Question E4.2- Probability with Martingales by Williams

Question

For $s > 1$, define the zeta function as $$\zeta(s):= \sum_{n=1}^{\infty} \frac{1}{n^s}.$$

Given a probability space $(\Omega,F,P)$, let $X : \Omega \rightarrow {1,2,3,\ldots}$ and $Y : \Omega \rightarrow {1,2,3,\ldots}$ be two independent random variables such that

$$P(X =n)=P(Y =n)= \frac{n^{-s}}{\zeta(s)} \quad \forall n=1,2,\ldots$$

Define $E_p = \left\{\omega \in \Omega: \frac{X(\omega)}{p} \in \{1,2,3,\ldots\}\right\}$, where $p$ is a prime number.

Facts:

• {$E_p: p \ge 2 \text{ is prime}$} is an independent family of events, and $P(E_p) = \frac{1}{p^s}$.
• $\frac{1}{\zeta(s)} = \prod_{p\ge2 \text{ prime}} \left(1 - \frac{1}{p^s}\right)$.

I'd like to prove that, if H is the highest common factor of X and Y, then

$$P(H = n) = n^{-2s} / \zeta(2s).$$

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What I came up with so far is that if $n$ is the h.c.f of X and Y, $np$, for any prime $p$, can not be the common factor of X and Y. In other words

$$\{H = n\} \equiv \left\{ \{E^X_n \cap E^Y_n\} \cap \left\{\bigcap_{p \ge 2 \text{ prime}}\{E^X_{np} \cap E^Y_{np}\}^c \right\}\right\},$$

where $E^X_n$ and $E^Y_n$ are the divisibility sets for X and Y respectively, as defined above.

This equivalence, however does not yield the correct answer. Something is missing and I appreciate any help.

Answer 1

Let $p_n=P(H=n)$ and $q_n=P(n\textrm{ divides }H)$. Then $$q_n=P(n\textrm{ divides } X\textrm{ and }n\textrm{ divides } Y) =P(n\textrm{ divides } X)^2 =\left(\sum_{m:n\mid m}\frac{m^{-s}}{\zeta(s)}\right)^2 =\frac{n^{-2s}\zeta(s)^2}{\zeta(s)^2}=n^{-2s}.$$ We can now use Mobius inversion to obtain $p_n$. $$\sum_{d=1}^\infty\mu(d)q_{nd}=\sum_{d,e=1}^\infty\mu(d)p_{nde} =\sum_{f=1}^\infty p_{nf}\sum_{d\mid f}\mu(d)=p_n$$ as $\sum_{d\mid f}\mu(d)=0$ for $f\ge2$. Therefore $$p_n=\sum_{d=1}^\infty\mu(d)(nd)^{-2s}=n^{-2s}\sum_{d=1}^\infty\mu(d)d^{-2s} =\frac{n^{-2s}}{\zeta(2s)}.$$

Answer 2

$$\newcommand{\N}{\mathbb{N}}$$

Those events are no longer independent, is where the issue comes into play. Here's one way to look at it:

First note that $$P(X/n = k \, | \, X/n \in \N) = \frac{(nk)^{-s}}{\sum_{j\geq1} (nj)^{-s}} = \frac{k^{-s}}{\zeta(s)}$$

implying that $(X/n \, | \, X/n \in \N)$ has the same law as $X$. From here, We can compute: \begin{align*} P(H = n) &= P(X/n \in \N, Y/n \in \N, \text{gcd}(X/n,Y/n) = 1)\\ &=P(X/n \in \N, Y/n \in \N) \cdot P(\text{gcd}(X/n,Y/n) = 1 \, | \,X/n \in \N, Y/n \in \N) \\ &= n^{-s}\cdot n^{-s}\cdot P\left( \text{gcd}(X,Y) = 1 \right)\,. \end{align*} Your decomposition gives that this last probability is exactly $1/\zeta(2s)$.