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Let $U,H$ be a separable $\mathbb R$-Hilbert spaces, $M$ be a $U$-valued square-integrable martingale on a filitered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\in[0,\:T]},\operatorname P)$ with martingale covariance $Q$, $\Phi:\Omega\times[0,T]\to\mathfrak L(U,H)$ be a predictable process with $$\operatorname E\left[\int_0^T\left\|\Phi_sQ_s^{1/2}\right\|_{\operatorname{HS}(U,\:H)}^2\:{\rm d}[M]_s\right]<\infty$$ and $\Phi\cdot M$ denote the Itō integral process.

Above $[M]$ denotes the (scalar) quadratic variation of $M$, i.e. the unique nondecreasing adapted process such at $\left\|M\right\|_H^2-[M]$ is a square-integrable martingale. Moreover, if $[\![M]\!]$ denotes the tensor-quadratic variation of $M$, i.e. the unique adapted process with values in the space of nuclear, nonnegative and self-adjoint operators on $U$ such that $M\otimes M-[\![M]\!]$ is a square-integrable martingale, then $$[\![M]\!]_t=\int_0^tQ_s\:{\rm d}[M]_s\;\;\;\text{for all }t\in[0,T]\;.$$

Now, if $N$ is another $U$-valued square-integrable martingale, can we find an expression for the covariation $[\Phi\cdot M,N]$?

I'm also interested in an expression for the tensor-covariation $[\![\Phi\cdot M,N]\!]$ where $N$ might take values in another separable $\mathbb R$-Hilbert space $\tilde H$.

All I was able to find in books are the classical results $$[\Phi\cdot M,\Psi\cdot M]_t=\int_0^t\langle\Phi_sQ_s^{1/2},\Phi_sQ_s^{1/2}\rangle_{\operatorname{HS}(U,\:H)}\:{\rm d}[M]_s$$ and $$[\![\Phi\cdot M]\!]_t=\int_0^t\Phi_sQ_s\Phi_s^\ast\:{\rm d}[M]_s\;.$$

Remark: Cross-posted on mathoverflow.

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  • Hi, I asked a few months ago a similar question: https://math.stackexchange.com/questions/2496441/hilbert-valued-martingales-help-with-reference . Unfortunately none answered in that case neither, but many of the formulas you use already partially answer my question. In which books have you found them? I'm quite new to stochastic integration in infinite dimension, and many of these formulas do not appear on Da Prato's book (at least as far as I know) – Lucio Feb 11 '18 at 20:29

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