(New page: Given X~exp(<math>lambda</math>) CDF of X = 1 - exp(-lambda*x) = P[X<=x] P[X>x] = exp(-lambda*x) P[Y>k] = P[X>k] = exp(-lambda*k) P[Y>k] = P[Y=k+1] + P[Y=k+2] + ... -- (1) P[Y>k-1] = P[...) |
(No difference)
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Revision as of 16:33, 15 October 2008
Given X~exp($ lambda $) CDF of X = 1 - exp(-lambda*x) = P[X<=x] P[X>x] = exp(-lambda*x) P[Y>k] = P[X>k] = exp(-lambda*k)
P[Y>k] = P[Y=k+1] + P[Y=k+2] + ... -- (1) P[Y>k-1] = P[Y=k] + P[Y=k+1] + P[Y=k+2] + ... -- (2)
= P[Y=k] + P[Y>k]
then find P[Y=k] ...
Alternatively, without using the hint given, we can also approach the problem using the PDF of X PDF of X = lambda*exp(-lambda*x) For Y to have a value k (k is integer), X has to fall within the range of k-1 to k P[Y=k] = P[k-1<X<k]
= (integ:k-1 to k) lambda*exp(-lambda*x) dx = ...
Both methods should lead to the same answer.
