# Quick Brownian Motion question

• Apr 13th 2010, 11:51 AM
Quick Brownian Motion question
How do calculate something like...

$\mathbb{E}[W_{\tfrac{1}{2}} W_{\tfrac{3}{4}}]$ ..?

Where $\{W_t\}_{t \geq 0}$ is a standard Brownian motion.

$\frac{1}{\sqrt{2 \pi 1/2} \sqrt{2 \pi 3/4}} \int_{-\infty}^{\infty} e^{-x^2}e^{-\frac{2x^2}{3}} dx$ ..?
• Apr 13th 2010, 12:23 PM
Laurent
Quote:

How do calculate something like...

$\mathbb{E}[W_{\tfrac{1}{2}} W_{\tfrac{3}{4}}]$ ..?

Where $\{W_t\}_{t \geq 0}$ is a standard Brownian motion.

By writing $E[W_{1/2}W_{3/4}]=E[W_{1/2}(W_{1/2}+(W_{3/4}-W_{1/2}))]=E[(W_{1/2})^2]+E[W_{1/2}]E[W_{3/4-1/2}]$ using independence and stationarity of increments.
• Apr 13th 2010, 12:40 PM
Quote:

Originally Posted by Laurent
By writing $E[W_{1/2}W_{3/4}]=E[W_{1/2}(W_{1/2}+(W_{3/4}-W_{1/2}))]=E[(W_{1/2})^2]+E[W_{1/2}]E[W_{3/4-1/2}]$ using independence and stationarity of increments.

Cool, but also grim since I have like, 6 of these to calculate in a question...

I hope this isn't dumb but...

$\mathbb{E}[(W_{1/2})^2]+\mathbb{E}[W_{1/2}]\mathbb{E}[W_{3/4-1/2}] = \frac{1}{2} + 0$?

So the other ones I have which include...

$\mathbb{E}[W_{1/2}W_1]$ and $\mathbb{E}[W_{3/4}W_1]$ equal $\frac{1}{2}$ and $\frac{3}{4}$ respectively..?
• Apr 13th 2010, 12:51 PM
Laurent
Quote:

Cool, but also grim since I have like, 6 of these to calculate in a question...

I hope this isn't dumb but...

$\mathbb{E}[(W_{1/2})^2]+\mathbb{E}[W_{1/2}]\mathbb{E}[W_{3/4-1/2}] = \frac{1}{2} + 0$?

So the other ones I have which include...

$\mathbb{E}[W_{1/2}W_1]$ and $\mathbb{E}[W_{3/4}W_1]$ equal $\frac{1}{2}$ and $\frac{3}{4}$ respectively..?

That looks correct, well done.
• Apr 13th 2010, 01:00 PM
Allow me one further 'make or break my reputation with Laurent' conjecture...

For $b > a > 0$.

$\mathbb{E}[W_a W_b] = a$
• Apr 13th 2010, 01:06 PM
Laurent
Quote:

Allow me one further 'make or break my reputation with Laurent' conjecture...

For $b > a > 0$.

$\mathbb{E}[W_a W_b] = a$

No worry for your reputation :).

By the way, one of the definitions of Brownian motion is: a continuous centered Gaussian process ( $\forall t$, $W_t$ is a centered Gaussian r.v.) with covariance $E[W_aW_b]=\min(a,b)$ for all $a,b\geq 0$.
• Apr 13th 2010, 01:10 PM
By the way, one of the definitions of Brownian motion is: a continuous centered Gaussian process ( $\forall t$, $W_t$ is a centered Gaussian r.v.) with covariance $E[W_aW_b]=\min(a,b)$ for all $a,b\geq 0$.