# How can I prove that these events are independent?

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.

Thank you!

• Before you can speak of dependence you have to have a probability distribution, but you haven't specified one. – lulu May 21 at 13:50
• @lulu: I believe that $P(x\in{A})=\frac{1}{p}$, $P(x\in{B})=\frac{1}{q}$ and $P(x\in{A})\land P(x\in{B})=\frac{1}{qp}$. – goodvibration May 21 at 13:51
• 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? – lulu May 21 at 13:52
• @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. – goodvibration May 21 at 13:53
• 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. – lulu May 21 at 13:54

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.
• 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. – goodvibration May 21 at 15:26
• @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 $1$)? – goodvibration May 21 at 15:41