# Basic Questions about Brownian Motion

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))

• Since $B_t-B_s$ is independent of $\mathcal F_s$, $\mathbb E[B_t-B_s\mid \mathcal F_s]=\mathbb E[B_t-B_s]$. – Math1000 Apr 30 at 22:30
• 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) – saz May 1 at 5:07
• 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 – user56628 May 1 at 15:14