# Given $S$ is a Geometric Brownian Motion, how to show that $S^n$ is also a Geometric Brownian Motion?

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 $S^n_t$

$E[S(t)|S(u),0\leq u \leq s]= E[e^{Y(t)}|Y(u),0\leq u \leq s]$

$L.H.S=E[e^{Y(s)+Y(t)-Y(s)}|Y(u),0\leq u\leq s]$

$L.H.S.=e^{Y(s)}E[e^{Y(t)-Y(s)}|Y(u),0\leq u\leq s]$

$L.H.S.=S(s)E[e^{Y(t)-Y(s)}]$

• do you know ito's lemma? – Gordon Jan 18 '17 at 14:51
• Yes, I know Ito's lemma.I have also studied how to derive black-scholes-merton formula using ito's lemma. But i think this question relate to multi-variate ito's lemma. – Dhamnekar Winod Jan 18 '17 at 14:57
• Then why can not apply ito's lemma to $S^n$? – Gordon Jan 18 '17 at 14:59
• Please show how you get your answer and then we can assess whether your approach is fine. The answer from the internet is not important – Gordon Jan 26 '17 at 14:24
• Why $E[S{(t)}|S{(u)}, 0\leq u \leq s]=S(s)E[e^{S(t)-S(s)]}$? – Gordon Jan 26 '17 at 15:10

As stock price $S$ follows Geometric Brownian motion,we can use Ito's lemma to determine the process followed by $S^a$. We obtain

$dS^n=nS^{n-1}dS + \frac12 n(n-1)S^{n-2}(\sigma S)^2dt$

$L.H.S.=nS^n\frac{dS}{S}+\frac12 n(n-1)S^n\sigma^2dt$

Dividing by $S^n$,we get

$\frac{dS^n}{S^n}=[n(\alpha-\delta) +\frac12 n(n-1)\sigma^2]dt + n\sigma dZ$

Thus $S^n$ follows same process as $S$ with drift $n(\alpha-\delta) + \frac12 n(n-1)\sigma^2$ and risk $n\sigma dZ$

$E[S(T)^n]=S(t)^n e^{[n(\alpha-\delta)+0.5n(n-1)\sigma^2][T-t]}$

• What are $\alpha$ and $\delta$? – Carmen González Apr 23 at 22:10
• @CarmenGonzález,$\alpha$ is the continuously compounded expected return on the stock and $\delta$ is the dividend yield on the stock – Dhamnekar Winod Apr 24 at 4:00