Proof of The Conditional Law of the Head Given the Tail for a Measure with a Density Relative to a Diagonal Gaussian Measure on a Hilbert Space

Each result cited is universally quantified over the data in its own statement.

Write $\gamma=\gamma_{c^{(n)}}$, a probability measure on $\mathcal{B}(\mathbb{R}^{n})$ by Diagonal Gaussian Measures on Euclidean Space §measure, and $\nu^{0}=(Q_{n})_{\#}\gamma_{c}$, a probability measure on $\mathcal{B}(X)$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables, $Q_{n}$ being Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity. Let $d_{\times}$ be the product metric on $X\times\mathbb{R}^{n}$, a metric by claim 1 of The Product Metric is a Metric, so that $d_{\times}((w,u),(w',u'))=\max\{|w-w'|,\lVert u-u'\rVert\}$, using $d_{E}(u,u')=\lVert u-u'\rVert$ from Euclidean Space and Lebesgue Measure: Standing Notation §space; its Borel $\sigma$-algebra is $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$, as recalled in the statement. That $f$ is a density of $\mu$ with respect to $\gamma_{c}$ means, by The Radon-Nikodym Theorem for a Finite Measure and a Sigma-Finite Measure, and Uniqueness of Densities, that $\mu(C)=\int_{X}\mathbf{1}_{C}\,f\,d\gamma_{c}$ for every $C\in\mathcal{B}(X)$.

Step 1 ($k_{n}$ is continuous; the law of $k_{n}$ under $\gamma_{c}$). For $x,x'\in X$, Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity gives $|Q_{n}x-Q_{n}x'|\le|x-x'|$ and $\lVert p_{n}(x)-p_{n}(x')\rVert\le|x-x'|$, so $d_{\times}(k_{n}(x),k_{n}(x'))\le|x-x'|$; hence $k_{n}$ is continuous (take $\delta=\varepsilon$) and Borel by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. This is the first assertion of claim 1. Let $h_{n}$ be the map of Head and Tail of a Diagonal Gaussian Measure on a Hilbert Space are Independent, $h_{n}(x)=(p_{n}(x),Q_{n}x)$. For $A\in\mathcal{B}(X)$ and $B\in\mathcal{B}(\mathbb{R}^{n})$ we have $k_{n}^{-1}(A\times B)=Q_{n}^{-1}(A)\cap p_{n}^{-1}(B)=h_{n}^{-1}(B\times A)$, so by Head and Tail of a Diagonal Gaussian Measure on a Hilbert Space are Independent §head-tail,

$$(k_{n})_{\#}\gamma_{c}(A\times B)=(h_{n})_{\#}\gamma_{c}(B\times A)=\gamma(B)\,\nu^{0}(A)=\nu^{0}(A)\,\gamma(B).$$

The measure $(k_{n})_{\#}\gamma_{c}$ on $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ thus has the defining rectangle values of the product of the probability (hence $\sigma$-finite) measures $\nu^{0}$ and $\gamma$, so by the uniqueness in Existence and Uniqueness of the Product Measure,

$$(k_{n})_{\#}\gamma_{c}=\pi,\qquad\pi=\nu^{0}\otimes\gamma .$$

Let $\kappa^{0}:X\times\mathcal{B}(\mathbb{R}^{n})\to\mathbb{R}$, $\kappa^{0}(w,B)=\gamma(B)$. Each $\kappa^{0}(w,\cdot)=\gamma$ is a probability measure and each $\kappa^{0}(\cdot,B)$ is constant, hence measurable by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; so $\kappa^{0}$ is a probability kernel from $(X,\mathcal{B}(X))$ to $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, and by Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §constant the composite measure $\nu^{0}\otimes\kappa^{0}$ is $\pi$. Hence Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §nonnegative applies to $\pi$ with $\kappa^{0}_{w}=\gamma$ for every $w\in X$.

Step 2 (the function $F$). Define $\Theta:X\times\mathbb{R}^{n}\to X$ by $\Theta(w,u)=p_{n}^{*}(u)+w$. Since $p_{n}^{*}(u)-p_{n}^{*}(u')=p_{n}^{*}(u-u')$ by the definition of $p_{n}^{*}$ in Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §coordinates and $|p_{n}^{*}(u-u')|=\lVert u-u'\rVert$ by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §continuity, the triangle inequality gives $|\Theta(w,u)-\Theta(w',u')|\le|w-w'|+\lVert u-u'\rVert\le2d_{\times}((w,u),(w',u'))$; so $\Theta$ is continuous (take $\delta=\varepsilon/2$), hence Borel by claim 3 of Borel Measurability and Bounded Integration on a Metric Space. Let $F=f\circ\Theta:X\times\mathbb{R}^{n}\to\mathbb{R}$, $F(w,u)=f(p_{n}^{*}(u)+w)$; it is nonnegative and measurable with respect to $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ by claim 4 of Borel Measurability and Bounded Integration on a Metric Space. For each $w\in X$ the section $F(w,\cdot)$, which is the function $u\mapsto f(p_{n}^{*}(u)+w)$ of the statement, is measurable with respect to $\mathcal{B}(\mathbb{R}^{n})$ by claim 3 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable. For $x\in X$, since $P_{n}=p_{n}^{*}\circ p_{n}$ and $Q_{n}x=x-P_{n}x$ (Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §coordinates),

$$\Theta(k_{n}(x))=p_{n}^{*}(p_{n}(x))+Q_{n}x=P_{n}x+(x-P_{n}x)=x,\qquad\text{so}\qquad f=F\circ k_{n}.$$

Step 3 (the image of $\mu$ as an integral against $\pi$). Let $E\in\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$. The function $\mathbf{1}_{E}F$ is nonnegative and measurable by claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $\mathbf{1}_{k_{n}^{-1}(E)}f=(\mathbf{1}_{E}F)\circ k_{n}$ by Step 2. By the density property of $f$ and claim 2 of Image Measures, Measures with Densities, and Change of Variables applied to $k_{n}$ and $\gamma_{c}$, together with Step 1,

$$(k_{n})_{\#}\mu(E)=\mu\bigl(k_{n}^{-1}(E)\bigr)=\int_{X}(\mathbf{1}_{E}F)\circ k_{n}\,d\gamma_{c}=\int_{X\times\mathbb{R}^{n}}\mathbf{1}_{E}F\,d\pi.\tag{1}$$

With $E=X\times\mathbb{R}^{n}$ this gives $\int F\,d\pi=\mu(X)=1$, so $F$ is integrable with respect to $\pi$ (Measure Spaces and the Lebesgue Integral: Standing Notation §integral).

Let $N$ be the set of $w\in X$ for which $F(w,\cdot)$ is not integrable with respect to $\gamma$. By Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §nonnegative applied to $F$: $N\in\mathcal{B}(X)$, and since $F$ is integrable, $\nu^{0}(N)=0$. For $B\in\mathcal{B}(\mathbb{R}^{n})$ define $J_{B}:X\to\mathbb{R}$ by

$$J_{B}(w)=\int_{\mathbb{R}^{n}}\mathbf{1}_{B}(u)\,F(w,u)\,\gamma(du)\quad(w\notin N),\qquad J_{B}(w)=0\quad(w\in N);$$

for $w\notin N$ the integrand is measurable and satisfies $0\le\mathbf{1}_{B}F(w,\cdot)\le F(w,\cdot)$, so it is integrable by monotonicity of the integral (Linearity and Monotonicity of the Lebesgue Integral, in force by Measure Spaces and the Lebesgue Integral: Standing Notation §background), and $0\le J_{B}(w)\le J_{\mathbb{R}^{n}}(w)$. Write $\tilde{Z}=J_{\mathbb{R}^{n}}$.

Step 4 (rectangles). We show: for every $B\in\mathcal{B}(\mathbb{R}^{n})$ the function $J_{B}$ is Borel, and for every $A\in\mathcal{B}(X)$,

$$(k_{n})_{\#}\mu(A\times B)=\int_{X}\mathbf{1}_{A}\,J_{B}\,d\nu^{0}.\tag{2}$$

Let $H=\mathbf{1}_{A\times B}F$, nonnegative and measurable, with $H\le F$, so $H$ is integrable with respect to $\pi$. By Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §nonnegative applied to $H$, the set $N_{H}$ of $w$ for which $H(w,\cdot)$ is not $\gamma$-integrable lies in $\mathcal{B}(X)$ and has $\nu^{0}(N_{H})=0$, the function $g_{H}$ equal to $\int H(w,u)\,\gamma(du)$ off $N_{H}$ and to $0$ on $N_{H}$ is Borel, and $\int H\,d\pi=\int g_{H}\,d\nu^{0}$. For $w\notin N$, $H(w,\cdot)=\mathbf{1}_{A}(w)\mathbf{1}_{B}F(w,\cdot)$ is integrable with integral $\mathbf{1}_{A}(w)J_{B}(w)$, by Step 3 and linearity of the integral; hence $w\notin N_{H}$ and $g_{H}(w)=\mathbf{1}_{A}(w)J_{B}(w)$. Taking $A=X$ shows $J_{B}=\mathbf{1}_{X\setminus N}\,g_{X\times B}$ everywhere (both vanish on $N$), so $J_{B}$ is Borel by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and so is $\mathbf{1}_{A}J_{B}$. For general $A$, $g_{H}$ is nonnegative and integrable with respect to $\nu^{0}$: it vanishes on $N_{H}$, off $N_{H}$ its value is the integral of the nonnegative integrable function $H(w,\cdot)$, which is nonnegative because the negative part of $H(w,\cdot)$ in Integrable Function and the Lebesgue Integral is the zero function, and it is integrable by Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §nonnegative, $H$ being integrable with respect to $\pi$. The nonnegative Borel functions $g_{H}$ and $\mathbf{1}_{A}J_{B}$ agree outside the $\nu^{0}$-null set $N$, so $\mathbf{1}_{A}J_{B}$ is integrable and $\int g_{H}\,d\nu^{0}=\int\mathbf{1}_{A}J_{B}\,d\nu^{0}$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison. Combined with (1) for $E=A\times B$, this gives (2).

Step 5 (claim 1). For $A\in\mathcal{B}(X)$, $Q_{n}^{-1}(A)=k_{n}^{-1}(A\times\mathbb{R}^{n})$, so by (2) with $B=\mathbb{R}^{n}$,

$$\nu_{n}(A)=\mu\bigl(Q_{n}^{-1}(A)\bigr)=(k_{n})_{\#}\mu(A\times\mathbb{R}^{n})=\int_{X}\mathbf{1}_{A}\,\tilde{Z}\,d\nu^{0}.\tag{3}$$

Thus $\nu_{n}$ is the measure with density $\tilde{Z}:X\to[0,\infty)$ with respect to $\nu^{0}$, in the sense of claim 3 of Image Measures, Measures with Densities, and Change of Variables. By the definitions of $N$ and $\tilde{Z}$, a point $w\in X$ belongs to $W_{0}$ exactly when $w\notin N$ and $\tilde{Z}(w)>0$, and then $Z(w)=\tilde{Z}(w)$. Hence

$$W_{0}=(X\setminus N)\cap\tilde{Z}^{-1}\bigl((0,\infty)\bigr)\in\mathcal{B}(X),\qquad X\setminus W_{0}=N\cup\{w\in X:\tilde{Z}(w)=0\},$$

using $\tilde{Z}\ge0$. By (3), $\nu_{n}(N)=\int_{X}\mathbf{1}_{N}\tilde{Z}\,d\nu^{0}=0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral, since $\mathbf{1}_{N}\tilde{Z}$ vanishes off the $\nu^{0}$-null set $N$; and $\nu_{n}(\{\tilde{Z}=0\})=\int_{X}\mathbf{1}_{\{\tilde{Z}=0\}}\tilde{Z}\,d\nu^{0}=0$, the integrand being identically $0$. By subadditivity $\nu_{n}(X\setminus W_{0})=0$, and as $\nu_{n}(X)=\mu(X)=1$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables, $\nu_{n}(W_{0})=1$. This proves claim 1.

Step 6 (claim 2). Let $w\in W_{0}$. The function $u\mapsto F(w,u)/Z(w)$ is a nonnegative measurable function on $\mathbb{R}^{n}$ (Step 2 and claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions), so by claim 3 of Image Measures, Measures with Densities, and Change of Variables the formula $B\mapsto\int\mathbf{1}_{B}(u)\,(F(w,u)/Z(w))\,\gamma(du)$ defines a measure on $\mathcal{B}(\mathbb{R}^{n})$; by linearity of the integral of the integrable function $\mathbf{1}_{B}F(w,\cdot)$ this measure is $K(w,\cdot)$, and $K(w,\mathbb{R}^{n})=Z(w)/Z(w)=1$. For $w\notin W_{0}$, $K(w,\cdot)=\gamma$ is a probability measure. So every $K(w,\cdot)$ is a probability measure on $\mathcal{B}(\mathbb{R}^{n})$.

For measurability in $w$, let $R:X\to\mathbb{R}$ be $R(w)=1/\tilde{Z}(w)$ if $\tilde{Z}(w)>0$ and $R(w)=0$ otherwise. For real $t$, the set $\{R>t\}$ is $X$ if $t<0$, is $\{\tilde{Z}>0\}$ if $t=0$, and is $\{0<\tilde{Z}<1/t\}$ if $t>0$; each lies in $\mathcal{B}(X)$ since $\tilde{Z}$ is Borel, so $R$ is Borel by the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable. For $w\in W_{0}$ we have $w\notin N$, so $K(w,B)=J_{B}(w)/Z(w)=R(w)J_{B}(w)$. Hence, for every $B\in\mathcal{B}(\mathbb{R}^{n})$,

$$K(\cdot,B)=\mathbf{1}_{W_{0}}\,R\,J_{B}+\gamma(B)\,\mathbf{1}_{X\setminus W_{0}},$$

which is Borel by claims 1, 2 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, $W_{0}$ and $J_{B}$ being Borel by Steps 5 and 4. Thus $K$ is a probability kernel from $(X,\mathcal{B}(X))$ to $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ in the sense of Probability Kernels Between Measurable Spaces §kernel, which proves claim 2.

Step 7 (claim 3). Fix $B\in\mathcal{B}(\mathbb{R}^{n})$. We claim that $K(w,B)\tilde{Z}(w)=J_{B}(w)$ for every $w\in X$. For $w\in W_{0}$ this is $K(w,B)Z(w)=J_{B}(w)$, shown in Step 6. For $w\in N$ both sides are $0$, as $\tilde{Z}(w)=J_{B}(w)=0$ by definition. For $w\notin N$ with $\tilde{Z}(w)=0$, the left side is $0$ and $0\le J_{B}(w)\le\tilde{Z}(w)=0$ by Step 3. Now let $A\in\mathcal{B}(X)$. The function $\mathbf{1}_{A}K(\cdot,B)$ is Borel with values in $[0,1]$, so by (3), claim 3 of Image Measures, Measures with Densities, and Change of Variables (integration against a measure with density), the identity just shown, and (2),

$$\int_{X}\mathbf{1}_{A}(w)\,K(w,B)\,\nu_{n}(dw)=\int_{X}\mathbf{1}_{A}\,K(\cdot,B)\,\tilde{Z}\,d\nu^{0}=\int_{X}\mathbf{1}_{A}\,J_{B}\,d\nu^{0}=(k_{n})_{\#}\mu(A\times B).$$

Here $\nu_{n}$ is a probability measure on $\mathcal{B}(X)$, $K$ is a probability kernel by Step 6, and $(k_{n})_{\#}\mu$ is a measure on $\mathcal{B}(X)\otimes\mathcal{B}(\mathbb{R}^{n})$ by claim 1 of Image Measures, Measures with Densities, and Change of Variables. By Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral §rectangles, $(k_{n})_{\#}\mu=\nu_{n}\otimes K$, which proves claim 3.

Step 8 (claim 4). Let $B\in\mathcal{B}(\mathbb{R}^{n})$ with $\lambda_{n}(B)=0$. By Diagonal Gaussian Measures on Euclidean Space §measure, $\gamma(B)=\int\mathbf{1}_{B}\,\rho_{c^{(n)}}\,d\lambda_{n}$ with the nonnegative Borel density $\rho_{c^{(n)}}$ named there; the integrand vanishes off the $\lambda_{n}$-null set $B$, so $\gamma(B)=0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral. Let $w\in X$. If $w\notin W_{0}$, then $K(w,B)=\gamma(B)=0$. If $w\in W_{0}$, the nonnegative measurable function $\mathbf{1}_{B}F(w,\cdot)$ vanishes off the $\gamma$-null set $B$, so its $\gamma$-integral is $0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-integral, and $K(w,B)=0$. This proves claim 4.