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 D5143: Set partition] of $\Omega$
|
| (iii) |
$\mathcal{G} = \sigma \langle E_0, E_1, E_2, \ldots \rangle$ is a D318: Generated sigma-algebra on $\Omega$ with generators $E_0, E_1, E_2, \ldots$
|
| (iv) |
$X : \Omega \to \mathbb{R}$ is a D3161: Random real number on $P$
|
| (v) |
\begin{equation}
\mathbb{E} |X| < \infty
\end{equation}
|
Let $n \in \mathbb{N}$ be a
D996: Natural number such that
| (i) |
\begin{equation}
\mathbb{P}(E_n) > 0
\end{equation}
|
We show that the basic real number $\frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)}$ equals the conditional expectation $\mathbb{E}(X \mid \mathcal{G})$ on the event $E_n$. If $\frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)}$ is understood as a constant map, then result
R1177: Constant map is always measurable shows that it is measurable in $\mathcal{G}$, which is the first condition required of a conditional expectation.
Next, applying results
we have
\begin{equation}
\int_{E_n} \frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)} \, d \mathbb{P} = \frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)} \int_{E_n} \, d \mathbb{P} = \frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)} \mathbb{P}(E_n) = \mathbb{E}(X I_{E_n}) = \int_{E_n} X \, d \mathbb{P}
\end{equation}
This finishes the proof. $\square$
