Stochastic Calculus Final Exam With Solutions

Stochastic Calculus — Final Examination SolutionsJune 17, 2005 There are 12 problems and 10 points each. 1. (Property of Brownian Bridge) Xt dt + dBt , Let Bt0 = {Bt , 0 ≤ t ≤ 1|B1 = 0} be a Brownian bridge, and define dXt = − 1−t with X0 = 0. a) Show that {Bt0 , 0 ≤ t ≤ 1} and {Xt , 0 ≤ t ≤ 1} have the same distribution. b) Show that Yt = (1 + t)B 0 t is a standard Brownian motion. 1+t sol: (a) (1)Bt0 = Bt − tB1 , ∵ Bt ∼ N (0, t), B1 ∼ N (0, 1), tB1 ∼ N (0, t2 ) ∴ linear combination of normal distribution is still normal. Compute mean and variance. EBt0 = 0, EBs0 Bt0 = s(1 − t)ifs < t ⇒ VarBt0 = t(1 − t) Bt0 ∼ N (0, t(1 − t)) ∀0 ≤ t ≤ 1. Rt (2) According to the solution of OU process, Xt = (1 − t) 0 1 dBs 1−s Xt is the linear combination of N.D., therefore it is still a N.D. Compute mean and variance Xt = 0, Ifs ≤ t : EXs Xt Rs 1 Rt 1 = (1 − s)(1 − t)E[ 0 1−u dBu 0 1−v dBv ] Rs 1 Rs 1 Rt 1 = (1 − s)(1 − t)E[ 0 1−u dBu ( 0 1−v dBv + s 1−v dBv )] Rs 1 = (1 − s)(1 − t)E[( 0 1−u dBu )2 ] (independent increment) Rs 1 2 = (1 − s)(1 − t)E[ 0 ( 1−u ) ds] (Ito’s isometry) s = s(1 − t) ⇒ VarXt = t(1 − t) = (1 − s)(1 − t) 1−s Xt ∼ N (0, t(1 − t)) (b) ∵B t 1+t ∼ N (0, t s ), B 1+s 1+t t EXt = E(1 + t)(B 1+t − ∼ N (0, t B) 1+t 1 s ), B1 1+s =0 1 ∼ N (0, 1) ∴ Yt ∼N.D. . < tn = 1 : Yt1 − Yt0 . Ys ) = t + s − 2s = t − s Yt − Ys ∼ N (0. Ytj − Ytj−1 ) = 0 (3)stationary increment E(Yt − Ys ) = 0 Var(Yt − Ys ) = Var(Yt ) + Var(Ys ) − 2Cov(Yt . 2.. t) (5)Bt is continuous in t ⇒ Yt is also continuous in t From (1)∼(5). and find the limiting distribution. Yt is a SBM. t) (1)Y0 = 0 (2)Independent increment ∀0 = t0 < t1 < . t − s) (4)Yt ∼ N (0. as t → ∞. Yt2 − Yt1 . where X0 = x0 and α > 0. a) Solve the following stochastic differential equation dXt = (−αXt + β)dt + σdBt .t s < 1+t ≤ 1) : If 0 ≤ s < t ≤ 1 (0 ≤ 1+s s B t EXs Xt = (1 + s)(1 + t)[EB 1+s − 1+t VarYt = t t s B EB 1+s 1 1+t − s t B1 EB 1+t 1+s + s t EB12 ] 1+s 1+t =s⇒ Yt ∼ N (0.. Ytn − Ytn−1 Cov(Yti − Yti−1 . . b) Verify the solution can be written as Xt = e −αt β (x0 + (eαt − 1) + σ α Z t eαs dBs ). d) Find the covariance Cov(Xs ... Xt ) for s < t. sol: (a) dXt + αXt dt = βdt + σdBt multiply eαt for both sides d(eαt Xt ) = βE αt dt + σeαt dBt Rt Rt ⇒ eαs Xs |t0 = β 0 eαs ds + σ 0 eαs dBs 2 . 0 c) Show that Xt converges in distribution. where Mt is the P-mtgle defined by Mt = exp(µBt − (b) 3 µ2 t ) 2 . sol: (a) If the process {Bt } is a P-BM and Q is the measure on C[0. b) Let L be the line given by the equation y = a + bt with a > 0. Use the simplest Girsanov theorem (Brownian motion with drift) to derive the probability density function fτL (t) of τL . T ] satisfies EQ (W ) = EP (W MT ). a) Give a precise description of the simplest Girsanov theorem. T ] induced by the process Xt = Bt + µt. 2α 2 −α(t+s) Therefore Xt → (d) Cov(Xs . Define τL = inf{t : Bt = a + bt}. 0 eαu dBu ) 3.⇒ Xt = e−αt (x0 + αβ (eαt − 1) + σ (b) ∂Xt = −αXt + β ∂t Rt 0 eαs dBs ) ∂Xt =σ ∂Bt 2 ∂ Xt =σ ∂Bt2 By Ito’s formula: dXt = (−αXt + β)dt + σdBt (c) (1)EXt = x0 e−αt + αβ (1 − e−αt ) → αβ as t → ∞ Rt σ2 (2)VarXt = σ 2 E 0 e−2α(t−s) ds = 2α (1 − e−2αt ) → D σ2 ) as t N ( αβ . then every bounded Borel measurable function W on the space C[0. Xt ) = σ e R s∧t = σ 2 e−α(t+s) E 0 e2αu du σ 2 −α(t+s) 2α(s∧t) = 2α e (e − 1) σ2 2α as t → ∞ →∞ Rt Rs Cov( 0 eαu dBu . 4 . x) = f (x) where f : R → R is also bounded. If u(t. x) + q(x)u(t. 2 and ∃! φ(s. dβt = rβt dt. x) = 21 uxx (t. where {Ft } is the standard Brownian filtration. such that Xt = Z t 0 φ(s. x) u(0. Consider  ut (t. x) is the unique bounded solution of the function. ∀0 ≤ t ≤ T . ω)dBs. T ]. such that E(XT2 ) < ∞. φ( a+bt fτa (t) = ∂t P (τL ≤ t) = t3/2 t 4.τL = inf{t : Bt = a + bt}. x) = E[f (x + Bt )exp( Rt 0 q(x + Bs )ds)] 5. The Black-Scholes model is assumed to be dSt = µSt dt + σSt dBt . then u(t. a) By using arbitrage theory to show the procedure of deriving the Black-Scholes PDE for European call option. ω) ∈ H [0. if ∃a and T . τa = inf{t : Xt = a}. (b) Suppose that the function q : R → R is bounded. b) State the Feynman-Kac representation theorem for Brownian motion. Sol: (a) Xt is an {Ft }-martingale. whereXt = Bt − bt P (τL ≤ t) = Q(τa ≤ t) = EQ (1(τa ≤ t)) = EP (1(τa ≤ t)MT )(simplest Girsanov theorem) 2 whereMt = exp{−bBT − b2 T } = EP (1(τa ≤ t)Mt∧τa ) ({τa ≤ t} is Ft∧τa measurable) 2 = EP (1(τa ≤ t) exp(−ab − b2 τa )) Rt 2 = 0 exp(−ab − b2 s) a3 φ( √as )ds s2 ∂ a √ ) for t ≥ 0. a) State the Martingale representation theorem. St ) = 0 Y (T. z) = f (z) = (ez − K)+ − t) Set u(θ. z) = er(T −t)Y (t. Ito’s formula: 2 dY = ( ∂Y + µS ∂Y + 21 σ 2 S 2 ∂∂SY2 )dt + σS ∂Y dWt ∂t ∂S ∂S And dY = βt dBt + γt dSt = (rβt Bt + µγt St )dt + σγt St dWt Matching coefficient:  ∴ Yt = ⇒  1 ∂Y ( rBt ∂t γt = βt = 2 + 21 σ 2 S 2 ∂∂SY2 )Bt + ∂Y ∂S 1 ∂Y ( rBt ∂t 2 + 21 σ 2 S 2 ∂∂SY2 ) ∂Y S ∂S t 2 (T. St ) + 12 σ 2 (T. (continued) c) Use Fourier transform technique to derive the solution of the above heat equation.St ) . Sol: (a) Hedging price Yt = βt Bt + γt St . then ∂u ∂θ 2 5 .b) Show how to transform this PDE into a heat equation. St )ST2 ∂∂SY2 (T. d) Derive Black-Scholes formula for European call option via Feynman-Kac representation. S) = f (T. 6. S) = (S − K)+ ∂Y ∂t (b) Let   θ = σ 2 (T − t) σ2 )(T 2  z = ln S + (r −  = 12 ∂∂zu2 u(0. z) = √1 2π R∞ −∞ z2 1 2 fˆ(w)e− 2 w θ eiwz dw By transform of Gaussian. where {Wθ } is the standard Brownian motion. θ) u ˆ(0. θ) ∂t 1 2 ⇒u ˆ(θ. w) = fˆ(w) ∂ u ˆ(w. Derive the pricing formula for European put option by using the technique of change of numeraire and Girsanov theorem. payoff X = (K − S(T ))+ ˜ −rT (K − S(T ))+ ] p = E[e ˜ −rT (K − S(T ))I(k ≥ S(T )] = E[e ˜ = e−rT K P˜ (K ≥ S(T )) − e−rT E[S(T )I(K ≥ S(T ))] = (1) − (2) 6 . √ 1 − K)+ E(eWθ +z − K)+ = E(ez e θW √ θ θ = E(ez+ 2 e θW1 − 2 − K)+ θ θ √ K) ) − Kez+ 2 Φ( z−ln = ez+ 2 Φ( z−ln√K+θ θ θ 2 S = S0 Φ( ln( K0 )+(r+ σ2 )(T −t) √ ) σ T −t − Ke−rT Φ( S 2 ln( K0 )+(r− σ2 )(T −t) √ ) σ T −t 7. we have F (e− 2θ )(w) = 2 − z2θ = F ( √1θ e ⇒e Therefore.(c) Take Fourier transform:  = 12 (iw)2 u ˆ(w. w) = fˆ(w)e− 2 w θ Inverse Fourier transform: u(θ. z) = f ∗ gRθ (z) ∞ = √12π −∞ f (y)gθ (z − y)dy R∞ (z−y)2 = √12π −∞ f (y) √1θ e− 2θ dy R∞ (z−y)2 1 = √2πθ f (y)e− 2θ dy −∞ R∞ y (z−y)2 1 + − 2θ dy = √2πθ (e − K) e −∞ (d) y = Wtheta + z ⇒ dy = dWθ . θ) = − 21 uˆ(w. 2 − z2θ √ 1 θe− 2 w 2θ )(w) ≡ F (gθ (z))(w) u(θ. b) Let N (t) be a Poisson process with intensity λ. Barrier option pricing: Find the arbitrage price of a contingent claim that pays M if the stock price gets to a level K or higher during the time period [0. b = − σ1 (r − σ2 )t τ = inf{Bt = a + bt} (a+bt)2 a − 2t fτ = √2πt e 3 −rT ˜ u0 = e E[M 1( sup St > K)] = e−rT M P˜ ( sup St > K) t∈[0.T ] t∈[0.T ] ln SK0 2 Let a = σ1 ln SK0 . Show that exp{ln(1 − u)N (t) + uλt}. p = Ke−rT Φ(−d2 ) − S0 Φ(−d1 ) 8.T ] t∈[0. t∈[0. σ2 T 2 ¯t = W ˜ t − σt Let W √ σ2 σ2 ¯ k ≥ S0 e(r− 2 )T +σ T Z = S0 e(r+ 2 )T +σWT ⇒ z ≤ −d1 ) = S0 P¯ (z ≤ −d1 ) = S0 Φ(−d1 ) Therefore. In other words. t ≤ T |PT )) = e−rT M P (τL ≤ T ) −rT = e−rT M [1 − Φ( 2 ln K −(r− σ2 )T S0 √ σ τ 2 )+ − ln K σ22 (r− σ2 ) Φ( e S0 2 K −(r− σ2 )T S0 √ σ T )] 9. t ≤ T |P˜T ) = Law(r + σBt .T ] = e M P (τ ≤ T ) (Law(µ + σBt . 7 .σ2 2 √ )T +σ T Z σ2 )T ] 2 ⇔ z ≤ − σ√1 T [ln( SK0 ) + (r − 2 1 √ [ln( SK0 ) + (r − σ2 )T ] σ T (1) = Ke−rT P˜ (z ≤ −d2 ) = Ke−rT Φ(−d2 ) ˜ −rT S(T ) I(K ≥ S(T ))] (2) = S0 E[e K ≥ S0 e(r− where d2 = (Let dP¯ dP˜ = S(0) −rT S(T ) e S(0) ˜ = e σ WT − ≡ −d2 . find a formula for the arbitrage price of the claim X = M I{ sol: sup St > K ⇔ sup Bt + σ1 (r − t∈[0.T ] σ2 )t 2 ≥ 1 σ sup St >K} . a) State the Poisson process martingale property theorem. T ] and that pays zero otherwise. 0 < u < 1. Sol: (a) For European call option. St ). x) − r(t.is a martingale. Sol: (a) Let N (t) be a Poisson process with intensity λ. St )dt + σ(t. St ) is the solution of the terminal value problem 1 ft (t. σ(t. and r(t. x)fxx (t. where f (t. 2 f (T. St ) are given by explicit functions of the current time and the current stock price. x)xfx (t. the payout is h(St ) = (ST − K)+ . x)f (t. Consider the stock and bond model given by dSt = µ(t. Consider the replicating portfolio at time t. St )dBt and dβt = r(t. bt : the number of units of the bond. Total value of the portfolio at time t is V t = a t S t + b t βt We require that the restructuing of the portfolio be self-financing. x). E[exp{ln(1 − u)N (t + s) + uλ(t + s)}|Ft ] = exp{ln(1 − u)N (t) + uλt} 10. Vt Let at : the number of units of stock. St ) = − σ 2 (t. x) + r(t. (b) E[exp{ln(1 − u)N (t + s)}|Ft ] = E[exp{ln(1 − u)N (t) + ln(1 − u)(N (t + s) − N (t))}|Ft ] = exp{ln(1 − u)N (t)}E[exp{ln(1 − u)(N (t + s) − N (t))}|Ft ] = exp{ln(1 − u)N (t)}E[exp{ln(1 − u)(N (t + s) − N (t))}] = exp{ln(1 − u)N (t)} exp{−uλs} Therefore. Then M (t) = N (t)−λt is a martingale. So. St ). where all of the model coefficients µ(t. the requirement is dVt = at dSt + bt dβt 8 . x) = h(x). St )βt dt. St ). Use the coefficient matching method to show that arbitrage price at time t of a European option with terminal time T and payout h(ST ) is given by f (t. St )dBt ∵ Vt = f (t. The price of a security U is defined as the product of two asset prices R and S. Ut = Rt St . St ))]dt + fx (t. ⇒ dVt = ft (t. x)fxx (t. St ) and Ito formula for geometric Brownian Motion. St σ 2 )(t. St )dSt = [ft (t. St )σ 2 (t. x) which is the solution of the terminal value problem. St )σ 2 (t.e.. St ) + 1/2fxx (t. St )βt ]dt + at σ(t. x) f (T. St µ(t. x) = − 12 σ 2 (t.S1t )βt (ft (t.S1t )βt (ft (t. Suppose R and S are Ito processes given by dRt = 2(100 − Rt )dt + St dBt . St )dBt Compare (1) and (2) ⇒ at = fx (t. x)f (t. x) − r(t. St )dBt = [at u(t. St ) + 1/2fxx (t. St )) The arbitrage price at time t of European option with terminal time T is f (t. St ) = a t S t + b t βt = fx (t.Hence put (1) and (2) into (4) dVt = µdt + σ(t. St ) + bt r(t.  ft (t. St )dSt dSt + fx (t.S1t )βt (ft (t. 9 . St )σ 2 (t. i. St ) = at St + bt βt Hence they replicate h(ST ). St )St + r(t. x) = h(x) (b)from (a) at = fx (t. dSt = Rt St dt + St dBt . x)xfx (t. St )dt + (1/2)fxx (t. St )) Vt = f (t. 11. St ) + 1/2fxx (t. St ) + fx (t. x) + r(t. St ) + (1/2)fxx (t. St )) ∴ Vt = f (t. St )σ(t. St ) bt = r(t. St ) bt = r(t. 15·35 5. t E( dtU ) = dt = = = = Ut dt + 200tSt − 2tUt + tUt Rt + tSt2 U0 + 0 (t → ∞) R 0 S0 0. by Ito formula dtUt = ft dt + fU dU + 12 fU U (dU )2 = Ut dt + tdUt = Ut dt + (200tSt − 2tUt + tUt Rt + tSt2 )dt + (tSt2 + tRt St )dBt (b). and find its expected instantaneous rate of return at time 0 if R0 = 0. What is the value of the machine? Hint: By how much should the asset value change at the time the certificate is printed? Assume that V is linear in X. which pays a cash flow of X(t) determined by the Poisson process: V = V (X) By Ito’s lemma: dV = VX dX + 21 VXX (dX)2 = [αXVX + 12 σ 2 X 2 VXX ]dt + σXVX dW Expected cash gain (ECG) = E[dV ] = [αXVX 12 σ 2 X 2 VXX ]dt Expected cash flow (ECF) = Xλdt (a cash flow paid with probability λdt) Total Revenue (TR)=ECG+ECF=[αXVX + 21 σ 2 X 2 VXX + Xλ]dt Arbitrage theory: [αXVX + 21 σ 2 X 2 VXX + Xλ]dt = rV dt 10 . t ≥ 0} is also an Ito process. dUt = UR dR + US dS + URS dRdS = St (2(100 − Rt )dt + St dBt ) + Rt (Rt St dt + St dBt ) + St2 dt = (200St − 2St Rt + Rt2 St + St2 )dt + (St2 + Rt St )dBt Let f = tUt .Show that {tUt . The economy is risk-neutral. sol: dX = αXdt + σXdW To find the value V of the machine. Assume that X follows a geometric Brownian motion with drift α and volatility σ. A machine prints a certificate worth X(t) at random time t generated by a Poisson arrival process with intensity λ. and the risk-free rate of interest is r.25 12. Sol: (a).15 and S0 = 35. VXX = 0 r(AX + B) = αXA + 0 + Xλ ⇒ rA = αA + λrB = 0 λ A = r−α . V = λX r−α 11 . VX = A.B = 0 Hence.⇒ αXVX + 12 σ 2 X 2 VXX + Xλ = rV Let V = AX + B. Documents Similar To Stochastic Calculus Final Exam With SolutionsSkip carouselcarousel previouscarousel next01 Latihan Cuti Pertengahan Penggal Pertama 2013Kalman Pascal Stallings1105 SaumAnother Integer Algorithm To Solve Linear Equations (Using Congruences)2014 Abadir Distaso Giraitis Koul Asymptotic Normality for Weighted Sums of Linear Processes40 Tricks of Math Two Test PapersBlacketscholes Formule de ItoSolutions to HW2Chapter 08 Statfolland2Maths Olympiad 2009Chpt2 Text Answers_Stock_Watson_Introduction to EconometricsTopic 3-Quadratic FunctionsBlack Scholes DerivationRatio OooBrownian Motion[1]10miMathsAHBoxExpectation Integral PDFsol1Home Assignment 1things you must know for math staar9789814321303Year4 Term2 Test - EditedUltimate HW1SolsCNY2015 Hwork.docxNitsche Method Interface ProblemChapter 8 Probability Lecture 1673642McaMore From TrbvmSkip carouselcarousel previouscarousel nextLabVIEW 5.0farshAulainternacional4-LP(2).pdfGuida Connettori236454513 Solution Manual for Advanced Accounting 11th Edition by Beams236454513 Solution Manual for Advanced Accounting 11th Edition by BeamsThe Mathematics of Wonton Burrito FieldsLaboratorio de Control Automatico 1Auditing Ch 8Lab ViewAuditing Ch 3 HWManana 1 Libro del Alumno.pdfAuditing Ch 14BooksPruebas Del GicaSTI-H_enLivre Blanc CardiologieVarioMacqweb.a m.a(Ps) Prospectus 2014 15jNetNeans Shortcuts-74test 2New Microsoft Word DocumentPro Log TutorialtestsdtruiuEmperor of China Power SymbolsBusiness Process Re-EnggineeringChaosFooter MenuBack To TopAboutAbout ScribdPressOur blogJoin our team!Contact UsJoin todayInvite FriendsGiftsLegalTermsPrivacyCopyrightSupportHelp / FAQAccessibilityPurchase helpAdChoicesPublishersSocial MediaCopyright © 2018 Scribd Inc. .Browse Books.Site Directory.Site Language: English中文EspañolالعربيةPortuguês日本語DeutschFrançaisTurkceРусский языкTiếng việtJęzyk polskiBahasa indonesiaSign up to vote on this titleUsefulNot usefulMaster Your Semester with Scribd & The New York TimesSpecial offer for students: Only $4.99/month.Master Your Semester with a Special Offer from Scribd & The New York TimesRead Free for 30 DaysCancel anytime.Read Free for 30 DaysYou're Reading a Free PreviewDownloadClose DialogAre you sure?This action might not be possible to undo. Are you sure you want to continue?CANCELOK