(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[...) |
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Given X~exp(<math>lambda</math>) | Given X~exp(<math>lambda</math>) | ||
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CDF of X = 1 - exp(-lambda*x) = P[X<=x] | CDF of X = 1 - exp(-lambda*x) = P[X<=x] | ||
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P[X>x] = exp(-lambda*x) | P[X>x] = exp(-lambda*x) | ||
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P[Y>k] = P[X>k] = exp(-lambda*k) | P[Y>k] = P[X>k] = exp(-lambda*k) | ||
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P[Y>k] = P[Y=k+1] + P[Y=k+2] + ... -- (1) | P[Y>k] = P[Y=k+1] + P[Y=k+2] + ... -- (1) | ||
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P[Y>k-1] = P[Y=k] + P[Y=k+1] + P[Y=k+2] + ... -- (2) | P[Y>k-1] = P[Y=k] + P[Y=k+1] + P[Y=k+2] + ... -- (2) | ||
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= P[Y=k] + P[Y>k] | = P[Y=k] + P[Y>k] | ||
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then find P[Y=k] ... | then find P[Y=k] ... | ||
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Alternatively, without using the hint given, we can also approach the problem using the PDF of X | Alternatively, without using the hint given, we can also approach the problem using the PDF of X | ||
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PDF of X = lambda*exp(-lambda*x) | PDF of X = lambda*exp(-lambda*x) | ||
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For Y to have a value k (k is integer), X has to fall within the range of k-1 to k | For Y to have a value k (k is integer), X has to fall within the range of k-1 to k | ||
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P[Y=k] = P[k-1<X<k] | P[Y=k] = P[k-1<X<k] | ||
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= (integ:k-1 to k) lambda*exp(-lambda*x) dx = ... | = (integ:k-1 to k) lambda*exp(-lambda*x) dx = ... | ||
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Both methods should lead to the same answer. | Both methods should lead to the same answer. | ||
Latest revision as of 16:34, 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.
