# Quadratic Variation of Sum of Local Martingales

I have a question about calculating covariances of local martingales. Suppose $M_1$ and $M_2$ are local martingales. Put $M = M_1+M_2$. Is there a nice way to calculation $[M]$ in terms of $[M_1]$ and $[M_2]$?

I feel that if $B_1$ and $B_2$ are independent Brownian motion, and $M_1 = \int f(B_1(s),B_2(s),s) \, dB_1(s)$ and $M_2 = \int g(B_1(s),B_2(s),s) \, dB_2(s)$, then $[M] = \int f^2 + g^2 \, dt$.

-
@ Minke : your are implicitly supposing that $B_1$ and $B_2$ are independent, you need to add a "quadratic covariation term" to be general. Regards – TheBridge Feb 27 '12 at 8:38

To answer this question one must understand the covariation of $M_1$ and $M_2$ (see page 44 of these notes). We denote the covariation of continuous local martingales $L$ and $N$ by $[L,N]$ and, under suitable conditions, we have
$$[L,N]_t = \lim_{n\to\infty}\sum_{k=0}^{\lfloor 2^nt\rfloor}(L_{(k+1)2^{-n}}-L_{k2^{-n}})(N_{(k+1)2^{-n}}-N_{k2^{-n}})$$
So, in particular, $[M]_t=[M,M]_t$. Covariation of processes behaves in much the same way as the covariance of random variables. For example:
• $[M_1]_t\geq 0$, but $[M_1,M_2]_t$ can be negative.
• If $M_1$ and $M_2$ are independent then $[M_1,M_2]_t=0$, but the reverse needn't hold.
• $[,]$ is bilinear. So $[M]=[M_1]+[M_2]+2[M_1,M_2]$. So we need to know $[M_1,M_2]$ in addition to $[M_1]$ and $[M_2]$ to determine $[M]$.