ThmDex – An index of mathematical definitions, results, and conjectures.
Probability calculus expression for conditional expectation given disjoint non-null partition
Formulation 0
Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that
(i) $E_0, E_1, E_2, \ldots \in \mathcal{F}$ are each an D1716: Event in $P$
(ii) $E_0, E_1, E_2, \ldots $ is a D83: Proper set partition of $\Omega$
(iii) \begin{equation} \mathbb{P}(E_0), \mathbb{P}(E_1), \mathbb{P}(E_2), \ldots > 0 \end{equation}
(iv) $\mathcal{G} := \sigma \langle E_0, E_1, E_2, \ldots \rangle$ is a D318: Generated sigma-algebra on $\Omega$ with generators $E_0, E_1, E_2, \ldots$
(v) $X : \Omega \to \mathbb{R}$ is an D3161: Random real number on $P$
(vi) \begin{equation} \mathbb{E} |X| < \infty \end{equation}
Then \begin{equation} \forall \, n \in \mathbb{N} \text{ and } \omega \in \Omega \left( \omega \in E_n \quad \implies \quad \mathbb{E}(X \mid \mathcal{G}) (\omega) \overset{a.s.}{=} \frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)} \right) \end{equation}
Subresults
R4925: Probability calculus expression for conditional probability given disjoint non-null partition
Proofs
Proof 0
Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that
(i) $E_0, E_1, E_2, \ldots \in \mathcal{F}$ are each an D1716: Event in $P$
(ii) $E_0, E_1, E_2, \ldots $ is a D83: Proper set partition of $\Omega$
(iii) \begin{equation} \mathbb{P}(E_0), \mathbb{P}(E_1), \mathbb{P}(E_2), \ldots > 0 \end{equation}
(iv) $\mathcal{G} := \sigma \langle E_0, E_1, E_2, \ldots \rangle$ is a D318: Generated sigma-algebra on $\Omega$ with generators $E_0, E_1, E_2, \ldots$
(v) $X : \Omega \to \mathbb{R}$ is an D3161: Random real number on $P$
(vi) \begin{equation} \mathbb{E} |X| < \infty \end{equation}
Fix $n \in \mathbb{N}$ such that $\omega \in E_n$ for some outcome $\omega \in \Omega$. We show that $\frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)}$ satisfies the properties required of a conditional expectation of $X$ given $\mathcal{G}$. Interpreting the real number $\frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)}$ as a constant map, result R1177: Constant map is always measurable guarantees that it is measurable in $\mathcal{G}$. Next, applying results
(i) R4652: Real-linearity of real expectation
(ii) R2089: Unsigned basic expectation is compatible with probability measure

we have \begin{equation} \mathbb{E} \left( \frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)} I_{E_n} \right) = \frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)} \mathbb{E} \left( I_{E_n} \right) = \frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)} \mathbb{P}(E_n) = \mathbb{E} (X I_{E_n}) \end{equation} This finishes the proof. $\square$