Markov inequality question












0












$begingroup$


Let $x$ be a random variable and let $m = sqrt[3]{Bbb E[X^3]}.$
Show that, for each $a gt 0$, then $Bbb P[X geq acdot m] leq frac{1}{a^3}.$



If anyone could help I'll appreciate.






probability probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Notice $mathbf{E}(X) = mathbf{E}(X; X leq a m) + mathbf{E}(X; X > am),$ where the $;$ stands for "integrating on the set" (that is to say, multiplication by an indicator function). Realise that $X > c$ and $X^3 > c^3$ are equivalent relations.
    $endgroup$
    – Will M.
    Jan 1 at 20:30






  • 1




    $begingroup$
    Do we need $X$ to be a nonnegative random variable?
    $endgroup$
    – angryavian
    Jan 1 at 20:30










  • $begingroup$
    Yes X should be a nonnegative random variable
    $endgroup$
    – user610402
    Jan 1 at 20:37
















0












$begingroup$


Let $x$ be a random variable and let $m = sqrt[3]{Bbb E[X^3]}.$
Show that, for each $a gt 0$, then $Bbb P[X geq acdot m] leq frac{1}{a^3}.$



If anyone could help I'll appreciate.






probability probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Notice $mathbf{E}(X) = mathbf{E}(X; X leq a m) + mathbf{E}(X; X > am),$ where the $;$ stands for "integrating on the set" (that is to say, multiplication by an indicator function). Realise that $X > c$ and $X^3 > c^3$ are equivalent relations.
    $endgroup$
    – Will M.
    Jan 1 at 20:30






  • 1




    $begingroup$
    Do we need $X$ to be a nonnegative random variable?
    $endgroup$
    – angryavian
    Jan 1 at 20:30










  • $begingroup$
    Yes X should be a nonnegative random variable
    $endgroup$
    – user610402
    Jan 1 at 20:37














0












0








0





$begingroup$


Let $x$ be a random variable and let $m = sqrt[3]{Bbb E[X^3]}.$
Show that, for each $a gt 0$, then $Bbb P[X geq acdot m] leq frac{1}{a^3}.$



If anyone could help I'll appreciate.






probability probability-theory







share|cite|improve this question











$endgroup$




Let $x$ be a random variable and let $m = sqrt[3]{Bbb E[X^3]}.$
Show that, for each $a gt 0$, then $Bbb P[X geq acdot m] leq frac{1}{a^3}.$



If anyone could help I'll appreciate.






probability probability-theory




probability probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 20:33









Namaste

1




1










asked Jan 1 at 20:12









user610402user610402

215




215








  • 1




    $begingroup$
    Notice $mathbf{E}(X) = mathbf{E}(X; X leq a m) + mathbf{E}(X; X > am),$ where the $;$ stands for "integrating on the set" (that is to say, multiplication by an indicator function). Realise that $X > c$ and $X^3 > c^3$ are equivalent relations.
    $endgroup$
    – Will M.
    Jan 1 at 20:30






  • 1




    $begingroup$
    Do we need $X$ to be a nonnegative random variable?
    $endgroup$
    – angryavian
    Jan 1 at 20:30










  • $begingroup$
    Yes X should be a nonnegative random variable
    $endgroup$
    – user610402
    Jan 1 at 20:37














  • 1




    $begingroup$
    Notice $mathbf{E}(X) = mathbf{E}(X; X leq a m) + mathbf{E}(X; X > am),$ where the $;$ stands for "integrating on the set" (that is to say, multiplication by an indicator function). Realise that $X > c$ and $X^3 > c^3$ are equivalent relations.
    $endgroup$
    – Will M.
    Jan 1 at 20:30






  • 1




    $begingroup$
    Do we need $X$ to be a nonnegative random variable?
    $endgroup$
    – angryavian
    Jan 1 at 20:30










  • $begingroup$
    Yes X should be a nonnegative random variable
    $endgroup$
    – user610402
    Jan 1 at 20:37








1




1




$begingroup$
Notice $mathbf{E}(X) = mathbf{E}(X; X leq a m) + mathbf{E}(X; X > am),$ where the $;$ stands for "integrating on the set" (that is to say, multiplication by an indicator function). Realise that $X > c$ and $X^3 > c^3$ are equivalent relations.
$endgroup$
– Will M.
Jan 1 at 20:30




$begingroup$
Notice $mathbf{E}(X) = mathbf{E}(X; X leq a m) + mathbf{E}(X; X > am),$ where the $;$ stands for "integrating on the set" (that is to say, multiplication by an indicator function). Realise that $X > c$ and $X^3 > c^3$ are equivalent relations.
$endgroup$
– Will M.
Jan 1 at 20:30




1




1




$begingroup$
Do we need $X$ to be a nonnegative random variable?
$endgroup$
– angryavian
Jan 1 at 20:30




$begingroup$
Do we need $X$ to be a nonnegative random variable?
$endgroup$
– angryavian
Jan 1 at 20:30












$begingroup$
Yes X should be a nonnegative random variable
$endgroup$
– user610402
Jan 1 at 20:37




$begingroup$
Yes X should be a nonnegative random variable
$endgroup$
– user610402
Jan 1 at 20:37










1 Answer
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$begingroup$

With Markov you have
$$P[Xgeq a]leq frac{E[X]}{a}$$
and so
$$P[Xgeq acdot m]leq P[X^3geq a^3m^3]leq frac{E[X^3]}{a^3cdot (E[X^3])}=frac{1}{a^3}.$$






share|cite|improve this answer









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    active

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    1












    $begingroup$

    With Markov you have
    $$P[Xgeq a]leq frac{E[X]}{a}$$
    and so
    $$P[Xgeq acdot m]leq P[X^3geq a^3m^3]leq frac{E[X^3]}{a^3cdot (E[X^3])}=frac{1}{a^3}.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      With Markov you have
      $$P[Xgeq a]leq frac{E[X]}{a}$$
      and so
      $$P[Xgeq acdot m]leq P[X^3geq a^3m^3]leq frac{E[X^3]}{a^3cdot (E[X^3])}=frac{1}{a^3}.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        With Markov you have
        $$P[Xgeq a]leq frac{E[X]}{a}$$
        and so
        $$P[Xgeq acdot m]leq P[X^3geq a^3m^3]leq frac{E[X^3]}{a^3cdot (E[X^3])}=frac{1}{a^3}.$$






        share|cite|improve this answer









        $endgroup$



        With Markov you have
        $$P[Xgeq a]leq frac{E[X]}{a}$$
        and so
        $$P[Xgeq acdot m]leq P[X^3geq a^3m^3]leq frac{E[X^3]}{a^3cdot (E[X^3])}=frac{1}{a^3}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 20:29









        model_checkermodel_checker

        4,34121931




        4,34121931






























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