Existence of a "quasi-uniform" probability distribution on $\mathbb{Z}$

Question

Does there exist a probability distribution on $\mathbb{Z}$ such that for every integer $n\geq 1$, the probability that a random integer $x$ is divisible by $n$ equals $1/n$?

Henry Cohn has an argument why this is not possible, but it is not completely rigorous. First, it is easy to see that we can assume that the distribution is supported on the positive integers. Let $p_n$ be the probability of $n$. For any function $f$ on the positive integers for which we get convergence, we have (by the assumption on $p_n$) $$ \sum_k p_k\sum_{n|k}f(n) = \sum_n \frac{f(n)}{n}. $$ Let $g(k)=\sum_{n|k} f(n)$. By Möbius inversion, $f(n) = \sum_{k|n}g(k)\mu(n/k)$. Writing $n=mk$, the first equation becomes $$ \sum_k p_k g(k) =\sum_k g(k)\sum_m \frac{\mu(m)}{mk}. $$ This should hold for all $g$ for which $\sum_n f(n)/n$ converges absolutely, so it should follow that $$ p_k = \frac 1k\sum_m \frac{\mu(m)}{m}. $$ This is nonsense since first of all, $\sum_m \mu(m)/m =0$ (equivalent to the prime number theorem), and even if we didn't know that, there's no way $p_k$ can be proportional to $1/k$ since $\sum 1/k=\infty$.

This argument is not completely rigorous since we have interchanged sums and equated coefficients of $g(k)$ without justification. It's also a problem that $\sum_m \mu(m)/m$ is conditionally convergent.

Answer 1

No. Let's restrict our attention to $\mathbb{N}$. The hypotheses imply that if $q$ is a prime, then the probability that a random positive integer is not divisible by $q$ is $1 - \frac{1}{q}$. They also imply that these events are independent. Now let $n$ be a positive integer. If $q_1, q_2, \dots$ is an enumeration of the primes not dividing $n$ it follows that

$$p_n \le \prod_{i=1}^m \left( 1 - \frac{1}{q_i} \right)$$

for all $m$. But taking $m \to \infty$ the RHS approaches $0$; contradiction. Note that this argument does not require the prime number theorem; we just need to know that the harmonic series diverges.

Edit: Here is a generalization which more completely rescues Henry Cohn's argument. Generalize the condition to being that the probability of a positive integer being divisible by $n$ is $\frac{1}{n^s}$, for some real number $s > 0$. This is equivalent to requiring that the probability is 1) multiplicative in $n$ and 2) monotonically decreasing.

It follows that if $q$ is prime, then the probability that the exponent of $q$ in the prime factorization of a random positive integer is exactly $k$ is

$$\frac{1}{q^{ks}} - \frac{1}{q^{(k+1)s}} = \frac{1}{q^{ks}} \left( 1 - \frac{1}{q^s} \right).$$

We again have that for different primes $q$ these events are independent. Now, if $q_1, q_2, \dots$ is an enumeration of the primes and $n = \prod q_i^{k_i}$ is a positive integer, it follows that

$$p_n \le \prod_{i=1}^m \frac{1}{q_i^{k_i s}} \left( 1 - \frac{1}{q_i^s} \right).$$

for all $m$. If $s \le 1$ then the RHS converges to $0$ as $m \to \infty$ (this, again, does not require the prime number theorem) and we get a contradiction. If $s > 1$ then the RHS converges to

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

and this is an equality because the RHS is the probability that a random positive integer has the same prime factorization as $n$.

There is a straightforward generalization where $\mathbb{N}$ is replaced by the set of nonzero ideals in the ring of integers $\mathcal{O}_K$ of a number field $K$ and the probability of an ideal being divisible by an ideal $I$ is $\frac{1}{N(I)^s}$, where we get the Dedekind zeta function instead.

Answer 2

Let $i, j$ be positive integers such that $p_i \ne p_j$. Assume without loss of generality that $p_i > p_j$, and let $$\varepsilon = p_i - p_j.$$

Since $\sum_{k=1}^{\infty}{p_k}$ is a convergent series, $\lim_{N \to \infty}{\sum_{k=N}^{\infty}{p_k}} = 0$. Let $N$ be a positive integer large enough that $\sum_{k=N}^{\infty}{p_k} < \varepsilon$, and let us also require $N > i$ and $N > j$. By assumption, $$\sum_{k=0}^{\infty}{p_{i + N k}} = \frac{1}{N} = \sum_{k=0}^{\infty}{p_{j + N k}},$$ and therefore $$\sum_{k=1}^{\infty}{p_{j + N k}} - \sum_{k=1}^{\infty}{p_{i + N k}} = \left(\frac{1}{N} - p_j\right) - \left(\frac{1}{N} - p_i\right) = p_i - p_j = \varepsilon.$$ But this contradicts the definition of $N$. Indeed, $$\sum_{k=1}^{\infty}{p_{j + N k}} - \sum_{k=1}^{\infty}{p_{i + N k}} \le \sum_{k=1}^{\infty}{p_{j + N k}} \le \sum_{k=N}^{\infty}{p_k} < \varepsilon.$$

Answer 3

What is the problem if $p_n$ is 0 for every $n \in \mathbb{N}$? It would be a problem if $P(S) = 0$ for every $S \subset \mathbb{N}$. Perhaps you are worried because $\sum_{i=1}^{\infty} p_i = 0$, but this is not a problem without $\sigma$-additivity, as Bruno de Finetti suggested.

A curious paper (unluckily in italian)