Let $\{W_x\}_{x \in [0,t]}$ be a Wiener process and $f(x)$ and $g(x)$ exponential functions (with different coefficients inside). Let $s$ be a positive number, with $0<s<t$.
I would like to calculate the following expectation value:
$\mathbb{E}\left[\int_0^t \int_0^s f(x)f(y)g(W_x)g(W_y)\, dW_x \, dW_y \right]$
Since the two integrals are not defined on the same interval, I guess I can't use the Ito isometry, at least in its most common form. I am wondering if there are any other useful tricks to do that.
$\int_0^t \int_0^s f(x)f(y)g(W_x)g(W_y)\, dW_x \, dW_y$ can be rewritten as $\int_0^t \int_0^t 1_{y<s}f(x)f(y)g(W_x)g(W_y)\, dW_x \, dW_y$
Then you can apply the isometry on $X_t$ and $Y_t$ where
$X_t=f(x)g(W_x)$
and $Y_t=1_{t<s}f(y)g(W_y)$