5
$\begingroup$

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Var}{\operatorname{Var}}$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.

$\endgroup$
2
  • $\begingroup$ @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

1
$\begingroup$

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 $$P(Y=y) = \frac{1}{y(y-1)}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.