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Brownian Motion

  1. Apr 3, 2007 #1
    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.

  2. jcsd
  3. Apr 20, 2007 #2


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    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?
  4. Apr 29, 2007 #3
    Break it up step-by-step. First,

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

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

    [tex]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][/tex]

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

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

    [tex]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)].[/tex]
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