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Using Ito, I am trying to show that $M_t$ = $\mathbb{E}[(f(X_1))|F_t]$ is a martingale. (I know that it's a martingale by definition of it, but this is an exam question, which stipulates use of Ito.)

Here $f:\mathbb{R}^d\to\mathbb{R}$ is a function of class $C^2$ on $\mathbb{R}^d$, such that $||\nabla f(x)||\leq K$ for some $K$.

First I have shown that $M_t=P_{1-t}f(X_t) \text{ a.s.},$ where $P_sf(x)=\mathbb{E}_x[f(X_s)]$.

Now I define $g(t,x)=\int_{\mathbb{R}^d}f(x+\sqrt{1-t}z)\phi(dz):(\mathbb{R},\mathbb{R}^d)\to\mathbb{R}$. Where $\phi$ is the density of $Z \sim \mathcal{N}_d(0,I)$.

Then $M_t(X_t)=g(t,X_t)$, which puts me in a position to apply Ito.

I found it straightforward to find $\frac{\partial g}{\partial x_i}$, $\frac{\partial^2 g}{\partial x_i^2}$.

Please help me with finding $\frac{\partial g}{\partial t}(t,x)$.

I am trying to apply this:

directional derivative

However, $\sqrt{1-t}$ is in the way.

For people who came here by multivariable-calculus tag, this should hold: $$\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 g}{\partial x_i^2}(t,x)+\frac{\partial g}{\partial t}(t,x)$$

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up vote 1 down vote accepted

No Itô there. Differentiating under the integral sign and integrating by parts yields $$\frac{\partial g}{\partial t}(t,x)=\frac1{2(1-t)}\int_{\mathbb R^d}f\left(x+\sqrt{1-t}z\right)\cdot(d-\|z\|^2)\cdot\phi(\mathrm dz). $$

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