The time convergence of stochastic integral and Doob's convergence.

Question

Consider the process $$X_{t}=\int_{0}^{t}e^{-s}dW_{s},$$ where $e^{-s}$ is deterministic.

I am wondering if $\lim_{t\rightarrow\infty}X_{t}$ exists almost surely... I understand that $X_{t}$ in the case is a martingale, so we can use Doob's martingale convergence theorem. However, I have no idea about how to show $$\sup_{t}\mathbb{E}X_{t}^{-}<\infty.$$ If we can show this, then yes, $X_{t}$ converges almost surely to a limit..

Also, is there any way to know the distribution of the limit? I know that the distribution converges since it converges almost surely, but I am not sure how to compute the limiting distribution.. using central limit theorem??

Thanks!

Answer 1

$E|X_t|^{2}=\int_0^{t} e^{-2s} ds=\frac 1 2(1-e^{-t}) <1$ for all $t$ and this implies $E|X_t|$ is bounded.

The limiting distribution is $N(0,\int_0^{\infty} e^{-2s} ds)$ i.e. $N(0, \frac 1 2) $.

Answer 2

Regarding the distribution: It holds $$\int_{0}^{t}f(\tau)dW_{\tau}\sim N(0,\int_{0}^{t}|f(\tau)|^{2}d\tau),$$ since $f(t)=e^{-t}$ is a square integrable deterministic function. As you now know the distribution of $X_t$ and it converges in distribution to $X_\infty$ you can calculate its distribution.