1. ## Brownian Motion

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

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

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

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

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

I have this...

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

which is apparently equal to $\displaystyle t-s$. Why is this?
This results possibly from your definition of Brownian motion : for $\displaystyle 0\leq s\leq t$, $\displaystyle W_t-W_s$ is a Gaussian random variable with mean 0 and variance $\displaystyle t-s$, that furthermore is independent of $\displaystyle W_s$. 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 $\displaystyle 0\leq s\leq t$, $\displaystyle W_t-W_s$ is a Gaussian random variable with mean 0 and variance $\displaystyle t-s$, that furthermore is independent of $\displaystyle W_s$. 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...

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

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

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