What is Countable probability space? — Mathematics

Definition 21.1 University Mathematics — Year 2 · Chapter 21 — Probability on Countable Spaces

Let $\Omega$ be a nonempty finite or countable set (the sample space). A probability measure on $\Omega$ is a map $\P$ from the set $\mathcal{P}(\Omega)$ of all subsets of $\Omega$ (events) to $[0, 1]$ such that:

$\P(\Omega) = 1$ ;
( $\sigma$ -additivity) for every sequence $(A_n)_{n\in\N}$ of pairwise disjoint events,

$\P\Bigl(\,\bigcup_{n \in \N} A_n\Bigr) = \sum_{n=0}^{\infty} \P(A_n) .$

The pair $(\Omega, \P)$ is a (countable) probability space.

Examples

Example 21.5 (Geometric model: waiting for the first head)

Toss a coin with head probability $p \in \intoo{0}{1}$ repeatedly, and let $\Omega = \N^* \cup \{\infty\}$ record the rank of the first head. The natural weights are

\P(\{k\}) = (1 - p)^{k-1}p \quad (k \in \N^*), \qquad \P(\{\infty\}) = 0 ,

a probability measure since $\sum_{k\geq1}(1-p)^{k-1}p = \frac{p}{1 - (1-p)} = 1$ : with probability $1$ the game ends — but the sample space must still contain the possibility that it does not. Countable additivity is what lets us assert $\P(\text{the game ends}) = \sum_k \P(\{k\})$ .

Example 21.8 (The union bound: crude but indestructible)

Subadditivity with finitely many events — the union bound — trades precision for universality. For the birthday problem with $23$ people, bounding the collision probability by the sum over pairs gives

\P(\text{collision}) \leq \binom{23}2\cdot\frac1{365} = \frac{253}{365} \approx 0.693 ,

against the true $0.507$ : off by a wide margin, because collisions overlap. Yet the bound needs no independence, no joint law, nothing but the pair probabilities — which is why, in the weekend problem and throughout Chapter 22, the union bound is the first tool drawn: when it happens to be small, the matter is settled with no further modelling.

Example 21.9 (A six comes, eventually)

Roll a fair die forever and let $B_n = {}$ “at least one six among the first $n$ rolls”, an increasing sequence of events with $\P(B_n) = 1 - (5/6)^n$ . Monotone continuity gives

\P(\text{a six eventually appears}) = \P\Bigl(\bigcup_nB_n\Bigr) = \lim_n\bigl(1 - (5/6)^n\bigr) = 1 .

The point is not the (obvious) limit but the logical step: “eventually” is an event about infinitely many rolls, outside the reach of finite additivity, and monotone continuity — that is, $\sigma$ -additivity — is precisely the axiom that assigns it a probability. Every almost-sure statement in the rest of this book passes through this same narrow door.

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Examples

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