1. ## Brownian Motion

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. Originally Posted by 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
Hi. Your original idea is correct, but the notation makes it difficult to see what to do. I suggest this:

Let X = B(u), Y = B(u+v) - B(u), Z = B(u+v+w) - B(u+v).

Then you want to evaluate

E[X(X+Y)(X+Y+Z)] = E[X^3] + 2E[X^2 Y] + E[X Y^2] + E[X^2 Z] + E[XYZ].

Which of the three properties defining Brownian Motion allows you to factor E[XYZ] and the terms with squares into simpler expressions that can be immediately evaluated using the other two properties of Brownian Motion?

3. Originally Posted by JakeD
Hi. Your original idea is correct, but the notation makes it difficult to see what to do. I suggest this:

Let X = B(u), Y = B(u+v) - B(u), Z = B(u+v+w) - B(u+v).

Then you want to evaluate

E[X(X+Y)(X+Y+Z)] = E[X^3] + 2E[X^2 Y] + E[X Y^2] + E[X^2 Z] + E[XYZ].

Which of the three properties defining Brownian Motion allows you to factor E[XYZ] and the terms with squares into simpler expressions that can be immediately evaluated using the other two properties of Brownian Motion?
I wasn't sure that X^2 Y, X Y^2 and X^2 Z were independent of each other (i.e. X and Y in the first 2 cases and X and Z in the last case) because of the squares. So if the expression can be factorized into

E[X(X+Y)(X+Y+Z)] = E[X^3] + 2E[X^2] E[Y] + E[X] E[Y^2] + E[X^2] E[Z] + E[X]E[Y]E[Z]

then I should get E[X(X+Y)(X+Y+Z)] = 0 since
E[X^3] = E[X] = E[Y] = E[Z] = 0?

4. Originally Posted by lancer6238
I wasn't sure that X^2 Y, X Y^2 and X^2 Z were independent of each other (i.e. X and Y in the first 2 cases and X and Z in the last case) because of the squares. So if the expression can be factorized into

E[X(X+Y)(X+Y+Z)] = E[X^3] + 2E[X^2] E[Y] + E[X] E[Y^2] + E[X^2] E[Z] + E[X]E[Y]E[Z]

then I should get E[X(X+Y)(X+Y+Z)] = 0 since
E[X^3] = E[X] = E[Y] = E[Z] = 0?
That is the answer, but you do not need that X^2 and Y and the like are independent. What matters is that the independence of X and Y implies E[X^2 Y] = E[X^2] E[Y]. You should be able to show that easily just by evaluating the two sides using the definitions of expectation and independence.