FE 610 hw1_step 2

FE 610 hw1_step 2 - state one because is the stock price is...

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Class: FE 610 Problem 1- I- The matrix as defined for the problem is as follows = Step 1= The set up of the initial matrix is like this: [1] = [1+r 1+r] [S(t)=36.2] [36 40] [C(t)] [2 0] [D(t)] [0 2] As shown in he examples, the first row shows that the bond price (1) = 1+ the LIBOR rate in both states of the world. The second row show the stock price (S(t)=36.2) and the two states of the world where the price can go to either 36 in state 1 or 40 in state 2. The put option shows the price of the put option C(t) = 2 in state 1. This 2 was derived because the strike price is 38 and in state 1 the price of the stock goes to 36 and we have a 2 dollar profit, making it 38-36=2. In state 2 it is assume that the option is not exercised and the price goes to 0. The last line of the matrix shows that the call option D(t) = 0 in
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Unformatted text preview: state one because is the stock price is lower than the strike price there is no need to exercise that option. In state 2 we show that the D(t)= 2 because if the strike price is 38, and the stock goes to 40 we make $2 profit. Now if we set a separate vector L as follows: [1] = [1+.052632 1+.052632] [L1] [36.2] [36 40] [L2] [C(t)] [2 0] [D(t)] [0 2] As shown in An Introduction to the Mathematics of Financial Derivatives by Neftci, if we can solve for vector L as positive probabilities, which we can in this problem because the matrices are solvable then there is no arbitrage possibilities in the system. [page 19] p1 = .44996 , p2= .500003 II- P1=p1(1+.052632), P2=p2(1+.052632), so the risk adjusted probabilities P1 and P2 are .473642 and .526348...
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This note was uploaded on 09/01/2011 for the course FE 610 taught by Professor Prasad during the Summer '10 term at Stevens.

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