ThmDex – An index of mathematical definitions, results, and conjectures.
Expectation of conditional expectation for a random real number
Formulation 0
Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that
(i) $\mathcal{G} \subseteq \mathcal{F}$ is a D470: Subsigma-algebra of $\mathcal{F}$ on $\Omega$
(ii) $X : \Omega \to \mathbb{R}$ is a D3161: Random real number on $P$
(iii) \begin{equation} \mathbb{E} |X| < \infty \end{equation}
Then \begin{equation} \mathbb{E}(\mathbb{E}(X \mid \mathcal{G})) = \mathbb{E}(X) \end{equation}
Proofs
Proof 0
Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that
(i) $\mathcal{G} \subseteq \mathcal{F}$ is a D470: Subsigma-algebra of $\mathcal{F}$ on $\Omega$
(ii) $X : \Omega \to \mathbb{R}$ is a D3161: Random real number on $P$
(iii) \begin{equation} \mathbb{E} |X| < \infty \end{equation}
Since $\mathcal{G}$ is a sigma-algebra on $\Omega$, then $\Omega \in \mathcal{G}$. Since $\mathbb{E}(X \mid \mathcal{G})$ is the conditional expectation of $X$ given $\mathcal{G}$, then \begin{equation} \mathbb{E}(\mathbb{E}(X \mid \mathcal{G}) I_G) = \mathbb{E}(X I_G) \end{equation} for all $G \in \mathcal{G}$. Thus, in particular, we have \begin{equation} \mathbb{E}(\mathbb{E}(X \mid \mathcal{G})) = \mathbb{E}(\mathbb{E}(X \mid \mathcal{G}) I_{\Omega}) = \mathbb{E}(X I_{\Omega}) = \mathbb{E}(X) \end{equation} $\square$