Show that $X_t = \int_0^t B_s^2 dB_s$ is a martingale.

Question

In this post Is this Stochastic integral a martingale ? it is claimed that the process $(X_t)_{t \geq 0}$ defined by $$X_t = \int_0^t B_s^2 dB_s$$ is a martingale.

I want to be able to show this directly, without any machinery higher than the Ito formula. I get to here: For $s < t$ consider that \begin{eqnarray*} \mathbb{E} (X_t - X_s \vert \mathscr{F}_s) &=& \mathbb{E} \left( \int_0^t B_r^2 dB_r - \int_0^s B_r^2 dB_r \Bigg \vert \mathscr{F}_s \right) \\ &=& \mathbb{E} \left( \int_s^t B_r^2 dB_r \Bigg \vert \mathscr{F}_s \right) \end{eqnarray*}

before getting stuck. Any assistance is appreciated.

Is it possible to show that $(X_t)_{t \geq 0}$ is a martingale directly by computing the conditional expectation?

Answer 1

Here's an approach using Ito: By Ito's formula, applied to $f(x)={1\over 3} x^3$, you have $$ \int_0^t B_s^2\,dB_s = {1\over 3} B_t^3-\int_0^t B_s\,ds. $$ It's not hard to take conditional expectations of the objects on the right side of this equality.