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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.

Alexander K.
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  • Doesn't the answer to the question in your link explain where the $2k+1$ comes from? – angryavian Jan 16 '19 at 18:10
  • Please, what link are you referring to? The only link, namely http://stats.stackexchange.com/a/90591/25936 , points to an application using above expression. – Alexander K. Jan 16 '19 at 20:00

2 Answers2

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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.

angryavian
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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).

Alexander K.
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