Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Probability axioms ▷ Probability of the empty set

Theorem: The probability of the empty set is zero:

\[\label{eq:prob-emp} P(\emptyset) = 0 \; .\]

Proof: Let $E_i = \emptyset$ for $i = 1,2,\ldots$ Then, $E_i \cap E_j = \emptyset \cap \emptyset = \emptyset$ for $i,j \geq 1$ and $\bigcup_{i=1}^\infty E_i = \bigcup_{i=1}^\infty \emptyset = \emptyset$. Thus, $E_1, E_2, \ldots$ is a countable sequence of disjoint events so that, with the third axiom of probability, it holds that:

\[\label{eq:prob-emp-qed} \begin{split} P\left(\bigcup_{i=1}^\infty E_i \right) &= \sum_{i=1}^\infty P(E_i) \\ P(\emptyset) &= \sum_{i=1}^\infty P(\emptyset) \; . \end{split}\]

Since, by the first axiom of probability, probabilities are non-negative, i.e. $P(\emptyset) \geq 0$, we are searching for a non-negative number which, when added to itself infinitely, is equal to itself. The only such number is zero, i.e. $P(\emptyset) = 0$.

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Metadata: ID: P464 | shortcut: prob-emp2 | author: JoramSoch | date: 2024-08-08, 11:41.