Lyapunov's inequality in Probability












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$begingroup$


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)










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$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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)










share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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)










share|cite|improve this question









$endgroup$




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






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asked Nov 9 '17 at 4:14









C. ZhihaoC. Zhihao

474213




474213












  • $begingroup$
    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






  • 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












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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 $.






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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 $.






    share|cite|improve this answer









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      0












      $begingroup$

      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 $.






      share|cite|improve this answer









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        0





        $begingroup$

        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 $.






        share|cite|improve this answer









        $endgroup$



        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 $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 25 '18 at 9:30









        Radu TRadu T

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