1
$\begingroup$

I was hoping someone could answer a few basic Brownian Motion questions.

A Brownian Motion, $B_t$, (let us assume defined on the continuous path space with the Brownian Measure starting at $0$) has the following properties:

1) For $0 \leq t_0 < t_1, ..., t_n$ we have $B_{t_1} - B_{t_0}, ..., B_{t_n} - B_{t_{n-1}}$ are independent.

2) For $t \geq s, B_t - B_s$ is $N(0,t-s)$.

3) The map $t \rightarrow B_t$ is almost surely continuous.

4) $B_0 = 0$

Let $\mathcal{F}_t$ be the "infinitesimal peak into the future" filtration. I have seen the following claims.

a) $B_t - B_s$ is independent of $\mathcal{F}_s$

b) $\mathbb{E}(B_t - B_s| \: \mathcal{F}_s$) is $N(0,t-s)$. EDIT: This is incorrect

c) $B_t$ is a martingale wrt $\mathcal{F}_t$.

Could someone explain a) and b) for me? I believe a) follows from 1) but I am not 100% certain. b) is stating that the unconditional distribution (property 1)) and the conditional distribution are the same, which I find confusing.

My problem with c) is the following: If $B_t$ is a martingale wrt $\mathcal{F}_t$, why is it not the case that

$\mathbb{E}(B_t - B_s| \: \mathcal{F}_s) = \mathbb{E}(B_t|\: \mathcal{F}_s) -\mathbb{E}(B_s| \: \mathcal{F}_s) = B_s - B_s = 0$? (thus contradicting b))

$\endgroup$
  • $\begingroup$ Since $B_t-B_s$ is independent of $\mathcal F_s$, $\mathbb E[B_t-B_s\mid \mathcal F_s]=\mathbb E[B_t-B_s]$. $\endgroup$ – Math1000 Apr 30 at 22:30
  • $\begingroup$ b) is wrong. It holds that $B_t-B_s \sim N(0,t-s)$ (... which is nothing but 2)) and $\mathbb{E}(B_t-B_s \mid \mathcal{F}_s)=0$ (... which you correctly proved in your question) $\endgroup$ – saz May 1 at 5:07
  • $\begingroup$ I see that now. I believe there was a post somewhere that incorrectly claimed b). I will leave this up in case anyone else runs into this $\endgroup$ – user56628 May 1 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.