Heuristic justification for Goldbach's conjecture

Question

On the Wikipedia page of Goldbach's conjecture, a heuristic justification is given, which did not completely satisfy me. It roughly goes as follows:

• randomly define a subset integers in accordance with the prime number theorem
• Let $K_n$ be the random variable, counting the number of ways the natural number $2n$, can be written as a sum of two members of this set.

Then $E[K_n]\rightarrow \infty$ .

The problem is that, although the mean goes to infinity, it still might be true that the probability that $K_n>0$ for all $n$ is zero.

So I thought of a different heuristic, and I am curious about whether anything is known about it:

Let $\mathcal P$ be the collection of all subsets of odd numbers whose density agrees with the prime number theorem, and let $\mathcal G$ be the collection of subsets for which Goldbach's property holds (i.e. every even number can be written in at least one way with two members of the set). Let $\mu$ be the uniform product measure of the space $\{0,1\}^{\mathbb > N}$. Then the quantity $$ > \frac{\mu(\mathcal P \cap \mathcal > G)}{\mu(\mathcal P)} $$ is (significantly) greater than zero.

Edit: As pointed out in the comments, $\mu(\mathcal P) = 0$, so this quantity is meaningless as it is, but I think it can be formalized in some way.

I do not know if this is easy or almost as difficult as the original problem. But it would be a very convincing heuristic for me in that, it would tell me how much of Goldbach's conjecture is already explained by the prime number theorem.

I would appreciate answers, or references to any known results, or reasons if this kind of heuristic is not relevant, if that is the case.

Answer 1

I'm not even sure that your heuristic is as easy as Goldbach. On one hand it allows exceptions, but on the other it requires that only the density be used, not other properties of the primes.

I prefer to justify the conjecture by looking at the expected number of exceptions (again, using only the density of the primes). If there are $n/\log n$ primes $p$ between $n$ and $2n$ and the chance that $2n-p$ is prime is $1/\log n$ then the chance that $2n$ is a Goldbach exception is $$\left(1-\frac{1}{\log n}\right)^{n/\log n}$$ which is $$\left(\left(1-\frac{1}{\log n}\right)^{\log n}\right)^{n/\log^2 n}$$ which is asymptotically $$\exp(-n/\log^2 n)$$ so the expected number of exceptions for $2n>a$ is about $$\int_a^\infty\exp(-n/\log^2 n)$$ which is about 5.7 for $a=1$ and about $10^{-1011269}$ for $a=10^9$. Since $10^{-1011269}$ is small, Goldbach seems likely, given that there are no small exceptions.

Answer 2

Let me add a historical remark. J.J. Sylvester was probably the first to come up with a heuristic argument. See his (somewhat naive) approach in this short paper and this Nature note.

P.S. I found these two while working on this mildly related paper.

Answer 3

A few thoughts.

1. The $\mu$-ratio statement, assuming it can be formalized, does not lead to quantitative predictions. In this sense it is not comparable to the density heuristics leading to Goldbach and a host of other asymptotic predictions in number theory.

2. It seems overwhelmingly likely that the $\mu$-ratio, if it exists in a suitable sense, is 1 or 0. I would guess that the value is $1$ and that this is an elaboration of the existing probability arguments for Goldbach.

3. For the state of the art in Goldbach heuristics, see Andrew Granville's paper "Refinements of Goldbach's conjecture,and the generalized Riemann hypothesis" and subsequent corrigendum.