Let's consider a random variable $X$ on $\mathbb{N}^*$ such as $\mathbb{P}[X=n]=n^{-s}\zeta(s)$.
Thanks to that random variable we can prove that $\zeta(s)= \underset{i=1}{\overset{\infty}{\prod}}\frac{1}{1-p_i^{-s}}$, where $p_i$ is the $i^\text{th}$ prime number, by calculating the probability of the event $\mathbb{P}[k \mid X]$. Then there a question I couldn't solve.
Let's take a random variable $Y$ defined on $\mathbb{N}^*$, which has the same distribution of $X$. $X$ and $Y$ are independent. Calculate the distribution of $Z = \gcd(X,Y)$ where $\gcd$ represents the greatest common divisor.
I would be grateful for any suggestions or ideas.
Let we compute the probability that $Z=1$, for first. Given two random integers $X,Y$, they are coprime if for every prime $p$ such a prime divides at most one of them. So, what is the probability that $X$ is even? It is:
$$ \mathbb{P}[X\text{ is even}]=\frac{1}{\zeta(s)}\sum_{n\geq 1}\frac{1}{(2n)^s} = \frac{1}{2^s}$$ In the same way we have that the probability that $p$ divides $X$ is exactly $\frac{1}{p^s}$. Now it is important to check that given two different primes $p,q$, the events $X\equiv 0\pmod{p}$ and $X\equiv 0\pmod{q}$ are independent. The unique factorization theorem grants that.
So we have: $$ \mathbb{P}[Z=1] = \prod_{p}\left(1-\frac{1}{p^{2s}}\right) = \frac{1}{\zeta(2s)}.$$ Now I bet you can prove from this argument that: $$ \mathbb{P}[Z=m] = \frac{1}{\zeta(2s)}\cdot\frac{1}{m^{2s}}.$$