Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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: $$\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 g}{\partial x_i^2}(t,x)+\frac{\partial g}{\partial t}(t,x)$$

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.