Question about calculate expected value

Question

Assume $X(t)$ is a Brownian motion.

Find $E[X(u)X(u+v)X(u+v+w)]$, where $0<u<u+v<u+v+w$

I have an idea to solve this problem, as follows:

$E[X(u)X(u+v)X(u+v+w)]=E\{X(u)X(u+v)[X(u+v+w)-X(u+v)+X(u+v)]\}$

=$E\{X(u)X(u+v)[X(u+v+w)-X(u+v)]+X(u)X(u+v)X(u+v)\}$

=$E\{X(u)X(u+v)[X(u+v+w)-X(u+v)]+X(u)X(u+v)^2\}$

=$E\{X(u)[X(u+v)-X(u)+X(u)][X(u+v+w)-X(u+v)]+X(u)X(u+v)^2\}$

=$E\{X(u)[X(u+v)-X(u)][X(u+v+w)-X(u+v)]+X(u)^2[X(u+v+w)-X(u+v)]+X(u)X(u+v)^2\}$

=$E\{0+X(u)^2[X(u+v+w)-X(u+v)]+X(u)X(u+v)^2\}$ "Since $X(t)$ is Brownian motion"

=$E[X(u)^2X(u+v+w)-X(u)^2X(u+v)+X(u)X(u+v)^2]$

Now I get a problem, Can I say that $E[X(u)^2X(u+v+w)]$=0?

And is there have any better way to solve this problem? Thanks!!!

Answer 1

The distributions of $(X(u),X(u+v),X(u+v+w))$ and $(-X(u),-X(u+v),-X(u+v+w))$ coincide hence $E[X(u)X(u+v)X(u+v+w)]=-E[X(u)X(u+v)X(u+v+w)]=0$.