0
$\begingroup$

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 $\mathbb{E[x^{2}_s]} = s$

$\endgroup$
6
  • $\begingroup$ 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 $x_s$. $\endgroup$
    – Kurt G.
    Oct 20, 2022 at 15:36
  • $\begingroup$ 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. $\endgroup$
    – nachofest
    Oct 20, 2022 at 16:12
  • $\begingroup$ @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. $\endgroup$
    – nachofest
    Oct 20, 2022 at 16:20
  • $\begingroup$ 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 ? $\endgroup$
    – Kurt G.
    Oct 20, 2022 at 18:27
  • 1
    $\begingroup$ It is as simple as that. $\endgroup$
    – Kurt G.
    Oct 21, 2022 at 11:49

0

You must log in to answer this question.

Browse other questions tagged .