independent normal random variables. We follow the simplification in the hint and consider I(tk ) − I(tl ) with tl < tk . Then tm ≤ s < tm+1 ≤ tk ≤ t < tk+1 .2. tm+1 ) for some m. m < k. 33 . So k−1 I(t) − I(s) = j=m ∆tj (Mtj+1 − Mtj ) + ∆tk (Ms − Mtk ) − ∆tm (Ms − Mtm ) k−1 = j=m+1 ∆tj (Mtj+1 − Mtj ) + ∆tk (Mt − Mtk ) + ∆tm (Mtm+1 − Ms ). E[I(t) − I(s)|Fs ] = E[I(t) − I(s)] = 0. Combined. we assume s ∈ [tm . t s k−1 j=l ∆2j V ar(Wtj+1 − Wtj ) = t k−1 j=l ∆2j (tj+1 − tj ) = t tk tl ∆2 du. Since ∆t is a non-random process and Wtj+1 − Wtj ⊥ Ftj ⊃ Ftl for j ≥ l. (ii) Proof. we conclude I(t) is a martingale. we must have I(tk ) − I(tl ) ⊥ Ftl . For s < t. (iii) E[I(t) − I(s)|Fs ] = Ws E[Wt − Ws |Fs ] = 0. Proof.4. Øksendal [3]. (Cf.) We first note that B tj +tj+1 (Btj+1 − Btj ) j 2 = j B tj +tj+1 (Btj+1 − B tj +tj+1 ) + Btj (B tj +tj+1 − Btj ) + 2 2 2 (B tj +tj+1 − Btj )2 . k E B tj +tj+1 − Btj 2 − tj+1 − tj 2 2 B tk +tk+1 − Btk 2 − tk+1 − tk 2 = j E B 2j+1 −tj − t 2 tj+1 − tj 2 = j 2· tj+1 − tj 2 2 ≤ T max |tj+1 − tj | → 0.3. Since E[(I(t) − I(s))4 ] = 3E[(I(t) − I(s))2 ]. Proof. j 2 The first term converges in L2 (P ) to   E  j 2 T 0 Bt dBt . since Ws ∈ Fs . (iv) t s E[I 2 (t) − 0 ∆2 du − (I 2 (s) − u 0 ∆2 du)|Fs ] u t 2 Wu du|Fs ] = E[(I(t) − I(s))2 + 2I(t)I(s) − 2I 2 (s) − 2 = E[Ws (Wt − Ws )2 + 2Ws (Wt − Ws ) − 2 2 = Ws (t − s) − Ws (t − s) s 2 Ws (t − s)|Fs ] 2 2 = Ws E[(Wt − Ws )2 ] + 2Ws E[Wt − Ws |Fs ] − Ws (t − s) = 0. For the second term. I(t) − I(s) is not normally distributed. I(t) − I(s) = ∆0 (Wt1 − W0 ) + ∆t1 (Wt2 − Wt1 ) − ∆0 (Wt1 − W0 ) = ∆t1 (Wt2 − Wt1 ) = Ws (Wt − Ws ). Exercise 3.9. (i) I(t) − I(s) is not independent of Fs . 2 1≤j≤n 34 . we note   2 2 B tj +tj+1 − Btj t  −   2 2  tj+1 − tj    2 2   = E  j 2 B tj +tj+1 − Btj 2 − j 2 = j.4. 4 (ii) E[(I(t) − I(s))4 ] = E[Ws ]E[(Wt − Ws )4 ] = 3s · 3(t − s) = 9s(t − s) and 3E[(I(t) − I(s))2 ] = 2 2 3E[Ws ]E[(Wt − Ws ) ] = 3s(t − s). 4. 2 2St 2 (ii) Proof. d ln St = dSt 1d S t 2St dSt − d S − 2 = 2 St 2 St 2St t = 2 2 2St (αt St dt + σt St dWt ) − σt St dt 1 2 = σt dWt + (αt − σt )dt. we assume p = 1.2 4 2 2 since E[(Bt − t)2 ] = E[Bt − 2tBt + t2 ] = 3E[Bt ]2 − 2t2 + t2 = 2t2 .8. (iii) 6 Proof.7. (i) Proof. 2 4. Without loss of generality. dWt6 = 6Wt5 dWt + 1 · 6 · 5Wt4 dt. So WT = 6 2 3 15T . (i) Proof. 2 2 4. (xp ) = p(p − 1)xp−2 . t 0 σs dWs + t (αs 0 − 1 2 2 σs )ds}. So T B tj +tj+1 (Btj+1 − Btj ) → j 2 Bt dBt + 0 T 1 2 = BT in L2 (P ). 2 So St = S0 exp{ 4. Since (xp ) = pxp−1 . T 0 Wt5 dWt + 15 T 0 6 Wt4 dt. 35 . dWt4 = 4Wt3 dWt + (ii) 4 Proof. So WT = 4 T 0 Wt3 dWt + 6 T 0 Wt2 dt. E[WT ] = 6 T 0 1 2 · 4 · 3Wt2 d W t 4 = 4Wt3 dWt + 6Wt2 dt. we have p d(St ) 1 p−1 p−2 = pSt dSt + p(p − 1)St d S t 2 1 p−1 p−2 2 = pSt (αSt dt + σSt dWt ) + p(p − 1)St σ 2 St dt 2 1 p = St [pαdt + pσdWt + p(p − 1)σ 2 dt] 2 p−1 2 p = St p[σdWt + (α + σ )dt]. t t ln St = ln S0 + 0 σs dWs + 0 1 2 (αs − σs )ds. Hence E[WT ] = 15 T 0 3t2 dt = 4. Proof.6.5. tdt = 3T 2 . x) − rKe−r(T −t) N (d− ) − xN (d+ ) d+ (T − t. √ σ 2 (T −t) −r(T −t) σ T −td+ − 2 (ii) Proof. x) ∂x ∂x = N (d+ ). x) ∂x ∂x ∂ ∂ = N (d+ ) + xN (d+ ) d+ (T − t. (i) Proof. d(eβt Rt ) = βeβt Rt dt + eβt dRt = eβt(αdt+σdWt ) . Hence t eβt Rt = R0 + 0 eβs (αds + σdWs ) = R0 + t −β(t−s) e dWs . 0 and Rt = R0 e−βt + α (1 − e−βt ) + σ β 4.9.Proof. ct = xN (d+ ) ∂ ∂ d+ (T − t. −r(T −t) Ke N (d− ) = = = = = Ke Ke Ke −r(T −t) e − d2 − 2 √ 2π (d+ −σ √ 2 T −t)2 − −r(T −t) e √ 2π e e N (d+ ) x (r+ σ2 )(T −t) − σ2 (T −t) 2 2 Ke−r(T −t) e e N (d+ ) K xN (d+ ). = N (d+ ) + xN (d+ ) (iii) Proof. 2 T −t (iv) 36 . x) ∂x ∂t ∂ ∂ σ = xN (d+ ) d+ (T − t. 0 α βt (e − 1) + σ β t eβs dWs . x) − xN (d+ ) d+ (T − t. x) − Ke−r(T −t) N (d− ) d− (T − t. cx ∂ ∂ d+ (T − t. x) + √ ∂t ∂t 2 T −t σx −r(T −t) = −rKe N (d− ) − √ N (d+ ). x) − rKe−r(T −t) N (d− ) − Ke−r(T −t) N (d− ) d− (T − t. x) > 0 and limt↑T d+ (T − t. d+ (T − t. (i) 37 . limx↓0 c(t. Similarly. x) = xN (lim d+ ) − KN (lim d− ) = t↑T t↑T t↑T x − K. x) = √ √ 1 x 1 limτ ↓0 d− (τ. 0.Proof. 4. K). if x ≤ K (vi) Proof. x) = limτ ↓0 d+ (τ. Also it’s clear that limt↑T d± = 0 for x = K. we get x = K exp{σ T − td+ − (T − t)(r + 2 σ 2 )}. limt↑T d± = −∞ for x ∈ 2 (0. So lim c(t. (vii) Proof. For t ∈ [0. x) = limτ ↓0 σ√τ ln K + σ (r + 1 σ 2 ) τ − σ τ = ∞. if x > K = (x − K)+ . So for t ∈ [0. (v) Proof. So we have −x e− 2 −Keσ lim x(N (d+ ) − 1) = lim N (d+ ) √ = lim √ x→∞ x→∞ σ T − t d+ →∞ 2π Therefore x→∞ d2 + √ T −td+ −(T −t)(r+ 1 σ 2 ) 2 √ σ T −t = 0. limt↑T d− (T − t. lim [c(t. T ]. x) = ∞. x) − (x − e−r(T −t) K)] lim [xN (d+ ) − Ke−r(T −t)N (d− ) − x + Ke−r(T −t) ] = = = = x→∞ x→∞ x→∞ lim [x(N (d+ ) − 1) + Ke−r(T −t) (1 − N (d− ))] lim x(N (d+ ) − 1) + Ke−r(T −t) (1 − N ( lim d− )) x→∞ 0. It is easy to see limx↓0 d± = −∞.10. x) = −rKe−r(T −t) N (d− ) − √ 2 ∂x 2 T −t σx 1 1 = rc − √ N (d+ ) + σ 2 x2 N (d+ ) √ 2 2 T −t σ T − tx = rc. it is clear limx→∞ d± = ∞. T ]. 1 ct + rxcx + σ 2 x2 cxx 2 1 σx ∂ N (d+ ) + rxN (d+ ) + σ 2 x2 N (d+ ) d+ (T − t. Note ∂ N (d+ ) σ√1 −t N (d+ ) ∂x d+ T lim x(N (d+ ) − 1) = lim = lim . x)) = 0. x→∞ x→∞ x→∞ −x−2 −x−1 √ 1 By the expression of d+ . x)) − Ke−r(T −t) N (limx↓0 d− (T − t. For x > K. x) = limx↓0 xN (limx↓ d+ (T − t. 3). Finally. cx (t. we clarify the problems by stating explicitly the given conditions and the result to be proved. St )St Nt Xt − ∆t St = = . St ) + (α − r)cx (t. So dXt = ∆t dSt +Γt dMt ⇐⇒ St d∆t + dSt d∆t + Mt dΓt + dMt dΓt = 0. St ) = Xt (by the calculations in Subsection 4. St )dWt 2 1 2 2 2 rc(t.10. dXt = ∆t dSt +Γt dMt +(St d∆t +dSt d∆t +Mt dΓt +dMt dΓt ). St )dt + cx (t. St )dt + cx (t.5.9) ⇐⇒ (4. assuming X has the representation Xt = ∆t St + Γt Mt . by the self-financing property and ∆t = cx (t. Indeed. St )σ 2 St dt + [cx (t.15). St ) dt + σ2 St cx (t. St ) + cxx (t. Proof. St ). 2 where Nt = c(t.10.Proof. St )dt + Γt dMt = ∆t dSt + Γt dMt = dXt .1-4.9) ⇐⇒ (4. St )dSt + Γt dMt − dXt ].10.10. St )σ 2 St dt + Mt dΓt + dMt dΓt + [cx (t. St ) = = = . We show (4.e. x) solves the Black-Scholes-Merton PDE with volatility σ1 : 2 ∂ ∂ 1 2 ∂ + rx + x2 σ1 2 − r c(t. This uniquely determines Γt as Γt = Moreover. Mt Mt Mt By self-financing property. (4. First.10.11. St )dWt . Indeed.10. We let c(t. St ) dt + σ2 St cx (t. St ) = 0.16) + (4. St ).e. St ) + cxx (t.10.5.15). St ) − cx (t. St )(αSt dt + σ2 St dWt ) + cxx (t. St ) + σ1 St cxx (t.15). (4. we note c(t. “continuous-time self-financing condition” has two equivalent formulations. St )St + St (σ2 − σ1 )cxx (t.16) + (4. St )σ 2 St dt = dNt − Mt dΓt − dMt dΓt = Γt dMt = Γt rMt dt = rNt dt. St ) − rc(t. St )dSt − d(Xt − Γt Mt )] 2 1 2 ct (t. St ) + σ 2 St cxx (t. we assume the portfolio is self-financing. St ) + αcx (t.9) or (4. St )d St t − d(∆t St ) 2 1 2 ct (t. 2 1 2 2 ct (t. St ) + cxx (t. 2 c(t. 2 38 and dc(t. St ) − ∆t St . St ) denote the price of call option at time t and set ∆t = cx (t. so 1 2 ct (t. First. St )St + σ2 St cxx (t. i. We assume we have a portfolio Xt = ∆t St + Γt Mt . The problem is to show 1 2 rNt dt = ct (t. St ) dt. x) = 0. St )σ2 St dt 2 1 2 2 ct (t. ∂t ∂x 2 ∂x So 1 2 2 ct (t.10. dNt = = = 1 ct (t. (ii) Proof. 2 4. St )dSt + cxx (t. St ) + rSt cx (t. we have c(t. i. = This implies Xt = X0 ert . By the put-call parity. Solve the equation for α(t) and β(t). We suppose (W1 . For more details. he should short the underlying stock and put p(t. St ) + rSt cx (t. 4. 4. we conclude Xt = 0 for all t ∈ [0. since ∆t = px (t. St ) − cx (t. x)) and pt (t. we have β(t) = √ 1 1−ρ2 (t) (1) (2) . x)) − 1. St )αSt dt + [σ2 St cx (t.12. St ) + rXt − rc(t. see Wilmott [8]. x)) − √ 2 T −t (ii) Proof. x) = cxx (t. So px (t. x) = x − Ke−r(T −t) satisfies the Black-Scholes-Merton partial differential equation. x) = cx (t. (i) Proof. For an agent hedging a short position in the put. So and α(t) = − √ ρ(t)2 1−ρ (t) W1 (t) = B1 (t) t W2 (t) = 0 √−ρ(s) dB1 (s) + 2 1−ρ (s) t 0 √ 1 dB2 (s) 1−ρ2 (s) (3) 39 . T ].5. x) − 1 = N (d+ (T − t. x)) − √ N (d+ (T − t. x)). x) + re−r(T −t) K σx = −rKe−r(T −t) N (d− (T − t. x) = σx√T −t N (d+ (T − t. Proof. x) = −rKe−r(T −t) + σ 2 x2 · 0 + rx · 1 − r(x − Ke−r(T −t) ) = 0. x)) + rKe−r(T −t) 2 T −t σx N (d+ (T − t. 1 1 ∂ ∂2 ∂ + σ 2 x2 2 + rx − r f (t. We want to find α(t) and β(t) such that (dW2 (t))2 = [α2 (t) + β 2 (t) + 2ρ(t)α(t)β(t)]dt = dt dW1 (t)dW2 (t) = [α(t) + β(t)ρ(t)]dt = 0. = rKe−r(T −t) N (−d− (T − t. 1 pxx (t. ∂t 2 ∂x ∂x 2 Remark: The Black-Scholes-Merton PDE has many solutions. St ) + (α − r)cx (t. By X0 .29). W2 ) is a pair of local martingale defined by SDE dW1 (t) = dB1 (t) dW2 (t) = α(t)dB1 (t) + β(t)dB2 (t). x) < 0.Therefore dXt = 1 2 2 2 rc(t. Indeed. Proper boundary conditions are the key to uniqueness. c(t. St )σ2 St ]dWt 2 rXt dt. St ) 2 1 2 2 2 − (σ2 − σ1 )St cxx (t. x) − p(t. St )St + St (σ2 − σ1 )σxx (t. By (4. St ) − cx (t. St ) − px (t. x) = ct (t.13. it suffices to show f (t. x) = x − e−r(T −t) K. (iii) Proof. St )St (> 0) cash in the money market account. 15. E[ (iii) Proof. since n−1 j=0 E[(f (Wtj ))2 ](tj+1 − tj ) → T 0 E[(f (Wt ))2 ]dt < ∞. (ii) Proof. (i) 40 .14. we have 2 E[Zj |Ftj ] = = = = [f (Wtj )]2 E[(Wtj+1 − Wtj )4 − 2(tj+1 − tj )(Wtj+1 − Wtj )2 + (tj+1 − tj )2 |Ftj ] [f (Wtj )]2 (E[Wt4 −tj ] − 2(tj+1 − tj )E[Wt2 −tj ] + (tj+1 − tj )2 ) j+1 j+1 [f (Wtj )]2 [3(tj+1 − tj )2 − 2(tj+1 − tj )2 + (tj+1 − tj )2 ] 2[f (Wtj )]2 (tj+1 − tj )2 . we have B1 (t) = W1 (t) t B2 (t) = 0 ρ(s)dW1 (s) + (4) t 0 1 − ρ2 (s)dW2 (s). Equivalently. Moreover E[Zj |Ftj ] = f (Wtj )E[(Wtj+1 − Wtj )2 − (tj+1 − tj )|Ftj ] = f (Wtj )(E[Wt2 −tj ] − (tj+1 − tj )) = 0. Clearly Zj ∈ Ftj+1 . 4. (i) Proof. where we used the independence of Browian motion increment and the fact that E[X 4 ] = 3E[X 2 ]2 if X is Gaussian with mean 0. Finally. V ar[ j=0 Zj ] = E[( j=0 n−1 Zj )2 ] 2 Zj + 2 = E[ j=0 n−1 Zi Zj ] 0≤i<j≤n−1 = j=0 n−1 2 E[E[Zj |Ftj ]] + 2 0≤i<j≤n−1 E[Zi E[Zj |Ftj ]] = j=0 n−1 E[2[f (Wtj )]2 (tj+1 − tj )2 ] 2E[(f (Wtj ))2 ](tj+1 − tj )2 j=0 n−1 = ≤ 2 0≤j≤n−1 max |tj+1 − tj | · j=0 E[(f (Wtj ))2 ](tj+1 − tj ) → 0. 4.is a pair of independent BM’s. n−1 n−1 n−1 j=0 Zj ] = E[ n−1 j=0 E[Zj |Ftj ]] = 0 by part (i). j+1 since Wtj+1 − Wtj is independent of Ftj and Wt ∼ N (0. t). dBm (t))tr = A(t)(dW1 (t). i = 1. This motivates us to define A as the square root of C. Bi is a local martingale with  (dBi (t))2 =  j=1 d 2 σij (t) dWj (t) = σi (t) d j=1 2 σij (t) dt = dt. dWm (t)) = A(t)−1 (dB1 (t). dBm (t))(A(t)−1 )tr dt = Im×m dt. where Im×m is the m × m unit matrix. 4. · · · . dBm (t)) = C(t). · · · . dWm (t))tr . we get (dB1 (t). Proof. We will try to solve all the sub-problems in a single. So A(t)−1 C(t)(A(t)−1 )tr = Im×m . 2 σi (t) So Bi is a Brownian motion. and (dW1 (t). Reverse the above analysis. dBm (t))tr . dBm (t))tr (dB1 (t). 4. which gives C(t) = A(t)A(t)tr .  dBi (t)dBk (t) =  j=1 d  σij (t) dWj (t) σi (t) d l=1 σkl (t) dWl (t) σk (t) = 1≤j. dBm (t))tr (dB1 (t).16. · · · . l≤d d σij (t)σkl (t) dWj (t)dWl (t) σi (t)σk (t) = j=1 σij (t)σkj (t) dt σi (t)σk (t) = ρik (t)dt. · · · . · · · . · · · . We start with the general Xi : t t Xi (t) = Xi (0) + 0 θi (u)du + 0 σi (u)dBi (u). dWm (t))tr (dW1 (t). we need to find A(t) = (aij (t)) so that (dB1 (t). · · · . Proof. Wm (t). · · · . we obtain a formal proof. dWm (t))tr = A(t)−1 (dB1 (t). · · · . · · · . long solution. To find the m independent Brownion motion W1 (t). or equivalently (dW1 (t). By the condition dBi (t)dBk (t) = ρik (t)dt. · · · . 41 .17.Proof. (ii) Proof. 2. E t0 t + (Θi (u) − Θi (t0 ))du|Ft0 ≤E t0 t + |Θi (u) − Θi (t0 )|du|Ft0 Since 1 t00 |Θi (u) − Θi (t0 )|du ≤ 2M and lim →0 1 t00 |Θi (u) − Θi (t0 )|du = 0 by the continuity of Θi . 2). 2. t0 + t0 + E[Xi (t0 + ) − Xi (t0 )|Ft0 ] t0 + t0 + E t0 Θi (u)du + t0 t0 + σi (u)dBi (u)|Ft0 Θi (t0 ) + E t0 (Θi (u) − Θi (t0 ))du|Ft0 . This proves (iii). So E[(Xi (t0 + ) − Xi (t0 ))(Xj (t0 + ) − Xj (t0 ))|Ft0 ] t0 + t0 + = E Yi (t0 + ) − Yi (t0 ) + t0 Θi (u)du Yj (t0 + ) − Yj (t0 ) + t0 t0 + Θj (u)du |Ft0 t0 + = E [(Yi (t0 + ) − Yi (t0 )) (Yj (t0 + ) − Yj (t0 )) |Ft0 ] + E t0 t0 + Θi (u)du t0 Θj (u)du|Ft0 t0 + +E (Yi (t0 + ) − Yi (t0 )) t0 Θj (u)du|Ft0 + E (Yj (t0 + ) − Yj (t0 )) t0 Θi (u)du|Ft0 = I + II + III + IV. By an argument similar to that involved in the proof of part (iii). for i = 1. we conclude I = σi (t0 )σj (t0 )ρij (t0 ) + o( ) and t0 + t0 + t0 + II = E t0 (Θi (u) − Θi (t0 ))du t0 Θj (u)du|Ft0 + Θi (t0 ) E t0 Θj (u)du|Ft0 = o( ) + (Mi ( ) − o( ))Mj ( ) = Mi ( )Mj ( ) + o( ). the Dominated Convergence Theorem under Conditional Expectation implies 1 lim E →0 t0 + |Θi (u) − Θi (t0 )|du|Ft0 = E lim t0 1 t0 + →0 |Θi (u) − Θi (t0 )|du|Ft0 = 0. First. t0 + I = E[Yi (t0 + )Yj (t0 + ) − Yi (t0 )Yj (t0 )|Ft0 ] = E t0 σi (u)σj (u)ρij (t)dt|Ft0 . t0 So Mi ( ) = Θi (t0 ) + o( ). 42 . t To calculate the variance and covariance. we note Yi (t) = 0 σi (u)dBi (u) is a martingale and by Itˆ’s o t formula Yi (t)Yj (t) − 0 σi (u)σj (u)du is a martingale (i = 1. we have Mi ( ) = = = By Conditional Jensen’s Inequality.The goal is to show lim ↓0 C( ) V1 ( )V2 ( ) = ρ(t0 ). t0 + III ≤ E |Yi (t0 + ) − Yi (t0 )| t0 |Θj (u)|du|Ft0 ≤ M ≤ M ≤ M E[(Yi (t0 + ) − Yi (t0 ))2 |Ft0 ] E[Yi (t0 + )2 − Yi (t0 )2 |Ft0 ] t0 + E[ √ t0 Θi (u)2 du|Ft0 ] ≤ M ·M = o( ) Similarly. we get dζt = −θζt dWt − rζt dt. Since d(ert ζt ) = ζt dXt + Xt dζt + dXt dζt ζt (rXt dt + ∆t (α − r)St dt + ∆t σSt dWt ) + Xt (−θζt dWt − rζt dt) +(rXt dt + ∆t (α − r)St dt + ∆t σSt dWt )(−θζt dWt − rζt dt) ζt (∆t (α − r)St dt + ∆t σSt dWt ) − θXt ζt dWt − θ∆t σSt ζt dt ζt ∆t σSt dWt − θXt ζt dWt . (ii) Proof. Bt is a local martingale with [B]t = motion. (This can be seen as a version of risk-neutral pricing. X0 = ζ0 X0 = E[ζT Xt ] = E[ζT VT ]. where for the second “=”. we have E[(Xi (t0 + ) − Xi (t0 ))(Xj (t0 + ) − Xj (t0 ))|Ft0 ] = Mi ( )Mj ( ) + σi (t0 )σj (t0 )ρij (t0 ) + o( ) + o( ). lim ↓0 C( ) V1 ( )V2 ( ) = lim ↓0 ρ(t0 )σ1 (t0 )σ2 (t0 ) + o( ) 2 2 (σ1 (t0 ) + o( ))(σ2 (t0 ) + o( )) = ρ(t0 ). only that the pricing is carried out under the actual probability measure. d(ert ζt ) = (de−θWt − 2 θ t ) = −e−θWt − 2 θ t θdWt = −θ(ert ζt )dWt . sign(Ws )2 ds = t. we used the fact that e−θWt − 2 θ rert ζt dt + ert dζt . 4. (i) Proof.19. d(ζt Xt ) = = = = So ζt Xt is a martingale. Finally. IV = o( ). This proves part (iv) and (v). Bt is a Brownian e 43 . So by L´vy’s theorem. Part (i) and (ii) are consequences of general cases.By Cauchy’s inequality under conditional expectation (note E[XY |F] defines an inner product on L2 (Ω)). This proves part (vi). (iii) Proof. t 0 1 2 1 2 1 2 t solves dXt = −θXt dWt . By part (ii). (i) Proof.) 4. In summary.18. d(Bt Wt ) = Bt dWt + sign(Wt )Wt dWt + sign(Wt )dt. if x = K undefined. Bt and Wt are not independent.42) cannot hold. If we equal to 0. f (x) = (ii) Proof. the expectation of RHS of (4. o d(Bt Wt2 ) = = = So t t Bt dWt2 + Wt2 dBt + dBt dWt2 Bt (2Wt dWt + dt) + Wt2 sign(Wt )dWt + sign(Wt )dWt (2Wt dWt + dt) 2Bt Wt dWt + Bt dt + sign(Wt )Wt2 dWt + 2sign(Wt )Wt dt.10. if x < K. we get t t E[Bt Wt ] = 0 E[sign(Ws )]ds = 0 E[1{Ws ≥0} − 1{Ws <0} ]ds = 1 1 t − t = 0. Proof. E[f (WT )] = ∞ (x K e 2T − K) √2πT dx = 2 −x standard normal random variable. 2 2 (iii) Proof. if x ≥ K So f (x) = undefined. o (iv) Proof. By Itˆ’s formula. dWt2 = 2Wt dWt + dt.42) is 44 . (i)  1. Integrate both sides of the resulting equation and the expectation.(ii) Proof. So (4. By Itˆ’s formula. E[Bt Wt2 ] = E[ 0 t Bs ds] + 2E[ 0 t sign(Ws )Ws ds] E[sign(Ws )Ws ]ds = 0 E[Bs ]ds + 2 0 t = = = = 2 0 t (E[Ws 1{Ws ≥0} ] − E[Ws 1{Ws <0} ])ds ∞ 0 t 4 0 e− 2s dxds x√ 2πs x2 s ds 2π 0 0 = E[Bt ] · E[Wt2 ].10. (iii) T −K 2T 2π e T suppose 0 2 K − KΦ(− √T ) where Φ is the distribution function of f (Wt )dt = 0. if x = K. 4 Since E[Bt Wt2 ] = E[Bt ] · E[Wt2 ]. 4. if x < K 0.20.  0. if x = K and f (x) =  0. if x > K  x − K. 2. (i) Proof. max0≤t≤T Wt (ω) < K − 2n .10. so that for any n ≥ n0 . So T LK (T )(ω) = lim n n→∞ 0 1 1 1(K− 2n . (vi) Proof. 4. Fix ω. see Hull [2]. 45 1 f (Xt )dt + f (Xt )d X 2 1 f (Xt )(dXt + d X t ) 2 t 1 2 1 2 f (Xt ) σt dWt + (αt − Rt − σt )dt + σt dt 2 2 f (Xt )(αt − Rt )dt + f (Xt )σt dWt . so its expectation is 0.21. x ≤ K − 2n . First. Similarly.. if x < K  (5) lim fn (x) = 1 . so that Wt (ω) < K for any t ∈ [0. for n large enough. the purchases and sales cannot be made at exactly the same price K. if x > K.26) is a martingale. limn→∞ fn (x) = 8n = 0. T ]. so limn→∞ fn (x) = limn→∞ 0 = 0. if x < K. x ≥ K + 2n . Risk-Neutral Pricing 5. T ]. So XT = (ST − K)+ . (iv) 1 1 Proof.Proof.10. If x = K. we can show  0. For more details. if x > K. There are two problems.K+ 2n ) (Wt (ω))dt = 0.20). (ii) Proof.s. Since Wt (ω) can obtain its maximum on [0. for n large enough. 5. In summary. the transaction cost could be big due to active trading.1. second. df (Xt ) = = = = This is formula (5. The RHS of (4. So we cannot have LK (T ) = 0 a. This is trivial to check.45). . there 1 exists n0 . so limn→∞ fn (x) = 1 limn→∞ (x − K) = x − K. (v) Proof. Take expectation on both sides of the formula (4. we have E[LK (T )] = E[(WT − K)+ ] > 0. (i) Proof. if x = K n→∞ 2  1. But E[(ST − K)+ ] > 0. No. limn→∞ fn (x) = (x − K)+ . This is formula (5.2.2. x) T = N (d+ (T.x)} dz 2π √ T )2 (z−σ 1 2 √ e− 2π −∞ N (d+ (T. If we set ZT = eσWT − 2 σ T and Zt = E[ZT |Ft ].4. by Girsanov’s Theorem.x)} dz (ii) Proof.1. then P is a probability measure equivalent to P . ST = xeσWT +(r− 2 σ 1 2 1 2 t (−σ)du 0 = Wt − σt is a P -Brownian motion (set Θ = −σ )T = xeσWT +(r+ 2 σ 1 2 1 2 )T . cx (0. Moreover. Therefore (5.2. x)).2. P (ST > K) = P (xeσWT +(r+ 2 σ )T > K) = P 46 . x) = = = = = = = = 1 2 d E[e−rT (xeσWT +(r− 2 σ )T − K)+ ] dx 1 2 d E e−rT h(xeσWT +(r− 2 σ )T ) dx DT VT ZT Zt |Ft . E[DT VT |Ft ] = E E[DT VT ZT |Ft ]. (i) Proof. and cx (0.3. then Z is a P -martingale.) (iii) Proof. Wt = Wt + in Theorem 5. d(Dt St ) = St dDt + Dt dSt + dDt dSt = −St Rt Dt dt + Dt αt St dt + Dt σt St dWt = Dt St (αt − Rt )dt + Dt St σt dWt . 5. 1{z−σ√T >−d+ (T. x)).. Proof. x) = E[ZT 1{ST >K} ] = P (ST > K). Zt > 0 and E[ZT ] = 1 2 E[eσWT − 2 σ T ] = 1. So if we define P by dP = ZT dP on FT .20).2.(ii) Proof. 5. By Lemma 5. So WT √ > −d+ (T.30) is equivalent to Dt Vt Zt = E e−rT eσWT +(r− 2 σ e e 1 − 2 σ2 T 1 2 )T 1 {xeσWT +(r− 2 σ 1 2 )T >K} E e σ WT √ σ T 1{WT > 1 (ln K −(r− 1 σ2 )T )} σ x 2 WT √ T 1 − 2 σ2 T E e 1 √ W { √T −σ T > T σ 1 √ T (ln K x √ −(r− 1 σ 2 )T )−σ T } 2 e− 2 σ ∞ 1 2 ∞ T −∞ √ z2 1 √ e− 2 eσ T z 1{z−σ√T >−d+ (T. 5.4. First, a few typos. In the SDE for S, “σ(t)dW (t)” → “σ(t)S(t)dW (t)”. In the first equation for c(0, S(0)), E → E. In the second equation for c(0, S(0)), the variable for BSM should be   1 T 2 1 T r(t)dt, σ (t)dt . BSM T, S(0); K, T 0 T 0 (i) Proof. d ln St = X = − is a Gaussian with X ∼ N ( (ii) Proof. For the standard BSM model with constant volatility Σ and interest rate R, under the risk-neutral measure, we have ST = S0 eY , where Y = (R− 1 Σ2 )T +ΣWT ∼ N ((R− 1 Σ2 )T, Σ2 T ), and E[(S0 eY −K)+ ] = 2 2 eRT BSM (T, S0 ; K, R, Σ). Note R = 1 T (rt 0 T T dSt 1 2 1 1 2 2 St − 2St d S t = rt dt + σt dWt − 2 σt dt. So ST = S0 exp{ 0 (rt − 2 σt )dt + 0 T 1 2 2 σt )dt + 0 σt dWt . The first term in the expression of X is a number and the T 2 random variable N (0, 0 σt dt), since both r and σ ar deterministic. Therefore, T T 2 2 (rt − 1 σt )dt, 0 σt dt),. 2 0 σt dWt }. Let second term ST = S0 eX , 1 T (E[Y ] + 1 V ar(Y )) and Σ = 2 T, S0 ; K, 1 T 1 T V ar(Y ), we can get 1 V ar(Y ) . T E[(S0 eY − K)+ ] = eE[Y ]+ 2 V ar(Y ) BSM So for the model in this problem, c(0, S0 ) = = e− e− T 0 1 E[Y ] + V ar(Y ) , 2 rt dt E[(S0 eX − K)+ ] e BSM T, S0 ; K, 1 T T 0 T 0 1 rt dt E[X]+ 2 V ar(X) 1 T  1 E[X] + V ar(X) , 2 1 V ar(X) T  = 1 BSM T, S0 ; K, T 0 T rt dt, 2 σt dt . 5.5. (i) 1 1 Proof. Let f (x) = x , then f (x) = − x2 and f (x) = 2 x3 . Note dZt = −Zt Θt dWt , so d 1 Zt 1 1 1 2 2 2 Θt Θ2 t = f (Zt )dZt + f (Zt )dZt dZt = − 2 (−Zt )Θt dWt + 3 Zt Θt dt = Z dWt + Z dt. 2 Zt 2 Zt t t (ii) Proof. By Lemma 5.2.2., for s, t ≥ 0 with s < t, Ms = E[Mt |Fs ] = E Zs Ms . So M = Z M is a P -martingale. (iii) Zt Mt Zs |Fs . That is, E[Zt Mt |Fs ] = 47 Proof. dMt = d Mt · 1 Zt = 1 1 1 Γt M t Θt M t Θ2 Γ t Θt t dMt + Mt d + dMt d = dWt + dWt + dt + dt. Zt Zt Zt Zt Zt Zt Zt (iv) Proof. In part (iii), we have dMt = Let Γt = 5.6. Proof. By Theorem 4.6.5, it suffices to show Wi (t) is an Ft -martingale under P and [Wi , Wj ](t) = tδij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale under P if and only if Wi (t)Zt is an Ft -martingale under P , since Wi (t)Zt E[Wi (t)|Fs ] = E |Fs . Zs By Itˆ’s product formula, we have o d(Wi (t)Zt ) = Wi (t)dZt + Zt dWi (t) + dZt dWi (t) = Wi (t)(−Zt )Θ(t) · dWt + Zt (dWi (t) + Θi (t)dt) + (−Zt Θt · dWt )(dWi (t) + Θi (t)dt) d Γt M t Θt M t Θ2 Γt Θt Γt M t Θt t dWt + dWt + dt + dt = (dWt + Θt dt) + (dWt + Θt dt). Zt Zt Zt Zt Zt Zt then dMt = Γt dWt . This proves Corollary 5.3.2. Γt +Mt Θt , Zt = Wi (t)(−Zt ) j=1 d Θj (t)dWj (t) + Zt (dWi (t) + Θi (t)dt) − Zt Θi (t)dt = Wi (t)(−Zt ) j=1 Θj (t)dWj (t) + Zt dWi (t) This shows Wi (t)Zt is an Ft -martingale under P . So Wi (t) is an Ft -martingale under P . Moreover, · · [Wi , Wj ](t) = Wi + 0 Θi (s)ds, Wj + 0 Θj (s)ds (t) = [Wi , Wj ](t) = tδij . Combined, this proves the two-dimensional Girsanov’s Theorem. 5.7. (i) Proof. Let a be any strictly positive number. We define X2 (t) = (a + X1 (t))D(t)−1 . Then P X2 (T ) ≥ X2 (0) D(T ) = P (a + X1 (T ) ≥ a) = P (X1 (T ) ≥ 0) = 1, and P X2 (T ) > X2 (0) = P (X1 (T ) > 0) > 0, since a is arbitrary, we have proved the claim of this problem. D(T ) Remark: The intuition is that we invest the positive starting fund a into the money market account, and construct portfolio X1 from zero cost. Their sum should be able to beat the return of money market account. (ii) 48 Proof. We define X1 (t) = X2 (t)D(t) − X2 (0). Then X1 (0) = 0, P (X1 (T ) ≥ 0) = P X2 (T ) ≥ X2 (0) D(T ) = 1, P (X1 (T ) > 0) = P X2 (T ) > X2 (0) D(T ) > 0. 5.8. The basic idea is that for any positive P -martingale M , dMt = Mt · sentation Theorem, dMt = Γt dWt for some adapted process Γt . So martingale must be the exponential of an integral w.r.t. Brownian motion. Taking into account discounting factor and apply Itˆ’s product rule, we can show every strictly positive asset is a generalized geometric o Brownian motion. (i) Proof. Vt Dt = E[e− 0 Ru du VT |Ft ] = E[DT VT |Ft ]. So (Dt Vt )t≥0 is a P -martingale. By Martingale Represent tation Theorem, there exists an adapted process Γt , 0 ≤ t ≤ T , such that Dt Vt = 0 Γs dWs , or equivalently, −1 t −1 t −1 Vt = Dt 0 Γs dWs . Differentiate both sides of the equation, we get dVt = Rt Dt 0 Γs dWs dt + Dt Γt dWt , i.e. dVt = Rt Vt dt + (ii) Proof. We prove the following more general lemma. Lemma 1. Let X be an almost surely positive random variable (i.e. X > 0 a.s.) defined on the probability space (Ω, G, P ). Let F be a sub σ-algebra of G, then Y = E[X|F] > 0 a.s. Proof. By the property of conditional expectation Yt ≥ 0 a.s. Let A = {Y = 0}, we shall show P (A) = 0. In∞ 1 1 deed, note A ∈ F, 0 = E[Y IA ] = E[E[X|F]IA ] = E[XIA ] = E[X1A∩{X≥1} ] + n=1 E[X1A∩{ n >X≥ n+1 } ] ≥ 1 1 1 1 1 P (A∩{X ≥ 1})+ n=1 n+1 P (A∩{ n > X ≥ n+1 }). So P (A∩{X ≥ 1}) = 0 and P (A∩{ n > X ≥ n+1 }) = 0, ∞ 1 1 ∀n ≥ 1. This in turn implies P (A) = P (A ∩ {X > 0}) = P (A ∩ {X ≥ 1}) + n=1 P (A ∩ { n > X ≥ n+1 }) = 0. ∞ Γt Dt dWt . T 1 Mt dMt . By Martingale RepreΓ dMt = Mt ( Mtt )dWt , i.e. any positive By the above lemma, it is clear that for each t ∈ [0, T ], Vt = E[e− t Ru du VT |Ft ] > 0 a.s.. Moreover, by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3.4)), we have the following stronger result: for a.s. ω, Vt (ω) > 0 for any t ∈ [0, T ]. (iii) 1 1 Proof. By (ii), V > 0 a.s., so dVt = Vt Vt dVt = Vt Vt Rt Vt dt + Γt Dt dWt Γt = Vt Rt dt + Vt Vt Dt dWt = Rt Vt dt + T σt Vt dWt , where σt = 5.9. Γt Vt Dt . This shows V follows a generalized geometric Brownian motion. Proof. c(0, T, x, K) = xN (d+ ) − Ke−rT N (d− ) with d± = then f (y) = −yf (y), cK (0, T, x, K) = xf (d+ ) 1 √ σ T x (ln K + (r ± 1 σ 2 )T ). Let f (y) = 2 y √1 e− 2 2π 2 , ∂d+ ∂d− − e−rT N (d− ) − Ke−rT f (d− ) ∂y ∂y −1 1 = xf (d+ ) √ − e−rT N (d− ) + e−rT f (d− ) √ , σ TK σ T 49 and cKK (0, T, x, K) x ∂d− e−rT 1 ∂d+ d− − √ − e−rT f (d− ) + √ (−d− )f (d− ) xf (d+ ) √ f (d+ )(−d+ ) 2 ∂y ∂y ∂y σ TK σ TK σ T x xd+ −1 −1 e−rT d− −1 √ √ − e−rT f (d− ) √ − √ f (d− ) √ f (d+ ) + √ f (d+ ) σ T K2 σ TK Kσ T Kσ T σ T Kσ T x d+ e−rT f (d− ) d− √ [1 − √ ] + √ f (d+ ) [1 + √ ] 2σ T K σ T Kσ T σ T e−rT x f (d− )d+ − 2 2 f (d+ )d− . Kσ 2 T K σ T = = = = 5.10. (i) Proof. At time t0 , the value of the chooser option is V (t0 ) = max{C(t0 ), P (t0 )} = max{C(t0 ), C(t0 ) − F (t0 )} = C(t0 ) + max{0, −F (t0 )} = C(t0 ) + (e−r(T −t0 ) K − S(t0 ))+ . (ii) Proof. By the risk-neutral pricing formula, V (0) = E[e−rt0 V (t0 )] = E[e−rt0 C(t0 )+(e−rT K −e−rt0 S(t0 )+ ] = C(0) + E[e−rt0 (e−r(T −t0 ) K − S(t0 ))+ ]. The first term is the value of a call expiring at time T with strike price K and the second term is the value of a put expiring at time t0 with strike price e−r(T −t0 ) K. 5.11. Proof. We first make an analysis which leads to the hint, then we give a formal proof. (Analysis) If we want to construct a portfolio X that exactly replicates the cash flow, we must find a solution to the backward SDE dXt = ∆t dSt + Rt (Xt − ∆t St )dt − Ct dt XT = 0. Multiply Dt on both sides of the first equation and apply Itˆ’s product rule, we get d(Dt Xt ) = ∆t d(Dt St ) − o T T Ct Dt dt. Integrate from 0 to T , we have DT XT − D0 X0 = 0 ∆t d(Dt St ) − 0 Ct Dt dt. By the terminal T T −1 condition, we get X0 = D0 ( 0 Ct Dt dt − 0 ∆t d(Dt St )). X0 is the theoretical, no-arbitrage price of the cash flow, provided we can find a trading strategy ∆ that solves the BSDE. Note the SDE for S −R gives d(Dt St ) = (Dt St )σt (θt dt + dWt ), where θt = αtσt t . Take the proper change of measure so that Wt = t θ ds 0 s + Wt is a Brownian motion under the new measure P , we get T T T Ct Dt dt = D0 X0 + 0 T 0 ∆t d(Dt St ) = D0 X0 + 0 ∆t (Dt St )σt dWt . T This says the random variable 0 Ct Dt dt has a stochastic integral representation D0 X0 + 0 ∆t Dt St σt dWt . T This inspires us to consider the martingale generated by 0 Ct Dt dt, so that we can apply Martingale Representation Theorem and get a formula for ∆ by comparison of the integrands. 50 dBi (t)dBk (t) = (dBi (t) + γi (t)dt)(dBj (t) + γj (t)dt) = dBi (t)dBj (t) = ρik (t)dt. t 0 Similarly. by part (iii). with X0 = M0 = E[ Ct Dt dt] solves the SDE dXt = ∆t dSt + Rt (Xt − ∆t St )dt − Ct dt XT = 0. by L´vy’s Theorem. so that Mt = M0 + + t 0 Γt dWt . This is the no-arbitrage price of the cash flow at time t. Integrate from t to T . we get T T 0 − Dt Xt = t ∆u d(Du Su ) − t Cu Du du. and we have justified formula (5. o t t t E[Bi (t)Bk (t)] = E[ 0 t Bi (s)dBk (s)] + E[ 0 t Bk (s)dBi (s)] + E[ 0 dBi (s)dBk (s)] = E[ 0 ρik (s)ds] = 0 ρik (s)ds.t. dSi (t) = = = R(t)Si (t)dt + σi (t)Si (t)dBi (t) + (αi (t) − R(t))Si (t)dt − σi (t)Si (t)γi (t)dt d d σij (t) σij (t) d d j=1 σi (t) Θj (t)dt = j=1 σi (t) dWj (t) + σij (t)2 d e j=1 σi (t)2 dt = dt.12. To check the terminal condition. Then by Martingale Representation Theot 0 rem.10) in the textbook. d(Dt Xt ) = ∆t d(Dt St ) − Ct Dt dt. we have found a trading strategy ∆. So Bi is a is a Brownian motion under R(t)Si (t)dt + σi (t)Si (t)dBi (t) + j=1 σij (t)Θj (t)Si (t)dt − Si (t) j=1 σij (t)Θj (t)dt R(t)Si (t)dt + σi (t)Si (t)dBi (t).(Formal proof) Let MT = Xt = −1 Dt (D0 X0 T 0 Ct Dt dt. Since dBi (t)dBi (t) = P. Bi σij (t) d j=1 σi (t) dWj (t). (i) Proof. we can check Cu Du du). and Mt = E[MT |Ft ]. Thus. 51 . dBi (t) = dBi (t) + γi (t)dt = martingale. we note T T T XT DT = D0 X0 + 0 ∆t Dt St σt dWt − 0 Ct Dt dt = M0 + 0 Γt dWt − MT = 0. we can show E[Bi (t)Bk (t)] = (v) ρik (s)ds.r. By Itˆ’s product rule and martingale property. so that the corresponding portfolio X replicates the cash flow and has zero T terminal value. we get T T −1 −Dt Xt = −E[ t Cu Du du|Ft ]. Remark: As shown in the analysis.6. 5. it is easy to see that X satisfies the first equation. (iv) Proof. Ft on both sides. So X0 = E[ 0 Ct Dt dt] is the no-arbitrage price of the cash flow at time zero. So Xt = Dt E[ t Cu Du du|Ft ]. Take conditional expectation w. Indeed. If we set ∆t = T 0 ∆u d(Du Su ) − t 0 Γt Dt St σt . (iii) Proof. (ii) Proof. So XT = 0. we can find an adapted process Γt . o t t E[B1 (t)B2 (t)] = E[ 0 sign(W1 (u))du] = 0 [P (W1 (u) ≥ 0) − P (W1 (u) < 0)]du = 0. 1 − P (W1 (u) < u) du 2 for any t > 0. 2 5. we are motivated to change the measure t in such a way that St −r 0 Su du−at is a martingale under the new measure. By Itˆ’s product formula. This is a good place to pause and think about the meaning of “martingale measure. T ].9. Cov[W1 (T ). we can find an equivalent probability measure P . So E[B1 (t)B2 (t)] = E[B1 (t)B2 (t)] for all t > 0. W2 (T )] = E[W1 (T )W2 (T )] T T t 0 W1 (u)du] = 0. Equation (5. Hence dSt −rSt dt−adt = [(αt −r)St −a]dt+σSt dWt = σSt Set θt = (αt −r)St −a σSt (αt −r)St −a dt σSt + dWt . and Wt = t θ ds 0 s + Wt .6) can be transformed into d(e−rt Xt ) = ∆t [d(e−rt St ) − ae−rt dt] = ∆t e−rt [dSt − rSt dt − adt]. to make the discounted portfolio value e−rt Xt a martingale.Proof. This is the rational for formula (5. under which S satisfies the SDE dSt = rSt dt + σSt dWt + adt and Wt is a BM.9. E[W1 (t)] = E[W1 (t)] = 0 and E[W2 (t)] = E[W2 (t) − (ii) Proof. t E[B1 (t)B2 (t)] = E[ 0 t sign(W1 (u))du [P (W1 (u) ≥ 0) − P (W1 (u) < 0)]du = 0 t = 0 t [P (W1 (u) ≥ u) − P (W1 (u) < u)]du 2 0 = < 0. for all t ∈ [0. = E 0 T W1 (t)dW2 (t) + 0 W2 (t)dW1 (t) T = E 0 W1 (t)(dW2 (t) − W1 (t)dt) + E 0 T W2 (t)dW1 (t) = −E 0 T W1 (t)2 dt tdt = − 0 1 = − T 2. Meanwhile. we note the SDE for S is dSt = αt St dt+σSt dWt .13.” What is to be a martingale? The new measure P should be such that the discounted value process of the replicating 52 .14.7). 5. So. To do this. (i) Proof. T ].9. we note when the portfolio is self-financing.9. where X is given by (5.7). by Martingale Representation rt t Theorem.9. when the underlying pays dividends. but self-financing with consumption. This would give us a theoretical representation of ∆ by comparison of integrands. However. So (e−rt Xt )t≥0 . If we let ∆t = Γt e t . This is not a coincidence. As indicated in the above analysis. Let VT be a payoff at time T . then the σS value of the corresponding portfolio X satisfies d(e−rt Xt ) = Γt dWt . To do so. Just like solving linear ODE. XT = VT . we first set Ut = e−rt St to remove the rSt dt term. we can find an adapted process Γt .7) under P . the discounted price process of the underlying is a martingale under P . So Dt Xt being a martingale under P is more or less equivalent to Dt St being a martingale under P . (i) Proof. XT = VT . (ii) 1 1 Proof. then for the martingale Mt = E[e−rT VT |Ft ]. then d(e−rt Xt ) = ∆t [d(e−rt St ) − ae−rt dt] = ∆t e−rt σSt dWt . as shown in formula (5. T ].6). we consider Vt = Ut e−σWt . In particular. t then e−rt Xt = E[e−rT VT |Ft ] := Mt . not the discounted price process of the underlying. so that Mt = M0 + 0 Γs dWs . Also note the risk-neutral measure is different from the one in case of no cost of carry. as in the classical Black-Scholes-Merton model without dividends or cost of carry. is a P -martingale. We need to solve the backward SDE dXt = ∆t dSt − a∆t dt + r(Xt − ∆t St )dt XT = VT for two unknowns. if St = S0 Yt + Yt 0 Ys ds. First. d(Dt Xt ) = ∆t d(Dt St ) no longer holds. The SDE for U is dUt = σUt dWt + ae−rt dt. to remove U in the dWt term. In order to avoid arbitrage. So by setting X0 = M0 = E[e−rT VT ]. By Itˆ’s formula. dYt = Yt [σdWt + (r − 2 σ 2 )dt] + 2 Yt σ 2 dt = Yt (σdWt + rdt). we must have Xt = Vt for any t ∈ [0. You can think of martingale measure as a calculational convenience. Ys This shows S satisfies (5. we want Dt Xt to be a martingale under P because we suppose that X is able to replicate the derivative payoff at terminal time. The portfolio is no longer self-financing. then t dSt = S0 dYt + 0 a dsdYt + adt = Ys t S0 + 0 a ds Yt (σdWt + rdt) + adt = St (σdWt + rdt) + adt. X and ∆. not that of Dt St . hence a perfect replication of VT . we have in this case the relation d(Dt Xt ) = ∆t d(Dt St ).9. The difficulty is how to calculate Xt and the magic is brought by the martingale measure in the following line of reasoning: −1 −1 Vt = Xt = Dt E[DT XT |Ft ] = Dt E[DT VT |Ft ]. we must have e−rt Xt = Mt . 2 53 . So d(e−rt Yt ) = o t a σe−rt Yt dWt and e−rt Yt is a P -martingale. That is all about martingale measure! Risk neutral is a just perception. Itˆ’s product formula yields o dVt = = e−σWt dUt + Ut e−σWt 1 (−σ)dWt + σ 2 dt + dUt · e−σWt 2 1 (−σ)dWt + σ 2 dt 2 1 e−σWt ae−rt dt − σ 2 Vt dt. Remark: To obtain this formula for S. referring to the actual effect of constructing a hedging portfolio! Second. What we still want to retain is the martingale property of Dt Xt . Martingale Representation Theorem gives Mt = M0 + 0 Γu dWu for some adapted process Γ.6). for all t ∈ [0. This justifies the risk-neutral pricing of European-type contingent claims in the model where cost of carry exists. if we have (5. Moreover. Another perspective for perfect replication is the following. Thus the portfolio perfectly hedges VT . This is how we choose martingale measure in the above paragraph. under which e−rt Xt is a martingale. we find a probability measure P . or there is cost of carry.portfolio is a martingale. Indeed. 2. E[ST ] = S0 erT − a (1 − erT ). XT = ST − S0 erT − a (erT − 1). we get Xt = St − S0 ert + a (1 − ert ) = St − S0 ert − a (ert − 1). 0 Ys E[ST |Ft ] = S0 E[YT |Ft ] + E YT 0 t a ds + YT Ys T t T a ds|Ft Ys E YT |Ft ds Ys E[YT −s ]ds t = S0 E[YT |Ft ] + 0 a dsE[YT |Ft ] + a Ys t t T = S0 Yt E[YT −t ] + 0 t a dsYt E[YT −t ] + a Ys T t = = S0 + 0 t a ds Yt er(T −t) + a Ys ads Ys er(T −s) ds S0 + 0 a Yt er(T −t) − (1 − er(T −t) ). As we have argued at the beginning of the solution. the portfolio has exactly zero value. After the agent delivers the commodity. (v) Proof. So St = Yt (S0 + t ads ). r (iv) Proof. risk-neutral pricing is valid even in the presence of cost of carry. 54 . d(e−rt Xt ) = d(e−rt St ) − ae−rt dt. (vi) Proof. we have d(St /Yt ) = adt/Yt . T ) = F utS (t. T ). So E[ST |Ft ] is a P -martingale. T ). t dE[ST |Ft ] = aer(T −t) dt + S0 + 0 t ads Ys a (er(T −t) dYt − rYt er(T −t) dt) + er(T −t) (−r)dt r = S0 + 0 ads Ys er(T −t) σYt dWt .Note V appears only in the dt term. F orS (t. We solve the equation E[e−r(T −t) (ST − K)|Ft ] = 0 for K. So F orS (0. r In particular. So F orS (t. t 1 2 1 2 t on both sides of the equation. First. r r r Meanwhile. In particular. )t . So by an argument similar to that of §5.6. we solve the SDE dXt = dSt − adt + r(Xt − St )dt X0 = 0. We follow the hint. Set Yt = eσWt +(r− 2 σ (iii) Proof. and get K = E[ST |Ft ]. and receives the forward price F orS (0. T ). whose value r is ST . T ) = r S0 erT − a (1 − erT ) and hence XT = ST − F orS (0. so multiply the integration factor e 2 σ we get 1 2 1 2 d(e 2 σ t Vt ) = ae−rt−σWt + 2 σ t dt. Integrate from 0 to t on both sides. T ) = E[ST |Ft ] = S0 + t ads 0 Ys Yt er(T −t) − a (1−er(T −t) ). By our analysis in part (i). T ) = F uts (t. the process E[ST |Ft ] is the futures price process for the commodity. it is easy to obtain dZu = bu Zu du + σu Zu dWu . then X satisfies the SDE ¯ ¯ ¯ dXu = au du + (¯u + σu Xu )dWu = (¯u du + γu dWu ) + σu Xu dWu . then Xt = Yt Zt = x · 1 = x and dXu = = = = Yu dZu + Zu dYu + dYu Zu au − σu γu γu du + dWu Zu Zu [Yu bu Zu + (au − σu γu ) + σu γu ]du + (σu Zu Yu + γu )dWu Yu (bu Zu du + σu Zu dWu ) + Zu (bu Xu + au )du + (σu Xu + γu )dWu . a ˆ ˆ Write everything back into the original X. Zt = 1 is obvious. au = e− ¯ u t bv dv au and γu = e− ¯ bv dv ¯ γu . we manipulate the equation (6. If Xu = Yu Zu (u ≥ t). Note the form of Z is similar to that of a geometric Brownian motion. + σu Z u γu du Zu Remark: To see how to find the above solution.2. we consider Xu = Xu e− ˆ dXu = e− u t u t σv dWv .1.2. we multiply on both sides of (6. a ˆ ˆ 2 where au = au e− ˆ ¯ ˆ 1 d Xu e 2 u t σv dWv 2 σv dv and γu = γu e− ˆ ¯ = e2 1 u t 2 σv dv u t σv dWv . to u remove the term bu Xu du. Finally. d u t bv dv− u t 1 σv dWv + 2 u t 2 σv dv = e2 1 u t 2 σv dv− u t σv dWv − u t bv dv [(au − σu γu )du + γu dWu ]. Then σv dWv σv dWv ¯ ¯ [(¯u du + γu dWu ) + σu Xu dWu ] + Xu e− a ¯ u t u t 1 (−σu )dWu + e− 2 u t σv dWv 2 σu du ¯ +(¯u + σu Xu )(−σu )e− γ σv dWv du 1 ˆ 2 ˆ ˆ ˆ = au du + γu dWu + σu Xu dWu − σu Xu dWu + Xu σu du − σu (ˆu + σu Xu )du ˆ ˆ γ 2 1 ˆ 2 = (ˆu − σu γu − Xu σu )du + γu dWu . we have u t 1 ˆ ˆ 1 2 (dXu + Xu · σu du) = e 2 2 [(ˆu − σu γu )du + γu dWu ]. use the integrating factor e u t 2 σv dv u 1 2 σ dv t 2 v .6. Connections with Partial Differential Equations 6. ¯ γ a ¯ ¯ ˆ ¯ To deal with the term σu Xu dWu .e. u t bv dv Xu . 6. (i) Proof. (i) 55 .2. a and γ.4) the integrating factor e− t bv dv . First. (ii) Proof. u ≥ t. we get d Xu e− i. Then d(Xu e− ¯ Let Xu = e− u t u t bv dv ) = e− u t bv dv (au du + (γu + σu Xu )dWu ).4) as follows. Zu This inspired us to try Xu = Yu Zu . Xu Zu = 1 [(au − σu γu )du + γu dWu ] = dYu . So by Itˆ’s o formula. Rt . Rt . 2 56 . T2 )]dWt . T1 )]fr (t. T ) + 2 γ (t. T2 )γ 2 (t. T2 )|dt. then d(Dt Xt ) = ∆(t)Dt −Rt f (t. T1 ) + ∆2 (t)df (t. T1 ) + ∆2 (t)f (t. Rt . Rt . under the assumption that fr (t.9. T2 ) + γ 2 (t. T1 ) + α(t. T1 ). ω. Rt . Rt . Rt )fr (t. This implies β(t. Rt . T1 )dRt + frr (t. Rt . T1 ) + ft (t. T ) dt. T1 ) + ∆2 (t)fr (t. T ) + γ 2 (t. Rt . T2 ) and ∆2 (t) = −St fr (t. T ) does not depend on T . To avoid arbitrage in this case. Rt )frr (t. Rt )frr (t. Rt . Rt . Rt )dt 2 1 +∆2 (t) ft (t. Rt . T2 ) + α(t. Rt . T ) = Rt f (t. In (6. then d(Dt Xt ) = Dt St [β(t. T1 )fr (t. T2 )]dt 2 +Dt γ(t.Proof. Rt )[∆1 (t)fr (t. Rt )frr (t. Rt )fr (t. Rt )frr (t. T ) + ft (t. T2 ))dt] 1 = ∆1 (t)Dt [−Rt f (t. β(t. T2 ) for some t ∈ [0. T1 )fr (t. T ) + 1 γ 2 (t. T2 ). T )dWt . Rt . Rt . T1 ) + ∆2 (t)fr (t. T1 )dt + ∆2 (t)Dt [α(t. Rt . T ]. Rt . T2 )]]dWt = ∆1 (t)Dt [α(t. T ]. we get T DT XT − D0 X0 = 0 Dt |[β(t. (iii) Proof. Rt )[∆1 (t)fr (t. Rt ) − β(t. Rt . Rt . we get d(Dt Xt ) = 1 ∆(t)Dt −Rt f (t. Rt . Rt . Rt . Rt . T2 )|dt. we must have for a. r. Rt . T2 )]fr (t. Rt . T1 )]fr (t. T ) + ft (t. Rt . Rt . Rt . Rt . Rt . Rt . Rt . Rt . T2 ) + ft (t. T1 ) − β(t. Rt . r. Rt . Exercise 5. Rt . T1 )fr (t. and d(Dt Xt ) = −Rt Dt Xt dt + Dt dXt = Dt [∆1 (t)df (t. If β(t. T ) dt 2 +Dt γ(t. T2 )dt + fr (t. r. or equivalently. T2 )dRt + frr (t. T1 ) + ∆2 (t)df (t. Rt )frr (t. Rt .7). r. T ). To avoid arbitrage (see. Rt . T ) = rf (t. Rt . Rt . We 1 2 choose ∆(t) = sign −Rt f (t. Rt ) − β(t. Integrate from 0 to T on both sides of the above equation. Rt )frr (t.s. Rt . Rt . T1 ) − β(t. r. The portfolio is self-financing. we must have ft (t. Rt )[Dt γ(t. T1 )dt + fr (t. Rt . for any r in the 2 range of Rt . Rt . Let ∆1 (t) = St fr (t. Rt . T2 )]fr (t. T ) . T1 )]dt 2 1 +∆2 (t)Dt [−Rt f (t. so for any t ≤ T1 . T ) + 2 γ 2 (t. Rt . 1 If fr (t. T ) + α(t. Rt )∆(t)fr (t. T1 ) + ∆2 (t)f (t.4). Rt . T1 )γ 2 (t. Rt . Rt .5). Rt . T ) + 1 γ 2 (t. Rt . T ) = 0 for all values of r and 0 ≤ t ≤ T . Rt . Rt )fr (t. Rt . T2 ))dt. T1 ) = β(t. T1 ) = β(t. T ) = 0. Rt . Rt . T2 )dt = Dt |[β(t. Rt )dt 2 −Rt (∆1 (t)f (t. we have dXt = ∆1 (t)df (t. T2 )]fr (t. Rt . T1 ) + γ 2 (t. T2 )dt +Dt γ(t. Rt . (ii) Proof. T2 ) + Rt (Xt − ∆1 (t)f (t. This is formula (6. Rt . Rt . Rt . T1 = T and ∆2 (t) = 0. T ) + ft (t. ft (t. Rt . Rt . Rt . for example. T1 ) − ∆2 (t)f (t. Rt . T2 ) − Rt (∆1 (t)f (t. let ∆1 (t) = ∆(t). T2 ) − β(t. DT XT − D0 X0 > 0.9. r)frr (t. T2 ))dt] 1 = Dt [∆1 (t) ft (t. T ). r. ∀t ∈ [0. T )(−bs ) + bs C(s. The characteristic equation of ϕ (t) − bϕ (t) − 2 σ 2 ϕ(t) = 0 is λ2 − bλ − 2 σ 2 = 0. T )] 2 2 1 4 1 σ ϕ(t)C 2 (t.5. It is then easy to see that we can choose appropriate constants c1 and c2 so that ϕ(t) = 1 2b 1 1 c1 c2 e−( 2 b+γ)(T −t) − 1 e−( 2 b−γ)(T −t) . By the definition of ϕ. we get − 2ϕ (t) 1 2 2 2ϕ (t) 1 + σ C (t.8) and (6. (iii) 1 1 Proof. C(u. T ) = 0. we have ϕ (t) = e 2 σ 1 2 T t T (a(s)C(s. T ) = − bv dv T t e− bv dv ds = T e t s bv dv ds. T ) = − ϕ(t)σ 2 C(t. Finally. Proof. T ). So integrate on both sides of the equation from t to T. by A (s. T ) − 1] = −e− s 0 bv dv . we obtain e− T 0 bv dv C(T. T ) = 6. T )ds. T ). So C (t. 4 2 2ϕ (t) σ 2 ϕ(t) . 2 2ϕ (t) 1 So C(t. T ) = − φ(t)σ2 . which gives two √ √ 1 1 1 roots 2 (b ± b2 + 2σ 2 ) = 2 b ± γ with γ = 2 b2 + 2σ 2 . T ) − σ 2 ϕ(t)C (t. T ) = 0. Plug formulas (6. +γ b−γ 2 57 . We note d − e ds s 0 bv dv C(s. T ))ds.4. T ) = (ii) 1 4 2 4 σ ϕ(t)C (t. Therefore by standard theory of ordinary differential 1 equations. t Since A(T. T ) = − t s 0 e− T t s 0 bv dv ds. Since C(T.9. T ) + ϕ(t)C (t. T ) + 2 σ 2 (s)C 2 (s. σ 2 ϕ(t) 2 σ ϕ(t) 2 1 i.9) into (6. (i) Proof.14). T ) − 1. T )] 2 1 1 − σ 2 [− ϕ(t)σ 2 C 2 (t.e. T ). T ) = b(−1) 2 + σ 2 C 2 (t. a general solution of ϕ is ϕ(t) = e 2 bt (a1 eγt + a2 e−γt ) for some constants a1 and a2 . we get A(T. T ) = e 1 −a(s)C(s. we have C(t. T ) − 2 Proof. T ) − e− t 0 t 0 T bv dv C(t. T ) + ϕ(t)C (t. we get ϕ (t) = = = 1 − σ 2 [ϕ (t)C(t. T ) = e− s 0 bv dv [C(s. T ) t − 1 2 2 2 σ (s)C (s. T )ds + t 1 2 σ 2 (s)C 2 (s. T ) − A(t. we have A(t.9. T ). ϕ (t) − bϕ (t) − 2 σ 2 ϕ(t) = 0.6.T )du 1 2 1 σ 2 (−1)C(t. T ) 1 − ϕ (t) / 1 ϕ(t)σ 2 = 2 σ 2 C 2 (t. Differentiate both sides of the equation ϕ (t) = − 2 ϕ(t)σ 2 C(t. T ) = T a(s)C(s.3. T ) = t 2aϕ (s) 2a ϕ(T ) ds = 2 ln . σ2 This implies C(t. and (cosh z) = sinh z. it is easy to see ϕ (t) = c1 e−( 2 b+γ)(T −t) − c2 e−( 2 b−γ)(T −t) . T ) = 2aϕ (t) σ 2 ϕ(t) . 0 = C(T.(iv) Proof.8). σ2 ϕ(t) σ γ cosh(γ(T − t)) + 1 b sinh(γ(T − t)) 2 58 . From part (iii). T ) = − (vi) Proof. T ) = − So c1 = c2 . 2ϕ (t) sinh(γ(T − t)) . A (t. We first recall the definitions and properties of sinh and cosh: sinh z = Therefore ϕ(t) = c1 e− 2 b(T −t) = c1 e− 2 b(T −t) = = and ϕ (t) = 1 2c1 − 1 b(T −t) [b sinh(γ(T − t)) + 2γ cosh(γ(T − t))] b· 2 e 2 2 σ 2c1 1 + 2 e− 2 b(T −t) [−γb cosh(γ(T − t)) − 2γ 2 sinh(γ(T − t))] σ 1 bγ bγ 2γ 2 b2 = 2c1 e− 2 b(T −t) sinh(γ(T − t)) + 2 cosh(γ(T − t)) − 2 cosh(γ(T − t)) − 2 sinh(γ(T − t)) 2 2σ σ σ σ 2 2 1 b − 4γ = 2c1 e− 2 b(T −t) sinh(γ(T − t)) 2σ 2 1 = −2c1 e− 2 b(T −t) sinh(γ(T − t)). T ) − A(t. cosh z = .15) and (6. In particular.9. By (6. σ 2 ϕ(T ) σ ϕ(T ) ez − e−z ez + e−z . 2 2 eγ(T −t) e−γ(T −t) − 1 1 2b + γ 2b − γ 1 2b 1 2 4b 1 − γ −γ(T −t) b + γ γ(T −t) e − 12 2 e 2 2 −γ 4b − γ 1 2c1 − 1 b(T −t) 1 e 2 −( b − γ)e−γ(T −t) + ( b + γ)eγ(T −t) 2 σ 2 2 2c1 − 1 b(T −t) e 2 [b sinh(γ(T − t)) + 2γ cosh(γ(T − t))]. 2 ϕ(s) σ σ ϕ(t) 1 and A(t. T ) = − 2a ϕ(T ) 2a γe 2 b(T −t) ln = − 2 ln .5. (sinh z) = cosh z. = σ 2 ϕ(t) γ cosh(γ(T − t)) + 1 b sinh(γ(T − t)) 2 1 1 1 1 2ϕ (T ) 2(c1 − c2 ) =− 2 . (v) Proof. Hence T A(T. The PDE for f can be similarly obtained. (ii) and (iii) Proof. X2 (t)) and e−rt f (t. X1 (t). X1 (t). (i) Proof.5. X2 (t)) 1 1 = gt dt + gx1 dX1 (t) + gx2 dX2 (t) + gx1 x2 dX1 (t)dX1 (t) + gx2 x2 dX2 (t)dX2 (t) + gx1 x2 dX1 (t)dX2 (t) 2 2 1 2 2 = gt + gx1 β1 + gx2 β2 + gx1 x1 (γ11 + γ12 + 2ργ11 γ12 ) + gx1 x2 (γ11 γ21 + ργ11 γ22 + ργ12 γ21 + γ12 γ22 ) 2 1 2 2 + gx2 x2 (γ21 + γ22 + 2ργ21 γ22 ) dt + martingale part. Since g(t. 2 1 So e 2 bt Xj (t) − Xj (0) = 1 t 1 2σ 0 e 2 bu dWj (u) and Xj (t) = e− 2 bt Xj (0) + 1 σ 2 1 1 1 t 0 e 2 bu dWj (u) . X2 (t)) ar both martingales. 2 Xj (t) d j=1 R(t) dt Let B(t) = then B is a local martingale with dB(t)dB(t) = = dt. we get 1 1 b 1 1 d(e 2 bt Xj (t)) = e 2 bt Xj (t) bdt + (− Xj (t)dt + σdWj (t) 2 2 2 1 1 = e 2 bt σdWj (t). By Theorem = σ2 −bt ). 6.6. j=1 0 R(s) d 2 σ − bR(t) dt + σ 4 R(t) j=1 Xj (t) R(t) dWj (t). Multiply e 2 bt on both sides of (6. X1 (t). Suppose R(t) = d j=1 2 Xj (t). 4b (1 − e 1 4. X1 (t).9. So by L´vy’s Theorem. iterated conditioning argument shows g(t. 59 R(t)dB(t) (a := d σ 2 ) 4 . Xj (t) is normally distributed with mean Xj (0)e− 2 bt and variance (ii) Proof. X2 (t)) = E[e−rT h(X1 (T ).4. then d e−bt 2 t bu e du 4 σ 0 dR(t) = j=1 d (2Xj (t)dXj (t) + dXj (t)dXj (t)) 1 2Xj (t)dXj (t) + σ 2 dt 4 1 2 −bXj (t)dt + σXj (t)dWj (t) + σ 2 dt 4 d = j=1 d = j=1 = t Xj (s) d √ dWj (s). X2 (T ))|Ft ]. 2 Taking ρ = 0 will give part (ii) as a special case.9. X2 (t)) = E[h(X1 (T ).15). Therefore dR(t) = (a − bR(t))dt + σ e and R is a CIR interest rate process. (i) Proof. X2 (T ))|Ft ] and e−rt f (t. 2 So we must have 1 2 2 gt + gx1 β1 + gx2 β2 + gx1 x1 (γ11 + γ12 + 2ργ11 γ12 ) + gx1 x2 (γ11 γ21 + ργ11 γ22 + ργ12 γ21 + γ12 γ22 ) 2 1 2 2 + gx2 x2 (γ21 + γ22 + 2ργ21 γ22 ) = 0. X1 (t). B is a Brownian motion.6. We note dg(t. St . e − 2v(t)/(1−2uv(t)) µ(t) (x− 1−2uv(t) ) 2 dx · e− µ2 (t)(1−2uv(t))−µ2 (t) 2v(t)(1−2uv(t)) 1 − 2uv(t) = e uµ2 (t) − 1−2uv(t) 1 − 2uv(t) (v) Proof. Vt )(a − bVt ) + css (t. Vt )rSt + cv (t. St . By R(t) = d j=1 2 Xj (t) and the fact X1 (t).(iii) Proof. Xd (t) are i.i.d. normal with the same mean µ(t) and variance v(t).16).d. 6. Xj (t) is dependent on Wj only and is normally distributed with mean e− 2 bt Xj (0) and 2 variance σ [1 − e−bt ]. (i) Proof. St . 2 (ii) 60 . This is equation (6. St . Since d(e−rt c(t. St . St . 2 d udµ2 (t) 2a e−bt uR(0) 1−2uv(t) E[euR(t) ] = (E[euX1 (t) ])d = (1 − 2uv(t))− 2 e 1−2uv(t) = (1 − 2uv(t))− σ2 e− . Xd (t) are i. · · · .. 4b (iv) Proof.7. Vt )σVt St ρ dt + martingale part. 2 1 we conclude rc = ct + rscs + cv (a − bv) + 1 css vs2 + 2 cvv σ 2 v + csv σsvρ. Vt )) 1 2 = e−rt c(t. St . Vt ) + cs (t.9. Vt )σ 2 Vt + csv (t.i. So X1 (t).26). e−rt c(t. Vt )(−r) + ct (t. St . · · · .9. By (6. E e 2 uXj (t) 1 ∞ = −∞ ∞ e ux2 e− (x−µ(t))2 2v(t) dx 2πv(t) (1−2uv(t))x2 −2µ(t)x+µ2 (t) − 2v(t) = −∞ ∞ e 2πv(t) 1 2πv(t) e− µ(t) (x− 1−2uv(t) ) 2 dx µ2 (t) µ2 (t) − 1−2uv(t) (1−2uv(t))2 2v(t)/(1−2uv(t)) + = −∞ ∞ dx = −∞ 1 − 2uv(t) 2πv(t) . Vt ) = E[e−rT (ST − K)+ |Ft ] is a martingale by iterated conditioning argument. Vt )Vt St + 2 1 cvv (t. St . log s. Vt ) = 1{XT ≥log K} . v). f (t. v) 2 + e−r(T −t) Kgs (t. v). v) − e−r(T −t) Kg(t.32) for f . c satisfies the PDE (6. log s. v) = sf (t. Vt ) + fx (t. x. so by differentiating f and setting the dt term as zero. 2 1 So we must have ft + (r + 2 v)fx + (a − bv + ρσv)fv + 1 fxx v + 1 fvv σ 2 v + σvρfxv = 0. v) − e−r(T −t) Kgs (t. s. v) − e−r(T −t) Kgvv (t. Vt )(a − bvt + ρσVt ) + fxx (t. 1 1 cs = f (t. First. log s. s s cv = sfv (t. Similar to (iii). log s. s s s s K csv = fv (t. log s.26). Vt ) is a martingale. Vt )Vt 2 2 1 + fvv (t. v) − e−r(T −t) Kgv (t. log s. s cvv = sfvv (t. log s. Suppose c(t. v) . log s. Vt ) = E[1{XT ≥log K} |Ft ]. v) − re−r(T −t) Kg(t. c(T. Vt ) = 1 1 ft (t. 61 . log s. log s. Vt )(r + Vt ) + fv (t. That is. 2 2 (iv) Proof. log s. Xt . Xt .9. then ct = sft (t. (iii) Proof.32). v) − e−r(T −t) Kgss (t.9. we have the PDE (6. This is (6. (v) Proof.9. v) 2 . log s. Xt . So f (T. v). Xt . log s.Proof. log s. log s. log s. v) = sf (T. log s. Xt . 6. 1 1 1 1 css = fs (t. Xt . v) = 1{x≥log K} for all x ∈ R. So 1 1 ct + rscs + (a − bv)cv + s2 vcss + ρσsvcsv + σ 2 vcvv 2 2 = sft − re−r(T −t) Kg − e−r(T −t) Kgt + rsf + rsfs − rKe−r(T −t) gs + (a − bv)(sfv − e−r(T −t) Kgv ) 1 1 gs K K 1 + s2 v − fs + fss − e−r(T −t) 2 gss + e−r(T −t) K 2 + ρσsv fv + fsv − e−r(T −t) gsv 2 s s s s s 1 2 + σ v(sfvv − e−r(T −t) Kgvv ) 2 1 1 1 1 = s ft + (r + v)fs + (a − bv + ρσv)fv + vfss + ρσvfsv + σ 2 vfvv − Ke−r(T −t) gt + (r − v)gs 2 2 2 2 1 1 2 +(a − bv)gv + vgss + ρσvgsv + σ vgvv + rsf − re−r(T −t) Kg 2 2 = rc. v) − e−r(T −t) Kgt (t. log s. v) + sfs (t. v) − e−r(T −t) gsv (t. Xt . Xt . v) + fss (t. by Markov property. log s. v). v ≥ 0. s. log s. f (t. Second. v). log s. Vt )σ 2 Vt + fxv (t. which implies f (T. v) = s1{log s≥log K} − K1{log s≥log K} = 1{s≥K} (s − K) = (s − K)+ . v) + fsv (t.8. v) − e−r(T −t) Kg(T. Vt )σVt ρ dt + martingale part. df (t. Indeed. Xt . Xt . By integration-by-parts formula.43). y)dy − h(x) 1 E t. y) + γ 2 (t. Xu ) + γ 2 (u. We first note dhb (Xu ) = hb (Xu )dXu + 2 hb (Xu )dXu dXu = hb (Xu )β(u. 2 This is (6. ∂y Plug these formulas into (6. x. By the arbitrariness of h and assuming β. Xu )hb (Xu ) du+ hb (Xu )γ(u. x. x. y)p(t. we have b b β(u. x. we get ∞ E t. x. x.x [hb (Xu )β(u.9. x) = 0 ∞ ∂ ∂t ∞ ∞ h(y)p(t. x. x. y)p(t. px . x)px (t. u. T. x. u. x) = gx (t.Proof. y)p(t.9. y)dy. x. Differentiate w. T. y) + γ 2 (u. x. u. We follow the hint. u. T. x. we have b hb (y) 0 ∂ p(t. we get (6.x [hb (XT ) − hb (Xt )] = −∞ T hb (y)p(t. Xu ) + 2 γ 2 (u. x.9. the integration range can be changed from (−∞. y))hb (y)dy = ∂y b 0 ∂2 2 (γ (u.45) implies 0 h(y) pt (t. u. x. we have T hb (XT ) − hb (Xt ) = t 1 hb (Xu )β(u. y)dy = 0 ∞ 0 h(y)pt (t. T. u. x)pxx (t. 2 Since hb vanishes outside (0. T. u.49). y)]hb (y)dy + ∂y 2 62 b 0 ∂2 2 [γ (T. y)p(t. Suppose h is smooth and compactly supported. y)p(t. y) + β(t. y)]hb (y)dy. y) = 0.48).t.9. y) dy = 0. T. Integrate on both sides of the equation. x) = ∞ 1 So (6. u. T on both sides of (6. y)dydu. x. x. 6. y)dy = − ∂T b 0 ∂ 1 [β(T. y)p(t. y))dy. x)px (t. Xu )hb (Xu ) du + martingale part. T. we have 1 pt (t. y) + 2 γ 2 (t. b). h(y)px (t. T. y) + β(t. x)pxx (t.48). T. y)hb (y) p(t.49). y)p(t. Xu )dWu . y)hb (y)dy = − 0 0 ∂ 2 (γ (u. b). ∂y and b b γ 2 (u. T.9. y)p(t. which gives (6. y)p(t. 1 1 Proof.9. v. y)dy. pxx are all continuous. ∞) to (0. x. x.r. 0 gxx (t. y))dy ∂y = − 0 ∂ hb (y) (β(u. x. T. x. x. y)dy. pt . T. T. y)hb (y)dy 0 = hb (y)β(u. 2 Take expectation on both sides. x.9. ∂y 2 . y))hb (y)dy. h(y)pxx (t. then it is legitimate to exchange integration and differentiation: gt (t. Xu ) + γ 2 (u. Xu )hb (Xu )]du 2 ∞ −∞ = t T = t 1 hb (y)β(u. y)|b − 0 0 b hb (y) ∂ (β(u. T. x. x. T.9. p(t. T. K)K 2 cKK (0. K). K) 2 ∞ 1 −rT = re K p(0. T. T. x. T. K) = = −rc(0.58). T. b hb (y) 0 ∂ ∂ 1 ∂2 2 (γ (T. T. y)dy 1 +e−rT σ 2 (T.that is.50) and the arbitrariness of hb . T. x. x. y)dy − e−rT ∂ (ryp(t. x. T. y))dy = K 2 ∂y ∂y K 1 = − (σ 2 (T. x. T. x. x. p(t. y)dy. y))|∞ − (σ (T. K) + e−rT K ∞ (y − K)pT (0.10. x. x. x. y))dy ∂y +e−rT K 1 ∂2 2 (y − K) (σ (T. If we further assume (6. x.9. y)dy = K K ryp(0. K) 2 K 1 2 = −rKcK (0. Under the assumption that limy→∞ (y − K)ryp(0. T. y)y 2 p(t. y) + (β(T. T.57) and (6. ∞ cT (0. y)p(t. T. x. ∂ ∂ 1 ∂2 2 (γ (T. y))dy = −(y−K)ryp(0. T. x.9. y)y 2 p(0. y) = 0. x. we have ∞ − K (y−K) ∂ (ryp(0. y)p(t. we have 1 ∞ ∂2 (y − K) 2 (σ 2 (T. T. y))dy 2 K ∂y ∞ ∂ 2 1 ∂ 2 (y − K) (σ (T. T. y)) dy = 0.50). x. x. we conclude for any y ∈ (0. T. K). T. ∞). 2 7. y)dy (y − K) K = −re−rT K ∞ (y − K)p(0. K)K 2 p(0. y)|∞ ) K 2 1 2 = σ (T. K) + σ (T. y) + (β(T. T. x. y)p(t. (i) 63 . x. Exotic Options 7. y)dy + e−rT σ 2 (T. x. x. T. y)p(t. T. y)y 2 p(0. x. y)) − ∂T ∂y 2 ∂y 2 6. T.9. x. T. T. y))dy 2 ∂y 2 ∞ ∞ = −re−rT K (y − K)p(0. T. x. Proof. y)|∞ + K ∂y ∞ ∞ ryp(0. y)dy ∞ −re−rT K ∞ (y − K)p(0. x. y)y 2 p(0. y)dy + e−rT K ryp(0. y)y 2 p(0. 2 Therefore. x.1. T. K)K 2 p(0. y)) − ∂T ∂y 2 ∂y 2 This is (6. T. y)dy + e−rT K ∞ (y − K)pT (0. T. x. By (6. x. then use integration-by-parts formula twice. T. y)) = 0. T. T. x. K)K 2 p(0. x. s) 4 1 1 2σ 2 τ . s)). Since δ± (τ. (log s+rτ )2 ±σ 2 τ (log s+rτ )+ 1 σ 4 τ 2 δ± (τ. s) − δ± (τ. s)) = √ e− 2 = √ e− 2π 2π Therefore 2σ 2 τ (log s+rτ ) N (δ+ (τ. ) = ∂x x ∂x σ τ σ τ 1 √ 1 √ log log x 1 + (r ± σ 2 )τ c 2 = 1 √ . s)). xσ τ 1 √ . s) = (vi) Proof. s 2r 2 −(log 1 +rτ )2 ±σ 2 τ (log s−log 1 ) s s 2σ 2 τ ] = e− 4rτ log s±2σ 2 τ log s 2σ 2 τ = e−( σ2 ±1) log s = s−( σ2 ±1) . σ τ . δ± (τ. s) ∂t r ± 1 σ 2 1 − 1 ∂τ log s 1 − 3 ∂τ 2 (− )τ 2 + τ 2 σ 2 ∂t σ 2 ∂t 1 r ± 2 σ2 √ 1 log s 1 √ (−1) − = − τ (−1) 2τ σ σ τ 1 1 1 = − · √ − log ss + (r ± σ 2 )τ ) 2τ σ τ 2 1 1 = − δ± (τ. ∂ x ∂ δ± (τ. xσ τ 1 c + (r ± σ 2 )τ x 2 =− (iii) Proof. σ ∂ δ± (τ. s) − δ− (τ. (iv) Proof. s−1 ) = (vii) Proof. δ+ (τ. s)) = sN (δ+ (τ. ). 2τ s (ii) Proof. ) = ∂x c ∂x c ∂ ∂ δ± (τ.Proof. s−1 )) = s( σ2 ±1) N (δ± (τ. 2r 2r So N (δ± (τ. 2 64 . N (y) = y √1 e− 2 2π 2 1 √ σ τ 1 log s + (r + 2 σ 2 )τ − 1 √ σ τ log s + (r − 1 σ 2 )τ = 2 1 √ σ2 τ σ τ √ = σ τ. s)) e−rτ 2σ 2 τ = = e− N (δ− (τ. 1 √ σ τ 1 log s + (r ± 2 σ 2 )τ − 1 √ σ τ log s−1 + (r ± 1 σ 2 )τ = 2 2 log s √ . so N (y) = y √1 e− 2 2π 2 (− y2 ) = −yN (y). s)) s and e−rτ N (δ− (τ. N (δ± (τ. s) = 1 √ [log s σ τ + (r ± 1 σ 2 )τ ] = 2 = log s − 1 2 σ τ + 1 r± 2 σ 2 √ τ. [(log s+rτ ) N (δ± (τ. (v) Proof. s)) = e− −1 )) N (δ± (τ. 4. supt≤u≤T (Wu − Wt ) is independent of Ft .1) or (8. L1∗ ]. rv2 (x) − rxv2 (x) − 1 2 2 2 σ x v2 (x) ≥ 0 for all x ≥ 0. = − σ2r (K − L).3.3. equality holds in either (8. This implies 8. x YT −t 1{ y ≤ YT −t } + y1{ y ≤ YT −t } )]. By Cauchy’s inequality and the monotonicity of Y . 2 So the linear complementarity conditions for v2 imply v2 (x) = (K2 − x)+ = K2 − x > K1 − x = (K1 − x)+ on [0. WT − Wt = (WT − Wt ) + α(T − t) is independent of Ft . If we increase the number of partition points to infinity and let the length of the longest subinterval m 2 max1≤j≤m |tj − tj−1 | approach zero.1. 8. y = Yt . YT )|Ft ] = E[f (x ST 0 . Proof. We note ST = S0 eσWT = St eσ(WT −Wt ) . then [S]T − [S]0 < ∞ and max1≤j≤m |Ytj − j=1 (Stj − Stj−1 ) → Ytj−1 | → 0 a. we can see v2 (x) ≥ (K2 − x)+ ≥ (K1 − x)+ .2) or both.. American Derivative Securities 8.3.8. Hence v2 (x) does not satisfy the third linear complementarity condition for v1 : for each x ≥ 0. YT )|Ft ] is a Borel function of (St . Solve for L.s. we get L = 2L 8. and for 0 ≤ x < L1∗ < L2∗ . by the continuity of Y . 7. 7. Therefore S S0 x S0 x S0 E[f (ST . Proof. x=L 2rK 2r+σ 2 . and YT = = = S0 eσMT S0 eσ supt≤u≤T Wu 1{Mt ≤sup St e σ supt≤u≤T (Wu −Wt ) t≤u≤T Wt } + S0 eσMt 1{Mt >sup } t≤u≤T Wu } } 1 { St ≤e t Y σ supt≤u≤T (Wu −Wt ) + Yt 1 { St ≤e t Y σ supt≤u≤T (Wu −Wt ) . vL (L+) = (K − L)(− σ2 )( L )− σ2 −1 L 2r m j=1 (Ytj − Ytj−1 )(Stj − Stj−1 ) → 0.. By the calculation in Section 8. 2r x 1 Proof. where x = St . we have m m | j=1 (Ytj − Ytj−1 )(Stj − Stj−1 )| ≤ j=1 |Ytj − Ytj−1 ||Stj − Stj−1 | m m ≤ j=1 (Ytj − Ytj−1 )2 j=1 (Stj − Stj−1 )2 m ≤ 1≤j≤m max |Ytj − Ytj−1 |(YT − Y0 ) j=1 (Stj − Stj−1 )2 .3. (i) 65 .To be continued . − σ2r (K − L) = −1.8. 1 rv2 (x) − rxv2 (x) − σ 2 x2 v2 (x) = rK2 > rK1 > 0.2. So vL (L+) = vL (L−) if and only if 2L Proof. Yt ). −t So E[f (ST . (v) Proof.20).3.4). So a general solution of (8. such that v(x) ≡ 0 satisfies (8. So v(0) = K.8. Contradiction. v(x0 ) = v (x0 ) = 0.8.4) has the form C1 x + C2 x− σ2 . If v satisfy (K − x)+ with equality for all x ≥ 0.3.3.8. This is a contradiction. This is already shown in our solution of part (iii): near 0. we get p = 1 or − σ2 . we have C1 = −1. so that v(x) = C1 x + C2 x− σ2 on [x1 . starting from first principles (8.” This is weaker than “v(x) = (K − x) in a right neighborhood of x2 . by the uniqueness of the solution of (8.e.19) with equality on [x1 . We have thus concluded simultaneously that v cannot satisfy (8. If in a right neighborhood of 0. (vi) 1 Note we have interpreted the condition “v(x) satisfies (8.3. This is a contradiction. Assume there is an interval [x1 .8.3. (iii) Proof. Suppose x takes its values in a domain bounded away from 0. Then v(0) = limx↓0 v(x) = 0 < (K − 0)+ . which is 0).18) with equality for x at and immediately to the left of x1 and 2r for x at and immediately to the right of x2 . and the only solution v that satisfies the specified conditions in the problem is the zero solution.20).19) with equality.3. v satisfies (8.8. If for some x0 ∈ [x1 .18) with equality for x at and immediately to the right of x ” 2 as “v(x2 ) = (K − x2 )+ and v (x2 ) =the right derivative of (K − x)+ at x2 .4) yields xp (r−pr− 2 σ 2 p(p−1)) = 0.19) with equality near 0 and v(0) = K.” 66 . then any solution of (8. i. we would conclude v ≡ 0. So such an x0 cannot exist.3.19) with equality. Therefore our initial assumption is incorrect. + v(x) = (K − x) near o. then we can find some C1 and C2 . x2 ] and satisfies (8. we have C1 = K−x1 = K−x2 . v2 (x) of (8. we would have x1 x2 x1 = x2 .3. (ii) Proof.4). v cannot satisfy (8. v(x2 ) = (K − x2 )+ = 0 and v (x2 )=the right derivative of (K − x)+ at x2 . the last two equations imply C2 = 0. Assume v(x) = xp for some constant p to be 1 1 determined. plug C2 = 0 into the last two equations. According to (8.18) will be 1 violated. if we can find two linearly independent solutions v1 (x). (iv) Proof. So it suffices to find two linearly independent special solutions of (8. then part (i) implies v(x) = C1 x + 2r C2 x− σ2 for some constants C1 and C2 . Plug C2 = 0 into the first two equations. So we must have rv − rxv − 2 σ 2 x2 v > 0 in a right neighborhood of 0.3. Since x1 = x2 . By the general theory of linear differential equations.3. (8. Solve the quadratic equation 0 = r−pr− 2 σ 2 p(p−1) = 2r 1 2r (− 2 σ 2 p − r)(p − 1).Proof. then v cannot have a continuous derivative as stated in the problem.4) can be represented in the form of C1 v1 +C2 v2 where C1 and C2 are constants.4). 1 Thus we have four equations for C1 and C2 :  2r C1 x1 + C2 x− σ2 = K − x1  1   2r  C x + C x − σ 2 = K − x 1 2 C −  1     C1 − 2 2 2r − σ2 −1 2r σ 2 C2 x1 2r − σ2 −1 2r σ 2 C2 x2 2 = −1 = −1. x2 ] where 0 < x1 < x2 < ∞.18)-(8.8. This implies 0 < x1 < x2 < K (if K ≤ x2 . x2 ]. Combined. x2 ]. (8. 5). both v and v are continuous. we are done. Then g(1) = 0 and g (θ) = 2r( σ2 + 1)θ σ2 − (σ 2 + 2r 2r 2r 2r 2r) σ2 θ σ2 −1 = σ2 (σ 2 + 2r)θ σ2 −1 (θ − 1) ≥ 0. ∞). L = 2rK 2r+σ 2 . This shows f (x) ≥ (K − x)+ for L ≤ x < K. So v satisfies (8.B= 2rK L(σ 2 +2r) − 1. Combined. f (x) − (K − x) = 2 σ + 2r L(σ 2 + 2r) (σ 2 + 2r)L 2r Let g(θ) = σ 2 + 2rθ σ2 +1 − (σ 2 + 2r)θ σ2 with θ ≥ 1. Since (K −x)+ is not differentiable at K.20). 2r 2r 2r 2r 2r 2r 2r σ 2 KL σ2 σ 2 +2r 2r . 2r (v) Proof. we get f (x) ≥ (K − x)+ for all x ≥ L. B = 0 if and only if 2r σ2 K x − σ2 σ 2 +2r ( L ) 2r 2rK L(σ 2 +2r) − 1 = 0. By the assumption of the problem. on the interval [L. x1 ] and v(x) = C1 x + C2 x− σ2 on [x1 . By (8.4. B > 0. but v is not the function vL∗ given by (8. (i) Proof. In summary.3. we must have C1 = 0 and the above equations only have two unknowns: C2 and x1 . Combined. (ii) Proof. B the equations AL− σ2 + BL = K − L 2r 2r − σ2 AL− σ2 −1 + B = −1.20).4. This is already shown in part (i) of Exercise 8. we can only obtain E[e−(r−a)τL ]. Since limx→∞ v(x) = limx→∞ f (x) = ∞ and limx→∞ vL∗ (x) = limx→∞ (K − L∗ )( L∗ )− σ2 = 0. and we obtain A = (iii) Proof.19). we can start with v(x) = (K − x)+ on [0.3. we must have x1 ≤ K.3. 8. v(x) = Ax− σ2 = 2r x = (K − L)( L )− σ2 = vL∗ (x). we also showed v satisfies (8.This gives us the equations  2r K − x = (K − x )+ = C x + C x− σ2 1 1 1 1 2 1 2r −1 = C − 2r C x− σ2 −1 .e.2. (iv) x Proof. v satisfies the linear complementarity condition (8. i. f (x) ≥ BK > 0 = (K − x)+ .18).Proof. So we have to start from Theorem 8.8.3. By the result of part (i). 1 2 2 rv − rxv − 2 σ x v ≥ 0 for x ≥ 0. Along the way. x x 2r KL σ2 σ 2 + 2r( L ) σ2 +1 − (σ 2 + 2r)( L ) σ2 2r 2rKx σ 2 KL σ2 − 2r − σ2 + σ2 + x −K =x . 8. So v satisfies (8. not E[e−rτL ]. For 0 ≤ x ≤ L. (i) 67 . rv − rxv − 2 σ x v = r(K − x) + rx = rK.3.13). The difficulty of the dividend-paying case is that from Lemma 8. ∞). So g(θ) ≥ 0 for any θ ≥ 1.3. By part (ii). Solve them for C2 and x1 . By part (iii). So for x ≥ K. In this case. v(x) ≥ (K − x) .3. v(x) + and vL∗ (x) are different. 1 σ2 2 1 2r Because v is assumed to be bounded. We solve for A. If L ≤ x < K.5. For x ≥ L.18)-(8.3. rv − rxv − 1 2 2 1 2 2 1 2 2 2 σ x v = rf − rxf − 2 σ x f = 0.3. Combined. x > L. 1 −r − (r − a)γ + σ 2 γ(γ + 1) .Proof. So 2 the risk-neutral expected discounted pay off of this strategy is vL (x) = K − x. (iii) Proof. 2 1 By the definition of γ. By Theorem 8. then St = L if and only if −Wt − σ (r − a − 1 − σ log x L 1 1 2 σ (r−a− 2 σ )+ 1 σ2 (r−a− 1 σ 2 )2 +2r 2 .2. we get d e−rt vL∗ (St ) = −e−rt 1{St <L∗ } (rK − aSt )dt + e−rt vL∗ (St )σSt dWt . if we define u = r − a − 2 σ 2 . u 1 + 2 σ σ 2u u2 + 3 σ4 σ u2 + 2r σ2 2 + u 1 + 2 σ σ u2 + 2r σ2 u2 + 2r u σ2 + u2 u + σ2 σ u2 + 2r σ2 u2 + 2r σ2 1 u2 + 2r + 2 σ2 σ u2 u − 2 2σ σ u2 1 + 2r − 2 σ 2 u2 u2 u + 2r + 2 + 2 σ σ σ 1 If x < L∗ . we have L∗ = γK γ+1 . we have 1 r + (r − a)γ − σ 2 γ(γ + 1) 2 1 2 2 1 = r − σ γ + γ(r − a − σ 2 ) 2 2 1 = r − σ2 2 1 = r − σ2 2 = r− = 0. 68 . By (8.3. Assume S0 = x. L Set ∂ ∂L vL (x) = 0 and solve for L∗ . By Itˆ’s formula. E[e−rτL ] = e 1 2 )t 1 . 0≤x≤L x −γ (K − L)( L ) . −rvL∗ (x) + vL∗ (x)(r − a)x + 2 vL∗ (x)σ 2 x2 = −r(K − x) + (−1)(r − a)x = −rK + ax. 1 −rvL∗ (x) + vL∗ (x)(r − a)x + vL∗ (x)σ 2 x2 2 −γ x 1 x−γ−2 x−γ−1 = −r(K − L∗ ) + (r − a)x(K − L∗ )(−γ) −γ + σ 2 x2 (−γ)(−γ − 1)(K − L∗ ) −γ L∗ 2 L∗ L∗ = (K − L∗ ) x L∗ −γ ∂ ∂L vL (x) x = −( L )−γ (1 − γ(K−L) ). St = S0 eσWt +(r−a− 2 σ 1 x 1 2 2 σ )t = σ log L . we have o 1 2 d e−rt vL∗ (St ) = e−rt −rvL∗ (St ) + vL∗ (St )(r − a)St + vL∗ (St )σ 2 St dt + e−rt vL∗ (St )σSt dWt . (ii) Proof. we can write E[e−rτL ] as e−γ log L = ( L )−γ .9). x 1 1 1 x 1 If we set γ = σ2 (r − a − 1 σ 2 ) + σ σ2 (r − a − σ2 )2 + 2r.8. 2 If x > L∗ . then necessarily r < a. since L∗ = γ+1 ≤ K. where we take the convention that e−rτ (Sτ −K)+ = 0 when τ = ∞. Theorem 8. by γ ≥ 0. So for L∗ ≤ x ≤ K. rK − aL∗ < 0. Define θ = 1 σ (θ r−a σ . Indeed. Plug L∗ = γ+1 into the expression and 1 1 1 note γ ≥ σ σ2 (r − a − 1 σ 2 )2 + σ2 (r − a − 1 σ 2 ) ≥ 0.3. Since θσ 2 < 0. r. As shown before. For x ≥ K > L∗ .T . The proof is similar to that of Corollary 8. Part (iii) already proved the supermartingale-martingale property.7. and vL∗ (x) ≥ (K −x)+ .5. we only need to show 1{x<L∗ } (rK − ax) ≥ 0 to finish γK the solution. 8. this means r(γ + 1) − aγ < 0. the inequality is further reduced to r(γ + 1) − aγ ≥ 0. e−rt∧τL∗ vL∗ (St ∧τL∗ ) is a martingale.1 implies E[e−rT (ST − K)+ ] ≥ E[e−rτ ∧T (Sτ ∧T − K)+ ] ≥ E[e−rτ (Sτ − K)+ 1{τ <∞} ] = E[e−rτ (Sτ − K)+ ].6.3. vL∗ (x) = K − x = (K − x)+ . For any τ ∈ Γ0. This is further equivalent to proving rK − aL∗ ≥ 0.3. (iv) Proof. Proof. dx γ + 1 γ+1 L∗ L∗ and (vL∗ (x) − (K − x))|x=L∗ = 0. Note the only properties used in the proof of Corollary 8. E[e−rT (ST − K)+ ] ≥ maxτ ∈Γ0. we have vL∗ (x) ≥ (K − x)+ ≥ 0 for all x ≥ 0. for 0 ≤ x < L∗ . 2 2 We prove this inequality as follows.8.6.Following the reasoning in the proof of Theorem 8.T . a and σ (K and σ are assumed to be strictly positive. r and a are assumed to be γK non-negative). Combined. for L∗ ≤ x ≤ K. So our initial assumption is incorrect. The other direction “≤” is trivial since T ∈ Γ0. −γ−1 x−γ−1 1 d L∗ γK (vL∗ (x) − (K − x)) = −γ(K − L∗ ) −γ + 1 ≥ −γ(K − L∗ ) −γ + 1 = −γ(K − ) γK + 1 = 0. Since τ is arbitrarily chosen. By Lemma 8. so it suffices to show vL∗ (x) ≥ (K − x)+ γK in our problem. vL∗ (x) > 0 = (K − x)+ . vL∗ (x) − (K − x)+ ≥ 0. then θ < 0 and γ = 1 σ 2 (r − a − 1 σ2 ) + 2 1 σ 1 σ 2 (r 1 − a − 2 σ 2 )2 + 2r = − 1 σ) + 2 1 σ (θ − 1 σ)2 + 2r. Xt = e−rt (St − K)+ is a submartingale. finally.1.T E[e−rτ (Sτ − K)+ ].6 are that e−rt vL∗ (St ) is a supermartingale. Assume for some K. We have 2 ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ (r − a)γ + r < 0 (r − a) 1 1 1 (θ − σ) + σ 2 σ 1 (θ − σ)2 + 2r + r < 0 2 r(γ + 1) − aγ < 0 1 1 θ(θ − σ) + θ (θ − σ)2 + 2r + r < 0 2 2 1 2 1 θ (θ − σ) + 2r < −r − θ(θ − σ)(< 0) 2 2 1 2 1 1 2 2 2 θ [(θ − σ) + 2r] > r + θ (θ − σ)2 + 2θr(θ − σ 2 ) 2 2 2 0 > r2 − θrσ 2 0 > r − θσ 2 . L∗ = γ+1 < K. 69 . 8. we have obtained a contradiction.5. and rK − aL∗ ≥ 0 must be true. Foreign risk-neutral measure P f is such that is a P f -martingale. Suppose λ ∈ [0. If we e assume the market is arbitrage-free and complete. if the market is arbitrage-free. no-arbitrage and completeness properties of market are independent of the choice of num´raire (see. Here B and ¯ ¯ B are both one-dimensional while S could be a vector price process. the foreign risk-neutral measure is dP f = f QT MT f Q0 M0 M0 dP = MT E f QT DT MT dP Q0 on FT . S) as in [1] Definition 2.1 or [5] page 383.1.Proof. if fT is a European contingent claim with maturity N and the market is both arbitrage-free and complete. M f Q.1. g((1 − λ)x1 + λx2 ) ≤ (1 − λ)h(x1 ) + λh(x2 ). B). then corresponding to B (resp. P ). Suppose B and B are both strictly positive. such that B B B B both arbitrage-free and complete. Shiryaev [5] page 413 Remark and page 481). That is. S). the traded assets are (M. We consider the following market: a domestic money market account M (M0 = 1). Under the above set-up. BT BT That is. where M f Q can be seen as the price process of one unit foreign currency. (resp. Therefore its foreign price is E f domestic currency. First. for a European contingent claim fT . Dt The RHS is exactly the price of fT in domestic market if we apply risk-neutral pricing. 1] and 0 ≤ x1 ≤ x2 . (i) 70 . 9. foreign currency exchange rate is Q. So h((1 − λ)x1 + λx2 ) = max{f ((1 − λ)x1 + λx2 ). a foreign money market account f M f (M0 = 1). Use the relation between P f and P on FT and the Bayes f DT fT f Dt QT Qt E f |Ft = E DT fT |Ft . Denominated by foreign currency. for example. ¯ ) is a martingale under P (resp. P ). Finally. e The above theoretical results can be applied to market involving foreign money market account. from M to M f Q. Domestic riskneutral measure P is such that assets are S Mf. a (vector) asset price process S called stock. there is an equivalent probability ¯ B S B S ¯ ¯ . if the market is P (resp. Convert this price into |Ft . Q Q Mf Q S M . Third. Denominated by domestic currency. g((1 − λ)x1 + λx2 )} ≤ (1 − λ)h(x1 ) + λh(x2 ). Note Q is not a traded asset. B. This summary is based on Shiryaev [5]. then both of them can be chosen as num´aire. Similarly. then fT ¯ ¯ fT Bt E ¯ |Ft = Bt E |Ft . we get f DT fT f Dt QT f DT fT f Dt QT |Ft . the traded M QM f S . M . we have f ((1 − λ)x1 + λx2 ) ≤ (1 − λ)f (x1 ) + λf (x2 ) ≤ (1 − λ)h(x1 ) + λh(x2 ). M is a P -martingale. e ¯ Consider a model of financial market (B. e ¯ Second. its payoff in foreign currency is fT /QT . e Several results hold under this model. we have the relation ¯ dP = ¯ BT BT E 1 ¯ B0 B0 dP . ¯ . This is a change of num´raire in the market denominated by domestic currency. Suppose the domestic vs. QM f . Change of Num´raire e To provide an intuition for change of num´raire. we give a summary of results for change of e num´raire in discrete case. h is also convex. we have Qt E f formula. denominated in domestic currency. 9. the price of fT is independent of the choice of num´raire. 2.2. Let M1 (t) = Dt St and M2 (t) = Dt Nt /N0 .Proof. (i) −1 Proof. we need to assume −1 < ρ < 1. Hence M1 (t) = Nt N0 is a martingale under P (N ) . which implies St = Nt is a martingale M2 (t) t t under P (N ) .3.5. For any 0 ≤ t ≤ T .2.2. 9. To avoid singular cases. Nt = N0 eν W3 (t)+(r− 2 ν dNt−1 1 2 )t . we have 1 1 [−νdWt − (r − ν 2 )dt + ν 2 dt] = Nt−1 (−νdWt − rdt). So 1 2 −1 = d(N0 e−ν W3 (t)−(r− 2 ν −1 = N0 e )t ) 1 −ν W3 (t)−(r− 2 ν 2 )t 1 1 −νdW3 (t) − (r − ν 2 )dt + ν 2 dt 2 2 = Nt−1 [−νdW3 (t) − (r − ν 2 )dt]. M2 (T ) M2 (t) M2 (T ) M2 (t) M2 (t) is a martingale under P M2 . d Mt = Mt d 1 Nt + 1 dMt + d Nt 1 Nt dMt = Mt (−νdWt − rdt) + rMt dt = −ν Mt dWt . 71 . by Lemma 5. (ii) Proof. Xt Nt dMt 9. dXt 1 1 1 + dXt + d dXt Nt Nt Nt 1 1 1 = (∆t St + Γt Mt )d + (∆t dSt + Γt dMt ) + d (∆t dSt + Γt dMt ) Nt Nt Nt 1 1 1 1 1 1 + dSt + d dSt + Γt Mt d + dMt + d = ∆t St d Nt Nt Nt Nt Nt Nt = d = Xt d = ∆t dSt + Γt dMt . (i) Proof. E (M2 ) So M1 (t) M2 (t) M2 (T ) M1 (T ) E[M1 (T )|Ft ] M1 (T ) M1 (t) Ft = E Ft = = .2.5. Remark: This can also be obtained directly from Theorem 9. Since Nt−1 = N0 e−ν Wt −(r− 2 ν 1 1 2 )t .6) is P (M2 ) as defined in (N ) S S Remark 9. Then P (N ) as defined in (9. (iii) Proof.2. 2 2 −1 d(Nt−1 ) = N0 e−ν Wt −(r− 2 ν 2 )t (ii) Proof. ν 9. (W1 . (N ) ρdt (N ) St (ν 2 (N ) St (σdW1 (t) σ γ W1 (t) σ 2 − 2ρσν + ν 2 and W4 (t) = [W4 ]t = − ν γ W3 (t). dW2 (t) 1 − ρ2 ). 9. consistent with the result of part (i). So under P . St e (ii) has volatility γ = σ 2 − 2ρσν + ν 2 . So t 0 σ2 (s)dW3 (s)− t 2 (Rs − 1 σ2 (s))ds 2 0 72 . 0). Proof. By Theorem 1 − ρ2 . and the volatility vector for N is (νρ. f Proof. From (9. γ2 γ r (N ) By L´vy’s Theorem. the volatility vector for S (N ) is (v1 . −ν the volatility of S (N ) is 2 2 v1 + v2 = (σ − νρ)2 + (−ν 1 − ρ2 )2 = σ 2 − 2νρσ + ν 2 .2. we define W2 (t) = √−ρ 2 W1 (t) + √ 1 1−ρ 1−ρ2 W3 (t). we have Mtf Qt = M0 Q0 e f Dt f = D0 Q−1 e− 0 Qt t 0 σ2 (s)dW3 (s)+ t 2 (Rs − 1 σ2 (s))ds 2 0 . with (dW2 (t)) = 2 − ρ 1 − ρ2 1−ρ2 dW1 (t) + 1 1 − ρ2 2 dW3 (t) = 1 2ρ2 ρ2 + − 1 − ρ2 1 − ρ2 1 − ρ2 dt = dt. then W4 is a martingale with quadratic σν ν2 σ2 t − 2 2 ρt + 2 t = t. and dW2 (t)dW1 (t) = − √ ρ dt + √ ρ 1−ρ2 dt = 0. then W2 is a martingale. v2 ) = (σ − νρ.15). the volatility vector for S is (σ.2.and dSt (N ) = Nt−1 dSt + St dNt−1 + dSt dNt−1 = Nt−1 (rSt dt + σSt dW1 (t)) + St Nt−1 [−νdW3 (t) − (r − ν 2 )dt] (N ) = St = Define γ = variation (rdt + σdW1 (t)) + St − σρ)dt + (N ) [−νdW3 (t) − (r − ν 2 )dt] − σSt − νdW3 (t)).13.3. and     dSt = rSt dt + σSt dW1 (t) = rSt dt + St (σ. W4 is a BM and therefore. W2 ) is a two-dimensional BM. and dNt = rNt dt + νNt dW3 (t) = rNt dt + νNt [ρdW1 (t) + (iii) 1 − ρ2 dW2 (t)]. This problem is the same as Exercise 4. Under P . So W2 is a BM independent of W1 . ν   dW1 (t) dW2 (t) 1 − ρ2 ) · dW1 (t) . 0) ·    dNt = rNt dt + νNt dW3 (t) = rNt dt + Nt (νρ. In particular. under the measure P (N ) . Proof.4. σ4 σ4 Then W4 is a martingale with [W4 ]t = So W4 is a Brownian motion under P .14) and (9. Qt 9. Qt To get (9. So 2 if we set a = r − rf + ρσ1 σ2 − σ2 . By risk-neutral pricing formula.3.3.and d f Dt Qt = f Dt 1 2 1 2 Df 2 [−σ2 (t)dW3 (t) − (Rt − σ2 (t))dt + σ2 (t)dt] = t [−σ2 (t)dW3 (t) − (Rt − σ2 (t))dt]. we note d f Mt D t Qt = = = = Mt d f Dt Qt + f Dt dMt + dMt d Qt f Dt Qt f f Mt Dt Rt Mt Dt 2 [−σ2 (t)dW3 (t) − (Rt − σ2 (t))dt] + dt Qt Qt − − f Mt Dt 2 (σ2 (t)dW3 (t) − σ2 (t)dt) Qt f Mt Dt f σ2 (t)dW3 (t). The payoff of a S quanto call is ( QT − K)+ units of domestic currency at time T . we have St = S0 e Qt = Q0 eσ2 W3 (t)+(r−r √ 2 f 2 − 1 σ2 )t 2 2 σ1 W1 (t)+(r− 1 σ1 )t 2 and √ = Q0 eσ2 ρW1 (t)+σ2 . Qt 2 2 Qt To get (9.3. we have 2 St S0 σ4 W4 (t)+(r−a− 1 σ4 )t 2 = e and d Qt Q0 St Qt = St [σ4 dW4 (t) + (r − a)dt]. We combine the solutions of all the sub-problems into a single solution as follows. we note d f Dt St Qt = = f Dt dSt + St d Qt f Dt Qt + dSt d f Dt Qt f f St Dt Dt 2 St (Rt dt + σ1 (t)dW1 (t)) + [−σ2 (t)dW3 (t) − (Rt − σ2 (t))dt] Qt Qt +St σ1 (t)dW1 (t) = = f Dt (−σ2 (t))dW3 (t) Qt f Dt St 2 [σ1 (t)dW1 (t) − σ2 (t)dW3 (t) + σ2 (t)dt − σ1 (t)σ2 (t)ρt dt] Qt f Dt St f f [σ1 (t)dW1 (t) − σ2 dW3 (t)]. Define 2 1−ρ2 W2 (t)+(r−r f − 1 σ2 )t 2 .22). So St Qt = S0 (σ1 −σ2 ρ)W1 (t)−σ2 Q0 e 2 2 1−ρ W2 (t)+(r f + 1 σ2 − 1 σ1 )t 2 2 σ4 = 2 (σ1 − σ2 ρ)2 + σ2 (1 − ρ2 ) = 2 2 σ1 − 2ρσ1 σ2 + σ2 and W4 (t) = 2 (σ1 −σ2 ρ)2 t + σ2 (1−ρ ) t + t.5.3.16). 2 2 σ4 σ4 σ1 − σ2 ρ σ2 1 − ρ2 W1 (t) − W2 (t). Qt 73 . its price at T S time t is E[e−r(T −t) ( QT − K)+ |Ft ]. So we need to find the SDE for T St Qt under risk-neutral measure P . Proof.23). By formula (9. (i) Proof. So d− (t) = d+ (t) − σ T − t.T ) 2 K ] e−d+ (t)/2 [ForS (t.12) gives us the desired price formula for quanto call option. d+ (t) − d− (t) = (ii) Proof. −t (viii) 74 .5. T ) 2 σ T − t ForS (t.T ) .T ) . T ))2 + σ (T − t)]dt + √ − − σdt 1σ (T − t)3 [log 2 2 K 2 2ForS (t. T ) − Kelog 0. + − K K = = = e−d+ (t)/2 [ForS (t. By (iv) and (v). dd− (t) = dd+ (t) − d(σ T − t) = dd+ (t) + (vi) Proof. T ) 1 ForS (t. log dt − √ K 2σ(T − t)3/2 4 T −t T −t = = = (v) √ Proof. (dd− (t))2 = (dd+ (t))2 = (vii) 1 Proof. dN (d+ (t)) = N (d+ (t))dd+ (t)+ 2 N (d+ (t))(dd+ (t))2 = √1 e− 2π d2 (t) + 2 σdt √ .6. 2 T −t dt T −t . Qt behaves like dividend-paying stock and the price of the quanto call option is like the t price of a call option on a dividend-paying stock. T ) − Ked+ (t)/2−d− (t)/2 ] ForS (t.S Therefore. 9. T ) 1 2 dForS (t. T )e−d+ (t)/2 − Ke−d− (t) 2 2 √1 σ 2 (T σ T −t √ √ − t) = σ T − t. T ) 3σ dt + √ . d+ (t)+d− (t) = (iii) Proof. dd+ (t) 1 1 1√ ForS (t. Thus formula (5. T ) (dForS (t. So d2 (t)−d2 (t) = (d+ (t)+d− (t))(d+ (t)−d− (t)) = 2 log ForS (t. T ) σ 1 1 2 1 2 log dt + √ dt + √ (σdW T (t) − σ dt − σ dt) K 2 2 4 T −t σ T −t 2σ (T − t)3 dW T (t) 1 ForS (t. 2 2 2 (iv) Proof. ForS (t. dd+ (t)+ 1 √1 e− 2 2π d2 (t) + 2 (−d+ (t)) Tdt . √2 σ T −t log ForS (t. under P . I2 (t). T )e−d+ (t)/2 √ dForS (t. T )e−d+ (t)/2 − Ke−d− (t)/2 dd+ (t) 2π = 0. dN (d− (t)) 1 = N (d− (t))dd− (t) + N (d− (t))(dd− (t))2 2 = d2 (t) − 1 √ e− 2 2π σdt dd+ (t) + √ 2 T −t 2 dt 1 e− 2 √ (−d− (t)) + 2 T −t 2π d2 (t)(σ − √ T −t−d+ (t)) d2 (t) − = 2 2 1 e− σe−d− (t)/2 √ e−d− (t)/2 dd+ (t) + √ dt + 2π 2(T − t) 2π 2 2π(T − t) 2 1 σe−d− (t)/2 d+ (t)e− 2 √ dt. T )dN (d+ (t)) + dForS (t. T −t 2π 2π(T − t) T 2 2 (x) Proof. ForS (t. if λ1 = λ2 . which implies e−d− (t)/2 = ForS (t. we can write Y1 (t) and Y2 (t) as Y1 (t) = e−λ1 t Y1 (0) + e−λ1 t I1 (t) and Y2 (t) = λ21 λ21 −λ1 t − e−λ2 t )Y1 (0) + e−λ2 t Y2 (0) + λ1 −λ2 e−λ1 t I1 (t) − e−λ2 t I2 (t) − e−λ2 t I3 (t). T )e−d+ (t)/2 √ e−d+ (t)/2 dt + = ForS (t.Proof. T )e−d+ (t)/2 √ √ e−d− (t)/2 dt = e + − − + 2(T − t) 2π 2(T − t) 2π 2π(T − t) 2π(T − t) 2 2 1 +√ ForS (t. T )dN (d+ (t)) = σForS (t. T )d+ (t) −d2 (t)/2 σForS (t. 75 . Using the notation I1 (t). T )dW (t) √ dW T (t) = dt. (i) Proof.1. T ) √ e−d+ (t)/2 dd+ (t) − dt 2π 2(T − t) 2π 2π(T − t) 2 −K e−d− (t)/2 √ dd+ (t) + 2π 2 σ 2π(T − t) e−d− (t)/2 dt − 2 2 2 d+ (t) √ e−d− (t)/2 dt 2(T − t) 2π 2 2 Kσe−d− (t)/2 Kd+ (t) ForS (t. √ e−d− (t)/2 dd+ (t) + dt − 2π 2(T − t) 2π 2π(T − t) 2 d2 (t) − dt = (ix) Proof.T ) e−d+ (t)/2 . Term-Structure Models 10. 1 e−d+ (t)/2 σForS (t. 2 2 The last “=” comes from (iii). K 10. T )dN (d+ (t)) − KdN (d− (t)) 2 2 1 d+ (t) σForS (t. λ1 −λ2 (e −λ1 t −λ1 t −λ1 t −λ21 te Y1 (0) + e Y2 (0) − λ21 te I1 (t) − e−λ1 t I4 (t) + e−λ1 t I3 (t). if λ1 = λ2 . I3 (t) and I4 (t) introduced in the problem. Y1 (t). then Xt Yt − [X. Y1 (t). Y2 (t))(σ21 Y1 (t) + α + βY1 (t))]dt + martingale part. if λ1 = λ2 . Y1 (t). E[I1 (t)I2 (t)] = 2λ1 t t e(λ1 +λ2 )u du = 0 e(λ1 +λ2 )t − 1 .Since all the Ik (t)’s (k = 1. we must have −(δ0 + δ1 y1 + δ2 y2 ) + ∂ ∂ ∂ 1 + (µ − λ1 y1 ) − λ 2 y2 + ∂t ∂y1 ∂y2 2 2σ21 y1 ∂2 ∂2 ∂2 2 + y1 2 + (σ21 y1 + α + βy1 ) 2 ∂y1 ∂y2 ∂y1 ∂y2 f = 0. Thus t 2 E[I1 (t)] = 0 e2λ1 u du = e2λ1 t − 1 . 2 T Since Dt B(t. Y2 (t))dt+ df (t. In particular. Y1 (t). The calculation relies on the following fact: if Xt and Yt are both martingales. if λ1 = λ2 . Assume B(t. Y2 (t)). Y2 (t))Y1 (t) 2 1 2 + fy2 y2 (t. E[Xt Yt ] = E{[X. E[I1 (t)I4 (t)] = 0 ue2λ1 u du = 1 e2λ1 t − 1 te2λ1 t − 2λ1 2λ1 and t 2 E[I4 (t)] = 0 u2 e2λ1 u du = t2 e2λ1 t te2λ1 t e2λ1 t − 1 − + . (ii) Proof. T ) is a martingale. Y1 (t). Y ]t }. Y2 (t)) + fy1 (t. Following the hint. λ1 + λ2 10. λ1 + λ2 E[I1 (t)I3 (t)] = 0. (i) Proof. Y2 (t))(µ − λ1 Y1 (t)) + fy2 (t. · · · . T )) = Dt [−Rt f (t. Y1 (t). Y1 (t). Y ]t is also a martingale. T ) = E[e− t Rs ds |Ft ] = f (t. 4) are normally distributed with zero mean. 2λ1 2λ2 4λ3 1 1 (iii) Proof. Y1 (t). we can conclude E[Y1 (t)] = e−λ1 t Y1 (0) and E[Y2 (t)] = λ21 −λ1 t − e−λ2 t )Y1 (0) + e−λ2 t Y2 (0). Y2 (t)) = [ft (t. Y2 (t))σ21 Y1 (t) + fy1 y1 (t. Y2 (t))]. λ1 −λ2 (e −λ21 te−λ1 t Y1 (0) + e−λ1 t Y2 (0).2. Y2 (t))(−λ2 )Y2 (t)] 1 +fy1 y2 (t. Then d(Dt B(t. By Itˆ’s formula. Y1 (t). we have t E[I1 (s)I2 (t)] = E[J1 (t)I2 (t)] = 0 e(λ1 +λ2 )u 1{u≤s} du = e(λ1 +λ2 )s − 1 . (ii) 76 . o df (t. Y1 (t). 4). So the PDE in part (i) becomes −(δ0 +δ1 y1 +δ2 y2 )+y1 C1 +y2 C2 +A −(µ−λ1 y1 )C1 +λ2 y2 C2 + 1 2 2 2 2σ21 y1 C1 C2 + y1 C1 + (σ21 y1 + α + βy1 )C2 = 0. Y2 (t))(−λ21 Y1 (t) − λ2 Y2 (t)) 1 1 + fy1 y1 (t. y2 ). Y2 (t))dt + df (t. Y2 (t))]dt + martingale part. T ) − y2 dt C2 (t. 10. T. T. ∂f ∂y2 = −C2 (T − t)f . check!  2 C = −λ2 C2 + δ2  2  1 2 A = µC1 − 2 αC2 + δ0 . T. T )f (t. T )) = Dt [−Rt f (t. Y1 (t). Y1 (t). 2 2 77 . we can obtain the ODEs for C1 . 2 In other words. T. y1 . y1 . then A (T − t)]f . y2 ) = e−y1 C1 (T −t)y2 C2 (T −t)−A(T −t) . T ) is a martingale under risk-neutral measure. T )f (t. y ). Y1 (t). y1 . y1 . T. T. T. Y1 (t). y2 ). T. T )C2 (t. y2 ) = −C1 (t. Y2 (t))(−λ1 Y1 (t)) + fy2 (t. Y1 (t). T. T. Y2 (t)) + fy1 (t. (i) Proof. T ) f (t. T ) + C1 (t. T. T. 2 Sorting out the LHS according to the independent variables y1 and y2 .T ) . 2 2 Since Dt B(t. ∂t ∂y1 ∂y2 2 ∂y1 2 Suppose f (t. y . Y2 (t))] and df (t.    fy y (t. y2 ) = C1 (t. ∂ f ∂y1 ∂y2 2 ∂ ∂t f = [y1 C1 (T − t) + y2 C2 (T − t) + ∂2f 2 ∂y1 2 = C1 (T − t)f .Proof. T. T.T )−A(t. T. y . y1 . y1 . T ) − A(t. y1 . Y1 (t). T. y2 ) = e−y1 C1 (t. T )f (t. Y2 (t)) = [ft (t. y2 ) = C2 (t. y2 ) = −y1 dt C1 (t. So the PDE becomes −(δ0 (t) + δ1 y1 + δ2 y2 ) + −y1 d dt A(t. y2 ) = C 2 (t. y1 . C2 and A as follows  1 2 2 2 C1 = −λ1 C1 − σ21 C1 C2 − 2 C1 − 1 (σ21 + β)C2 + δ1 different from (10. then  d d ft (t.7. y ) = −C (t. Y2 (t)) + fy2 y2 (t. T ) −    fy1 (t. y1 . T. T )f (t. y2 ). T. y1 . T )f (t. Y1 (t). T. and ∂ f 2 ∂y2 2 ∂ ∂y1 f = −C1 (T − t)f . y2 ). y1 . y2 ). T. T ) + C2 (t.T )−y2 C2 (t. y2 ) = 0.    f (t. T ) dt dt dt 1 2 1 2 +(λ21 y1 + λ2 y2 )C2 (t. = C1 (T − t)C2 (T − t)f .3. y1 . T. 2 = C2 (T − t)f . y1 . we get  1 1 2 2 2 −δ1 + C1 + λ1 C1 + σ21 C1 C2 + 2 C1 + 2 (σ21 + β)C2 = 0  −δ + C2 + λ2 C2 = 0  2  2 −δ0 + A − µC1 + 1 αC2 = 0. T. we have the following PDE: −(δ0 (t) + δ1 y1 + δ2 y2 ) + 1 ∂ ∂ ∂ ∂ 1 ∂2 − λ1 y1 − (λ21 y1 + λ2 y2 ) + 2 + 2 ∂y 2 f (t. d(Dt B(t. If we suppose f (t. T ) = 0. d d d C1 (t. Y1 (t). T ) − y2 C2 (t. T ) + λ1 y1 C1 (t.  1  1 1   2 fy2 y2 (t. y2 1 2 2 1 2 fy1 y2 (t. From the ODE d dt A(t. T ) = 2 C1 (t. 2 (iv) Proof. T ) = −δ2 t e−λ2 s ds = λ2 (e−λ2 T − e−λ2 t ).T )−A(0. we have T δ δ 0 − e−λ2 t C2 (t. T ) − δ1  d C (t. T ) + C2 (t. (iii) Proof. T ) + 2 C2 (t. T ) + C1 (T. T ) = −Y1 (0) C1 (0. 2 1 Taking derivative w. T ) = λ21 δ2 −λ1 (T −t) λ2 λ1 (e λ21 δ2 −λ1 (T −t) λ2 λ1 (e δ2 − λ21λ1 (e−λ1 T − e−λ1 t ) − λ2 δ2 − λ21λ1 (e−λ1 T − e−λ1 t ) − λ2 λ21 δ2 −λ2 T e(λ2 −λ1 )T −e(λ2 −λ1 )t δ + λ1 (e−λ1 T λ2 e λ2 −λ1 1 λ21 δ2 −λ2 T δ (T − t) + λ1 (e−λ1 T − e−λ1 T ) λ2 e 1 − e−λ1 T ) if λ1 = λ2 if λ1 = λ2 . For C1 . − 1) + − 1) + λ21 δ2 e−λ1 (T −t) −e−λ2 (T −t) δ − λ1 (e−λ1 (T −t) − 1) λ2 λ2 −λ1 1 λ21 δ2 −λ2 T +λ1 t δ (T − t) − λ1 (e−λ1 (T −t) − 1) λ2 e 1 if λ1 = λ2 . T ) − δ0 (T ). Y1 (0). T ) − δ0 (t). Integrate from t to T . T ) − δ2  dt 2 d 1 2 1 2 dt A(t. T )) ds. T ) = λ2 C2 (t. we note 2 2 λ21 δ2 −λ1 t d −λ1 t (e C1 (t. T ) = −Y1 (0)C1 (0.t. T ) + 1 C1 (t. T ) + λ1 C1 (t. T ) = 0. we get 2 T A(t.T ) = B(0. T ) + C2 (T. T ) + λ2 C2 (t. T ) + 2 C2 (t. ∂T ∂T ∂T 2 2 78 . T ) + C2 (s. T ) 2 2 = 1 (C1 (t. T ) − Y2 (0)C2 (0. T ) = λ2 (1 − e−λ2 (T −t) ). Y2 (0)) = e−Y1 (0)C1 (0. we get  1 2 d 2 −δ0 (t) − dt A(t. we get T log B(0. T ) + C2 (s. we note dt [e−λ2 t C2 (t. Take logarithm on both sides and plug in the expression of A(t. (ii) d Proof. That is  d  dt C1 (t. T ) + 0 1 2 2 (C (s. T ) for all T > 0. we have ∂ ∂ ∂ 1 2 1 2 log B(0. For C2 . T ) = t 1 2 2 δ0 (s) − (C1 (s. T ) − δ1 )e−λ1 t = (e − e−λ2 T +(λ2 −λ1 )t ) − δ1 e−λ1 t . T ). T )) = (λ21 C2 (t. T )) − δ0 (t). T ) + λ21 C2 (t. if λ1 = λ2 .r.T )−Y2 (0)C2 (0. T ) = λ1 C1 (t. T ) = 0  1 dt 1  d −δ2 − dt C2 (t. We want to find δ0 so that f (0. T ) = So C1 (t. dt λ2 Integrate from t to T . T )] = −e−λ2 t δ2 from the ODE in (i). T )) − δ0 (s) ds.Sorting out the terms according to independent variables y1 and y2 . T. T ) − Y2 (0) C2 (0. T. T ) = 0  2 −δ − d C (t. T ) + λ21 C2 (t. So C2 (t. we get −e−λ1 t C1 (t. (i) Proof. W 2 1−ρ2 = −√ ρ 1−ρ2 B1 + √ 1 = = t. Wt = CΣBt = 1 σ1 − √ρ 2 σ1 1−ρ t 0 √1 σ2 1−ρ2 t σ1 0 0 Bt = σ2 1−ρ2 t 1 −√ ρ 1−ρ2 0 √1 t Bt . T ) λ2 e ∂ λ21 δ2 e−λ2 T T − Y2 (0)δ2 e−λ2 T − ∂T log B(0. dYt = CdXt = −CK Xt dt+CΣdBt = −CKC Yt dt+dWt = −ΛYt dt+dWt . Therefore W is a two-dimensional BM. T ) ∂T ∂T ∂T −λ1 T −Y1 (0) δ1 e−λ1 T − −Y1 (0) δ1 e − λ21 δ2 −λ2 T ∂ − Y2 (0)δ2 e−λ2 T − ∂T log B(0. 0 . t t Xt = = = Xt + e−Kt 0 eKu Θ(u)du = C −1 Yt + e−Kt 0 eKu Θ(u)du Θ(u)du σ1 ρσ2 σ2 0 1 − ρ2 Y1 (t) + e−Kt Y2 (t) t e 0 Ku σ1 Y1 (t) + e−Kt ρσ2 Y1 (t) + σ2 1 − ρ2 Y2 (t) 79 t eKu Θ(u)du. (iii) Proof. T ) − log B(0.4. dXt = dXt + Ke−Kt 0 eKu Θ(u)dudt − Θ(t)dt t = −KXt dt + ΣdBt + Ke−Kt 0 eKu Θ(u)dudt = −K Xt dt + ΣdBt . ρ t + 1−ρ2 −2 1−ρ2 ρt = 1−ρ2 So W is a martingale with W 1 ρ2 +1−2ρ2 t 1−ρ2 = B1 t = t. and W 1 . W 2 t = B1. where   1 √1 0 1  σ2 1−ρ2 0  λ1 0 σ1 · Λ = CKC −1 = − √ρ ρ 1 √1 −1 λ2 |C| σ √1−ρ2 σ1 σ1 1−ρ2 σ2 1−ρ2 1 = ρλ − √1 σ1 λ1 σ1 1−ρ2 − √1 σ2 1−ρ2 0 λ √2 σ2 1−ρ2 σ1 ρσ2 σ2 0 1 − ρ2 = λ1 ρσ2 (λ2 −λ1 )−σ1 √ σ2 1−ρ2 0 λ2 . T ) − Y2 (0) C2 (0.So δ0 (T ) = −Y1 (0) = 10. T ) if λ1 = λ2 if λ1 = λ2 . Moreover. (ii) Proof. − √ ρ B1 + √ 1 B2 1−ρ2 ρt −√ 2 + 1−ρ B2 = ρ2 t 1−ρ2 √ ρt −1 1−ρ2 = 0. t ∂ ∂ ∂ C1 (0. τ )]. If δ2 = 0. By (9. t + τ ) aare constants ¯ ¯ when τ is fixed. So C(t. 10.2. −C2 (T − t)) dW1 (t) .5).2. T ) under P is (−C1 (T − t). ¯ ¯ ¯ ¯ it is easy to verify that their correlation is 1. 10. where δ0 (t) is the second coordinate of e−Kt and can be derived explicitly by Lemma 10. We use the canonical form of the model as in formulas (10. 2 2 δ1 + δ2 and Bt = √ δ1 2 2 δ1 +δ2 W1 (t) + √ δ2 2 2 δ1 +δ2 So the volatility vector of B(t.20). −C2 (T − t)). Then δ1 = ρσ2 and δ2 = σ2 1 − ρ2 . t + τ ) ¯ ¯ 1 [C(t. T ) are dependent only on T − t. We note C(t. WjT (t) = u)du + Wj (t) (j = 1. So a = δ0 λ1 and b = λ1 . τ )R (t) + A(0. T )(−C1 (T − t). dB(t. σ = 10. t + τ )] ¯ ¯ τ ¯ 1 [C(0. By (10. 2) form a two-dimensional P T −BM. Y1 (t). b = λ2 . t Ku e Θ(u)du 0 Proof.5.2. t + τ )] ¯ ¯ ¯ τ B(t.3.So Rt = X2 (t) = ρσ2 Y1 (t)+σ2 1 − ρ2 Y2 (t)+δ0 (t). δ2 2 δ1 2 + δ2 dW2 (t) . ¯ ¯ τ ¯ 1 1 Hence L(t2 )−L(t1 ) = τ C(0.2. t + τ )R (t) − A(t. t + τ ) and A(t.7. So ¯ d Lt dt = − = = B(t. t + τ )[−C(t. T ) = df (t. (i) δ0 Proof. τ )[R(t2 )−R(t1 )]+ τ A(0.6).6. τ )(t2 −t1 ). dW2 (t) t 0 2 2 δ1 + δ2 δ1 2 δ1 + 2 δ2 dW1 (t) + W2 (t). dRt = δ1 dY1 (t) + δ2 dY2 (t) = −δ1 λ1 Y1 (t)dt + λ1 dW1 (t) − δ2 λ21 Y1 (t)dt − δ2 λ2 Y2 (t)dt + δ2 dW2 (t) = −Y1 (t)(δ1 λ1 + δ2 λ21 )dt − δ2 λ2 Y2 (t)dt + δ1 dW1 (t) + δ2 dW2 (t) = −Y1 (t)λ2 δ1 dt − δ2 λ2 Y2 (t)dt + δ1 dW1 (t) + δ2 dW2 (t) = −λ2 (Y1 (t)δ1 + Y2 (t)δ2 )dt + δ1 dW1 (t) + δ2 dW2 (t) = −λ2 (Rt − δ0 )dt + So a = λ2 δ0 . (ii) 80 Cj (T − . T )[−C1 (T − t)dW1 (t) − C2 (T − t)dW2 (t)] = dt term + B(t. (i) Proof. Since L(t2 )−L(t1 ) is a linear transformation. then dRt = δ1 dY1 (t) = δ1 (−λ1 Y1 (t)dt + dW1 (t)) = δ1 ( δ1 − Rt δ1 )λ1 dt + dW1 (t) = (δ0 λ1 − λ1 Rt )dt + δ1 dW1 (t).2. (ii) Proof.4)-(10. t + τ )R (t) + A(t. T ) and A(t. Y2 (t)) = de−Y1 (t)C1 (T −t)−Y2 (t)C2 (T −t)−A(T −t) = dt term + B(t. Tj ) − B(0. Its no-arbitrage price at time 0 is δ(KB(0.Proof. T ) B(T. 10. So the value of the swap is n+1 n+1 n+1 δ[KB(0.4) and (10. T )(eµ N (d+ ) − KN (d− )). Tj−1 )) by Theorem 10. We can rewrite (10. we get (10. Tj )L(0. Tj )L(0. we recall the Black-Scholes formula for call options: if dSt = µSt dt + σSt dWt . the payoff of this swap contract is δ(K − L(Tj−1 . T ).4. Y2 ) is jointly Gaussian and X is therefore Gaussian. we have d B(0. Tj ) − δ j=1 B(0. Under the T-forward measure. Tj ) − B(0. T ) = 1. S0 = 1 and ξ = σW1 + (µ − 1 σ 2 ). 81 . (iv) Proof. Tj−1 ). Tj−1 )] = δK j=1 j=1 B(0. Check!).7. at time zero the risk-neutral price V0 of the option satisfies V0 1 ¯ ¯ ¯ = ET (e−C1 (T −T )Y1 (T )−C2 (T −T )Y2 (T )−A(T −T ) − K)+ .2. 1 1 where d± = σ (− log K + (µ ± 2 σ 2 )) (different from the problem. Since under P T .19). (iii) Proof. Let T = 1.12. By risk-neutral pricing. T ) Note B(T. the numeraire is B(t. B(0. then ξ = N (µ − 2 σ 2 .2. Then Y1 (t) = Y1 (0)e−λ1 t + Y2 (t) = Y0 e−λ2 t − λ21 t λ1 (s−t) t T e dW1 (s) − 0 C1 (T − s)eλ1 (s−t) ds 0 t t t Y (s)eλ2 (s−t) ds + 0 eλ2 (s−t) dW2 (s) − 0 0 1 C2 (T − s)eλ2 (s−t) ds. σ 2 ) 2 2 E[(eξ − K)+ ] = eµ N (d+ ) − KN (d− ). Tj )L(0. First. σ ). On each payment date Tj . So (Y1 . Tj−1 )). T )E T [(eX − K)+ ] = B(0. 10. Proof. X = 1 2 2 N (µ − 2 σ .11.5) as T dY1 (t) = −λ1 Y1 (t)dt + dW1 (t) − C1 (T − t)dt T dY2 (t) = −λ21 Y1 (t)dt − λ2 Y2 (t)dt + dW2 (t) − C2 (T − t)dt. then E[e−µT (S0 eσWT +(µ− 2 σ with d± = and 1 √ σ T 1 2 )T − K)+ ] = S0 N (d+ ) − Ke−µT N (d− ) d (log S0 K 1 + (µ ± 1 σ 2 )T ). That is. T + δ) 1 − B(T. 2 E[Mt2 − λt|Fs ] = Ms − λs. we have E Su Ft St = = = = = = So St = E[Su |Ft ] and S is a martingale. Proof. (i) Proof. Introduction to Jump Processes 11. T + δ) δB(T. f (x) = x2 is a convex function.4. T ).3.T +δ) ∈ FT . We note Mt has independent and stationary increment. Mt2 = Nt2 − 2λtNt + λ2 t2 . So ∀s ≤ t. (ii) 2 Proof. T + δ) = δ = B(0. E[f (Mt )|Fs ] ≥ f (E[Mt |Fs ]) = f (Ms ). T ) = 1−B(T. we have = E[E[D(T + δ)L(T. P (Ns+t = k|Ns = k) = P (Ns+t − Ns = 0|Ns = k) = P (Nt = 0) = e−λt = 1 − λt + O(t2 ).4)) e 1. So E[Mt2 ] < ∞. T )] 11. Similarly. T + δ) D(T ) − D(T )B(T. T + δ) = E E[D(T + δ)|FT ] δB(T. E[Mt2 − Ms |Fs ] = E[(Mt − 2 2 Ms ) |Fs ] + E[(Mt − Ms ) · 2Ms |Fs ] = E[Mt−s ] + 2Ms E[Mt−s ] = V ar(Nt−s ) + 0 = λ(t − s). T )|FT ]] 1 − B(T. First. E[(σ + 1)Nt −Nu e−λσ(t−u) |Ft ] e−λσ(t−u) E[(σ + 1)Nt−u ] e−λσ(t−u) E[eNt−u log(σ+1) ] e−λσ(t−u) eλ(t−u)(e e −λσ(t−u) λσ(t−u) log(σ+1) ∞ (λt)k −λt k=2 k! e = O(t2 ). 82 .Proof. So by conditional Jensen’s inequality.2. 1 we have P (Ns+t = k + 1|Ns = k) = P (Nt = 1) = (λt) e−λt = λt(1 − λt + O(t2 )) = λt + O(t2 ). Proof. ∀s ≤ t. For any t ≤ u. Since L(T.T +δ) δB(T.3. So Mt2 is a submartingale. T + δ) = E D(T )B(T. T + δ)L(0. and 1! P (Ns+t ≥ k + 2|N2 = k) = P (Nt ≥ 2) = 11. 11. E[D(T + δ)L(T. T + δ) = E δ B(0. T ) − B(0. −1) (by (11. 11.1. · · · . N2 have no simultaneous jump.r. by Itˆ’s product formula. This shows N1 (t) − N1 (s) is independent of N2 (t) − N2 (s). Meanwhile. we consider Xt = u1 (N1 (t) − N1 (s)) + u2 (N2 (t) − N2 (s)) − λ1 (eu1 − 1)(t − s) − λ2 (eu2 − 1)(t − s). N2 (t2 ) − N2 (t1 ). Working by induction.t. Then by independence E[M1 (t)M2 (t)] = E[M1 (t)]E[M2 (t)] = 0. Suppose N1 and N2 are independent. M1 (t)M2 (t) = o t t t t M1 (s−)dM2 (s) + 0 M2 (s−)dM1 (s) + [M1 . Since 0<s≤t ∆N1 (s)∆N2 (s) ≥ 0 a. 83 . Both 0 M1 (s−)dM2 (s) and 0 M2 (s−)dM1 (s) are mar0 tingales. then E[e = = = E[e E[e E[e n i=1 ui (N1 (ti )−N1 (ti−1 ))+ ui (N1 (ti )−N1 (ti−1 ))+ ui (N1 (ti )−N1 (ti−1 ))+ ui (N1 (ti )−N1 (ti−1 ))+ n j=1 vj (N2 (tj )−N2 (tj−1 )) ] E[eun (N1 (tn )−N1 (tn−1 ))+vn (N2 (tn )−N2 (tn−1 )) |Ftn−1 ]] ]E[eun (N1 (tn )−N1 (tn−1 ))+vn (N2 (tn )−N2 (tn−1 )) ] ]E[eun (N1 (tn )−N1 (tn−1 )) ]E[evn (N2 (tn )−N2 (tn−1 )) ]. · · · . 2. we would guess the condition should be that N1 and N2 are independent. M2 ]t . Now. we have E[e n n i=1 ui (N1 (ti )−N1 (ti−1 ))+ n n j=1 vj (N2 (tj )−N2 (tj−1 )) ] = i=1 E[eui (N1 (ti )−N1 (ti−1 )) ] j=1 n i=1 E[evj (N2 (tj )−N2 (tj−1 )) ] ]E[e n j=1 = E[e ui (N1 (ti )−N1 (ti−1 )) vj (N2 (tj )−N2 (tj−1 )) ]. Proof. 2) relative to Ftn−1 and the third equality comes from the result obtained in the above paragraph. 0<u≤t Since ∆Xt = u1 ∆N1 (t)+u2 ∆N2 (t) and N1 . N1 (t2 ) − N1 (t1 ). t > s. n−1 i=1 n−1 i=1 n−1 i=1 n−1 j=1 n−1 j=1 n−1 j=1 vj (N2 (tj )−N2 (tj−1 )) vj (N2 (tj )−N2 (tj−1 )) vj (N2 (tj )−N2 (tj−1 )) where the second equality comes from the independence of Ni (tn ) − Ni (tn−1 ) (i = 1. So taking expectation on both sides. N1 (tn )) is independent of (N2 (t1 ). · · · . we can find a set Ω0 of probability 1. (Ft )t≥s . · · · . According to page 524.s. E[eu1 (N1 (t)−N1 (s))+u2 (N2 (t)−N2 (s)) ] = eλ1 (e u1 −1)(t−s) eλ2 (e u2 −1)(t−s) = E[eu1 (N1 (t)−N1 (s)) ]E[eu2 (N2 (t)−N2 (s)) ]. we get 0 = 0 + E{[M1 . M2 ]t } = E[ 0<s≤t ∆N1 (s)∆N2 (s)]. Therefore N1 and N2 can have no simultaneous jump. So E[eXt ] ≡ 1. following the scheme suggested by page 489. Define M1 (t) = N1 (t) − λ1 t and M2 (t) = N2 (t) − λ2 t. paragraph 1. · · · . eXu −eXu− = eXu− (e∆Xu −1) = eXu− [(eu1 − 1)∆N1 (u) + (eu2 − 1)∆N2 (u)]. 11.Proof.. Let t0 = 0. The problem is ambiguous in that the relation between N1 and N2 is not clearly stated. paragraph 2. Fix s ≥ 0. This shows (eXt )t≥s is a martingale w. N2 (tn ) − N2 (tn−1 )). By letting t = 1. We shall prove the whole path of N1 is independent of the whole path of N2 . Then by Itˆ’s formula for jump process.e. we conclude 0<s≤t ∆N1 (s)∆N2 (s) = 0 a. i.5. So eXt − 1 t = s t eXu− [−λ1 (eu1 − 1) − λ2 (eu2 − 1)]du + s<u≤t eXu− [(eu1 − 1)∆N1 (u) + (eu2 − 1)∆N2 (u)] = s eXu− [(eu1 − 1)d(N1 (u) − λ1 u) − (eu2 − 1)d(N2 (u) − λ2 u)]. 0<s≤t ∆N1 (s)∆N2 (s) = 0 for all t > 0. then the vector (N1 (t1 ). N2 (tn )) if and only if (N1 (t1 ).s. suppose we have 0 ≤ t1 < t2 < t3 < · · · < tn . N1 (tn ) − N1 (tn−1 )) is independent of (N2 (t1 ). so that ∀ω ∈ Ω0 . we have o t e Xt − e Xs = s t c eXu dXu + 1 2 t c c eXu dXu dXu + s (eXu − eXu− ) s<u≤t = s eXu [−λ1 (eu1 − 1) − λ2 (eu2 − 1)]du + (eXu − eXu− ). 2004. Springer-Verlag. Cambridge University Press. Singapore. [5] A. futures. x) = E[h(QT −t + x)]. Qt ). Options. and other derivatives. New York. Itˆ’s formula for jump process yields o t eXt − 1 = 0 1 1 eXs (u1 dWs − u2 ds − λ(ϕ(u2 ) − 1)ds) + 1 2 2 t eXs u2 ds + 1 0 (eXs − eXs− ). 1995. Wilmott. The mathematics of financial derivatives: A student introduction. 1 This shows eXt is a martingale and E[eXt ] ≡ 1.. 2003. models. where g(t. Øksendal.6. Berline. 2004. [4] D. Hull. 2000. then Ht − λE[eu2 YNt − 1]t = Ht − λ(ϕ(u2 ) − 1)t is a martingale. Berlin. So eXt − eXt− = Nt eXt− (e∆Xt − 1) = eXt− (eu2 YNt − 1)∆Nt .This shows the whole path of N1 is independent of the whole path of N2 . Continous martingales and Brownian motion. [6] S. Stochastic calculus for finance II. 0<s≤t Note ∆Xt = u2 ∆Qt = u2 YNt ∆Nt . Let Xt = u1 Wt − 1 u2 t + u2 Qt − λt(ϕ(u2 ) − 1) where ϕ is the moment generating function of the jump 2 1 size Y . Shreve. [7] S. References [1] F.7. 1999. theory. Prentice-Hall International Inc. Springer-Verlag. Springer. Springer-Verlag. 2006. [2] J. Sixth edition. Proof. The mathematics of arbitrage. The binomial asset pricing model. Delbaen and W. eXt − eXt− = eXt− ∆Ht and t eXt − 1 = 0 t 1 1 eXs (u1 dWs − u2 ds − λ(ϕ(u2 ) − 1)ds) + 1 2 2 t t t eXs u2 ds + 1 0 0 eXs− dHs = 0 eXs u1 dWs + 0 eXs− d(Hs − λ(ϕ(u2 ) − 1)s). Revuz and M. Fourth Edition. E[h(QT )|Ft ] = E[h(QT −Qt +Qt )|Ft ] = E[h(QT −t +x)]|x=Qt = g(t. New Jersey. Third edition. So E[eu1 Wt +u2 Qt ] = e 2 u1 t eλt(ϕ(u2 )−1)t = E[eu1 Wt ]E[eu2 Qt ]. SpringerVerlag. [8] P. 11. Shiryaev. Shreve. This shows Wt and Qt are independent. Proof. 84 . N. Continuous-time models. where Nt is the Poisson process associated with Qt . Stochastic differential equations: An introduction with applications. Yor. 11. [3] B. Schachermayer. Essentials of stochastic finance: facts. 1998. World Scientific. Stochastic calculus for finance I. New York. Consider the compound Poisson process Ht = i=1 (eu2 Yi − 1).