Use MGF of standard Brownian motion to compute its expectation

Question

I have a standard Brownian motion $B(t)$, $B(0)=0$ and I have to compute $\mathbb{E}(B(t))$ and Var($B(t)$), using $\mathbb{E}(e^{-sB(t)})$ (the moment generating function). I thought this is equal to $e^{-s \mu + 1/2 \sigma^2 s^2} = e^{ts^2/2}$, since the mean is zero and variance is t.

So now I have to compute $\mathbb{E}(B(t))$, which I think will be zero because we have a BM. But how do you do this, because I do not see the link between $\mathbb{E}(B(t))$ and $\mathbb{E}(e^{-sB(t)})$.

Thanks in advance!

Answer 1

$\frac d {ds} Ee^{-sB(t)}=-EB(t)e^{-sB(t)}$. So $tse^{ts^{2}/2}=-EB(t)e^{-sB(t)}$. Put $s=0$ to see that $E(B(t))=0$. Differentiate one more time and put $s=0$ to find the value of $E(B(t))^{2}$. Finally $var (B(t))=E(B(t))^{2}-(E(B(t)))^{2}$.