# Given $dX_t = W_t^2dW_t$, and $Y_t= X_t^2$, is $Y_t$ a martingale?

Suppose $X_t$ satisfy the following stochastic differential:

$dX_t = W_t^2 dW_t$

Let $Y_t = X_t^2$

Is $Y_t$ a martingale?

I first thought to find $X_t$ from the differential using Ito's differential and comparing coefficients then see if that was a martingale using the definition, then square it to see if $X_t^2$ is a martingale.

However, its impossible to express $X_t$ as a function of Brownian motion. Does this then infer that $X_t$ is not a martingale? If not, I'm not sure what other method to take?

Note that $d\langle X\rangle_t=W_t^4dt$. By Ito's lemma, $dY_t=2X_tdX_t+d\langle X\rangle_t=2X_tW_t^2dW_t+W_t^4dt$. The first term integrates to a local martingale, which by directly computing quadratic variation can be shown to be a martingale. The second term $W_t^4dt$ integrates to a process with non-zero finite variation, hence is not a martingale. This shows that $Y_t$ is not a martingale.
• Thank you! Could you please explain how you get $dX_t = W_t^4$ Mar 19 '18 at 16:12
• You're welcome. It's a standard result that $dX_t=Z_tdW_t$ implies $d\langle X\rangle_t=Z_t^2dt$. You may find this proposition in your textbook. Mar 20 '18 at 10:43