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Let $X=\mathbb{N}\times \mathbb{N}$ be the product of positive integers. I wonder how to establish the sigma-algebra and the probability measure on $X$ in a formal way so that the probability of the set

$$\{(m,n):\gcd(m,n)=1\}$$

is equal to $\frac{6}{\pi^2}$.

If we start with the probability $\mathbb P$ on $\mathbb N$ and take the product, then this structure should make the probability of $\{m:p|m\}$ equal to $\frac{1}{p}$. Then what are the probabilities of the singletons $\{n\}$? (If the sigma algebra is discrete, then they shouldn't be all zero otherwise there would be a contradiction... So maybe we can use nondiscrete sigma algebra )

Reference: Probability that two random numbers are coprime is $\frac{6}{\pi^2}$

No One
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2 Answers2

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This is not a rigorous statement about probability in a fixed probability space. You could make it rigorous by saying that if $X = \{1,2,\dots,N\} \times \{1,2,\dots,N\}$ and we sample an ordered pair $(m,n)$ uniformly from $X$, then $$\lim_{N \to \infty} \Pr[\gcd(m,n) = 1] = \frac{6}{\pi^2}.$$

Misha Lavrov
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The $\frac6{\pi^2}$ you reference is not really a probability but the limit of a sequence of probabilities, where the pairs are chosen uniformly from a finite set for each experiment in the sequence. So you're asking whether there's a single distribution over $\Bbb N\times\Bbb N$ that "does the same thing".

There is no uniform discrete distribution on an infinite set, since the probabilities must sum to $1$. If you allow non-uniform distributions, then many "silly" solutions are possible that give the desired probability, e.g.

$$P(X=x)=\begin{cases}\frac6{\pi^2}&\text{if }x=(1,1)\\1-\frac6{\pi^2}&\text{if }x=(2,2)\\0&\text{otherwise.}\end{cases}$$

Karl
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