X random variable in $\mathbb{N}$ independence of events If I have a random variable $X$ with values in $\mathbb{N}$,
$$\mathbb{P}(X=n)=\frac{1}{n^s\zeta(s)}$$
where $s>1$ and $\zeta$ the Riemann zeta function, then how can I show that $$A_i=E_{p_i^2}=\left\{X\text{ is divisible for } p_i^2\right\}$$ events are independent?
Maybe with $$1=\frac{X}{X}=\frac{m_1p_1^2}{m_2p_2^2}\Rightarrow m_1p_1^2=m_2p_2^2\,?$$
 A: Suppose that $p_i$ and $p_j$ are prime.  Then
$$
A_i\cap A_j=\{X=p_i^2p_j^2k\mid k\in\mathbb{N}\}.
$$
So,
$$
\begin{align*}
P(A_i\cap A_j)&=\sum_{k=1}^{\infty}P(X=p_i^2p_j^2k)\\
&=\frac{1}{\zeta(s)p_i^{2s}p_j^{2s}}\sum_{k=1}^{\infty}\frac{1}{k^s}\\
&=\frac{1}{p_i^{2s}p_j^{2s}}.
\end{align*}
$$
On the other hand,
$$
\begin{align*}
P(A_i)&=P\{X=p_i^{2}k\mid k\in\mathbb{N}\}\\
&=\frac{1}{\zeta(s)p_i^{2s}}\sum_{k=1}^{\infty}\frac{1}{k^s}\\
&=\frac{1}{p_i^{2s}},
\end{align*}
$$
and similarly
$$
P(A_j)=\frac{1}{p_j^{2s}}.
$$
Combining these, we see that
$$
P(A_i\cap A_j)=P(A_i)P(A_j),
$$
as desired.
Now, you might be wondering: where did we use the fact that $p_i$ and $p_j$ are prime?  Well, we made the assumption that a number $n$ is divisible by both $p_i^2$ and $p_j^2$ if and only if it is of the form $p_i^2p_j^2k$, $k\in\mathbb{N}$.  This is only true because $p_i$ and $p_j$ are relatively prime, and therefore $p_i^2$ and $p_j^2$ are relatively prime.  We could extend this result to any collection of numbers which are pairwise relatively prime; the primes themselves are the most accessible such set of numbers.
