Dealing with a term coming from Ito formula.

Question

Say I have a d-dimensional SDE $$dX_t=\sigma(X_t) dW_t,$$ here $\sigma \in \mathbb{R}^{d\times d}$ and $W_t$ is a d-dimensional Brownian motion. After using Ito formula on the function $x\mapsto\|x\|^2$ and taking expectations ( $\mathbb{E}$ ), among other things, I have to deal with the following term

$$ 2\mathbb{E} \Big[\int_0^t \langle X_s,\sigma(X_s)dW_s \rangle \Big]. $$

Now when can I say this term is zero? Or can how can I use a Cauchy Schwarz inequality to start bounding this term, I want to bound it by terms like $\mathbb{E} \Big[\int_0^t \|X_s\|^2ds \Big] $, assume $\sigma$ is uniformly Lipschitz.

$\mathbb E[\int_0^t f(X_s),\mathrm d W_s]=0$ only if $\int_{0}^t\mathbb E[f(X_s)^2,\mathrm d s]<\infty $. Do you have such information ? — Surb, Jan 12 '22 at 13:21

score 1 · Answer 1 · answered Aug 15 '23 at 03:25

As mentioned in an answer here and in the blog here,

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

$f(\omega,s)$ is adapted, measurable in s, and

$\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.

If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. Also, by taking $f=x^{p}$ for some $0<p<1/2$, we then deal with rough-integrals.

Dealing with a term coming from Ito formula.

1 Answers1

Linked