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

Result R3404 on D2012: Conditional probability

Bayes' theorem in the case of two events

Formulation 0

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

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

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

Formulation 1

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

(i)	$\text{Evidence}, \text{Hypothesis} \in \mathcal{F}$ are each an D1716: Event in $P$
(ii)	\begin{equation} \mathbb{P} (\text{Evidence}), \mathbb{P} (\text{Hypothesis}) > 0 \end{equation}

Then \begin{equation} \mathbb{P}(\text{Hypothesis} \mid \text{Evidence}) \mathbb{P}(\text{Evidence}) = \mathbb{P}(\text{Evidence} \mid \text{Hypothesis}) \mathbb{P}(\text{Hypothesis}) \end{equation}

Formulation 2

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

(i)	$\text{Observation}, \text{Assumption} \in \mathcal{F}$ are each an D1716: Event in $P$
(ii)	\begin{equation} \mathbb{P} (\text{Observation}), \mathbb{P} (\text{Assumption}) > 0 \end{equation}

Then \begin{equation} \mathbb{P}(\text{Assumption} \mid \text{Observation}) \mathbb{P}(\text{Observation}) = \mathbb{P}(\text{Observation} \mid \text{Assumption}) \mathbb{P}(\text{Assumption}) \end{equation}

Formulation 3

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

(i)	$\text{Measurement}, \text{Hypothesis} \in \mathcal{F}$ are each an D1716: Event in $P$
(ii)	\begin{equation} \mathbb{P} (\text{Measurement}), \mathbb{P} (\text{Hypothesis}) > 0 \end{equation}

Then \begin{equation} \mathbb{P}(\text{Hypothesis} \mid \text{Measurement}) \mathbb{P}(\text{Measurement}) = \mathbb{P}(\text{Measurement} \mid \text{Hypothesis}) \mathbb{P}(\text{Hypothesis}) \end{equation}

Proofs

Proof 0

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

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

Using the definition of an D5811: Event-conditional probability, we have \begin{equation} \mathbb{P}(E \mid F) = \frac{\mathbb{P}(E \cap F)}{\mathbb{P}(F)} \end{equation} and thus $\mathbb{P}(E \cap F) = \mathbb{P}(E \mid F) \mathbb{P}(F)$. But now we can switch the places of $E$ and $F$ to obtain \begin{equation} \mathbb{P}(E \mid F) \mathbb{P}(F) = \mathbb{P}(E \cap F) = \mathbb{P}(F \mid E) \mathbb{P}(E) \end{equation} This is precisely what was required to be shown. $\square$