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Simple example of a distribution on +ve natural numbers with $sum_{n=1}^infty n^2 p_n < infty$ and...
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DISCLAIMER: This is probably too elementary and I haven't spent much time thinking about it (as I should).
Question
What is a simple example of a distribution $(p_1,p_2,ldots,p_n,ldots)$ on +ve natural numbers (meaning that $p_n ge 0;forall n =1,2,ldots$ and $sum_{n=1}^infty p_n = 1$) satisfying simultaneously
finite-second moment: $sum_{n=1}^infty n^2 p_n < infty$
divergent square root of probabilities: $sum_{n=1}^infty sqrt{p_n} = infty$ ?
Edit
User Stephan Lafon has given a really simple proof which deserves to be expanded upon. In fact one can proof a stronger statement.
For any $alpha in (1, infty)$, there is no distribution $p:=(p_1,ldots,p_n,ldots)$ with fintite $alpha$-moment such that $sum_n p_n^{1/alpha} = infty$.
Proof. Let $beta := alpha / (alpha-1) > 1$ be the harmonic conjugate of $alpha$. Define the functions $f,g: {1,2,ldots} rightarrow mathbb R_+$ by $f(n) := 1/n$ and $g(n) := np_n^{1/alpha}$. If $sum_n n^alpha p_n < infty$, then one computes
$$
begin{split}
sum_n p_n^{1/alpha} &= sum_n (1/n) np^{1/alpha} = langle f, grangle
overset{text{Cauchy-Schwarz}}{le}|f|_beta|g|_alpha = left(sum_n 1/n^betaright)^{1/beta}left(sum_n n^alpha p_nright)^{1/alpha}\
&= zeta(beta)^{1/beta}left(sum_n n^alpha p_nright)^{1/alpha} < infty.
end{split}
$$
probability-distributions random-variables elementary-probability
asked Jan 29 at 19:32
dohmatobdohmatob
3,727629
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add a comment |
$begingroup$
DISCLAIMER: This is probably too elementary and I haven't spent much time thinking about it (as I should).
Question
What is a simple example of a distribution $(p_1,p_2,ldots,p_n,ldots)$ on +ve natural numbers (meaning that $p_n ge 0;forall n =1,2,ldots$ and $sum_{n=1}^infty p_n = 1$) satisfying simultaneously
finite-second moment: $sum_{n=1}^infty n^2 p_n < infty$
divergent square root of probabilities: $sum_{n=1}^infty sqrt{p_n} = infty$ ?
Edit
User Stephan Lafon has given a really simple proof which deserves to be expanded upon. In fact one can proof a stronger statement.
For any $alpha in (1, infty)$, there is no distribution $p:=(p_1,ldots,p_n,ldots)$ with fintite $alpha$-moment such that $sum_n p_n^{1/alpha} = infty$.
Proof. Let $beta := alpha / (alpha-1) > 1$ be the harmonic conjugate of $alpha$. Define the functions $f,g: {1,2,ldots} rightarrow mathbb R_+$ by $f(n) := 1/n$ and $g(n) := np_n^{1/alpha}$. If $sum_n n^alpha p_n < infty$, then one computes
$$
begin{split}
sum_n p_n^{1/alpha} &= sum_n (1/n) np^{1/alpha} = langle f, grangle
overset{text{Cauchy-Schwarz}}{le}|f|_beta|g|_alpha = left(sum_n 1/n^betaright)^{1/beta}left(sum_n n^alpha p_nright)^{1/alpha}\
&= zeta(beta)^{1/beta}left(sum_n n^alpha p_nright)^{1/alpha} < infty.
end{split}
$$
probability-distributions random-variables elementary-probability
asked Jan 29 at 19:32
dohmatobdohmatob
3,727629
$endgroup$
add a comment |
$begingroup$
DISCLAIMER: This is probably too elementary and I haven't spent much time thinking about it (as I should).
Question
What is a simple example of a distribution $(p_1,p_2,ldots,p_n,ldots)$ on +ve natural numbers (meaning that $p_n ge 0;forall n =1,2,ldots$ and $sum_{n=1}^infty p_n = 1$) satisfying simultaneously
finite-second moment: $sum_{n=1}^infty n^2 p_n < infty$
divergent square root of probabilities: $sum_{n=1}^infty sqrt{p_n} = infty$ ?
Edit
User Stephan Lafon has given a really simple proof which deserves to be expanded upon. In fact one can proof a stronger statement.
For any $alpha in (1, infty)$, there is no distribution $p:=(p_1,ldots,p_n,ldots)$ with fintite $alpha$-moment such that $sum_n p_n^{1/alpha} = infty$.
Proof. Let $beta := alpha / (alpha-1) > 1$ be the harmonic conjugate of $alpha$. Define the functions $f,g: {1,2,ldots} rightarrow mathbb R_+$ by $f(n) := 1/n$ and $g(n) := np_n^{1/alpha}$. If $sum_n n^alpha p_n < infty$, then one computes
$$
begin{split}
sum_n p_n^{1/alpha} &= sum_n (1/n) np^{1/alpha} = langle f, grangle
overset{text{Cauchy-Schwarz}}{le}|f|_beta|g|_alpha = left(sum_n 1/n^betaright)^{1/beta}left(sum_n n^alpha p_nright)^{1/alpha}\
&= zeta(beta)^{1/beta}left(sum_n n^alpha p_nright)^{1/alpha} < infty.
end{split}
$$
probability-distributions random-variables elementary-probability
asked Jan 29 at 19:32
dohmatobdohmatob
3,727629
$endgroup$
DISCLAIMER: This is probably too elementary and I haven't spent much time thinking about it (as I should).
Question
What is a simple example of a distribution $(p_1,p_2,ldots,p_n,ldots)$ on +ve natural numbers (meaning that $p_n ge 0;forall n =1,2,ldots$ and $sum_{n=1}^infty p_n = 1$) satisfying simultaneously
finite-second moment: $sum_{n=1}^infty n^2 p_n < infty$
divergent square root of probabilities: $sum_{n=1}^infty sqrt{p_n} = infty$ ?
Edit
User Stephan Lafon has given a really simple proof which deserves to be expanded upon. In fact one can proof a stronger statement.
For any $alpha in (1, infty)$, there is no distribution $p:=(p_1,ldots,p_n,ldots)$ with fintite $alpha$-moment such that $sum_n p_n^{1/alpha} = infty$.
Proof. Let $beta := alpha / (alpha-1) > 1$ be the harmonic conjugate of $alpha$. Define the functions $f,g: {1,2,ldots} rightarrow mathbb R_+$ by $f(n) := 1/n$ and $g(n) := np_n^{1/alpha}$. If $sum_n n^alpha p_n < infty$, then one computes
$$
begin{split}
sum_n p_n^{1/alpha} &= sum_n (1/n) np^{1/alpha} = langle f, grangle
overset{text{Cauchy-Schwarz}}{le}|f|_beta|g|_alpha = left(sum_n 1/n^betaright)^{1/beta}left(sum_n n^alpha p_nright)^{1/alpha}\
&= zeta(beta)^{1/beta}left(sum_n n^alpha p_nright)^{1/alpha} < infty.
end{split}
$$
probability-distributions random-variables elementary-probability
probability-distributions random-variables elementary-probability
asked Jan 29 at 19:32
dohmatobdohmatob
3,727629
asked Jan 29 at 19:32
dohmatobdohmatob
3,727629
edited Jan 29 at 23:36
dohmatob
asked Jan 29 at 19:32
dohmatobdohmatob
3,727629
asked Jan 29 at 19:32
dohmatobdohmatob
3,727629
asked Jan 29 at 19:32
dohmatobdohmatob
3,727629
3,727629
add a comment |
add a comment |
1 Answer
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There is no such distribution. Proof: By Cauchy-Schwarz:
$$sum_{n=1}^infty sqrt{p_n} = sum_{n=1}^infty frac{n}{n}sqrt{p_n} leq left(sum_{n=1}^infty n^2p_nright)^{frac1 2}left(sum_{n=1}^inftyfrac 1 {n^2}right)^{frac1 2}$$
answered Jan 29 at 19:38
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Great. Thanks, just chewed out something similar :)
$endgroup$
– dohmatob
Jan 29 at 19:39
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Updated the question with a proof to a slightly more general claim. Yes, still Cauchy-Schwarz at work.
$endgroup$
– dohmatob
Jan 29 at 21:22
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There is no such distribution. Proof: By Cauchy-Schwarz:
$$sum_{n=1}^infty sqrt{p_n} = sum_{n=1}^infty frac{n}{n}sqrt{p_n} leq left(sum_{n=1}^infty n^2p_nright)^{frac1 2}left(sum_{n=1}^inftyfrac 1 {n^2}right)^{frac1 2}$$
answered Jan 29 at 19:38
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Great. Thanks, just chewed out something similar :)
$endgroup$
– dohmatob
Jan 29 at 19:39
$begingroup$
Updated the question with a proof to a slightly more general claim. Yes, still Cauchy-Schwarz at work.
$endgroup$
– dohmatob
Jan 29 at 21:22
add a comment |
$begingroup$
There is no such distribution. Proof: By Cauchy-Schwarz:
$$sum_{n=1}^infty sqrt{p_n} = sum_{n=1}^infty frac{n}{n}sqrt{p_n} leq left(sum_{n=1}^infty n^2p_nright)^{frac1 2}left(sum_{n=1}^inftyfrac 1 {n^2}right)^{frac1 2}$$
answered Jan 29 at 19:38
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
Great. Thanks, just chewed out something similar :)
$endgroup$
– dohmatob
Jan 29 at 19:39
$begingroup$
Updated the question with a proof to a slightly more general claim. Yes, still Cauchy-Schwarz at work.
$endgroup$
– dohmatob
Jan 29 at 21:22
add a comment |
$begingroup$
There is no such distribution. Proof: By Cauchy-Schwarz:
$$sum_{n=1}^infty sqrt{p_n} = sum_{n=1}^infty frac{n}{n}sqrt{p_n} leq left(sum_{n=1}^infty n^2p_nright)^{frac1 2}left(sum_{n=1}^inftyfrac 1 {n^2}right)^{frac1 2}$$
answered Jan 29 at 19:38
Stefan LafonStefan Lafon
3,005212
$endgroup$
There is no such distribution. Proof: By Cauchy-Schwarz:
$$sum_{n=1}^infty sqrt{p_n} = sum_{n=1}^infty frac{n}{n}sqrt{p_n} leq left(sum_{n=1}^infty n^2p_nright)^{frac1 2}left(sum_{n=1}^inftyfrac 1 {n^2}right)^{frac1 2}$$
answered Jan 29 at 19:38
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 19:38
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 19:38
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 19:38
Stefan LafonStefan Lafon
3,005212
3,005212
$begingroup$
Great. Thanks, just chewed out something similar :)
$endgroup$
– dohmatob
Jan 29 at 19:39
$begingroup$
Updated the question with a proof to a slightly more general claim. Yes, still Cauchy-Schwarz at work.
$endgroup$
– dohmatob
Jan 29 at 21:22
add a comment |
$begingroup$
Great. Thanks, just chewed out something similar :)
$endgroup$
– dohmatob
Jan 29 at 19:39
$begingroup$
Updated the question with a proof to a slightly more general claim. Yes, still Cauchy-Schwarz at work.
$endgroup$
– dohmatob
Jan 29 at 21:22
$begingroup$
Great. Thanks, just chewed out something similar :)
$endgroup$
– dohmatob
Jan 29 at 19:39
$begingroup$
Great. Thanks, just chewed out something similar :)
$endgroup$
– dohmatob
Jan 29 at 19:39
$begingroup$
Updated the question with a proof to a slightly more general claim. Yes, still Cauchy-Schwarz at work.
$endgroup$
– dohmatob
Jan 29 at 21:22
$begingroup$
Updated the question with a proof to a slightly more general claim. Yes, still Cauchy-Schwarz at work.
$endgroup$
– dohmatob
Jan 29 at 21:22
add a comment |
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