For a brownian motion, $W_t$ with scale $\sigma^2$ and for $t<s<u$, I want to find: $$ E(W_s|W_t=a,W_u=b) $$ I have the solution, but I am confused about two of the following statements:
First $$ E(W_s|W_t,W_u) = E(W_s|W_t,W_u-W_t) $$ does this follow due to the fact that $(W_u,W_t) \overset{d}{=} (W_u, W_t-W_u)$? The solution then goes on to show that we may simplify this down to: $$ E(W_s-W_t|W_u-W_t) + W_t $$ which makes sense. The solution then says, using the "Best linear prediction"
$$ E(W_s | W_t,W_u) = \frac{cov(W_s - W_t, W_u - W_t) ~~(W_u - W_t)}{Var (W_u - W_t)} $$ I'm not really sure where this formula comes from either? Is it simply some sort of linear interpolation between the points? Does anyone know a derivation of this, this might even be a typo since the solution simplifies down $E(W_s|W_t,W_u)$ to $E(W_s-W_t|W_u-W_t) + W_t$ and then just uses the linear prediction..