Let $W_1,$ $W_2$ be to Wiener processes under the martingale measure $Q$. What can be said about $dW_1*dW_2$? I know that $$(dW_i)^2=dt$$ but what about the case with two different wiener processes?

  • $\begingroup$ It doesn't matter that the measure is the martingale measure. If two processes are both BMs in some probability space, then their covariation process is just $\rho t$ where $\rho$ is the correlation between them. $\endgroup$ – Calculon Jan 25 '18 at 17:22

The following is not a proof but some reasoning:

If you consider the L2-limits then you see something along the lines: $$ dW^2 = \lim_{n \rightarrow \infty}\sum_{j=1}^n (W_{t_{j+1}} - W_{t_{j}})^2 \rightarrow t $$ For $W_1, W_2$ with correlation $\rho$ this transaltes to $$ dW^1 dW^2 = \lim_{n \rightarrow \infty}\sum_{j=1}^n (W_{t_{j+1}}^1 - W_{t_{j}}^1)(W_{t_{j+1}}^2 - W_{t_{j}}^2) \rightarrow \rho t, $$ which can be seen by considering that $\left((W_{t_{j+1}}^1 - W_{t_{j}}^1),(W_{t_{j+1}}^2 - W_{t_{j}}^2)\right)$ is bivariate Gaussian.

You find details in Introduction to Stochastic Differential Equations (SDEs) for Finance on page 19 Correlated Brownian Motion.

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