Probabilistic Riemann hypothesis under Cramér random model

Question

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.

Answer 1

It turns out that a little bit of stretch completes the argument above, which I shall post below, verifications are welcomed.

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 $\sum_{2 \leq n \leq x}1 / \log n = \textrm{Li}(x) + O(1)$.