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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?

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    $\begingroup$ 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. $\endgroup$
    – Calculon
    Oct 17, 2016 at 13:44
  • $\begingroup$ Yes I need everything to be $L^2$. $\endgroup$
    – user355356
    Oct 17, 2016 at 13:52
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    $\begingroup$ 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. $\endgroup$
    – Calculon
    Oct 17, 2016 at 13:53

1 Answer 1

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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 :)

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