Concentration around the mean in Cramer random model

Question

This question is related to: Probabilistic prime number theorem

The Cramér random model is a random model ${{\mathcal P}}$ that is a random subset of the natural numbers, suh that each natural number ${n > 2}$ has an independent probability of ${\frac{1}{\log n}}$ of lying in ${{\mathcal P}}$, show that almost surely, the quantity $\frac{1}{x/\log x} |\{n \leq x: n \in {\mathcal P}\}|$ converges to one as ${x \rightarrow \infty}$.

The hint is to use Chebyshev’s inequality to get $S_x = |\{n \leq x: n \in {\mathcal P}\}|$ close to $\sum_{n \leq x} \frac{1}{\log n}$, which one can show in turn to be somewhat close to $x /\log x$.

Question: Let $S_x = |\{n \leq x: n \in {\mathcal P}\}| = 1 + \sum_{2 < n \leq x} 1_{{\mathcal P}}(n)$, where the $1_{{\mathcal P}}(n)$ are independent Bernoulli random variables with parameters $1 / \log n$. By Chebyshev's inequality, we have

$\displaystyle {\bf P}(|\sum_{2 < n \leq x} 1_{{\mathcal P}}(n) - \sum_{2 < n \leq x} \frac{1}{\log n}| \geq \varepsilon) \leq \frac{1}{\varepsilon^2} \sum_{2 < n \leq x} \frac{1}{\log n}(1 - \frac{1}{\log n})$.

Also, it can be shown from integration by parts that

$\displaystyle \sum_{2 < n \leq x} \frac{1}{\log n} = \int_{3}^x \frac{dt}{\log t} + O(1) = \frac{x}{\log x} + \frac{x}{\log^2 x} + O(\frac{x}{\log^3 x}) + O(1)$.

I am not sure how to proceed with these bounds though, any help will be appreciated.

Answer 1

You don't need to do any very precise calculations, simply note that the variance of any Bernoulli random variable is at most its expectation, and hence the same applies to any sum of independent Bernoulli random variables.

Thus, if $S_n=\sum_{i=1}^n X_i$ with $X_i$ independent Bernoulli, then $$P(|S_n-E(S_n)|\geq\delta E(S_n))\leq \frac{1}{\delta^2E(S_n)}.$$ In your case this is essentially tight: the ratio of $E(S_n):\operatorname{Var}(S_n)$ tends to $1$, so you can't get a meaningfully better bound from Chebyshev.

Unfortunately this isn't good enough: it gives you convergence in probability (since for any fixed $\delta$ the bound tends to $0$) but not almost sure convergence, which is what you were asked for. To get almost sure convergence you want the bound to be summable, for Borel-Cantelli, but $\sum_{n=2}^{\infty}\frac{\log n}{\delta^2n}=\infty$.

Hoeffding's inequality is strong enough to give almost sure convergence.

Answer 2

Edit: I managed to write up a solution using the moment method which I post below, verifications and suggestions are welcomed.

For all $n > 2$, let $X_n = 1_{{\mathcal P}}(n)$ be independent Bernoulli random variables with parameters $1 / \log n$. Take $x$ to be a natural number, and let $(Y_{n,x})_{n,x > 2: n \leq x}$ be the triangular array given by $Y_{n, x} := \frac{1}{x / \log x} X_n$.

Consider the partial sums $S_x = \frac{\sum_{2 < n \leq x} X_n}{x / \log x} = Y_{3,x} + \dots + Y_{x,x}$. For convenience, we normalise each $X_n$ to have zero mean, by replacing $X_n$ by $X_n - 1 / \log n$, so that $S_x$ also gets replaced by

$\displaystyle S_x - \frac{\sum_{2 < n \leq x} 1 / \log n}{x / \log x} = S_x - \frac{\int_{3}^x dt / \log t + O(1)}{x / \log x} = S_x - (1 + O(1 / \log x))$,

where we use integration by parts for the last inequality. We can then expand

$\displaystyle {\bf E}|S_x|^4 = {\bf E}|Y_{3,x} + \dots + Y_{x,x}|^4 = \sum_{2 < i, j, k, l \leq x} {\bf E}|Y_{i,x}Y_{j,x}Y_{k,l}Y_{l,x}|^4$.

The independence hypothesis leaves only a few quadruples ${(i,j,k,l)}$ for which ${{\bf E} Y_{i,x} Y_{j,x} Y_{k,x} Y_{l,x}}$ could be non-zero: the three cases ${i=j \neq k=l}, {i = k \neq j = l}, {i=l \neq j=k}$ where each of the indices ${i,j,k,l}$ is paired up with exactly one other index; and the diagonal case ${i=j=k=l}$. If for instance ${i=j \neq k=l}$, then

$\displaystyle {\bf E}Y_{i,x} Y_{j,x} Y_{k,x} Y_{l,x} = {\bf E}Y_{i,x}^2Y_{k,x}^2 = ({\bf E}Y_{i,x}^2)({\bf E}Y_{k,x}^2) \leq (\frac{\log x}{x})^2 \times (\frac{\log x}{x})^2$.

Similarly for the cases ${i=k \neq j=l}$ and ${i=l \neq j=k}$, which gives a total contribution of at most ${3x(x-1) (\frac{\log x}{x})^4}$ to ${{\bf E} |S_x|^4}$. Finally, when ${i=j=k=l}$, then ${\bf E}Y_{i,x} Y_{j,x} Y_{k,x} Y_{l,x} \leq (\frac{\log x}{x})^4$, and there are at most ${x}$ contributions of this form to ${{\bf E} |S_x|^4}$. We conclude that

$\displaystyle {\bf E} |S_x|^4 \leq {3x(x-1) (\frac{\log x}{x})^4} + x(\frac{\log x}{x})^4 = \frac{(3x - 2)\log^4 x}{x^3}$

and hence by Markov's inequality,

$\displaystyle {\bf P}(|S_x| > \varepsilon) \leq \frac{(3x - 2)\log^4 x}{\varepsilon^4{x^3}}$ for any $\varepsilon > 0$.

Remove the normalisation, we conclude that

$\displaystyle {\bf P}(|S_x - (1 + O(1 / \log x))| > \varepsilon) \leq \frac{(3x - 2)\log^4 x}{\varepsilon^4{x^3}}$ for any $\varepsilon > 0$.

By the comparison test, the right-hand side is summable in $x$. The Borel-Cantelli lemma thus implies that $S_x$ converges almost surely to one as $x \to \infty$, as desired.