Probabilistic Goldbach conjecture 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 $${{\bf P}(n \in {\mathcal P}) = \frac{1}{\log n}}$$.

(Probabilistic Goldbach conjecture) Show that almost surely, all but finitely many natural numbers $${n}$$ are expressible as the sum of two elements of $${{\mathcal P}}$$.

Question: Consider the r.v $$S_x := \sum_{3 \leq n \leq x} 1_{\mathcal{P}}(n)1_{\mathcal{P}}(x-n)$$, which we interpret as the number of ways $$x$$ (taken to be a natural number) is expressible as the sum of two Cramér random model primes, scaled by a factor of two. One can deduce as in this paper https://arxiv.org/pdf/1508.05702 that $${\bf E}S_x \sim x / \log^2 x$$, thus it seems natural to show a strong concentration of the quantity $$S_x$$ around its mean (which tends to infinity). However, without using Chernoff bounds, I'm not clear about how to establish this (having tried to calculate the $$k$$th moment $${\bf E}(S_x - {\bf E}S_x)^k$$ for $$k = 2, 4, etc$$. Or should one use the Borel-Cantelli lemma to show that $${\bf P}(S_x = 0)$$ is summable in $$x$$?

Edit: An attempted solution by the moment method is posted below, verification are welcomed.

Take the r.v $$\sum_{3 \leq n \leq x} 1_{\mathcal{P}}(n)1_{\mathcal{P}}(x-n)$$, which we interpret as the number of ways $$x$$ (taken to be a natural number) is expressible as the sum of two Cramér random model primes, scaled by a factor of two. Consider only integers greater than $$2$$ in $${\mathcal P}$$, by independence:

$$\displaystyle {\bf E}S_x = \sum_{3 \leq n \leq x-3}\frac{1}{\log n \log (x-n)} \sim 2 \int_{3}^{x/2} \frac{dt}{\log t \log (x-t)} := 2I_x$$.

Moreover, since $$\displaystyle \frac{2}{\log (x-3)} \int_{3}^{x/2} \frac{dt}{\log t} \leq 2I_x \leq \frac{2}{\log x - \log 2} \int_{3}^{x/2} \frac{dt}{\log t}$$,

one may deduce from $$\displaystyle \int_{3}^x \frac{dt}{\log t} \sim x / \log x$$ that $$\displaystyle {\bf E}S_x \sim x / \log^2 x$$.

We normalise each $$X_n = 1_{\mathcal P}(n)1_{\mathcal P}(x - n)$$ to have mean zero by replacing it with $$Y_n := \displaystyle 1_{\mathcal P}(n)1_{\mathcal P}(x - n) - \frac{1}{\log n \log (x-n)}$$, so that $$S_x$$ also gets replaced by $$\displaystyle S_x' = \sum_{3 \leq n \leq x-3} Y_n$$. As the $$Y_n$$ are independent, have mean zero and of size $$O(1)$$, one can calculate the fourth moment $${\bf E}(\sum_{3 \leq n \leq x-3} Y_n)^4 = O(x^2)$$. By Markov's inequality, $$\forall \varepsilon > 0$$, this implies

$$\displaystyle {\bf P}(|\frac{S_x'}{x / \log^2 x}| > \varepsilon) \leq \frac{\log^8 x O(x^2)}{\varepsilon^4 x^4} = O(\frac{\log^8 x}{x^2})$$,

which are summable in $$x$$ by the comparison test. By the Borel-Cantelli lemma, $$\displaystyle \frac{S_x'}{x / \log^2 x}$$ thus converges almost surely to zero. Equivalently, we have

$$\displaystyle {\bf P}(\lim_{x \to \infty} \frac{S_x}{x / \log^2 x} = 1) = 1$$. i.e, $$S_x \stackrel{\text{a.s}}{\sim} x / \log^2 x$$, and the desired claim follows.