Covariation of product of local martingales

Question

Let $(M_t)$ and $(N_t)$ be continuous local martingales, and let $[M,N]_t$ be their continuous covariation. Show that $(MN - [M,N])$ is a local martingale.

I have tried the following. Since $M_tN_t = M_0N_0 + \int_{[0,t]} M_s dN_s + \int_{[0,t]} N_s dM_s + [M_t,N_t]$, we have

$$M_tN_t - [M_t,N_t] = M_0N_0 + \int_{[0,t]} M_s dN_s + \int_{[0,t]} N_s dM_s$$

However, I do not know how to show that the sum of these integrals is a local martingale. Any help is appreciated!

Answer 1

Hints. The following identity plays a crucial role: $$xy = \frac{1}{4} ((x+y)^2-(x-y)^2). \tag{1}$$

1. Using $(1)$ and the very definition of the quadratic (co)variation show that $$[M,N]_t = \frac{1}{4} ([M+N,M+N]_t - [M-N,M-N]_t).$$
2. Recall (or check) that for any continuous local martingale $(X_t)_{t \geq 0}$ the process $X_t^2 - [X]_t$ is a local martingale.
3. By Step 1 and another application of $(1)$, \begin{align*} M_t N_t - [M,N]_t &= \frac{1}{4} ((M_t+N_t)^2-(M_t-N_t)^2) \\ &\quad - \frac{1}{4}([M+N,M+N]_t - [M-N,M-N]_t).\end{align*} Rearrange the terms on the right-hand side in a suitable way and apply Step 2 (twice) to prove that it is a local martingale.