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 $$\langle I(g),I(g) \rangle=\int_{0}^t\int_{T}^S\int_T^S g(s,x)g(s,y)Q(x,y)dxdydt.$$

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 $x.$

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 $x.$

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 $$\langle I(g),I(g) \rangle=\int_{s}^t\int_{T}^S\int_T^S g(s,x)g(s,y)Q(x,y)dxdydt.$$

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 $x.$

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 $$\langle I(g),I(g) \rangle=\int_{0}^t\int_{T}^S\int_T^S g(s,x)g(s,y)Q(x,y)dxdydt.$$

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 $x.$

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

$$I(g)=\int_0^t\int_T^Sg(s,x)dW_s^x dx.$$ How can we find the quadratic variation $\langle I(g),I(g) \rangle$ of this integral?

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

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!

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 $$\langle I(g),I(g) \rangle=\int_{s}^t\int_{T}^S\int_T^S g(s,x)g(s,y)Q(x,y)dxdydt.$$

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 $x.$