Posterior-predictive mean and variance decomposition

Let $s_2(\theta )=\int z^2{Q}_{\theta }(dz)$. It is measurable by [F4]. Since $\int s_{2}d{\mathrm{\Pi }}_y<\mathrm{\infty }$, the set $E_y=\{s_2<\mathrm{\infty }\}$ is measurable and has ${\mathrm{\Pi }}_y(E_y)=1$. On $E_y$, $\mathrm{\mid }z\mathrm{\mid }\le 1+z^2$ gives a finite first absolute moment, so $\mu (\theta )$ is finite and measurable by [F4]. Expanding the square in the variance definition [F3] gives $v(\theta )=s_2(\theta )-\mu (\theta {)}^2\ge 0$, hence $\mu (\theta {)}^2\le s_2(\theta )$; therefore $\mu$ is square-integrable under ${\mathrm{\Pi }}_y$ and $v$ is integrable.

For the joint-model hypotheses in the Statement, let $B$ be a measurable future event and $G\in \mathcal{G}$. The function $\theta \mapsto Q_{\theta }(B)$ is measurable by [F9]. Apply [F5] to the posterior conditional law and this bounded function to obtain ${\int }_G{Q}_Y^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}(B)dP={\int }_G{Q}_X(B)dP$. Since $G\in \mathcal{G}\subseteq \mathcal{H}$, apply [F5] again to the conditional law $Q_X$ to get ${\int }_G{Q}_X(B)dP=P(G\cap \{Z\in B\})$. By [F2], the mixture has probability-measure sections and measurable evaluations; these event identities therefore make $Q_Y^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}$ a regular conditional distribution of $Z$ given $\mathcal{G}$ under [F6].

By [F1], $Q_y^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}$ is the mixture defined from $Q_{\theta }$ and ${\mathrm{\Pi }}_y$. Apply [F2] to $g(z)=z^2$ to obtain $\int z^2{Q}_y^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}(dz)=\int s_{2}d{\mathrm{\Pi }}_y<\mathrm{\infty }$. Apply [F2] to $\mathrm{\mid }z\mathrm{\mid }$ and use $\mathrm{\mid }z\mathrm{\mid }\le 1+z^2$ to see that $Q_y^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}$ has finite first absolute moment. Applying [F2] to the nonnegative positive and negative parts of $z$ and subtracting their finite integrals gives $\int z{Q}_y^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}(dz)=\int \mu d{\mathrm{\Pi }}_y$.

Apply [F5] to $z$ and $z^2$ under $Q_X$. Writing $m:=\mu (X)$ and $s:=s_2(X)$, their conditional-integration identities give $\mathbb{E}[1_{A}Z]=\mathbb{E}[1_{A}m]$ and $\mathbb{E}[1_A{Z}^2]=\mathbb{E}[1_{A}s]$ for every $A\in \mathcal{H}$, and $\mathbb{E}[s]=\mathbb{E}[Z^2]<\mathrm{\infty }$. Applying [F5] to $s_2$ under ${\mathrm{\Pi }}_Y$ gives a conditional version of $s$ with expectation $\mathbb{E}[s]$; hence $D:=\{y:\int s_2(\theta ){\mathrm{\Pi }}_y(d\theta )<\mathrm{\infty }\}$ has full probability. Take the zero-filled representative $\bar{s}(Y):=\int s_2(\theta ){\mathrm{\Pi }}_y(d\theta )$ on $D$ and zero elsewhere. Then $\mathbb{E}[\bar{s}(Y)]=\mathbb{E}[s]=\mathbb{E}[Z^2]<\mathrm{\infty }$ and ${\mathrm{\Pi }}_Y(E)=1$ almost surely for $E=\{\theta :s_2(\theta )<\mathrm{\infty }\}$. The conditioning identity for $X$ then gives $P(X\in E)=1$. By step 1.1, $m^2\le s$ and $0\le v(X)\le s$ almost surely; hence $m\in L^2$ and $v(X)$ is integrable, with $v(X)=s-m^2$ almost surely. Because $X$ is $\mathcal{H}$-measurable and these moment functions are measurable, $m,s,v(X)$ are $\mathcal{H}$-measurable. The first event identity extends from bounded simple $\mathcal{H}$-measurable multipliers to $1_{A}m$ by simple approximation and truncation; $2\mathrm{\mid }mZ\mathrm{\mid }\le m^2+Z^2$ and $m^2\in L^1$ justify the limits. Thus $\mathbb{E}[1_{A}mZ]=\mathbb{E}[1_A{m}^2]$, and expansion gives $\mathbb{E}[1_A(Z-m{)}^2]=\mathbb{E}[1_A(s-m^2)]=\mathbb{E}[1_{A}v(X)]$ for every $A\in \mathcal{H}$. This proves eventwise that $m$ is the conditional-mean version of $Z$ and $v(X)$ is the conditional-variance version given $\mathcal{H}$.

On $E_y$, $s_2=v+{\mu }^2$, and off $E_y$ the set is ${\mathrm{\Pi }}_y$-null. By [F3] and step 2.1, ${\mathrm{Var}}_{Q_y^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}}(Z)=\int s_{2}d{\mathrm{\Pi }}_y-(\int \mu d{\mathrm{\Pi }}_y{)}^2=\int vd{\mathrm{\Pi }}_y+\int {\mu }^{2}d{\mathrm{\Pi }}_y-(\int \mu d{\mathrm{\Pi }}_y{)}^2=\int vd{\mathrm{\Pi }}_y+{\mathrm{Var}}_{{\mathrm{\Pi }}_y}(\mu )$. This proves both displayed kernel identities without choosing conditional-expectation representatives.

On the full-measure data domain where $b(Y):=\int s_{2}d{\mathrm{\Pi }}_Y<\mathrm{\infty }$, set $c(Y):=\int \mu d{\mathrm{\Pi }}_Y$, $d(Y):=\int {\mu }^{2}d{\mathrm{\Pi }}_Y$, and $e(Y):=\int vd{\mathrm{\Pi }}_Y$, and extend $b,c,d,e$ by zero on its null complement, using the signed zero-filled extension for $c$. By [F5] and the kernel identities in steps 2.1 and 3.1, $c,b$ are conditional first- and second-moment versions of $Z$ given $\mathcal{G}$, while $c,d,e$ are conditional moments of $m,m^2,v(X)$ given $\mathcal{G}$. Since $Y$ is $\mathcal{G}$-measurable, these functions of $Y$ are $\mathcal{G}$-measurable. Also $\mathbb{E}[b(Y)]=\mathbb{E}[Z^2]<\mathrm{\infty }$, $c^2\le b$ almost surely since $b-c^2={\mathrm{Var}}_{Q_Y^{\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}}}(Z)\ge 0$, and $d\le b<\mathrm{\infty }$ almost surely since ${\mu }^2\le s_2$.

For each $A\in \mathcal{G}$, the same simple-multiplier and truncation argument as in step 2.2, now applied to $m$ with center $c(Y)$ and to $Z$ with that same center, gives $\mathbb{E}[1_A(m-c(Y){)}^2]=\mathbb{E}[1_A(d-c(Y{)}^2)]$ and $\mathbb{E}[1_A(Z-c(Y){)}^2]=\mathbb{E}[1_A(b(Y)-c(Y{)}^2)]$; the bounds $2\mathrm{\mid }cm\mathrm{\mid }\le c^2+m^2$ and $2\mathrm{\mid }cZ\mathrm{\mid }\le c^2+Z^2$ justify the respective truncations. Thus $d-c^2$ and $b-c^2$ are the conditional-variance versions of $m$ and $Z$ given $\mathcal{G}$. The kernel identity in step 3.1 gives $b-c^2=e+(d-c^2)$, so the conditional variance of $Z$ given $\mathcal{G}$ is the expected conditional variance given $\mathcal{H}$ plus the conditional variance of the conditional mean. These event-identity versions are choice-free; under [A1] they may also be read as conditional-expectation equivalence classes by [F5], while [F8] gives the eventwise version convention. No value on a null data fiber is thereby made canonical.