Given prob space $(\Omega, \mathscr{F}, P)$ and a Wiener process $(W_t)_{t \geq 0}$, define filtration $\mathscr{F}_t = \sigma(W_u : u \leq t)$
Let $(B_t)_{t \geq 0}$ where $B_t = W_t^3 - 3tW_t$. Show that $E[B_t|\mathscr{F}_s] = B_s$ whenever $s < t$.
I think this all comes down to manipulation since there are martingales somewhere
My attempt:
Splitting up into $E[W_t^3|\mathscr{F}_s] - 3E[tW_t|\mathscr{F}_s]$ doesn't do anything since those guys aren't martingales? So, I tried splitting it up into:
$E[W_t(W_t^2 - 3t)|\mathscr{F}_s]$
$= E[W_t(W_t^2 - t -2 t)|\mathscr{F}_s]$
$= E[W_t(W_t^2 - t) -2 tW_t)|\mathscr{F}_s]$
$= E[W_t(W_t^2 - t)|\mathscr{F}_s] -2E[ tW_t|\mathscr{F}_s]$
$W_t$ is not $\mathscr{F}_s$-measurable, so we can't take that out...
$tW_{1/t}$ is Brownian and thus a martingale, but I don't know about $tW_t$...
$cW_{t/c^2}$ is Brownian and thus a martingale, but I don't think we can set c = t...
Help please?