Proving $E[max(X^2,Y^2)]le 1+sqrt{1-rho^2}$
$begingroup$
The question is
Let $X$ and $Y$ be random variables with mean $0$, and variances $1$ and correlation coefficient $rho$. Show that $E[max(X^2,Y^2)]le 1+sqrt{1-rho^2}$.
My attempt:
$$E[max(X^2,Y^2)]=
Eleft(frac{X^2+Y^2-|X^2-Y^2|}{2}right)=
1-frac{E|X^2-Y^2|}{2}$$
which is certainly lesser than the rhs. Is there anything wrong in my reasoning?
probability-theory statistics inequality expected-value
edited Dec 8 '18 at 10:02
StubbornAtom
5,97311238
asked Dec 8 '18 at 7:42
user587126user587126
155
$endgroup$
add a comment |
$begingroup$
The question is
Let $X$ and $Y$ be random variables with mean $0$, and variances $1$ and correlation coefficient $rho$. Show that $E[max(X^2,Y^2)]le 1+sqrt{1-rho^2}$.
My attempt:
$$E[max(X^2,Y^2)]=
Eleft(frac{X^2+Y^2-|X^2-Y^2|}{2}right)=
1-frac{E|X^2-Y^2|}{2}$$
which is certainly lesser than the rhs. Is there anything wrong in my reasoning?
probability-theory statistics inequality expected-value
edited Dec 8 '18 at 10:02
StubbornAtom
5,97311238
asked Dec 8 '18 at 7:42
user587126user587126
155
$endgroup$
add a comment |
$begingroup$
The question is
Let $X$ and $Y$ be random variables with mean $0$, and variances $1$ and correlation coefficient $rho$. Show that $E[max(X^2,Y^2)]le 1+sqrt{1-rho^2}$.
My attempt:
$$E[max(X^2,Y^2)]=
Eleft(frac{X^2+Y^2-|X^2-Y^2|}{2}right)=
1-frac{E|X^2-Y^2|}{2}$$
which is certainly lesser than the rhs. Is there anything wrong in my reasoning?
probability-theory statistics inequality expected-value
edited Dec 8 '18 at 10:02
StubbornAtom
5,97311238
asked Dec 8 '18 at 7:42
user587126user587126
155
$endgroup$
The question is
Let $X$ and $Y$ be random variables with mean $0$, and variances $1$ and correlation coefficient $rho$. Show that $E[max(X^2,Y^2)]le 1+sqrt{1-rho^2}$.
My attempt:
$$E[max(X^2,Y^2)]=
Eleft(frac{X^2+Y^2-|X^2-Y^2|}{2}right)=
1-frac{E|X^2-Y^2|}{2}$$
which is certainly lesser than the rhs. Is there anything wrong in my reasoning?
probability-theory statistics inequality expected-value
probability-theory statistics inequality expected-value
edited Dec 8 '18 at 10:02
StubbornAtom
5,97311238
asked Dec 8 '18 at 7:42
user587126user587126
155
edited Dec 8 '18 at 10:02
StubbornAtom
5,97311238
asked Dec 8 '18 at 7:42
user587126user587126
155
edited Dec 8 '18 at 10:02
StubbornAtom
5,97311238
edited Dec 8 '18 at 10:02
StubbornAtom
5,97311238
edited Dec 8 '18 at 10:02
StubbornAtom
5,97311238
5,97311238
asked Dec 8 '18 at 7:42
user587126user587126
155
asked Dec 8 '18 at 7:42
user587126user587126
155
asked Dec 8 '18 at 7:42
user587126user587126
155
155
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Actually, the correct expression is
$$max(X^2,Y^2)=frac{1}{2}left[X^2+Y^2color{red}+|X^2-Y^2|right]tag{1}$$
Now simply note that
begin{align}
E|X^2-Y^2|&=E(|(X+Y)(X-Y)|)
\&=Eleft[sqrt{(X+Y)^2(X-Y)^2}right]
\&le sqrt{Eleft[(X+Y)^2(X-Y)^2right]}tag{2}
\&lesqrt{E,[(X+Y)^2]E,[(X-Y)^2]}tag{3}
end{align}
In $(2)$, Jensen's inequality was used since $xmapstosqrt x$ is a concave function for all $xge0$.
In $(3)$, Cauchy-Schwarz inequality was used.
So when you take expectation on both sides of $(1)$, use the above inequality and substitute the values of the expectations, you would get the desired result.
answered Dec 8 '18 at 9:57
StubbornAtomStubbornAtom
5,97311238
$endgroup$
$begingroup$
Thank you very much! I messed up the formula pretty badly!
$endgroup$
– user587126
Dec 8 '18 at 11:46
$begingroup$
Actually the inequality (3) directly follows from Cauchy-Scharwz; no need for Jensen. Previously asked and answered: math.stackexchange.com/questions/1395820/….
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:49
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Actually, the correct expression is
$$max(X^2,Y^2)=frac{1}{2}left[X^2+Y^2color{red}+|X^2-Y^2|right]tag{1}$$
Now simply note that
begin{align}
E|X^2-Y^2|&=E(|(X+Y)(X-Y)|)
\&=Eleft[sqrt{(X+Y)^2(X-Y)^2}right]
\&le sqrt{Eleft[(X+Y)^2(X-Y)^2right]}tag{2}
\&lesqrt{E,[(X+Y)^2]E,[(X-Y)^2]}tag{3}
end{align}
In $(2)$, Jensen's inequality was used since $xmapstosqrt x$ is a concave function for all $xge0$.
In $(3)$, Cauchy-Schwarz inequality was used.
So when you take expectation on both sides of $(1)$, use the above inequality and substitute the values of the expectations, you would get the desired result.
answered Dec 8 '18 at 9:57
StubbornAtomStubbornAtom
5,97311238
$endgroup$
$begingroup$
Thank you very much! I messed up the formula pretty badly!
$endgroup$
– user587126
Dec 8 '18 at 11:46
$begingroup$
Actually the inequality (3) directly follows from Cauchy-Scharwz; no need for Jensen. Previously asked and answered: math.stackexchange.com/questions/1395820/….
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:49
add a comment |
$begingroup$
Actually, the correct expression is
$$max(X^2,Y^2)=frac{1}{2}left[X^2+Y^2color{red}+|X^2-Y^2|right]tag{1}$$
Now simply note that
begin{align}
E|X^2-Y^2|&=E(|(X+Y)(X-Y)|)
\&=Eleft[sqrt{(X+Y)^2(X-Y)^2}right]
\&le sqrt{Eleft[(X+Y)^2(X-Y)^2right]}tag{2}
\&lesqrt{E,[(X+Y)^2]E,[(X-Y)^2]}tag{3}
end{align}
In $(2)$, Jensen's inequality was used since $xmapstosqrt x$ is a concave function for all $xge0$.
In $(3)$, Cauchy-Schwarz inequality was used.
So when you take expectation on both sides of $(1)$, use the above inequality and substitute the values of the expectations, you would get the desired result.
answered Dec 8 '18 at 9:57
StubbornAtomStubbornAtom
5,97311238
$endgroup$
$begingroup$
Thank you very much! I messed up the formula pretty badly!
$endgroup$
– user587126
Dec 8 '18 at 11:46
$begingroup$
Actually the inequality (3) directly follows from Cauchy-Scharwz; no need for Jensen. Previously asked and answered: math.stackexchange.com/questions/1395820/….
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:49
add a comment |
$begingroup$
Actually, the correct expression is
$$max(X^2,Y^2)=frac{1}{2}left[X^2+Y^2color{red}+|X^2-Y^2|right]tag{1}$$
Now simply note that
begin{align}
E|X^2-Y^2|&=E(|(X+Y)(X-Y)|)
\&=Eleft[sqrt{(X+Y)^2(X-Y)^2}right]
\&le sqrt{Eleft[(X+Y)^2(X-Y)^2right]}tag{2}
\&lesqrt{E,[(X+Y)^2]E,[(X-Y)^2]}tag{3}
end{align}
In $(2)$, Jensen's inequality was used since $xmapstosqrt x$ is a concave function for all $xge0$.
In $(3)$, Cauchy-Schwarz inequality was used.
So when you take expectation on both sides of $(1)$, use the above inequality and substitute the values of the expectations, you would get the desired result.
answered Dec 8 '18 at 9:57
StubbornAtomStubbornAtom
5,97311238
$endgroup$
Actually, the correct expression is
$$max(X^2,Y^2)=frac{1}{2}left[X^2+Y^2color{red}+|X^2-Y^2|right]tag{1}$$
Now simply note that
begin{align}
E|X^2-Y^2|&=E(|(X+Y)(X-Y)|)
\&=Eleft[sqrt{(X+Y)^2(X-Y)^2}right]
\&le sqrt{Eleft[(X+Y)^2(X-Y)^2right]}tag{2}
\&lesqrt{E,[(X+Y)^2]E,[(X-Y)^2]}tag{3}
end{align}
In $(2)$, Jensen's inequality was used since $xmapstosqrt x$ is a concave function for all $xge0$.
In $(3)$, Cauchy-Schwarz inequality was used.
So when you take expectation on both sides of $(1)$, use the above inequality and substitute the values of the expectations, you would get the desired result.
answered Dec 8 '18 at 9:57
StubbornAtomStubbornAtom
5,97311238
answered Dec 8 '18 at 9:57
StubbornAtomStubbornAtom
5,97311238
answered Dec 8 '18 at 9:57
StubbornAtomStubbornAtom
5,97311238
answered Dec 8 '18 at 9:57
StubbornAtomStubbornAtom
5,97311238
5,97311238
$begingroup$
Thank you very much! I messed up the formula pretty badly!
$endgroup$
– user587126
Dec 8 '18 at 11:46
$begingroup$
Actually the inequality (3) directly follows from Cauchy-Scharwz; no need for Jensen. Previously asked and answered: math.stackexchange.com/questions/1395820/….
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:49
add a comment |
$begingroup$
Thank you very much! I messed up the formula pretty badly!
$endgroup$
– user587126
Dec 8 '18 at 11:46
$begingroup$
Actually the inequality (3) directly follows from Cauchy-Scharwz; no need for Jensen. Previously asked and answered: math.stackexchange.com/questions/1395820/….
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:49
$begingroup$
Thank you very much! I messed up the formula pretty badly!
$endgroup$
– user587126
Dec 8 '18 at 11:46
$begingroup$
Thank you very much! I messed up the formula pretty badly!
$endgroup$
– user587126
Dec 8 '18 at 11:46
$begingroup$
Actually the inequality (3) directly follows from Cauchy-Scharwz; no need for Jensen. Previously asked and answered: math.stackexchange.com/questions/1395820/….
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:49
$begingroup$
Actually the inequality (3) directly follows from Cauchy-Scharwz; no need for Jensen. Previously asked and answered: math.stackexchange.com/questions/1395820/….
$endgroup$
– StubbornAtom
Dec 18 '18 at 19:49
add a comment |
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