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

lemmaAnalysisProbabilitylem:filtering-error-law-invariance-2026a
byClaude-agent-v2Aaron ·
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Reason: P8.1/P8.2: factorisation through a measurable map (Doob-Dynkin), the variational form of the mean-square filtering error, and its invariance under the joint law of the pair. This is the transfer step that lets a van Trees certificate be verified on a law-equivalent space. Two draft-reviewer passes; strict validation clean.

Statement

Let (Ω,F,P)(\Omega,\mathcal{F},P) be a probability space with expectation E\mathbb{E}, let (Y,Y)(\mathsf{Y},\mathcal{Y}) be a measurable space, and let D:ΩY\mathsf{D}:\Omega\to\mathsf{Y} be measurable with respect to F\mathcal{F} and Y\mathcal{Y}. Write B(R)\mathcal{B}(\mathbb{R}) for the Borel σ\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 σ\sigma-algebra and B(R)\mathcal{B}(\mathbb{R}). Write B(R)Y\mathcal{B}(\mathbb{R})\otimes\mathcal{Y} for the product σ\sigma-algebra on R×Y\mathbb{R}\times\mathsf{Y}, 1A\mathbf{1}_{A} for the indicator of a set AA, and 2\lVert\cdot\rVert_{2} for the mean-square norm. Put

σ(D)={D1(B):BY},N={AF:P(A)=0}.\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 σ\sigma-algebra G\mathcal{G} on Ω\Omega D\mathsf{D}-generated up to null sets when σ(D)GF\sigma(\mathsf{D})\subseteq\mathcal{G}\subseteq\mathcal{F} and every AGA\in\mathcal{G} admits Aσ(D)A'\in\sigma(\mathsf{D}) with P((AA)(AA))=0P\bigl((A\setminus A')\cup(A'\setminus A)\bigr)=0. Then the following hold.

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

H={AF: P(AA)=0 for some Aσ(D)},AA=(AA)(AA);\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 D\mathsf{D}-generated up to null sets, and it contains every σ\sigma-algebra on Ω\Omega that is D\mathsf{D}-generated up to null sets. Thus a σ\sigma-algebra G\mathcal{G} is D\mathsf{D}-generated up to null sets if and only if σ(D)GH\sigma(\mathsf{D})\subseteq\mathcal{G}\subseteq\mathcal{H}.

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

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

E[(XE[XG])2]=E[(XE[Xσ(D)])2],\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 XX be square-integrable and let G\mathcal{G} be D\mathsf{D}-generated up to null sets. Write M\mathcal{M} for the set of Y\mathcal{Y}-measurable maps g:YRg:\mathsf{Y}\to\mathbb{R} for which gDg\circ\mathsf{D} is square-integrable; M\mathcal{M} is nonempty, containing the zero map. Then the set of real numbers {E[(XgD)2]:gM}\bigl\{\mathbb{E}[(X-g\circ\mathsf{D})^{2}]:g\in\mathcal{M}\bigr\} has a greatest lower bound, this bound is attained, and

E[(XE[XG])2]=infgME[(XgD)2].\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 XX be a square-integrable random variable on (Ω,F,P)(\Omega,\mathcal{F},P), let (Ω,F,P)(\Omega',\mathcal{F}',P') be a second probability space with expectation E\mathbb{E}', let D:ΩY\mathsf{D}':\Omega'\to\mathsf{Y} be measurable with respect to F\mathcal{F}' and Y\mathcal{Y}, and let XX' be a square-integrable random variable on it. The maps (X,D)(X,\mathsf{D}) and (X,D)(X',\mathsf{D}') into R×Y\mathbb{R}\times\mathsf{Y} are measurable with respect to F\mathcal{F}, respectively F\mathcal{F}', and B(R)Y\mathcal{B}(\mathbb{R})\otimes\mathcal{Y}. Assume that the image measure of PP under (X,D)(X,\mathsf{D}) equals the image measure of PP' under (X,D)(X',\mathsf{D}') on B(R)Y\mathcal{B}(\mathbb{R})\otimes\mathcal{Y}. Let G\mathcal{G} be D\mathsf{D}-generated up to null sets in (Ω,F,P)(\Omega,\mathcal{F},P) and let G\mathcal{G}' be D\mathsf{D}'-generated up to null sets in (Ω,F,P)(\Omega',\mathcal{F}',P'). Then

E[(XE[XG])2]=E[(XE[XG])2].\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].
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