ThmDex – An index of mathematical definitions, results, and conjectures.
Conditional expectation of known random complex 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) $Z : \Omega \to \mathbb{C}$ be a D4877: Random complex number on $P$
(iii) \begin{equation} \mathbb{E} |Z| < \infty \end{equation}
(iv) \begin{equation} Z \in \mathcal{G} \end{equation}
Then \begin{equation} \mathbb{E}(Z \mid \mathcal{G}) \overset{a.s.}{=} Z \end{equation}
Formulation 1
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) $Z : \Omega \to \mathbb{C}$ be a D4877: Random complex number on $P$
(iii) \begin{equation} \mathbb{E} |Z| < \infty \end{equation}
(iv) \begin{equation} \sigma_{\text{pullback}} \langle Z \rangle \subseteq \mathcal{G} \end{equation}
Then \begin{equation} \mathbb{P}( \mathbb{E}(Z \mid \mathcal{G}) = Z) = 1 \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) $Z : \Omega \to \mathbb{C}$ be a D4877: Random complex number on $P$
(iii) \begin{equation} \mathbb{E} |Z| < \infty \end{equation}
(iv) \begin{equation} Z \in \mathcal{G} \end{equation}
We show that under these assumptions, $Z$ itself is a version of the conditional expectation $\mathbb{E}(Z \mid \mathcal{G})$ by confirming that the required properties hold. We have assumed that $Z$ is measurable in $\mathcal{G}$ which takes care of the the first condition. Next, fixing an event $G \in \mathcal{G}$ and applying the results
(i) R2150: Expectation of conditional expectation for a random euclidean real number
(ii) R2549: Conditional expectation of random complex product when factor is known

we have \begin{equation} \begin{split} \mathbb{E}(Z I_G) = \mathbb{E}(\mathbb{E}(Z I_G \mid \mathcal{G})) = \mathbb{E}(\mathbb{E}(Z \mid \mathcal{G}) I_G) \end{split} \end{equation} This completes the proof. $\square$