I have a real number $x$, and $W$ is a standard Brownian motion. Let $0 < s < t$. How to find $$ \mathsf E[W_s | W_t = x] $$

Please provide me with a step by step answer as I want to understand your steps and the concept.

Many thanks in advance.


1 Answer 1


Let $U_{s,t}=W_s-(s/t)W_t$ and $V_{s,t}=(s/t)W_t$. The process $W$ is centered gaussian, $\mathbb E(W_t^2)=t$ and $\mathbb E(W_sW_t)=s$, hence $U_{s,t}$ and $W_t$ are independent. Thus, $W_s=U_{s,t}+V_{s,t}$ where $U_{s,t}$ is centered and independent of $W_t$ and $V_{s,t}$ is measurable with respect to $W_t$.

Thus, $\mathbb E(W_s\mid W_t)=\mathbb E(U_{s,t}\mid W_t)+\mathbb E(V_{s,t}\mid W_t)=$ $____$ $+$ $____$ $=$ $____$.

  • $\begingroup$ Is true that since $U_{s,t}$ is independent of $W_{t}$ then the corresponding conditional expectation $\mathbb{E}(U_{s, t} | W_{t})$ reduces to zero and $\mathbb{E}(\frac{s}{t}W_{t} | W_{t}) = \frac{s}{t}W_{t}$ due to the basic properties of an expectation?? $\endgroup$ Apr 6, 2017 at 13:16
  • $\begingroup$ @Arteom Yes, if $X$ is independent of $Y$ then $E(X\mid Y)=E(X)$ while if $X$ is $\sigma(Y)$-measurable then $E(X\mid Y)=X$. Which source on conditional expectations does not explain this? $\endgroup$
    – Did
    Apr 6, 2017 at 14:41
  • $\begingroup$ How does this imply $U_{s,t}$ and $W_t$ are independent? $\endgroup$
    – user428487
    Apr 3, 2018 at 4:22
  • 2
    $\begingroup$ @user428487 "This" does not imply that they are independent, rather one uses the fact that they are to deduce other facts. $\endgroup$
    – Did
    Apr 3, 2018 at 7:57
  • $\begingroup$ Why U and W_t are independent? $\endgroup$
    – ggkkll
    Nov 5, 2021 at 17:09

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