Is $X(t)=B_1(t)\,B_2(t)$ a martingale?

Question

I want to check if $X(t)=B_1(t)\,B_2(t)$ is a martingale with respect to the filtration $F(t)$, where $B_1(t)$ and $B_2(t)$ are two independent Brownian motions.

For this I need to see if: $\mathbb{E}[X(t)|F(s)]=X(s)$. Do you have any suggestion?

Answer 1

I finally knew how to do it:

\begin{align} \mathbb{E}[X(t)|F(s)] &= \mathbb{E}[B_1(t)B_2(t)|F(s)]\\ &=\mathbb{E}[(B_1(t)-B_1(s)+B_1(s))\,(B_2(t)-B_2(s)+B_2(s)|F(s)]\\ &=\mathbb{E}[(B_1(t)-B_1(s))(B_2(t)-B_2(s))+B_2(s)(B_1(t)-B_1(s))+B_1(s)(B_2(t)-B_2(s))+B_1(s)B_2(s)|F(s)]\\ &=B_2(s)B_1(s)\\ &=X(s) \end{align} Thus it is a martingale.

We have used the fact that:

$$\mathbb{E}[B(s)|F(s)]=B(s)\mathbb{E}[1|F(s)]=B(s),$$ because $B(s)$ is $F(s)-$measurable. And:

$$\mathbb{E}[B(t)-B(s)|F(s)]=\mathbb{E}[B(t)-B(s)]=0,$$ since, for $0\leq s\leq t$:

\begin{align} B(t)-B(s)&=\sum_{i=s+1}^tY_i, \end{align} where Y is a random variable, so $B(t)-B(s)$ is independent of filtration $F(s)$, and by definition the expected value of brownian motion increments is equal to zero.