Conditional distribution of Ito's integral

Question

Suppose $\sigma=\sigma(t)$ is a deterministic process and $X(t)=\int_0^t \sigma(u)dW(u)$. I would like to compute the conditional distribution of $x(t)-x(s)$ given $\mathcal{F}_s$, where $0\leq s<t$. My intuition is that since $$X(t)-X(s)=\int_s^t \sigma(u)dW(u),$$ it should be that $$\mathbb{E}[X(t)-X(s)|\mathcal{F}_s]=0$$ $$Var(X(t)-X(s)|\mathcal{F}_s)=\int_s^t\sigma(u)^2du$$ How can I prove such claim (if it is correct)?

Answer 1

$X(t)-X(s)$ is independent of $\mathcal F_s$. $$ E[(X(t)-X(s))^2] = E[X(t)^2-2X(t)X(s)+ X(s)^2]= E[X(t)^2] -2E[X(s)X(t)] +E[X(s)^2]= \int_0^t \sigma^2(u)du -2\int_0^s \sigma^2(u)du +\int_0^s \sigma^2(u)du=\int_s^t\sigma^2(u)du. $$ Another way: since $X(t)-X(s)$ and $X(s)$ are independent $$D^2(X(t))=D^2(X(t)-X(s)+X(s))= D^2(X(t)-X(s)) + D^2(X(s))$$ from which you get $$D^2(X(t)-X(s))= D^2(X(t))-D^2(X(s)).$$