What the expectation of S^2 is from GBM? [closed]

Question

I was at an interview and was asked to write down the SDE for GBM.

$$ dS = S\mu dt + S\sigma dX $$

Then I was asked how I would compute the expectation of S^2. I didn't know where to start. Any ideas?

$$ {\mathbb{E}} [S_t^2] $$

Answer 1

As Sanjay said, you can apply Itô's Lemma to $f(t,x)=x^2$ and obtain \begin{align*} \mathrm{d} S^2_t=\left(2\mu S_t^2+\sigma^2S_t^2\right)\mathrm{d}t+\left(2\sigma S_t^2\right)\mathrm{d}W_t. \end{align*} Thus, $(S_t^2)$ is again a geometric Brownian motion and hence, for each time point $t$ log-normally distributed with drift $2\mu+\sigma^2$ and volatility $2\sigma$. Then, \begin{align*} S_t^2 &= S_0^2\cdot\exp\left(\left(2\mu+\sigma^2-\frac{1}{2}\cdot 4\sigma^2\right)t+2\sigma W_t\right) \\ &= S_0^2\cdot\exp\left(\left(2\mu-\sigma^2\right)t+2\sigma W_t\right). \end{align*}

Alternatively, you can use that you know $(S_t)$ explicitly and directly obtain \begin{align*} S_t &= S_0\cdot\exp\left(\left(\mu-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right)\\ \implies S_t^2 &= S_0^2 \cdot\exp\left(\left(2\mu-\sigma^2\right)t+2\sigma W_t\right). \end{align*}

In either case, you get that \begin{align*} \mathbb{E}[S_t^2]=S_0^2e^{(2\mu+\sigma^2)t}. \end{align*}

You can of course easily generalise this to the process $(S_t^\gamma)$ for some $\gamma>0$.