# Computing cross variation of independent brownian motions

I am familiar with computing the quadratic variation of Brownian motion, but was confused when the text I'm working through introduced cross variation of independent Brownian motions. the notation is as follows:

$$\langle X,Y\rangle_t = \lim_{||\Delta||\to 0} \sum_i(X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i})$$

Where $X_t$ and $Y_t$ are independent Brownian motions and $\Delta$ is a partition of $[0,t]$. I believe to proceed I should try to calculate the $L^2$ limit (as hinted at in the text), but I am not sure where to start here. The issue is that the only way I know to prove that $X_n\to X$ in $L^2$ is by showing that $E[| X_n-X|]\to0$, but I don't know what to use for $X$ here since I'm trying to compute the limit. Any help is appreciated.

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Compute $E(\langle X,Y\rangle_t^2)$ by expanding the square of the RHS. A lot of terms have zero expectation... – Did Feb 23 '14 at 9:09

Let $U^\Delta_i=(X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i})$, then $E((U^\Delta_i)^2)=(t_{i+1}-t_i)^2$ and $E(U^\Delta_iU^\Delta_j)=0$ for every $i\ne j$ hence the square of the $L^2$ norm of the RHS for subdivision $\Delta$ is $$\sum_i(t_{i+1}-t_i)^2\leqslant\|\Delta\|\cdot t.$$ Edit: The processes $X$ and $Y$ are independent hence, for every $i$, $$E((U^\Delta_i)^2)=E((X_{t_{i+1}}-X_{t_i})^2)\cdot E((Y_{t_{i+1}}-Y_{t_i})^2)=(t_{i+1}-t_i)\cdot(t_{i+1}-t_i).$$ Likewise, $X$ and $Y$ are independent, the increments of $X$ are independent and the increments of $Y$ are independent hence, for every $i\ne j$, $$E(U^\Delta_iU^\Delta_j)=E(X_{t_{i+1}}-X_{t_i})\cdot E(Y_{t_{i+1}}-Y_{t_i})\cdot E(X_{t_{j+1}}-X_{t_j})\cdot E(Y_{t_{j+1}}-Y_{t_j})=0\cdot0\cdot0\cdot0.$$
I think it might be my lack of familiarity with the subject but can you clarify why we can say that $E((U^\Delta_i)^2)=(t_{i+1}-t_i)^2$ and $E(U^\Delta_iU^\Delta_j)=0$? It intuitively makes sense to me but I don't know how to justify that step. – thorspinkhammer Feb 23 '14 at 9:48
See Edit.   – Did Feb 23 '14 at 10:07
So with this we can prove that $E[(\langle X,Y\rangle_t)^2]=0$, which would imply that $\langle X,Y\rangle_t=0$ since for any random variable $W$ we have $E(W^2)=var(W)+E(W)^2$. Is this sufficient to prove the probability limit even without using the definition of the $L^2$ limit? – thorspinkhammer Feb 23 '14 at 21:04