Problem
Let $\Omega = {0, 1, ...}$
Prove that there does not exist a uniform distribution on $\Omega$, that is, if $P(A) = P(B)$ whenever $|A| = |B|$ then $P$ cannon satisfy the axioms of probability.
Solution
First, let's review the axioms of probability
- $P(A) \geq 0$ for every $A \subseteq \Omega$
- $P(\Omega) = 1$
- If $P_1, P_2, ... P_n$ are disjoint, then $P(\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} P(A_i) $
If $P$ is a uniform distribution, $P(\Omega) = 1 = \bigcup_{i=0}^{\infty} \{i\} = \sum_{i=0}^{\infty} \{i\} = \sum_{i=0}^{\infty} c$
for some $c$. If $c=0$ then $P(\Omega) = 0$, and if $c>=$ then $P(\Omega) = \infty$, both breaking axiom #2.
Sources
- 36-705 hw1 problem 4