Cramér's Model - "The Prime Numbers and Their Distribution" - Part 1

I was reading "The Prime Numbers and Their Distribution" by Gérald Tenenbaum and Michel Mendès France, the section about Cramér's Model, and I couldn't prove a couple of results. I would like to start with an introduction of the topic.

Introduction:

Cramér's Model consists in considering a sequence of independent random variables $\{X_n\}_{n=2}^\infty$ which take values 0 and 1, such that $\Bbb P (X_n=1)=1/\log n$ for $n \ge 3$. The definition of $\Bbb P (X_2=1)$ is considered irrelevant for the purposes of the model.

The idea of this is to give a probabilistic model for the behaviour of prime numbers: according to the prime numbers theorem, the number of primes less or equal than $n$, with $n$ sufficiently large, is approximately $n/\log n$, so the "probability that $n$ is prime" would be $1/\log n$.

Now we define $S:=\{n\ge2:X_n=1\}$. The usual way of working with this model seems to be, proving that some property holds almost surely for $S$, and then conjecturing that property holds for $\mathcal P$, the sequence of prime numbers.

Question:

By analogy to the usual notation, we define $\pi_S(x):=\sum_{n\le x} X_n$ (note that $\pi_\mathcal P(x)=\pi(x)$). We also define: $$\upsilon_S(x):=\frac{\pi_S(x)-\mathrm {li}(x)}{\sqrt{2x\frac{\log\log x}{\log x}}}$$ The book states that it's easy to show that $\upsilon_S(x)$ oscillates asymptotically between -1 and 1, but a paper by Cramér only states that it's shown in another paper (which I couldn't find) that $\limsup_{x\to \infty}|\upsilon_S(x)|=1$ holds almost surely. I understand that the statement of the book is that both, $\limsup_{x\to \infty}\upsilon_S(x)=1$ and $\liminf_{x\to \infty}\upsilon_S(x)=-1$, hold almost surely, but I could be wrong since I haven't found the definition of "asymptotic oscillation".

Solution 1:

This statement is a corollary to a theorem, proved by Andrey Kolmogorov in “Über das Gesetz des iterierten Logarithmus”:

Suppose that $\{Y_n\}_{n = 1}^\infty$ are independent zero-mean random variables, $S_n:=\Sigma_{i = 1}^n Y_i$, $B_n:=Var\, S_n\to\infty$, $|Y_n|\le M_n\in(0,\infty)$, and $M_n=o((\frac{B_n}{\ln\ln B_n})^{\frac{1}{2}})$. Then $\limsup_{n \to \infty}\frac{S_n}{\sqrt{2B_n\ln\ln B_n}}=1$ almost surely.