Question
Question 1 Consider the binomial model with T = 3, u = .2, d = .4, r = .05, p = .8, S0 = 100,
Question 1
Consider the binomial model with T = 3, u = .2, d = .4, r = .05, p = .8, S0 = 100, and the following portfolio: begin with 10 shares of stocks, and no bonds; if the stock prices become larger than 110, trade all stocks for bonds and wait until the terminal time; otherwise no change until the terminal time. (a) Express 1, 1. (b) Express 2, 2 in our notations in the in the path-space point of view. That is, find (2(1), 2(1)) = ( 1 = 1 = . (c) Express 3, 3 accordingly (you should have four different cases to discuss here).
Question 2
Consider {Xn}nN i.i.d. random variables such that P(X1 = 1) = 1/2. Consider Mn := (X1 + + Xn) 2 n, n N, and Mn = 0. Show that E(Mn+1 | M0, . . . , Mn) = Mn, n N0. That is {Mn}nN0 is a martingale.
Question 3
. Consider the following so-called urn model. There is an urn (box) with w white balls and b black balls at the beginning, with w, b > 0. At each round, one takes a ball from the urn randomly, put back the same ball, and put back another ball with the same color. The balls with the same color are indistinguishable. For example, the first pick has chance w/(w + b) to be white, and in this case, after first round there will be w + 1 white and b black balls in the urn. Let Xn denote the proportion of white balls in the run right after the first n rounds, and X0 := w/(w + b). Let Yn denote the number of white balls right after the first n rounds of sampling. A key observation is that P(Yn+1 = Yn + 1 | X1, . . . , Xn) = P(Yn+1 = Yn | Xn) = Xn. Note also that E(Xn+1 | X1, . . . , Xn) = E(Xn+1 | Xn). (a) Compute P(Y2 = w), P(Y2 = w + 1), P(Y2 = w + 2). (b) Find a relation between Yn and Xn. (c) Show that E(Xn+1 | X1, . . . , Xn) = Xn, for all n = 0, 1, 2, . . . . (The above tells that {Xn}nN0 is a martingale.)
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