ThmDex – An index of mathematical definitions, results, and conjectures.
Real conditional covariance partition into conditional moments
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) $X, Y : \Omega \to \mathbb{R}$ are each a D3161: Random real number on $P$
(iii) \begin{equation} \mathbb{E} |X|^2, \mathbb{E} |Y|^2 < \infty \end{equation}
Then \begin{equation} \text{Cov}(X, Y \mid \mathcal{G}) = \mathbb{E}(X Y \mid \mathcal{G}) + \mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) \end{equation}
Subresults
R4787
R3562: Real conditional variance partition into conditional moments
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) $X, Y : \Omega \to \mathbb{R}$ are each a D3161: Random real number on $P$
(iii) \begin{equation} \mathbb{E} |X|^2, \mathbb{E} |Y|^2 < \infty \end{equation}
By definition, $\mathbb{E}(X \mid \mathcal{G})$ and $\mathbb{E}(Y \mid \mathcal{G})$ are measurable in $\mathcal{G}$. Since we have \begin{equation} \begin{split} (X - \mathbb{E}(X \mid \mathcal{G})) (Y - \mathbb{E}(Y \mid \mathcal{G})) = X Y - X \mathbb{E}(Y \mid \mathcal{G}) - Y \mathbb{E}(X \mid \mathcal{G}) + \mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) \end{split} \end{equation} then, applying results
(i) R4784: Basic real linearity of basic real conditional expectation
(ii) R4781: Conditional expectation of known random real number

we have \begin{equation} \begin{split} \text{Cov}(X, Y \mid \mathcal{G}) & = \mathbb{E}( (X - \mathbb{E}(X \mid \mathcal{G})) (Y - \mathbb{E}(Y \mid \mathcal{G})) \mid \mathcal{G}) \\ & = \mathbb{E}(X Y \mid \mathcal{G}) - \mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) - \mathbb{E}(Y \mid \mathcal{G}) \mathbb{E}(X \mid \mathcal{G}) + \mathbb{E}(\mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) \mid \mathcal{G}) \\ & = \mathbb{E}(X Y \mid \mathcal{G}) - 2 \mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) + \mathbb{E}(\mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) \mid \mathcal{G}) \\ & = \mathbb{E}(X Y \mid \mathcal{G}) - 2 \mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) + \mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) \\ & = \mathbb{E}(X Y \mid \mathcal{G}) - \mathbb{E}(X \mid \mathcal{G}) \mathbb{E}(Y \mid \mathcal{G}) \\ \end{split} \end{equation} This is what was required to be shown. $\square$