Finding $cov(B(t,x), B(t,y))$

Question

Suppose we have a stochastic process $B(t,x)$ so process so that for every $x$ fixed, $\{B(t,x)\}_t$ is a Brownian motion. Also, define some correlation $\rho_{xy}$ so that $$dB(t,x)dB(t,y)=\rho_{xy} dt$$

In other words, we can thing of the correlation structure of this process as $$corr(dB(t,x),dB(t,y))=\rho_{xy}.$$

I suppose it's not incorrect to think of the covariance as $$cov(dB(t,x),dB(t,y))=\rho_{xy} dt.$$

Now my question is, can we find what $cov(B(t,x),B(t,y))$ is based on this?

Answer 1

Sure, we just need the product formula:

\begin{align*} d(B(t,x)B(t,y)) &= B(t,x)dB(t,y) + B(t,y)dB(t,x) + dB(t,y)dB(t,x) \\ &= B(t,x)dB(t,y) + B(t,y)dB(t,x) + \rho_{xy}dt. \end{align*}

Now we just integrate and take expectation. Since the stochastic integrals will have expectation $0$, we'll end up with just $\mathbb{E}[B(t,x)B(t,y)] = \rho_{xy}t$ and since each is a Brownian motion (and hence has mean $0$) this also gives the covariance is $\rho_{xy}t$.