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On set $\Omega = \{ 1,2,\dots,n\}$, $n \in \mathbb{N}$ and $\sigma$-algebra, which is its power set, we denote $P$ is probability measure, where $P(\{ k \}) = \frac{1}{n}$ for all $k = 1,2,\dots,n$. Let $\mathbf{P}$ is set of prime numbers. We denote $\phi(n)$ is number of all distinct numbers of $n$. We define events $A_p := \{p,2p,3p,\dots \} \cap \{1,2,\dots, n \}$ where p is prime.

How to prove that $A_p$ are independent, where p is prime divisor of n? And then i need to prove $\frac{\phi(n)}{n} = \prod_{p\in \mathbf{P}, p|n} (1-\frac{1}{p}).$

Thank you for any help.

  • 1
    In this setting, what will be $P(\{1,2 \})$ if $n>2$? – 2017-02-20
  • 0
    Possible duplicate of [A simple way to obtain $\prod_{p\in\mathbb{P}}\frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty}\frac{1}{n^s}$](http://math.stackexchange.com/questions/427910/a-simple-way-to-obtain-prod-p-in-mathbbp-frac11-p-s-sum-n-1-in) – 2017-03-07

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$A_p$'s are not independent.

Counter-example:

Let $\Omega = \{ 1,2,3\}$, $A_2 = \{2\}$, $A_3 = \{ 3 \}$. So $$0 = P(\{\phi\}) = P(A_2,A_3) \neq P(A_2)*P(A_3) = \frac{1}{3}*\frac{1}{3}. $$

  • 0
    Is the same if $p$ is all prime divisors of number $n$? – 2017-02-20
  • 1
    if $p$ is all prime divisors of number $n$ then the counter examples fails. – 2017-02-20
  • 1
    How to prove independence in that case? – 2017-02-20