Proving the expected value of the square root of X is less than the square root of the expected value of X
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How do I show that $E(sqrt{X}) leq sqrt{E(X)}$ for a positive random variable $X$?
I may be intended to use the Cauchy-Schwarz Inequality, $[E(XY)]^2 leq E(X^2)E(Y^2)$, but I'm not sure how.
probability inequality expectation
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add a comment |
$begingroup$
How do I show that $E(sqrt{X}) leq sqrt{E(X)}$ for a positive random variable $X$?
I may be intended to use the Cauchy-Schwarz Inequality, $[E(XY)]^2 leq E(X^2)E(Y^2)$, but I'm not sure how.
probability inequality expectation
$endgroup$
add a comment |
$begingroup$
How do I show that $E(sqrt{X}) leq sqrt{E(X)}$ for a positive random variable $X$?
I may be intended to use the Cauchy-Schwarz Inequality, $[E(XY)]^2 leq E(X^2)E(Y^2)$, but I'm not sure how.
probability inequality expectation
$endgroup$
How do I show that $E(sqrt{X}) leq sqrt{E(X)}$ for a positive random variable $X$?
I may be intended to use the Cauchy-Schwarz Inequality, $[E(XY)]^2 leq E(X^2)E(Y^2)$, but I'm not sure how.
probability inequality expectation
probability inequality expectation
edited Nov 12 '14 at 3:23
asked Nov 12 '14 at 2:45
user191823
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4 Answers
4
active
oldest
votes
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$sqrt{x}$, $xgeq 0$ is a concave function so Jensen's inequality gives us the result without further effort.
answered Nov 12 '14 at 3:35
Suzu HiroseSuzu Hirose
4,17021228
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add a comment |
$begingroup$
The Cauchy-Schwarz inequality tells us that if $A=sqrt{X}$ and $B=1$, then
$$E[sqrt{X}]^2=(E[AB])^2leq E[A^2]E[B^2]=E[X]$$
so
$$E[sqrt{X}]leq sqrt{E[X]}$$
edited Jul 27 '15 at 20:55
Daniel Fischer♦
174k16167287
answered Nov 12 '14 at 3:22
Matt SamuelMatt Samuel
38.7k63769
$endgroup$
add a comment |
$begingroup$
Consider the function $f(x)=sqrt{x}$.
As this is clearly a concave function, Jensen's inequality states
$$
mathbb{E}Big[f(X)Big]leq f Big(mathbb{E}[X] Big)
$$
and there you go
answered Mar 6 '18 at 12:54
EmGEmG
688
$endgroup$
$begingroup$
$f$ is concave, not convex. Otherwise the inequality would be the other way around.
$endgroup$
– 0xbadf00d
Dec 20 '18 at 17:49
add a comment |
$begingroup$
Let us assume a variable x and its probability p. Then we calculate (x^2)p - (x^2)p^2. Since p is a fraction it is greater than p^2. So (x^2){p - p^2} is positive. Summing over all variables we get by definition E[X^2] - (E[X])^2 >= 0 or E[X^2] >= (E[X])^2. Taking square root and putting X = X^0•5 gives us the result. Since probability is a fraction and becomes very small as number of variables are large summing over (x^2)p^2 can be approximated as summation of product of x and p whole square.
answered Aug 18 '15 at 17:11
dev kitdev kit
11
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$sqrt{x}$, $xgeq 0$ is a concave function so Jensen's inequality gives us the result without further effort.
answered Nov 12 '14 at 3:35
Suzu HiroseSuzu Hirose
4,17021228
$endgroup$
add a comment |
$begingroup$
$sqrt{x}$, $xgeq 0$ is a concave function so Jensen's inequality gives us the result without further effort.
answered Nov 12 '14 at 3:35
Suzu HiroseSuzu Hirose
4,17021228
$endgroup$
add a comment |
$begingroup$
$sqrt{x}$, $xgeq 0$ is a concave function so Jensen's inequality gives us the result without further effort.
answered Nov 12 '14 at 3:35
Suzu HiroseSuzu Hirose
4,17021228
$endgroup$
$sqrt{x}$, $xgeq 0$ is a concave function so Jensen's inequality gives us the result without further effort.
answered Nov 12 '14 at 3:35
Suzu HiroseSuzu Hirose
4,17021228
answered Nov 12 '14 at 3:35
Suzu HiroseSuzu Hirose
4,17021228
answered Nov 12 '14 at 3:35
Suzu HiroseSuzu Hirose
4,17021228
answered Nov 12 '14 at 3:35
Suzu HiroseSuzu Hirose
4,17021228
4,17021228
add a comment |
add a comment |
$begingroup$
The Cauchy-Schwarz inequality tells us that if $A=sqrt{X}$ and $B=1$, then
$$E[sqrt{X}]^2=(E[AB])^2leq E[A^2]E[B^2]=E[X]$$
so
$$E[sqrt{X}]leq sqrt{E[X]}$$
edited Jul 27 '15 at 20:55
Daniel Fischer♦
174k16167287
answered Nov 12 '14 at 3:22
Matt SamuelMatt Samuel
38.7k63769
$endgroup$
add a comment |
$begingroup$
The Cauchy-Schwarz inequality tells us that if $A=sqrt{X}$ and $B=1$, then
$$E[sqrt{X}]^2=(E[AB])^2leq E[A^2]E[B^2]=E[X]$$
so
$$E[sqrt{X}]leq sqrt{E[X]}$$
edited Jul 27 '15 at 20:55
Daniel Fischer♦
174k16167287
answered Nov 12 '14 at 3:22
Matt SamuelMatt Samuel
38.7k63769
$endgroup$
add a comment |
$begingroup$
The Cauchy-Schwarz inequality tells us that if $A=sqrt{X}$ and $B=1$, then
$$E[sqrt{X}]^2=(E[AB])^2leq E[A^2]E[B^2]=E[X]$$
so
$$E[sqrt{X}]leq sqrt{E[X]}$$
edited Jul 27 '15 at 20:55
Daniel Fischer♦
174k16167287
answered Nov 12 '14 at 3:22
Matt SamuelMatt Samuel
38.7k63769
$endgroup$
The Cauchy-Schwarz inequality tells us that if $A=sqrt{X}$ and $B=1$, then
$$E[sqrt{X}]^2=(E[AB])^2leq E[A^2]E[B^2]=E[X]$$
so
$$E[sqrt{X}]leq sqrt{E[X]}$$
edited Jul 27 '15 at 20:55
Daniel Fischer♦
174k16167287
answered Nov 12 '14 at 3:22
Matt SamuelMatt Samuel
38.7k63769
edited Jul 27 '15 at 20:55
Daniel Fischer♦
174k16167287
edited Jul 27 '15 at 20:55
Daniel Fischer♦
174k16167287
edited Jul 27 '15 at 20:55
Daniel Fischer♦
174k16167287
174k16167287
answered Nov 12 '14 at 3:22
Matt SamuelMatt Samuel
38.7k63769
answered Nov 12 '14 at 3:22
Matt SamuelMatt Samuel
38.7k63769
answered Nov 12 '14 at 3:22
Matt SamuelMatt Samuel
38.7k63769
38.7k63769
add a comment |
add a comment |
$begingroup$
Consider the function $f(x)=sqrt{x}$.
As this is clearly a concave function, Jensen's inequality states
$$
mathbb{E}Big[f(X)Big]leq f Big(mathbb{E}[X] Big)
$$
and there you go
answered Mar 6 '18 at 12:54
EmGEmG
688
$endgroup$
$begingroup$
$f$ is concave, not convex. Otherwise the inequality would be the other way around.
$endgroup$
– 0xbadf00d
Dec 20 '18 at 17:49
add a comment |
$begingroup$
Consider the function $f(x)=sqrt{x}$.
As this is clearly a concave function, Jensen's inequality states
$$
mathbb{E}Big[f(X)Big]leq f Big(mathbb{E}[X] Big)
$$
and there you go
answered Mar 6 '18 at 12:54
EmGEmG
688
$endgroup$
$begingroup$
$f$ is concave, not convex. Otherwise the inequality would be the other way around.
$endgroup$
– 0xbadf00d
Dec 20 '18 at 17:49
add a comment |
$begingroup$
Consider the function $f(x)=sqrt{x}$.
As this is clearly a concave function, Jensen's inequality states
$$
mathbb{E}Big[f(X)Big]leq f Big(mathbb{E}[X] Big)
$$
and there you go
answered Mar 6 '18 at 12:54
EmGEmG
688
$endgroup$
Consider the function $f(x)=sqrt{x}$.
As this is clearly a concave function, Jensen's inequality states
$$
mathbb{E}Big[f(X)Big]leq f Big(mathbb{E}[X] Big)
$$
and there you go
answered Mar 6 '18 at 12:54
EmGEmG
688
edited Dec 22 '18 at 19:00
answered Mar 6 '18 at 12:54
EmGEmG
688
answered Mar 6 '18 at 12:54
EmGEmG
688
answered Mar 6 '18 at 12:54
EmGEmG
688
688
$begingroup$
$f$ is concave, not convex. Otherwise the inequality would be the other way around.
$endgroup$
– 0xbadf00d
Dec 20 '18 at 17:49
add a comment |
$begingroup$
$f$ is concave, not convex. Otherwise the inequality would be the other way around.
$endgroup$
– 0xbadf00d
Dec 20 '18 at 17:49
$begingroup$
$f$ is concave, not convex. Otherwise the inequality would be the other way around.
$endgroup$
– 0xbadf00d
Dec 20 '18 at 17:49
$begingroup$
$f$ is concave, not convex. Otherwise the inequality would be the other way around.
$endgroup$
– 0xbadf00d
Dec 20 '18 at 17:49
add a comment |
$begingroup$
Let us assume a variable x and its probability p. Then we calculate (x^2)p - (x^2)p^2. Since p is a fraction it is greater than p^2. So (x^2){p - p^2} is positive. Summing over all variables we get by definition E[X^2] - (E[X])^2 >= 0 or E[X^2] >= (E[X])^2. Taking square root and putting X = X^0•5 gives us the result. Since probability is a fraction and becomes very small as number of variables are large summing over (x^2)p^2 can be approximated as summation of product of x and p whole square.
answered Aug 18 '15 at 17:11
dev kitdev kit
11
$endgroup$
add a comment |
$begingroup$
Let us assume a variable x and its probability p. Then we calculate (x^2)p - (x^2)p^2. Since p is a fraction it is greater than p^2. So (x^2){p - p^2} is positive. Summing over all variables we get by definition E[X^2] - (E[X])^2 >= 0 or E[X^2] >= (E[X])^2. Taking square root and putting X = X^0•5 gives us the result. Since probability is a fraction and becomes very small as number of variables are large summing over (x^2)p^2 can be approximated as summation of product of x and p whole square.
answered Aug 18 '15 at 17:11
dev kitdev kit
11
$endgroup$
add a comment |
$begingroup$
Let us assume a variable x and its probability p. Then we calculate (x^2)p - (x^2)p^2. Since p is a fraction it is greater than p^2. So (x^2){p - p^2} is positive. Summing over all variables we get by definition E[X^2] - (E[X])^2 >= 0 or E[X^2] >= (E[X])^2. Taking square root and putting X = X^0•5 gives us the result. Since probability is a fraction and becomes very small as number of variables are large summing over (x^2)p^2 can be approximated as summation of product of x and p whole square.
answered Aug 18 '15 at 17:11
dev kitdev kit
11
$endgroup$
Let us assume a variable x and its probability p. Then we calculate (x^2)p - (x^2)p^2. Since p is a fraction it is greater than p^2. So (x^2){p - p^2} is positive. Summing over all variables we get by definition E[X^2] - (E[X])^2 >= 0 or E[X^2] >= (E[X])^2. Taking square root and putting X = X^0•5 gives us the result. Since probability is a fraction and becomes very small as number of variables are large summing over (x^2)p^2 can be approximated as summation of product of x and p whole square.
answered Aug 18 '15 at 17:11
dev kitdev kit
11
edited Aug 19 '15 at 10:44
answered Aug 18 '15 at 17:11
dev kitdev kit
11
answered Aug 18 '15 at 17:11
dev kitdev kit
11
answered Aug 18 '15 at 17:11
dev kitdev kit
11
11
add a comment |
add a comment |
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