Let $\{W_t\}_{t\ge0}$ be a Brownian Motion and $\{F_t\}_{t\ge0}$ its natural Filtration.
Let $0<r<s$. I want to determine $\mathbb E[(W_s-W_r)^2|F_t],\ \forall t \ge 0$.
I looked at the following cases:
1) $t\ge s$:
$E[(W_s-W_r)^2|F_t]=(W_s-W_r)^2$
(using that everything is measurable)
2) $t \in [r,s)$:
$E[(W_s-W_r)^2|F_t]=E[W_s^2-2W_sW_r+W_r^2|F_t]=s-t+W_t^2-2W_rW_t+W_r^2$
using that $W_r$ is measurable and using the martingale property of $W_s^2$ and $W_s$.
Does that look correct so far? And how can I do the 3rd case
$t<r$:
$E[(W_s-W_r)^2|F_t]=E[W_s^2-2W_sW_r+W_r^2|F_t]$. I thought about using again Martingale property for $W_s^2$ and $W_r^2$ but what can I do with the term $2W_sW_r$
Your answer is correct in the first two cases. If $t <r<s$ then $W_s-W_r$ is independent of $F_t$. [This can be proved from the fact that BM has independent increments]. Hence the answer in this case is $E(W_s-W_r)^{2}=s-r$.