0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.