Proof of Localized Filtering Lower Bound from a Mixture-Weight van Trees Certificate

Claim 1. The indicator $\mathbf{1}_{\mathcal{H}}$ is $\mathcal{G}$-measurable, because $\mathcal{H}\in\mathcal{G}$, and bounded by $1$. By claim 1 of the conditional expectation properties lemma, applied finitely many times, $c\cdot M$ is a conditional expectation of $c\cdot X$ given $\mathcal{G}$. By claim 4 of the same lemma, applied with $Z=\mathbf{1}_{\mathcal{H}}$, the random variables $\mathbf{1}_{\mathcal{H}}(c\cdot X)$ and $\mathbf{1}_{\mathcal{H}}(c\cdot M)$ are square-integrable and $\mathbf{1}_{\mathcal{H}}(c\cdot M)$ is a conditional expectation of $\mathbf{1}_{\mathcal{H}}(c\cdot X)$ given $\mathcal{G}$.

Since $\mathbf{1}_{\mathcal{H}}^{2}=\mathbf{1}_{\mathcal{H}}$ and $c\cdot X-c\cdot M=c\cdot\varepsilon$ at every point of $\Omega$,

at every point of $\Omega$. Each of the finitely many products $\mathbf{1}_{\mathcal{H}}\varepsilon^{\gamma}\varepsilon^{\delta}$ is integrable: each $\varepsilon^{\gamma}$ is square-integrable, so each product $\varepsilon^{\gamma}\varepsilon^{\delta}$ is integrable by the closure properties of the square-integrability definition, and multiplying by the bounded $\mathbf{1}_{\mathcal{H}}$ preserves integrability, by the monotonicity of the integral applied to the pointwise bound $|\mathbf{1}_{\mathcal{H}}\varepsilon^{\gamma}\varepsilon^{\delta}|\le|\varepsilon^{\gamma}\varepsilon^{\delta}|$. Consequently all three expressions have the same expectation; the equality of claim 1 is the equality of the expectations of the first two, obtained from the linearity of the integral.

Claim 2. Write $T=\mathbf{1}_{\mathcal{H}}(c\cdot X)$ and $\hat{T}=\mathbf{1}_{\mathcal{H}}(c\cdot M)$, so that by claim 1

The random variable $\hat{T}$ is $\mathcal{G}$-measurable and square-integrable, so by (C1) there is a measurable $g:\mathsf{Y}\to\mathbb{R}$ with $g(\mathsf{D})$ square-integrable and $\hat{T}=g(\mathsf{D})$ almost surely. Then $(T-\hat{T})^{2}=(T-g(\mathsf{D}))^{2}$ almost surely, and almost surely equal integrable random variables have equal expectations, so

Let $G:\mathsf{Y}\to\mathbb{R}$ be the measurable map furnished by (C3). The difference $h=g-G:\mathsf{Y}\to\mathbb{R}$ is measurable with respect to $\mathcal{Y}$ and the Borel $\sigma$-algebra, being a difference of two such maps (claim 2 of the arithmetic of measurable functions, applied on the measurable space $(\mathsf{Y},\mathcal{Y})$), and $h(\mathsf{D})=g(\mathsf{D})-G(\mathsf{D})$ is square-integrable by the closure properties of square-integrability, both $g(\mathsf{D})$ and $G(\mathsf{D})$ being square-integrable. By (C2$'$) the setting of Score Identities and the Mixture-Weight Directional van Trees Inequality, that is, the setting and the hypotheses (i), (ii), (iii) of The Multivariate van Trees Inequality together with the hypothesis (iv$'$) of the former lemma, holds for the certificate data, with $l$ there replaced by $d$, the probability space $(\Omega,\mathcal{F},\mathbb{P})$, the data space $(\mathsf{Y},\mathcal{Y})$ with the measure $\varrho_{0}$ in the roles of $(Y,\mathcal{G})$ and $\mu$ there, the random variables $\Theta_{1},\dots,\Theta_{d}$, the map $\mathsf{D}$ in the role of $D$ and the function $p$ in the role of the density. The map $h$ is measurable with $h(\mathsf{D})$ square-integrable, so it is admissible as the map $m$ of that lemma. Claim 2 of that lemma, applied with the shifts $a_{1},\dots,a_{n}$, the direction $u=z$, the map $m=h$ and the vector $\alpha$, gives

with $\mathcal{I}_{z}\in[0,\infty)$ the mixture-weight information of (C2$'$). By (C4$'$), $\alpha\cdot z>0$, so the left-hand side $|\alpha\cdot z|$ equals $\alpha\cdot z$ and is strictly positive; hence so is the right-hand side, which forces $\sqrt{\mathcal{I}_{z}}>0$ (for $\sqrt{\mathcal{I}_{z}}=0$ would make the right-hand side $0$), and dividing by it,

the second step because the inequality of (C4$'$), divided by $\sqrt{\mathcal{I}_{z}}>0$, reads $\alpha\cdot z/\sqrt{\mathcal{I}_{z}}\ge\max_{1\le j\le n}|\alpha\cdot a_{j}|+\sqrt{\varkappa}$. Thus $\lVert h(\mathsf{D})-\alpha\cdot\Theta\rVert_{2}\ge\sqrt{\varkappa}$.

At every point of $\Omega$,

so by claim 2 of the triangle inequality for the mean-square norm and by (C3), which reads $\lVert\alpha\cdot\Theta+G(\mathsf{D})-T\rVert_{2}\le\epsilon$,

the last step using $\lVert -U\rVert_{2}=\lVert U\rVert_{2}$, which is immediate from the definition of the mean-square norm. Hence $\lVert T-g(\mathsf{D})\rVert_{2}\ge\sqrt{\varkappa}-\epsilon$, and the right-hand side is nonnegative by the hypothesis $\sqrt{\varkappa}\ge\epsilon$. Squaring an inequality between nonnegative real numbers preserves it, so

which is claim 2.

Independence of the choice of conditional expectations. Any two conditional expectations of $X^{\gamma}$ given $\mathcal{G}$ are almost surely equal, by the uniqueness part of the existence and uniqueness theorem. Replacing the $M^{\gamma}$ therefore changes $c\cdot\varepsilon$ only on a set of probability zero, which leaves the expectations displayed in claims 1 and 2 unchanged.