Limit of Itô-Integral

Question

We consider the following Itô-Integral $$Z_t := \int_0^t \exp(-\mu s) W_s ds$$ for $\mu\geq0$. I wonder if I could calculate the limit as $t\to\infty$ in some convergence, but how should I start? Some ideas?

Edit: Is it possible to get convergence almost surely to some random variable? Or in $L^2$?

Answer 1

Apply Ito's lemma to $-\frac{1}{\mu}\exp(-\mu t)W_t$. You get

$$d\left(-\frac{1}{\mu}\exp(-\mu s)W_s\right)=-\frac{1}{\mu}\exp(-\mu s)dW_t+\exp(-\mu t)W_sds$$

Taking integrals and setting $W_0=0$

$$-\frac{1}{\mu}\exp(-\mu t)W_t=\int_0^t-\frac{1}{\mu}\exp(-\mu t)dW_s+\int_0^t\exp(-\mu s)W_sds$$

Rearranging

$$\int_0^t\exp(-\mu s)W_sds=\int_0^t\frac{1}{\mu}\exp(-\mu s)dW_s-\frac{1}{\mu}\exp(-\mu t)W_t$$

The 2nd expression converges to zero a.s. because $W_t/t$ tends to zero a.s.. The 1st term is a normal random variable with mean 0 and variance $\int_0^t \frac{1}{\mu^2}\exp(-2\mu s)ds$, which converges to $\frac{1}{2\mu^3}$. Hence $\int_0^t\exp(-\mu s)W_sds$ converges a.s. to $\int_0^t\frac{1}{\mu}\exp(-\mu s)dW_s$. It also converges in distribution to a $N(0,\frac{1}{2\mu^3})$ random variable.

Answer 2

As shown by @fesman, $Z_t$ differs from the stochastic integral $\int_0^t e^{-\mu s}\,dW_s$ by an amount that converges a.s (and also in $L^2$) to $0$. It's not hard to show that $\int_0^t e^{-\mu s}\,dW_s$ converges in $L^2$ to $\int_0^\infty e^{-\mu s}\,dW_s$ (the $L^2$ norm of their difference is ${e^{-2\mu t}\over t}$). In fact, by the martingale convergence theorem, the convergence is also almost sure, because $$ \int_0^t e^{-\mu s}\,dW_s = E\left[\int_0^\infty e^{-\mu s}\,dW_s\mid\mathcal F_t\right]. $$ It follows that $Z_t$ converges to $\int_0^\infty e^{-\mu s}\,dW_s$ a.s. and in $L^2$.