# Probabilistic Riemann hypothesis under Cramér random model

The Cramér random model for the primes is a random subset $${{\mathcal P}}$$ of the natural numbers with $${1 \not \in {\mathcal P}}, {2 \in {\mathcal P}}$$, and the events $${n \in {\mathcal P}}$$ for $${n=3,4,\dots}$$ being jointly independent with $${{\bf P}(n \in {\mathcal P}) = \frac{1}{\log n}}$$ (the restriction to $${n \geq 3}$$ is to ensure that $${\frac{1}{\log n}}$$ is less than $${1}$$).

Show that if $${\varepsilon > 0}$$, then the quantity $$\displaystyle \frac{1}{x^{1/2+\varepsilon}} (|\{ n \leq x: n \in {\mathcal P}\}| - \int_2^x \frac{dt}{\log t} )$$ converges almost surely to zero as $${x \rightarrow \infty}$$.

Attempt: Interpreting the number of Cramer random model primes up to $$x$$ as the sum up to $$x$$ of a sequence of independent Bernoulli random variables $$X_n$$ each of parameter $$1 / \log n$$, we want to show the quantity $$\displaystyle \frac{\sum_{n \leq x} X_n - \textrm{Li}(x)}{x^{1/2 + \varepsilon}}$$ converges almost surely to zero as $$x \to \infty$$. As $$\sum_{2 \leq n \leq x} \frac{1}{\log n} = \textrm{Li}(x) + O(1)$$, it suffices to show the quantity $$\displaystyle \frac{\sum_{2 < n \leq x} X_n - \frac{1}{\log n}}{x^{1/2 + \varepsilon}}$$ converges almost surely to zero as $$x \to \infty$$. Consider the partial sums $$\displaystyle S_x = \sum_{2 < n \leq x} Y_{n,x}$$, where $$\displaystyle Y_{n,x} := \frac{X_n - \frac{1}{\log n}}{x^{1/2 + \varepsilon}}$$, one therefore aims to show that $$S_x$$ converges almost surely to its mean (which is zero).

However, working as in this post Concentration around the mean in Cramer random model with the fourth moment $${\bf E}|S_x|^4$$, the concentration of measure still seems not strong enough for one to apply the Borel-Cantelli lemma. Some suggestions for fixing this will be appreciated.

Let $$X_n$$ be as in the OP, and $$(Y_{n,x})_{n,x > 2: n \leq x}$$ be the triangular array given by $$Y_{n, x} := \frac{X_n - 1 / \log n}{x^{1/2 + \varepsilon}}$$, the $$Y_{n,x}$$ have mean zero and are independent row-wise, thus
$$\displaystyle \mathop{\bf E} (\sum_{2 < n \leq x} Y_{n,x})^k = \frac{1}{x^{k/2 + k\varepsilon}}{\bf E}(\sum_{2 < n \leq x}X_n - 1 / \log n)^k = O(x^{-k\varepsilon})$$
for any fixed even number $${k}$$ (the only terms in $$(\sum_{2 < n \leq x}X_n - 1 / \log n)^k$$ that give a non-zero contribution are those in which each $$Z_n = X_n - 1 / \log n$$ appears at least twice, so there are at most $${k/2}$$ distinct indices of $${n}$$ that arise, and so there are only $${O_k(x^{k/2})}$$ such terms, each contributing $${O(1)}$$). By Markov's inequality, $$\displaystyle {\bf P}(|\sum_{2 < n \leq x} Y_{n,x}| > \delta) \leq O(1 / \delta^k x^{k\varepsilon})$$ for any $$\delta > 0$$ and $$k$$ even. Setting $${k}$$ sufficiently large depending on $${\varepsilon}$$, we see the LHS is summable in $$x$$, and the claim follows from the fact that $$\sum_{2 \leq n \leq x}1 / \log n = \textrm{Li}(x) + O(1)$$.