The fourth moment of a centered random variable is at least equal to the square of its variance












3















Let X be a random variable with mean µ and variance $sigma^2$. Show that



$E(X-mu)^4geq sigma^4$



and use this to show that the kurtosis of $X$ is at least $-2$.




This looks like a form of Chebyshev's equation:



$P(|X-mu|geq a) leq frac{sigma^2}{a^2}$



But does not relate the inequality in the probability.



I tried expanding $E(X-mu)^4$ out, but that seemed like a dead end.



Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?






random-variables expected-value







share|cite|improve this question





























    3















    Let X be a random variable with mean µ and variance $sigma^2$. Show that



    $E(X-mu)^4geq sigma^4$



    and use this to show that the kurtosis of $X$ is at least $-2$.




    This looks like a form of Chebyshev's equation:



    $P(|X-mu|geq a) leq frac{sigma^2}{a^2}$



    But does not relate the inequality in the probability.



    I tried expanding $E(X-mu)^4$ out, but that seemed like a dead end.



    Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?






    random-variables expected-value







    share|cite|improve this question



























      3












      3








      3








      Let X be a random variable with mean µ and variance $sigma^2$. Show that



      $E(X-mu)^4geq sigma^4$



      and use this to show that the kurtosis of $X$ is at least $-2$.




      This looks like a form of Chebyshev's equation:



      $P(|X-mu|geq a) leq frac{sigma^2}{a^2}$



      But does not relate the inequality in the probability.



      I tried expanding $E(X-mu)^4$ out, but that seemed like a dead end.



      Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?






      random-variables expected-value







      share|cite|improve this question
















      Let X be a random variable with mean µ and variance $sigma^2$. Show that



      $E(X-mu)^4geq sigma^4$



      and use this to show that the kurtosis of $X$ is at least $-2$.




      This looks like a form of Chebyshev's equation:



      $P(|X-mu|geq a) leq frac{sigma^2}{a^2}$



      But does not relate the inequality in the probability.



      I tried expanding $E(X-mu)^4$ out, but that seemed like a dead end.



      Can someone guide me on how to solve this? If it does deal with Chebyshev, can someone explain in detail how to get the inequality out of the absolute value probability?






      random-variables expected-value




      random-variables expected-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 at 17:06









      Did

      246k23220454




      246k23220454










      asked Dec 9 at 16:14









      user603569

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          2 Answers
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          3














          Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.



          This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$






          share|cite|improve this answer





























            2














            You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
              2






              active

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              active

              oldest

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              active

              oldest

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              3














              Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.



              This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$






              share|cite|improve this answer


























                3














                Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.



                This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$






                share|cite|improve this answer
























                  3












                  3








                  3






                  Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.



                  This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$






                  share|cite|improve this answer












                  Setting $Y=X-mu$ it must actually be shown that $mathbb EY^4geq(mathbb EY^2)^2$.



                  This is a direct consequence of: $$mathbb EY^4-(mathbb EY^2)^2=mathsf{Var}Y^2geq0$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 at 16:20









                  drhab

                  97.3k544128




                  97.3k544128























                      2














                      You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.






                      share|cite|improve this answer


























                        2














                        You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.






                        share|cite|improve this answer
























                          2












                          2








                          2






                          You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.






                          share|cite|improve this answer












                          You don't need Chebyshev. Note that $E(X-mu)^4=operatorname{Var}(X-mu)^2+E^2(X-mu)^2$. The first term $ge 0$; the second is $sigma^4$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 9 at 16:19









                          J.G.

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