# Expectation and variance of standard brownian motion

Assuming that the price of the stock follows the model

$$S(t) = S(0) exp ( mt − (σ^2/ 2) t + σW(t) ) ,$$ where W(t) is a standard Brownian motion; σ > 0, S(0) > 0, m are some constants.

What is the expectation and variance of S(2t)?

Expectation:

$$E[S(2t)]=E[S(0)exp(2mt-(t\sigma^2)+\sigma W(2t)] =$$

$$S(0)E[exp(2mt-(t\sigma^2)+\sigma W(2t))] = S(0)exp(2mt-\sigma^2 t)E[exp(\sigma W(2t)]$$

using that $$W(2t)$$ is $$N(0,2t)$$ I get that $$W(2t)=\sqrt{2t} Z$$, where $$Z$$ is $$N(0,1)$$.

$$S(0)exp(2mt-\sigma^2 t)E[exp(\sigma \sqrt{2t} Z)]$$ =

$$S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})E[e^{Z}] =$$

$$S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})$$

Is this solution correct?

Variance: Assuming that the expectation is correctly solved I could just use that $$Var(S(2t))= E[(S(2t))^2] - E[S(2t)]^2$$ ?

No because $$E(e^Z)=e^{\frac{1}{2}}\neq1$$
More generally: $$N \sim \mathcal N(\mu,\sigma^2)\\ E(e^{Nt})=MGF_{\mathcal N(\mu, \sigma^2)}(t)=e^{\mu t+\frac{1}{2}\sigma^2t^2}$$
The last lines should be: $$S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})E[e^{Z}] =\\ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})e^{\frac{1}{2}}$$