# Expectation of Brownian Motion adapted for Filtration

I just started learning about Brownian Motions and martin gales and have the following issue.

If $$X_t$$ is a brownian motion, I cannot understand how the results below are different when adapting for filtration $$\mathcal{F}_s$$.

$$\mathbb{E[x^{2}_s | \mathcal{F}_s]} = x^{2}_s$$ whereas $$\mathbb{E[x^{2}_s]} = s$$

• Before you started learning about BM have you learned about conditional expectation ? The first result is then trivial. The second is nothing else than the variance of $x_s$. Oct 20, 2022 at 15:36
• I understand conditional expectation as $\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X|Y]]$ but I cannot see how to go from there to the first. For the second, I believe the result comes from $\mathbb{E}[X^2_s] = Var(X_s)+\mathbb{E}[X_s]^2$ and since $X_s$ is centered then its just $Var(X)$ which is $s$? @KurtG. Oct 20, 2022 at 16:12
• @KurtG. Also I understand that since $X_m$ is a brownian motion its therefore a martingale and $\mathbb{E}[X_m|\mathcal{F}_s] = X_s$. But I dont know whether that is generalisable to $X^2_s$, and if so why. Oct 20, 2022 at 16:20
• Correct: $\operatorname{Var}(X_\color{red}{s})=s$ . A hint about $\mathbb E[X_s^2|{\cal F}_s]=X_s^2\,:$ we have an $s$ everywhere. What did you learn about the conditional expectation of an ${\cal F}_s$-measurable random variable ? Oct 20, 2022 at 18:27
• It is as simple as that. Oct 21, 2022 at 11:49