# Brownian Motion

• Mar 17th 2010, 11:00 AM
Brownian Motion
I don't understand this one small point about Brownian motion...

Here's the question, my answer and the part I always miss out...

Let $\displaystyle \{W_t\}_{t \geq 0}$ be a standard Brownian motion. Show that.
(a) $\displaystyle \{X_t\}_{t \geq0}$ is a standard Brownian motion, where $\displaystyle X_t = -W_t$

Question is easy enough. Can show that $\displaystyle X_0 = 0$ almost surely and get that for $\displaystyle 0 \leq s \leq t$, $\displaystyle X_t - X_s = -(W_t - W_s)$ is distributed with mean $\displaystyle 0$ and variance $\displaystyle t-s$ and is independent of $\displaystyle F_s = \sigma (W_u: 0 \leq u \leq s)$. (Not ENTIRELY sure why this is though)

Then comes this small statement that I don't know how to show (and hence don't know if it's true).

$\displaystyle X_t$ has continuous paths.

How do I show that? It's only ever stated in answers so I don't need some in depth proof, just a simple, it has continuous paths because...
• Mar 17th 2010, 11:13 AM
Laurent
Quote:

it has continuous paths because...

because if a function $\displaystyle f$ is continuous, then so is $\displaystyle -f$. That's all! Now you get why it's not extensively proved in the answers... (Wink)

If $\displaystyle t\mapsto W_t$ is continuous, then so is $\displaystyle t\mapsto -W_t$.
• Mar 17th 2010, 02:11 PM
Focus
Quote:

Can show that $\displaystyle X_0 = 0$ almost surely and get that for $\displaystyle 0 \leq s \leq t$, $\displaystyle X_t - X_s = -(W_t - W_s)$ is distributed with mean $\displaystyle 0$ and variance $\displaystyle t-s$ and is independent of $\displaystyle F_s = \sigma (W_u: 0 \leq u \leq s)$. (Not ENTIRELY sure why this is though)
The reason for the law is because the normal distribution is symmetric. The independence follows rather easily if you use the weak definition (see below), that is for any continuous bounded function f and $\displaystyle A \in \mathcal{F}_s$ you have that (by independent increments of BM)
$\displaystyle \mathbb{E}[f(W_t-W_s)\mathbf{1}_{A}]=\mathbb{E}[f(W_t-W_s)]\mathbb{P}(A)$.
Now if you consider any f that is bounded and continuous then $\displaystyle g(x)=f(-x)$ is bounded and continuous hence
$\displaystyle \mathbb{E}[f(-(W_t-W_s))\mathbf{1}_A]=\mathbb{E}[g(W_t-W_s)\mathbf{1}_A]=\mathbb{E}[g(W_t-W_s)]\mathbb{P}(A)$
$\displaystyle =\mathbb{E}[f(-(W_t-W_s))]\mathbb{P}(A)$.