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What does $\langle dW_1, d W_2\rangle = \rho$ mean?

It shows up in stochastic differential equations involving brownian motions. I am told it's the "correlation", but why, how, and what does it mean?

In particular, I don't know what the scalar product of stuff like $"dW_1"$ is.

Jaood
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  • FYI, $\langle dW_1, d W_2\rangle = \rho$ never "shows up" but $\langle dW_1, d W_2\rangle_t = \rho dt$ sometimes does although one should prefer $d\langle W_1, W_2\rangle_t = \rho dt$ or even, $\langle W_1, W_2\rangle_t = \rho t$. – Did Mar 27 '17 at 10:07

1 Answers1

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Given a continuous square integrable martingale $X_t$, there is a unique predictable process $\langle X\rangle_t$ such that $X^2_t-\langle X\rangle_t$ is a martingale. We call $\langle X\rangle$ the quadratic variation of $X$. If $X$ and $Y$ are two continuous square integrable martingales, then the quadratic covariation is defined as

$$\langle X,Y\rangle_t:=\tfrac14(\langle X+Y\rangle_t-\langle X-Y\rangle_t)$$

and you can easily check this is equal to $\langle X\rangle_t$ when $X=Y$. Then $\langle dX,dY\rangle:=d\langle X,Y\rangle$, that is, it is simply (Ito) integration with respect to the process $\langle X,Y\rangle$.

Jason
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