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Unit 01 /
Problem 2:


C probability function properties
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Easy


Assume an outcome space consisting of all positive integers, $\Omega = {1,2,3,...}$, and a probability function $P$ defined on the power set. Prove that it is not possible for the probability of every integer to be equal. That is it is not possible that $P({a}) = P({b})$ for every pair of integers $a,b \in {1,2,3,...}$.

Solution:

Assume for contradiction that there is a number $c \ge 0$ such that $P({a}) = c$ for every integer $a$. The positive integers are countable, so by the countable additivity axiom, we know,

$$P(\Omega) = P( {1} \cup {2} \cup {3} \cup ...) = P( {1} ) + P( {2} ) + P( {3} ) + ... = c + c + c + ...$$

$P(\Omega) = 1$ by the unitarity axiom, so the infinite sum on the right must equal 1. However, if $c=0$ then the infinite sum equals zero, and if $c>0$ then the infinite sum us unbounded (and certainly bigger than 1). This is a contradiction, so it is not possible that all integers have the same probability.