Suppose that a stock price $S$ follows Geometric Brownian Motion with expected return $\mu$ and volatility $\sigma:$
$$dS = \mu S dt +\sigma S dz$$
How to find out the process followed by variable $S^n$?
How to prove that $S^n$ also follows geometric brownian motion?
The expected value of $S_T,$ the stock price at time $T,$.is $Se^{\mu(T-t)}$.
What is the expected value of $S^n_T$?
Answer:- The expected value of $S^n_T$ is $S(t)^n e^{[n(r-\delta)+\frac12 n^2\sigma^2]T}$
But i found tha answer in some study material on internet as $S(t)^n e^{[n(r-\delta)+\frac12 n(n-1)\sigma^2]T}$
Would anyone explain me why the difference occurred between my answer and answer provided by study material on internet? r is risk-free interest rate.$\delta$ is dividend yield on the stock. $S(t)=e^{Y(t)}$
For a geometric Brownian motion ${S_t}$, the expected value of the process at time t given the history of the process up to time s, for $s < t$
$E[S{(t)}|S{(u)}, 0\leq u \leq s]=S(s)E[e^{Y(t)-Y(s)]}$
Now the mgf of a normal random variable $W$ is given by
$E[e^{nW}]=\exp[nE(W)+n^2 Var(W)/2]$
Hence, since $S(t)-S(s)$ is normal with mean $(r-\delta)(t-s)$ and variance $\sigma^2 (t-s)$ it follows that
$E[e^{S(t)-S(s)}]=e^{n(r-\delta)+\frac12(t-s)n^2\sigma^2}$
Thus we will get final answer to expected value of $S^n_t$
$E[S(t)|S(u),0\leq u \leq s]= E[e^{Y(t)}|Y(u),0\leq u \leq s]$
$L.H.S=E[e^{Y(s)+Y(t)-Y(s)}|Y(u),0\leq u\leq s]$
$L.H.S.=e^{Y(s)}E[e^{Y(t)-Y(s)}|Y(u),0\leq u\leq s]$
$L.H.S.=S(s)E[e^{Y(t)-Y(s)}]$