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I have a pair of sets:

  • $A=\{n\in\mathbb{N}\mid p\cdot n\}$
  • $B=\{n\in\mathbb{N}\mid q\cdot n\}$

Where $p$ and $q$ are two different prime numbers.

And the following event definitions:

  • $X_n$: $n\in{A}$
  • $Y_n$: $n\in{B}$

Does it follow that the events $X_n$ and $Y_n$ are independent for every $n\in\mathbb{N}$ and every pair of different primes?

I believe that $P(X_n\land Y_n)=P(X_n)\cdot P(Y_n)=\frac1p\cdot\frac1q$ for every $n\in\mathbb{N}$ and every pair of different primes, hence the answer is true, but I'm finding it hard to prove this.

Can anyone please help with this?

Thank you!

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  • $\begingroup$ Before you can speak of dependence you have to have a probability distribution, but you haven't specified one. $\endgroup$ – lulu May 21 at 13:50
  • $\begingroup$ @lulu: I believe that $P(x\in{A})=\frac{1}{p}$, $P(x\in{B})=\frac{1}{q}$ and $P(x\in{A})\land P(x\in{B})=\frac{1}{qp}$. $\endgroup$ – goodvibration May 21 at 13:51
  • $\begingroup$ Please use Mathjax. And it's not a question of what you believe, but of what you define. What is the sample space? What is the probability distribution on it? $\endgroup$ – lulu May 21 at 13:52
  • $\begingroup$ @lulu: I thought it was pretty obvious that since we're dealing with sequences, the sample space is $\mathbb{N}$. I can mention that explicitly, but I've figured less text <==> higher level of clarity. $\endgroup$ – goodvibration May 21 at 13:53
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    $\begingroup$ It's also not a question of what you think is obvious. There is no uniform distribution on $\mathbb N$ so the critical issue is what probability distribution you had in mind. $\endgroup$ – lulu May 21 at 13:54
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Since the author doesn't know the probability distribution, I will assume exponential distribution: $$ P(n=k) = \frac122^{-k}, \qquad k=0,1,2... $$

With that assumption, $P(X_n)=\frac{1/2}{1-2^{-p}}$, $P(Y_n)=\frac{1/2}{1-2^{-q}}$, $P(X_n\wedge Y_n)=\frac{1/2}{1-2^{-qp}}$. One can see that these events are not independent.

Can one come up with the distribution, so these events are independent? It is quite another story.

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  • $\begingroup$ Sure you can. The limit distribution of the uniform on bounded sets works. But your point is correct, of course. The choice of distribution is the critical thing. $\endgroup$ – lulu May 21 at 15:24
  • $\begingroup$ The distribution is uniform, i.e., choose a random positive integer $n$, then $X_n$ is the event $x\in A$ and $Y_n$ is the event $x\in B$. I know that $P(X_n)=\frac1p$ and $P(Y_n)=\frac1q$, and I'm pretty sure that the events are independent, but I'm not sure how to come up with a formal proof. $\endgroup$ – goodvibration May 21 at 15:26
  • $\begingroup$ @goodvibration As has been pointed out to you repeatedly, there is no uniform distribution on an infinite set. That's the entire problem here. $\endgroup$ – lulu May 21 at 15:36
  • $\begingroup$ @lulu: So given the two infinite (but enumerable) sets $A$ and $\mathbb{N}$, I cannot state that when selecting an element from $\mathbb{N}$, the probability of it being in $A$ is well defined (and of course, a value between $0$ and $1$)? $\endgroup$ – goodvibration May 21 at 15:41
  • $\begingroup$ @goodvibration No. Not without saying what you mean by probability. Infinite sets are complicated. You can't just wave your hands. That's what everyone has been trying to tell you. $\endgroup$ – lulu May 21 at 15:47

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