- #1
tyler_T
- 17
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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
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