Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $i\ne j$, $$ E(U^\Delta_iU^\Delta_j)=E(X_{t_{i+1}}-X_{t_i})\cdot E(Y_{t_{i+1}}-Y_{t_i})\cdot E(X_{t_{j+1}}-X_{t_j})\cdot E(Y_{t_{j+1}}-Y_{t_j})=0\cdot0\cdot0\cdot0. $$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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