Computation of basic stochastic integral.

Question

I am trying to compute the covariance of a 1 dimensional Ornstein-Uhlenbeck process $dx_t=-\theta x_t dt+ \sigma dW_t$, $\theta>0$ and I am at the stage, $$\text{Cov }(x_s,x_t)=\sigma^2 e^{-\theta(t+s)} \mathbb{E}\left[ \int_0^s e^{\theta u}dW_u \int_0^t e^{\theta v} dW_v\right].$$

Is it possible to evaluate the stochastic integrals explicitly and if not how does one go about simplifying this. On Wikipedia, they say this is equal to $\frac{\sigma^2}{2\theta} e^{-\theta(t+s)} (e^{2\theta s\wedge t } -1)$ but I cannot see how they reach this conclusion.

Thanks.

Answer 1

The trick you shall learn when dealing with Ito integrals is that $$ \int_a^bf_s\mathrm d W_s\quad\text{ and }\quad \int_b^cf_s\mathrm d W_s $$ are independent whenever $a<b<c$, it follows from independence of increments of $W_t$. As a result, if you assume that in your case $s\leq t$ then $$ \mathsf E\left[\int_0^s f_u\mathrm dW_u \int_0^t f_v\mathrm dW_v\right] = \mathsf E\left[\left(\int_0^s f_u\mathrm dW_u \right)^2\right] + \mathsf E\left[\int_0^s f_u\mathrm dW_u\int_s^t f_v\mathrm dW_v\right] $$ $$ = \int_0^s\mathsf E[f^2_u]\mathrm du + \mathsf E\left[\int_0^s f_u\mathrm dW_u\right]\cdot\mathsf E\left[\int_s^t f_v\mathrm dW_v\right] = \int_0^s\mathsf E[f^2_u]\mathrm du. $$ Now you only have to choose $f$ appropriately.