Brownian Motion: Calculate E[B(u) B(u+v) B(u+v+w)] using...

[lancer6238]

Hi all, I need help with a question.

Let B(t), t>= 0 be a standard Brownian motion and let u, v, w > 0. Calculate E[B(u) B(u+v) B(u+v+w)], using the fact that for a zero mean normal random variable Z, E[Z^3] = 0.

I tried to do this question by breaking up the brownian motions, i.e. B(u+v) = B(u) + (B(u+v) - B(u)) and B(u+v+w) = B(u) + (B(u+v) - B(u)) + (B(u+v+w) - B(u+v)), and then putting them into the expectation.

I got

E[B(u) B(u+v) B(u+v+w)] = E[ B(u) (B(u) + (B(u+v) - B(u))) (B(u) + (B(u+v) - B(u)) + (B(u+v+w) - B(u+v))) ]

and then expanding the terms, I got

E[B(u)^3] + E[ B(u)^2 (B(u+v) - B(u)) ] + E[ B(u)^2 (B(u+v+w) - B(u+v)) ] +
E[ B(u)^2 (B(u+v) - B(u)) ] + E[ B(u) (B(u+v) - B(u))^2 ] + E[B(u)] E[B(u+v) - B(u)] E[B(u+v+w) - B(u+v)]
= 2 E[ B(u)^2 (B(u+v) - B(u)) ] + E[ B(u)^2 (B(u+v+w) - B(u+v)) ] + E[ B(u) (B(u+v) - B(u))^2 ]

since the brownian motions in the last term are independent of one another and E[B(u)] = 0.

Up to here, I'm stuck as I'm not sure how to handle the square terms.

Am I correct up to this point? If I am, how should I continue?

Thank you.

Regards,
Rayne

 

[royhaas]

Don't forget to apply independence all the way through. B(x+h)-B(x) is independent of B(x).

 

[lancer6238]

I wasn't sure that B(u)^2 (B(u+v) - B(u)), B(u) (B(u+v) - B(u))^2 and B(u)^2 (B(u+v+w) - B(u+v)) were independent of each other (i.e. B(u) and B(u+v) - B(u) in the first 2 cases and B(u) and B(u+v+w) - B(u+v) in the last case) because of the squares. So if the expression can be factorized into

E[B(u) B(u+v) B(u+v+w)] = E[B(u)^3] + 2E[B(u)^2] E[B(u+v) - B(u)] + E[B(u)] E[(B(u+v) - B(u))^2] + E[B(u)^2] E[B(u+v+w) - B(u+v)] + E[B(u)]E[B(u+v) - B(u)]E[B(u+v+w) - B(u+v)]

then I should get E[B(u) B(u+v) B(u+v+w)] = 0 since
E[B(u)^3] = E[B(u)] = E[B(u+v) - B(u)] = E[B(u+v+w) - B(u+v)] = 0?