1. 2 marks each True/False: briefly explain (no formal proofs/derivations are required for full mark).

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1 The University of Toronto ACT460/STA2502 Stochastic Methods for Actuarial Science Fall 2016 Midterm Test You must show your steps or no marks will be awarded 1 Name Student # 1. 2 marks each True/False: briefly explain (no formal proofs/derivations are required for full mark). (a) An arbitrage-free market has one risky asset and a money market account. Any contingent claim payable at time n can be perfectly replicated by a self-financing portfolio of these two securities, if the risky asset follows a n-period binomial model. Ans: T (b) Consider a market that has one risky asset and a money market account. The risky asset is modelled by a 4-period annual time step recombining binomial tree with S 0 = 100,u = 1.3,d = 1.1. The continuously compounding annual interest rate r is given by r = ln(1.05). This market does not admit arbitrage. Ans: F d = 1.1 > e r = e ln(1.05) = (c) The Delta of a European put option is positive. Ans: F Keeping others fixed, Put price decreases in stock price.

2 2 (d) An American call option on a non-dividend paying stock is always exercised at the expiration date. Ans: T (e) The put-call parity holds for American put and call options with the same strike price. Ans: F Put-call parity holds for European type. (f) The annual insurance charges of a variable annuity as a percentage of the subaccount value vary from year to year. Ans: F Usually it is a fixed percentage. (g) If {W t,t 0} denotes a standard Brownian motion, then {e σwt,t 0} is a martingale. Ans: F Easy to check using the definition. (h) The mean of an Ito integral depends on its integrand. Ans: F Ito integral has zero mean.

3 3 2. You are given the following VBA codes for computing an option price. In this function, mu and sigma are the annualized drift and volatility, period represents the duration of the option contract and δ t is the time step. Function Option_price(mu, sigma, r, period, delta_t, K, S) r interest rate (annual continuously compounding) K strike price S initial stock price ********* represents hidden codes up = Exp(Sqr(muˆ2*delta_tˆ2+sigmaˆ2*delta_t)) down = ********* q_up = ******** q_down = 1-q_up Option_price = 0 N = period/delta_t For Index = 0 To N Option_price = Option_price + _ Application.Combin(N, Index) * q_up ˆ Index * q_down ˆ (N - Index) * _ (Application.Max(S * (up ˆ Index) * (down ˆ (N - Index)) - K, 0) _ +2 * Application.Combin(N, Index) * q_up ˆ Index * q_down ˆ (N - Index) * _ (Application.Max(0.5*K - S * (up ˆ Index) * (down ˆ (N - Index)), 0) Next Index Option_price = Exp(-period * r) * Option_price End Function

4 4 a) 1 marks According to the expression for up, what is the underlying real world assumption in this Binomial tree model. Ans: ud=1. b) 2 marks Recover the VBA codes for down and q up. down = Exp(-Sqr(muˆ2*delta_tˆ2+sigmaˆ2*delta_t)) q_up = (Exp(r*delta_t)-down)/(up-down) c) 2 marks Sketch the payoff (as a function of S T ) of this option.

5 5 3. You are building a binomial model for stock price S t,s 0 = 100, over a three-years period. You are using 1 year (δt = 1) as its time step and for each period the probabilities of price going up and going down are equal under the real world probability measure. As defined in the lecture notes, the expected annual log-return µ = 0.10 and the annualized volatility on the log-price σ = a) 5 marks Determine the values of u and d, the total returns of the stock in one period when the price goes up and goes down, respectively. Ans: u = e µδt+σ δt = e = d = e µδt σ δt = e =

6 6 b) 5 marks Determine the real world probability that the stock price is greater than 110 at the end of three years. Ans: Accordingtothecomputation,theterminalstockpricesatdifferentstatesareS 3 = , , and Therefore the probability of S 3 > 100 is 3(1/2) 3 +(1/2) 3 = 1/2

7 7 c) 5 marks An at the money European put option on the stock expires in three years. Assume that the interest rate r compounded continuously is 0. Determine the option price at time 0. Ans: The risk neutral probabilities The Put option price is q u = e = q d = 1 q u = e 3r ( )q 3 d =

8 8 d) 5 marks If you use a replicating portfolio to hedge the option, determine the number of shares of the underlying stock in the portfolio at time 0. Ans: First compute the discounted expected payoff at t = 1. The number of shares in stock at time 0 can be computed as =

9 9 4. The price S t of a stock is assumed to follow the process S t = 100e µt+0.4 W t, t 0. Here W t denotes a standard Brownian motion under the risk-neutral probability measure Q, and µ is a constant. The interest rate r is 8% compounded continuously. a) 5 marks Today is t = 0. Find the risk-neutral probability that three months from now, the stock will have outperformed the money market account. Ans: Let W t denote a Brownian motion under risk-neutral measure Q. Then under the risk-neutral measure, the stock price dynamic should satisfy This implies ds t /S t = rdt+0.4dw t S t = S 0 e (r 0.08)t+0.4dWt The risk-neutral probability that stock will outperform the m.m. account is Q(S 3 /S 0 > e 3r ) = Q(0.4W 3 > 0.24) Where W 3 N(0,3) under Q. Q(0.4W 3 > 0.24) = Q(Z > 0.6/ 3) =

10 10 b) 5 marks Express S t as a solution of a SDE under Q. Ans: As previously stated S t = 100e (r 0.08)t+0.4dWt

11 11 5. The price S t of a stock is assumed to follow the stochastic process ds t = 0.2S t dt+e 0.5t S t dw t, S 0 = 100. Here W t denotes a standard Brownian motion under the real world probability measure P, and µ and σ are positive constants. t is measured in years. Today is t = 0. a) 5 marks Solve the equation for S t. Ans: According to Ito s lemma dlns t = ds t S t ds tds t 2S 2 t Substitute the expression for ds t and ds t ds t = e t S 2 tdt we get Integrate both sides from s = 0 to s = t get Therefore Where S 0 = 100. dlns t = 0.2dt+e 0.5t dw t e t 2 dt = (0.2 e t /2)dt+e 0.5t dw t lns t lns 0 = 0.2t 1 e t 2 + t S t = S 0 e 0.2t 1 e t 2 + t 0 e 0.5s dw s 0 e 0.5s dw s

12 12 b) 5 marks Determine the mean E(S 2 ) of S 2. Ans: Note that t 0 e 0.5s dw s N(0, t 0 e s ds), hence E(S 2 ) = S 0 e e 2 2 E(e t 0 e 0.5s dw s ) = S 0 e e 2 2 e 1 e 2 2 = S 0 e 0.4 = Another way is taking expectation on both sides of SDE ds t = 0.2S t dt + e 0.5t S t dw t to get E(dS t ) = de(s t ) = 0.2E(S t )dt. Solve the ODE for f t = E(S t ) with f 0 = 100.