Sim Exercises



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¶ x4 − 15000 x4 − 15000 ⇒ = 0.8 ⇒ x4 = 16600 U4 = 0.7881 = Φ 2000 2000 Because of the deductible amount 10000 the insurer’s simulated claim costs for the four months are 5200, 2600, 0, 6600 respectively. Hence the total claim costs are 14400. The correct choice is b. µ 1.8 EXERCISES: 1. Using the Rejection Method generate a random sample from f (x) = 2x, 0 < x < 1. Y are ou given U1 = 1/3 and U2 = 1/4 as random numbers from U(0, 1). a)1/3 b)1/4 c)1/5 d)1/6 e)1/2 2. Y are given: ou ( −1/6 + x/12 for 2≤x<6 f (x) = 4/3 − x/6 for 6≤x≤8 0 otherwise Y are also given U1 = 1/3 and U2 = 1/4 as random numbers from U(0, 1). Generate a random ou sample from f (x) using the Rejection Method. a)3 b)4 c)2 d)5 e)7 x − 10 , 10 < x < 20. 50 Y are given the following random numbers from U(0, 1) : U1 = 0.096, U2 = 0.966, U3 = ou 0.648, U4 = 0.492. 3. Using the Rejection Method generate a random sample from: f (x) = a)12.48 b)14.48 c)16.48 d)18.48 e)19.48 4. Y are to generate a random sample from: f (x) = Cx2 , 1 < x < 5, using the Rejection ou Method. Y are given U1 = 0.156, U2 = 0.1 as random numbers from U(0, 1). ou a)2.5 b)4.624 c)3.624 d)1.624 e)2.624 5. Y are to generate a random observation from P (2) using the Composition Method. Use the ou following sequence of random numbers from U(0, 1) : 0.64, 0.50, 0.25, 0.75, 0.50, 0.67, 0.05, 0.15. What is your random observation? a)6 b)5 c)4 d)3 e)2 6. Y are to draw a random sample from B(8, 0.5) where you are given the following 8 independent ou observations from U (0, 1) : 0.64, 0.50, 0.25, 0.75, 0.50, 0.67, 0.05, 0.15. a)1 b)2 c)3 d)4 e)5 7. Using the Inverse Transform Method, generate random samples from B(4, 0.5) where you are given the following random numbers from U (0, 1) : U1 = 0.965, U2 = 0.539, U3 = 0.303, U4 = 0.731, U5 = 0.662. 8. Using the Inverse Transform Method, generate random samples from P (3) where you are given the following random numbers from U (0, 1) : U1 = 0.64, U2 = 0.25. x2 9. Using the Inverse Transform Method, you are to generate 3 observations from f (x) = , 0 < 9 x < 3. Y may use the following random numbers from U(0, 1) : U1 = 0.008, U2 = 0.729, U3 = ou 0.125. II - 17 Y are given the following distribution function F (x) = x2 . 9 for the group using the following random numbers ou from U(0.8. . 0.250. . 1).216 e)0. Determine the average amount per person paid by the insurance company.01 Y are to simulate the yi .995. 2.92. 0 ≤ x ≤ 1. if any.036 c)0. A random number U from U (0. 0. . Using the 5 simulated examination grades estimate the percentage of passing grades. a)14444 b)1444 c)144444 d)10000 e)70000 16. The distribution of yi is given by: y: 0 100 50000 100000 P (Y = y) : 0. 0.640. x. 0. Y are ou ou 0 otherwise to simulate 2 observations from f (x) using the Inverse Transform Method and the following two random numbers from U(0.10. 0. 0. Y are to use simulation to analyze the medical claims.18 a)30 b)25 c)20 d)15 .027 ≤ U < k 1 .256 14. 0. e)10 12.3.22 d)0. from a B(n.780. 0. Y are given the following probability function: f (x) = 1 − |x| −1 ≤ x ≤ 1 . 1) : 0.910. the repair time.300. n a)0. .80 0. i = 1.19 c)0. A grade of 6 or above is a passing grade.7. is U (0. 0. .64. Customers arrive at a store according to the Poisson Process at a mean rate of λ = ln2. Y are to use ou (10 − x)/25 for 5 < x < 10 the following 5 random numbers from U(0. 0. 9 of each person ou in a group of nine.66.3. 10). Y are to esou timate X. 1): 0.02 and 0.10 0.3 c)0 d)0. It generated the following observations from F (x) : 0.49.7. a)0. 1) to simulate repair times:0. i = 1. 0. The following table gives the correspondence between the values of U and values of x: Values of U Values of x 0 ≤ U < 0. Determine X. .6. Y ou II . b) − 0.350.5. 0. 2. a)20% b)25% c)30% d)35% e)40% 11.68. 3.1 e)0. . 0. The number needing repair in each week has a Binomial distribution with p = 0. 0. For each machine.657 ≤ U < 1 n Determine k. 0.027 0 0.1. Examination grades are distributed according to the following probability density function: ½ x/25 for 0 < x < 5 f (x) = .49. Insurance coverage pays for the excess. Using the two simulated observations estimate the mean of the distribution. 0.3. Four machines are in a shop. . yi . Use the following numbers from U(0.16 b)0. 0. The Inverse Transou form Method was applied to three random numbers from U(0. 1) to simulate 5 examination grades: 0.09 0. Determine the mean of the uniform random numbers used.02. . . the total repair time (in hours) for a three weeks period.002. . of each claim over 10000. 0. 1) is used to generate a random observation.31 15. .77. 0. p).1 13. 0. using the simulation.5.030 b)0. 0. Use the following numbers from U(0.6.26 e)0.890. in hours.1.116 d)0. 3. .09. 0. 0.3 a) − 0. 0. 1) to simulate the number of machines needing repair during each of the three weeks : 0. I. . i = 1. Y are devising a procedure for generating a random value.973 ≤ U. . Generate the random values U1 . .216 ≤ U < 0. from B(3.784 ≤ U < 0. using the Inverse Transform Method. . . Generate xi from the Mixed Congruential generator xn+1 = 5xn + 7 modulo 8. a)I only b)II only c)III only d)I and III e)II and III 18. . S = 3. S = 3 if 7/8 ≤ U1 ≤ 1. II. Let S represent the number of heads in 3 tosses of a fair coin. otherwise P set si = 0.2-1 with a = 3.027 ≤ U < 0. 2. x0 = 3. Let Ui . n be random variables from U(0. U2 . . s = 2. 3.3. if U < 0. . n. S = 2 if 4/8 ≤ U1 < 7/8. . III.657. . S = 2. II. II. Let Yk be a function of Uk . III. U5 . Yi = P − ln(1 − Ui ) for i = 1. i = 1. if Z = 5 or 6. . U4 . x2 . represent independent observations from U (0. .657 ≤ U. . xi .343 ≤ U < 0. Generate one random value. . if 0. . Let Z be the number of Ui s less than or equal to 1/2 among {U1 . 1). Which of the following represent valid procedures for simulating an observation on S. 1). III. if 0.343. ½ 1/2 I. Set s = 3. U. 3. y2 . The average of the numbers xi . . U6 }. III e)None of the above. s = 3. where I is an integer satisfying: I/8 ≤ Ui < (I + 1)/8. if Z = 0 or 1. Set s = 0. . a)I only b)II only c)III only d)I. 2. . U2 . 1). . III. Determine the simulated value of Y corresponding to U = 0. if Z = 3 or 4. with x0 = 4. Set xi = 1 + INT [8Ui ]. from U (0.19 .216. . from U(0. Yk = 1 if Uk ≤ 1/2 . Which of the ou following are valid procedures? I. 1). Set S = Y1 + Y2 + Y3 . if 0.973. where for all k. Set s = 3 si . III e)None of the above a)I only 19. and U. . s = 2 if 0. 2. Repeat Example 1. n is equal to the average of the numbers yi . if U < 0. 3. Z is a random number from Gamma distribution. . s = 1. yn are random numbers from the same exponential distribution II. . U2 . .3). . The numbers x1 . i=1 II. For each Ui set si = 1 if Ui ≤ 0. 0. 0 if Uk > II. Define Xi = − ln Ui . . an observation from U (0. s. S = 1 if 1/8 ≤ U1 < 4/8.027. n. if 0 ≤ U1 < 1/8. U3 from U(0. 3. c = 2. U. Which of the following is true? i=1 I.are to simulate one observation of Y . erify that now the cycle length is 21. if 0. s = 0. . and Z = n Xi . Generate one random value. . . of a random variable X that is uniformly distributed on the set {1. Which of the following methods are valid for the purpose of simulating observations. m = 17. II. 1). . 1). 2. 1). 8}. 20. i = 1. Determine S as follows: S = 0. the time until the next arrival. S = 1 if Z = 2. U3 . V II . III e)None of the above. 2. Then set S = 0. . Let U1 . s = 1. a)e−1/2 b) ln 2 c)1 d)e1/2 e)2 17. b)II only c)III only d)I. xn and y1. Generate random values Ui from U(0. 3. Set xi = I. if 0. . U3 . 1). Generate a random number Ui from U (0.784. a)I only b)II only c)III only d)I.5. once expenses occur. multiplier and starting value. III only c)II. X2 . has the following distribution. III only d)I. U3 = 1/4. x < 0. Y are given the following multiplicative congruential generator: x0 = 1. Y are also given U1 = 0. of medical expenses for any person. ⎧ x<0 ⎨ 0 x/8 0≤x<2 24. 1) random numbers.01. . Y are given the following uniform numbers ou in (0. Use the following U(0.23 b)0. (iii) F (x) = x. 1). X3 are simulated from the distribution of X by using the Inverse Transform Method to three observations from U(0.876 from U(0. c)5100 d)5200 e)10200 a)0 b)5000 28.41 e)0. The random variable X has distribution function given by: F (x) = (x + 1)/4 2 ≤ x < 3 . Use e−2. 0.80.20.95.124 and ou U2 = 0. U2 = 1/2. The observations are: 0..9 0. ou 22. in the order given. A random variable X has distribution function F (x) with (i) F (x) = 0 for. a)0. What are the corresponding random numbers from the given probability function? 23. .082 to determine Z. 0. 1).5 = 0.80. II.70. the number with medical expense during a year is distributed according to a B(3.16.625 and 0. 0. First Cup: U = 0. Determine the sample mean.The maximum possible cycle length is 12. to generate the amount of each claim: 0. .. Use the U(0. 1. .25. . III e)None of the above 27.01. II. Y are given a probability distribution with the following probability function:The Inverse ou Transform II . and 0. in the order given.35 d)0. j = 1.52. 0. Three samples X1 . II only b)I. III. Y are given P (X = j) = (0.1 Each year. . Calculate the total amount that the insurance company pays for the two years. . a)I. Y are to ou simulate Z.29 c)0. ⎩ 1 3≤x Y are given U1 = 3/4.96. (ii) F (0) = 0 0 1/2. Using the Inverse Transform Method ou obtain the corresponding random values of X and calculate X. 3.9). the number of flies in 2 cups of soup.x4 = 9. 0.54. xn+1 = 7xn modulo ou 13. 2. 3. The number of flies in a cup of soup has a Poisson distribution with mean of 2. The amount X. 1) random numbers 0. n = 0. a)19/64 b)1 c)2 d)17/8 e)3 25.40.47 26.5. U4 = 7/8. . 2. 0. 1). the insurance company pays the total medical expenses for the group in excess of 5000.6)j−1 . to generate the number of claims for each of two years.4)(0. Second Cup: U = 0. 0. Of a group of 3 independent lives with medical expense insurance.. Which of the following are true? I.20 a)2 b)3 c)4 d)5 e)6 . x 100 10000 P (x = x) 0.The above generator is faster than a linear congruential generator with nonzero increment using the same modulus. for 0 < x < 1 where F (x) is the derivative of F (x). 86.3. 1. . a uniform random number between 0 and 1.072 c)0. 1]: 0. 0. Use. a)0. the Inverse Transform Method. Determine the value of U. 2. 0. x2 .5 0. in order.620 c)9.03. in order. 0.2 0. The liability loss from an individual suit is a random variable with the following cumulative distribution function ( 0 x<0 1 x=0 F (x) = 2 1 x + 1 0 < x ≤ 10. a)1/6 b)1/3 c)3/5 d)5/12 e)5/6 29. 20 2 Y generate uniform random numbers from the interval [0. 0. a)1 b)3 c)5 d)7 e)8 32.8.Method was used to generate a random observation x1 = 5. 0. 0.8 e)14. Let U be a uniform random number between 0 and 1.2 0. and then divide by 1000 to obtain three uniform random numbers over [0. you simulate the number of suits per year by generating one uniform random number from the interval (0.01. 0.150 b)9.51. Use this generator to obtain x1 . are: 0. Y are given the following Multiplicative Congruential random number generator: xn+1 = ou 17xn (modulo 1001) with x0 = 30. 0. The number of suits for a given year is distributed as follows: # of suits: 0 1 2 3 Probability: 0. .0 b)0.858 31. as many of these uniform random numbers as needed to simulate the company’s total loss due to liability suits over the three year period.69.7.7 e)10.1.915 d)10. 0. Determine the number of unique random numbers (excluding the seed) that can be generated.59. . A company is insured against liability suits. Use the 0 otherwise Inverse Transform Method to determine a random observation from the distribution of X. 0. a)9. ou Using these numbers.6 d)5. 0. The first nine uniform random numbers. 0.250 d)0. 0.1 For each of these three years. Then apply the Central Limit Theorem to use these uniform random numbers to obtain one observation from an approximately normal distribution with mean 10 and standard deviation 5.325 e)0.19. 1] and the applying the Inverse Transform Method. Y three our random numbers are 0.83. 0.125.78.17.21 .5 from this distribution by using U.375 30. a)0.4 33.2 c)5. 1]. Y are given a random variable X with the following probability density function: ou n 2 f (x) = 24x 0 ≤ x ≤ 1/2 .003 b)0. .82. generate as many random observations as possible from the uniform distribution over the II . given U = 0.74. x3 . and the method of complementary random numbers. n = 0. Y are given the following five uniform random numbers over the interval [0. 1] and apply the Inverse Transform ou method to simulate the amounts of the individual losses. ou xn+1 = 6xn + 7 modulo 8.4. Y are given the following mixed congruential random number generator: x0 = 2 (seed). Let α = 0. Use U = 0. H(y) = 0. U2 = 0. U4 = 0. H(y) that of P (3).5. F (y) = αG(y) + (1 − α)H(y) where 0 < α < 1.7 c)4.interval [2. which is expressed as a weighted sum of distribution functions G(y) and H(y) of two other random variables Y1 and Y2 .62. The Composition Method for simulating a random sample from the distribution of Y can be carried out as follows: STEP 1: Generate U1 from U(0. 36.3. 1). 35. Repeat Example 1. Do Exercise 36 where G(y) is the distribution function of Exp(2). STEP 3: If U1 < α.0 e)5. Let G(y) be the distribution function of B(5. U2 = 0.6-8 with T = 2 and λ = 1.5) and H(y) that of P (2). a)1 b)3 c)5 d)7 ( e)9 0 f or y < −3 (y + 3)/6 f or −3 ≤ y < 3 .30 5)e 16)c 26)d 36)c c)0. 0.3 b)2. i. U1 = 0.10. α = 1 f or 3 ≤ y e)0.62. Y may use U1 = 0. STEP 2: Generate U2 from U(0.8. U1 = 0. U5 = 0. Determine the sample average of these random observations.6-7 with T = 2 and λ = 1.12. Generate a random sample from F (y).29 to simulate the value of N(T ). a)2.15.5. respectively. U3 = 0.3.15.22 .75 to generate the times of occurrence of the Poisson events. 1). U1 = 0.24 ANSWERS: 1)a 2)b 12)b 13)d 20)b 23)e 30)a 31)b 3)c 14)c 24)d 32)d 4)d 15)a 25)a 33)d b)0. Do Exercise 36 where G(y) = 1 − e−2y . ou U4 = 0. U6 = 0.34 6)e 17)d 27)c 37)b d)0. otherwise generate a random sample from Y2 .8 d)5. Use U1 = 0. U5 = 0. F (y).3.2.2.62.2 34.58 38. U2 = 0.1.e. a)0.. U2 = 0. then generate a random sample from Y1 using U2 . a)1 b)2 c)3 d)4 e)5 37. α = 0. Let the random variable Y have distribution function. Repeat Example 1. Thus Y has the probability density function of Y1 with probability α and that of Y2 with probability 1 − α.48 10)e 18)d 28)d 38)d 11)a 19)e 29)c II . where α is known as the Mixing Parameter and Y is known as the Mixture Random Variable.3. U2 = 0.15.15. 8]. U3 = 0. Now U3 = 0. we select g(x) = 1 1 = . 8−2 6 which is a uniform random variable defined over the same interval as f (x). Observe that cg(y) 2(1) 3 that X = y = 1/3 is a random sample from f (x). Let g(x) = 1. 0 < x < 1.648.096. We are given that U = U2 = 1/4 which is < 1/2.23 . Since U = 1/4 ≤ 1/3. 3. (ii) U1 U2 = 0. Hence c = 2.6-5. 2 < x < 8. 2. Now.156.1 < = = cg(y) 25 (1. If (−1/6 + y/12) f (y) = = 1/2.105. hence we have X = 4 is indeed a random sample from f (x). Thus f (x)/g(x) = ½ −1 + x/2 for 2 ≤ x ≤ 6 . Now U1 = 0. We are given U1 = 1/3 is from g(x) and hence y = 1/3. Taking g(x) = 1/4. Following Example 1. if U ≤ cg(y) random sample from g(x) and U is a random number from U (0. The correct 25 choice is d.135. Here f (x) = 2x.48 is a random U = U4 = 0. 1).96. The correct choice is c. we have f (x)/g(x) = 31 y2 f (y) U1 = 0. 5. Also U = U2 = 0. It follows 8−x for 6 ≤ x ≤ 8 that c = 2.08 < 0. 4. Hence c = 75/31 and for 1 < x < 5. 8]. (i) U1 = 0. 1). Now U1 = 1/3. it follows that y = 2 + (8 − 2)U1 = 4 is a random sample from g(x). Hence y = 10 + (20 − 10)U3 = 16. The correct choice is a. 5).135.135. 3 2 x ≤ 75/31 for all x ∈ (1. Since here f (x) is defined over the interval [2. 10 < x < 20. 1).966.64 > e−2 = 0. we conclude that X = y = 16. then it will follow that X = y = 4 is a random sample from U≤ cg(y) 2(1/6) f (x). we proceed further and generate another random number from U(0. The correct choice is e.624)2 = 0. e−2 = 0. and since U £ = = cg(y) 10 0. 20) so that c = 2. It follows that n = 3 and X = n − 1 = 2 is a random observation from P (2). which has its maximum value of 2 for all x ∈ [2.492 ≤ cg(y) 10 sample from f (x).648 is the next uniform random number that is given.32 > 0. 0 < x < 1. Now U = U2 = 0.096 so 5 f (y) y − 10 that y = 10 + (20 − 10)U1 = 10. we have y = 1 + (5 − 1)U1 = 1. 20).624 is a random sample from f (x). II . We are given that f (x) = Cx2 . we select g(x) = 20 − 10 10 x − 10 Consequently. (iii) U1 U2 U3 = 0. 1 < x < 5.135. according to the Rejection Method. then it follows that X = y is a random sample from f (x). 8]. 1 1 = . f (x)/g(x) = ≤ 2 for all x ∈ (10. it follows that X = y = 1.48. Since y − 10 f (y) = = 0. The correct choice is b.624. it follows that C = 3/124. Since f (x) is defined over the interval (10. it follows given that U = U2 = 1/4. We are also f (y) 2y 1 = = y = . so that f (x)/g(x) = 2x ≤ 2.SOLUTIONS TO EXERCISES: 1. for all f (y) where y is a x ∈ (0. 6 ⇒ x4 = 6.6875 0.5).423 2 3 3 p2 = 0.000 V erify that U1 = 0.7. U6 = 0.008 = 1 ⇒ x1 = 27 27 x3 x3 2 3 0. 0. Hence U1 = 0.5 ⇒ x5 = 5. U2 = 0. 3.0625 F (j) 0. As in Exercise 7 we have j 0 1 2 3 4 pj 0.25 ⇒ X2 = 2. we observe that of the 8 uniform random numbers given only 5 of them are ≤ p = 0.68 will give a pass grade 25 of 6 or better. U2 = 0. 40%.25 0. pj = P (X = j) = e−3 3i . Since F (x) = P (X ≤ x) = 0 (x /9)dx = .2240418 0.0625 0.303 ⇒ X3 = 1. 2.5)4 . U3 = 0.0497 1 3p0 = 0.125 = ⇒ x3 = 1.539 ⇒ X2 = 2. U5 = 0..965 ⇒ X1 = 4. where pj+1 = pj = j!(4 − j)! j+11−p 4−j pj . We have pj = P (X = j) = 4! n−j p (0.125 ⇒ 0. i. 0. Thus values of uniform random numbers which are at least 0..0625 0.25 0. The correct choice is e. p0 = e−3 = 0. j = 0. U3 = 0. Only 2 out of 5 values satisfy that condition hence the percentage of passing grades 2 is . F (x) = 1 − (10 − x)2 /50 5 ≤ x < 10 . x3 x3 9. 3. 1. 3.729 = ⇒ x2 = 2. U4 = 0. U2 = 0.. 4.0497 0. II .1 ⇒ x2 = 1. U2 = 0.375 0.6. 1. 8.199 2 3 p1 = 0.9375 1. The correct choice is e.662 ⇒ X5 = 2. Hence X = 5 is a random sample from B(8.25 0.375 0.0625 0.6875 0.64 ⇒ X1 = 3. U4 = 0.3 ⇒ x1 = 10(0. 2.3125 0.00 V erify that U1 = 0. The repair times for the 6 machines are obtained as follows: If Ui ∼ U(0. Since you are drawing only one random number from B(8.7 ⇒ y3 = 3. 10).0497 j! F (j) j pj 0 0. e.2240418 0. U5 = 0. Hence X = 3 + 1 + 7 + 6 + 5 + 8 = 30 hours in all. Hence the repair times are U1 = 0.25 0. The correct choice is a.0625.6 ⇒ y2 = 2 and U3 = 0.731 ⇒ X4 = 3. 27 27 ( 2 x /50 0<x≤5 16 34 (10 − 6)2 10.3) = 3. 2..3125 0.8 ⇒ x6 = 8. j+1 j 0 1 2 3 4 pj 0. 1) then 10Ui ∼ U(0. U3 = 0. .0625 0.5.7 ⇒ x3 = 7. j = 1. Now F (6) = 1 − = 1− = = 50 50 50 1 10 ≤ x 17 = 0.815 4 V erify that U1 = 0.008 ⇒ 0.0625 F (j) 0.647 3 4 3 p3 = 0..1680314 0. j = 0.729 ⇒ 0.1493612 0. Thus the total number of machines for the 3-week period is y1 + y2 + y3 = 1 + 2 + 3 = 6 machines.5) and we are given 8 uniform random numbers.6.5. 4 and p0 = (1 − p)4 = (1/2)4 = 0. 5 2 Rx 11.68. U2 = 0.3 ⇒ y1 = 1. 7.24 .9375 1. True: This procedure amounts to setting s = number of Ui ≤ p = 0.995 0.01 100000 1. III. The correct choice is d.68 = 1 − (1 − x2 )2 /2 ⇒ x2 = 0. and F (x) = 1 − (1 − x) /2 for 0 ≤ x < 1 1 − x for 0 ≤ x ≤ 1 ⎩ 1 for 1 ≤ x Also note that F (0) = 1/2.02.350 Claim amount 100 0 50000 100 100000 100 100 100 100 Amount paid 0 0 40000 0 90000 0 0 0 0 40000 + 90000 130000 Hence the average amount paid = = ' 14444.216.2 ⇒ x1 = −0.1 0 0. 9 9 16.027 (0. The correct choice is c.3)3 = ⇒ n = 3. I.3.343 (0. we note that P (X = j) = 14.657 = 0. p = 0.002 0.09)2 = 0. II. set 0. The inter arrival time is Exp(λ) if the customers are arriving according to a Poisson Process with rate λ.3-1. Hence P (X = 0) + P (X = 1) = k ⇒ k = = pn 0.8 ⇒ 0. U2 = F (x2 ) = x2 = (0. Since F (x) = x2 . 15.68. II . U3 = F (x3 ) = x3 = (0. The correct choice is b.99 40000 3 0.⎧ for − 1 ≤ x < 0 ⎨ (1 + x)2 /2 1 + x for − 1 ≤ x < 0 2 . Observe that f (x) = . The correct choice is a. in general. 2 2 ½ n! pj (1 − p)n−j so that P (X = j!(n − j)! P (X = 0) = 0) = (1 − p)n = 0..5) = λ ln 2 = 1.3 as the random sample from B(3. It −0. The correct choice is c.8 + 0.5 = 1 − e−λy or 0. Then Y ∼ Exp(λ) where λ = ln 2.250 0.6-2).09 50000 0.2401..2192667 ' 0..True: Since the sum of n independent and identically distributed exponential random variables is indeed a Gamma random variable.6578 2 2 = 0.8 or −1.2 x1 + x2 follows x= = = −0.7)3 (0. Thus P (X = n) (1 − p)n 0. For U2 = 0. Hence for U1 = 0. The correct choice is d.0081.7) = 0. (See Example 1. Hence (U1 +U2 +U3 ) = 3 3 0. set 0. 12. 17.3). j pj Claim amount F (j) Amount paid 0 0. 18. Since F (y) = P (Y ≤ y) = 1 − e−λy .4096. Following Example 1. 1 1 for U = 0.False: Two random samples of size n do not. we have 0.22 is mean of the uniform random numbers.343 = pn .780 0. I.6-1).640 0.49) = 0. 13.25 .8.64)2 = 1 2 1 0. set U1 = F (x1 ) = x2 = (0.5.890 0.910 0.7. yield the same sample mean.True.1 0 1 0.00 90000 U 0.2.3)2 (0.027 and P (X = n) = 1 − 0..3)3 + 3(0.02 = (1 + x1 )2 /2 ⇒ x1 = −0.8 100 0.9 0 2 0. 0. it follows.300 0.2 or 1.5 = e−λy ⇒ y = − λ ln(0. (See Example 1. 784 0.. x9 = 3. x14 = 7.216 ⇒ 3 − X = 1 ⇒ X = 2 Y = 2 if 0..II.3)j (0. 2.6) 23.189 0. x8 = 4.027 0. The correct choice is b.876) 22. Here n = 3. I .973 1.7)3−j . 3.657 ≤ U < 1. Let Xi be the random variable representing the number of flies in the ith cup of soup.3 where the procedure for obtaining a random number from a discrete uniform distribution requires setting X = j if and only if j − 1 ≤ nU < j ⇔ X = I INT [nU] + 1. Clearly P (S = 0) 6= P (Z = 0 or 1).027 ⇒ 3 − X = 0 ⇒ X = 3 Y = 1 if 0.124) = INT [4. 3. N = INT = INT = INT [0. it is clear that I and II are true. p) then Y = n−X ∼ B(n. x7 = 9.True: Observe that if P (X = j) = pj = 3! (0. x4 = 0. i = 1. Thus we have and pj = P (Y = j) = j!(3 − j)! j 0 1 2 3 pj 0. x2 = 1.False: V erify that x1 = 3.216 0. x11 = 10. ∙ II . x7 = 1. x4 = 0. then j!(3 − j)! j 0 1 2 3 pj 0. 1/2). x13 = 13. x3 = 5.189 0. Note that the value 8 will never be obtained by this procedure despite P (X = 8) = 1/8. V erify that for U1 = 0. On the other hand P (X = 0) = 0 but will be obtained as a random sample.3 3! (0.441 0. x6 = 2. 2.000 Hence set Y = 0 if 0 ≤ U < 0. x16 = 3.000 ⇒ 3 − X = 3 ⇒ X = 0 The correct choice is d.124.657 1. 1−p). p = 0. x5 = 7. 20.027 F (j) 0. x3 = 5.216 ≤ U < 0. Recall that if Z = number of Ui ≤ 1/2 then Z ∼ B(6. j = 0. Here n = 8 and therefore X = I + 1 if and only if I ≤ 8U < I + 1 ⇔ ≤ U < 8 I +1 .True: If X ∼ B(n. Hence III is false.7)j (0. x6 = 8. Let us examine if P (S = 0) = P (Z = 0 or 1). 1.26 . ¸ ∙ ¸ ln(1 − U ) ln(0. 1/2).3)3−j .4-1). (See Result 1. 1.027 0. x8 = 12. x15 = 6. V erify that x1 = 11.343 0. 21. Now S ∼ B(3.343 0.True III .259] = ln(1 − p) ¸ ln(0. x9 = 4.027 ≤ U < 0.000 An application of the Inverse Transform method immediately shows the statement given is true. on the other hand P (Z = 0 or 1) = P (Z = 0) + P (Z = 1 6 7 1) = (1/2)6 + 6(1/2)6 = + = . 8 II . x2 = 6. Following the solution for Exercise 18. x12 = 15. N = INT ln(0. x5 = 2.657 ⇒ 3 − X = 2 ⇒ X = 1 Y = 3 if 0. Since P (S = 0) = (1/2)3 = 1/8. j = 0.False: Recall Note-2 of Section 1. U2 = 0.876.6) ∙ ln(0. III.343 F (j) 0. 64 64 64 The correct choice is e.08] = 4 ⇒ X = N +1 = 5. 19. x10 = 14.441 0. 2. 0 ⇒ X = N +1 = 1. we solve for x2 where x2 /2 + 1/2 = 0.80. The correct choice is e.625. II .40.5 p = 0.True: Since we need to carry out one extra addition operation to calculate each xi .04 ⇒ x3 = 0. Hence x = 3 3 30 26. we observe that 1/4 = F (2− ) ≤ U2 = 1/2 < F (2) = 3/4.25. .2. II . 3 3 x1 + x2 + x3 0 + 0.757 0. we observe that 1/4 = F (2− ) ≤ U3 = 1/4 < F (2) = 3/4. 3/4 = (x + 1)/4. 25. 4 8 x1 + x2 + x3 + x4 2 + 2 + 2 + 5/2 Hence x = ¯ = = 17/8.23. x4 + 1 7 For U4 = 7/8.52. Hence x2 = 2 is the simulated value of X. The correct choice is d. III .2 7 = = = 0. 4 4 NOTE: This is an example of a random variable which has a pdf which is discontinuous at X = 2 but continuous for all other values. x4 = 9.287 0. For U2 = 0. x12 = 1 = x0 .1334636 4 3 . x9 = 8. .543 0.5 + 0.True: V erify that x1 = 7.5 2 2 For U3 = 0.125) = 0.25625 2 1 2. 24. . For U = 0. x8 = 3. we obtained x4 = 9. The graph of F (x) is given here. .205 2. we have 0 = F (0 − ²) ≤ 0. we solve for x3 where x2 /2 + 1/2 = 0. .5p0 = 0. . Hence x1 = 2 is the simulated value of X.5 F (j) 0.082 e 2. we solve = ⇒ x4 = 5/2. x5 = 11.5 p = 0.2333 ' 0. Solve for x.02) = 0.5 p = 0. For U2 = 1/2.27 . pj = 0. For U1 = 3/4. Clearly the cycle length is 12. The correct choice is d. The correct choice is a.2135417 3 2 2. F (x) = 2 ⎪ 0 tdt + 1/2 = x /2 + 1/2 for 0 ≤ x < 1 ⎪ ⎩ 1 for 1 ≤ x Its graph is given here For U1 = 0. x7 = 6. x3 = 5.082 0. ⎧ for x < 0 ⎪ 0 ⎪ ⎨ 1/2 for x = 0 Rx . x10 = 4.625 ⇒ x2 = 2(0.True: In I.25 ⇒ x2 = 0. x6 = 12. x11 = 2. Hence x3 = 2 is the simulated value of X.890 . −2.25 < F (0) = 1/2 ⇒ x1 = 0. x2 = 10.52 ⇒ x2 = 2(0. X2 = 2. Thus Z = X1 + X2 = 6. Observe that P (X = 2) = F (2) − F (2− ) = (2 + 1)/4 − 2/8 = 3/4 − 1/4 = 1/2. X1 = 4 and for U = 0. For U3 = 1/4. I .j 0 1 2 3 4 . second year expense. U2 = 0. we conclude N2 = 2 is the number of claims in year two..0 Thus for U1 = 0. the pdf is given by the equation of the line through (2. f (x) = (x − 2)/12. 1/6) and (8. the pdf is given by the equation of the line through (8. f (x) = (10 − x)/12.271 1.8.20. 12 (NOTE: The above procedure was followed to illustrate the process in general. For 4 ≤ x < 8. 1/6). Second year expenses amount is 10000 + 100 = 10100 and the payment amount will be 5100. Thus 1 = (area A) + (area B) + (area C) = 1 (2)h + 4h + 1 (2)h = h + 4h + h = 6h =⇒ h = 1/6 2 2 Note that the pdf of X is non-zero only for 2 ≤ x ≤ 10.95. the company pays 5100. The correct choice is c.01.001 0. i.5) = Area A + Area II . i.243 0. 3! (0.1)3−j . x3 = 100 is the second. 1/6) and (10.027 0. For 8 ≤ x < 10. For 2 ≤ x < 4.5 − 3)/6 = . the pdf is given by the equation of the line through (4.9 1. U3 = 0.729 pj F (j) 0.1 P (X ≤ x) = F (x) 0.e.5) = (5. Similarly the amount.. f (x) = 1/6. Thus ⎧ x<2 ⎪ 0 ⎪ (x − 2)/12 ⎨ 2≤x<4 1/6 4≤x<8 f (x) = ⎪ (10 − x)/12 8 ≤ x ≤ 10 ⎪ ⎩ 0 10 < x Since f (x) is continuous for all x. X.001 0.e. 3. of medical expenses for any person has the following distribution: X=x 100 10000 P (X = x) 0. 1/6). 1.000 For U1 = 0.. we conclude N1 = 1 is the number of claims in year one For U2 = 0.28 . i. x1 = 100 is the first year expense. The correct choice is d. Area under the curve equals the areas of the two triangular regions A and C plus the area of the rectangle B.9 0. Let N be the number of medical claims where P (N = j) = 0.9)j (0. For the given problem we have P (2 < X ≤ 5.70. 0).5) = P (2 < X ≤ 4) + P (4 < X ≤ 5. j = j!(3 − j)! j 0 1 2 3 0. Being a probability density function the area under the curve must equal 1. second year expense. 2. 28. Thus for the two years. 0) and (4. x2 = 10000 is the first. we have ⎧ x<2 ⎪ 0 ⎪ (x − 2)2 /24 ⎨ 2≤x<4 F (x) = (x − 3)/6 4≤x<8 ⎪ [24 − (x − 10)2 ]/24 8 ≤ x ≤ 10 ⎪ ⎩ 1 10 < x 5 Now U = F (5. First year expenses amount is 100 and the payment amount is 0.028 0.27.e. It follows. 1). 1/4). Thus the total loss for the three year period is y1 + y2 + y3 = 0.6 = 5. Here a = 2 and b = 8. 30.125 = 8x3 ⇒ x3 = 0. and 5. Then E(Y ) = 3(1/2) = 3/2 and V (Y ) = 3(1/12) = 1/4 and using the Y − 3/2 ' N(0. 1) then Y = a + (b − a)X ∼ U(a. b) (Note 4 of Section 1. The correct choice is d. 0 ≤ x ≤ 1/2. Since x1 = 510. For U3 = 0. If X ∼ U(0. 1) so is 1 − U (called complementary random number. Since F (x) = P (X ≤ x) = 8x3 .51 ⇒ y1 = 0. E(U) = 1/2 and V (U) = 1/12.415 − 3/2 W − 10 p = Hence ⇒ W = 9. 33. Recall that for U ∼ U(0.1 G(j) 0.2 0.9 1. Note 3 of Section 1. 1/4).) 29. 52 ) Central Limit Theorem we have p 1/4 W − 10 ∼ N(0. 1.51 ⇒ 1/2 + y1 /20 = 0.2).415 is a random sample from N(3/2. Let Y = U1 + U2 + U3 .59 ⇒ x2 = 1 suits in the second year. U2 = 0.01 ⇒ y2 = 0. The correct 5 1/4 choice is a. 31.662. 1) ⇔ Y ' N(3/2.78 ⇒ 1/2 + y3 /20 = 0.250. Now the cumulative distribution function for the liability loss. 3 suits in all for the three-year period. 1). x2 = 662 and x3 = 243. x2 = 1.of a rectangle with base 1.6. 32. If W ∼ N(10.5)(1/6) = 5/12.15 is a random sample from N(10.5 and height 1/6 = (1/2)(2)(1/6) + (1.2 0. Thus we have II .78 ⇒ y3 = 5. x3 = 5.19 ⇒ x3 = 0.29 . Y.5 0. The cumulative distribution function G(j) for the number of suits for the given year is given by: j: 0 1 2 3 pj : 0. The correct choice is b. U3 = 0. U3 = 0. is given by ⎧ y<0 ⎨ 0 1/2 y=0 F (y) = ⎩ 1/2 + y/20 0 < y ≤ 10 V erify the following: U1 = 0.015625 ⇒ x = 0. The correct choice is c.8. U1 + U2 + U3 = 1. In fact the expression for F (x) could also be obtained using the areas of the associated regions.2.2 + 0 + 5. then 5 1. U2 = 0. Recall if U ∼ U(0.83 ⇒ x1 = 2 suits in the first year.2). x4 = 5 we conclude that the number of unique random numbers (excluding the seed) is just three and they are 3. For U2 = 0. 25). Since x1 = 3.243. suits in the third year.7 0.510.5 0. we have 0.0 V erify that the Inverse Transform method yields the following: For U1 = 0. Thus we have. the uniform random numbers are U1 = 0. 38.5). 2) and its time of occurrence is S1 = 1. . Following Example 1. λ = 1.1353352.4-3 it follows that U2 = 0. 36.4 + 6. Here T = 2.15. S2 = T U(2) = 0. Step-2: U2 = 0. Step-4: I = 1.2 3. Here U1 = 0.4 The sample average of the 10 values of Y obtained is 1 (3.8 = 5. For U1 = 0..48.609. we generate a random sample from G(y). U5 = 0.6 Y = 2 + 6(1 − U) : 6. U2 = 0. Since t = 3. We proceed to generate the time of occurrence of the second Poisson event. we do not stop and go to the next step. 1. Since t = 1. II . 34. 0. The correct choice is b..2 7. The values of the uniform random numbers are U1 = 0. Thus only one Poisson event happens during the time interval (0.2 + 3.4 0. p1 = 2e−2 = 0.6.8 6.15 > 0.6 50 10 0. for j = 0.609 − ln 0.2.U: 0.1. U(2) = 0. The correct choice is d. ..29.1 1−U : 0.10 = α.7 0.3. Hence U1 = 0.15 < 0.6 4. S1 = 1. I = 0 Step-2: U1 = 0.3 and S1 = T U(1) = 0. Since U1 = 0. we conclude that Y is a random sample from P (3). we verify that N(2) = 2( since pj = e−2 2j /j!. so that p0 = e−2 = 0.5.7 0. 0.38) = 0. 2].e.6 + 4.4 5. Exp(2).3 = α. Since T = 2.2 3.30 . From Example 1. i.8) = 10 The correct choice is d.9 Y = 2 + 6U : 3. Thus the time at which the first Poisson event occurs is S1 = 1.8 0. .15. 2.483792 ≈ 0.2 0.62 yields Y = 3 as a random sample from F (y). 37. 3. Step-3: t = t − (1/λ) ln U2 = 1. Hence the random sample is Y = −(1/λ) ln(1 − U2 ) = −(1/2) ln(0.2706706.2 Step-3: t = t − (1/λ) ln U1 = 0 − ln 0.3 0.1.2. U2 = 0. Y = 3 is the required random sample from F (y).609.506 > 2 = T. we stop.609.6-7 we have: Step-1: t = 0. U4 = 0. N(T ) ∼ P (2λ) = P (2).). 35.2 + 6.506.2 = 1. .3 = α.8 + 6.3 ⇔ U(1) = 0.4 + 5. The second Poisson event does not occur in the interval (1. U3 = 0. we conclude that Y is a random sample from B(5.8 2.609.8 + 2.2 + 7.3-3 it follows that for U2 = 0.609 < 2 = T.609.3 6.62.6 + 3. The correct choice is c.12.15 < 0.15 = 3. Since U1 = 0. From Example 1. The four random numbers from U(0. #32/May 2000 Insurance for a city’s snow removal costs covers 4 winter months. 0 < x < 1.0013. The insurer assumes that the city’s monthly costs are independent and normally distributed with mean 15000 and standard deviation 2000. 0.0013 = Φ 2000 ¡ x4 −15000 ¢ =⇒ x4 = 15000 + 0. The denominations of the coins found are independent. 20002 ) we have X−15000 ∼ N(0. The expected number of iterations needed to generate a random variable from f (x). For each coin found.78 c) 1.. II .e. 12(1 − 4x + 3x2 = 0 =⇒ x = 1 and 1/3. #43/November 2000 Lucky Tom finds coins on his 60 minute walk to work at a Poisson rate of 2 per minute.11 Solution f (x) = 12(x − 2x2 + x3 ) is to be minimized over 0 < x < 1. 60% of the coins are 1 each.86 d) 2. It is clear that £ ¤ f (1) f (1/3) 1 = 0 and c = = 12 1 − 2( 1 ) + 27 = 16 ' 1. There is deductible of 10000 per month. where f (x) = 12(x − 2x2 + x3 ). Hence the total claim costs are 14400. 20% are worth 5 each.52 b) 1.5389. #14/May2000 The Rejection Method is used to generate a random Variable with pdf f (x) using the pdf of g(x) and constant c as the basis. (i) The first actuary begins by simulating the number of coins found. 1. 1) are 0. 2600. he simulates its denomination.2(2000) = 12600 U1 = 0.8(2000) = 16600 U4 = 0.and g(x) = 1. To simulate 4 months claim costs. The expected number of random number he needs for one simulation of the is F.7881.1151. a) 1. The correct choice is b.1(2000) = 15200 µ 2000 ¶ x2 − 15000 =⇒ x2 = 15000 − 1.31 . Calculate n. Hence using 2000 the Inverse Transform method ¶ µ x1 − 15000 U1 = 0. 20% are worth 10 each. the insurer uses the Inverse Transform method where small random numbers correspond to low costs.5398 = Φ =⇒ x1 = 15000 + 0. using the Inverse Transform algorithm. 1). 0 < x < 1. until the length of the walk is exceeded.05 2.1151 = Φ µ 2000 ¶ x3 − 15000 =⇒ x3 = 15000 − 3(2000) = 9000 U2 = 0. 0. a) 13400 b) 14400 c) 17800 d) 20000 e) 26600 Solution Since X = Claim cost is distributed N(15000. Two actuaries are simulated Tom’s 60 minute walk to work. CHAPTER 3 3. 0. Calculate the insurer’s simulated claim costs.Problems from sample tests issued by the Society of Actuaries. 3 4 9 g(1) g(1/3) 2. 6600 respectively.78. The constant c has been chosen so as to minimize n. using the procedure of repeatedly simulating the time until the next coin is found. 0.7881 = Φ 2000 Because of the deductible amount 10000 the insurer’s simulated claim costs for the four months are 5200. This is done by setting the first g(x) derivative of the ratio equal to 0. i. The correct choice is b. 4 b) 0. then simulates finding coins worth 10.32 . 1) are 0. The correct choice is d. After a coin is found we need to obtain its denomination (1. or 10) and that requires one random number to ascertain that. (ii) Here N(T ) = N1 (T ) + N5 (T ) + N10 (T ) where N1 (T ) ∼ P (0.46 < 2 = T λ 5 STEP-4: I = 1. I = 0 STEP-2: U1 = 0.10 1 1 STEP-3: t = t − lnU1 = 0 − ln(0. G 123 5. the first actuary’s is not d) Both methods are valid. In Chapter 3.6-7. The expected number of random numbers she needs for one complete simulation of Tom’s walk is G. Thus we will need on the average λT + 1 random numbers to generate times of the coins found. she first simulates finding coins worth 1. λ = 5. and 10 respectively. but they may produce different results from the same sequence of random numbers e) Both methods are valid and they produce identical results from the same sequence of random numbers Solution (i) Here T = 60 minutes and λ = 2 per minute. 4.6 e) Solution F 241 = = 1.20λT ) = P (24). Following Example 1. 2).20λT ) = P (24) are the mutually independent Poisson process corresponding to coins of denominations 1. then simulates finding coins worth 5.01. N(T ) ∼ P (λT ) = P (120). However. Same as #3 Determine which of the following ranges contain the rate F/G. Which of the following statements is true? a) Neither is a valid method for simulating the process b) The first actuary’s method is valid. Thus G = 73 + 50 = 123. here F = 121 + 120 = 241. Thus we need 120 more random numbers. 121 random numbers.2 1. N5 (T ) ∼ P (0. 5. we have shown that both the procedures are valid but will not yield the same results.24.2 < F/G ≤ 1. Similarly for N5 (T ) and N10 (T ) we will require 2(24 + 1) = 50 random numbers.50. SI = 0. 0. the second actuary’s is not c) The second actuary’s method is valid.8 < F/G ≤ 1. we have STEP-1: t = 0. #37/November 2000 (Modified to conform to the current syllabus) Arrival times of a Poisson process are simulated from Exp(5).10. 0. 0.46 We go to step 2: U2 = 0. i.e.8 c) 0. The random numbers from U (0.96. in each case simulating until the 60 minutes are exceeded. Accordingly for N1 (T ) we will require 72 + 1 = 73 uniform random numbers. N10 (T ) ∼ P (0. a) 0 < F/G ≤ 0. First actuary is following Example 1. 5.6-7. Thus the correct choice is e.? a) 1 b) 2 c) 3 d) 4 e) 5 Solution Here T = 2..6 < F/G d) 1.4 < F/G ≤ 0. How many Poisson events happen in the interval (0.60λT ) = P (72)..05 II .(ii) The second actuary uses the same algorithm for simulating the times between events.10) = 0. 27 > 2 = T λ 5 So we stop.24) = 1.06 < 2 = T λ 5 STEP-4: I2 = 2.33 .01) = 2.60 = 1. U4 = 0.06 − ln(0. Thus 3 Poisson events happen during (0.e.1 1 STEP-3: t = t − lnU2 = 0.35 < 2 = T λ 5 STEP-4: I = 3.06 We go to step 2 1 1 STEP-3: t = t − lnU3 = 1.46 − ln(0.35 We go to step 2. i. S3 = 1.. The correct choice is c.46 + 0.01 1 1 STEP-3: t = t1 − lnU4 = 1. II . 2).35 − ln(0. S2 = 1.05) = 0.
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