ThmDex – An index of mathematical definitions, results, and conjectures.
Real conditional expectation is absolutely integrable
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} \left| \mathbb{E}(X \mid \mathcal{G}) \right| < \infty \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}
Result R2150: Expectation of conditional expectation for a random euclidean real number shows that $\mathbb{E} (\mathbb{E}(X \mid \mathcal{G})) = \mathbb{E} X$. Thus, applying result R3582: Signed expectation finite iff absolutely integrable random basic number twice, we have the chain of equivalencies \begin{equation} \begin{split} \mathbb{E}|\mathbb{E}(X \mid \mathcal{G})| < \infty \quad & \iff \quad \mathbb{E} (\mathbb{E}(X \mid \mathcal{G})) = \mathbb{E} X \in (-\infty, \infty) \quad & \iff \quad \mathbb{E} |X| < \infty \end{split} \end{equation} We have assumed the rightmost expression and hence the proof is complete. $\square$