Financial Economics. Practice Exam #1. Exam MFE

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1 Financial Economics Practice Exam #1 Exam MFE These practice exams should be used during the last month prior to your exam. This exam has 30 questions, corresponding to an exam of about 3 hours. Each problem is similar to a problem in my study guide, sold separately. Solutions to problems are at the end of each practice exam. prepared by Howard C. Mahler, FCAS Copyright 2016 by Howard C. Mahler. Howard Mahler hmahler@mac.com

2 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page 1 MFE, Financial Economics Practice Exam #1 1. On January 1, 2012, a common stock is priced at $78. You are given the following: (i) Dividends of equal amounts will be paid on January 31, April 30, July 31, and October 31. (ii) A 9-month 70-strike European call option on this stock has a premium of $ (iii) A 9-month 70-strike European put option on this stock has a premium of $3.86. (iv) The continuously compounded risk-free interest rate is 4.8%. Calculate the amount of each dividend. (A) $0.57 (B) $0.58 (C) $0.59 (D) $0.60 (E) $ S(t) is the price of a stock at time t. The stock pays no dividends X(t) = S(t) 5 exp[ (b r + c σ 2 ) (T - t) ], for 0 < t < T, for constants b and c. For any values of r and σ, X satisfies the Black-Scholes partial differential equation. Determine b + c. A. 11 B. 12 C. 13 D. 14 E. 15

3 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page 2 3. Define: (i) W(t) = 0.02 t 2. Let w be the total variation of W over the interval 0 to 3. 0 for t < 0.7 1/ 4 for 0.7 t < 1.2 (ii) X(t) = 1/ 2 for 1.2 t < / 4 for 2.4 t < for 4.5 t Let x be the quadratic variation of X over the interval 0 to 3. (iii) Y(t) = 0.1(t ) Z(t), where {Z(t): t 0} is a standard Brownian motion. Let y be the quadratic variation of Y over the interval 0 to 3. Which of the following if true? (A) w < x < y (B) w < y < x (C) x < w < y (D) x < y < w (E) None of A, B, C, or D. 4. Currently one can buy 1 U.S. Dollar for 122 Yen. Dollar and Yen interest rates are 4.5% and 2.0%, respectively. The price of a 120 Yen strike 2-year put option to sell one U.S. dollar is 16 Yen. What is the price in Yen of a similar call? A. Less than 13 B. At least 13, but less than 14 C. At least 14, but less than 15 D. At least 15, but less than 16 E. At least 16

4 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page 3 5. S 0 = 200. α = 12% per year. δ = 2% per year. σ = 30% per year. Simulate the path of this stockʼs price over each of the next 3 months. Use the following three random numbers from (0, 1): , , Determine the average simulated stock price: (S 1/12 + S 2/12 + S 3/12 ) / 3. A. 200 B. 202 C. 204 D. 206 E Consider the following information about an at-the-money European put option on stock ABC: The current stock price is $100. The time to expiration is 2 years. The continuously compounded risk-free rate is 9% annually. The stock pays dividends at a continuous rate of 2%. The price is calculated using a 2-step Binomial Model where each step is one year in length. The stock price tree is shown below: Calculate the price of the put on stock ABC. A. Less than 0.30 B. At least 0.30, but less than 0.35 C. At least 0.35, but less than 0.40 D. At least 0.40, but less than 0.45 E. At least 0.45

5 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page 4 7. Assume the Black-Scholes framework. Suppose the spot exchange rate is $1.40 per Euro. Assume r = 0.05, r Euros = 0.035, volatility = What is the premium in dollars for an at the money put option that expires in 10 months? A. Less than 0.05 B. At least 0.05, but less than 0.06 C. At least 0.06, but less than 0.07 D. At least 0.07, but less than 0.08 E. At least You are given the following information on European Calls: S K σ r T δ Option Premium % 6% 3 years 2% % 6% 3 years 2% % 6% 3 years 2% % 6.5% 3 years 2% % 6% 3.5 years 2% % 6% 3 years 1% Estimate rho (per percent), the option Greek, for a 120 strike 3 year call with S 0 = 100, σ = 20%, δ = 2%, and r = 6%. A. 0.3 B. 0.5 C. 0.7 D. 0.9 E The future price of a stock follows a LogNormal Distribution with parameters m = 5 and v = 0.8. Which of the following is a 95% confidence interval for the price of this stock? A. [40, 553] B. [31, 712] C. [23, 954] D. [19, 1165] E. None of the above

6 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page The short-rate process {r(t)} in a Cox-Ingersoll-Ross model follows: dr(t) = { r(t)} dt r(t) dz(t), where {Z(t)} is a standard Brownian motion under the true probability measure. For t T, let P(r,t,T) denote the price at time t of a zero-coupon bond that pays 1 at time T, if the short-rate at time t is r. lim ln[ P(r,t,T) ] / T = for each r > 0. T Determine the Sharpe ratio when the short term interest rate is 9%. (A) 0.04 (B) 0.07 (C) 0.10 (D) 0.13 (E) Consider the stochastic differential equation dx(t) = 8 dt - 4 X(t) dt + 3 dz(t), where {Z(t)} is a standard Brownian motion. You are given that a solution is X(t) = A + B e -Ct + D e -Ct t e Ls dz(s), 0 where A, B, C, D, and L are constants. Calculate the product: A C D L. (A) 48 (B) 96 (C) 120 (D) 144 (E) Barrier call option prices are shown in the table below. Each option has the same underlying asset. Option Strike Barrier Price down-and-out Call up-and-out Call down-and-in Call X up-and-in Call down-and-out Put up-and-out Put Y down-and-in Put up-and-in Put Determine X - Y. A. Less than 9 B. At least 9, but less than 11 C. At least 11, but less than 13 D. At least 13, but less than 15 E. At least 15

7 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page Let S(t) be the price of a stock at time t. S(0) = 90. σ = 35%. r = 5%. δ = 1%. At time 1, a contract pays Max[Min[S(1), 110], 85]. Determine the time-zero premium of this contract. (A) 90 (B) 91 (C) 92 (D) 93 (E) The underlying stock for a European exchange option has S = $110, continuous rate of dividends = 1.2%, volatility = The strike stock has S = $140, continuous rate of dividends = 1.8%, volatility = The two stocks have a correlation coefficient of If the exchange option expires in 2 years, what is the price of the call using the Black-Scholes approach? A. Less than 4 B. At least 4, but less than 5 C. At least 5, but less than 6 D. At least 6, but less than 7 E. At least ds(t) / S(t) = 0.1dt + 0.3dZ(t). G(t) = S(t) t. Based on Itoʼs Lemma, which of the following stochastic differential equations is satisfied by G(t)? (A) dg(t)/g(t) = {0.1t + ln[g(t)]/t} dt dz (B) dg(t)/g(t) = {0.055t t 2 + ln[g(t)]/t} dt dz (C) dg(t)/g(t) = {0.1t + ln[g(t)]/t} dt + 0.3t dz (D) dg(t)/g(t) = {0.055t t 2 + ln[g(t)]/t} dt + 0.3t dz (E) None of A, B, C, or D

8 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page For a Vasicek model, you are given the following information: a = 0.12 b = 0.06 r = 0.08 σ = 0.04 Calculate the expected change in the interest rate over the next 6 months. A. Less than B. At least but less than C. At least but less than D. At least but less than E. At least Assume the Black-Scholes framework. For a stock that pays dividends continuously at a rate proportional to its price, you are given: (i) The current stock price is 58. (ii) The stockʼs volatility is (iii) The continuously compounded expected rate of stock-price appreciation is 15%. Let X be the arithmetic average of stock prices in six months and twelve months. Calculate the variance of X. (A) 210 (B) 230 (C) 250 (D) 270 (E) 290

9 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page Which of the following statements is false? A. The control variate method uses the insight that for every simulated realization, there is an opposite and equally likely realization. B. The continuation value is the value of leaving an option unexercised. C. Latin hypercube sampling is a generalization of stratified sampling when the payoff depends on more than one random variable. D. When valuing options using true probabilities, the discount rates are not the same along the different paths of future stock prices. E. None of A. B. C. or D is false. 19. The following tree was produced by the Black-Derman-Toy Model: Year 0 Year 1 Year % 12.6% 9.0% 9.3% 10.6% Determine the yield to maturity of a 3 year zero-coupon bond. A. 11.0% B. 11.2% C. 11.4% D. 11.6% E. 11.8% 20. For a lognormally distributed stock, S 0 = 150, α = 10%, δ = 3%, and σ = 30%. What is the probability that the stock price at time 5 is between 200 and 250? (A) 12% (B) 14% (C) 16% (D) 18% (E) 20%

10 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page The following five charts are profit diagrams for option strategies. Each strategy is constructed with the purchase or sale of 1-year European options. In each chart, the profit is shown with one month to expiration. Profit 10 Chart1 Profit 15 Chart S S Profit Chart3 Profit Chart S S Profit 10 Chart S Which of the following correctly matches charts with option strategies? A. Chart 1 = Bull Spread, Chart 3 = Written Straddle. B. Chart 2 = Straddle, Chart 3 = Butterfly Spread. C. Chart 4 = Butterfly Spread, Chart 5 = Bull Spread. D. Chart 1 = Straddle, Chart 4 = Bear Spread. E. Chart 2 = Written Straddle, Chart 5 = Bear Spread.

11 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page You own a one-year 80 strike American call option on a dividend paying stock. The stock currently sells at 83, and is expected to sell at either 81 or 90 one year from now. The stock pays dividends at a continuously compounded rate of 2.5% per year. What annual continuously compounded risk-free interest rate would equate the value of exercising the call immediately with the present value of waiting until the expiration to exercise? A. Less than 3.0% B. At least 3.0%, but less than 3.5% C. At least 3.5%, but less than 4.0% D. At least 4.0%, but less than 4.5% E. At least 4.5% 23. Warren owns one thousand shares of a stock. Because he worries that its price may drop over the next three years, he decides to employ a rolling insurance strategy, which entails obtaining one year European put options on the stock every year, with the first one being bought immediately. Use the following information: (i) The continuously compounded risk-free interest rate is 10%. (i) The continuously compounded dividend rate is 3%. (iii) The stockʼs volatility is 20%. (iv) The current stock price is 80. (v) The strike price for each option is equal to the then-current stock price. His broker will sell him the options but will charge him for their total cost now. Under the Black-Scholes framework, how much does Warren now pay his broker? (A) 10,700 (B) 10,800 (C) 10,900 (D) 11,000 (E) 11, Use the following information: In one year the economy will be in one of two states, high or low. The value of Oscorp stock one year from now in the high state is 110. The value of Oscorp stock one year from now in the low state is 70. The current price of Oscorp stock is 85. Oscorp stock pays no dividends. The premium for a 120-strike one-year put on Oscorp stock is 24. Determine the risk neutral probability of the low state. A. 33% B. 35% C. 37% D. 39% E. 41%

12 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page Assume the Black-Scholes framework. A stock has S 0 = 81, α = 12%, σ = 33%, δ = 2.5%. r = 5.7%. A market maker, sells ten thousand 2-year 70-strike European puts on this stock. The market maker will delta hedge his position by using 2-year 70-strike European calls on this stock. How many calls should he sell? A. less than 4000 B. at least 4000 but less than 5000 C. at least 5000 but less than 6000 D. at least 6000 but less than 7000 E. at least S(t) = exp[ t Z(t)], t 0, where {Z(t)} is a standard Brownian motion. Which of the following is true? A. ds(t) = S(t) dt S(t) dz(t) B. ds(t) = S(t) dt S(t) dz(t) C. ds(t) = S(t) dt S(t) dz(t) D. ds(t) = S(t) dt S(t) dz(t) E. None of A, B, C, or D 27. You are given the following regarding the foreign currency Bongo Bucks: (i) One Bongo Buck currently sells for $0.06. (ii) Six months from now a Bongo Buck will sell for either $0.05 or $0.07. (iii) The continuously compounded risk-free interest rate on Bongo Bucks is 7%. (iv) The continuously compounded risk-free interest rate on U.S. dollars is 4%. While reading the Bongo Congo Times, Leonardo notices that a six-month $0.065 strike European call written on Bongo Bucks is selling for $ Use the binomial option pricing model to determine if an arbitrage opportunity exists. What transactions should Leonardo enter into to exploit the arbitrage opportunity (if one exists)? (A) No arbitrage opportunity exists. (B) Short Bongo Bucks, lend at the $ risk-free rate, and buy the call. (C) Buy Bongo Bucks, borrow at the $ risk-free rate, and buy the call. (D) Buy Bongo Bucks, borrow at the $ risk-free rate, and short the call. (E) Short Bongo Bucks, lend at the $ risk-free rate, and short the call.

13 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page The price of a stock is $120. The stock pays dividends at the continuous rate of 1.5%. The price of a European call with a strike price of $125 is $9.72 and the price of a European put with a strike price of $125 is $ Both options expire in six months. Calculate the annual continuously compounded risk-free rate on a synthetic T-Bill created using these options. A. Less than 1% B. At least 1%, but less than 2% C. At least 2%, but less than 3% D. At least 3%, but less than 4% E. At least 4% 29. For a non-dividend stock you are given the following information: The current stock price is 70. The volatility of the return on the stock is 30%. The continuously compounded risk-free interest rate is 5%. Using the Black-Scholes formula, calculate the cost of an at-the-money one year straddle. A. Less than 16 B. At least 16, but less than 17 C. At least 17, but less than 18 D. At least 18, but less than 19 E. At least The current price of a stock is 90. r = 5%. continuous rate of dividends = 1.5%. volatility = 0.3. Use a Binomial Tree with 4 month time periods, in order to determine the premium of a 100-strike one-year European call option. A. Less than 8 B. At least 8, but less than 9 C. At least 9, but less than 10 D. At least 10, but less than 1 E. At least 11 END OF PRACTICE EXAM

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15 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page 14 Solutions: 1. D. By put-call parity, C = P + S - PV[Div] - K e -rt. Let x be the size of each dividend = x{e -(0.048)(1/12) + e -(0.048)(4/12) + e -(0.048)(7/12) } - 70 e -(0.048)(9/12). x = / = Comment: Similar to Q.4.46 (MFE, 5/07, Q.1). In put-call parity we only include those dividends paid from the time when we buy the option to when it expires; in this case we only include the dividends paid in January, April, and July. 2. D. Treating S as the variable and t as a constant, X = S 5 times constants: X S = 5 S(t)4 exp[(b r + c σ 2 )(T - t)] = 5X/S. 2 X S 2 = 20 S(t)3 exp[(b r + c σ 2 )(T - t)] = 20X/S 2. Treating t as the variable and S as a constant, X = exp[-(b r + c σ 2 ) t] times constants: X t = -(b r + c σ 2 ) S(t) 5 exp[(b r + c σ 2 )(T - t)] = -(b r + c σ 2 ) X. The Black-Scholes partial differential equation, when there are no dividends: σ 2 S 2 2 X 1 S r S X S + X t = r X. σ 2 20X/2 + r 5X - (b r + c σ 2 )X = rx. (10 - c)σ 2 + (4 - b)r = 0. This equation has to be satisfied for all values of r and σ. The only way this can be true is if the coefficients multiplying σ 2 and r are each zero. Therefore, b = 4 and c = 10. b + c = 14. Alternately, F P t,t [S T a ] = e -r(t-t) S t a exp[{a(r - δ) + a(a-1)σ 2 /2}(T - t)]. For a = 5 and δ = 0, the prepaid forward price is: e -r(t-t) S t 5 exp[{5r + 20σ 2 /2}(T - t)] = S t 5 exp[{4r + 10σ 2 }(T - t)]. This prepaid forward price must satisfy the Black-Scholes partial differential equation. Matching the form of this prepaid forward price to X(t), b = 4 and c = 10. b + c = 14. Comment: Similar to Q The Black-Scholes equation holds for the prepaid forward prices of options, in other words option premiums, rather the forward prices (cash on delivery.).

16 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page E. (i) W(t) = 0.02 t 2. dw = 0.04 t dt. Total variation over the time interval 0 to 3 is: 3 3 dw(t) = 0.04 t dt = (0.02) (3 2 ) = for t < 0.7 1/ 4 for 0.7 t < 1.2 (ii) X(t) = 1/ 2 for 1.2 t < / 4 for 2.4 t < for 4.5 t X(t) jumps from 0 to 1/4 at t = 0.7, from 1/4 to 1/2 at t = 1.2, and from 1/2 to 3/4 at t = 2.4. (The jump at t = 4.5 is outside the interval from 0 to 3.) Thus for small time intervals, all of the increments are zero, except for those time intervals that include t = 0.7, t = 1.2, or t = 2.4, in which case the increment is 1/4. n Therefore, {X(j3 / n) - X((j 1)3 / n)} 2 = 1/ / /4 2 = 3/16 = j=1 Taking the limit as n approaches infinity, the quadratic variation of X is iii) Y(t) = 0.1(t ) Z(t). dy = 0.1dt dZ(t). dy 2 = 0.01dt dt dz(t) dz(t) 2 = dt. Quadratic variation over the time interval 0 to 3 is: 3 3 {dy(t)} 2 = dt = (3)(0.0576) = = y < 0.18 = w < = x. Comment: Similar to Q (MFE Sample Exam Q. 63). 4. A. C P (x 0, K, T) = P P (x 0, K, T) + x 0 exp[-tr $ ] - K exp[-tr Yen ]. = 16 + (122)e -(2)(0.045) - (120)e -(2)(0.02) = 12.2 Yen. Comment: Similar to Q. 6.8.

17 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page D. Φ 1 [0.7324] = , Φ 1 [0.4681] = , Φ 1 [0.1515] = The return over one month is Normal with parameters: ( /2)/12 = , and 0.3 1/ 12 = S 1/12 = (200) exp[ ( )(0.0866)] = S 2/12 = (212.00) exp[ ( )(0.0866)] = S 3/12 = (211.50) exp[ ( )(0.0866)] = ( )/3 = Comment: Similar to Q The stock price in month two is dependent on the stock price in month one. If R 1 is the simulated return in month 1, and R 2 is the simulated return in month 2, then S 1/12 = S 0 exp[r 1 ], and S 2/12 = S 0 exp[r 1 + R 2 ] = S 1/12 exp[r 2 ]. 6. E. p* = exp[h(r - δ)] - d u - d The premium for the put is: = exp[(1)( )] = 86.25%. (0.8625) 2 (0) + (2)(0.8625)( )(100-99) + ( ) 2 (100-81) exp[(2)(0.09)] = $ Comment: Similar to Q (CAS3, 5/07, Q.14). 7. C. At the money means that K = S = d 1 = ln(1.4 / 1.4) + ( / 2)(10 / 12) (0.15) 10 / 12 = d 2 = (0.15) 10 / 12 = N[ ] = N[ ] = Value of this put option is: = (1.4) exp[-(0.05)(10/12)]( ) - exp[-(0.035)(10/12)] (1.4)( ) = $ Comment: Similar to Q Euros act as the asset; replace δ everywhere by r Euros. 8. E. rho = change in option premium change in r = = Comment: Similar to Q

18 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page B. A 95% confidence interval for lny is: [5 - (1.960)(0.8), 5 + (1.960)(0.8)] = [3.432, 6.568]. A 95% confidence interval for Y is: [e 3.432, e ] = [31, 712]. Comment: Similar to Q D. For the CIR Model, as T, the yield on the bond approaches: r = 2 a b a - φ + γ. γ = (a φ ) 2 + 2σ 2. dr = a(b- r)dt + σ r dz. Thus, from the given form of dr: ab = 0.014, a = 0.2, and σ = b = 7%. We are given that for very large T, ln[p(t)]/t = P(T) = e T. The (continuously compounded) yield on the long-lived bond is = r = 2 a b a - φ + γ. (0.076) (0.2 - φ + γ) = (2)(0.014). γ = φ. (a - φ ) 2 + 2σ 2 = ( φ ) 2. (0.2 - φ ) 2 + 2(0.1 2 ) = ( φ ) φ + φ = φ + φ 2. φ = / = When the short term interest rate is 9%, the Sharpe ratio is: φ = φ r / σ = (0.0429) 0.09 / 0.10 = Comment: Similar to Q (MFE Sample Exam, Q.60).

19 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page B. X(t) = A + Be -Ct + D e -Ct e Ls dz(s). Therefore, t 0 dx(t) = -CBe -Ct dt - Cdt D e -Ct t e Ls dz(s) + D e -Ct e Lt dz(t). 0 = -C(X - A) dt + D e (L-C)t dz(t) = CA dt - CX dt + D e (L-C)t dz(t) But we are given that dx(t) = 8dt - 4X(t)dt + 3dZ(t). Matching terms in the three equations: CA = 8, C = 4, and 3 = D e (L-C)t. Thus L = C = 4, A = 8/4 = 2, and D = 3. A C D L = (2)(4)(3)(4) = 96. Alternately, for the Ornstein-Uhlenbeck Process: dx(t) = λ {α - X(t)} dt + σ dz(t), for λ > 0. Since we are given that dx(t) = 8dt - 4X(t)dt + 3dZ(t), matching terms we have in this case: σ = 3, λα = 8, and λ = 4. α = 2. The Ornstein-Uhlenbeck Process X(t) satisfies the following stochastic integral equation: X(t) = X(0) e -λt + α(1 - e -λt ) + σ e -λt t e λs dz(s). 0 We are given that in this case the solution is: X(t) = A + Be -Ct + D e -Ct t e Ls dz(s). 0 Therefore, matching terms: A = α = 2, C = λ = 4, D = σ = 3, L = λ = 4. A C D L = (2)(4)(3)(4) = 96. Comment: Similar to Q (MFE Sample Exam, Q.68). B = X(0)e -λt - α.

20 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page E. Working with calls with a barrier of 200, the value of the regular Call with K = 80 is: up-and-out + up-and-in = = Working with calls with a barrier of 85: = down-and-out 80 strike Call + down-and-in 80 strike Call = X. X = Working with puts with a barrier of 35, the value of the regular Put with K = 70 is: down-and-out + down-and-in = = Working with puts with a barrier of 110: 7.14 = up-and-out 70 strike Put + up-and-in 70 strike Put = Y Y = X - Y = = Comment: Similar to Q (CAS3, 5/07, Q.34). Based on the Black-Scholes framework, with S 0 = 90, σ = 0.4, r = 5%, T = 2, δ = 1%.

21 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page if S(1) > A. Max[Min[S(1), 110], 85] = S(1) if 110 S(1) if S(1) < 85 By checking all three cases, one can verify that, Max[Min[S(1), 110], 85] = 85 + (S(1) - 85) + - (S(1) - 110) +. Thus, taking the actuarial present value in the risk-neutral environment, the premium is: 85e (premium of a 1-year 85-strike call) - (premium of a 1-year 110-strike call). Working at 85: d 1 = ln(90/ 85) + (1)( / 2) = d 2 = d = Premium of a 1-year 85-strike call is: 90e Φ[ ] - 85e Φ[ ] = (89.10)( ) - (80.85)( ) = Working at 110: d 1 = ln(90/ 110) + (1)( / 2) = d 2 = d = Premium of a 1-year 110-strike call is: 90e Φ[ ] - 110e Φ[ ] = (89.10)( ) - (104.64)( ) = Thus the premium of this contract is: 85e = Alternately, Max[Min[S(1), 110], 85] = (85 - S(1)) + - (110 - S(1)) +. Thus, taking the actuarial present value in the risk-neutral environment, the premium is: 110e (premium of a 1-year 85-strike put) - (premium of a 1-year 110-strike put). Premium of a 1-year 85-strike put = 85e Φ[ ] - 90e Φ[ ] = (80.85)( ) - (89.10)( ) = Premium of a 1-year 110-strike put = 110e Φ[ ] - 90e Φ[ ] = (104.64)( ) - (89.10)( ) = Thus the premium of this contract is: 110e = Comment: Similar to Q If S(1) > 110, then max [min [S(1),110], 85] = (S(1) - 85) + - (S(1) - 110) + = 85 + {S(1) - 85} - {S(1) - 110} = 110. Try S(1) = 115. If 110 > S(1) > 85, then max [min [S(1),110], 85] = S(1) (S(1) - 85) + -(S(1) -110) + = 85 + {S(1) - 85} + 0 = S(1). Try S(1) = 100. If 85 > S(1), then max [min [S(1),110], 85] = (S(1)-85) + - (S(1)-110) + = = 85. Try S(1) = 80. Since the two expressions are equal for all three cases, they are indeed equal.

22 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page E. volatility = (2)(0.6)(0.24)(0.32) = d 1 = ln(110/ 140) + ( / 2)(2) = d 2 = ( ) 2 = N[ ] = N[ ] = Premium for this call option is: exp[-(0.012)(2)] (110) ( ) - (140) exp[-(0.018)(2)] ( ) = $7.15. Comment: Similar to Q D. G S = G S = t St-1. G SS = 2 G S 2 = t(t-1)st-2. G t = G t = ln(s) S t. ln(g) = t ln(s). Therefore, G t = S t ln(g)/t. By Itoʼs Lemma, dg = G S ds + G SS ds 2 /2 + G t dt = t S t-1 S(0.1dt + 0.3dZ) + t(t-1)s t-2 S 2 (0.1dt + 0.3dZ) 2 /2 + S t ln(g)/t dt = 0.1t G dt + 0.3t G dz + t(t-1)g 0.045dt + (ln(g)/t) G dt. dg/g = {0.1t t(t-1) + ln(g)/t}dt + 0.3t dz = {0.055t t 2 + ln(g)/t} dt + 0.3t dz. Comment: Similar to Q C. For the Vasicek model, dr = a (b - r) dt + σ dz. A (Standard) Brownian Motion has an expected value of 0; E[dZ] = 0. E[dr] = a (b - r) dt. E[dr] is the expected change in the interest rate, over a very short period of time dt. If r = 0.04, then E[dr] = a (b - r) dt = (0.12)( ) dt = dt. For Δt = 1/2, then the expected change in interest rate would be: (1/2)( ) = Comment: Similar to Q (CAS3, 5/07, Q.36).

23 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page C. The continuously compounded expected rate of stock-price appreciation α - δ = 15%. X = S(1/ 2) + S(1) 2. Var[X] = Var[S(1/2) + S(1)]/2 2 = Var[S(1/2)]/4 + Var[S(1)]/4 + Cov[S(1/2), S(1)]/2. S(1/2) is LogNormal with parameters: m = ln[58] + ( /2)(1/2) = , and v = 0.3 1/ 2 = E[S(1/2)] = exp[ /2] = E[S(1/2) 2 ] = exp[(2)(4.1129) + (2)( )] = Var[S(1/2)] = = S(1) is LogNormal with parameters: m = ln[58] + ( /2)(1) = , and v = = 0.3. E[S(1)] = exp[ /2] = E[S(1) 2 ] = exp[(2)(4.1654) + (2)(0.3 2 )] = Var[S(1)] = = For s < t, Cov[S s, S t ] = E[S s ] E[S t ] {exp[sσ 2 ] - 1}. Cov[S(1/2), S(1)] = E[S 1/2 ] E[S 1 ] {exp[σ 2 /2] - 1} = (62.51)(67.38){exp[0.3 2 /2] - 1} = Var[X] = Var[S(1/2)]/4 + Var[S(1)]/4 + Cov[S(1/2), S(1)]/2 = 180.5/ / /2 = 249. Alternately, S(1)/S(1/2) is LogNormal with parameters: ( /2)(1/2) = , and 0.3 1/ 2 = E[S(1)/S(1/2)] = exp[ /2] = Since S(1/2) and S(1)/S(1/2) are independent, E[S(1/2) S(1)] = E[S(1/2) 2 S(1)/S(1/2)] = E[S(1/2) 2 ] E[S(1)/S(1/2)] = (4088.0)( ) = Cov[S(1/2), S(1)] = E[S(1/2) S(1)] - E[S(1/2)] E[S(1)] = (62.51)(67.38) = Proceed as before. Alternately, Cov[X, Y] = Cov[X, E[Y X] ]. E[S(1) S(1/2)] = S(1/2) exp[(1-1/2) (α - δ)] = S(1) e Thus, Cov[S(1/2), S(1)] = Cov[S(1/2), E[S(1) S(1/2)] ] = Cov[S(1/2), S(1/2) e ] = e Cov[S(1/2), S(1/2)] = e Var[S(1/2)] = e (180.5) = Proceed as before. Comment: Similar to Q (MFE Sample Exam Q.56). E[S t ] = S 0 exp[t(α - δ)]. Var[S t ] = E[S t ] 2 {exp[σ 2 t] - 1}.

24 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page A. A is false, see page 632 of McDonald. A would be true if it said antithetic variate method. B is true. C is true, see page 633 of McDonald. D is true. Comment: See Sections 10, 17, and 67 of Mahlerʼs Guide to Financial Economics. 19. B. Using the geometric progression of the values in a column of the BDT tree, the missing value is: (0.172)(0.106) = Year 0 Year 1 Year % 12.6% 9.0% 13.5% 9.3% 10.6% Assume for simplicity that the bond pays $1 upon maturity. In order to get the price of the 3 year bond, we discount along of each of four possible paths: 1 { (1.09)(1.126)(1.172) + 1 (1.09)(1.126)(1.135) + 1 (1.09)(1.093)(1.135) + 1 (1.09)(1.093)(1.106) } / 4 = Yield to maturity is: (1/0.7279) 1/3-1 = 11.17%. Comment: Similar to Q

25 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page A. S 5 is a LogNormal with parameters: ln150 + ( /2)(5) = , and = F(250) - F(200) = Φ[ ln(250) ln(200) ] - Φ[ ] = Φ[ ] - Φ[ ] = = %. Comment: Similar to Q

26 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page E. The profit at one month until expiration should approach the profit at expiration, which is the payoff at expiration minus the original cost to set up the position accumulated for interest. Thus, the profit at expiration should look like the payoff graph, translated either up or down. Bull Spread: The purchase of an option together with the sale of an otherwise identical option with a higher strike price. If we use calls with strikes of 90 and 110, then the payoff is: (S - 90) + - (S - 110) +. This is: 0 if S < 90, S - 90 if 90 S 110, and 20 if S > 110. Chart 1 appears to be approaching, other than a constant, the payoff for a Bull Spread. Written Straddle: Sell a put and a call, with the same strike and time until expiration. If the strike is 100, then the payoff is: -(100 - S) + - (S - 100) + = - S Chart 2 appears to be approaching, other than a constant, the payoff for a Written Straddle.

27 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page 26 Butterfly Spread: Buying a K strike option, selling two K + ΔK strike options, and buying a K + 2ΔK strike option. If we use calls with strikes of 90, 100, and 110, then the payoff is: (S - 90) + + (S - 110) (S - 100) +. This is: 0 if S < 90, S - 90 if 90 S 100, S if 90 S 110, and 0 if S > 110. Chart 3 appears to be approaching, other than a constant, the payoff for a Butterfly Spread. Straddle: Buy a put and a call, with the same strike and time until expiration. If the strike is 100, then the payoff is: (100 - S) + + (S - 100) + = S Chart 4 appears to be approaching, other than a constant, the payoff for a Straddle.

28 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page 27 Bear Spread: The purchase of an option together with the sale of an otherwise identical option with a lower strike price. If we use calls with strikes of 90 and 110, then the payoff is: (S - 110) + - (S - 90) +. This is: 0 if S < 90, 90 - S if 90 S 110, and -20 if S > 110. Chart 5 appears to be approaching, other than a constant, the payoff for a Bear Spread. Comment: Similar to Q (MFE Sample Exam Q.40). If we buy a Bull Spread, we are hoping that the stock price goes up, which is chart 1. If we buy a Bear Spread, we are hoping that the stock price goes down, which is chart 5. Graphs were constructed for S 0 = 100, σ = 30%, r = 6%, and δ = 1%. 22. A. If we exercise right away, we get = 3. If the stock price goes up the payoff is 10, while if it goes down the payoff is 1. The continuation value is: {10p* + 1(1 - p*)} / e r = (9p* + 1) / e r. p* = 83 exp[r ] = e r We want: 3 = (9p* + 1) / e r. 3e r = 9p* + 1 = 83e r e r = 80 / (83e ) = r = 2.6%. Comment: Similar to Q (5B, 5/99, Q.34). p* = 83 exp[ ] = 23.14%. The continuation value is: {(10)(0.2314) + (1)( )} / e = 3.00.

29 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page B. For a forward start put, with forward start date t 1, expiration time T, with strike price equal to the current stock price, the premium is: S 0 exp[-δ t 1 ] {exp[-r(t- t 1 )] Φ[-d 2 ] - exp[-δ(t - t 1 )] Φ[-d 1 ]}, where d 1 = -ln[1] + (r - δ + σ2 / 2)(T - t 1 ) σ T - t 1, and d 2 = d 1 - σ T - t 1. d 1 = -ln[1] + (10% - 3% / 2)(1) = d 2 = = 0.25 Φ[0.45] = Φ[0.25] = Premium for the first forward start put is: 80 {exp[-10%]( ) - exp[-3%]( )} = For the second forward start put, T- t 1 = 2-1 = 1. Thus d 1 and d 2 are the same as before. The premium for the second option is: S 0 exp[-δ] {exp[-r] Φ[-d 2 ] - exp[-δ] Φ[-d 1 ]}. This is the premium for the first put times exp[-δ]: e = Similarly, the third put has premium: S 0 exp[-2δ] {exp[-r] Φ[-d 2 ] - exp[-δ] Φ[-d 1 ]}. This is the premium for the first put times exp[-2δ]: e = Therefore, the premium for the rolling insurance strategy is: (1,000)( ) = 10,807. Comment: Similar to Q (MFE Sample Exam, Q.33). The first forward start put is just an ordinary put. 24. E. Let x = pu H, and y = (1-p)U L. Then, stock price = 85 = 110x + 70y. put premium = 24 = 10x + 50y. x = and y = p* = y/(x+y) = /( ) = Comment: Similar to Q

30 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page A. d 1 = ln(s / K) + (r - δ + σ2 / 2)T σ T = ln[81/ 70] + ( / 2)(2) = Φ[ ] = Δ call = e -δt Φ[d 1 ] = exp[-(0.025)(2)] ( ) = Δ put = -e -δt Φ[-d 1 ] = -exp[-(0.025)(2)] ( ) = Having sold 10,000 puts, his delta is: (-10,000)( ) = In order to get his delta to be zero, he needs to sell calls: / = Comment: Similar to Q C. S(t) = exp[ t Z(t)]. ln[s(t)] = t Z(t). ln[s(t)] is an Arithmetic Brownian Motion with µ = α - δ - σ 2 /2 = , and σ = α - δ = /2 = ds(t) = S(t) dt S(t) dz(t). Comment: Similar to Q In order to go from the exponential form to the stochastic differential equation we need to add σ 2 /2. S(0) = e = 87.

31 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page D. This is a $ denominated call, and therefore dollars act as money and Bongo Bucks act as the asset. r = 4%. δ = 7%. Δ = e -δh C u - C d S 0 (u - d) = e-0.07/ = B = e -rh u C d - d C u u - d = e-0.04/2 (7 / 6)(0) - (5 / 6)(0.005) 7 / 6-5 / 6 = -$ Therefore, the call premium should be: ($0.06)(0.241) - $ = $ Therefore, the call is overpriced at $ We can create a synthetic call by replicating its cashflows; buy Δ Bongo Bucks and sell bonds. We want to buy a synthetic call and sell one of the real overpriced call. Therefore, Leonardo can exploit the arbitrage opportunity by selling (shorting) a call, buying Bongo Bucks, and selling risk-free bonds (borrowing money.) exp[( )(1/ 2)] - 5 / 6 Alternately, p* = = / 6-5 / 6 Therefore, the call premium should be: ($0.005) e -0.04/2 (0.4553) = $ Therefore, the call is overpriced at $ Proceed as before. Comment: Similar to Q (MFE/3F, 5/09, Q.3). If u is too large or d is too small, then there is arbitrage inherent in the binomial tree itself. However, here d < exp[(r-δ)h] < u. Even if the values of u and d are fine, as they are here, one can not pick the price of an option out of thin air as was done in this question. We were given enough information to price the option. If the option is selling for other than this fair market price, then there is an opportunity for arbitrage. (We are implicitly assuming that the model given in the question is correct.) 28. E. Using put call parity, C = P - Ke -rt + Se -δt = e -r/ e /2. e -r/2 = /125. r = 4.5%. Comment: Similar to Q. 9.7 (CAS3, 5/07, Q.13).

32 2016-MFE-Ex1 Financial Economics Practice Exam #1 HCM 11/29/15, Page B. Straddle: Purchase a call and the otherwise identical put. In this case both have K = S 0 = 70. d 1 = ln(s / K) + (r - δ + σ2 / 2)T σ T = ln(1) + ( / 2)(1) = d 2 = d 1 - σ T = (0.3) 1 = Φ[ ] = Φ[ ] = Call premium is: e -δt S Φ[d 1 ] - K e -rt Φ[d 2 ] = exp[-(0)(1)] (70) ( ) - (70) exp[-(0.05)(1)] ( ) = Put premium is: K e -rt Φ[-d 2 ] - e -δt S Φ[-d 1 ] = = (70) exp[-(0.05)(1)] ( ) - exp[-(0)(1)] (70) ( ) = Straddle Premium = = Comment: Similar to Q B. u = exp[(5% - 1.5%)/ / 3 ] = d = exp[(5% - 1.5%)/ / 3 ] = p* = exp[(5% - 1.5%)/ 3] = S uuu = (90)( ) = S duu = (90)(0.8508)( ) = S ddu = (90)( )(1.2031) = S ddd = (90)( ) = The Binomial probabilities are: = , (3)( )( ) = , (3)(0.4568)( ) 2 = , ( ) 3 = Discounting back in the risk-neutral environment, the premium of the call is: ( )(0.0953) + ( )(0.3400) = 8.6. exp[0.05] Comment: Similar to Q Send any corrections or comments to: Howard Mahler, hmahler@mac.com