local martingales/ Ito formula

Question

I have a problem with following task. Find a process $(A_t)_{t\ge0}$ of bounded variation on bounded intervals, such that $A_0=0$ and process $M_t=W_t\sin(\int^t_0W_s^3dW_s)-A_t$ is a local martingale. I know that I should probably use two-dimensional Ito formula but I don't know how to deal with this integral $\int^t_0W_s^3dW_s$. Any help appreciated.

Answer 1

Let $X_t = \int_0^t W_s^3\,dW_s$. Note that $[W,X]_t = \int_0^t W_s^3\,ds$ and $[X,X]_t = \int_0^t W_s^6\,ds$. Also define $N_t := W_t\sin(X_t)$. We will apply the two dimensional version of Ito's lemma to the process $N$.

\begin{align}N_t = N_0 &+ \int_0^t \sin(X_s)\,dW_s + \int_0^t W_s\cos(X_s)\,dX_s\\ &-\frac{1}{2}\int_0^t W_s\sin(X_s)\,d[X,X]_s + \int_0^t \cos(X_s)\,d[W,X]_s\end{align}

Performing the substitutions mentioned in the beginning we get

\begin{align}N_t = N_0 &+ \int_0^t \left(\sin(X_s) - \cos(X_s)W_s^4\right)\,dW_s \\ &\int_0^t \left(-\frac{1}{2}W_s^7\sin(X_s) + \cos(X_s)W_s^3\right)\,ds\end{align}

$N_0 = 0$. We have $M_t = N_t - A_t$. If we choose $$A_t = \int_0^t \left(-\frac{1}{2}W_s^7\sin(X_s) + \cos(X_s)W_s^3\right)\,ds$$ then we get $$M_t = \int_0^t \left(\sin(X_s) - \cos(X_s)W_s^4\right)\,dW_s$$ The integrand is adapted and almost surely continuous, which is sufficient to conclude that the $M_t$ is a local martingale. The idea behind ensuring that such a process is a local martingale is to cancel out the "drift terms" in it, which is what we did.