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Based on this question, I want to ask:

Suppose $q$ is drawn uniformly from $[0,1]$. What is the probability that the reduced form of $q$ has the form $\frac{n}{n+1}$ conditioned on $q\in\Bbb{Q}$?

(I'm reminded that one can't have a probability distribution on a countably infinite discrete set; but I don't recall offhand how to prove that there are no Borel probability measures on any countably infinite set. So I think the modification above will make sense but am happy to contemplate reasons why it doesn't and/or alternative probability measures to use on $\Bbb{Q}\cap [0,1]$.)

Here's my quick answer attempt. It will become evident that it's flawed but I don't have clarity on how exactly to fix the argument -- mostly I don't have good intuition for the function $$n\mapsto \#\{\mbox{integers $k<n$ that are coprime to $n$}\}.$$

I'm going to write $$ \Bbb{Q}\cap [0,1] = \bigcup_{n=0}^\infty Q_n $$ where $$ Q_n = \bigg\{ \frac{k}{n+1}\ \bigg|\ k=0,1,\ldots,n+1 \bigg\}.$$ For each $n$ we have $|Q_n| = n+1$. As there is precisely one element of $Q_n$ of the form $\frac{n}{n+1}$ we have for each $N$ the probability of randomly drawing such an element from $\cup_{n=0}^N Q_n$ is no less than $$ \frac{N+1}{\sum_{j=0}^N Q_j} = \frac{N+1}{1+2+\cdots + N+1} = \frac{2(N+1)}{(N+1)(N+2)} = \frac{2}{N+2} $$ However: This goes to zero, so: great, the probability is no less than zero

Notice that for composite $n$, many elements of $Q_n$ are not in reduced form, so I'm overcounting a lot of fractions.

Am I overcounting enough fractions to push the probability that $q = \frac{n}{n+1}$ above zero?

Neal
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    You can't have a uniform probability distribution on a countably infinite set. – paw88789 Jul 06 '17 at 17:06
  • @paw88789 Good call! Then I'll modify the question. – Neal Jul 06 '17 at 17:07
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    The probability is zero. Consider how many numbers are of the form $\frac nk$ with $k>n+1$. – Simply Beautiful Art Jul 06 '17 at 17:10
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    I'd say it is "not even zero" – Hagen von Eitzen Jul 06 '17 at 17:11
  • @SimplyBeautifulArt I don't think we know that without a distribution. There are countably infinite rationals of the form $\frac{n}{n+1}$ and countably many not of this form. So the two sets are equinumerous; so perhaps the probability is $\frac{1}{2}$. – paw88789 Jul 06 '17 at 17:12
  • @paw88789 Sorry, I meant the density is zero. – Simply Beautiful Art Jul 06 '17 at 17:13
  • @HagenvonEitzen $ 0 > Pr[q=n/(n+1)] > -\epsilon$ for all $\epsilon > 0$? :P – Neal Jul 06 '17 at 17:14
  • It seems to me that if we had a distribution for the numerator and for the denominator separately and independently, then we'd have a positive probability for our answer. But perhaps this is reading too much into the question. – paw88789 Jul 06 '17 at 17:15
  • @paw88789 Also, your suggestion that the probability would be $\frac12$ has false logic. While it is true that the cardinality of the naturals and the rationals are equal, the rationals have a' higher probability' of being picked than the naturals. – Simply Beautiful Art Jul 06 '17 at 17:19
  • @SimplyBeautifulArt I wasn't seriously proposing that the 'correct' answer is $\frac12$. :-) – paw88789 Jul 06 '17 at 17:20
  • @paw88789 I'm definitely open to revising the question further. – Neal Jul 06 '17 at 17:22
  • This kind of questions do make sense. The probability of taking two integers randomly which are coprime is known, so why not? Actually I just made an experiment and got $\frac{91}{10000}$ consecutive pairs of random numbers. To be fair I restricted the range to $10^6$. Increasing the range to $10^8$ I got only one on $10,000$ I just realized that the experiment is not right. I considered only the irreducible fractions. I am going to redo it properly – Raffaele Jul 06 '17 at 17:30
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    Can you condition on a null event? – Ranc Jul 06 '17 at 17:30
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    As has been said in the comments above : there is no uniform probability on a countable set. Therefore the answer will completely depend on the probability measure. It could be any $x\in [0;1]$, depending on the measure you're willing to put. According to the measurable sets you allow, it is possibly not defined. So the answer is "as you wish" – Maxime Ramzi Jul 06 '17 at 17:36
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    One might note that $\frac n{n+1}=1-\frac1{n+1}$, so it comes down to "what's the probability that I happen to choose a number of the form $\frac1n$". – Simply Beautiful Art Jul 06 '17 at 17:40
  • @Ranc Why not? c.f. https://math.stackexchange.com/questions/110112/probability-conditional-on-a-zero-probability-event – Neal Jul 06 '17 at 17:41
  • @Neal From the link: You can "condition on an individual event of probability zero, if that event admits a natural approximation by events of positive probability". I don't see how that applies here. – leonbloy Jul 06 '17 at 19:25

2 Answers2

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I don't think that the conditioning solves the problem of defining a uniform probability measure over the rationals.

If, for$X\sim U[0,1]$, we were allowed to condition on the (zero probability) event $A \equiv X \in \mathbb{Q}$ (I don't think we can) then, for any $q\in \mathbb{Q}$ we'd should be able to calculate $P(X=q \mid A)=\alpha$ ; but again, this cannot work, because of additivity: no value of $\alpha$ can give us $\sum_{q\in \mathbb{Q}} P(X=q \mid A) =1$

We'd need some non-uniform (and not simple) measure, for example.

So, I'm afraid it's hard to make sense of the question.

leonbloy
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For every $n$ such that

$$x_n=\frac n{n+1}$$

There exists infinitely many rationals of the form

$$y_n=\frac nk,k>n+1$$

Thus, the 'probability' of choosing a number of the form $x_n$ is $0$.

Note a probability of $0$ does not mean it is impossible for you to choose a number of the form of $x_n$, but rather that it is improbable.

A classic example is the probability you'd choose $1$ out of all the real numbers in $[0,1]$.

Simply Beautiful Art
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  • Sorry to be dense - does this imply an answer to my revised question, conditioning $q = \frac{n}{n+1}$ on $q\in\Bbb{Q}$? – Neal Jul 06 '17 at 17:46
  • @Neal I'm unsure what you mean. Are you taking $n\in\mathbb Z\setminus-1$? Either way, yes, the density of your $q$ is zero in $\mathbb Q$. – Simply Beautiful Art Jul 06 '17 at 17:48
  • Let $R = {\frac{n}{n+1}\ |\ n = 0, 1, 2, \ldots}$ and $Q = \Bbb{Q}\cap[0,1]$. My revised question is: what is $\operatorname{Pr}[R|Q]$ with respect to the Lebesgue measure on $[0,1]$? – Neal Jul 06 '17 at 17:54
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    Pretty sure it's still zero. – Simply Beautiful Art Jul 06 '17 at 18:10
  • Does this mean that the probability of each one specific rational is $0$? Because if so, then this is not a (conditional) probability distribution. (The (conditional) probabilities don't sum to $1$.) – paw88789 Jul 06 '17 at 22:23
  • @paw88789 Indeed, that is the case. It's like the probability of choosing 1 out of all the natural numbers, to give an analogy. – Simply Beautiful Art Jul 06 '17 at 22:27
  • When you say "the probability of choosing 1 out of all the natural numbers," what do you mean? – Cameron Buie Jul 07 '17 at 11:05
  • @CameronBuie something along the lines of density, like how the even numbers and off numbers have 1/2 density in the natural and the primes have 0 density in the naturals. – Simply Beautiful Art Jul 07 '17 at 11:15