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Unformatted text preview: MATH 238 WINTER 2009 PROBLEM SET 1 - SOLUTIONS Problem 1 : Let S be the current stock price, K the strike price of the option, T the expiration time of the option, t the current time, S T the stock price at time T , r the risk-free interest rate, c the price of a European call option and p the price of a European put option. Explain why the two portfolios (a) one European call option plus cash equal to Ke- r ( T- t ) (b) one European put option plus one share have the same payoff at time T : max( S T ,K ). Deduce that the value of the portfolios today must be the same, so that (1) c + Ke- r ( T- t ) = p + S Explain this put-call parity relation with figures. (Ch. 8, Hull.) Solution : At any time t ≤ T , we are considering the two portfolios (a): one European call option plus an amount of Ke- r ( T- t ) cash put in the bank (b): one European put option and one share of the underlying asset S t . Then the value of portfolio (a) at time T equals c T + K = max( S T- K, 0) + K = max( S T ,K ) while the value of portfolio (b) at time T equals p T + S T = max( K- S T , 0) + S T = max( S T ,K ) Therefore both portfolios have indeed the same payoff max( S T ,K ) at the expiration time T of the options. We now claim that this implies that the values of these portfolios today, i.e. at any time t ≤ T , must be the same, so that we have the put-call parity relation c t + Ke- r ( T- t ) = p t + S t for all t ≤ T . One way to see this is from the assumption that there is no arbitrage which we now present. Assume that at some time t < T portfolio (a) is cheaper than portfolio (b) (i.e. c t + Ke- r ( T- t ) < p t + S t ). Then we buy (a) (i.e. we buy a call option and we put an amount of Ke- r ( T- t ) in the bank ) and sell (b) (i.e. write a put option and short sell a share of the stock). This creates a positive cash flow at time t, and no other flow afterwords (i.e. an arbitrage). To see this assume that at time T it happens that S T < K . The call option is not 1 2 MATH 238 WINTER 2009 PROBLEM SET 1 - SOLUTIONS S T K-K Payoff Figure 1. Payoff: One Call minus one Put equals the Stock- K exercised. The cash in the bank is now worth K and we use it to buy the one share of the stock (for K) from the person to whom we wrote the put option. Finally we take this one share and return to the broker who made the initial short sale possible. Similarly if S T ≥ K we are able at time T to fulfill all our obligations with zero cost. If portfolio (b) is cheaper than (a) we buy (b), sell (a) and we show that this is again an arbitrage. Mathematically, we simply have to take conditional expectations of the discounted payoffs in the equation e- r ( T- t ) max( K- S T , 0)+ Ke- r ( T- t ) = e- r ( T- t ) max( K- S T , 0)+ e- r ( T- t ) S T with respect to the risk neutral probability law, given S t = S . This gives C ( t,S ) + Ke- r ( T- t ) = P ( t,S ) + S for t < T . We have used here the fact that the discounted price of the risky asset is a martingale under the risk neutral probability:...
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This note was uploaded on 11/08/2009 for the course MATH 238 taught by Professor Papanicolaou during the Winter '08 term at Stanford.
- Winter '08