Brownian Motion

**Deadstar** · April 1st 2010, 07:29 AM

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.

**Laurent** · April 1st 2010, 09:21 AM

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.

**Deadstar** · April 1st 2010, 09:47 AM

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...) $\mathbb{E}[W_t - W_s] = 0$ ..?

**Laurent** · April 1st 2010, 10:32 AM

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).

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