Given a Standard Brownian motion $(B_t)_{t\in\mathbf{R}_{+}}$, define:
$$E(e^{\int_0^tudB_u})=?$$ $$ Var(e^{\int_0^tudB_u})=?$$
I started off assuming (!) that $X_t=$ $\int_0^tudB_u \sim N(\mu,\sigma^2)$; so that I could apply:
$E(e^{X_t})=\int_Re^{X_t}.(pdf) dx = e^{(\mu+\frac[2\sigma^2)}$
And went on to find $\mu$ and $\sigma$ of $X_t$:
1) $E(X_t)=E(\int_0^tudB_u)$
2)Since $X_t$ has no drift, and only a diffusion term, it should be a Martingale assuming (!) that
$E(X_t)^2<\infty$ (is bounded)
Which is since:
$E(X_t)=E(\int_0^tudB_u)^2 = E(\int_0^tu^2dt) = (\frac{t^3}{3}) < \infty$
So since it is a Martingale and $B_0=0$:
3)$E(X_t)=E(\int_0^tudB_u)=0$
4) $Var(X_t)=E(X_t^2) - [E(X_t)]^2$$= E(\int_0^tudB_u)^2-0$ $=(\frac{t^3}{3})$
5)At this point I can state:
$$X_t=\int_0^tudB_u \sim N(0,\frac{t^3}{3})$$
and apply
$E(e^{\int_0^tudB_u})=e^{0+\frac{t^3}{6}}$
6) For the variance:
$Var(e^{X_t})=E(e^{2N(0; \frac{t^3}{3}})-[E(e^{X_t})]^2 = e^{0+ \frac{1}{2} \frac{t^3}{3}} -[e^{0+\frac{t^3}{6}}]^2$
PS: Fixed thanks to comments