Convexity of $x^{2q}$ implies $mathbb{E} big[ left(X-X^{'}right)^{2q} big]leq 2^{2q-1} left( mathbb{E}X^{2q}+...
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This problem is from page 25 of concentration inequalities a nonasymptotic theory of independence
$X$ is a random variable
and $X^{'}$ is independent copy of $X$
q is integer and $qgeq1$
It is said that the following inequality was implied by the convexity of $x^{2q}$
$$mathbb{E} big[ left(X-X^{'}right)^{2q} big]leq 2^{2q-1} left( mathbb{E}X^{2q}+ mathbb{E}X^{'2q} right) $$
However, I do not know how to deduce it. The first guess is using Minkowski inequality to get terms of $mathbb{E}X^{2q}$ and $mathbb{E}X^{'2q}$. The Jensen ineqaulity will give inequality of reverse direction, I am wondering how convexity is applied. Any suggestion is welcomed.
probability inequality convexity-inequality
asked Nov 16 at 23:33
Rikeijin
878
add a comment |
up vote
0
down vote
favorite
This problem is from page 25 of concentration inequalities a nonasymptotic theory of independence
$X$ is a random variable
and $X^{'}$ is independent copy of $X$
q is integer and $qgeq1$
It is said that the following inequality was implied by the convexity of $x^{2q}$
$$mathbb{E} big[ left(X-X^{'}right)^{2q} big]leq 2^{2q-1} left( mathbb{E}X^{2q}+ mathbb{E}X^{'2q} right) $$
However, I do not know how to deduce it. The first guess is using Minkowski inequality to get terms of $mathbb{E}X^{2q}$ and $mathbb{E}X^{'2q}$. The Jensen ineqaulity will give inequality of reverse direction, I am wondering how convexity is applied. Any suggestion is welcomed.
probability inequality convexity-inequality
asked Nov 16 at 23:33
Rikeijin
878
2
math.stackexchange.com/questions/1102727/…
– d.k.o.
Nov 16 at 23:38
Thx. I am wondering how $$(a+b)^rleqslant a^r+b^rmbox{ if }0lt rleqslant 1 $$ is deduced from concavity of $ x^r$. I can see it throguh Bernoulli inequality but it's fine to find a unified way to do it through Jensen's inequality.
– Rikeijin
Nov 17 at 10:17
math.stackexchange.com/questions/318649/…
– d.k.o.
Nov 17 at 16:44
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This problem is from page 25 of concentration inequalities a nonasymptotic theory of independence
$X$ is a random variable
and $X^{'}$ is independent copy of $X$
q is integer and $qgeq1$
It is said that the following inequality was implied by the convexity of $x^{2q}$
$$mathbb{E} big[ left(X-X^{'}right)^{2q} big]leq 2^{2q-1} left( mathbb{E}X^{2q}+ mathbb{E}X^{'2q} right) $$
However, I do not know how to deduce it. The first guess is using Minkowski inequality to get terms of $mathbb{E}X^{2q}$ and $mathbb{E}X^{'2q}$. The Jensen ineqaulity will give inequality of reverse direction, I am wondering how convexity is applied. Any suggestion is welcomed.
probability inequality convexity-inequality
asked Nov 16 at 23:33
Rikeijin
878
This problem is from page 25 of concentration inequalities a nonasymptotic theory of independence
$X$ is a random variable
and $X^{'}$ is independent copy of $X$
q is integer and $qgeq1$
It is said that the following inequality was implied by the convexity of $x^{2q}$
$$mathbb{E} big[ left(X-X^{'}right)^{2q} big]leq 2^{2q-1} left( mathbb{E}X^{2q}+ mathbb{E}X^{'2q} right) $$
However, I do not know how to deduce it. The first guess is using Minkowski inequality to get terms of $mathbb{E}X^{2q}$ and $mathbb{E}X^{'2q}$. The Jensen ineqaulity will give inequality of reverse direction, I am wondering how convexity is applied. Any suggestion is welcomed.
probability inequality convexity-inequality
probability inequality convexity-inequality
asked Nov 16 at 23:33
Rikeijin
878
asked Nov 16 at 23:33
Rikeijin
878
asked Nov 16 at 23:33
Rikeijin
878
asked Nov 16 at 23:33
Rikeijin
878
asked Nov 16 at 23:33
Rikeijin
878
878
2
math.stackexchange.com/questions/1102727/…
– d.k.o.
Nov 16 at 23:38
Thx. I am wondering how $$(a+b)^rleqslant a^r+b^rmbox{ if }0lt rleqslant 1 $$ is deduced from concavity of $ x^r$. I can see it throguh Bernoulli inequality but it's fine to find a unified way to do it through Jensen's inequality.
– Rikeijin
Nov 17 at 10:17
math.stackexchange.com/questions/318649/…
– d.k.o.
Nov 17 at 16:44
add a comment |
2
math.stackexchange.com/questions/1102727/…
– d.k.o.
Nov 16 at 23:38
Thx. I am wondering how $$(a+b)^rleqslant a^r+b^rmbox{ if }0lt rleqslant 1 $$ is deduced from concavity of $ x^r$. I can see it throguh Bernoulli inequality but it's fine to find a unified way to do it through Jensen's inequality.
– Rikeijin
Nov 17 at 10:17
math.stackexchange.com/questions/318649/…
– d.k.o.
Nov 17 at 16:44
2
2
math.stackexchange.com/questions/1102727/…
– d.k.o.
Nov 16 at 23:38
math.stackexchange.com/questions/1102727/…
– d.k.o.
Nov 16 at 23:38
Thx. I am wondering how $$(a+b)^rleqslant a^r+b^rmbox{ if }0lt rleqslant 1 $$ is deduced from concavity of $ x^r$. I can see it throguh Bernoulli inequality but it's fine to find a unified way to do it through Jensen's inequality.
– Rikeijin
Nov 17 at 10:17
Thx. I am wondering how $$(a+b)^rleqslant a^r+b^rmbox{ if }0lt rleqslant 1 $$ is deduced from concavity of $ x^r$. I can see it throguh Bernoulli inequality but it's fine to find a unified way to do it through Jensen's inequality.
– Rikeijin
Nov 17 at 10:17
math.stackexchange.com/questions/318649/…
– d.k.o.
Nov 17 at 16:44
math.stackexchange.com/questions/318649/…
– d.k.o.
Nov 17 at 16:44
add a comment |
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math.stackexchange.com/questions/1102727/…
– d.k.o.
Nov 16 at 23:38
Thx. I am wondering how $$(a+b)^rleqslant a^r+b^rmbox{ if }0lt rleqslant 1 $$ is deduced from concavity of $ x^r$. I can see it throguh Bernoulli inequality but it's fine to find a unified way to do it through Jensen's inequality.
– Rikeijin
Nov 17 at 10:17
math.stackexchange.com/questions/318649/…
– d.k.o.
Nov 17 at 16:44