Brownian motion or not?

Question

Suppose that $(X_t , t\in [0;1])$ are independent normal r.v with mean 0 and variance $\sigma^2 _{t}$. Is this process brownian motion?

Got something from an answer below? – Did Jul 05 '13 at 07:44 — Did, Jul 05 '13 at 07:44

user81566 · Answer 1 · 2013-06-12T16:56:55.640

1

You first need $\sigma_t=t$, since the Brownian motion should have variance $t$ at time $t$.

A condition for $X$ to be a Brownian motion is that for any $s<t$, $$X_t-X_s\sim N(0,t-s)\qquad (1).$$

From the independence of $X_t$ and $X_s$, you get $$X_t-X_s\sim N(0,t+s),$$ which is a contradiction to the condition $(1)$.

edited Jun 12 '13 at 16:56

answered Jun 12 '13 at 16:36

user81566

337

Why it variance is t+s? – Mex Jun 12 '13 at 17:36
This comes from the independence of the normal random variables $X_t$ and $X_s$. See link for example. – user81566 Jun 12 '13 at 17:49

score 0 · Answer 2 · answered Jun 14 '13 at 06:06

0

The answer is never, because, if $(W_t)_{t\geqslant0}$ is a Brownian motion, then $W_{1/2}$ and $W_1$ are not independent while you assumed that $X_{1/2}$ and $X_1$ are independent.

answered Jun 14 '13 at 06:06

Did

279,727

It is the increments that are independent. – GEdgar Jul 04 '13 at 00:33
@GEdgar The increments of a Brownian motion: yes. The increments of the process $(X_t)$ in this question: no. Which is the point of this answer. Is this also the point of your (rather cryptic) comment? – Did Jul 04 '13 at 06:17
Yes, the (cryptic) comment was to help Mex understand better. – GEdgar Jul 04 '13 at 12:07

Brownian motion or not?

2 Answers2