Quadratic variation of the sum

Question

Let $X_t$ and $Y_t$ be two independent martingales whose quadratic variations are given by $\langle X\rangle_t$ and $\langle Y\rangle_t$ respectively...

If we define $Z_t=a X_t+ bY_t$ can we say anything about the quadratic variation of $Z_t$?

$a$ and $b$ are two constants.

Answer 1

All càdlàg martingales, and local martingales have well defined quadratic variation, which follows from the fact that such processes are examples of semi- martingales.

However, let $I=\{t_0,t_1,\cdots,t_n\}$ is a sequence of partitions of $[0,t]$ with mesh going to zero, then the Quadratic variation of $Z_t$ is defined as $$[Z,Z]([0,t])=\lim_{\delta_n\to 0}\sum_{i=1}^{n}(\,Z(t_i)-Z(t_{i-1})\,)^2$$ where the limit is taken over all shrinking partitions of $[0, t]$, with $\delta_n=\max\{t_i-t_{i-1}\}$ as $n\to \infty$. We have $$[Z,Z]([0,t])=\lim_{\delta_n\to 0}\sum_{i=1}^{n}(\,aX(t_i)-aX(t_{i-1})+bY(t_i)-bY(t_{i-1})\,)^2$$ thus $$[Z,Z]([0,t])=a^2\lim_{\delta_n\to 0}\sum_{i=1}^{n}(\,X(t_i)-X(t_{i-1}))^2+b^2\lim_{\delta_n\to 0}\sum_{i=1}^{n}(\,Y(t_i)-Y(t_{i-1}))^2\\ +2ab\lim_{\delta_n\to 0}\sum_{i=1}^{n}(\,X(t_i)-X(t_{i-1}))(\,Y(t_i)-Y(t_{i-1}))$$ Therefore $$[Z,Z]([0,t])=a^2[X]([0,t])+b^2[Y]([0,t])+2ab[X,Y]([0,t])$$

Answer 2

Two complements to the answer of @BehrouzMaleki:

1. Because $X$ and $Y$ are independent, the product $X_tY_t$ is also a martingale; the semimartingale product rule then tells us that $[X,Y]_t$ is a martingale with initial value $0$; if one or the other of $X$ or $Y$ is continuous, then $[X,Y]$ is predictable, hence constant, hence identically $0$.

2. If $X$ and $Y$ are locally square integrable, then they admit predictable quadratic variations $\langle X\rangle$ and $\langle Y\rangle$. The predictable covariation $\langle X,Y\rangle$ then exists, and now the independence of $X$ and $Y$ implies that $\langle X,Y\rangle$ is a predictable local martingale, hence constant, hence identically $0$.