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Given $X_t=\int_0^t s W_s dW_s$ and the process $M_t=X_t^3-\int_0^t X_sY_s ds$. Find $Y_t$ such that $M_t$ is a martingale.

I started thinking that $X_t$ can be seen as: $dX_t=tW_tdW_t$ then by applying Ito's lemma: $d(X_t^3)=3X_t^2dX_t+3X_t(dX_t)^2=3X_t^2tW_tdW_t+3X_t t^2 W_t^2 dt$.

Integrating the latter between s and t yields: $X_t^3-X_s^3=3\int_s^tX_t^2tW_tdW_t+3\int_s^tX_t t^2W_t^2dt$

$M_t$ is a martingale if $\mathbb{E}[M_t|F_s]=M_s$.

$\mathbb{E}[M_t|F_s]=\mathbb{E}[X_t^3-\int_0^tX_sY_sds|F_s]$.

Subtracting and adding $X_s^3$ and breaking the integral leads to:$\mathbb{E}[X_t^3-X_s^3+X_s^3-\int_0^sX_sY_sds-\int_s^tX_sY_sds|F_s]=M_s+\mathbb{E}[X_t^3-X_s^3-\int_s^tX_sY_sds|F_s]$.

I still don't know how to treat the latter expectation.

thanks for any help.

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    $\begingroup$ that was my idea $\endgroup$ – riccardo giussani Dec 30 '15 at 14:09
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I started thinking that $X_t$ can be seen as: $dX_t=tW_tdW_t$ then by applying Ito's lemma: $$d(X_t^3)=3X_t^2dX_t+3X_t(dX_t)^2=3X_t^2tW_tdW_t+3X_t t^2 W_t^2 dt.$$

Yeah, that's a good idea. It follows directly from this identity that

$$dM_t = d(X_t^3)-X_t Y_t \, dt = 3 X_t^2 t W_t \, dW_t + (3t^2 W_t^2-Y_t) X_t \, dt. \tag{1}$$

Since the process $(X_t)_{t \geq 0}$ is nicely integrable, the process

$$N_t :=3 \int_0^t X_s^2 s W_s \, dW_s$$

is a martingale (as a stochastic integral with respect to Brownian motion). Therefore, $(1)$ shows that $(M_t)_{t \geq 0}$ is a martingale if

$$(3t^2 W_t^2-Y_t) X_t \, dt = 0$$

i.e.

$$Y_t = 3 t^2 W_t^2.$$

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