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..

  • $\begingroup$ First question: the sigma-algebras generated by $(X,Y)$ and by $(X,Y-X)$ coincide since each random vector is a function of the other. Second question: a general property of gaussian families is that conditional expectations are linear combinations, in particular, for every gaussian vector $(X,Y)$, $E(X\mid Y)=aY+b$ where $(a,b)$ is uniquely determined by $E(X)=aE(Y)+b$ and $E(XY)=aE(Y^2)+bE(Y)$. $\endgroup$ – Did Oct 19 '15 at 10:20
  • $\begingroup$ Note furthermore that the second remark allows to bypass some steps in the computations you present in your question since $E(W_s\mid W_t,W_u)=aW_t+bW_u+c$ where $$E(W_s)=aE(W_t)+bE(W_u)+c\quad E(W_sW_t)=aE(W_t^2)+bE(W_tW_u)+cE(W_t)\quad E(W_sW_u)=aE(W_tW_u)+bE(W_u^2)+cE(W_u),$$ that is, since $t<s<u$, $$0=c\quad t=at+bt\quad s=at+bu,$$ from which $(a,b,c)$ follows readily. $\endgroup$ – Did Oct 19 '15 at 10:26

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