# Ito Isometry against non-Brownian SDE

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 $$\mathbb{E}[H\cdot W_T^2] = \mathbb{E}\bigg[\int_0^TH_t^2dt\bigg],$$ 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?

• It is then equal to $E[\int_0^TH_t^2d[X]_t]$ if $X_t$ is an $L^2$ martingale. Otherwise, no such isometry exists as far as I know. Oct 17, 2016 at 13:44
• Yes I need everything to be $L^2$.
– user355356
Oct 17, 2016 at 13:52
• Actually more importantly, the expectation you define with $X_t$ being a semi-martingale doesn't necessarily exist. So you need to cover that base before you think about isometries etc. Oct 17, 2016 at 13:53

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...
$$\mathbb{E}\left[\left( \int_0^T H_t dX_t\right)^2 \right] = \mathbb{E}\left[ \int_0^T H_t^2 d[X]_t \right],$$ 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 :)