# Is the cross variation (of stochastic processes) bilinear?

I have not been able to find whether for three stochastic processes (adapted with respect to one constant filtration): $(X_t)_{t\geq 0}$, $(Y_t)_{t\geq 0}$ and $(Z_t)_{t\geq 0}$ we have the following equality:

$\langle X+Y, Z\rangle_t= \langle X,Z\rangle_t+\langle Y,Z\rangle_t$

I ask this because I saw a statement along the lines:

"If $\langle L,N\rangle =\langle L',N\rangle$ for all $N$ (all three processes are continuous second order martingale), then $\langle L-L', L-L'\rangle=0$."

This lead me to the more general question, but I assume that if it is true then it's a famous result, and if not there are several classical counterexamples. I would appreciate any hints or answer.

• This is true. By definition, $\langle X+Y, Z\rangle$ is the unique process such that $(X+Y)Z - \langle X+Y, Z\rangle$ is a martingale. It is easy to see that $(X+Y)Z - (\langle X,Z\rangle+ \langle Y,Z\rangle)$ is a martingale because $XZ-\langle X,Z\rangle$ and $YZ-\langle Y,Z\rangle$ are martingales and sum of martingales is still a martingale. Aug 6, 2018 at 13:42
• @JamesYang You should probably post that comment as an answer. Aug 6, 2018 at 13:43

This is true. By definition, $\langle X+Y, Z\rangle$ is the unique process such that $(X+Y)Z - \langle X+Y, Z\rangle$ is a martingale. It is easy to see that $(X+Y)Z - (\langle X,Z\rangle+ \langle Y,Z\rangle)$ is a martingale because $XZ-\langle X,Z\rangle$ and $YZ-\langle Y,Z\rangle$ are martingales and sum of martingales is still a martingale.
• The definition I know for cross variation is of $\langle X,Y\rangle= \frac{1}{4} \Big( \langle X+Y \rangle - \langle X+Y \rangle \Big)$ where $\langle U \rangle$ is the quadratic variation process given by the Doob-Meyer decomposition theorem. Does it follow from the decomposition theorem? Aug 6, 2018 at 13:46
• This is a consequence of what I give. The point is that $XY = \frac{1}{4} ((X+Y)^2 - (X-Y)^2)$. Aug 6, 2018 at 13:51