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Example of a non-negative discrete distribution where the mean (or another moment) does not exist?
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I was doing some work in scipy and a conversation came up w/a member of the core scipy group whether a non-negative discrete random variable can have a undefined moment. I think he is correct but do not have a proof handy. Can anyone show/prove this claim? (or if this claim is not true disprove)
I don't have an example handy if the discrete random variable has support on $mathbbZ$ but it seems that some discretized version of the Cauchy distribution should serve as an example to get an undefined moment. The condition of non-negativity (perhaps including $0$) is what seems to make the problem challenging (at least for me).
mathematical-statistics expected-value
asked 51 mins ago
Lucas Roberts
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up vote
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I was doing some work in scipy and a conversation came up w/a member of the core scipy group whether a non-negative discrete random variable can have a undefined moment. I think he is correct but do not have a proof handy. Can anyone show/prove this claim? (or if this claim is not true disprove)
I don't have an example handy if the discrete random variable has support on $mathbbZ$ but it seems that some discretized version of the Cauchy distribution should serve as an example to get an undefined moment. The condition of non-negativity (perhaps including $0$) is what seems to make the problem challenging (at least for me).
mathematical-statistics expected-value
asked 51 mins ago
Lucas Roberts
2,013429
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I was doing some work in scipy and a conversation came up w/a member of the core scipy group whether a non-negative discrete random variable can have a undefined moment. I think he is correct but do not have a proof handy. Can anyone show/prove this claim? (or if this claim is not true disprove)
I don't have an example handy if the discrete random variable has support on $mathbbZ$ but it seems that some discretized version of the Cauchy distribution should serve as an example to get an undefined moment. The condition of non-negativity (perhaps including $0$) is what seems to make the problem challenging (at least for me).
mathematical-statistics expected-value
asked 51 mins ago
Lucas Roberts
2,013429
I was doing some work in scipy and a conversation came up w/a member of the core scipy group whether a non-negative discrete random variable can have a undefined moment. I think he is correct but do not have a proof handy. Can anyone show/prove this claim? (or if this claim is not true disprove)
I don't have an example handy if the discrete random variable has support on $mathbbZ$ but it seems that some discretized version of the Cauchy distribution should serve as an example to get an undefined moment. The condition of non-negativity (perhaps including $0$) is what seems to make the problem challenging (at least for me).
mathematical-statistics expected-value
mathematical-statistics expected-value
asked 51 mins ago
Lucas Roberts
2,013429
asked 51 mins ago
Lucas Roberts
2,013429
asked 51 mins ago
Lucas Roberts
2,013429
asked 51 mins ago
Lucas Roberts
2,013429
asked 51 mins ago
Lucas Roberts
2,013429
2,013429
add a comment |Â
add a comment |Â
3 Answers
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3
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Here's a famous example: Let $X$ take value $2^k$ with probability $2^-k$, for each integer $kge1$. Then $X$ takes values in (a subset of) the positive integers; the total mass is $sum_k=1^infty 2^-k=1$, but its expectation is
$$E(X) = sum_k=1^infty 2^k P(X=2^k) = sum_k=1^infty 1 = infty.
$$
This random variable $X$ arises in the St. Petersburg paradox.
answered 16 mins ago
grand_chat
1,70724
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up vote
2
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Let the CDF $F$ equal $1-1/n$ at the integers $n=1,2,ldots,$ piecewise constant everywhere else, and subject to all criteria to be a CDF. The expectation is
$$int_0^infty d(1-F(x)) = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
If you're uncomfortable with this notation, note that for $n=1,2,3,ldots,$
$$Pr_F(n) = frac1n - frac1n+1.$$
This defines a probability distribution since each term is positive and $$sum_n=1^infty Pr_F(n) = sum_n=1^infty left(frac1n - frac1n+1right) = lim_nto infty 1 - frac1n+1 = 1.$$
The expectation is
$$sum_n=1^infty n,Pr_F(n) = sum_n=1^infty nleft(frac1n - frac1n+1right) =sum_n=1^infty frac1n+1 = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
answered 45 mins ago
whuber♦
195k31417778
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up vote
1
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The zeta distribution is a fairly well-known discrete distribution on the positive integers that doesn't have finite mean (for $1<thetaleq 2$) .
$P(X=x|theta)=frac 1zeta (theta)x^-theta,,: x=1,2,...,:theta>1$
where the normalizing constant involves $zeta(cdot)$, the Riemann zeta function
(edit: The case $theta=2$ is very similar to whuber's answer)
Another distribution with similar tail behaviour is the Yule-Simon distribution.
Another example would be the beta-negative binomial distribution with $0<alphaleq 1$:
$P(X=x|alpha ,beta ,r)=frac Gamma (r+x)x!;Gamma (r)frac mathrmB (alpha +r,beta +x)mathrmB (alpha ,beta ),,:x=0,1,2...:alpha,beta,r > 0$
answered 33 mins ago
Glen_b♦
202k22381707
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Here's a famous example: Let $X$ take value $2^k$ with probability $2^-k$, for each integer $kge1$. Then $X$ takes values in (a subset of) the positive integers; the total mass is $sum_k=1^infty 2^-k=1$, but its expectation is
$$E(X) = sum_k=1^infty 2^k P(X=2^k) = sum_k=1^infty 1 = infty.
$$
This random variable $X$ arises in the St. Petersburg paradox.
answered 16 mins ago
grand_chat
1,70724
add a comment |Â
up vote
3
down vote
Here's a famous example: Let $X$ take value $2^k$ with probability $2^-k$, for each integer $kge1$. Then $X$ takes values in (a subset of) the positive integers; the total mass is $sum_k=1^infty 2^-k=1$, but its expectation is
$$E(X) = sum_k=1^infty 2^k P(X=2^k) = sum_k=1^infty 1 = infty.
$$
This random variable $X$ arises in the St. Petersburg paradox.
answered 16 mins ago
grand_chat
1,70724
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here's a famous example: Let $X$ take value $2^k$ with probability $2^-k$, for each integer $kge1$. Then $X$ takes values in (a subset of) the positive integers; the total mass is $sum_k=1^infty 2^-k=1$, but its expectation is
$$E(X) = sum_k=1^infty 2^k P(X=2^k) = sum_k=1^infty 1 = infty.
$$
This random variable $X$ arises in the St. Petersburg paradox.
answered 16 mins ago
grand_chat
1,70724
Here's a famous example: Let $X$ take value $2^k$ with probability $2^-k$, for each integer $kge1$. Then $X$ takes values in (a subset of) the positive integers; the total mass is $sum_k=1^infty 2^-k=1$, but its expectation is
$$E(X) = sum_k=1^infty 2^k P(X=2^k) = sum_k=1^infty 1 = infty.
$$
This random variable $X$ arises in the St. Petersburg paradox.
answered 16 mins ago
grand_chat
1,70724
answered 16 mins ago
grand_chat
1,70724
answered 16 mins ago
grand_chat
1,70724
answered 16 mins ago
grand_chat
1,70724
1,70724
add a comment |Â
add a comment |Â
up vote
2
down vote
Let the CDF $F$ equal $1-1/n$ at the integers $n=1,2,ldots,$ piecewise constant everywhere else, and subject to all criteria to be a CDF. The expectation is
$$int_0^infty d(1-F(x)) = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
If you're uncomfortable with this notation, note that for $n=1,2,3,ldots,$
$$Pr_F(n) = frac1n - frac1n+1.$$
This defines a probability distribution since each term is positive and $$sum_n=1^infty Pr_F(n) = sum_n=1^infty left(frac1n - frac1n+1right) = lim_nto infty 1 - frac1n+1 = 1.$$
The expectation is
$$sum_n=1^infty n,Pr_F(n) = sum_n=1^infty nleft(frac1n - frac1n+1right) =sum_n=1^infty frac1n+1 = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
answered 45 mins ago
whuber♦
195k31417778
add a comment |Â
up vote
2
down vote
Let the CDF $F$ equal $1-1/n$ at the integers $n=1,2,ldots,$ piecewise constant everywhere else, and subject to all criteria to be a CDF. The expectation is
$$int_0^infty d(1-F(x)) = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
If you're uncomfortable with this notation, note that for $n=1,2,3,ldots,$
$$Pr_F(n) = frac1n - frac1n+1.$$
This defines a probability distribution since each term is positive and $$sum_n=1^infty Pr_F(n) = sum_n=1^infty left(frac1n - frac1n+1right) = lim_nto infty 1 - frac1n+1 = 1.$$
The expectation is
$$sum_n=1^infty n,Pr_F(n) = sum_n=1^infty nleft(frac1n - frac1n+1right) =sum_n=1^infty frac1n+1 = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
answered 45 mins ago
whuber♦
195k31417778
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let the CDF $F$ equal $1-1/n$ at the integers $n=1,2,ldots,$ piecewise constant everywhere else, and subject to all criteria to be a CDF. The expectation is
$$int_0^infty d(1-F(x)) = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
If you're uncomfortable with this notation, note that for $n=1,2,3,ldots,$
$$Pr_F(n) = frac1n - frac1n+1.$$
This defines a probability distribution since each term is positive and $$sum_n=1^infty Pr_F(n) = sum_n=1^infty left(frac1n - frac1n+1right) = lim_nto infty 1 - frac1n+1 = 1.$$
The expectation is
$$sum_n=1^infty n,Pr_F(n) = sum_n=1^infty nleft(frac1n - frac1n+1right) =sum_n=1^infty frac1n+1 = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
answered 45 mins ago
whuber♦
195k31417778
Let the CDF $F$ equal $1-1/n$ at the integers $n=1,2,ldots,$ piecewise constant everywhere else, and subject to all criteria to be a CDF. The expectation is
$$int_0^infty d(1-F(x)) = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
If you're uncomfortable with this notation, note that for $n=1,2,3,ldots,$
$$Pr_F(n) = frac1n - frac1n+1.$$
This defines a probability distribution since each term is positive and $$sum_n=1^infty Pr_F(n) = sum_n=1^infty left(frac1n - frac1n+1right) = lim_nto infty 1 - frac1n+1 = 1.$$
The expectation is
$$sum_n=1^infty n,Pr_F(n) = sum_n=1^infty nleft(frac1n - frac1n+1right) =sum_n=1^infty frac1n+1 = 1/2 + 1/3 + 1/4 + cdots$$
which diverges.
answered 45 mins ago
whuber♦
195k31417778
answered 45 mins ago
whuber♦
195k31417778
answered 45 mins ago
whuber♦
195k31417778
answered 45 mins ago
whuber♦
195k31417778
195k31417778
add a comment |Â
add a comment |Â
up vote
1
down vote
The zeta distribution is a fairly well-known discrete distribution on the positive integers that doesn't have finite mean (for $1<thetaleq 2$) .
$P(X=x|theta)=frac 1zeta (theta)x^-theta,,: x=1,2,...,:theta>1$
where the normalizing constant involves $zeta(cdot)$, the Riemann zeta function
(edit: The case $theta=2$ is very similar to whuber's answer)
Another distribution with similar tail behaviour is the Yule-Simon distribution.
Another example would be the beta-negative binomial distribution with $0<alphaleq 1$:
$P(X=x|alpha ,beta ,r)=frac Gamma (r+x)x!;Gamma (r)frac mathrmB (alpha +r,beta +x)mathrmB (alpha ,beta ),,:x=0,1,2...:alpha,beta,r > 0$
answered 33 mins ago
Glen_b♦
202k22381707
add a comment |Â
up vote
1
down vote
The zeta distribution is a fairly well-known discrete distribution on the positive integers that doesn't have finite mean (for $1<thetaleq 2$) .
$P(X=x|theta)=frac 1zeta (theta)x^-theta,,: x=1,2,...,:theta>1$
where the normalizing constant involves $zeta(cdot)$, the Riemann zeta function
(edit: The case $theta=2$ is very similar to whuber's answer)
Another distribution with similar tail behaviour is the Yule-Simon distribution.
Another example would be the beta-negative binomial distribution with $0<alphaleq 1$:
$P(X=x|alpha ,beta ,r)=frac Gamma (r+x)x!;Gamma (r)frac mathrmB (alpha +r,beta +x)mathrmB (alpha ,beta ),,:x=0,1,2...:alpha,beta,r > 0$
answered 33 mins ago
Glen_b♦
202k22381707
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The zeta distribution is a fairly well-known discrete distribution on the positive integers that doesn't have finite mean (for $1<thetaleq 2$) .
$P(X=x|theta)=frac 1zeta (theta)x^-theta,,: x=1,2,...,:theta>1$
where the normalizing constant involves $zeta(cdot)$, the Riemann zeta function
(edit: The case $theta=2$ is very similar to whuber's answer)
Another distribution with similar tail behaviour is the Yule-Simon distribution.
Another example would be the beta-negative binomial distribution with $0<alphaleq 1$:
$P(X=x|alpha ,beta ,r)=frac Gamma (r+x)x!;Gamma (r)frac mathrmB (alpha +r,beta +x)mathrmB (alpha ,beta ),,:x=0,1,2...:alpha,beta,r > 0$
answered 33 mins ago
Glen_b♦
202k22381707
The zeta distribution is a fairly well-known discrete distribution on the positive integers that doesn't have finite mean (for $1<thetaleq 2$) .
$P(X=x|theta)=frac 1zeta (theta)x^-theta,,: x=1,2,...,:theta>1$
where the normalizing constant involves $zeta(cdot)$, the Riemann zeta function
(edit: The case $theta=2$ is very similar to whuber's answer)
Another distribution with similar tail behaviour is the Yule-Simon distribution.
Another example would be the beta-negative binomial distribution with $0<alphaleq 1$:
$P(X=x|alpha ,beta ,r)=frac Gamma (r+x)x!;Gamma (r)frac mathrmB (alpha +r,beta +x)mathrmB (alpha ,beta ),,:x=0,1,2...:alpha,beta,r > 0$
answered 33 mins ago
Glen_b♦
202k22381707
edited 8 mins ago
answered 33 mins ago
Glen_b♦
202k22381707
answered 33 mins ago
Glen_b♦
202k22381707
answered 33 mins ago
Glen_b♦
202k22381707
202k22381707
add a comment |Â
add a comment |Â
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