ThmDex – An index of mathematical definitions, results, and conjectures.

Result R4340 on D2012: Conditional probability
Subresult of R3404: Bayes' theorem in the case of two events

Bayes' theorem in the case of event and complement

Formulation 0

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$E \in \mathcal{F}$ is an D1716: Event in $P$
(ii)	\begin{equation} \mathbb{P}(E) \in (0, 1) \end{equation}

Then \begin{equation} \mathbb{P}(E \mid E^{\complement}) \mathbb{P}(E^{\complement}) = \mathbb{P}(E^{\complement} \mid E) \mathbb{P}(E) \end{equation}

Proofs

Proof 0

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$E \in \mathcal{F}$ is an D1716: Event in $P$
(ii)	\begin{equation} \mathbb{P}(E) \in (0, 1) \end{equation}

Since $\mathbb{P}(E) \in (0, 1)$ ensures that $\mathbb{P}(E) > 0$ and $\mathbb{P}(E^{\complement}) > 0$, this result is a particular case of R3404: Bayes' theorem in the case of two events. $\square$