ThmDex – An index of mathematical definitions, results, and conjectures.
Probability calculus expression for probability conditioned on event of nonzero probability
Formulation 0
Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space.
Let $E, F \in \mathcal{F}$ each be an D1716: Event in $P$ such that
(i) $\mathbb{P}(F) > 0$
Then \begin{equation} \mathbb{P}(E \mid F) = \frac{\mathbb{P}(E \cap F)}{\mathbb{P}(F)} \end{equation}
Proofs
Proof 0
Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space.
Let $E, F \in \mathcal{F}$ each be an D1716: Event in $P$ such that
(i) $\mathbb{P}(F) > 0$
Since $\mathbb{P}(F) > 0$, then $F \neq \emptyset$. If $F = \Omega$, then result R4338: Conditional probability of almost surely true event yields the claim, so we may further assume that $F \neq \Omega$. Under this assumption, the pair $F, \Omega \setminus F$ is a disjoint partition of $\Omega$ and applying results
(i) R2089: Unsigned basic expectation is compatible with probability measure
(ii) R4333: Binary product of indicator functions equals indicator of intersection
(iii) R3401: Probability calculus expression for conditional expectation given disjoint non-null partition

we have \begin{equation} \begin{split} \frac{\mathbb{P}(E \cap F)}{\mathbb{P}(F)} = \frac{\mathbb{E}(I_{E \cap F})}{\mathbb{P}(F)} & = \frac{\mathbb{E}(I_E I_F)}{\mathbb{P}(F)} \\ & = \mathbb{E}(I_E \mid \sigma \langle F \rangle) & = \mathbb{P}(E \mid \sigma \langle F \rangle) = \mathbb{P}(E \mid F) \end{split} \end{equation} where the last two equalities hold by definition. This is what was required to be shown. $\square$