Consider stochastic processes $X_t$, $\sigma_t$, and Wiener process $W_t$, with $X_t$ independent of $\sigma_s$ $\forall t, s$, and $W_t$ independent of $\sigma_s$ $\forall t, s$, but where $X_t$ and $W_s$ are independent $\forall t \neq s$. Also, $\mathbb{E} X_t = 0$. What is the value of the expression:
\begin{equation}
\mathbb{E} X_T \int_0^T \sigma_t dW_t
\end{equation}
Clearly, the only part of the integral exhibiting dependency with $X_T$ is the end-point $W_T$. Does this prevent me from separating the expression into $\mathbb{E} X_T \mathbb{E} \int_0^T \sigma_t dW_t$?
Best Answer
Two random variables $X$ and $W$ are independent if the sigma algebras $\sigma(X)$ and $\sigma(W)$ are independent sigma algebras. Then under your assumptions we could say that $\sigma(X_s)$ is independent of $\sigma(W_t)$ for all $t<s$, then $\sigma(X_s)$ must be independent of $\bigcap_{t<s} \sigma(W_t)$.
But by continuity of the Wiener processes we have that $\bigcap_{t<s} \sigma(W_t)= \sigma(W_s)$.