# Brownian Motion

1. Apr 3, 2007

### wu_weidong

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

2. Apr 20, 2007

### EnumaElish

Is there a way to break up B(u+v) into, say, B(u) + B(v) + "other terms" in a meaningful way?

Other than this, can't the square terms be expressed in terms of a distribution parameter, e.g. σ2?

3. Apr 29, 2007

### Jason Swanson

Break it up step-by-step. First,

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

By independence, the first term on the right-hand side is zero. Next,

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

Again by independence, the first and second terms are zero.

Another way to get the result is to note that $$W(t):=-B(t)$$ is also a Brownian motion. So $$E[W(u)W(u+v)W(u+v+w)]$$ should be the same number. But

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