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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
asked 5 hours ago
Richard
356
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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
asked 5 hours ago
Richard
356
I think you are right.
– Kavi Rama Murthy
5 hours ago
add a comment |
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
asked 5 hours ago
Richard
356
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
asked 5 hours ago
Richard
356
asked 5 hours ago
Richard
356
edited 5 hours ago
asked 5 hours ago
Richard
356
asked 5 hours ago
Richard
356
asked 5 hours ago
Richard
356
356
I think you are right.
– Kavi Rama Murthy
5 hours ago
add a comment |
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
add a comment |
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I think you are right.
– Kavi Rama Murthy
5 hours ago