Expectation and variance of standard brownian motion

Question

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 $Var(S(2t))= E[(S(2t))^2] - E[S(2t)]^2$ ?

Answer 1

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: $$ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})E[e^{Z}] =\\ S(0)exp(2mt-\sigma^2 t)exp(\sigma \sqrt{2t})e^{\frac{1}{2}} $$

For the rest it is correct.