(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Brownian Motion 2 (probability)

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