Chapter 4: E4.2 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Let $s>1$ , and define $\zeta(s):=\sum_{n\in\mathbb{N}}n^{-s}$ , as usual. Let $X$ and $Y$ be independent $\mathbb{N}$ -valued random variables with $\mathbb{P}(X=n)=\mathbb{P}(Y=n)=n^{-s}/\zeta(s).$ Prove that the events $(E_{p}:p\mbox{ prime})$ , where $E_{p}=\{X\mbox{ is divisible by }p\}$ , are independent. Explain Euler’s formula $1/\zeta(s)=\prod_{p}(1-1/p^{s})$ probabilistically. Prove that $\mathbb{P}(\mbox{no square other than 1 divides }X)=1/\zeta(2s).$ Let $H$ be the highest common factor of $X$ and $Y$ . Prove that $\mathbb{P}(H=n)=n^{-2s}/\zeta(2s).$

$\mathbb{P}(E_{n})=n^{-s}$ . Hence, for any $\{n_{k}\}_{k=1}^{m}$ , $\begin{eqnarray*} \mathbb{P}(E_{\prod_{k=1}^{m}n_{k}}) & = & (\prod_{k=1}^{m}n_{k})^{-s}\\ & = & \prod_{k=1}^{m}\mathbb{P}(E_{n_{k}}). \end{eqnarray*}$ And for different prime numbers $\{p_{k}\}_{k=1}^{m}$ and any powers $\{r_{k}\}_{k=1}^{m}$ , $E_{\prod_{k=1}^{m}p_{k}^{r_{k}}}=\cap_{k=1}^{m}E_{p_{k}^{r_{k}}},$ so that events $(E_{p^{r_{p}}})$ are independent.

Euler’s formula expresses the probability of $1$ on both sides of the equality, because all $E_{p}$ are independent.

$X$ is not divisible by a square iff it is not divisible by a square of any prime number, and, using independence of $(E_{p^{2}})$ and Euler’s formula, we obtain the probability.

$\mathbb{P}(H=1)=$ $\prod_{p}(1-\mathbb{P}(E_{p})^{2})=$ $1/\zeta(2s)$ . Now, note that if $X\in E_{n}$ , then $\begin{eqnarray*} \mathbb{P}(X=kn) & = & (kn)^{-s}/\zeta(s)\\ & = & \mathbb{P}(E_{n})\mathbb{P}(X=k), \end{eqnarray*}$ hence, $\begin{eqnarray*} \mathbb{P}(H=n) & = & \sum_{p,q\mbox{ relatively prime}}\mathbb{P}(X=pn,Y=qn)\\ & = & \mathbb{P}(E_{n})^{2}\sum_{p,q\mbox{ relatively prime}}\mathbb{P}(X=p,Y=q)\\ & = & \mathbb{P}(E_{n})^{2}\mathbb{P}(H=1). \end{eqnarray*}$

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Chapter 4: E4.2 Solution

Chapter 4: E4.2 Solution