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 $A$ and $B$ be two events fulfilling $A \subseteq B$. Set $E_1 = A$, $E_2 = B \setminus A$ and $E_i = \emptyset$ for $i \geq 3$. Then, the sets $E_i$ are pairwise disjoint and $E_1 \cup E_2 \cup \ldots = B$. Thus, from the third axiom of probability, we have:

\[\label{eq:pB} P(B) = P(A) + P(B \setminus A) + \sum_{i=3}^\infty P(E_i) \; .\]

Assume that the probability of the empty set is not zero, i.e. $P(\emptyset) > 0$. Then, the right-hand side of \eqref{eq:pB} would be infinite. However, by the first axiom of probability, the left-hand side must be finite. This is a contradiction. Therefore, $P(\emptyset) = 0$.

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Metadata: ID: P244 | shortcut: prob-emp | author: JoramSoch | date: 2021-07-30, 11:58.