I reading the proof of R. Schilling in his book : Brownian motion, introduction to stochastic process, and his proof is really unclear for me. I want to prove that if $u$ is bounded, then $$\mathbb E[u(B_{t+s})\mid \mathcal F_s]=\mathbb E[u(B_t)\mid B_s],$$ where $(\mathcal F_t)$ is an admissible filtration to $(B_t)$ (i.e. $B_t-B_s$ is independent of $\mathcal F_s$ for all $0\leq s\leq t$, and $(B_t)$ is a Brownian motion.

The proof goes as follow :

$$\mathbb E[u(B_{t+s})\mid \mathcal F_s]=\mathbb E[u(B_{t+s}-B_s+B_s)\mid \mathcal F_s]=\mathbb E[u(B_t+x)]|_{x=B_s}=\mathbb E[u(B_t)\mid B_s],$$ but I neither understand the notation $\mathbb E[u(B_t+x)]|_{x=B_s}$ nor the two last inequalities. I thought that $$\mathbb E[u(B_t+x)]|_{x=B_s}:=\mathbb E[u(B_t+B_s)\mid B_s=x],$$ but it doesn't really make sense. Any help is welcome.

  • 1
    $\begingroup$ $$\mathbb{E}[u(B_t+x)] \bigg|_{x=B_s}$$ is a short-hand for the function $$f(x) := \mathbb{E}[u(B_t+x)]$$ evaluted at $x=B_s$, i.e. $$\mathbb{E}[u(B_t+x)] \bigg|_{x=B_s} = f(B_s).$$ $\endgroup$ – saz Apr 6 at 9:00
  • $\begingroup$ @saz: So, it's exactly $\mathbb E[u(B_t+B_s)]$, no ? So why doe he use this notation ? $\endgroup$ – user657324 Apr 6 at 9:20
  • $\begingroup$ No, it's not the same as the expression in your comment. This is already indicated by the fact that $f(B_s)$ is a random variable whereas $\mathbb{E}u(B_t+B_s)$ is just a constant. $\endgroup$ – saz Apr 6 at 9:23
  • $\begingroup$ Oh, I see. Just to look through my notation, could it be something as $\mathbb E[u(B_t+B_s)\mid B_s]$ ? This is a more comon notation, no ? @saz $\endgroup$ – user657324 Apr 6 at 9:26
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    $\begingroup$ Forget for the moment about conditional expectations. If $X$ and $Y$ are two random variables, then $$\mathbb{E}[u(x,Y)] \bigg|_{x=X}$$ is defined as $$\mathbb{E}[u(x,Y)] \bigg|_{x=X} := f(X)$$ where $$f(x) := \mathbb{E}(u(x,Y)).$$ There are no conditional expectations around. Note that, in general, $$\mathbb{E}[u(x,Y)] \bigg|_{x=X} \neq \mathbb{E}(u(X,Y))$$ and $$\mathbb{E}[u(x,Y)] \bigg|_{x=X} \neq \mathbb{E}(u(X,Y) \mid X).$$ $\endgroup$ – saz Apr 6 at 10:32

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