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.
Best Answer
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.