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Brownian Motion 2 (probability)

  1. Oct 23, 2011 #1
    Problem:

    Let M(t) = max X(s), 0<=s<=t

    Show that P{ M(t)>a | M(t)=X(t)} = exp[-a^2/(2t)]

    Attempt at solution:

    It seems this should equal P(|X(t)| > a), but evaluating the normal distribution from a to infinity cannot be expressed in closed form as seen in the solution (unless this is somehow a special case).

    Note: X(t) ~ N(0,t)
    |X(t)| ~ N(sqrt(2t/pi),t(1-2/pi))

    Anybody help?

    -Tyler
     
  2. jcsd
  3. Oct 24, 2011 #2
    The reflection principle might be useful here: take a mirror image of the sample path about the line y=M(t), for segment of the path after it first hits the maximum
     
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