Lyapunov's inequality in Probability
$begingroup$
I have a question about the proof of the inequality. The well known result stats
Let $Z$ be a RV and let $0<s<t$. Then
$$E(|Z|^s)^{1/s} leq E(|Z|^t)^{1/t}$$
The proof follows almost immediately from the Holder Inequality
$$E(|XY|)leq E(|X|^p)^{1/p} E(|Y|^q)^{1/q}$$
for $1<p,q$ such that $1/p+1/q =1$.
Taking $Z=|X|^s, Y=1, p= t/s $ we can derive the Lyapunov's inequality.
even though it's not mentioned , I think one must assume that $E(Z) =E(|X|^s)<infty$ to apply the inequality . However, I'm not assuming that in my hypothesis.
My question is, what happens if $E(Z)=infty$? Does the inequality still holds? In other words, $E(|X|^t)=infty$ as well? (If $1<s<t $ it follows easyly but I don't know how to argue in other cases)
probability-theory random-variables expectation integral-inequality
asked Nov 9 '17 at 4:14
C. ZhihaoC. Zhihao
489214
$endgroup$
add a comment |
$begingroup$
I have a question about the proof of the inequality. The well known result stats
Let $Z$ be a RV and let $0<s<t$. Then
$$E(|Z|^s)^{1/s} leq E(|Z|^t)^{1/t}$$
The proof follows almost immediately from the Holder Inequality
$$E(|XY|)leq E(|X|^p)^{1/p} E(|Y|^q)^{1/q}$$
for $1<p,q$ such that $1/p+1/q =1$.
Taking $Z=|X|^s, Y=1, p= t/s $ we can derive the Lyapunov's inequality.
even though it's not mentioned , I think one must assume that $E(Z) =E(|X|^s)<infty$ to apply the inequality . However, I'm not assuming that in my hypothesis.
My question is, what happens if $E(Z)=infty$? Does the inequality still holds? In other words, $E(|X|^t)=infty$ as well? (If $1<s<t $ it follows easyly but I don't know how to argue in other cases)
probability-theory random-variables expectation integral-inequality
asked Nov 9 '17 at 4:14
C. ZhihaoC. Zhihao
489214
$endgroup$
$begingroup$
Holder's inequality as stated by you holds always! Integrals of non-negative measurable functions are always defined, but the values may be $infty$.
$endgroup$
– Kavi Rama Murthy
Nov 9 '17 at 9:04
1
$begingroup$
You should not be afraid to write $text{something} leq infty$. It's true.
$endgroup$
– Gabriel Romon
Nov 9 '17 at 19:47
$begingroup$
What I mean, if $E(Z) =E(|X^p|)= infty$, I get $infty leq$ **something **, following your notation. But working on @KaviRamaMurthy comment I think that the proof still remains valid under the assumption of $E(Z) = infty$
$endgroup$
– C. Zhihao
Nov 9 '17 at 21:39
add a comment |
$begingroup$
I have a question about the proof of the inequality. The well known result stats
Let $Z$ be a RV and let $0<s<t$. Then
$$E(|Z|^s)^{1/s} leq E(|Z|^t)^{1/t}$$
The proof follows almost immediately from the Holder Inequality
$$E(|XY|)leq E(|X|^p)^{1/p} E(|Y|^q)^{1/q}$$
for $1<p,q$ such that $1/p+1/q =1$.
Taking $Z=|X|^s, Y=1, p= t/s $ we can derive the Lyapunov's inequality.
even though it's not mentioned , I think one must assume that $E(Z) =E(|X|^s)<infty$ to apply the inequality . However, I'm not assuming that in my hypothesis.
My question is, what happens if $E(Z)=infty$? Does the inequality still holds? In other words, $E(|X|^t)=infty$ as well? (If $1<s<t $ it follows easyly but I don't know how to argue in other cases)
probability-theory random-variables expectation integral-inequality
asked Nov 9 '17 at 4:14
C. ZhihaoC. Zhihao
489214
$endgroup$
I have a question about the proof of the inequality. The well known result stats
Let $Z$ be a RV and let $0<s<t$. Then
$$E(|Z|^s)^{1/s} leq E(|Z|^t)^{1/t}$$
The proof follows almost immediately from the Holder Inequality
$$E(|XY|)leq E(|X|^p)^{1/p} E(|Y|^q)^{1/q}$$
for $1<p,q$ such that $1/p+1/q =1$.
Taking $Z=|X|^s, Y=1, p= t/s $ we can derive the Lyapunov's inequality.
even though it's not mentioned , I think one must assume that $E(Z) =E(|X|^s)<infty$ to apply the inequality . However, I'm not assuming that in my hypothesis.
My question is, what happens if $E(Z)=infty$? Does the inequality still holds? In other words, $E(|X|^t)=infty$ as well? (If $1<s<t $ it follows easyly but I don't know how to argue in other cases)
probability-theory random-variables expectation integral-inequality
probability-theory random-variables expectation integral-inequality
asked Nov 9 '17 at 4:14
C. ZhihaoC. Zhihao
489214
asked Nov 9 '17 at 4:14
C. ZhihaoC. Zhihao
489214
asked Nov 9 '17 at 4:14
C. ZhihaoC. Zhihao
489214
asked Nov 9 '17 at 4:14
C. ZhihaoC. Zhihao
489214
asked Nov 9 '17 at 4:14
C. ZhihaoC. Zhihao
489214
489214
$begingroup$
Holder's inequality as stated by you holds always! Integrals of non-negative measurable functions are always defined, but the values may be $infty$.
$endgroup$
– Kavi Rama Murthy
Nov 9 '17 at 9:04
1
$begingroup$
You should not be afraid to write $text{something} leq infty$. It's true.
$endgroup$
– Gabriel Romon
Nov 9 '17 at 19:47
$begingroup$
What I mean, if $E(Z) =E(|X^p|)= infty$, I get $infty leq$ **something **, following your notation. But working on @KaviRamaMurthy comment I think that the proof still remains valid under the assumption of $E(Z) = infty$
$endgroup$
– C. Zhihao
Nov 9 '17 at 21:39
add a comment |
$begingroup$
Holder's inequality as stated by you holds always! Integrals of non-negative measurable functions are always defined, but the values may be $infty$.
$endgroup$
– Kavi Rama Murthy
Nov 9 '17 at 9:04
1
$begingroup$
You should not be afraid to write $text{something} leq infty$. It's true.
$endgroup$
– Gabriel Romon
Nov 9 '17 at 19:47
$begingroup$
What I mean, if $E(Z) =E(|X^p|)= infty$, I get $infty leq$ **something **, following your notation. But working on @KaviRamaMurthy comment I think that the proof still remains valid under the assumption of $E(Z) = infty$
$endgroup$
– C. Zhihao
Nov 9 '17 at 21:39
$begingroup$
Holder's inequality as stated by you holds always! Integrals of non-negative measurable functions are always defined, but the values may be $infty$.
$endgroup$
– Kavi Rama Murthy
Nov 9 '17 at 9:04
$begingroup$
Holder's inequality as stated by you holds always! Integrals of non-negative measurable functions are always defined, but the values may be $infty$.
$endgroup$
– Kavi Rama Murthy
Nov 9 '17 at 9:04
1
1
$begingroup$
You should not be afraid to write $text{something} leq infty$. It's true.
$endgroup$
– Gabriel Romon
Nov 9 '17 at 19:47
$begingroup$
You should not be afraid to write $text{something} leq infty$. It's true.
$endgroup$
– Gabriel Romon
Nov 9 '17 at 19:47
$begingroup$
What I mean, if $E(Z) =E(|X^p|)= infty$, I get $infty leq$ **something **, following your notation. But working on @KaviRamaMurthy comment I think that the proof still remains valid under the assumption of $E(Z) = infty$
$endgroup$
– C. Zhihao
Nov 9 '17 at 21:39
$begingroup$
What I mean, if $E(Z) =E(|X^p|)= infty$, I get $infty leq$ **something **, following your notation. But working on @KaviRamaMurthy comment I think that the proof still remains valid under the assumption of $E(Z) = infty$
$endgroup$
– C. Zhihao
Nov 9 '17 at 21:39
add a comment |
1 Answer
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$begingroup$
I believe you could argue thus under the above assumptions: First $ |X|^s > 1 Rightarrow |X| > 1 Rightarrow |X|^s < |X|^t $. So one could write:
$$ infty = int |X|^s = int_{X>1} |X|^s + int_{X leq 1} |X|^s leq int_{X>1} |X|^s + 1 leq int_{X>1} |X|^t + 1 leq int |X|^t + 1, $$
from which it follows that $ E(|X|^t) = infty $.
answered Mar 25 '18 at 9:30
Radu TRadu T
12
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add a comment |
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$begingroup$
I believe you could argue thus under the above assumptions: First $ |X|^s > 1 Rightarrow |X| > 1 Rightarrow |X|^s < |X|^t $. So one could write:
$$ infty = int |X|^s = int_{X>1} |X|^s + int_{X leq 1} |X|^s leq int_{X>1} |X|^s + 1 leq int_{X>1} |X|^t + 1 leq int |X|^t + 1, $$
from which it follows that $ E(|X|^t) = infty $.
answered Mar 25 '18 at 9:30
Radu TRadu T
12
$endgroup$
add a comment |
$begingroup$
I believe you could argue thus under the above assumptions: First $ |X|^s > 1 Rightarrow |X| > 1 Rightarrow |X|^s < |X|^t $. So one could write:
$$ infty = int |X|^s = int_{X>1} |X|^s + int_{X leq 1} |X|^s leq int_{X>1} |X|^s + 1 leq int_{X>1} |X|^t + 1 leq int |X|^t + 1, $$
from which it follows that $ E(|X|^t) = infty $.
answered Mar 25 '18 at 9:30
Radu TRadu T
12
$endgroup$
add a comment |
$begingroup$
I believe you could argue thus under the above assumptions: First $ |X|^s > 1 Rightarrow |X| > 1 Rightarrow |X|^s < |X|^t $. So one could write:
$$ infty = int |X|^s = int_{X>1} |X|^s + int_{X leq 1} |X|^s leq int_{X>1} |X|^s + 1 leq int_{X>1} |X|^t + 1 leq int |X|^t + 1, $$
from which it follows that $ E(|X|^t) = infty $.
answered Mar 25 '18 at 9:30
Radu TRadu T
12
$endgroup$
I believe you could argue thus under the above assumptions: First $ |X|^s > 1 Rightarrow |X| > 1 Rightarrow |X|^s < |X|^t $. So one could write:
$$ infty = int |X|^s = int_{X>1} |X|^s + int_{X leq 1} |X|^s leq int_{X>1} |X|^s + 1 leq int_{X>1} |X|^t + 1 leq int |X|^t + 1, $$
from which it follows that $ E(|X|^t) = infty $.
answered Mar 25 '18 at 9:30
Radu TRadu T
12
answered Mar 25 '18 at 9:30
Radu TRadu T
12
answered Mar 25 '18 at 9:30
Radu TRadu T
12
answered Mar 25 '18 at 9:30
Radu TRadu T
12
12
add a comment |
add a comment |
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$begingroup$
Holder's inequality as stated by you holds always! Integrals of non-negative measurable functions are always defined, but the values may be $infty$.
$endgroup$
– Kavi Rama Murthy
Nov 9 '17 at 9:04
1
$begingroup$
You should not be afraid to write $text{something} leq infty$. It's true.
$endgroup$
– Gabriel Romon
Nov 9 '17 at 19:47
$begingroup$
What I mean, if $E(Z) =E(|X^p|)= infty$, I get $infty leq$ **something **, following your notation. But working on @KaviRamaMurthy comment I think that the proof still remains valid under the assumption of $E(Z) = infty$
$endgroup$
– C. Zhihao
Nov 9 '17 at 21:39