TITLE: Ito Isometry against non-Brownian SDE
QUESTION [6 upvotes]: Suppose $X_t$ is a Semi-martingale and $H_t$ is $X_t$-predictable.
I know that if $X_t=W_t$ is a Wiener process then
$$
\mathbb{E}[H\cdot W_T^2] = \mathbb{E}[\int_0^TH_t^2dt],
$$
where $H\cdot W_T$ denotes the stochastic integral of $H_t$ against $W_t$ up to time $T$.  
My question is if $W_t$ is not a Weiner process then what is 
$$
\mathbb{E}[H\cdot W_T^2]
$$
equal to?

REPLY [5 votes]: If we assume that $X_t$ is $L^2$ and that $H_t$ is also $L^2$, in their respective senses then we may proceed as follows...
The result is again called the Ito isometry and given your setting is as follows:
Itô Isometry
$$
\mathbb{E}\left[\left(
\int_0^T H_t dX_t\right)^2
\right]
=
\mathbb{E}\left[
\int_0^T H_t^2 d[X]_t
\right],
$$
where $[X]_t$ denotes the quadratic variation of $X$.  Theorem 5 in this blog shows the details of the result.  
In particular if $X_t$ is an Ito process, that is $X_t$ satisfies the SDE
$$
dX_t= \mu_tdt +\Sigma_tdW_t,
$$
then $[X_t]=\Sigma^{\star}\Sigma_t$.  In this case the Ito isomtery simplifies to
$$
\mathbb{E}\left[\left(
\int_0^T H_t dX_t\right)^2
\right]
=
\mathbb{E}\left[
\int_0^T H_t^2 \Sigma^{\star}\Sigma dt
\right].  
$$
Hope this helped :)