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: Monday, December 14, Examination hours: 09:00 1:00. This problem set consists of 5 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 model with three scenarios and two stocks and one bond with interest rate r. S 1 1 = 2 S 2 1 = 1/2 ω 1 S 1 0 = 1 S 2 0 = 1 S 1 1 = S 2 1 = 1 ω 2 S 1 1 = 0 S 2 1 = ω B 0 = 1 B 1 = 1+r 1a Show that the model is not viable if r = 0, and find an arbitrage opportunity in this case. Possible solution: An implicit arbitrage opportunity is found by solving y S(ω i ) 0 for i = 1, 2,. These equations are y y 2 0, 2y 1 0, y 1 + 2y 2 0. (Continued on page 2.)

2 Examination in MAT2700, Monday, December 14, Page 2 One solution is z = y 1 = y 2 > 0, which gives the arbitrage opportunity ϕ = ( 2z, z, z) with V ϕ 1 (ω 1) = z/2, V ϕ 1 (ω 2) = 2z, V ϕ 1 (ω ) = z. Since we have arbitrage opportunities, the model is not viable. 1b Show that the model is viable and complete for r 0 < r < r 1, and find r 0 and r 1. Possible solution: The model is viable and complete if we have a unique risk free probability measure Q = (q 1, q 2, q ), the equations for Q are Solving this equation, we get q q 2 = 1 + r q 1 + r q 1 = 8 10r 7, q 2 = 9r, q = r For this to be a probability measure, we must have 0 < q i < 1 for i = 1, 2,. For i = 1, this gives 1/10 < r < 4/5, for i = 2, 1/ < r < 10/9, and for i =, 2 < r < 5. Therefore the model is viable and complete if r 0 = 1 < r < 4 5 = r 1. 1c In the rest of this exercise we assume that r = 5/8, then the model is viable and complete (you do not have to show this). Find the fair price of the indicator derivatives D j, with payoff at t = 1 { D j 1 i = j, (ω i ) = for j = 1, 2,. 0 i j, Possible solution: measure is If r = 5/8, then the unique risk neutral probability ( 1 Q = 4, 8, ). 8 The fair price of D j is E Q ( D j ) = q j /(1 + r), i.e., E Q ( D 1 ) = = 2 1, E Q( D 2 ) = = 1, E Q( D ) = = 1. (Continued on page.)

3 Examination in MAT2700, Monday, December 14, Page 1d Suppose a third stock was valued at t = 1 as follows S 1(ω 1 ) = 1, S 1(ω 2 ) = 2, S 1(ω ) =. What must its price be at t = 0 for the model with three stocks to be viable? Possible solution: We can view S as a derivative, and the unique fair price is the one which excludes arbitrage, hence S 0 = E Q ( S 1) = = 17 1 Problem 2 Consider the following binary two period model with one stock and one bond. S 2 = 140 ω 1 S 1 = 120 S 0 = 100 S 2 = 100 ω 2 S 1 = 90 S 2 = 110 ω S 2 = 70 ω 4 Assume that the interest rate is 0 and that the initial bond price is B 0 = 1. 2a Find the unique equivalent martingale measure Q. Possible solution: The risk neutral conditional probabilities are q uu = 1/2 q u = 1/ q ud = 1/2 q d = 2/ q du = 1/2 q dd = 1/2 From this we get the equivalent martingale measure Q(ω 1 ) = 1 6 Q(ω 2 ) = 1 6 Q(ω ) = 1 Q(ω 4 ) = 1. The model is viable and complete since we have a unique equivalent martingale measure. (Continued on page 4.)

4 Examination in MAT2700, Monday, December 14, Page 4 2b Let X be the random variable defined by X(ω 1 ) = 0, X(ω 2 ) = 1, X(ω ) = 1, X(ω 4 ) = 0. and let F t be the filtration defined by the stock prices. List F 0, F 1 and F 2. For which t is X measurable with respect to F t? Possible solution: Set Ω = {ω 1, ω 2, ω, ω 4 }, Ω u = {ω 1, ω 2 }, Ω d = {ω, ω 4 }, then F 0 = {, Ω}, F 1 = {, Ω, Ω u, Ω d }, F 2 = all subsets of Ω. 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 = 2. 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 2 ) Q(ω ) = = 1 6. Y 1 (u) = 1 1 Q(ω 2 ) = 1 2, Y 1 (d) = 1 2 ( Q(ω )) = d Find the fair price of the European call option with payoff max {S 2 100, 0}. Possible solution: We find the payoff of the derivative for each scenario, Hence the fair price is D(ω 1 ) = 40, D(ω 2 ) = 0, D(ω ) = 10, D(ω 4 ) = 0. D 0 = E Q (D) = ( 40 1 ) = 10. (Continued on page 5.)

5 Examination in MAT2700, Monday, December 14, Page 5 2e Assume that the real world probability of ω i is P(ω 1 ) = 1/6, P(ω 2 ) = 1/, P(ω ) = 1/, P(ω 4 ) = 1/6. Given an initial wealth V 0, find a self financing trading strategy Φ = {(x t, y t )} 2 t=1 which maximizes for the utility function u(v) = ln(v). E P [ u(v Φ 2 ) ], given that V Φ 0 = V 0, Possible solution: We use the martingale method to find this trading strategy. Since the utility function is the logarithm, the optimal derivative is 1, i = 1, D(ω i ) = P(ω i) Q(ω i ) V 2, i = 2, 0 = V 0 1, i =, 1/2, i = 4. We must find a portfolio for each of the two submodels at u, d. At u we must solve the equations This gives x + 140y = V 0 x + 100y = 2V 0. x 2 (u) = (9/2)V 0, y 2 (u) = (1/40)V 0, V Φ 2 (u) = (/2)V 0. Similarly for the node d x 2 (d) = (/8)V 0, y 2 (d) = (1/80)V 0, V Φ 2 (d) = (/4)V 0. The equations for the root node are x + 120y = (/2)V 0 x + 90y = (/4)V 0, giving x 1 = (/2)V 0, y 1 = (1/40)V 0, V Φ 1 = V 0. THE END