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Math Help - Brownian Motion

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by lancer6238 View Post
    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?
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  3. #3
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    Quote Originally Posted by JakeD View Post
    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?
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  4. #4
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    Quote Originally Posted by lancer6238 View Post
    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.
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