# What does it mean for a theorem to be “almost surely true”, in a probabilistic sense? (Note: Not referring to “the probabilistic method”)

I recently came across this paper where the Goldbach conjecture is explored probabilistically. I have seen this done with other unsolved theorems as well (unfortunately, I cant find a link to them anymore). The author purports to bound the probability of the Goldbach Conjecture being false as $\approx 10^{–150,000,000,000}$

What is mathematics to make of such a statement? Since a theorem is either true or false, how can a theorem be true with $99.999..\%$ certainty? In the past, I have seen another prime theorems (forgot which one) where it was shown that it was "almost surely true".

So, lets say I assume some (theoretically plausible) probability distribution over the instances $t\in T_n$ of a general theorem I am studying, where $|T_n|=n$. Lets call this measure $P_{T_n}(t)$, and lets say I conclude $\lim\limits_{n\to\infty}E_{T_n}[\mathbf{1}_{\perp}(T_n)]=0$. This seems to say that $P(\{\mathbf{1}_{\perp}(T)=1\;\; i.o\})=0$

Has anyone used a probabilistic heuristic like in the paper to demonstrate "almost surely true". Or have you also run across it. If so, how is it used to as part of mathematical research, apart from a "gee whiz...that's interesting...I guess." Is there a domain of "Theorems we are "almost sure" about, but can't prove" where this method takes the place of certain/non probabilistic deduction?

## 2 Answers

First of all, the phrase "almost surely true" or its equivalent "true with probability 1" has a technical and well defined meaning for results with some probability aspect. It means that although there may be examples involving specification of an infinite set of items which violate the statement, the probability measure of the collection of such sets is zero. A good example is the statement that a balanced random walk in 1 dimension will return to the origin almost certainly -- you can of course construct sequences (e.g., all +1 steps) that do not, but with probability 1 a random sequence will return to 0.

Next, there are some statements in number theory, typically involving distributions of primes, which have been proven in a limit/probabilistic sense. These statements general sound like "there exists some sufficiently large $N$ that for all intervals of length greater than $N$ the occurence of is rarer than . Some authors then say that the proposition is "almost surely true."

The statement you have encountered is weaker yet. It says that given some known distribution properties (again, of the primes) and other small-number information, the author can prove that the probability of a violation of the conjecture is less than that tine $\epsilon$, *assuming the pseudo-random distribution of the primes is not correlated with the properties that would affect Goldbach's conjecture."

For example, if we were to substitute the "conjecture" that there are no nontrivial tuples such that $a^3 + b^3 + c^3$ we might be able to say, based on knowing the conjecture is true for numbers below a million, and assuming the cubes are distributed randomly with a density $1/3n^2$, that the probability of that conjecture being false is less than 0.1.

Obviously, the interesting thing about this author's work is the incredibly tiny heuristic probability.

• Thanks! The instances I've run into fall into your "number theory" examples...basically they are saying that they know something is true for the vast majority of numbers, therefore, the statement "X is true w.p. P" is a pointwise statement, not a statement about the general theorem, which is false if a single counterexample is found. – user76844 Oct 27 '14 at 18:54

"Almost surely" is not being used in the mathematical sense, but in the common sense that the probability is very small. We use the idea is the the primes are distributed "randomly" with the probability that $$n$$ is prime being $$\frac 1{\log n}$$. If we then verify the Goldbach conjecture up to some large number $$M$$ we can compute the chance that no even number above $$M$$ fails to have a decomposition into two primes. Since there are lots of primes below $$M$$, the chance that any given number fails to have a decomposition is very small. We can then sum up the chance over all the even numbers greater than $$M$$. If this sum converges, we can claim that the probability the conjecture fails is less than this sum. Note that we are assuming that for any two numbers the chance that they are prime is independent.

Given an even number $$m$$ the potential decompositions are $$(a,m-a)$$. The chance that both numbers are prime is then $$\frac 1{\log a\cdot \log (m-a)}$$ The chance that $$m$$ is a number where the conjecture fails is then $$\prod_{a=2}^{m/2}\left(1-\frac 1{\log a\cdot \log (m-a)}\right)$$ Then the chance that the Goldbach conjecture fails is $$\sum_{m=M}^\infty\prod_{a=2}^{m/2}\left(1-\frac 1{\log a\cdot \log (m-a)}\right)$$ You can convert both of these to integrals and evaluate them. The result is a very small chance the conjecture fails.