Heuristic justification for Goldbach's conjecture

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.

• If I understand this right, I think $\mu(\mathcal{P}) = 0$. Indeed, any collection of sets of odd numbers has $\mu$ measure zero. Commented Jul 12, 2010 at 17:45
• Thanks, I think you are right! Could you help me correct the expression? There must be a way to write the expression as a generalization of conditional probabilities, so that $\mu(\mathcal P) = 0$ would not matter.
– AgCl
Commented Jul 12, 2010 at 17:51
• Moreover, $\mu$-almost every set of integers has density 1/2, by the strong law of large numbers. I think this is not the right heuristic. Commented Jul 12, 2010 at 17:55
• To correct your heuristic, the standard procedure in such countable situations is to cut at a finite step: denote by $\mathcal{P}_n$ the collection of all subsets of $\{1,\ldots,n\}$ with the right density, $\mathcal{G}_n$ those for which Goldbach's property holds for all integer less than or equal to $n$, and estimate the probability of $P\in\mathcal{G}_n$ for $P$ drawn uniformly in $\mathcal{P}_n$. The answer of Charles should apply to this setting. Commented Jul 12, 2010 at 20:39

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.

• Charles, the $n/log(n)$ could have been any asymptotically nearby density function, and using a different choice can raise the expected number of counterexamples. To get meaningful numbers out of this argument requires some analysis of the sensitivity to particular models of the prime distribution.
– T..
Commented Jul 12, 2010 at 21:47
• ...and the Rosser and Schoenfeld estimates may be useful for such analyses.
– T..
Commented Jul 12, 2010 at 21:48
• @T: Absolutely. Mine is a very rough heuristic, based only on a very simple form of the PNT. Commented Jul 13, 2010 at 12:44

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.

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.