Is the product of a random variable and martingale still a martingale?

Question

Assuming we have a Brownian motion $(B_t, \mathcal{F}_t)$, we define $M_t = M_0(B_t^2 - t)$, if $M_0$ is a constant, then I know $M_t$ is a martingale, however, if $M_0$ is a random variable which is independent of $\mathcal{F}_t$ and $M_0 < \infty$, is it still a martingale?

Answer 1

If $M_0 (B_t^2-t)$ is integrable for each $t$, then $(M_t)_{t \geq 0}$ is a martingale with respect to the augmented filtration $\mathcal{G}_t := \sigma(B_s; s \leq t; M_0)$.

Clearly, $M_t$ is measurable with respect to $\mathcal{G}_t$ and using the pull out property of the conditional expectation we find $$\mathbb{E}(M_t \mid \mathcal{G}_s) = M_0 \mathbb{E}((B_t^2-t) \mid \mathcal{G}_s) =M_0 \mathbb{E}((B_t^2-t) \mid \mathcal{F}_s) = M_0 (B_s^2-s) = M_s$$ for all $s \leq t$. For the second "$=$" we have used that $M_0$ is independent of $\mathcal{F}_t$ for each $t \geq 0$.