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
UNIVERSITY OF OSLO. Faculty of Mathematics and Natural Sciences
UNIVERSITY OF OSLO Faculty of Mathematics and Natural Sciences Examination in MAT2700 Introduction to mathematical finance and investment theory. Day of examination: November, 2015. Examination hours:??.????.??
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