Question

$S(n)\text{ is a stock price where } K(1)=u \text{ with probability } p \\ \text{ and } K(1)=d \text{ with probability } 1-p \text{ and } K(1)...K(n) \text{ are independent identically distributed one step returns.}$

5. Consider a binomial tree model for a stock price, S(n) as above. Find a probability value p, in the case when the risk free assest has a continuous compounding rate of r. What are the bounds on e', that is, what is the smallest and largest value it can be in terms of u and d which prevent arbitrage?

Answer 1

"To achieve the no of arbitrage condition, we need to improve the marginal prices followed by the stock prices therefore if we observe the prices one step ahead then it will have ""2"" possible outcomes either ""u or d"" E(s_0) = E(s(1))/(1 + r) therefore E(s_0) = p cl s_0 + (1 - p) d s_0/e^r [e^r almostequalto 1 + v for r rightarrow 0] As at t = 0 E(s_0) = s_0 therefore s_0 = (p u + d - pd)/e^r s_0 therefore 1 = pu + d - pd/e^r e^r = pu + d - pd e^r - d = p(u - d) p^* = e^r - d/u - d To prevent arbitrage we need to achieve risk neutral probability p^* A that should be p^* elementof[0, 1] if u > d: e^r > d u > e^r > d"