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NAME: M339W/389W Financial Mathematics for Actuarial Applications University of Texas at Austin In-Term Exam II Instructor: Milica Čudina Notes: This is a closed book and closed notes exam. The maximum number of points on this exam is 100. Time: 50 minutes TRUE/FALSE 1 (2) TRUE FALSE 2 (2) TRUE FALSE 2.14 (5) a b c d e 3 (2) TRUE FALSE 2.15 (5) a b c d e 4 (2) TRUE FALSE 2.16 (5) a b c d e 5 (2) TRUE FALSE 2.17 (5) a b c d e 2.18 (5) a b c d e MULTIPLE CHOICE 6 (2) TRUE FALSE 7 (2) TRUE FALSE 2.19 (5) a b c d e 8 (2) TRUE FALSE 2.20 (5) a b c d e 9 (2) TRUE FALSE 2.21 (5) a b c d e 10 (2) TRUE FALSE 2.22 (5) a b c d e 11 (2) TRUE FALSE FOR GRADER’S USE ONLY: T/F 2.12 2.13 M.C. Σ 2 2.1. TRUE/FALSE QUESTIONS. Please note your answers on the front page. Problem 2.1. (2 pts) In the setting of the Black-Scholes stock-price model, let S = {S(t), t ≥ 0} denote the stock price whose drift is α and volatility is σ. Define the new stochastic process X(t) = ln(S(t)), for every t ≥ 0. Then, E[X(t + h) | X(t)] = X(t), for every t ≥ 0 and h > 0. Solution: FALSE Since {X(t)} is an (arithmetic) Brownian motion, its increment X(t + h) − X(t) is independent from X(t) and it is normally distributed with mean αh and variance σ 2 h. So, E[X(t + h) | X(t)] = E[X(t + h) − X(t) | X(t)] + X(t) = αh + X(t), for every t ≥ 0 and h > 0. Problem 2.2. (2 pts) Let the movement of a stock price follow a geometric Brownian motion. The, the realized (rate of) return on that stock follows the normal distribution. Solution: TRUE Problem 2.3. (2 pts) The total variation of a standard Brownian motion on the interval from 0 to T is infinite. Solution: TRUE Problem 2.4. (2 pts) In the setting of the Black-Scholes stock-price model, let {S(t), t ≥ 0} denote the price of a non-dividend-paying stock. Then, 1 S(t) = S(0) exp{(α − σ 2 )t + σZ(t)} 2 for every t ≥ 0 and where Z denotes the “driving” standard Brownian motion under the “true” probability measure P (consistent with the mean rate of return α). Solution: TRUE Problem 2.5. (2 points) Asset S pays no dividends and satisfies the following SDE dS(t) = S(t)(µS dt + σ dZ̃t ), S(0) = s0 where Z̃ is a standard Brownian motion under the risk-neutral probability measure P∗ and with µS , σ, s0 > 0 constant. Assume that the continuously compouned, risk-free interest rate equals r = 0.05. Then, µS < 0.05. True or false? Solution: FALSE In fact, µS = r. 3 Problem 2.6. (2 points) In the Black-Scholes stock-price model, the value function of the K−strike European put option must satisfy the following terminal condition: v(T, s) = max(s, K). Solution: FALSE It should be v(T, s) = max(0, K − s). Problem 2.7. (2 pts) The Black-Scholes option pricing formula can always be used for pricing both European and American type options. Solution: FALSE Problem 2.8. (2 pts) Let the exchange rate for GBP to USD be modeled by a geometric Brownian motion. Then, the exchange rate from the USD to GBP is also a geometric Brownian motion. Solution: TRUE Problem 2.9. (2 pts) An investor uses a proportional investment strategy to build his portfolio. More precisely, he always invests a fixed proportion of his wealth W in the risky asset and the remainder at the risk-free interest rate. Assume that the price of the risky asset is modeled by a geometric Brownian motion. Then, the investor’s wealth process is a geometric Brownian motion itself. Solution: TRUE Problem 2.10. (2 pts) Let Z = {Z(t), t ≥ 0} be a standard Brownian motion. Define the process Y as Z t 3 Y (t) = (Z(t)) − 3 Z(s) ds. 0 Then, the process Y has zero drift. Solution: TRUE dY (t) = 3(Z(t))2 dZ(t) + 3Z(t)dt − 3Z(t) dt = 3(Z(t))2 dZ(t). Problem 2.11. (2 pts) Let the process S = {S(t), t ≥ 0} satisfy the following SDE dS(t) = S(t)[µ dt + σ dZ(t)] where Z stands for a standard Brownian motion while µ and σ are constant. Then the process S(t)e−µt has zero drift. 4 Solution: TRUE d[S(t)e−µt ] = −µe−µt S(t) dt + e−µt dS(t) = −µe−µt S(t) dt + e−µt (S(t)[µ dt + σ dZ(t)]) = e−µt S(t)σ dZ(t). 5 2.2. FREE-RESPONSE PROBLEMS. Please, explain carefully all your statements and assumptions. Numerical results or single-word answers without an explanation (even if they’re correct) are worth 0 points. Problem 2.12. (15 points) Let S = {S(t), t ≥ 0} denote the stock-price process following a geometric Brownian motion. The mean stock price at time−1 equals 120 and the median stock price 115. What is the probability that the time−1 stock price exceeds 100? Solution: Since S = {S(t), t ≥ 0} is the stock-price process modeled by a geometric Brownian motion, the stock price at time−1 is lognormally distributed. In fact, using our usual parameters, we can rewrite it as 1 S(1) = S(0)e(α−δ− 2 σ (α−δ− 21 σ 2 ) Recall that the median of S(1) equals S(0)e as P[S(1) > 100] = P[115e σZ(1) 2 )+σZ(1) . . So, the required probability can be expressed 100 1 > 100] = P Z(1) > ln σ 115 1 115 115 1 = P Z(1) < ln ln =N . σ 100 σ 100 Since the mean of S(1) equals S(0)e(α−δ) , we have p 1 2 120 ⇒ σ = 2 ln(1.04348) = 0.2918. e2σ = 115 So, our final answer is P[S(1) > 100] = N (0.48) = 0.6844. The required probability is: P[S(1) > 100] = 6 Problem 2.13. (20 points) Let Z = {Z(t), t ≥ 0} be a standard Brownian motion. According to Itô’s lemma, it is possible to express the product Y (t) = et sin(Z(t)) as a sum of a stochastic and a classical integral. Please, provide this representation. More precisely, according to Itô’s lemma, we can write Z t Z t σ(s) dZ(s), µ(s) ds + Y (t) = Y (0) + 0 0 for every t ≥ 0, and for appropriate stochastic processes µ and σ. Please, determine µ and σ as functions of the time variables t and the standard Brownian motion Z. We get: µ(s) = and σ(s) = Solution: Set F (t, x) = et sin(x). Then Ft (t, x) = et sin(x), Fx (t, x) = et cos(x), and Fxx (t, x) = −et sin(x). So, by Itô’s lemma, Z t Z t 1 s s Y (t) = (e sin(Z(s)) − e sin(Z(s))) ds + es cos(Z(s)) dZ(s) 2 0 0 Z t Z t 1 = es sin(Z(s)) ds + es cos(Z(s)) dZ(s) . 2 0 0 7 2.3. MULTIPLE CHOICE QUESTIONS. Please note your asnwers on the front page. Problem 2.14. (5 points) Let {Z(t), t ≥ 0} be a standard Brownian motion. Define the stochastic process Y (t) = tZ(t). Then, (a) (b) (c) (d) (e) Rt Rt Y (t) = 0 Z(s) ds + 0 Z(s) dZ(s) Rt Rt Y (t) = 0 Z(s) ds + 0 s dZ(s) Rt Y (t) = 0 sZ(s) ds Rt Y (t) = 0 sZ(s) dZ(s) None of the above. Solution: (b) Problem 2.15. (5 pts) The current price of a share of stock is given to be S(0) = 45. The stock is assumed to follow a geometric Brownian motion. Let σ = 0.4. The stock is projected to pay a single $5.00 dividend in 1 month. Assume that r = 0.08. Consider an at-the-money call on that stock with expiry in 3 months. Find the price VC (0) of this call. (a) 1.233 (b) 1.255 (c) 1.697 (d) 3.013 (e) None of the above. Solution: (c) The prepaid forward price on this stock is P F0,T (S) = S(0) − De−rTD = 45 − 5e− 0.08 12 = 40.0332. ≈ P VC (0) = F0,T (S)N (d1 ) − Ke−rT N (d2 ) with 1 1 P d1 = √ [ln(F0,T (S)/K) + (r + σ 2 )T ] = · · · = −0.38, 2 σ T √ d2 = d1 − σ T = −0.58. We get that the price is VC (0) = 40.0332N (−0.38) − 45e−0.08/4 N (−0.58) = 40.0332(1 − 0.6480) − 45e−0.02 (1 − 0.719) = 1.697. 8 Problem 2.16. (5 points) The current price of a continuous-dividend-paying stock is given to be $92. The stock’s volatility is 0.35 and its dividend yield is 0.02. The continuously-compounded, risk-free interest rate is 0.05. Consider a $90-strike European call option on the above stock with exercise date in a quarteryear. What is the Black-Scholes price of this call option? (a) 5.05 (b) 7.66 (c) 7.71 (d) 7.89 (e) None of the above. Solution: (d) d1 = 0.26, d2 = 0.08. So, VC (0) = 92e−0.02/4 × 0.6026 − 90e−0.05/4 × 0.5319 ≈ 7.89. 9 Problem 2.17. (5 points) Consider a non-dividend-paying stock whose price S = {S(t), t ≥ 0} is modeled using a geometric Brownian motion. Suppose that the current stock price equals $100 and that its volatility is given to be 0.25. The continuously compounded, risk-free interest rate is assumed to equal 0.04. Consider a derivative security which entitles its owner to obtain a European call option on the above stock six months from today, i.e., at time t∗ = 1/2. The call option is to be 3−month to expiration at time of delivery and have the strike equal to 105% of the time−t∗ price of the underlying asset. This contract is called a forward start option. What is the price of the forward start option? (a) 3.13 (b) 8.65 (c) 10.51 (d) 13.55 (e) None of the above. Solution: (a) or (e) At time t∗, the required Black-Scholes price of the call option equals ∗ VC (t∗ ) = S(t∗ )N (d1 ) − 1.05S(t∗ )e−r(T −t ) N (d2 ) = S(t∗ )(N (d1 ) − 1.05e−0.01 N (d2 )) with 1 0.252 1 d1 = − ln(1.05) + 0.04 − × = −0.2478, 0.125 2 4 √ d2 = d1 − σ T − t∗ = −0.2478 − 0.125 = −0.3728. So, N (d1 ) = 1 − N (0.25) = 1 − 0.5987 = 0.4013 and N (d2 ) = 1 − N (0.37) = 1 − 0.6443 = 0.3557 and Hence, VC (t∗ ) = S(t∗ )(0.4013 − 1.05e−0.01 × 0.3557) = S(t∗ )0.31531. So, one would need to buy exactly 0.031531 shares of stock to be able to buy the call option in question at time−t∗ . This amount of shares costs $3.1531. Problem 2.18. (5 points) A pair of stock prices is modeled using the following system of SDE: dS(t) = c(0.01 dt + 0.1 dZ(t)) S(t) dQ(t) = 0.03 dt + 0.2 dZ(t) Q(t) with Z a standard Brownian motion. The asset S does not pay dividends, while the asset Q is assumed to pay dividends continuously with the dividend yield equal to 0.02. The continuously-compounded, risk-free interest rate equals 0.04. What is the value of the constant c? (a) 3 10 (b) (c) (d) (e) 8 10 13 None of the above. Solution: (b) The Sharpe ratios of the two assets needs to be the same since they have a common source of uncertainty Z. We have 0.01c − 0.04 0.05 − 0.04 c−4 5−4 1 = ⇒ = = ⇒ 2(c − 4) = c ⇒ c = 8. 0.1c 0.2 c 2 2 Problem 2.19. We use the Black-Scholes model for the evolution of the stock price S. The current stock price is S(0) = $60, and its volatility given to be σ = 0.20 The stock pays no dividends. The continuously-compounded, risk-free interest rate equals 0.06. What is the price of a two-month, $60-strike put option on the above stock? (a) 1.62 (b) 2.55 (c) 2.84 (d) 3.37 (e) None of the above. Solution: (a) In our usual notation, the price is VP (0) = Ke−r·T N (−d2 ) − S(0)N (−d1 ) with √ 60 (0.2)2 1 ln + 0.06 + = (0.3 + 0.1)/ 6 = 0.16, 60 2 6 √ d2 = d1 − 0.2/ 6 = 0.08. 1 p d1 = 0.2 1/6 The standard normal table gives us N (d1 ) = 0.5636 ⇒ N (−d1 ) = 0.4364, N (d2 ) = 0.5319 ⇒ N (−d2 ) = 0.4681. So, VP (0) = 60 e−0.01 × 0.4681 − 0.4364 = 1.6225. Problem 2.20. Alice wagers to deliver one share of stock S to Bob in one year if the stock price at that time exceeds $70.00. Assume that the current price of the said stock equals S(0) = 60.00, and that its volatility parameter is given to be 0.25. Moreover, the stock pays dividends continuously with the dividend yield of 0.01. The continuously-compounded, risk-free interest rate equals 0.04. What is the Black-Scholes price of Alice’s bet? (a) 16.72 (b) 21.13 11 (c) 38.27 (d) 39.51 (e) None of the above. Solution: (b) We recognize that Alice’s bet is actually a written asset call. So, its price is the Black-Scholes price of an asset call, i.e., in our usual notation VAC (0) = S(0)e−δT N (d1 ) with 1 d1 = 0.25 60 1 2 ln + 0.04 − 0.01 + (0.25) = −0.37. 70 2 So, our answer is VAC (0) = 60e−0.01 N (−0.37) = 60e−0.01 (1 − 0.6443) ≈ 21.13. Problem 2.21. The stochastic process S = {S(t), t ≥ 0} satisfies the stochastic differential equation dS(t) = S(t)(α dt + σ dZ(t)) with α and σ constant and Z = {Z(t), t ≥ 0} a standard Brownian motion. Define the stochastic process Y as p Y (t) = S(t) for every t ≥ 0. Then, V ar[ln(Y (t)] equals . . . 2 (a) σ4 t (b) 2σ 2 t 2 (c) σ √t (d) σt (e) None of the above. Solution: (a) Problem 2.22. Source: Sample MFE Problem #16. Assume that the Black-Scholes framework holds. Let S = {S(t), t ≥ 0} denote the price of a nondividend-paying stock. The stocks volatility is given to be 20%. The continuously compounded risk-free interest rate is 4%. You are interested in contingent claims with payoff being the stock price raised to some power. For 0 ≤ t ≤ T, consider the equation P F0,T [Sa ] = [S(t)]a where the left-hand side stands for the prepaid forward price at time t of a contingent claim that pays the amount (S(T ))a at time T . An obvious solution to the above equation is a = 1. It corresponds to the stock itself. The other solution to the above equation is . . . (a) −4 12 (b) −2 (c) 0 (d) 2 (e) None of the above. Hint: You might find it handy to know that P F0,T [Sa ] = [S(t)]a e(r(a−1)+a(a−1)σ Solution: (b) 2 /2)(T −t)