# $B_t^3 - 3t B_t$ is a $L^2$ martingale ($B_t$ being a standard Brownian motion)

By Itô's formula I get that \begin{align} d(B_t^3 - 3t B_t) &= (3 B_t^2 dB_t + 3 B_t dt) - 3(B_t dt + 3 t d B_t) \\ &= (3 B_t^2 + 3t) d B_t \end{align} which seems related to martingale representation theorem, but I am missing the argument to conclude that expressing its "differential" as a process times $$d B_t$$ is enough to be a martingale.

Additionally, would there be a direct proof that it's a local martingale (without showing it to be a martingale) ?

For a direct proof of it being a martingale, I can use:

$$\mathbb{E}[B_t^2| \mathcal{F}_s] = (t -s) + B_s^2$$ and \begin{align} \mathbb{E}[B_t^3 - B_s^3| \mathcal{F}_s] &= \mathbb{E}[(B_t - B_s)^3] + \mathbb{E}[3 B_t B_s (B_t - B_s) | \mathcal{F}_s] \\ &= 0 + 3 B_s \mathbb{E}[B_t^2| \mathcal{F}_s] - 3 B_s^2 \\ &= 3(t -s) B_s \end{align}

but I was hoping for a knockout argument following the result of the Itô formula.

• What do you mean by "you are misssing the argument to conclude that expressing its differential as a process tims $dB_t$ is enough to be a martingale"...? It's simply a well-known fact that the stochastic integral (with respect to BM) is a martingale provided that the integrand is suitably measurable and integrable. Both assumptions (measurability and integrability) are easily verified here since $B$ is nicely measurable and has moments of arbitrary order. – saz Mar 1 at 6:55
• @saz well I simply missed this simple fact then... – P. Camilleri Mar 1 at 7:26