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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 $$1=\frac{X}{X}=\frac{m_1p_1^2}{m_2p_2^2}\Rightarrow m_1p_1^2=m_2p_2^2\,?$$

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    $\begingroup$ Who is $p_i$? :-) $\endgroup$ – Ant Jun 25 '16 at 9:43
  • $\begingroup$ sorry, p is a prime number @Ant $\endgroup$ – Bombadil Jun 25 '16 at 9:47
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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.

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