Bound on weighted probability distribution

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Let $p(x)$ be a probability distribution (say, on some finite space, $xin calX$), $f(x)>0$ a positive function on $calX$, with no further assumptions of $f$. I'd like to consider when can the product $p(x)f(x)$ be a probability distribution? For that end, I'm trying to bound

$sum_x p(x)f(x)$

either from above, or from below.

If my product were, say, $p^t(x)f(x)$ for some $0<t<1$, then using Hölder inequality for the conjugate exponents $frac1t$ and $frac11-t$ we would have had,

$sum_x p^t(x)f(x) leq left(sum_x p(x)right)^t left(sum_x f^frac11-t(x)right)^(1-t) = left(sum_x f^frac11-t(x)right)^(1-t)$

Where in the last equality I've exploited the fact that $p(x)$ is a probability distribution.

Questions:

1. How to bound $sum_x p(x)f(x)$, either from above or from below, using the fact that $p(x)$ is a probability distribution?




2. Any conditions under which the product $p(x)f(x)$ can or cannot be a probability distribution?

real-analysis probability inequality probability-distributions integral-inequality

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edited Aug 2 at 19:14

asked Aug 2 at 19:08

Shlomi A

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down vote

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Let $p(x)$ be a probability distribution (say, on some finite space, $xin calX$), $f(x)>0$ a positive function on $calX$, with no further assumptions of $f$. I'd like to consider when can the product $p(x)f(x)$ be a probability distribution? For that end, I'm trying to bound

$sum_x p(x)f(x)$

either from above, or from below.

If my product were, say, $p^t(x)f(x)$ for some $0<t<1$, then using Hölder inequality for the conjugate exponents $frac1t$ and $frac11-t$ we would have had,

$sum_x p^t(x)f(x) leq left(sum_x p(x)right)^t left(sum_x f^frac11-t(x)right)^(1-t) = left(sum_x f^frac11-t(x)right)^(1-t)$

Where in the last equality I've exploited the fact that $p(x)$ is a probability distribution.

Questions:

1. How to bound $sum_x p(x)f(x)$, either from above or from below, using the fact that $p(x)$ is a probability distribution?




2. Any conditions under which the product $p(x)f(x)$ can or cannot be a probability distribution?

real-analysis probability inequality probability-distributions integral-inequality

share|cite|improve this question

edited Aug 2 at 19:14

asked Aug 2 at 19:08

Shlomi A

258210

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $p(x)$ be a probability distribution (say, on some finite space, $xin calX$), $f(x)>0$ a positive function on $calX$, with no further assumptions of $f$. I'd like to consider when can the product $p(x)f(x)$ be a probability distribution? For that end, I'm trying to bound

$sum_x p(x)f(x)$

either from above, or from below.

If my product were, say, $p^t(x)f(x)$ for some $0<t<1$, then using Hölder inequality for the conjugate exponents $frac1t$ and $frac11-t$ we would have had,

$sum_x p^t(x)f(x) leq left(sum_x p(x)right)^t left(sum_x f^frac11-t(x)right)^(1-t) = left(sum_x f^frac11-t(x)right)^(1-t)$

Where in the last equality I've exploited the fact that $p(x)$ is a probability distribution.

Questions:

1. How to bound $sum_x p(x)f(x)$, either from above or from below, using the fact that $p(x)$ is a probability distribution?




2. Any conditions under which the product $p(x)f(x)$ can or cannot be a probability distribution?

real-analysis probability inequality probability-distributions integral-inequality

share|cite|improve this question

edited Aug 2 at 19:14

asked Aug 2 at 19:08

Shlomi A

258210

Let $p(x)$ be a probability distribution (say, on some finite space, $xin calX$), $f(x)>0$ a positive function on $calX$, with no further assumptions of $f$. I'd like to consider when can the product $p(x)f(x)$ be a probability distribution? For that end, I'm trying to bound

$sum_x p(x)f(x)$

either from above, or from below.

If my product were, say, $p^t(x)f(x)$ for some $0<t<1$, then using Hölder inequality for the conjugate exponents $frac1t$ and $frac11-t$ we would have had,

$sum_x p^t(x)f(x) leq left(sum_x p(x)right)^t left(sum_x f^frac11-t(x)right)^(1-t) = left(sum_x f^frac11-t(x)right)^(1-t)$

Where in the last equality I've exploited the fact that $p(x)$ is a probability distribution.

Questions:

1. How to bound $sum_x p(x)f(x)$, either from above or from below, using the fact that $p(x)$ is a probability distribution?




2. Any conditions under which the product $p(x)f(x)$ can or cannot be a probability distribution?

real-analysis probability inequality probability-distributions integral-inequality

share|cite|improve this question

edited Aug 2 at 19:14

asked Aug 2 at 19:08

Shlomi A

258210

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 19:14

edited Aug 2 at 19:14

edited Aug 2 at 19:14

asked Aug 2 at 19:08

Shlomi A

258210

asked Aug 2 at 19:08

Shlomi A

258210

asked Aug 2 at 19:08

Shlomi A

258210

258210

add a comment | 

add a comment | 

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