$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Var}{\operatorname{Var}}$We define the discrete random variable $X$ as having the probability mass function $$f_{X}(k) = \Pr(X=k) = \zeta(k)-1, $$ for $k \geq 2 $.

Here, $\zeta(\cdot)$ is the Riemann zeta function, defined as $$\zeta(s) = \sum_{n=1}^{\infty} n^{-s} $$ for $\Re(s) >1 $.

Now, $X$ is indeed a discrete RV, as we have $$\sum_{k=2}^{\infty} p_k = \sum_{k=2}^{\infty} (\zeta(k)-1) = 1,$$ (which we can find, for example, here) and it is clear that for all $k$ it holds that $$0 \leq p_k \leq 1 .$$

Furthermore, we can find the first and second moments of $X$. The mean amounts to $$E[X] = \sum_{k=2}^{\infty} k \big(\zeta(k)-1\big) = 1+\frac{\pi^{2}}{6} .$$

Moreover, we have $$E[X^{2}] = \sum_{k=2}^{\infty} k^{2} \big( \zeta(k)-1 \big) = 1 + \frac{\pi^{2}}{2} + 2 \zeta(3), $$ so we obtain \begin{align*} \Var(X) &= E[X^2] - E[X]^{2} \\ &= \frac{\pi^2}{6} +2 \zeta(3) - \frac{\pi^4}{36} \\ &= \zeta(2) + 2 \zeta(3) - \frac{5}{2} \zeta(4) \\ &= \zeta(2) \left( 1 - \zeta(2) \right) + 2 \zeta(3) \end{align*}

Question: does this discrete random variable involving curtailed Riemann zeta values come up in the literature on probability theory and/or statistics? Does it have any applications?

Note: please note that this RV differs from the Zeta distribution.

  • $\begingroup$ @whuber How is this a truncated version of the Zeta distribution? I don't see the simple relation you describe. It appears to me that the Zeta distribution has a different probability mass function than the one above $\endgroup$
    – Max Muller
    Commented Feb 7, 2023 at 21:03
  • 1
    $\begingroup$ Sorry, I misread your initial formula. +1 for the question -- and I'll delete that comment. FWIW, this would arise as a Pareto$(2)$ mixture of Geometric variables. I have not seen an application of such a model, but maybe others have. $\endgroup$
    – whuber
    Commented Feb 7, 2023 at 21:08

1 Answer 1


To illustrate Whuber's comment

$$\begin{array}{c|ccccccccc} & Y = 2 & Y =3& Y=4 & Y=5\\ \hline X =2 & \frac{1}{2^2} & \frac{1}{3^2} & \frac{1}{4^2} &\frac{1}{5^2} & \dots\\ X =3 & \frac{1}{2^3} & \frac{1}{3^3} & \frac{1}{4^3} & \frac{1}{5^3} & \dots\\ X =4 & \frac{1}{2^4} &\frac{1}{3^4} &\frac{1}{4^4} & \frac{1}{5^4} &\dots\\ \vdots& \\\text{etc.}& \end{array}$$

And those terms can be seen as the product of a product $P(X=x|Y=y)P(Y=y)$ with a shifted geometric distribution $$P(X=x|Y=y) = \left(\frac{1}{y}\right)^{x} y(y-1)$$ and some sort of variant of a Zipf distribution $$P(Y=y) = \frac{1}{y(y-1)}$$


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