Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Probability axioms ▷ Monotonicity of probability

Theorem: Probability is monotonic, i.e. if $A$ is a subset of or equal to $B$, then the probability of $A$ is smaller than or equal to $B$:

\[\label{eq:prob-mon} A \subseteq B \quad \Rightarrow \quad P(A) \leq P(B) \; .\]

Proof: When $A \subseteq B$, then $B$ is equal to the union of $A$ and the intersection of $B$ with the complement of $A$:

\[\label{eq:A-cup} B = A \cup (B \cap A^\mathrm{c}) \; .\]

Moreover, the intersection of A and the intersection of $B$ with the complement of $A$ is equal to the empty set:

\[\label{eq:A-cap} A \cap (B \cap A^\mathrm{c}) = \emptyset \; .\]

Thus, the third axiom of probability implies:

\[\label{eq:prob-mon-qed} \begin{split} P(B) &= P(A) + P(B \cap A^\mathrm{c}) \\ P(A) &= P(B) - P(B \cap A^\mathrm{c}) \; . \end{split}\]

Since $P(B \cap A^\mathrm{c}) \geq 0$ by the first axiom of probability, it must hold that $P(A) \leq P(B)$.

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Metadata: ID: P465 | shortcut: prob-mon2 | author: JoramSoch | date: 2024-08-08, 11:46.