multivariate Ito isometry

Question

I wonder whether there exists a straightforward extension of the Ito isometry to multidimensional processes.

In the one-dimensional case the Ito isometry can be written as

$\mathbb{E}[ (\int_0^T X_t \; \mathrm{d}W_t)^2 ] = \mathbb{E}[ (\int_0^T X_t^2 \;\mathrm{d}_t) ]$.

If now $X_t$ is a vector of random variables instead, do I get something along these lines:

$\mathbb{E}[ (\int_0^T X_t \; \mathrm{d}W_t) (\int_0^T X_t^\top \; \mathrm{d}W_t^\top) ] = \mathbb{E}[ (\int_0^T X_t X_t^\top \;\mathrm{d}_t) ]$

????

Answer 1

Short answer: Yes, what you wrote is basically correct.

Longer anser:

1. See the following for a more general case, which is Lemma D.1 on the page 19 of http://www.shuangjian.info/Wasserstein_Control_HRLMC.pdf

${\bf Statement.}$ Let ${\bf B}: [0, T]\times \Omega \rightarrow \mathbf{R}^p$ be the standard $p$-dimensional Brownian motion and ${\bf M}: [0, T]\times \Omega \rightarrow \mathbf{R}^{p\times p}$ be a matrix-valued stochastic process adapted to the natural filtration of the Brownian motion. Then $$\mathbf{E} \left[ \left\Vert\int_0^T {\bf M}_t d{\bf B}_t\right\Vert_2^2\right] = \mathbf{E} \left[ \int_0^T \left\Vert{\bf M}_t\right\Vert_F^2 dt\right],$$ where the norm in the right hand side is the Frobenius norm of the matrix-valued process.

2. The implication of the above statement is your case: take ${\bf X}_t$ as the first row of ${\bf M}_t$ and let the rest of the matrix be zero all the time, then you can recover the case for $p$-dimensional vector-valued random variables: $$\mathbf{E} \left[ \left(\int_0^T {\bf X}^{T}_t d{\bf B}_t\right)^2\right] = \mathbf{E} \left[ \int_0^T \left\Vert{\bf X}_t\right\Vert_2^2 dt\right].$$ This is basically what you wrote down. [1]: https://i.stack.imgur.com/KNvI4.png