Thread: Brownian Motion

1. Brownian Motion

Show that a standard Brownian motion $\{W_t\}_{t \geq 0}$ is a martingale with respect to the filtration $\{F_t \}_{t \geq 0}$, where $F_t = \sigma(W_s : 0 \leq s \leq t)$. Moreover, show that $\{ W_t^2 + t \}_{t \geq 0}$ is a martingale with respect to the filtration $\{F_t \}_{t \geq 0}$.

First part I have done, but for the second part I arrive at a point where I have this...

$\mathbb{E}[(W_t - W_s)^2]$

which is apparently equal to $t-s$. Why is this?

(More info if needed)
$\mathbb{E}[(W_t^2 - t |F_s] = \mathbb{E}[(W_t - W_s + W_s)^2 - t|F_s]$ and continuing I arrive at the above result I'm stuck at.

2. Originally Posted by Deadstar
I have this...

$\mathbb{E}[(W_t - W_s)^2]$

which is apparently equal to $t-s$. Why is this?
This results possibly from your definition of Brownian motion : for $0\leq s\leq t$, $W_t-W_s$ is a Gaussian random variable with mean 0 and variance $t-s$, that furthermore is independent of $W_s$. If that doesn't answer your question, please give us your definition of Brownian motion.

3. Originally Posted by Laurent
This results possibly from your definition of Brownian motion : for $0\leq s\leq t$, $W_t-W_s$ is a Gaussian random variable with mean 0 and variance $t-s$, that furthermore is independent of $W_s$. If that doesn't answer your question, please give us your definition of Brownian motion.
That seems to be right relative to what we've been told!

So does that mean...

$\mathbb{E}[(W_t - W_s)^n] = t-s$ $\forall n \geq 2$?
Or perhaps you take $t-s$ to the (n-1)th power..?

I say 2 since (correct me if I'm wrong which is likely...) $\mathbb{E}[W_t - W_s] = 0$..?

4. Originally Posted by Deadstar
So does that mean...

$\mathbb{E}[(W_t - W_s)^n] = t-s$ $\forall n \geq 2$?
Or perhaps you take $t-s$ to the (n-1)th power..?
No! It says that $E[(W_t-W_s)^n]$ is the n-th moment of a Gaussian random variable of mean 0 and variance $t-s$, that's all. For $n=2$, $E[(W_t-W_s)^2]$ is the variance itself (mean = 0), so it is $t-s$. We can also write $W_t-W_s=\sqrt{t-s}Z$, so that $Z$ is a standard Gaussian random variable (if $Z\sim \mathcal{N}(0,1)$ then $\lambda Z\sim \mathcal{N}(0,\lambda^2$)); then $E[(W_t-W_s)^n]=(t-s)^{n/2} E[Z^n]$ and there are formulas for $E[Z^n]$. By the way, $E[(W_t-W_s)^n]=0$ if $n$ is odd (by symmetry).