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Variance and covariance inequality
Minimizing the variance of weighted sum of two random variables with respect to the weightsFinding covariance given varianceVariance and covarianceVariance, Covariance, and Correlation answer checkConditional Expectation and Variance QuestionVariance and covariance between 2 variablesHow to prove an equality envolving variance and covarianceFinding the covariance of X,Y using variance?Covariance between given iid random variable and sample averageVariance of ratio of mean value of functions of a random variable
$begingroup$
Given a real-valued random variable $X$, is
$$2mathbb E[X] mathrmVar(X) geq mathrmCov(X, X^2)$$
true?
Any pointers for how to tackle this problem would be immensely helpful.
probability inequality variance covariance
asked 5 hours ago
sk1ll3rsk1ll3r
455
$endgroup$
add a comment |
$begingroup$
Given a real-valued random variable $X$, is
$$2mathbb E[X] mathrmVar(X) geq mathrmCov(X, X^2)$$
true?
Any pointers for how to tackle this problem would be immensely helpful.
probability inequality variance covariance
asked 5 hours ago
sk1ll3rsk1ll3r
455
$endgroup$
$begingroup$
One trick is to write a little computer program to generate examples randomly, which you can use to find counterexamples. Such as $P(X=a)=p, P(X=b)=1-p$, where you use a random number generator to pick the parameters $a,b,p$ and exact formulas in terms of $a,b.p$ for the relevant moments of $X$.
$endgroup$
– kimchi lover
4 hours ago
$begingroup$
I expect that the answer will follow quickly after applying the well-known identities $textVar(X)=mathbb E[X^2]-mathbb E[X]^2$ and $textCov(X,Y)=mathbb E[XY]-mathbb E[X]mathbb E[Y]$
$endgroup$
– M. Nestor
4 hours ago
add a comment |
$begingroup$
Given a real-valued random variable $X$, is
$$2mathbb E[X] mathrmVar(X) geq mathrmCov(X, X^2)$$
true?
Any pointers for how to tackle this problem would be immensely helpful.
probability inequality variance covariance
asked 5 hours ago
sk1ll3rsk1ll3r
455
$endgroup$
Given a real-valued random variable $X$, is
$$2mathbb E[X] mathrmVar(X) geq mathrmCov(X, X^2)$$
true?
Any pointers for how to tackle this problem would be immensely helpful.
probability inequality variance covariance
probability inequality variance covariance
asked 5 hours ago
sk1ll3rsk1ll3r
455
asked 5 hours ago
sk1ll3rsk1ll3r
455
asked 5 hours ago
sk1ll3rsk1ll3r
455
asked 5 hours ago
sk1ll3rsk1ll3r
455
asked 5 hours ago
sk1ll3rsk1ll3r
455
455
$begingroup$
One trick is to write a little computer program to generate examples randomly, which you can use to find counterexamples. Such as $P(X=a)=p, P(X=b)=1-p$, where you use a random number generator to pick the parameters $a,b,p$ and exact formulas in terms of $a,b.p$ for the relevant moments of $X$.
$endgroup$
– kimchi lover
4 hours ago
$begingroup$
I expect that the answer will follow quickly after applying the well-known identities $textVar(X)=mathbb E[X^2]-mathbb E[X]^2$ and $textCov(X,Y)=mathbb E[XY]-mathbb E[X]mathbb E[Y]$
$endgroup$
– M. Nestor
4 hours ago
add a comment |
$begingroup$
One trick is to write a little computer program to generate examples randomly, which you can use to find counterexamples. Such as $P(X=a)=p, P(X=b)=1-p$, where you use a random number generator to pick the parameters $a,b,p$ and exact formulas in terms of $a,b.p$ for the relevant moments of $X$.
$endgroup$
– kimchi lover
4 hours ago
$begingroup$
I expect that the answer will follow quickly after applying the well-known identities $textVar(X)=mathbb E[X^2]-mathbb E[X]^2$ and $textCov(X,Y)=mathbb E[XY]-mathbb E[X]mathbb E[Y]$
$endgroup$
– M. Nestor
4 hours ago
$begingroup$
One trick is to write a little computer program to generate examples randomly, which you can use to find counterexamples. Such as $P(X=a)=p, P(X=b)=1-p$, where you use a random number generator to pick the parameters $a,b,p$ and exact formulas in terms of $a,b.p$ for the relevant moments of $X$.
$endgroup$
– kimchi lover
4 hours ago
$begingroup$
One trick is to write a little computer program to generate examples randomly, which you can use to find counterexamples. Such as $P(X=a)=p, P(X=b)=1-p$, where you use a random number generator to pick the parameters $a,b,p$ and exact formulas in terms of $a,b.p$ for the relevant moments of $X$.
$endgroup$
– kimchi lover
4 hours ago
$begingroup$
I expect that the answer will follow quickly after applying the well-known identities $textVar(X)=mathbb E[X^2]-mathbb E[X]^2$ and $textCov(X,Y)=mathbb E[XY]-mathbb E[X]mathbb E[Y]$
$endgroup$
– M. Nestor
4 hours ago
$begingroup$
I expect that the answer will follow quickly after applying the well-known identities $textVar(X)=mathbb E[X^2]-mathbb E[X]^2$ and $textCov(X,Y)=mathbb E[XY]-mathbb E[X]mathbb E[Y]$
$endgroup$
– M. Nestor
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use the variance and covariance identities
$$textVar(X)=mathbb E[X^2] − mathbb E[X]^2$$
and
$$textCov(X,Y) = mathbb E[XY] − mathbb E[X] mathbb E[Y]$$
Then the given inequality is equivalent to
$$ 2 mathbb E[X] Big( mathbb E[X^2] - mathbb E[X]^2 Big)
overset?geq
mathbb E[X^3] - mathbb E[X] mathbb E[X^2] tag1$$
For a counterexample, let $X sim textExp(lambda)$ be the exponential distribution with parameter $lambda in (0, infty)$ and probability mass function
$$ f_X(x) = begincases
lambda e^-lambda x & textif x geq 0 \
0 & textif x < 0
endcases $$
Its $n$-th moment is given by $mathbb E[X^n]=fracn!lambda^n$, so the inequality $(1)$ becomes
$$2 cdot frac1lambda Big( frac2lambda^2 - frac1lambda^2 Big)
geq
frac6lambda^3 - frac1lambda cdot frac2lambda^2$$
which reduces to
$$frac2lambda^3 geq frac4lambda^3$$
This is false for all $lambda in (0, infty)$, hence we found a counterexample.
answered 4 hours ago
M. NestorM. Nestor
890115
$endgroup$
$begingroup$
hi! thanks a lot for this. do you think the inequality is false for a real-valued $X$ (as specified in the question) as well? i haven't been able to find a counter-example. for $X sim mathrmNormal(mu, sigma^2)$, both left and right hand side reduce to $2musigma^2$.
$endgroup$
– sk1ll3r
2 hours ago
1
$begingroup$
Sorry I did not notice that detail. I edited my answer to include an example where $X$ has continuous support.
$endgroup$
– M. Nestor
2 hours ago
add a comment |
$begingroup$
Alternate approach / observation...
Assume $2 E[X] Var(X) ge Cov(X, X^2)$ for any real-valued r.v. $X.$ Then in particular it would also be true for $-X$, so that:
$$
beginalign
2E[-X] Var(-X) &ge Cov(-X, (-X)^2)\
-2E[X] Var(X) &ge -Cov(X, X^2)\
2E[X] Var(X) &le Cov(X, X^2)
endalign
$$
which, combined with the original inequality, means: $2E[X]Var(X) = Cov(X, X^2)$ for all $X$.
At this point, it would seem one should be able to find a counter-example easily, without using e.g. $Cov(X, X^2) = E[X^3] - E[X]E[X^2]$ etc, but I just couldn't think of a quick one. :(
answered 1 hour ago
antkamantkam
4,489415
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use the variance and covariance identities
$$textVar(X)=mathbb E[X^2] − mathbb E[X]^2$$
and
$$textCov(X,Y) = mathbb E[XY] − mathbb E[X] mathbb E[Y]$$
Then the given inequality is equivalent to
$$ 2 mathbb E[X] Big( mathbb E[X^2] - mathbb E[X]^2 Big)
overset?geq
mathbb E[X^3] - mathbb E[X] mathbb E[X^2] tag1$$
For a counterexample, let $X sim textExp(lambda)$ be the exponential distribution with parameter $lambda in (0, infty)$ and probability mass function
$$ f_X(x) = begincases
lambda e^-lambda x & textif x geq 0 \
0 & textif x < 0
endcases $$
Its $n$-th moment is given by $mathbb E[X^n]=fracn!lambda^n$, so the inequality $(1)$ becomes
$$2 cdot frac1lambda Big( frac2lambda^2 - frac1lambda^2 Big)
geq
frac6lambda^3 - frac1lambda cdot frac2lambda^2$$
which reduces to
$$frac2lambda^3 geq frac4lambda^3$$
This is false for all $lambda in (0, infty)$, hence we found a counterexample.
answered 4 hours ago
M. NestorM. Nestor
890115
$endgroup$
$begingroup$
hi! thanks a lot for this. do you think the inequality is false for a real-valued $X$ (as specified in the question) as well? i haven't been able to find a counter-example. for $X sim mathrmNormal(mu, sigma^2)$, both left and right hand side reduce to $2musigma^2$.
$endgroup$
– sk1ll3r
2 hours ago
1
$begingroup$
Sorry I did not notice that detail. I edited my answer to include an example where $X$ has continuous support.
$endgroup$
– M. Nestor
2 hours ago
add a comment |
$begingroup$
Use the variance and covariance identities
$$textVar(X)=mathbb E[X^2] − mathbb E[X]^2$$
and
$$textCov(X,Y) = mathbb E[XY] − mathbb E[X] mathbb E[Y]$$
Then the given inequality is equivalent to
$$ 2 mathbb E[X] Big( mathbb E[X^2] - mathbb E[X]^2 Big)
overset?geq
mathbb E[X^3] - mathbb E[X] mathbb E[X^2] tag1$$
For a counterexample, let $X sim textExp(lambda)$ be the exponential distribution with parameter $lambda in (0, infty)$ and probability mass function
$$ f_X(x) = begincases
lambda e^-lambda x & textif x geq 0 \
0 & textif x < 0
endcases $$
Its $n$-th moment is given by $mathbb E[X^n]=fracn!lambda^n$, so the inequality $(1)$ becomes
$$2 cdot frac1lambda Big( frac2lambda^2 - frac1lambda^2 Big)
geq
frac6lambda^3 - frac1lambda cdot frac2lambda^2$$
which reduces to
$$frac2lambda^3 geq frac4lambda^3$$
This is false for all $lambda in (0, infty)$, hence we found a counterexample.
answered 4 hours ago
M. NestorM. Nestor
890115
$endgroup$
$begingroup$
hi! thanks a lot for this. do you think the inequality is false for a real-valued $X$ (as specified in the question) as well? i haven't been able to find a counter-example. for $X sim mathrmNormal(mu, sigma^2)$, both left and right hand side reduce to $2musigma^2$.
$endgroup$
– sk1ll3r
2 hours ago
1
$begingroup$
Sorry I did not notice that detail. I edited my answer to include an example where $X$ has continuous support.
$endgroup$
– M. Nestor
2 hours ago
add a comment |
$begingroup$
Use the variance and covariance identities
$$textVar(X)=mathbb E[X^2] − mathbb E[X]^2$$
and
$$textCov(X,Y) = mathbb E[XY] − mathbb E[X] mathbb E[Y]$$
Then the given inequality is equivalent to
$$ 2 mathbb E[X] Big( mathbb E[X^2] - mathbb E[X]^2 Big)
overset?geq
mathbb E[X^3] - mathbb E[X] mathbb E[X^2] tag1$$
For a counterexample, let $X sim textExp(lambda)$ be the exponential distribution with parameter $lambda in (0, infty)$ and probability mass function
$$ f_X(x) = begincases
lambda e^-lambda x & textif x geq 0 \
0 & textif x < 0
endcases $$
Its $n$-th moment is given by $mathbb E[X^n]=fracn!lambda^n$, so the inequality $(1)$ becomes
$$2 cdot frac1lambda Big( frac2lambda^2 - frac1lambda^2 Big)
geq
frac6lambda^3 - frac1lambda cdot frac2lambda^2$$
which reduces to
$$frac2lambda^3 geq frac4lambda^3$$
This is false for all $lambda in (0, infty)$, hence we found a counterexample.
answered 4 hours ago
M. NestorM. Nestor
890115
$endgroup$
Use the variance and covariance identities
$$textVar(X)=mathbb E[X^2] − mathbb E[X]^2$$
and
$$textCov(X,Y) = mathbb E[XY] − mathbb E[X] mathbb E[Y]$$
Then the given inequality is equivalent to
$$ 2 mathbb E[X] Big( mathbb E[X^2] - mathbb E[X]^2 Big)
overset?geq
mathbb E[X^3] - mathbb E[X] mathbb E[X^2] tag1$$
For a counterexample, let $X sim textExp(lambda)$ be the exponential distribution with parameter $lambda in (0, infty)$ and probability mass function
$$ f_X(x) = begincases
lambda e^-lambda x & textif x geq 0 \
0 & textif x < 0
endcases $$
Its $n$-th moment is given by $mathbb E[X^n]=fracn!lambda^n$, so the inequality $(1)$ becomes
$$2 cdot frac1lambda Big( frac2lambda^2 - frac1lambda^2 Big)
geq
frac6lambda^3 - frac1lambda cdot frac2lambda^2$$
which reduces to
$$frac2lambda^3 geq frac4lambda^3$$
This is false for all $lambda in (0, infty)$, hence we found a counterexample.
answered 4 hours ago
M. NestorM. Nestor
890115
edited 1 hour ago
answered 4 hours ago
M. NestorM. Nestor
890115
answered 4 hours ago
M. NestorM. Nestor
890115
answered 4 hours ago
M. NestorM. Nestor
890115
890115
$begingroup$
hi! thanks a lot for this. do you think the inequality is false for a real-valued $X$ (as specified in the question) as well? i haven't been able to find a counter-example. for $X sim mathrmNormal(mu, sigma^2)$, both left and right hand side reduce to $2musigma^2$.
$endgroup$
– sk1ll3r
2 hours ago
1
$begingroup$
Sorry I did not notice that detail. I edited my answer to include an example where $X$ has continuous support.
$endgroup$
– M. Nestor
2 hours ago
add a comment |
$begingroup$
hi! thanks a lot for this. do you think the inequality is false for a real-valued $X$ (as specified in the question) as well? i haven't been able to find a counter-example. for $X sim mathrmNormal(mu, sigma^2)$, both left and right hand side reduce to $2musigma^2$.
$endgroup$
– sk1ll3r
2 hours ago
1
$begingroup$
Sorry I did not notice that detail. I edited my answer to include an example where $X$ has continuous support.
$endgroup$
– M. Nestor
2 hours ago
$begingroup$
hi! thanks a lot for this. do you think the inequality is false for a real-valued $X$ (as specified in the question) as well? i haven't been able to find a counter-example. for $X sim mathrmNormal(mu, sigma^2)$, both left and right hand side reduce to $2musigma^2$.
$endgroup$
– sk1ll3r
2 hours ago
$begingroup$
hi! thanks a lot for this. do you think the inequality is false for a real-valued $X$ (as specified in the question) as well? i haven't been able to find a counter-example. for $X sim mathrmNormal(mu, sigma^2)$, both left and right hand side reduce to $2musigma^2$.
$endgroup$
– sk1ll3r
2 hours ago
1
1
$begingroup$
Sorry I did not notice that detail. I edited my answer to include an example where $X$ has continuous support.
$endgroup$
– M. Nestor
2 hours ago
$begingroup$
Sorry I did not notice that detail. I edited my answer to include an example where $X$ has continuous support.
$endgroup$
– M. Nestor
2 hours ago
add a comment |
$begingroup$
Alternate approach / observation...
Assume $2 E[X] Var(X) ge Cov(X, X^2)$ for any real-valued r.v. $X.$ Then in particular it would also be true for $-X$, so that:
$$
beginalign
2E[-X] Var(-X) &ge Cov(-X, (-X)^2)\
-2E[X] Var(X) &ge -Cov(X, X^2)\
2E[X] Var(X) &le Cov(X, X^2)
endalign
$$
which, combined with the original inequality, means: $2E[X]Var(X) = Cov(X, X^2)$ for all $X$.
At this point, it would seem one should be able to find a counter-example easily, without using e.g. $Cov(X, X^2) = E[X^3] - E[X]E[X^2]$ etc, but I just couldn't think of a quick one. :(
answered 1 hour ago
antkamantkam
4,489415
$endgroup$
add a comment |
$begingroup$
Alternate approach / observation...
Assume $2 E[X] Var(X) ge Cov(X, X^2)$ for any real-valued r.v. $X.$ Then in particular it would also be true for $-X$, so that:
$$
beginalign
2E[-X] Var(-X) &ge Cov(-X, (-X)^2)\
-2E[X] Var(X) &ge -Cov(X, X^2)\
2E[X] Var(X) &le Cov(X, X^2)
endalign
$$
which, combined with the original inequality, means: $2E[X]Var(X) = Cov(X, X^2)$ for all $X$.
At this point, it would seem one should be able to find a counter-example easily, without using e.g. $Cov(X, X^2) = E[X^3] - E[X]E[X^2]$ etc, but I just couldn't think of a quick one. :(
answered 1 hour ago
antkamantkam
4,489415
$endgroup$
add a comment |
$begingroup$
Alternate approach / observation...
Assume $2 E[X] Var(X) ge Cov(X, X^2)$ for any real-valued r.v. $X.$ Then in particular it would also be true for $-X$, so that:
$$
beginalign
2E[-X] Var(-X) &ge Cov(-X, (-X)^2)\
-2E[X] Var(X) &ge -Cov(X, X^2)\
2E[X] Var(X) &le Cov(X, X^2)
endalign
$$
which, combined with the original inequality, means: $2E[X]Var(X) = Cov(X, X^2)$ for all $X$.
At this point, it would seem one should be able to find a counter-example easily, without using e.g. $Cov(X, X^2) = E[X^3] - E[X]E[X^2]$ etc, but I just couldn't think of a quick one. :(
answered 1 hour ago
antkamantkam
4,489415
$endgroup$
Alternate approach / observation...
Assume $2 E[X] Var(X) ge Cov(X, X^2)$ for any real-valued r.v. $X.$ Then in particular it would also be true for $-X$, so that:
$$
beginalign
2E[-X] Var(-X) &ge Cov(-X, (-X)^2)\
-2E[X] Var(X) &ge -Cov(X, X^2)\
2E[X] Var(X) &le Cov(X, X^2)
endalign
$$
which, combined with the original inequality, means: $2E[X]Var(X) = Cov(X, X^2)$ for all $X$.
At this point, it would seem one should be able to find a counter-example easily, without using e.g. $Cov(X, X^2) = E[X^3] - E[X]E[X^2]$ etc, but I just couldn't think of a quick one. :(
answered 1 hour ago
antkamantkam
4,489415
answered 1 hour ago
antkamantkam
4,489415
answered 1 hour ago
antkamantkam
4,489415
answered 1 hour ago
antkamantkam
4,489415
4,489415
add a comment |
add a comment |
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$begingroup$
One trick is to write a little computer program to generate examples randomly, which you can use to find counterexamples. Such as $P(X=a)=p, P(X=b)=1-p$, where you use a random number generator to pick the parameters $a,b,p$ and exact formulas in terms of $a,b.p$ for the relevant moments of $X$.
$endgroup$
– kimchi lover
4 hours ago
$begingroup$
I expect that the answer will follow quickly after applying the well-known identities $textVar(X)=mathbb E[X^2]-mathbb E[X]^2$ and $textCov(X,Y)=mathbb E[XY]-mathbb E[X]mathbb E[Y]$
$endgroup$
– M. Nestor
4 hours ago