The heuristic justification section of the Wikipedia article about Goldbach's conjecture says that the argument that suggests that

the number of twin primes below $x$ should be roughly $\dfrac{x}{\log^{2} x}$ up to a multiplicative constant

is just heuristics, because of the lack of independence of the "random variables" considered therein. But what if we considered them as just uncorrelated instead of truly independent?

Could the argument be still used or not?

Edit: several people have said that the notion of probability here is rather irrelevant, cause the primes form a deterministic system. My question now is thus: in some dynamical system framework, would the set of congruences that are verified in the case of primes force any subset of the positive integers with the same natural density as the one given by the PNT to be the one of a subset of the primes? The idea is to make a random process evolve under constraints until it becomes entirely determined. I apologize for the vagueness of the question, I just try to make my intuitions as 'rigorizable' as possible. To make this more clear, let's consider the double sequence of random variables $X_{i,t}$ with $1\leq i\leq n$ and $t\in\mathbb{N}_{0}$ such that there exists $K_{i,t}>0$ such that $$P(X_{i,t}=1)=K_{i,t}\dfrac{\pi(n)}{n}\left(\prod_{k=1}^{t}(i \mod p_{k})\right)^{1_{t\gt 0}}$$ and $\forall t, \sum_{i=1}^{n}P(X_{i,t}=1)=1$.

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    $\begingroup$ The Wikipedia article does not use the word "proof" in that context. So the quotation mark is misleading here. Without the quotation mark, it would be equally misleading, so I suggest you change the first sentence. A more serious problem with your post is that it does not contain a well-formulated mathematical question. To clarify: no proof of Goldbach's conjecture would consider the primes random variables, because they are not random variables. $\endgroup$ – GH from MO Apr 16 '15 at 22:57
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    $\begingroup$ the probability that n is prime is ~ 1/log(n). The probability that n,n+2 are prime is therefore ~ 1/log(n)^2 $\endgroup$ – JMP Apr 16 '15 at 23:41
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    $\begingroup$ @JonMarkPerry, I hope you are being a little bit facetious, or else you are possibly misleading the questioner or other naive souls. $\endgroup$ – paul garrett Apr 17 '15 at 0:13
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    $\begingroup$ @JonMarkPerry, :) the point is that the notion of "probability" of primality is not well-founded, although an extremely intuitive/engaging idea. :) That it is not (at this time in history) something that generates causality, that is, that gives true proof, is disappointing, but a contingent fact. $\endgroup$ – paul garrett Apr 17 '15 at 0:40
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    $\begingroup$ I guess he had a problem with the second part of the statement, we have no idea what the conditional probability of $p+2$ being prime is, given $p$ prime... $\endgroup$ – Asvin Apr 17 '15 at 0:50

As I understand your question, we have random variables $X_n$, $n\in\mathbb N$ taking values in $\{0,1\}$.

We intuitively think of $X_n=1$ as ``$n$ is prime'', but other than that this has little to do with primes.

Letting $p_n:=\Pr(X_n=1)$, we have that heuristically $p_n\rightarrow 0$ at a certain rate, and we want to conclude something about the number of $X_k$'s that are $=1$ for $k\le n$.

Note that if $X_i$ and $X_j$ are uncorrelated, then they are (pairwise) independent. (This is true for any random variables that take values in $\{0,1\}$).

Moreover, the Law of Large Nunbers only requires pairwise independence.

[It may be that that fact breaks down for non-constant $\{p_n\}_{n\in\mathbb N}$ but I guess it should be okay.]

Thus, it seems the same informal argument does work with uncorrelatedness in place of independence.

  • $\begingroup$ I once saw a heuristic argument that the probability of two numbers being relatively prime is $6/\pi^2$, the computation involving calculating the value of Euler's zeta function value $\zeta(2)$. How does this gel with this? $\endgroup$ – P Vanchinathan Apr 17 '15 at 0:38
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    $\begingroup$ @PVanchinathan I don't see any canonical way that it gels yet $\endgroup$ – Bjørn Kjos-Hanssen Apr 17 '15 at 0:46

Although this is a good heuristic, I have to add it doesn't capture the number-theoretic nature of the problem. One way to see this is that the same heuristic would imply every odd number is the sum of two primes, which unfortunately has more counterexamples than examples. The reason is that the primes are not random at all in terms of congruences (or, using more jargon, in finite places). One extreme example is that almost all primes (except 2, to be precise) are odd, so if you want two primes to add to an odd number, one of them has to be 2, which screws up the probabilistic heuristic quite badly. Thus we need to take congruence relations into account even if we are heuristically guessing the number of representations of a number as a sum of two primes, which explains why our best guess is not $x/log^2x$, but a constant multiple of it (more precisely, the twin prime constant).


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