## Brownian Motion: Expectation and Variance

Consider the stochastic differential equation,

$\displaystyle dS_t = \mu S_tdt + \sigma S_tdW_t$, for all $\displaystyle t \in [0, T]$,

where $\displaystyle \{W_t\}_{t\geq0}$ is a standard Brownian motion, $\displaystyle \mu, \sigma$ and $\displaystyle S_0$ are positive constants. Derive the unique solution to the above SDE . Calculate $\displaystyle \mathbb{E}[S_t]$ and $\displaystyle Var[S_t]$.

Skipping working the unique solution is...

$\displaystyle S_t = S_0e^{(\mu - \tfrac{\sigma^2}{2})t - \sigma W_t}$.

Now to find the expectation and variance. I've realized that a lot of the things I do I do because I have seen them in lectures or done them in other questions but I now realize I have no idea why they are true.

$\displaystyle \mathbb{E}[S_t] = \mathbb{E}[S_0e^{(\mu - \tfrac{\sigma^2}{2})t - \sigma W_t}] = S_0e^{\mu t}\mathbb{E}[e^{-\tfrac{\sigma^2}{2}t - \sigma W_t}]$ $\displaystyle =S_0\mu$. Now why is that last step true?

My guess is something to do with...

$\displaystyle \mathbb{E}[W_t] = 0$

But I don't know what to do with the $\displaystyle \sigma$ although I assume that it eventually comes out to be $\displaystyle e^0 = 1$.

Now the variance...

The way my lecturer does it is to calculate $\displaystyle Var[S_t] = \mathbb{E}[(S_t - \mathbb{E}[S_t])^2]$ but I always calculate $\displaystyle \mathbb{E}[S_t^2]$ instead then do the usual subtraction of $\displaystyle (\mathbb{E}[S_t])^2$.

So, I get...

$\displaystyle \mathbb{E}[S_t^2] = \mathbb{E}[S_0^2e^{2(\mu - \tfrac{\sigma^2}{2})t + 2 \sigma W_t}]$.

$\displaystyle = S_0^2 e^{2\mu t}\mathbb{E}[e^{-\sigma^2 t + 2 \sigma W_t}]$ but now I am not sure how to justify the next step.

$\displaystyle = S_0^2 e^{2\mu t}e^{-\sigma^2 t}$ (actually should that be $\displaystyle e^{\sigma^2 t}$..?!?)