UNIVERSITY OF OSLO. Faculty of Mathematics and Natural Sciences

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1 UNIVERSITY OF OSLO Faculty of Mathematics and Natural Sciences Examination in MAT2700 Introduction to mathematical finance and investment theory. Day of examination: November, Examination hours:??.????.?? This problem set consists of 7 pages. Appendices: Permitted aids: None. None. Please make sure that your copy of the problem set is complete before you attempt to answer anything. Problem 1 Consider the following three period model with 10 scenarios, one stock and one bond, initially priced as B 0 = 1 and an interest rate of r = 0.1. S 1 = 90 S 2 = 70 S 2 = 110 S = 60 ω 1 S = 80 ω 2 S = 100 ω S = 10 ω 4 S 0 = 100 S 2 = 100 S = 90 ω 5 S = 120 ω 6 S 1 = 120 S 2 = 110 S = 100 ω 7 S = 140 ω 8 S 2 = 140 S = 120 ω 9 S = 160 ω 10 1a Find the equivalent martingale measures for this model and characterize those derivatives which are attainable. (Continued on page 2.)

2 Examination in MAT2700, November, Page 2 Possible solution: The conditional risk neutral probabilities are q u = 1/ q d = 2/ q uu = 11/40 q ud = 29/40 q du = λ q dm = 2/5 2λ q dd = /5 + λ q uuu = /20 q uud = 17/20 q udu = /10 q udd = 7/10 q duu = 1/ q dud = 2/ q dmu = 1/5 q dmd = 4/5 q ddu = /10 q ddd = 7/10, where λ is in the interval (0, 1/5). This gives an equivalent martingale measure Q given by Q(ω 1 ) = Q(ω 2 ) = 187 Q(ω ) = Q(ω 4 ) = 20 Q(ω 5 ) = λ 2 Q(ω 6 ) = λ 4 Q(ω 7 ) = λ Q(ω 8 ) = λ Q(ω 9) = λ Q(ω 10) = λ. For any derivative with payoff D i = D(ω i ) we have E Q (D) = D D D + 20 D D D D D 10 ( 2 + λ 9 D D D D D D 10 The attainable derivatives are those for which the last term in brackets is zero. ) 1b Find the range of fair prices at t = 0 of the put option with payoff max {100 S, 0}. Possible solution: For this put option we have D 1 = 40, D 2 = 20, D = 0, D 4 = 0, D 5 = 10, D 6 = 0, D 7 = 0, D 8 = 0, D 9 = 0, D 10 = 0. Hence any fair price is given by E Q ( D) = ( ) 9 λ. (Continued on page.)

3 Examination in MAT2700, November, Page Therefore πd b = 1 ( π a D = ( 40 ) ) The range of fair prices is the interval (π b D, πa D ). 1c Let D be the derivative with payoff at t = defined by { 1 i = 4 or i = 5, D(ω i ) = 0 otherwise, at define D t to be its conditional expectation, i.e., D t = E Q [ D Ft ], where F t is the filtration defined by stock price. Find D 1. Possible solution: Since D 1 is measurable with respect to F 1 it takes a value at the node u and another at the node d. These values are given by D 1 (u) = 1 1 D 1 (d) = i=1 10 i=5 D(ω i )Q(ω i ) = Q(ω 4 ) = D(ω i )Q(ω i ) = λ = λ. Problem 2 Consider the following binary three period model with one stock. (Continued on page 4.)

4 Examination in MAT2700, November, Page 4 S 2 = 14 S = 16 ω 1 S 1 = 12 S = 12 ω 2 S 2 = 10 S = 12 ω S 0 = 10 S = 9 ω 4 S 2 = 11 S = 1 ω 5 S 1 = 9 S = 10 ω 6 S 2 = 7 S = 8 ω 7 S = 6 ω 8 Assume that the interest rate is 0 and that the initial bond price is B 0 = 1. 2a Show that the model is complete and find the unique equivalent martingale measure Q. Possible solution: The risk neutral conditional probabilities are q u = 1/ q d = 2/ q uu = 1/2 q ud = 1/2 q du = 1/2 q dd = 1/2 q uuu = 1/2 q uud = 1/2 q udu = 1/ q udd = 2/ q duu = 1/ q dud = 2/ q ddu = 1/2 q ddd = 1/2. From this we get the equivalent martingale measure Q(ω 1 ) = 1 12 Q(ω 2 ) = 1 12 Q(ω ) = 1 18 Q(ω 4 ) = 1 9 Q(ω 5 ) = 1 9 Q(ω 6 ) = 2 9 Q(ω 7 ) = 1 6 Q(ω 8 ) = 1 6. The model is viable and complete since we have a unique equivalent martingale measure. (Continued on page 5.)

5 Examination in MAT2700, November, Page 5 2b Let X be the random variable defined by 1 i = 1, X(ω i ) = 1 i = 5, 0 otherwise, and let F t be the filtration defined by the stock prices. For which t is X measurable with respect to F t? Possible solution: X is measurable with respect to F t iff it takes a single value on each t-node. This is the case only for t =. 2c Find the conditional expectation Y t = E Q { X F t } for t = 0, 1. Possible solution: We have that also Y 0 = E Q (X) = Q(ω 1 ) Q(ω 5 ) = = 1 6, Y 1 (u) = 1 1 Q(ω 1 ) = 1 4, Y 1 (d) = 1 2 ( Q(ω 5 )) = d Find the fair price of the European call option with payoff max {S 10, 0}. Possible solution: We find the payoff of the derivative for each scenario, 6, i = 1, 2 i = 2,, D(ω i ) = 0 i = 4, 6, 7, 8, i = 5. Hence the fair price is D 0 = E Q (D) = 6 1 ( ) = (Continued on page 6.)

6 Examination in MAT2700, November, Page 6 2e Assume that the real world probability of ω i is P(ω i ) = 1/8 for each i = 1,..., 8. Given an initial wealth V 0, find a trading strategy Φ = {(x t, y t )} t=0 which maximizes the expected utility of the wealth at t =, u(v Φ ), for the utility function u(v) = ln(v). Possible solution: We use the martingale method to find this trading strategy. Since the utility function is the logarithm, the optimal derivative is /2 i = 1, D(ω i ) = V 0 P(ω i ) Q(ω i ) = V 0 /2 i = 2, 9/4 i =, 9/8 i = 4, 9/8 i = 5, 9/16 i = 6, /4, i = 7, /4 i = 8. To save ink, all numbers, i.e., the strategy and the values, are in multiples of V 0. We must find a portfolio for each of the four submodels at uu, ud, du and dd. At uu we must solve the equations This gives Similarly for the other 2-nodes For the 1-nodes we get At the root node we must solve x + 16y = 2 x + 12y = 2. Φ 2 (uu) = (/2, 0), V Φ 2 (uu) = /2. Φ 2 (ud) = ( 9/4, /8), V Φ 2 (ud) = /2, Φ 2 (du) = ( 21/16, /16), V Φ 2 (du) = /4, Φ 2 (dd) = (/4, 0), V Φ 2 (dd) = /4. Φ 2 (u) = (/2, 0), V Φ 1 (u) = /2, Φ 2 (d) = (/4, 0), V Φ 1 (d) = /4. x + 12y = /2, x + 9y = /4, which gives Φ 1 = ( /2, 1/4), V Φ 0 = 1. (Continued on page 7.)

7 Examination in MAT2700, November, Page 7 THE END