# Thread: Quick Brownian Motion question

1. ## Quick Brownian Motion question

How do calculate something like...

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

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

$\displaystyle \frac{1}{\sqrt{2 \pi 1/2} \sqrt{2 \pi 3/4}} \int_{-\infty}^{\infty} e^{-x^2}e^{-\frac{2x^2}{3}} dx$ ..?

How do calculate something like...

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

Where $\displaystyle \{W_t\}_{t \geq 0}$ is a standard Brownian motion.
By writing $\displaystyle E[W_{1/2}W_{3/4}]=E[W_{1/2}(W_{1/2}+(W_{3/4}-W_{1/2}))]=E[(W_{1/2})^2]+E[W_{1/2}]E[W_{3/4-1/2}]$ using independence and stationarity of increments.

3. Originally Posted by Laurent
By writing $\displaystyle E[W_{1/2}W_{3/4}]=E[W_{1/2}(W_{1/2}+(W_{3/4}-W_{1/2}))]=E[(W_{1/2})^2]+E[W_{1/2}]E[W_{3/4-1/2}]$ 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...

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

So the other ones I have which include...

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

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

I hope this isn't dumb but...

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

So the other ones I have which include...

$\displaystyle \mathbb{E}[W_{1/2}W_1]$ and $\displaystyle \mathbb{E}[W_{3/4}W_1]$ equal $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{3}{4}$ respectively..?
That looks correct, well done.

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

For $\displaystyle b > a > 0$.

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

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

For $\displaystyle b > a > 0$.

$\displaystyle \mathbb{E}[W_a W_b] = a$
No worry for your reputation .

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

7. Originally Posted by Laurent
No worry for your reputation .

By the way, one of the definitions of Brownian motion is: a continuous centered Gaussian process ($\displaystyle \forall t$, $\displaystyle W_t$ is a centered Gaussian r.v.) with covariance $\displaystyle E[W_aW_b]=\min(a,b)$ for all $\displaystyle a,b\geq 0$.
Ah cool, the more you know...

Cheers for the help.