Factorisation of Random Variables Through a Measurable Map, the Variational Form of the Mean-Square Filtering Error, and Its Invariance Under the Joint Law

Let $(\Omega,\mathcal{F},P)$ be a probability space with expectation $\mathbb{E}$, let $(\mathsf{Y},\mathcal{Y})$ be a measurable space, and let $\mathsf{D}:\Omega\to\mathsf{Y}$ be measurable with respect to $\mathcal{F}$ and $\mathcal{Y}$. Write $\mathcal{B}(\mathbb{R})$ for the Borel $\sigma$-algebra of the real line; a real-valued map on a measurable space is called measurable when it is measurable with respect to the named $\sigma$-algebra and $\mathcal{B}(\mathbb{R})$. Write $\mathcal{B}(\mathbb{R})\otimes\mathcal{Y}$ for the product $\sigma$-algebra on $\mathbb{R}\times\mathsf{Y}$, $\mathbf{1}_{A}$ for the indicator of a set $A$, and $\lVert\cdot\rVert_{2}$ for the mean-square norm. Put

$$\sigma(\mathsf{D})=\bigl\{\mathsf{D}^{-1}(B):B\in\mathcal{Y}\bigr\},\qquad \mathcal{N}=\{A\in\mathcal{F}:P(A)=0\} .$$

Call a $\sigma$-algebra $\mathcal{G}$ on $\Omega$ $\mathsf{D}$-generated up to null sets when $\sigma(\mathsf{D})\subseteq\mathcal{G}\subseteq\mathcal{F}$ and every $A\in\mathcal{G}$ admits $A'\in\sigma(\mathsf{D})$ with $P\bigl((A\setminus A')\cup(A'\setminus A)\bigr)=0$. Then the following hold.

1. (The generated $\sigma$-algebra.) $\sigma(\mathsf{D})$ is a $\sigma$-algebra on $\Omega$ with $\sigma(\mathsf{D})\subseteq\mathcal{F}$, and $\mathsf{D}$ is measurable with respect to $\sigma(\mathsf{D})$ and $\mathcal{Y}$. Moreover the $\sigma$-algebra generated by $\sigma(\mathsf{D})\cup\mathcal{N}$ is exactly the family

$$\mathcal{H}=\bigl\{A\in\mathcal{F}:\ P(A\triangle A')=0\ \text{for some }A'\in\sigma(\mathsf{D})\bigr\},\qquad A\triangle A'=(A\setminus A')\cup(A'\setminus A);$$

it is $\mathsf{D}$-generated up to null sets, and it contains every $\sigma$-algebra on $\Omega$ that is $\mathsf{D}$-generated up to null sets. Thus a $\sigma$-algebra $\mathcal{G}$ is $\mathsf{D}$-generated up to null sets if and only if $\sigma(\mathsf{D})\subseteq\mathcal{G}\subseteq\mathcal{H}$.

2. (Factorisation.) A map $Z:\Omega\to\mathbb{R}$ is measurable with respect to $\sigma(\mathsf{D})$ if and only if there is a $\mathcal{Y}$-measurable $g:\mathsf{Y}\to\mathbb{R}$ with $Z=g\circ\mathsf{D}$, that is, $Z(\omega)=g(\mathsf{D}(\omega))$ for every $\omega\in\Omega$.

3. (Conditioning on a $\mathsf{D}$-generated $\sigma$-algebra.) Let $\mathcal{G}$ be $\mathsf{D}$-generated up to null sets and let $X$ be a square-integrable random variable on $(\Omega,\mathcal{F},P)$. Then every conditional expectation of $X$ given $\sigma(\mathsf{D})$ is a conditional expectation of $X$ given $\mathcal{G}$, and any two conditional expectations, one given $\sigma(\mathsf{D})$ and one given $\mathcal{G}$, are almost surely equal. In particular

$$\mathbb{E}\Bigl[\bigl(X-\mathbb{E}[X\mid\mathcal{G}]\bigr)^{2}\Bigr]=\mathbb{E}\Bigl[\bigl(X-\mathbb{E}[X\mid\sigma(\mathsf{D})]\bigr)^{2}\Bigr],$$

both sides being independent of the choice of conditional expectations.

4. (Variational form of the mean-square filtering error.) Let $X$ be square-integrable and let $\mathcal{G}$ be $\mathsf{D}$-generated up to null sets. Write $\mathcal{M}$ for the set of $\mathcal{Y}$-measurable maps $g:\mathsf{Y}\to\mathbb{R}$ for which $g\circ\mathsf{D}$ is square-integrable; $\mathcal{M}$ is nonempty, containing the zero map. Then the set of real numbers $\bigl\{\mathbb{E}[(X-g\circ\mathsf{D})^{2}]:g\in\mathcal{M}\bigr\}$ has a greatest lower bound, this bound is attained, and

$$\mathbb{E}\Bigl[\bigl(X-\mathbb{E}[X\mid\mathcal{G}]\bigr)^{2}\Bigr]=\inf_{g\in\mathcal{M}}\mathbb{E}\bigl[(X-g\circ\mathsf{D})^{2}\bigr].$$

5. (Invariance under the joint law.) Let $X$ be a square-integrable random variable on $(\Omega,\mathcal{F},P)$, let $(\Omega',\mathcal{F}',P')$ be a second probability space with expectation $\mathbb{E}'$, let $\mathsf{D}':\Omega'\to\mathsf{Y}$ be measurable with respect to $\mathcal{F}'$ and $\mathcal{Y}$, and let $X'$ be a square-integrable random variable on it. The maps $(X,\mathsf{D})$ and $(X',\mathsf{D}')$ into $\mathbb{R}\times\mathsf{Y}$ are measurable with respect to $\mathcal{F}$, respectively $\mathcal{F}'$, and $\mathcal{B}(\mathbb{R})\otimes\mathcal{Y}$. Assume that the image measure of $P$ under $(X,\mathsf{D})$ equals the image measure of $P'$ under $(X',\mathsf{D}')$ on $\mathcal{B}(\mathbb{R})\otimes\mathcal{Y}$. Let $\mathcal{G}$ be $\mathsf{D}$-generated up to null sets in $(\Omega,\mathcal{F},P)$ and let $\mathcal{G}'$ be $\mathsf{D}'$-generated up to null sets in $(\Omega',\mathcal{F}',P')$. Then

$$\mathbb{E}\Bigl[\bigl(X-\mathbb{E}[X\mid\mathcal{G}]\bigr)^{2}\Bigr]=\mathbb{E}'\Bigl[\bigl(X'-\mathbb{E}'[X'\mid\mathcal{G}']\bigr)^{2}\Bigr].$$