How to prove that if $E(X^2) < infty$ then $E(X) < infty$?












-1












$begingroup$



How to prove that if $E(X^2) < infty$ then $E(X) < infty$?




Here is my attempt:



It's easy to show that if $E(X^2) < infty$ then $E(X) < infty$ when $X^2 ge X$ (by using the monotonicity of expectation).



But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as
$$
Eg(x) = int |g(x)|f(x) < infty?
$$






probability-theory expected-value







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
    $endgroup$
    – Yanko
    Dec 23 '18 at 18:34








  • 4




    $begingroup$
    $$|X|leqslant X^2+1quad text{(QED)}$$
    $endgroup$
    – Did
    Dec 23 '18 at 19:01


















-1












$begingroup$



How to prove that if $E(X^2) < infty$ then $E(X) < infty$?




Here is my attempt:



It's easy to show that if $E(X^2) < infty$ then $E(X) < infty$ when $X^2 ge X$ (by using the monotonicity of expectation).



But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as
$$
Eg(x) = int |g(x)|f(x) < infty?
$$






probability-theory expected-value







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
    $endgroup$
    – Yanko
    Dec 23 '18 at 18:34








  • 4




    $begingroup$
    $$|X|leqslant X^2+1quad text{(QED)}$$
    $endgroup$
    – Did
    Dec 23 '18 at 19:01
















-1












-1








-1





$begingroup$



How to prove that if $E(X^2) < infty$ then $E(X) < infty$?




Here is my attempt:



It's easy to show that if $E(X^2) < infty$ then $E(X) < infty$ when $X^2 ge X$ (by using the monotonicity of expectation).



But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as
$$
Eg(x) = int |g(x)|f(x) < infty?
$$






probability-theory expected-value







share|cite|improve this question











$endgroup$





How to prove that if $E(X^2) < infty$ then $E(X) < infty$?




Here is my attempt:



It's easy to show that if $E(X^2) < infty$ then $E(X) < infty$ when $X^2 ge X$ (by using the monotonicity of expectation).



But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as
$$
Eg(x) = int |g(x)|f(x) < infty?
$$






probability-theory expected-value




probability-theory expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 5:35







user587192

















asked Dec 23 '18 at 18:31









DianneDianne

62




62








  • 2




    $begingroup$
    $X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
    $endgroup$
    – Yanko
    Dec 23 '18 at 18:34








  • 4




    $begingroup$
    $$|X|leqslant X^2+1quad text{(QED)}$$
    $endgroup$
    – Did
    Dec 23 '18 at 19:01
















  • 2




    $begingroup$
    $X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
    $endgroup$
    – Yanko
    Dec 23 '18 at 18:34








  • 4




    $begingroup$
    $$|X|leqslant X^2+1quad text{(QED)}$$
    $endgroup$
    – Did
    Dec 23 '18 at 19:01










2




2




$begingroup$
$X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
$endgroup$
– Yanko
Dec 23 '18 at 18:34






$begingroup$
$X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
$endgroup$
– Yanko
Dec 23 '18 at 18:34






4




4




$begingroup$
$$|X|leqslant X^2+1quad text{(QED)}$$
$endgroup$
– Did
Dec 23 '18 at 19:01






$begingroup$
$$|X|leqslant X^2+1quad text{(QED)}$$
$endgroup$
– Did
Dec 23 '18 at 19:01












3 Answers
3






active

oldest

votes


















2












$begingroup$

The one reached onto you by the comment of Did is my favourite.



Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) A nice quantization of the difference using a common formula for the variance.
    $endgroup$
    – robjohn♦
    Dec 23 '18 at 23:06






  • 1




    $begingroup$
    @robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
    $endgroup$
    – drhab
    Dec 24 '18 at 6:00












  • $begingroup$
    Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
    $endgroup$
    – robjohn♦
    Dec 24 '18 at 9:17



















4












$begingroup$

By Jensen inequity for $g(x) = x^ 2$,
$$
g(mathbb{E}(X)) le E( g(X) ),
$$
hence,
$$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
$$
thus
$$
mathbb{E}(X) < infty
$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    (+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
    $endgroup$
    – robjohn♦
    Dec 23 '18 at 19:04



















2












$begingroup$

Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
$$
E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
$$






share|cite|improve this answer











$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The one reached onto you by the comment of Did is my favourite.



    Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (+1) A nice quantization of the difference using a common formula for the variance.
      $endgroup$
      – robjohn♦
      Dec 23 '18 at 23:06






    • 1




      $begingroup$
      @robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
      $endgroup$
      – drhab
      Dec 24 '18 at 6:00












    • $begingroup$
      Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
      $endgroup$
      – robjohn♦
      Dec 24 '18 at 9:17
















    2












    $begingroup$

    The one reached onto you by the comment of Did is my favourite.



    Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (+1) A nice quantization of the difference using a common formula for the variance.
      $endgroup$
      – robjohn♦
      Dec 23 '18 at 23:06






    • 1




      $begingroup$
      @robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
      $endgroup$
      – drhab
      Dec 24 '18 at 6:00












    • $begingroup$
      Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
      $endgroup$
      – robjohn♦
      Dec 24 '18 at 9:17














    2












    2








    2





    $begingroup$

    The one reached onto you by the comment of Did is my favourite.



    Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$






    share|cite|improve this answer











    $endgroup$



    The one reached onto you by the comment of Did is my favourite.



    Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 23 '18 at 20:29

























    answered Dec 23 '18 at 20:15









    drhabdrhab

    102k545136




    102k545136












    • $begingroup$
      (+1) A nice quantization of the difference using a common formula for the variance.
      $endgroup$
      – robjohn♦
      Dec 23 '18 at 23:06






    • 1




      $begingroup$
      @robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
      $endgroup$
      – drhab
      Dec 24 '18 at 6:00












    • $begingroup$
      Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
      $endgroup$
      – robjohn♦
      Dec 24 '18 at 9:17


















    • $begingroup$
      (+1) A nice quantization of the difference using a common formula for the variance.
      $endgroup$
      – robjohn♦
      Dec 23 '18 at 23:06






    • 1




      $begingroup$
      @robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
      $endgroup$
      – drhab
      Dec 24 '18 at 6:00












    • $begingroup$
      Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
      $endgroup$
      – robjohn♦
      Dec 24 '18 at 9:17
















    $begingroup$
    (+1) A nice quantization of the difference using a common formula for the variance.
    $endgroup$
    – robjohn♦
    Dec 23 '18 at 23:06




    $begingroup$
    (+1) A nice quantization of the difference using a common formula for the variance.
    $endgroup$
    – robjohn♦
    Dec 23 '18 at 23:06




    1




    1




    $begingroup$
    @robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
    $endgroup$
    – drhab
    Dec 24 '18 at 6:00






    $begingroup$
    @robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
    $endgroup$
    – drhab
    Dec 24 '18 at 6:00














    $begingroup$
    Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
    $endgroup$
    – robjohn♦
    Dec 24 '18 at 9:17




    $begingroup$
    Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
    $endgroup$
    – robjohn♦
    Dec 24 '18 at 9:17











    4












    $begingroup$

    By Jensen inequity for $g(x) = x^ 2$,
    $$
    g(mathbb{E}(X)) le E( g(X) ),
    $$
    hence,
    $$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
    $$
    thus
    $$
    mathbb{E}(X) < infty
    $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      (+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
      $endgroup$
      – robjohn♦
      Dec 23 '18 at 19:04
















    4












    $begingroup$

    By Jensen inequity for $g(x) = x^ 2$,
    $$
    g(mathbb{E}(X)) le E( g(X) ),
    $$
    hence,
    $$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
    $$
    thus
    $$
    mathbb{E}(X) < infty
    $$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      (+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
      $endgroup$
      – robjohn♦
      Dec 23 '18 at 19:04














    4












    4








    4





    $begingroup$

    By Jensen inequity for $g(x) = x^ 2$,
    $$
    g(mathbb{E}(X)) le E( g(X) ),
    $$
    hence,
    $$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
    $$
    thus
    $$
    mathbb{E}(X) < infty
    $$






    share|cite|improve this answer









    $endgroup$



    By Jensen inequity for $g(x) = x^ 2$,
    $$
    g(mathbb{E}(X)) le E( g(X) ),
    $$
    hence,
    $$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
    $$
    thus
    $$
    mathbb{E}(X) < infty
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 23 '18 at 18:42









    V. VancakV. Vancak

    11.1k2926




    11.1k2926








    • 1




      $begingroup$
      (+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
      $endgroup$
      – robjohn♦
      Dec 23 '18 at 19:04














    • 1




      $begingroup$
      (+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
      $endgroup$
      – robjohn♦
      Dec 23 '18 at 19:04








    1




    1




    $begingroup$
    (+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
    $endgroup$
    – robjohn♦
    Dec 23 '18 at 19:04




    $begingroup$
    (+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
    $endgroup$
    – robjohn♦
    Dec 23 '18 at 19:04











    2












    $begingroup$

    Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
    $$
    E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
    $$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
      $$
      E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
      $$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
        $$
        E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
        $$






        share|cite|improve this answer











        $endgroup$



        Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
        $$
        E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 23 '18 at 18:58

























        answered Dec 23 '18 at 18:49









        robjohn♦robjohn

        268k27308634




        268k27308634






























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