The expectation of a geometric random variable where its parameter is uniform











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First thanks for any help editing my text.



If a random variable $X$ has a geometric distribution with parameter $P$ where $P$ itself is a random variable and uniformly distributed from $0$ to $1-1/n$, I want to find $E[X]$.



I first learned that $E[X|P] = 1/P$, using the expectation of the geometric distribution.



But when it comes to $E[X]=E[E[X|P]]$, I tried integration and it seems to be



(integral from $0$ to $(n-1)/n$) , $(1/p)$ * $(n/(n-1))$ $dp$



but isn't the value infinite?






probability-distributions conditional-expectation uniform-distribution expected-value







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  • I think you are right.
    – Kavi Rama Murthy
    5 hours ago















up vote
0
down vote

favorite












First thanks for any help editing my text.



If a random variable $X$ has a geometric distribution with parameter $P$ where $P$ itself is a random variable and uniformly distributed from $0$ to $1-1/n$, I want to find $E[X]$.



I first learned that $E[X|P] = 1/P$, using the expectation of the geometric distribution.



But when it comes to $E[X]=E[E[X|P]]$, I tried integration and it seems to be



(integral from $0$ to $(n-1)/n$) , $(1/p)$ * $(n/(n-1))$ $dp$



but isn't the value infinite?






probability-distributions conditional-expectation uniform-distribution expected-value







share|cite|improve this question
























  • I think you are right.
    – Kavi Rama Murthy
    5 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











First thanks for any help editing my text.



If a random variable $X$ has a geometric distribution with parameter $P$ where $P$ itself is a random variable and uniformly distributed from $0$ to $1-1/n$, I want to find $E[X]$.



I first learned that $E[X|P] = 1/P$, using the expectation of the geometric distribution.



But when it comes to $E[X]=E[E[X|P]]$, I tried integration and it seems to be



(integral from $0$ to $(n-1)/n$) , $(1/p)$ * $(n/(n-1))$ $dp$



but isn't the value infinite?






probability-distributions conditional-expectation uniform-distribution expected-value







share|cite|improve this question















First thanks for any help editing my text.



If a random variable $X$ has a geometric distribution with parameter $P$ where $P$ itself is a random variable and uniformly distributed from $0$ to $1-1/n$, I want to find $E[X]$.



I first learned that $E[X|P] = 1/P$, using the expectation of the geometric distribution.



But when it comes to $E[X]=E[E[X|P]]$, I tried integration and it seems to be



(integral from $0$ to $(n-1)/n$) , $(1/p)$ * $(n/(n-1))$ $dp$



but isn't the value infinite?






probability-distributions conditional-expectation uniform-distribution expected-value




probability-distributions conditional-expectation uniform-distribution expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago

























asked 5 hours ago









Richard

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356












  • I think you are right.
    – Kavi Rama Murthy
    5 hours ago


















  • I think you are right.
    – Kavi Rama Murthy
    5 hours ago
















I think you are right.
– Kavi Rama Murthy
5 hours ago




I think you are right.
– Kavi Rama Murthy
5 hours ago















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