relation between moments of discrete random variable and CDF












0












$begingroup$


Consider a discrete positive random variable, say X. This link nicely shows that



begin{equation}
E[X] = sum_{k=0}^{infty} (1-F(k))
end{equation}



Moreover,



begin{equation}
E[X^2] = sum_{k=0}^{infty} (2k+1)(1-F(k))
end{equation}



I am puzzled on how to obtain the factor "(2k+1)" in above expression. It would be nice, when there were a relation between F(X) and X for higher order moments.






probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
    $endgroup$
    – angryavian
    Jan 16 at 18:10










  • $begingroup$
    Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
    $endgroup$
    – Alexander K.
    Jan 16 at 20:00


















0












$begingroup$


Consider a discrete positive random variable, say X. This link nicely shows that



begin{equation}
E[X] = sum_{k=0}^{infty} (1-F(k))
end{equation}



Moreover,



begin{equation}
E[X^2] = sum_{k=0}^{infty} (2k+1)(1-F(k))
end{equation}



I am puzzled on how to obtain the factor "(2k+1)" in above expression. It would be nice, when there were a relation between F(X) and X for higher order moments.






probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
    $endgroup$
    – angryavian
    Jan 16 at 18:10










  • $begingroup$
    Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
    $endgroup$
    – Alexander K.
    Jan 16 at 20:00
















0












0








0





$begingroup$


Consider a discrete positive random variable, say X. This link nicely shows that



begin{equation}
E[X] = sum_{k=0}^{infty} (1-F(k))
end{equation}



Moreover,



begin{equation}
E[X^2] = sum_{k=0}^{infty} (2k+1)(1-F(k))
end{equation}



I am puzzled on how to obtain the factor "(2k+1)" in above expression. It would be nice, when there were a relation between F(X) and X for higher order moments.






probability-distributions







share|cite|improve this question









$endgroup$




Consider a discrete positive random variable, say X. This link nicely shows that



begin{equation}
E[X] = sum_{k=0}^{infty} (1-F(k))
end{equation}



Moreover,



begin{equation}
E[X^2] = sum_{k=0}^{infty} (2k+1)(1-F(k))
end{equation}



I am puzzled on how to obtain the factor "(2k+1)" in above expression. It would be nice, when there were a relation between F(X) and X for higher order moments.






probability-distributions




probability-distributions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 17:53









Alexander K.Alexander K.

1




1












  • $begingroup$
    Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
    $endgroup$
    – angryavian
    Jan 16 at 18:10










  • $begingroup$
    Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
    $endgroup$
    – Alexander K.
    Jan 16 at 20:00




















  • $begingroup$
    Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
    $endgroup$
    – angryavian
    Jan 16 at 18:10










  • $begingroup$
    Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
    $endgroup$
    – Alexander K.
    Jan 16 at 20:00


















$begingroup$
Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
$endgroup$
– angryavian
Jan 16 at 18:10




$begingroup$
Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
$endgroup$
– angryavian
Jan 16 at 18:10












$begingroup$
Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
$endgroup$
– Alexander K.
Jan 16 at 20:00






$begingroup$
Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
$endgroup$
– Alexander K.
Jan 16 at 20:00












2 Answers
2






active

oldest

votes


















0












$begingroup$

The answer in your link gives you the explanation already...



If $G$ is the CDF of $X^2$ then the first formula yields
$$E[X^2] = sum_{k = 0}^infty (1 - G(k))
= (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    thank you for the hint. From this post, it is now clear that one has to rewrite
    begin{equation}
    E[X^2] = sum_k k^2 P(X=k)
    end{equation}
    as
    begin{equation}
    E[X^2] = sum_k H(k) P(X>k)
    end{equation}
    where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
    Setting
    begin{equation}
    sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
    end{equation}
    one can easily get the term H(k).






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      The answer in your link gives you the explanation already...



      If $G$ is the CDF of $X^2$ then the first formula yields
      $$E[X^2] = sum_{k = 0}^infty (1 - G(k))
      = (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
      Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
      Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        The answer in your link gives you the explanation already...



        If $G$ is the CDF of $X^2$ then the first formula yields
        $$E[X^2] = sum_{k = 0}^infty (1 - G(k))
        = (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
        Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
        Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          The answer in your link gives you the explanation already...



          If $G$ is the CDF of $X^2$ then the first formula yields
          $$E[X^2] = sum_{k = 0}^infty (1 - G(k))
          = (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
          Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
          Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.






          share|cite|improve this answer









          $endgroup$



          The answer in your link gives you the explanation already...



          If $G$ is the CDF of $X^2$ then the first formula yields
          $$E[X^2] = sum_{k = 0}^infty (1 - G(k))
          = (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
          Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
          Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 at 20:40









          angryavianangryavian

          42.5k23481




          42.5k23481























              0












              $begingroup$

              thank you for the hint. From this post, it is now clear that one has to rewrite
              begin{equation}
              E[X^2] = sum_k k^2 P(X=k)
              end{equation}
              as
              begin{equation}
              E[X^2] = sum_k H(k) P(X>k)
              end{equation}
              where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
              Setting
              begin{equation}
              sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
              end{equation}
              one can easily get the term H(k).






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                thank you for the hint. From this post, it is now clear that one has to rewrite
                begin{equation}
                E[X^2] = sum_k k^2 P(X=k)
                end{equation}
                as
                begin{equation}
                E[X^2] = sum_k H(k) P(X>k)
                end{equation}
                where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
                Setting
                begin{equation}
                sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
                end{equation}
                one can easily get the term H(k).






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  thank you for the hint. From this post, it is now clear that one has to rewrite
                  begin{equation}
                  E[X^2] = sum_k k^2 P(X=k)
                  end{equation}
                  as
                  begin{equation}
                  E[X^2] = sum_k H(k) P(X>k)
                  end{equation}
                  where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
                  Setting
                  begin{equation}
                  sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
                  end{equation}
                  one can easily get the term H(k).






                  share|cite|improve this answer









                  $endgroup$



                  thank you for the hint. From this post, it is now clear that one has to rewrite
                  begin{equation}
                  E[X^2] = sum_k k^2 P(X=k)
                  end{equation}
                  as
                  begin{equation}
                  E[X^2] = sum_k H(k) P(X>k)
                  end{equation}
                  where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
                  Setting
                  begin{equation}
                  sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
                  end{equation}
                  one can easily get the term H(k).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 16:48









                  Alexander K.Alexander K.

                  1




                  1






























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