# covariance of two stochastic integrals

I'm trying to evaluate the covariance between two stochastic integrals such as $$Cov(\int_0^t g_udW_u, \int_0^t h_udW_u) = \int_0^t E[g_uh_u]du$$So I am trying to prove this and I thought I would do so by using the regular covariance formula such as $$E[XY] - E[X]E[Y]$$. Using the fact of the martingale so that the expectations are 0 only the cross term remains to be evaluated. I don't really know how to proceed now just I assume that I have to use Ito isometry at some point.

• Use the identity $$X \cdot Y = \frac{1}{4} ((X+Y)^2-(X-Y)^2)$$ for $X:=\int_0^t g(u) \, dW_u$ and $Y:= \int_0^t h_u \, dW_u$, and apply Itô's isometry.
– saz
Apr 15 '19 at 18:39
• I get $\int_0^tg_uh_ud_u$ is that the same as $\int_0^t E[g_uh_u]d_u?$ Apr 15 '19 at 18:44
• How that? (Perhaps there are some typos in your comment? As it is currently stated, it's wrong.)
– saz
Apr 15 '19 at 18:51
• So I got $\frac{1}{4}E[(\int_0^t(g_u+h_u)dW_u)^2]-\frac{1}{4}E[(\int_0^t(g_u+h_u)dW_u)^2]$ and then applying Ito isometry I get $\frac{1}{4}E[\int_0^t(g_u+h_u)^2du]-\frac{1}{4}E[(\int_0^t(g_u+h_u)^2du]$ and then I got $\int_0^t\frac{1}{4}((\int_0^t(g_u+h_u)^2)du$ Apr 15 '19 at 18:55
• Well, the first part is correct, but why do you think that you can just omit the expectation value (at the end of your computation)? If $g$ and $h$ are deterministic, then you can do so, but in general you can't
– saz
Apr 15 '19 at 19:01

Itô isometry - The Itô integral respects the inner product: $$\mathbb{E}\left[\left(\int_0^t X_udW_u\right)\left(\int_0^t Y_udW_u\right)\right]=\mathbb{E}\left[\int_0^t X_uY_ud_u\right]$$
Therefore, \begin{align} Cov\left(\int_0^t g_udW_u, \int_0^t h_udW_u\right) &= \mathbb{E}\left[\left(\int_0^t g_udW_u\right)\left(\int_0^t h_udW_u\right)\right]-\mathbb{E}\left[\int_0^t g_udW_u\right]\mathbb{E}\left[\int_0^t h_udW_u\right]\\ &= \mathbb{E}\left[\int_0^t g_uh_ud_u\right]-0\times 0\\ &= \int_0^t E[g_uh_u]du \\ \end{align}