Conditional Ito's isometry

Question

I am looking for a formal proof of the following (if true):

$\mathbb E \left[ \int_0^1 g_1(s)\,dW_s \int_0^1 g_2(s) K_s\,dW_s \big|\mathscr F^K \right]=\int_0^1 g_1(s)g_2(s)K_s\,ds $,

where $g_1,g_2$ are deterministic functions, $W$ is the Wiener process, $K$ is a square integrable continuous stochastic process, $W$ and $K$ are independent and $\mathscr F^K$ is the sigma algebra generated by the process $K$.

Are there further conditions we need to assume for it to be true?

I was trying to use the known fact that for a pair of variables $X,Y$, a Borel function $h$ and a sigma algebra $\mathscr F$ such that $X$ is $\mathscr F$ measurable and $Y$ is $\mathscr F$ independent

$\mathbb E\left[ h(X,Y)\Bigg|\mathscr F\right]=H(X)$, where $H(x)=\mathbb E \left[h(x,Y)\right]$.

Regarding the processes $W$ and $K$ as random variables with values in the canonical space of continuous paths, I thought one could set:

$h(W,K)=\int_0^1 g_1(s)\,dW_s\int_0^1g_2(s)K_s\,dW_s$

The problem is that such a function $h$ need not exist since the stochastic integral cannot be constructed pathwise.

Any help is greatly appreciated.

Answer 1

Hint:

1. Case 1: $g_1$, $g_2$ and $K$ are simple functions. Then the claim follows from a straight-forward calculation (since the stochastic integrals can be calculated explicitly).
2. Case 2: $K$ is a simple function and $g_1,g_2$ are deterministic measurable (square integrable) functions. Approximate $g_1$ and $g_2$ by simple functions and use step 1.
3. Case 3: $K$ is an arbitrary function (satisfying all the assumptions in the OP). Approximate $K$ by simple functions and conclude.

(Don't hesitate to ask if you get stuck; the calculations are rather messy and technical but not that complicated.)