Proof: Probability of exhaustive events
Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Probability axioms ▷ Probability of exhaustive events
Metadata: ID: P249 | shortcut: prob-exh | author: JoramSoch | date: 2021-08-08, 04:10.
Theorem: Let $B_1, \ldots, B_n$ be mutually exclusive and collectively exhaustive subsets of a sample space $\Omega$. Then, their total probability is one:
\[\label{eq:prob-exh} \sum_i P(B_i) = 1 \; .\]Proof: Because all $B_i$ are mutually exclusive, we have:
\[\label{eq:B-exclusive} B_i \cap B_j = \emptyset \quad \text{for all} \quad i \neq j \; .\]Because the $B_i$ are collectively exhaustive, we have:
\[\label{eq:B-exhaustive} \cup_i \, B_i = \Omega \; .\]Thus, the third axiom of probability implies that
\[\label{eq:prob-exh-s1} \sum_i P(B_i) = P(\Omega) \; .\]and the second axiom of probability implies that
\[\label{eq:prob-exh-s2} \sum_i P(B_i) = 1 \; .\]∎
Sources: - Alan Stuart & J. Keith Ord (1994): "Probability and Statistical Inference"; in: Kendall's Advanced Theory of Statistics, Vol. 1: Distribution Theory, pp. 288-289; URL: https://www.wiley.com/en-us/Kendall%27s+Advanced+Theory+of+Statistics%2C+3+Volumes%2C+Set%2C+6th+Edition-p-9780470669549.
- Wikipedia (2021): "Probability axioms"; in: Wikipedia, the free encyclopedia, retrieved on 2021-08-08; URL: https://en.wikipedia.org/wiki/Probability_axioms#Axioms.
Metadata: ID: P249 | shortcut: prob-exh | author: JoramSoch | date: 2021-08-08, 04:10.