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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 evens can be written in at least one way$n$ 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.

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 all evens can be written in at least one way 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.

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.

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AgCl
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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 all evens can be written in at least one way 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.

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 all evens can be written in at least one way 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.

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.

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 all evens can be written in at least one way 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.

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