Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Cumulative distribution function ▷ Exceedance probability based on cumulative distribution function

Theorem: Let $X$ be a random variable with possible values $\mathcal{X}$ and cumulative distribution function $F_X(x)$. Then, the exceedance probability that $X$ is larger than some value $x$ is

\[\label{eq:cdf-prob-exc} \mathrm{Pr}(X > x) = 1 - F_X(x) \; .\]

Proof: Note that $\left\lbrace X \mid X > x \right\rbrace$ and $\left\lbrace X \mid X \leq x \right\rbrace$ are disjoint sets

\[\label{eq:intersection} \left\lbrace X \mid X > x \right\rbrace \cap \left\lbrace X \mid X \leq x \right\rbrace = \emptyset\]

and that they comprise the set of all outcomes, i.e. the sample space:

\[\label{eq:union} \left\lbrace X \mid X > x \right\rbrace \cup \left\lbrace X \mid X \leq x \right\rbrace = \mathcal{X} = \Omega \; .\]

Using the second axiom of probability, we have:

\[\label{eq:cdf-prob-exc-s1} \begin{split} P(\Omega) &= 1 \\ P\left( \left\lbrace X \mid X > x \right\rbrace \cup \left\lbrace X \mid X \leq x \right\rbrace \right) &= 1 \; . \end{split}\]

Using the third axiom of probability, we get:

\[\label{eq:cdf-prob-exc-s2} \begin{split} P\left( \left\lbrace X \mid X > x \right\rbrace \right) + P\left( \left\lbrace X \mid X \leq x \right\rbrace \right) &= 1 \\ P\left( \left\lbrace X \mid X > x \right\rbrace \right) &= 1 - P\left( \left\lbrace X \mid X \leq x \right\rbrace \right) \\ \mathrm{Pr}(X > x) &= 1 - \mathrm{Pr}(X \leq x) \; . \end{split}\]

Using the definition of the cumulative distribution function, we finally have:

\[\label{eq:cdf-prob-exc-qed} \mathrm{Pr}(X > x) = 1 - F_X(x) \; .\]
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Metadata: ID: P466 | shortcut: cdf-probexc | author: JoramSoch | date: 2024-09-06, 10:27.