$E(X!)$ for Poisson distribution [closed]
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For $Xsimtext{Pois}(λ)$, find $E(X!)$ (the average factorial of $X$), if it is finite.
Solution:
By LOTUS,
$$E(X!) =e^{−λ}sum_{k=0}^{infty} k!frac{lambda^k}{k!} = frac{e^{−lambda}}{1−lambda}$$
for $0<lambda<1$.
Question: Why is just the $k$ on the bottom of the fraction factorial, and not $lambda^{k!}$ ?
random-variables poisson-distribution expected-value
edited Dec 2 '18 at 7:29
Tianlalu
3,08621038
asked Dec 1 '18 at 23:32
user603569user603569
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closed as unclear what you're asking by Did, Paul Frost, Cesareo, Ali Caglayan, DRF Dec 2 '18 at 22:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
For $Xsimtext{Pois}(λ)$, find $E(X!)$ (the average factorial of $X$), if it is finite.
Solution:
By LOTUS,
$$E(X!) =e^{−λ}sum_{k=0}^{infty} k!frac{lambda^k}{k!} = frac{e^{−lambda}}{1−lambda}$$
for $0<lambda<1$.
Question: Why is just the $k$ on the bottom of the fraction factorial, and not $lambda^{k!}$ ?
random-variables poisson-distribution expected-value
edited Dec 2 '18 at 7:29
Tianlalu
3,08621038
asked Dec 1 '18 at 23:32
user603569user603569
728
$endgroup$
closed as unclear what you're asking by Did, Paul Frost, Cesareo, Ali Caglayan, DRF Dec 2 '18 at 22:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Because en.wikipedia.org/wiki/Poisson_distribution#Definition.
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– Did
Dec 2 '18 at 9:55
add a comment |
$begingroup$
For $Xsimtext{Pois}(λ)$, find $E(X!)$ (the average factorial of $X$), if it is finite.
Solution:
By LOTUS,
$$E(X!) =e^{−λ}sum_{k=0}^{infty} k!frac{lambda^k}{k!} = frac{e^{−lambda}}{1−lambda}$$
for $0<lambda<1$.
Question: Why is just the $k$ on the bottom of the fraction factorial, and not $lambda^{k!}$ ?
random-variables poisson-distribution expected-value
edited Dec 2 '18 at 7:29
Tianlalu
3,08621038
asked Dec 1 '18 at 23:32
user603569user603569
728
$endgroup$
For $Xsimtext{Pois}(λ)$, find $E(X!)$ (the average factorial of $X$), if it is finite.
Solution:
By LOTUS,
$$E(X!) =e^{−λ}sum_{k=0}^{infty} k!frac{lambda^k}{k!} = frac{e^{−lambda}}{1−lambda}$$
for $0<lambda<1$.
Question: Why is just the $k$ on the bottom of the fraction factorial, and not $lambda^{k!}$ ?
random-variables poisson-distribution expected-value
random-variables poisson-distribution expected-value
edited Dec 2 '18 at 7:29
Tianlalu
3,08621038
asked Dec 1 '18 at 23:32
user603569user603569
728
edited Dec 2 '18 at 7:29
Tianlalu
3,08621038
asked Dec 1 '18 at 23:32
user603569user603569
728
edited Dec 2 '18 at 7:29
Tianlalu
3,08621038
edited Dec 2 '18 at 7:29
Tianlalu
3,08621038
edited Dec 2 '18 at 7:29
Tianlalu
3,08621038
3,08621038
asked Dec 1 '18 at 23:32
user603569user603569
728
asked Dec 1 '18 at 23:32
user603569user603569
728
asked Dec 1 '18 at 23:32
user603569user603569
728
728
closed as unclear what you're asking by Did, Paul Frost, Cesareo, Ali Caglayan, DRF Dec 2 '18 at 22:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, Paul Frost, Cesareo, Ali Caglayan, DRF Dec 2 '18 at 22:07
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Because en.wikipedia.org/wiki/Poisson_distribution#Definition.
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– Did
Dec 2 '18 at 9:55
add a comment |
$begingroup$
Because en.wikipedia.org/wiki/Poisson_distribution#Definition.
$endgroup$
– Did
Dec 2 '18 at 9:55
$begingroup$
Because en.wikipedia.org/wiki/Poisson_distribution#Definition.
$endgroup$
– Did
Dec 2 '18 at 9:55
$begingroup$
Because en.wikipedia.org/wiki/Poisson_distribution#Definition.
$endgroup$
– Did
Dec 2 '18 at 9:55
add a comment |
2 Answers
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oldest
votes
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$EX!=sum P{X=k} k!$ and $P{X=k}=e^{-lambda} frac {lambda^{k}} {k!}$.
answered Dec 1 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
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add a comment |
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Here's the sum:
$${cal E}[n!] = sumlimits_{n=0}^infty {e^{-mu} mu^n over n!} n! = {e^{-mu} over 1 - mu}$$
answered Dec 1 '18 at 23:40
David G. StorkDavid G. Stork
10.7k31332
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$EX!=sum P{X=k} k!$ and $P{X=k}=e^{-lambda} frac {lambda^{k}} {k!}$.
answered Dec 1 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
add a comment |
$begingroup$
$EX!=sum P{X=k} k!$ and $P{X=k}=e^{-lambda} frac {lambda^{k}} {k!}$.
answered Dec 1 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
add a comment |
$begingroup$
$EX!=sum P{X=k} k!$ and $P{X=k}=e^{-lambda} frac {lambda^{k}} {k!}$.
answered Dec 1 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
$endgroup$
$EX!=sum P{X=k} k!$ and $P{X=k}=e^{-lambda} frac {lambda^{k}} {k!}$.
answered Dec 1 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 1 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 1 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
answered Dec 1 '18 at 23:34
Kavi Rama MurthyKavi Rama Murthy
55.4k42057
55.4k42057
add a comment |
add a comment |
$begingroup$
Here's the sum:
$${cal E}[n!] = sumlimits_{n=0}^infty {e^{-mu} mu^n over n!} n! = {e^{-mu} over 1 - mu}$$
answered Dec 1 '18 at 23:40
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
add a comment |
$begingroup$
Here's the sum:
$${cal E}[n!] = sumlimits_{n=0}^infty {e^{-mu} mu^n over n!} n! = {e^{-mu} over 1 - mu}$$
answered Dec 1 '18 at 23:40
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
add a comment |
$begingroup$
Here's the sum:
$${cal E}[n!] = sumlimits_{n=0}^infty {e^{-mu} mu^n over n!} n! = {e^{-mu} over 1 - mu}$$
answered Dec 1 '18 at 23:40
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
Here's the sum:
$${cal E}[n!] = sumlimits_{n=0}^infty {e^{-mu} mu^n over n!} n! = {e^{-mu} over 1 - mu}$$
answered Dec 1 '18 at 23:40
David G. StorkDavid G. Stork
10.7k31332
answered Dec 1 '18 at 23:40
David G. StorkDavid G. Stork
10.7k31332
answered Dec 1 '18 at 23:40
David G. StorkDavid G. Stork
10.7k31332
answered Dec 1 '18 at 23:40
David G. StorkDavid G. Stork
10.7k31332
10.7k31332
add a comment |
add a comment |
$begingroup$
Because en.wikipedia.org/wiki/Poisson_distribution#Definition.
$endgroup$
– Did
Dec 2 '18 at 9:55