Proof of Conditional Mean-Square Optimality Restricted to an Event of the Conditioning Sigma-Algebra

Throughout, $\mathcal{G}$-measurability of a real-valued function on $\Omega$ means measurability with respect to $\mathcal{G}$ and the Borel $\sigma$-algebra of the real line, and tuples are multiplied by $\mathbf{1}_{G}$ componentwise.

Step 1: pulling the indicator inside the conditional expectation. The indicator $\mathbf{1}_{G}$ is $\mathcal{G}$-measurable by claim 1 of the arithmetic lemma for measurable functions applied on $(\Omega,\mathcal{G})$, and $|\mathbf{1}_{G}|\le1$ everywhere. By claim 4 of the basic properties of conditional expectation (taking out what is known), for each $\gamma\in\{1,\dots,k\}$ the random variables $\mathbf{1}_{G}X^\gamma$ and $\mathbf{1}_{G}M^\gamma$ are square-integrable and $\mathbf{1}_{G}M^\gamma$ is a conditional expectation of $\mathbf{1}_{G}X^\gamma$ given $\mathcal{G}$. Set $X'=\mathbf{1}_{G}X$ and $M'=\mathbf{1}_{G}M$; then $M'$ is a tuple of conditional expectations of the components of $X'$ given $\mathcal{G}$, and its error tuple is $\varepsilon'=X'-M'=\mathbf{1}_{G}\varepsilon$.

Step 2: the tuple $Y'=\mathbf{1}_{G}Y$ is admissible. Fix $\gamma$. By admissibility of $Y$ there is a $\mathcal{G}$-measurable square-integrable random variable $\tilde{Y}^\gamma$ and an event $A_\gamma$ with $P(A_\gamma)=1$ such that $Y^\gamma=\tilde{Y}^\gamma$ on $A_\gamma$ (almost sure equality). Then $\mathbf{1}_{G}Y^\gamma$ is square-integrable, since $(\mathbf{1}_{G}Y^\gamma)^2\le(Y^\gamma)^2$ pointwise and expectation is monotone by monotonicity of the integral; it coincides with $\mathbf{1}_{G}\tilde{Y}^\gamma$ on $A_\gamma$; and $\mathbf{1}_{G}\tilde{Y}^\gamma$ is $\mathcal{G}$-measurable (claim 3 of the arithmetic lemma on $(\Omega,\mathcal{G})$) and square-integrable by the same comparison. Hence $Y'$ is admissible in the sense of the conditional mean-square optimality lemma.

Step 3: claims 1 and 2. Apply the cited lemma to the data $X'$, $M'$, $\varepsilon'$, $R$, and the admissible tuple $Y'$. Pointwise, $Y'-X'=\mathbf{1}_{G}(Y-X)$, $Y'-M'=\mathbf{1}_{G}(Y-M)$, and for any tuple $U$ of square-integrable random variables

$$(\mathbf{1}_{G}U)\cdot\big(R\,(\mathbf{1}_{G}U)\big)=\sum_{\gamma,\delta=1}^{k}R_{\gamma\delta}\,\mathbf{1}_{G}^2\,U^\gamma U^\delta=\mathbf{1}_{G}\,U\cdot(RU),$$

because $\mathbf{1}_{G}^2=\mathbf{1}_{G}$. Therefore claim 1 (orthogonal decomposition) and claim 2 (optimality) of the cited lemma for the primed data read exactly as claims 1 and 2 of the present lemma, and the integrability of all products involved is part of the setting of that lemma (each is a finite linear combination of products of two square-integrable random variables, integrable by the closure properties of the square-integrability definition; see also Step 4).

Step 4: trace form. Pointwise $\mathbf{1}_{G}\,\varepsilon\cdot(R\varepsilon)=\sum_{\gamma,\delta}R_{\gamma\delta}\,(\mathbf{1}_{G}\varepsilon^\gamma)\varepsilon^\delta$, each product being integrable as the product of the square-integrable random variables $\mathbf{1}_{G}\varepsilon^\gamma$ and $\varepsilon^\delta$; the displayed trace formula follows from linearity of the integral applied finitely many times.

Step 5: independence of the choice. Let $\tilde{M}^\gamma$ be any other conditional expectation of $X^\gamma$ given $\mathcal{G}$, and set $\tilde{\varepsilon}^\gamma=X^\gamma-\tilde{M}^\gamma$. By the uniqueness assertion of the existence and uniqueness theorem for conditional expectation, $P(\tilde{M}^\gamma=M^\gamma)=1$; the set $A'_\gamma=\{\tilde{M}^\gamma\neq M^\gamma\}$ is an event, being the set where the random variable $\tilde{M}^\gamma-M^\gamma$ (claim 2 of the arithmetic lemma) is nonzero, it has $P(A'_\gamma)=0$, and off it $\tilde{\varepsilon}^\gamma=\varepsilon^\gamma$. The event $A'=A'_1\cup\dots\cup A'_k$ satisfies $P(A')\le P(A'_1)+\dots+P(A'_k)=0$, by writing $A'$ as the disjoint union of the sets $A'_\gamma\setminus(A'_1\cup\dots\cup A'_{\gamma-1})$ and using additivity and monotonicity of the measure $P$. Off $A'$ all the $\tilde{\varepsilon}^\gamma$ agree with the $\varepsilon^\gamma$. Now fix $\gamma,\delta$ and use the polarization identity, valid pointwise because $\mathbf{1}_{G}^2=\mathbf{1}_{G}$:

$$\mathbf{1}_{G}\,\varepsilon^\gamma\varepsilon^\delta=\tfrac14\Big(\big(\mathbf{1}_{G}(\varepsilon^\gamma+\varepsilon^\delta)\big)^2-\big(\mathbf{1}_{G}(\varepsilon^\gamma-\varepsilon^\delta)\big)^2\Big),$$

and the same identity with $\tilde{\varepsilon}$ in place of $\varepsilon$. The random variables $\mathbf{1}_{G}(\varepsilon^\gamma\pm\varepsilon^\delta)$ and $\mathbf{1}_{G}(\tilde{\varepsilon}^\gamma\pm\tilde{\varepsilon}^\delta)$ are square-integrable and agree off the null event $A'$, so by the lemma on almost sure equality and second moments their second moments coincide. Taking expectations in the two polarization identities and using linearity of the integral gives $\mathbb{E}[\mathbf{1}_{G}\tilde{\varepsilon}^\gamma\tilde{\varepsilon}^\delta]=\mathbb{E}[\mathbf{1}_{G}\varepsilon^\gamma\varepsilon^\delta]$, which is claim 3. $\blacksquare$