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In an answer to a question on MU about the Riemann zeta function, I sketched a proof that the probability distribution on $\mathbb{N}$ which assigns $n$ the probability

$$\frac{ \frac{1}{n^s} }{\zeta(s)}$$

(henceforth called the zeta distribution with parameter $s$) where $s > 1$ is the unique family of probability distributions on $\mathbb{N}$ satisfying the following three requirements:

  • The exponent of $p$ and the exponent of $q$ in the prime factorization of $n$ are chosen independently for all pairs of primes $p \neq q$.
  • The exponent of a particular prime $p$ is geometrically distributed.
  • The probabilities are monotonically decreasing as a function of $n$.

The basic motivation for the first requirement is the Chinese Remainder Theorem. I can think of two motivations for the second requirement: first, that geometric distributions are the maximally entropic distributions on $\mathbb{N}_{\ge 0}$ with a given mean, and second, that (if I'm not mistaken) one naturally gets a geometric distribution from Haar measure on the $p$-adic integers.

In fact, the distribution one gets from Haar measure on the $p$-adic integers is the one in which a $p$-adic integer is divisible by exactly $p^k$ with probability $(1 - p^{-1}) p^{-k}$. This is essentially the $s \to 1$ limit of the zeta distribution above, which I gave as a reason one might deduce that this limit is important from first principles.

It seems like it should be possible to combine the motivations for the first two requirements into a statement about Haar measure on the profinite integers $\prod_p \mathbb{Z}_p$, except that I don't know exactly what kind of statement I'd be looking for, so I thought I'd ask here.

Question: Complete the following statement. It is natural to study the $s \to 1$ limit of the zeta distribution because (some statement about Haar measure on the profinite integers, maybe with "Tate's thesis" thrown in somewhere).

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This is also the unique nontrivial distribution on the positive integers that makes the events "$p$ divides $X$" and "$q$ divides $X$" independent whenever $p,q$ are distinct primes. – Kevin O'Bryant Aug 2 '10 at 2:57
Do you have a reference for that? As far as I can tell, it is not hard to construct distributions not of this form which don't satisfy the third requirement. I have read a statement like this in writing by Gian-Carlo Rota, but it was "subject to certain technical requirements" which I thought I had figured out. – Qiaochu Yuan Aug 2 '10 at 3:23
My source is also Gian-Carlo Rota (some popular lectures he gave). He also has a paper wherein this distribution arises as a profinite completion. It was a short paper (4 or 5 pages), and I think published in the 1990s. – Kevin O'Bryant Aug 2 '10 at 4:04
up vote 7 down vote accepted

...because for sets with logarithmic density, the logarithmic density and the $s\to1$ zeta-measure agree. And for sets with natural density, the logarithmic and natural densities are the same.

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I know I left out Tate & Haar, but I don't think there's any call to bring them along. – Kevin O'Bryant Aug 1 '10 at 23:20
This is a perfectly good answer, but it's a little unsatisfying for me. Presumably there are lots of notions of density that agree with and extend natural density. Why single out logarithmic density? – Qiaochu Yuan Aug 2 '10 at 3:26

"This is also the unique nontrivial distribution on the positive integers that makes the events "p divides X" and "q divides X" independent whenever p,q are distinct primes."

Kevin, consider $X = \prod_{i=1}^\infty p_i^{N_i}$ where the $N$'s are independent, and $N_i$ has a Poisson distribution with parameter $p_i^{-s}$. It turns out this distribution is related to the exponential of the prime zeta function.

In fact, whenever you construct a probability distribution by normalising the terms of a convergent Dirichlet series with positive, multiplicitive coefficients, this property shows up. It's the subject of my MPhil thesis, which I may manage to post up on the arxiv sometime this century.

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I agree. Rota's statement is a little unclear but I think I have found one way to make it precise here:… – Qiaochu Yuan Nov 20 '10 at 13:44
So you're using the memoryless property of the geometric distribution in your post? – Simon Lyons Nov 20 '10 at 17:19
I suppose so, yes. – Qiaochu Yuan Nov 21 '10 at 1:58

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