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Is it possible to define "uniform" probability on $\mathbb N$? I first thought that it is intuitively "clear" how the probability density is defined. Consider the following examples:

Example 1. Probability of choosing an even number has probability 1/2.

Example 2. Probability of choosing a multiple of 3 has probability 1/3.

However, more thoughts revealed that my initial postulation is wrong. By contradiction, if there is such probability density function, then $P(X \leq n) = 0$ for every $n \in \mathbb N$, which means that Density and cumulative distribution functions are 0. This doesn't make sense at all! (It is also possible to choose a set $A \subset \mathbb N$ with $|A| = \infty$ and $\frac{|[n] \cap A|}{[n]} \rightarrow 0$, where $[n] = \{1, \cdots, n\}$.)

So what went wrong? How can I define a sigma algebra and probability measure so that the event space behaves similar to uniform distribution so that

  1. $P$(Even numbers) = 1/2
  2. $P$(Multiples of 3) = 1/3?
James C
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1 Answers1

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Let $\mathbb{N}$ denote the set of positive integers, let $X$ be a random variable defined on $\mathbb{N}$, and for each $k\in N$, let $p_{\vphantom{\frac{1}{1}}k}=P(X=k)$.

It's clear that $X$ can't be truly uniform else we would have $$ 0 < p_1=p_2=p_3=\cdots $$ but then for all $n\in\mathbb{N}$ we would have $$ P(X\le n)=np_1 $$ which exceeds $1$ for sufficiently large $n$.

Moreover we can show

Claim:

It is not possible to define $p_1,p_2,p_3,...$ such that for all $n\in\mathbb{N}$, we have $$ P(A_n)=\frac{1}{n} $$ where $A_n$ is the event that $X$ is a multiple of $n$.

Proof:

Suppose otherwise.

Let $Q$ be the set of prime numbers, and for each $k\in\mathbb{N}$, let $Q_k=\{q\in Q{\,\mid\,}q > k\}$.

From the fact that $$ \prod_{\large{q\in Q}}\Bigl(1-\frac{1}{q}\Bigr)=0 $$ it follows that for all $k\in\mathbb{N}$ we have $$ \prod_{\large{q\in Q_k}}\Bigl(1-\frac{1}{q}\Bigr)=0 $$ Now if $m,n\in\mathbb{N}$ are such that $\gcd(m,n)=1$, then $$ P(A_m\cap A_n) = P(A_{mn}) = \frac{1}{mn} = \frac{1}{m}{\,\cdot\,}\frac{1}{n} = P(A_m)P(A_n) $$ so $A_m,A_n$ are independent.

Then for all $k\in N$ we have $$ p_k \le \prod_{\large{q\in Q_k}}\Bigl(1-\frac{1}{q}\Bigr) = 0 $$ contrary to $$ \sum_{k=1}^\infty p_{\vphantom{\frac{1}{1}}k}=1 $$ which completes the proof of the claim.

quasi
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  • I am unsure about the last two lines. By $p_k$, do you mean $p_k = P(X = k)$? – James C May 30 '21 at 22:25
  • $p_k = 0$, so $p_k \leq \prod_{q \in Q_k} (1-1/q)$ trivially. Or perhaps, you meant $p_k = P(A_k)$, where $A_k$ is the set of multiples of $k$. – James C May 30 '21 at 22:28
  • @Paul Pogba: I$,$ defined $,p_k = P(X = k)$, and then I$,$ proved $,$that $p_k=0$ for all $k$, which contradicts the requirement that $\sum p_k=1$. – quasi May 30 '21 at 23:23
  • Thanks for the comment. If we take out the singletons ${{n} \mid n \in \mathbb N}$ from the $\sigma$-algebra (more precisely, the event space $\mathcal F \subset 2^\mathbb N$ contains no finite set), then there is no contradiction since $p_n$ is undefined for every $n \in \mathbb N$. Note that $\mathcal F$ contains events such as "the set of all multiples of $n$" or "the set of all primes." – James C May 30 '21 at 23:49
  • @Paul Pogba: In order to have a distribution which allows for selecting random positive integers, the probabilities $p_1,p_2,p_3,...$ must be defined. – quasi May 31 '21 at 00:17
  • @Paul Pogba: Furthernore, if $A_n$ denotes the set of positive integers which are multiples of $n$, then any $\sigma$-algebra on $\mathbb{N}$ which contains $A_1,A_2,A_3,...,$ must $,$contain all singleton subsets of $\mathbb{N}$. – quasi May 31 '21 at 00:33
  • Yes, I get your idea, while I think a more interesting conversation can be made if we replace countably additivity by finite additivity. – James C May 31 '21 at 04:34
  • Paul Pogba: Sure, but the objective of my post was to show that there does not exist an actual probability distribution on $\mathbb{N}$ such that $P(A_n)={\large{\frac{1}{n}}}$ for all $n\in\mathbb{N}$.- – quasi May 31 '21 at 13:21