I need to describe a couple of integrals which are supposed to be evaluated in terms of Ito calculus.
$$ I_1 = \int_0^t e^{-2\tau}dW(\tau); \\ I_2 = \int_0^t e^{-3 W(\tau)} dW(\tau); $$
Here $W(\tau)$ is a Wiener process. What meaningful information can (should) be said here besides mean and variance (based on the nature of these processes I think the mean is zero in both cases)?
I surely was able to do this a year ago but my memory is very foggy now. Your help is appreciated.
As for the mean it appears to be the case that $$ \operatorname{E}\left( \int_0^t f(\tau,W(\tau)) dW(\tau) \right) = \operatorname{E}\left( \lim_{\Delta \tau \to 0} \sum_k f(\tau_k,W(\tau_k)) (W(\tau_{k+1}) - W(\tau_k)) \right) = 0 $$ because $\operatorname{E}(W(\tau_{k+1}) - W(\tau_k)) = 0$.
Note that $I_1$ is gaussian since the function $s\mapsto\mathrm e^{-2s}$ is deterministic, hence the distribution of $I_1$ is characterized by its mean and variance. The distribution of $I_2$ is more complicated but one can also compute its mean and variance.
Hint for both $I_1$ and $I_2$: $$I=\int_0^tH_s\,\mathrm dW_s\implies\left( E[I]=0,\ E[I^2]=\int_0^tE[H_s^2]\,\mathrm ds\right) $$ Hint for $I_2$: $$ E[\mathrm e^{aW_s}]=\mathrm e^{a^2s/2}$$ Edit: Note that $W_s=\sqrt{s}X$ where $X$ is standard normal and that $$ E[\mathrm e^{aX}]=\frac1{\sqrt{2\pi}}\int_\mathbb R\mathrm e^{ax}\mathrm e^{-x^2/2}\mathrm dx\stackrel{\large (t=x-a)}{=}\frac1{\sqrt{2\pi}}\int_\mathbb R\mathrm e^{a^2/2}\mathrm e^{-t^2/2}\mathrm dt=\mathrm e^{a^2/2}. $$