B.4 Solutions to Exam MFE/3F, Spring 2009

SOLUTIONS TO EXAM MFE/3F, SPRING 29, QUESTIONS 1 3 775 B.4 Solutions to Exam MFE/3F, Spring 29 The questions for this exam may be downloaded from http://www.soa.org/files/pdf/edu-29-5-mfe-exam.pdf 1. [Section 4.2] The multipliers for up and down moves and the risk-neutral probability are u = e (r δ+.5σ) = e.5.5+.3 = e.3 = 1.34986 d = e.3 =.7481 p = 1 1 + e = 1 =.425557 σ 1 + e.3 where we ve used equation (3.6) to calculate p, since this tree is the forward tree. The resulting stock prices are shown in Figure B.3. At the ending nodes, the option only pays off at the highest node. Pulling back one year: C tentative u = e.5 (.425557)(82.2119) = 33.2796 However, 33.2796 is less than the exercise value of 34.9859, so C u = 34.9859. Then C = e.5 (.425557)(34.9859) = 14.1623 (E) 2. [Sections 13.1 and 13.2] The average of the 12 stock prices is 15 + 12 + + 11 + 115 = 11 12 so the Asian call option pays 1. The up-and-out pays nothing since the 125 barrier was hit; the up-and-in pays 5 since the barrier of 12 was hit and 115 11 = 5. The answer is (B). 3. [Lesson 3] The risk-neutral probability of an up is p = e r δ d u = d = e.5.1.8 =.5498 1.1.8 134.9859 34.9859 182.2119 82.2119 1 14.1623 1 74.818 54.8812 Figure B.3: Binomial tree for S9:1

776 SOLUTIONS TO EXAM MFE/3F, SPRING 29, QUESTIONS 4 7 so the value of the call is e.5 (.5498)(55 5) = 2.3976 > 1.9, so buy the option and sell shares of WWW. When buying a call, shares must be sold since a call results in buying stock if it pays off. Lend the extra money. This is choice (B). The official solution also works this out with the replicating portfolio. 4. [Section 14.1] As listed in Table 14.1, the appropriate formula for this asset-or-nothing put option S S < K is Se δt N ( d 1 ). Multiplying by one million, the answer is (D). d 1 = ln(1/.6) +.25.2 +.5(.22 ) = 2.67913.2 N ( d 1 ) =.369 S S < K = 1e.2 (.369) = 3.61693 5. [Section 24.1] The zero-coupon bond is worth e.18 =.83527 at the top node, e.12 =.88692 at the two middle nodes, and e.6 =.94176 at the bottom node, so the option pays off at the top 3 nodes. Discounting with probabilities to the initial node: P = (.7 2 )e.12.15 (.9.83527) + (.7)(.3)e.12.15 (.9.88692) + (.7)(.3)e.12.9 (.9.88692) =.2421 +.21 +.222 =.29 (E) 6. [Lesson 18] The derivatives of Y (t ) are By Itô s Lemma, Y X = 1 X 2 Y X X = 2 X 3 Y t = dy (t ) = Y X dx +.5Y X X (dx ) 2 + Y t dt 8 2X (t ) = dt 8 64 dz (t ) + X (t ) 2 X (t ) 2 X (t ) dt 3 with the last term resulting from dx (t ) 2 2 = 8 dz (t ) = 64 dt. Ignore the dz (t ) term, since we don t need that for this question. The coefficient of dt is Y (t ) 2 8 2 + 64Y (t ) 3 = 8Y (t ) 2 + 2Y (t ) + 64Y (t ) 3 Y (t ) so α(y ) = 8y 2 + 2y + 64y 3, and α(1/2) = 8(.25) + 2(.5) + 64(.125) = 7. (D) 7. [Lesson 5] Tricky question: even though probabilities, assumptions, and payoffs don t change, the answer is not A because the risk-neutral probability changes. The second method in the official solution is easier, and that is what I will use here. From the information given, $1.13 is 2p discounted one year, or e r d e 1.13 = 2 e r r.8 = 2 e r = 5 4e r u d.4 e r = 5 1.13 =.9675 4 In the revised tree, e r.6 C = 2 e r = 1 2(.9675) = 1.3983.6 3 (D)

SOLUTIONS TO EXAM MFE/3F, SPRING 29, QUESTIONS 8 11 777 8. [Lesson 19] By the Black-Scholes Equation (19.1), For V = e r t ln S(t ),.5S 2 σ 2 V SS + V t + V S S(r δ) = r V V S = e r t S(t ) V SS = e r t S(t ) 2.5e r t σ 2 + (r δ) + r V = r V.5σ 2 + r δ = δ = r.5σ 2 V t = r V Plugging in the values we re given, δ =.55.5(.3 2 ) =.1. (B) The official solution has the following alternative. The risk-neutral expectation of the prepaid forward price of the security must equal the current value of the security. The current value (t = ) of the security is ln S(), while the prepaid forward price of it is e r t times its price, or ln S(t ). So ln S() = E ln S(t ) for all t. But ln S(t ) has a normal distribution with mean r δ.5σ 2 under the risk-neutral measure, so r δ.5σ 2 =. 9. [Subsection 1.2.5] This question requires a currency translation and put-call parity. If you do put-call parity first, then we calculate the value of a four-year dollar-denominated European call option on yen with strike price $.8:.5 C =.8e 4(.3).11e 4(.15) =.32645 C =.5 +.32645 =.37645 Then multiply by 125 since we need a put on 1 dollar or 125(.8) dollars, and divide by.11 to translate the currency into dollars:.37645(125/.11) = 42.7733. ((E)) If you want to do the currency translation first: a four-year put option to pay $.8 and get 1 is multiplied by 125 to make it pay $1 and get 125, and then the price is $125(.5). The price in yen is $125(.5)/.11 = 5.6818. By put-call parity, a put option on dollars to pay 125 and get $1 is worth: P 5.6818 = 125e 4(.15) (1/.11)e 4(.3) = 37.914 P = 5.6818 + 37.914 = 42.7733 1. [Section 2.2] As discussed in Example 2D, a risk-free portfolio is obtained by zeroing out the volatility, so if x i is the amount to invest in asset i in this question, then.2x 1.25x 2 = which implies x 1 = (5/4)x 2. Since x 1 + x 2 = 1, it follows that x 1 = (5/9)(1) = 555.56. (C) 11. [Lesson 22] The expected value of S(1) a is calculated from formula (22.1): E[S(1) a ] = S() a a (α δ)+.5a (a 1)σ2 e and is equal to 1.4. We are being asked for the prepaid forward price of S(1) a, or formula (22.3): F P,1 S(1) a = S() a r +a (r δ)+.5a (a 1)σ2 e

778 SOLUTIONS TO EXAM MFE/3F, SPRING 29, QUESTIONS 12 13 The quotient of the second formula over the first is F,1 P (Sa ) E r a (α r ) = e S(1) a so we need to back out a. The stochastic differential equation for S(t ) has α =.5 and σ =.2. E S(1) a = S() a a (α δ)+.5a (a 1)σ2 e ln 1.4 = a ln.5 +.3 + (.5a +.2a (a 1)).2a 2 + (ln.5 +.3)a ln 1.4 = a =.49985 The other solution to the quadratic is rejected since it is positive. Then, with r =.3, F P,1 (Sa ) = 1.4e.3+.49985(.5.3) = 1.372 (C) 12. [Subsection 1.2.1 and Lesson 2] I. C (5, T ) is worth more than C (55, T ) since calls decrease in value with increasing strike prices, and the difference in values is less than 5e r T because if you buy C (5, T ) and sell C (55, T ), the most you can get is 5 at time T, which is worth 5e r T at time. II. By put-call parity, P(5, T ) C (5, T ) = 5e r T S and P(45, T ) P(5, T ), so P(45, T ) C (5, T ) 5e r T, and is certainly also less than 55e r T, but the left side inequality is incorrect. III. We proved the right hand inequality in (II). The left hand inequality follows from and C (45, T ) C (5, T ). (E) P(45, T ) C (45, T ) = 45e r T S The answer choices are asymmetric, something which was not allowed on previous exams, but II and III cannot both be true, so they perhaps were forced to make them asymmetric. 13. [Section 9.1] By the Black-Scholes formula, the value of the option at time 8 months, with 4 months left to expiry, is d 1 = ln(85/75) +.5 +.5(.26 2 ) (1/3) = 1.1989.26 1/3 d 2 = 1.199.26 1/3 =.86978 N (d 1 ) = N (1.1989) =.84611 N (d 2 ) = N (.86978) =.8779 C = 85(.84611) 75e.5(1/3) (.8779) = 12.3365 The eight-month holding profit, taking into account the eight-month interest cost on the original investment of 8, is 12.3365 8e.5(2/3) = 4.653. Using the rounding rules for the normal CDF in effect for this exam, the answer would ve been 46, or (A).

SOLUTIONS TO EXAM MFE/3F, SPRING 29, QUESTIONS 14 16 779 14. [Subsection 24.2.2] First, in a Black-Derman-Toy tree, the vertical ratios between interest rates are constant, so.8/r u d = r u d /.2 and therefore r u d =.4. By formula (24.1), the forward s price is P(, 3)/P(, 2). We calculate these in terms of the initial interest rate r. In the following, P u (1, 3) is the price of a two-year bond at the upper node after one year, and P d (1, 3) is the price of a two-year bond at the lower node after one year. P(, 2) =.5 1 1 + r 1.6 + 1 =.697115 1.3 1 + r P u (1, 3) =.5 1 1.6 1.8 + 1 =.396825 1.4 P d (1, 3) =.5 1.3 1 1.4 + 1 1.2 =.595238 P(, 3) =.5 1 + r (.396825 +.595238) =.49632 1 + r The answer to the question is 1F,2 P(2, 3) = 1(.49632/.697115) = 711.55. (E) The question omitted r and used high interest rates to try to trick you. Some students mistakenly thought that the forward price is the probability-weighted average of the one-year bond prices at the end of two years, or and 74.37 is not one of the five answer choices. 1 1 1.25 +.5 +.25 =.7437 1.8 1.4 1.2 15. [Section 26.3] We recognize the model as a Vasicek model with a =.1 and σ =.5. The Sharpe ratio is deduced by comparing the risk-neutral process to the true process; the Sharpe ratio φ times σdt is added to go from the latter to the former, and the risk-neutral process is.5 dt more than the true process, so φ =.5/.5 =.1. Then 1 e a (T t ) B(t, T ) = a B(2, 5) = 1 e.1(3) = 2.59182.1 q(.4, 2, 5) = B(2, 5)σ = 2.59182(.5) α(.4, 2, 5).4 =.1 q(.4, 2, 5) α(.4, 2, 5) = 2.59182(.5)(.1) +.4 =.5296 (C) 16. [Lesson 16] The parameters m and v of the lognormal distribution of the stock price after 9 months are m = (α δ.5σ 2 )(t ) =.1.5(.3 2 ) (.75) =.4125 v = σ t =.3.75 =.25981 So the probability that the stock price is more than 125 is ln(125/1).4125 1 N = 1 N (.71) =.24193 (A).25981

78 SOLUTIONS TO EXAM MFE/3F, SPRING 29, QUESTIONS 17 19 17. [Subsection 1.1.1] The formula for a put s delta is e δt N (d 1 ) 1 (equation (1.3)), and since N ( d 1 ) = 1 N (d 1 ), this is the same as e δt N ( d 1 ). Since δ =, N ( d 1 ) =.4364 d 1 =.161 We ll use d 1 =.16, in accordance with the rounding rules for the normal distribution in effect for this exam. The answer is not significantly different if.161 is used. We set up the quadratic equation for σ. r +.5σ 2 =.16 σ.12 +.5σ 2 =.16σ.5σ 2.16 +.12 = The two solutions are σ =.12,.2. Higher volatility leads to higher put prices, so if 2% satisfies (i), 12% certainly does, and since the answer is unique, we know the answer has to be 12%. (A) The official solution shows how you can verify σ <.14 without explicitly valuing the put with Black-Scholes formula for σ =.2; you can express the Black-Scholes formula for the put over the stock price in terms of N (d 1 ) and N (d 2 ), set d 2 = d 1 σ, and then get an upper bound for σ. 18. [Lesson 2] The Sharpe ratios of the two stocks must equal. Let α i and σ i be the rates of return and volatilities of the stocks with prices S i. Then α 1 =.1 +.5(.2 2 ) =.12 α 2 =.125 +.5(.3 2 ) =.17.12 r =.17 r.2.3.36.3r =.34.2r.1r =.2 r =.2 (A) 19. [Section 9.1] The notation Var ln F P t,1 (S) looks a little weird, but it denotes the variance of the prepaid forward from time t to time 1, as of time, as a function of t. When t =, there is no volatility since the prepaid forward price is known, but as time goes on, the volatility, from the perspective of valuation at time, keeps growing. So σ 2 =.1 and σ =.1. We can now use the Black-Scholes formula on the prepaid forward of the stock, whose value at time is 5 5e r t = 5 5e.75(.12) = 45.4334 Plugging into the prepaid forward version of the Black-Scholes formula, equation (9.1), Using the rounding rules in effect at this exam, d 1 = ln(45.4334/45e.12 ) +.5(.1 2 ) = 1.34518.1 d 2 = 1.34518.1 = 1.24518 N ( d 1 ) = N ( 1.35) =.885 N ( d 2 ) = N ( 1.25) =.156

SOLUTIONS TO EXAM MFE/3F, SPRING 29, QUESTION 2 781 P = 45e.12 (.156) 45.4334(.885) =.1941 and 1 units have value 19.41. (D) Using 5-decimal precision for the normal CDF, and 1 units have value 19.58. N ( d 1 ) = N ( 1.34518) =.8928 N ( d 2 ) = N ( 1.24518) =.1653 P = 45e.12 (.1653) 45.4324(.8928) =.19575 2. [Section 12.2] The delta-gamma approximation is equation (12.2) without hθ. Thus we have ε +.5Γε 2 = 2.21 2.34 =.13.181ε +.5(.35)ε 2 =.13.175ε 2.181ε +.13 = ε =.181 ±.181 2 4(.175)(.13) = 9.5663,.7765.35 The original stock price is 86 ε. Using ε =.7765 gets an original stock price of S() = 85.2235. (C). Using 9.5663 gets a stock price below 8, violating (i). The delta-gamma approximation is invalid for such a large change, since the squared epsilon in the gamma term overwhelms delta in such a case.

782 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 1 3 B.5 Solutions to Sample Questions The SOA sample questions and solutions for Exam MFE/3F can be downloaded from http://www.beanactuary.org/exams/pdf/mfe_sampleqs1-76.pdf The proportion of these questions that are based on difficult topics such as Brownian motion is greater than the proportion of questions that a real exam would have. The solutions are lengthy and often go both beyond the question and the McDonald textbook; they tend to avoid using short cuts and instead derive everything from first principles, which makes them long. They are often not the way you would work out the question on an exam. The questions and their solutions are more an educational tool than a sample exam, and in fact, rather than providing the questions and solutions separately, each question is followed by its solution, which makes it hard to use these questions as a sample exam. The solutions provided here are the way I would solve these questions if they came up on an exam. I usually present only one method, my favorite method. 1. [Subsection 1.2.1] This is a straightforward put-call parity question. By put-call parity P(S, K, t ) C (S, K, t ) = K e r t Se δt.15 = 7e 4r 6 e 4r 6.15 = =.855 7 ln.855 r = =.3916 (A) 4 2. [Subsection 2.4.1] The call prices do not satisfy convexity, since 6 > (1/3)(11) + (2/3)(3) = 5 2. The put 3 prices, however, satisfy convexity. Mary s portfolio, if two calls with strike price 55 are longed, is a butterfly spread. A butterfly spread is the standard method for exhibiting arbitrage when convexity is violated. The portfolio costs (11) + 2(3) 3(6) = 1, and after lending 1, there is no initial cash flow. At expiry, the amount paid by Mary for the three call options shorted will always be no greater than the amount received on the three call options longed, so Mary will gain the repayment of the loan at the very least, with no risk of loss. In Peter s portfolio, every pair of calls longed and puts shorted with the same strike price will be worth, at expiry, S K. If three calls with strike price 5 are shorted and three puts with strike price 5 longed, Peter will receive a net of $2 initially, since 2(3 11) + (11 3) + 3(8 6) = 2. At expiry, the portfolio will be worth, since the pair of 4-strike options will yield S 4; the two pairs of 55-strike options will yield S 11; and the three pairs of 5-strike options will yield 15 3S, and these add up to. Peter will gain the repayment of the loan. Thus Mary s and Peter s portfolios exhibit arbitrage. (D) 3. [Section 13.3] The present value of the payout is equated to the premium π. We are given the price of a put that pays the maximum of and 13 S, so we would like to turn the max expression for the payoff into max(, 13 S). We do this by factoring out 1/S(): max S(T )/S(),(1 + g %) T = max S(1), 13 = S(1) + max, 13 S(1) 1 The present value of S(1) is S() or 1, since S(1) incorporates dividends. If it didn t incorporate dividends, the present value would be S()e δ, since F,T P (S) = Se δt. (See Table 1.2.) The present value of the maximum expression is therefore PV max S(T )/S(),(1 + g %) T 1 + 15.21 = = 1.1521 1 1

SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 4 5 783 Equating the present value of the payout to the premium, so y % = 1.86798 =.1322. π (1 y %) 1.1521 = π 1 y % = 1 1.1521 =.86798 4. [Section 4.1] Although not needed for the solution, u = e.25 and d = e.15, so this binomial tree is based on forward prices with σ =.2. Thus you can calculate p using the shortcut formula (3.6): 1/(1 + e σh ) = 1/(1 + e.2 ) =.452 if you wish. Since the stock pays no dividends, the American call option is worth the same as a European call option, so we have no need to induct on the tree. We just have to calculate the option payoffs at the ending nodes, weight them with risk-neutral probabilities, and discount them. The risk-neutral probability of an up-movement is p = e.5.867 1.284.867 =.452 The stock prices at the three ending nodes are S u u = 2(1.284 2 ) = 32.97, S u d = 2(1.284)(.867) = 22.1, and S d d = 2(.867 2 ) = 14.82. The call payoffs from top to bottom are 1.97,.1, and. Weighting and discounting: C = e.1 (.452 2 )(1.97) + 2(.452)(.5498)(.1) = 2.6 (C) 5. [Section 4.3] They didn t tell you which binomial tree to use. They intended that you use one based on forward prices. I think they would be more clear regarding which tree to use on a real exam. For the tree with forward prices, r is the domestic dollar interest rate or 8% and r f is 9%. u = e (r δ)h+σh = e (.8.9)(.25)+.3(.5) = 1.1589 d = e.25.3(.5) =.8586 p 1 = 1 + e = 1 =.4626 σ h 1 + e.3(.5) We then calculate the exchange rates and put payoffs at the ending nodes as indicated in Figure B.4 We compute the put prices as: P u u = P u d = e.8(.25) (.5374)(.3384) =.1783 P tentative d d = e.2.4626(.3384) +.5374(.655) =.4985 However, at the d d node, 1.56 1.541 =.559 >.4985, so the option is exercised early at that node. Continuing, P u = e.2 (.5374)(.1783) =.939 P d = e.2.4626(.1783) +.5374(.559) =.3473 P = e.2.4626(.939) +.5374(.3473) =.2255

784 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 6 7 1.927 2.2259 1.6573.939 1.649 1.43.2255 1.4229.1783 1.2277.3473 1.2216.3384 1.541.559.95.655 Figure B.4: Binomial tree for dollar/pound exchange rate question, sample question 5 6. [Section 9.1] The Black-Scholes formula gives: d 1 = ln(2/25) +.5.3 +.5(.24 2 ) (.25).24.25 d 2 = 1.7579.24.25 = 1.8779 N (d 1 ) = N ( 1.76) =.392 N (d 2 ) = N ( 1.88) =.31 C (S, K,σ, r, t,δ) = 2e.75 (.392) 25e.125 (.31) =.3499 = 1.7579 For 1 units, the price is 3.499. (C) 7. [Section 9.2] The question uses Brownian motion language to state that the Black-Scholes framework applies and to state the volatility, but only requires the Black-Scholes or Garman-Kohlhagen formula. Since the price of the option is denominated in dollars, the easier way to do this is from the American perspective. Yen must be sold for dollars, so the needed option is a put on yen. The domestic risk-free rate, which is used for r, is 3.5% and the foreign risk-free rate, which is used for δ, is 1.5%. The option is at-the-money, so K = x and ln(x /K ) =, where x is the spot exchange rate. The annual volatility is σ =.261712 365 =.5. We calculate N ( d 1 ) and N ( d 2 )..35.15 +.5(.5 2 ) (.25) d 1 =.5.25 =.2125 N (.2125) =.41586 d 2 =.2125.5.25 =.1875 N (.1875) =.42563 We want an option for 12 billion yen. The value of these is $1 billion and the strike price is $1 billion, so the cost of the options is 1,,, e.35(.25) (.42563) e.15(.25) (.41586) = 7,618,538 This answer should be rounded, since not all 7 digits are significant.

SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 8 9 785 The alternative is to calculate a yen-denominated call on dollars and then to translate the price into dollars using $1= 12. Then the domestic rate is.15 and the foreign rate is.35, leading to the following calculation:.15.35 +.5(.5 2 ) d 1 = =.1875 N (d 1 ) =.42563.25 d 2 =.1875.25 =.2125 N (d 2 ) =.41586 We calculate 12 billion calls, but then divide by 12 to convert to dollars, so effectively we multiply by 1 billion: 1,,, e.35(.25) (.42563) e.15(.25) (.41586) = 7,618,538 8. [Subsection 1.1.1] For a nondividend-paying stock, = N (d 1 ). Since = N (d 1 ) =.5, it follows that d 1 = and d 2 = σ t =.3.25 =.15. We will need to back out.25r. We have so d 1 = ln(4/41.5) + r +.5(.3 2 ) (.25).3.25 = ln(4/41.5) +.25r +.1125 =.25r =.1125 ln(4/41.5) =.25564 The price of the option is 4(.5) 41.5e.25r N (.15) = 2 4.453N (.15) We want to express this as one of the five choices, all of which have N (.15). 2 4.453N (.15) = 2 4.453 + 4.453N (.15) = 2 4.453 + 4.453.15 e x 2 /2 dx 2π.15 = 2.453 + 16.138 e x 2 /2 dx (D) 9. [Section 12.1] We must back out σ. Since = N (d 1 ) =.6179, d 1 =.3. (.1 +.5σ 2 )(.25) =.3 σ(.5).125σ 2.15σ +.25 = σ =.15 ±.15 2 4(.25)(.125).15 ±.1 = =.2, 1 2(.125).25 The solution σ = 1 is high and doesn t result in one of the five answer choices. On a real exam, they would tell you explicitly not to use the solution 1. Using σ =.2, the daily change resulting in no gain or loss is Sσ h = 5(.2) 1/365 =.5234. (B)

786 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 1 14 1. [Sections 17.1 and 23.2] (i) The Black-Scholes framework includes the hypothesis that X (t ) is an arithmetic Brownian motion. (ii) The Black-Scholes framework includes the hypothesis that volatility is constant. The variance of X (t + h) X (t ) is the square of the volatility times h, for any t and h. (iii) (E) If dx (t ) = µ dt +σ dz (t ), then dx (t ) 2 = σ 2 n dt by the multiplication rules, and lim n j =1 dt = T. 11. [Lesson 17] (i) (ii) (iii) (E) The Black-Scholes framework includes the hypothesis that volatility is constant and its square is σ 2 h over an interval of length h. We have ds(t ) = α dt + σ dz (t ) S(t ) dt is not stochastic, while the variance of dz (t ) is dt by the multiplication rules, since dz (t ) has mean and dz (t ) 2 = dt. It follows that the variance of α dt + σ dz (t ) is σ 2 dt. If we multiply (ii) by S(t ), we have Var ds(t ) S(t ) = S(t ) 2 σ 2 dt, and ds(t ) = S(t +dt ) S(t ). S(t ) is given and therefore does not change the variance. 12. [Lesson 2] Since d ln Y (t ) = µ dt +.85 dz (t ), it follows that B =.85; the volatility of any process is equal to the volatility of the logged process. We now equate Sharpe ratios of X (t ) and Y (t )..7.4 = A.4.12.85.255 =.12(A.4) A =.255.12 +.4 =.6125 (B) 13. See exercise 23.3, page 454. 14. [Section 26.3] The Sharpe ratio does not vary with α. In a Vasicek model, the Sharpe ratio is φ = α r q = α(r, t, T ) r B(t, T )σ where σ is constant. This question is asking the change in α as a result of changing r and T t. We must determine the change in B(t, T ). We can express B(t, T ) in terms of a =.6 and T t as ā T t a. Equating Sharpe ratios multiplied by σ: B(, 2) = 1 e 2(.6) = 1.164676.6 B(1, 4) = 1 e 3(.6) = 1.391169.6.4139761.4 α(.5, 1, 4).5 = 1.164676 1.391169 α(.5, 1, 4) =.5 + 1.391169(.139761) +.5 =.516694 1.164676

SOLUTIONS TO SAMPLE QUESTIONS, QUESTION 15 787 Year Year 1 Year 2 Year 3 17.2% 3.46317 16.8% 5.39384 12.6% 2.15876 13.594% 2.72384 9% 1.3264 13.5% 1.39836 9.3%.73285 11%.4545 1.596%.2365 8.99% Figure B.5: Interest rate tree, with caplet prices, for Sample Question 15 15. [Section 24.2] We back out the missing interest rates on the tree, using the fact that the ratios between vertically consecutive interest rates are constant. R x x x in the following refers to the effective annual interest rate at node x x x..168/.11 = 1.235829 R u u d =.168/1.235829 =.13594 R d d d =.11/1.235829 =.899.172/.135 = 1.27474 R d d =.135/1.27474 =.1596 The interest rate tree is shown in Figure B.5. The caplet prices C x x x at the ending nodes are: 16.8 1.5 C u u u = = 5.39384 1.168 13.594 1.5 C u u d = = 2.72384 1.13594 11 1.5 C u d d = =.4545 1.11 The other caplet prices are calculated by backwards induction on the tree: C u u =.5 (5.39384 + 2.72384) = 3.46317 1.172 C u d =.5 (2.72384 +.4545) = 1.39836 1.135 C d d =.5 (.4545) =.2365 1.1596

788 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 16 18 C u =.5 (3.46317 + 1.39836) = 2.15876 1.126 C d =.5 (1.39836 +.2365) =.73285 1.93 C =.5 (2.15876 +.73285) = 1.3264 1.9 The non-year 4 nodes for which the interest rate is greater than 1.5% do not contribute to the cost of the caplet. A caplet is effective in only one period; here, that period is year 4. In contrast, a cap is effective for all periods until it expires. The official solution s answer is slightly different since they keep less precision in the intermediate results. The official solution also has an alternative solution which is a longcut. Namely, discount the four ending prices along each path separately. There are eight paths to the ending four nodes, each with probability 1/8. The path d d d can be skipped since the value on that path is. This longcut is not needed for this question since the value of the caplet is path-independent, but would have to be used if the payoff were path-dependent, for example if the payoff were based on an average of interest rates. or 16. [Lesson 22] By formula (22.3) for the price of a prepaid forward of S(T ) a, we want x(x 1) exp (x 1)r + σ 2 = 1 2 x(x 1) (x 1)r + σ 2 = 2 Factoring out x 1, which generates the solution x = 1, we have r + xσ2 2 = x = 2r σ 2 For the parameters given by the question, x = 2(.4)/(.2 2 ) = 2. (B) 17. [Section 6.3] To make the calculation easy, the ending stock price equals the beginning stock price, so that µ =, and there are only two distinct ratios of stock prices,.8 and 1.25, whose logs are negative each other. In fact, ln 1.25 =.223144. So the unbiased sample variance over eight values is (8/7)(.223144 2 ) =.5696. This is the monthly variance; the annual variance is 12(.5696) =.68288. The annual volatility is.68288 =.82636. (A) 18. [Subsection 14.2.3] We express the gap call as a standard call plus all-or-nothing calls. C gap = C (S, 1, 1) (3 S > 1) = C (S, 1, 1) 3N d 2 (1) Delta for this option is then the derivative with respect to S, or = N d 1 (1) 3N d 2 (1) d 2 (1) ( )

SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 19 21 789 We calculate N d 1 (1), N d 2 (1), and d 2 (1). d 1 (1) =.5σ2 σ =.5 N d 1 (1) =.69146 d 2 (1) =.5 1 =.5 N d 2 (1) = e (.52 /2) 2π =.35265 d 2 (1) = lns ln 1 +.5 d 2 (1) = 1 S =.1 Substituting into ( ), we conclude that the number of shares needed to delta-hedge is 1 = 1.69146 3(.35265)(.1) = 585.84 (A) 19. [Subsection 14.4.2] One year from now, the call option value is as follows: d 1 =.8 +.5(.32 ) =.41667.3 N (d 1 ) = N (.41667) =.66154 d 2 =.41667.3 =.11667 N (d 2 ) = N (.11667) =.54644 C (S, K,.3,.8, 1, ) = S.65154 e.8 (.54644) =.15711 Therefore, the price now is.15711f P,1 (S). We are given that F,1(S) = $1, so F P,1 (S) = 1e.8 and the forward start option s value is 1e.8 (.15711) = 14.5. (C) 2. [Section 1.2] Let C and P be the call delta and put delta respectively. From the delta of Investor B s portfolio, we conclude 2 C 3 P = 3.4 ( ) Elasticity of a portfolio is S portfolio /C portfolio, where portfolio and C portfolio are delta and the price for the entire portfolio. Let C and P be the call and put prices respectively. From the elasticity of Investor A s portfolio, we conclude S(2 C + P ) = 5. 2C + P 45(2 C + P ) 2(4.45) + 1.9 = 5. 45 1.8 (2 C + P ) = 5. 2 C + P = 1.2 Subtracting equation ( ) from equation ( ), we get 4 P = 2.2, or P =.55. Then the put option elasticity is 45(.55)/1.9 = 13.3. (D) 21. [Section 26.3] In a Cox-Ingersoll-Ross model, the Sharpe ratio does not vary with T or t but is proportional to r. The volatility σ r is proportional to r as well. Also, q(r, t, T ) = B(t, T )σ. From the definition of the Sharpe ratio and from the values we are given, ( ) α(.5, 7, 9) r φ(.5) = q(r, t, T ).6.5 = B(t, T ) σ.5

79 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 22 25 and it follows that φ(.4) = Set this equal to the definition of φ(.4)..4.4.5 φ(.5) =.5.1 B(t, T ) σ.5 α(.4, 11, 13).4.4 B(t, T ) σ.1 =.4.5 B(t, T ) σ.5 Divide both sides by.4, and multiply both sides by B(t, T ) σ. α(.4, 11, 13).4.4 =.1.5 = 1 5 α(.4, 11, 13) =.4 +.4/5 =.48 (C) 22. [Example 26B] To go from the true process to the risk-neutral process, σ(r, t )φ(r, t ) is added to the drift term. We are given that σ =.3, so we can derive φ(r, t ): φ(r, t ) =.15.5r (t ).9.5r (t ).3 =.6.3 =.2 Since g r (t ), t depends on the same dz (t ), it must have the same Sharpe ratio. Let µ(r, g ) = m (r, g )g. Then m (r, g ) r =.2.4 m (r, g ) = r +.8 and µ(r, g ) = (r +.8)g. (D) 23. [Section 1.2, Lessons 12 and 17] Although the question is couched in Brownian motion terminology, it is an elasticity question. From (i) and (v), the risk premium for the stock is.1.4 =.6. To delta-hedge a call option one has written, one must buy shares, which costs S. The elasticity of the call option is S /C. From (iii) and (iv), we conclude that the elasticity of the call option is 9/6 = 1.5. By equation (1.7), the risk premium of the call option is Ω(.6) =.9. The rate of return on the call option is r plus the risk premium. Therefore, γ =.4 +.9 =.13. (C) 24. See Example 23G, page 451. 25. [Subsection 14.4.1] Let C (t, T ) be the price of a European call option at time t expiring at time T. The question s notation C (T ) is the same as our C (, T ). At time 1, the chooser option is worth max C (1, 3), P(1, 3) = C (1, 3) + max, P(1, 3) C (1, 3) = C (1, 3) + max, K S(1) where the last equality is by put-call parity, since r = δ =. Moving back to time, the first term becomes C (, 3) = C (3) and the second term is a put option expiring at time 1, so the chooser option is worth C (3) + P(1)

SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 26 28 791 By put-call parity, P(1) = C (1) + K S(). So we have C (3) + C (1) + K S() = 2 C (3) + 4 + 1 95 = 2 C (3) = 11 (B) 26. [Section 2.1] A call option is worth less than the stock price, and a put option is worth less than the strike price. A European call option is worth at least as much as implied by put-call parity against a put worth, or C Se δt K e r t. Here, e δt = 1 and K e r t = 1e.5 = 95.12. Thus graph II gives bounds for the value of a European call option. An American call option on a stock with no dividends is worth the same as a European call option, so graph II gives bounds for the value of an American call option as well. A European put option is worth at least as much as implied by put-call parity against a call worth, or P K e r t Se δt = 95.12 S, Thus graph IV gives bounds for the value of a European put option. An American put option can be exercised immediately so it must be worth at least its exercise value of K S = 1 S, so graph III gives bounds for the value of an American put option. (D) 27. [Section 3.2] We can either use a replicating portfolio or risk-neutral probabilities. Replicating Portfolio Let X and Y be the number of shares of X and Y respectively in the replicating portfolio, and B the amount lent. Then the three outcomes imply 2 X + B e.1 = 95 15 = 1 5 X + B e.1 = 95 3 Y + B e.1 = ( ) ( ) ( ) Subtracting ( ) from ( ), 15 X = 15, so X =.7 and B e.1 = 13. From ( ), 3 Y + 13 =, so Y = 13/3. Then P Y C X = 1( X + Y ) + B = 1(.7 13/3) + 13e.1 = 4.2955 (A) Risk-Neutral Probabilities Let p i be the risk-neutral probability of outcome i. For Y, we have 3p 3 = 1e.1, so p 3 = e.1 /3 =.36839. For X, we have 2p 1 + 5p.1 2 = 1e 2p 1 + 5(1 p 1 e.1 /3) = 1e.1 15p 35e.1 1 = 5 = 78.9366 3 p 1 = 78.9366 =.526244 15 so p 2 = 1.526244.36839 =.15366. Then P Y C X = e.1 1p 1 + 95p 2 = 4.2955. Note that the state prices in the official solution are the risk-neutral probabilities multiplied by e.1. 28. [Section 14.1] The option pays 1 if S(1) > 1, otherwise ; squaring S(1) plays no role. It is an all-ornothing option of the form 1 S(1) > 1. The price of such an option is C = 1e r t N d 2 (1). The number of shares to delta hedge is the derivative of this price, or 1e r t N (d 2 )d 2. ln(s/1) + r +.5σ2 d 2 = σ = ln(1/1) +.2.5(.22 ).2 =

792 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 29 3 N () = 1 2π =.3989 d 2 = 1 Sσ = 1 (1)(.2) =.5 C S = 1e.2 (.3989)(.5) = 19.55 (A) 29. [Section 24.2] First we back out r u d, using the fact that the ratio of 6% to r u d equals the ratio of r u d to 2%: r u d = (.6)(.2) =.3464 Then we compute the prices of the 3-year bond at the upper and lower Year 1 nodes, P u and P d respectively. P u =.5 1 1.5 1.6 + 1 =.995 1.3464 P d =.5 1 1.3 1.3464 + 1 =.9451 1.2 We convert these into effective annual interest rates, which we ll call R u and R d. 1 1/2 R u = 1 =.48573.995 1 1/2 R d = 1 =.28635.9451 The formula for volatility is R u /R d = e 2σh with h the time period. Here h = 1, and we get σ = ln(.48573/.28635) 2 =.2642 (D) 3. [Section 24.2] To get 1% volatility, we need r u = e.2 r d. To get a bond price of 88.5 for the two year bond, we need 1 1 (.5)(94.34) + = 88.5 1 + r d 1 + r d e.2 We solve this equation for r d. First multiply out the denominators (.5)(94.34) 2 + r d (1 + e.2 ) = 88.5 1 + r d (1 + e.2 ) + r 2 d e.2 Move all terms to one side to get a quadratic equation. 88.5e.2 r 2 d + 1 + e.2 88.5.5(94.34) r d + 88.5 94.34 = 18.9r 2 d + 91.81r d 5.84 = Solve r d = 91.81 + 1954 2(18.9) =.594 (A)

SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 31 32 793 31. [Subsections 1.1.1 and 1.1.7] Assume the bull spread is constructed by buying a 5-strike call and selling a 6-strike call. Then delta for the bull spread is N d 1 (5) N d 1 (6). Note that the same delta would result if the bull spread were constructed by buying a 5-strike put and selling a 6-strike put, since delta for a put on a stock with no dividends is N (d 1 ) 1, and N d 1 (5) 1 N d 1 (6) 1 is the same bull spread delta as above after canceling the 1 s. To work out this question, we must compute four N (d 1 ) s. Initially.5 +.5(.2 2 ) (.25) d 1 (5) = =.175 N d 1 (5) = N (.175) =.56946.1 d 1 (6) = ln(5/6) +.5 +.5(.2 2 ) (.25) = 1.64822 N d 1 (6) = N ( 1.64822) =.4965.1 After one month,.5 +.5(.2 2 ) (1/6) d 1 (5) =.2/ =.14289 N d 1 (5) = N (.14289) =.55681 6 d 1 (6) = ln(5/6) +.5 +.5(.2 2 ) (1/6).2/ 6 = 2.99 N d 1 (6) = N ( 2.99) =.183 The initial delta for the bull spread is.56946.4965 =.51981, and the delta after one month is.55681.183 =.58351. The difference is.53851.51981 =.187. (B) 32. [Lesson 21] The fund W has volatility ϕσ. Since it must have the same Sharpe ratio as S, its rate of return γ must satisfy giving us the geometric Brownian motion γ r ϕσ = α r σ γ = r + ϕ(α r ) dw (t ) W (t ) = ϕα + (1 ϕ)r dt + ϕσ dz (t ) which isn t quite choice (A). The solution of this differential equation for W (t ) is arrived at by subtracting half the volatility squared from the coefficient of dt and using that as an exponent, so W (t ) = W () exp ϕα + (1 ϕ)r.5ϕ 2 σ 2 t + ϕσz (t ) which isn t quite choices (B) or (C). So we ll also have to compute S(t )/S() ϕ. S(t ) ϕ = exp ϕα.5ϕσ 2 t + ϕσz (t ) S() Subtract the exponent of S(t )/S() ϕ from the exponent of W (t )/W (), and what remains is (1 ϕ)r.5σ 2 (ϕ 2 ϕ) t = (1 ϕ)r.5σ 2 ϕ(ϕ 1) t = (1 ϕ)(r +.5ϕσ 2 )t which is answer choice (E).

794 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 33 36 33. [Subsection 14.4.2] Each of the four put options is a forward start option; even the first option, purchased immediately, can be considered a forward start option with a delay of. The price P of each option at the time purchased is d 1 = ln(1/.9) +.8 +.5(.3 2 ) (.25).3 =.9174.25 N ( d 1 ) = N (.9174) =.18122 d 2 =.9174.3(.5) =.7674 N ( d 2 ) = N (.7674) =.22341 P = S.9e.2 (.22341).18122 =.1587S The sum of the cost of the four options if paid immediately is.1586 S() + F P,.25 S(.25) + F P,.5 S(.5) + F P,.75 S(.75) However, the prepaid forward price of a nondividend paying stock is the stock price. So the amount you pay your broker is.1587(45)(4) = 2.8562. (C) 34. [Section 23.2] The answer is built into the multiplication rules: the limit of (dz ) 3 = (dz ) 2 dz = dt dz =, making the answer (A). 35. [Lesson 23] Having seen Example 23G, this question is a breeze. Just imitate the steps. First, calculate dr(t ). As you know from Example 23G, the trick is to pull e t out of the integral, use the product rule, and then put most of the pieces of the result back together again as R(t ). t dr(t ) = R()e t dt +.5e t dt +.1d e t e s R(s )dz (s ) t = R()e t +.5e t.1e t e s R(s )dz (s ) dt +.1e t e t R(t ) dz (t ) where we evaluated d of the integral as the integrand with s replaced with t. Now, notice that the dt term is.5 R(t ). dr(t ) =.5 R(t ) dt +.1 R(t ) dz (t ) Now we move on to X (t ). Using Itô s lemma, dx (t ) = X (t ) 2 X (t ) dr(t 2 X (t ) dr(t ) +.5 ) + dt R(t ) R(t ) 2 t = 2R(t )dr(t ) + dr(t ) 2 and plugging in dr(t ), dx (t ) = 2R(t ).5 R(t ) dt +.2R(t ) R(t ) dz (t ) +.1R(t )dt =.11R(t ) 2R(t ) 2 dt +.2R(t ) 3/2 dz (t ) =.11 X (t ) 2X (t ) dt +.2X (t ) 3/4 dz (t ) (B) 36. [Lesson 19 or 22] Let C be the price of the derivative security. By the Black-Scholes equation, equation (19.1), 1 2 σ2 S 2 2 C C + (r δ)s S2 S + C t = r C

SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 37 38 795 Let C = S a. Then, with δ =, C S = a S(t ) a 1 2 C S = a (a 1) S(t ) a 2 2 C t =.5σ 2 a (a 1) S(t ) a a a + r a S(t ) = r S(t ) Divide out S(t ) a..5a (a 1)σ 2 + r a r = ( ) a = 1 is one solution. Dividing it out, we have.5a σ 2 + r =, or a = 2r /σ 2. In our case, a = k /σ 2 and r =.4, so a =.8/σ 2 and k =.8. (E) Another way to get the same answer is to realize that this security is S a for a = k /σ 2. What would the prepaid forward price of this security be? Well, whatever it is, it would not vary with t, since regardless of the time of maturity of the forward, you would receive a security with the same value. But in formula (22.3), the prepaid forward on S a has a value varying with T, namely S() a e u t, where u is an expression involving a, r, and δ. The only way to reconcile with that formula is to have u =. So we get the equation which with δ = is the same equation as ( ). u = a (r δ) +.5a (a 1)σ 2 r = 37. [Section 13.1] The trick is to express the S(t ) s in terms of Q(t ) s where Q(t ) = S(t )/S(t 1). So 1/3 G = S()Q(1) S()Q(1)Q(2) S()Q(1)Q(2)Q(3) = S()Q(1)Q(2) 2/3 Q(3) 1/3 The Q(t ) s are independent lognormal random variables with m =.3 and v.2. Logging them yields normal random variables with µ =.3.5(.2 2 ) =.1 and σ =.2. Then lng = lns() + lnq(1) + 2 lnq(2) + 1 lnq(3) 3 3 Var(lnG ) = Var lnq(1) + 4 Var lnq(2) + 1 9 9 lnq(3) =.4 1 + 4 9 + 1 9 =.6222 (D) 38. [Section 26.5] Delta is the first derivative of P with respect to r and gamma is the second derivative, and we plug in r =.5: P = Ae Br = BP Γ = B 2 P The delta-gamma approximation is (equation (26.21) without θ ) P(t, T, r ) = P(t, T, r ) + (r r ) +.5Γ(r r ) 2 P Est (, 3,.3) = 1 + (.2)( B) +.5(.2 2 )(B 2 ) P(, 3,.5) = 1 +.2(2) +.5(.2 2 )(2 2 ) = 1.48 (B)

796 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 39 42 39. [Lesson 3] This is a sophisticated question combining put-call parity, binomial trees, and interest rate binomial trees. From put-call parity, P(18) C (18) = F P (18) S. The prepaid forward price of 18 is the probabilityweighted discounted value of 18 paid two years from now. We must consider the two possibilities of u and d. To determine the probabilities, we use the stock prices. Since the stock pays no dividends, p = 1.5S S d 15 95 = S u S d 11 95 = 2 3 With 2/3 probability the two-year discounting rate is 1/(1.5)(1.6) and with 1/3 probability the two-year discounting rate is 1/(1.5)(1.4). Therefore F P (18) = 18 2 1 + 1 1 = 97.6571 3 (1.5)(1.6) 3 (1.5)(1.4) P(18) C (18) = 97.6571 1 = 2.34 (B) 4. [Subsection 1.1.3 and Section 11.1] The profit diagrams are interesting, but for answering the question, it suffices to look at the hatched light blue expiration graph. Portfolio I s expiration looks like a collar. Portfolio II s expiration has the signature look of a straddle, which gains with volatility. Portfolio III s expiration looks like a strangle, and Portfolio IV s expiration has the increasing Z shape of a bull spread. (D) See Subsection 1.1.3 for diagrams of the payoffs at expiration of these option strategies. 41. [Subsection 1.1.1 and Section 13.3] I did this question using the method in the official solution s remark rather than the main method, although they re both equally difficult. Let V be the value of the contingent claim at time. The payoff is min S(1), 42 = S(1) max,s(1) 42. Discounting to time-, this becomes V = Se δ C (S, 42, 1) = Se.3 C (S, 42, 1). Another way to derive this is to realize that this contingent claim is a bull spread with strikes and 42, and a call with strike price is the stock itself. Delta for this contingent claim is the derivative of V with respect to S, or contingent claim = e δ e δ N (d 1 ), since we know that for the call is e δ N (d 1 ) (equation (1.2)). Now let s compute C (S, 42, 1) and N (d 1 ). d 1 = ln(45/42) +.7.3 +.5(.252 ) =.5697.25 N (d 1 ) = N (.5697) =.71259 d 2 =.5697.25 =.3197 N (d 2 ) = N (.3197) =.6229 C (S, 42, 1) = 45e.3 (.71259) 42e.7 (.6229) = 6.75746 The elasticity is Ω = S contingent claim V = 45 e.3 e.3 (.71259) 45e.3 6.75746 =.343 (C) 42. [Section 13.2] For H =, an up-and-out call is a standard European call. The special option of this question is equivalent to two up-and-in calls with barrier 7 minus one up-and-in call with barrier 8. An up-and-in call is equal to a standard call minus an up-and-out call. If we let C (H) be the value of an up-and-out call with barrier H, then the value V of the special option is V = 2 C ( ) C (7) C ( ) C (8) = C ( ) C (7) + C (8) = 4.861 2(.1294) +.7583 = 4.5856 (D)

SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 43 44 797 3 39.73 375 14.463 21 9 468.75 262.5 41.1 147 153 585.9375 328.125 183.75 116.25 12.9 197.1 Figure B.6: Binomial tree for Sample Question 44 43. [Lesson 22] The fastest way to do this is to use the formula for the Itô process of S a, equation (22.4), with a = 1. Here, δ = r. So we have dy (t ) y (t ) = ( 1)(r r ) +.5( 1)( 2)σ 2 dt + ( 1)σ dz (t ) = (r r + σ 2 )dt σ dz (t ) (E) 44. [Section 4.1] In this tree, u = 1.25 and d =.7 at all nodes, although that is not needed to solve the question. The risk-neutral probability at all nodes is p = 3e.1.65 21 375 21 =.61218 At the ending nodes, the put payoffs are, from top to bottom,,, 116.25, 197.1. The tree is shown in Figure B.6. Pulling back to the third column of nodes: P u u = P u d = e.1 (.389782)(116.25) = 41.1 P tentative d d = e.1 (.61218)(116.25) + (.389782)(197.1) = 133.723 At the d d node, the exercise value 3 147 = 153 is greater than the calculated value, so early exercise is optimal. Pulling back to the second column of nodes: P u = e.1 (.389782)(41.1) = 14.463 P tentative d = e.1 (.61218)(41.1) + (.389782)(153) = 76.5997 At the d node, the exercise value 3 21 = 9 is greater than the calculated value, so early exercise is optimal. Initially, P = e.1 (.61218)(14.463) + (.389782)(9) = 39.73. (D)

798 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 45 46 96 1.7284 11 115.2 3.2 3.2 8 3.7838 3.8721 86.4 1.4 1.4 72.4551.4551 Figure B.7: Binomial tree for Sample Question 46. At each node, the first number is the future price, the second number is the European option price, and the third number is the American option price. The American option price is boldface if it is optimal to exercise early. 64.8 45. [Section 12.3] We calculate gamma by calculating delta at the two Year 1 nodes and then dividing the change in delta by the change in the stock price. The formula for delta is equation (3.1). Although the letter C is used in that formula, it applies equally well to puts. The formula for gamma is equation (12.5). u = e.65 41.1 468.75 262.5 =.186278.65 41.1 153 d = e 262.5 147 =.9867 (.186278) (.9867) Γ = =.4378 (C) 375 21 46. [Section 4.4] For a futures contract, the dividend rate is set equal to the risk-free rate in the formulas, so p = (1 d )/(u d ). Therefore, u and d are computed as follows: 1 3 = 1 d u d = 1 d d /3 The binomial tree is shown in Figure B.7. The calculation of the second column is: 3 3d = d 3 9 = 1d d =.9 u = 1.2 C European u = e.25 1 3 (3.2) + 2 3 (1.4) = 1.7284

SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 47 49 799 C d = e.25 1 3 (1.4) =.4551 At the u node, the exercise value is 96 85 = 11, so it is optimal to exercise. This is the only node where early exercise is optimal. We could therefore answer the question without calculating the initial option values by discounting the difference of the top node: e.25 1 (11 1.7284) =.883. (E) However, the tree shows all 3 of the option values. 47. [Section 12.1] Since they give you the put option price, you can be pretty sure that put-call parity will be used. The initial cash flow to the female investor for selling 1 call options and buying 79.4 (1 ) shares of stock is 1(8.88) 79.4(4) = 2288 The final cash flow at the close of the position, which means buying 1 call options and selling 79.4 shares of stock is 1(14.42) + 79.4(5) = 2528 To calculate interest, we use put-call parity. Let t 1 be the initial time, t 2 the time of closing the position, and T the expiry time for the options. Then Initially: 8.88 1.63 = 7.25 = 4 K e r (T t1) or K e r (T t1) = 32.75 At closing time: 14.42.26 = 14.16 = 5 K e r (T t2) or K e r (T t2) = 35.84 Dividing the first equation into the second yields e r (t2 t1) = 35.84/32.75 = 1.9435. The investor s profit is 2528 2288(1.9435) = 24.12. (B) Student reports indicated that this question appeared on the Fall 27 exam. 48. [Lesson 2] There is no need to calculate k, although you could calculate k by equating Sharpe ratios. If you are interested in a solution using Sharpe ratios, see the first sample solution that comes with the sample questions. A risk-free portfolio must have a risk premium of. The risk premium is.6.4 =.2 for S 1 and.3.4 =.1 for S 2. Let x be the number of shares of Stock 2 to purchase..2(1).1(5x) = 2.5x = x = 4 (E) Student reports indicated that this question appeared on the Fall 28 exam. 49. [Lesson 3] The binomial tree will have 1 p = u = e.4(.25)+.3.25 = e.16 d = e.1.15 = e.14 1 1 + e σ h = 1 1 + e.15 (Equation (3.7)) Note that 1u = 1e.16 = 117.35 < 118. When the option pays off at both nodes, it is optimal to exercise early since the option no longer has any risk. Let s assume the K < 117.35 so that the option does not pay at the upper node. For optimal early exercise, we need the current payoff to be worth more than the discounted payoff at the lower node, or K 1 e.1 1 p K 1e.14 K 1e = e.1.14 1 + e.15

8 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 5 53 K e.1 1e.15 1 + e.15 (K 1) 1 + e.15 K e.1 1e.15 K 1 + e.15 e.1 1 1 + e.15 e.15 = 1 K 1 1 + e.15 e = 1.1.87658 = 114.86 so the answer is 115. (B) 5. [Section 7.2] The parameters of the lognormal ratio S 1/2 /S are m =.5.15.5(.35 2 ) =.44375 v =.35.5 =.247487 The upper bound of the 9% confidence interval for the normal distribution is.44375 + 1.64485(.247487) =.451454. The upper bound of the 9% confidence interval for the stock price after six months is.25e.451454 =.39265. (A) 51. [Section 8.1] A statistical calculator may help for this question. We calculate ln(s t /S t 1 ) for times t = 2 through t = 7. Month 2 3 4 5 6 7 S t /S t 1.3637.15415.13613.871.339.6669 ln(s t /S t 1 ) is assumed to follow a normal distribution. We estimate the mean of the normal as the sample mean µ = x =.233 and the standard deviation as the sample standard deviation (dividing by n 1 = 5) σ =.1354. These are per month. The monthly return is therefore α/12 = µ+.5σ 2 =.233+.5(.1354 2 ) =.2839. The annual return is α = 12(.2839) =.346. (E) You can also reverse the order of these operations: first annualize µ (multiply by 12) and σ (multiply by 12) and then calculate α = µ +.5σ 2. It is strange that the answer is so far out of the ranges; they were thinking you d forget to add.5σ 2. 52. [Section 15.3] For 2 years, the mean and variance of the lognormal are m = 2.15.5(.3 2 ) =.21 v =.3 2 =.4243 Using the inversion method, the standard normal random numbers are N 1 (.983) = 2.12, N 1 (.384) = 1.77, and N 1 (.7794) =.77. The resulting ratios of S 1 /S are n i m + n i v e m+n i v 2.12 1.194 3.327 1.77.541.5822.77.5367 1.713 The average, multiplied by S = 5, is 5(3.327 +.5822 + 1.713)/3 = 88.75. (C) 53. [Section 14.1] We can express the gap option as a standard European call with strike price 4 minus a cash-or-nothing option paying 5 if the stock price is above 4. Now the two properties of gamma to use are: (i) Gamma is a linear function of the options, since it is a second derivative with respect to the stock price,

SOLUTIONS TO SAMPLE QUESTIONS, QUESTION 54 81 (ii) so a linear combination of the two options will have a gamma that is the linear combination of the gammas. Gamma for a put equals gamma for the corresponding call, the one with the same strike price and expiry. Thus gamma for a call plus gamma for a cash-or-nothing of 5 equals gamma for the gap call..8 + gamma cash-or-nothing =.7 gamma cash-or-nothing =.1 For cash-or-nothing of 1, gamma is (1/5)(.1) = 2. (B) Alternatively, write down all the information using our all-or-nothing notation: (ii) & (iv): 4 S < 4 S S < 4 Γ =.7 (iii) & (v): S S > 4 45 S > 4 Γ =.8 (*) and we need gamma for 1 S > 4. Since we need the condition S > 4, we will replace the first equation above with a call equation: S S > 4 4 S > 4 Γ =.7 (**) Subtracting (*) from (**), we get 5 S > 4 has Γ =.1, so 1 S > 4 has Γ = 2(.1) = 2. (B) 54. [Sections 1.2, 13.3, and 14.3] We want the value of an option with payoff max 17 min 2S 1 (1),S 2 (1),, since the option holder will sell the cheapest stock for 17. Let M be the value at time 1 of min 2S 1 (1),S 2 (1). Then the option given in (vi) call its value V 1 pays max(m 17, ) and the option we want to value call its value V 2 pays max(17 M, ). Since one option pays exactly when the other one doesn t, the difference, or V 1 V 2, is equal to the present value of M 17. The present value of 17 is 17e.5. The present value of M is an option paying the minimum of 2S 1 and S 2, which is S 2 C (S 2, 2S 1 ), where C (S 2, 2S 1 ) is an exchange option which allows one to buy S 2 in exchange for 2S 1. Let s value this exchange option. The volatility of the difference in returns between the two stocks is the square root of σ 2 1 + σ2 2 2ρσ 1σ 2 =.18 2 +.25 2 2(.4)(.18)(.25) =.139 and.139 =.3618. By Black-Scholes, since both dividend rates are, ln(2/2) +.5σ2 d 1 = =.5(.3618) =.189 σ d 2 =.189.3618 =.189 N (d 1 ) = N (.189) =.57178 N (d 2 ) = N (.188) =.42822 C (S 2, 2S 1 = 2(.57178.42822) = 2.87126 so the present value of M is 2 2.8712 = 17.1288. Then V 1 V 2 = 17.1288 17e.5 1.632 V 2 = 17.1288 17e.5 =.9579 V 2 = 1.632.9579 =.6741 (A)

82 SOLUTIONS TO SAMPLE QUESTIONS, QUESTIONS 55 56 55. [Section 9.3] We must back out σ. For an at-the-money future, d 1 = d 2. Therefore, 1.625 = 2e.75(.1) N ( d 2 ) N ( d 1 ) = 2e.75 N (d 1 ) 1 N (d 1 ) = 2e.75 2N (d 1 ) 1 1.625e.75 2N (d 1 ) = + 1 = 1.87578 2 N (d 1 ) =.54379 d 1 =.1999 =.5σ 3/4 σ =.21998 3/4 =.25411 Now we can value the option three months later using the Black formula. With six months to expiry: d 1 = ln(17.7/2) +.5(.254112 )(.5).25411.5 d 2 =.5937.25411.5 =.76998 N ( d 1 ) = N (.5937) =.72253 N ( d 2 ) = N (.76998) =.77934 =.5937 P(F, 2,.5) = 2e.5(.1) (.77934) 17.7e.5(.1) (.72253) = 2.66156 (D) 56. [Sections 7.1 and 13.1] Nothing about this question requires knowledge of average strike options. You just have to evaluate the variance of A(2). The stock price is 5 times a lognormal random variable with parameters m =.5.5(.2 2 ) t =.3t and v =.2 t. Therefore, A(2) = 1 2 S(1) +S(2) = S() 2 S(1) S() + S(2) S() While S(1)/S() and S(2)/S() are not independent because the periods (, 1) and (, 2) overlap, the variables X = S(1)/S() and Y = S(2)/S(1) are independent, so we set S(2)/S() = X Y, and A(2) = 5 2 (X + X Y ) where X and Y are independent lognormal random variables with parameters m =.3 and v =.2. We will use the two alternative methods as the official solution goes through: (1) calculating first and second moments, and (2) calculating variance of a sum directly. First let s calculate the variance as the second moment minus the first moment squared. We can hold off the 5/2 until the end. E[X + X Y ] = E[X ] + E[X Y ] = e.5 + e.1 E[(X + X Y ) 2 ] = E[X 2 ] + E[(X Y ) 2 ] + 2 E[X 2 Y ] X and Y are independent, so the expectations may be factored. Also, E[Z 2 ] = e 2m+2v 2 for a lognormal Z. So E[X 2 ] = e 2(.3)+2(.22) = e.14 E[(X Y ) 2 ] = E[X 2 ] E[Y 2 ] = e.14 e.14 = e.28 E[X 2 Y ] = E[X 2 ] E[Y ] = e.14 e.5 = e.19 (*)