# Brownian Motion

• Apr 3rd 2007, 09:46 PM
lancer6238
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
• Apr 4th 2007, 12:55 AM
JakeD
Quote:

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?
• Apr 5th 2007, 11:02 PM
lancer6238
Quote:

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?
• Apr 8th 2007, 10:18 PM
JakeD
Quote:

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.