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Consider the process $W^x_t$ which is a Brownian motion for every $x\geq 0$ such that $$d\langle W_t^x,W_t^y\rangle=Q(x,y)\,dt$$

where $Q$ is some non-negative definite function. Now consider the integral

$$I(g)=\int_{s=0}^t\int_{x=T}^Sg(s,x) \, dW_s^x \, dx.$$ How can we find the quadratic variation $\langle I(g),I(g) \rangle$ of this integral?

My guess is that $$\langle I(g),I(g) \rangle=\int_0^t\int_T^S\int_T^S g(s,x)g(s,y) Q(x,y) \, dx \, dy \, dt.$$

However, I do not know if this is correct or how this can be proved (or if this follows from some well-known result). Any help or references are much appreciated!

EDIT: For a fixed $t$ think of $W^x_t$ as some continuous function in $x.$

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  • $\begingroup$ Can you provide some context for the integral? I think it might help. I am pretty curious about this question too. $\endgroup$
    – Nate River
    Commented Jul 4, 2021 at 8:40
  • $\begingroup$ The inner integral can be thought of as a regular Reimann integral. The outer integral (in $s$) is w.r.t a Brownian motion. Therefore, it's like a stochastic integral of a Riemann integral. Does this make sense @NateRiver? $\endgroup$
    – Heisenberg
    Commented Jul 4, 2021 at 18:08
  • $\begingroup$ Right I was more thinking, how does this integral arise, what is it intuitively. Questions like that. $\endgroup$
    – Nate River
    Commented Jul 4, 2021 at 18:24
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    $\begingroup$ @NateRiver More like a specific case of a stochastic integral w.r.t a Brownian motion taking values in a Banach space (in this case, the space of a continuous functions). I came across this integral in this paper worldscientific.com/doi/abs/10.1142/9789812702852_0002. It is used in a finance context. I think you can think of this as an integral in the form $$\int_0^t \Sigma(s,T_1,T_2)dW(s)$$ where $W(s)\in C(R)$ and $\Sigma(s,T_1,T_2)\in C^*(R)$ such that $$\Sigma(s,T_1,T_2)f=\int_S^T \sigma(s,u)f(u)du.$$ $\endgroup$
    – Heisenberg
    Commented Jul 4, 2021 at 18:33

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