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

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?

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

• Apply Ito to $S^2$ to reach an expression for $d(S^2)$. You can easily compute $E[S^2]$ from there – Sanjay Oct 17 '19 at 14:33

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 $$\gamma>0$$.
• @Idonknow It would be in general $e^{\mu+0.5\sigma^2}$ for a lognormal distribution. We have a mean of $(2\mu-\sigma^2)t$ and a variance of $4\sigma^2t$. So this gives a plus in the exponential? – KeSchn Oct 18 '19 at 6:08