Proof of Localized Filtering Lower Bound from a 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 $\Omega$,

$$\mathbf{1}_{\mathcal{H}}\sum_{\gamma=1}^{l}\sum_{\delta=1}^{l}c^{\gamma}c^{\delta}\,\varepsilon^{\gamma}\varepsilon^{\delta}=\mathbf{1}_{\mathcal{H}}\,(c\cdot\varepsilon)^{2}=\bigl(\mathbf{1}_{\mathcal{H}}(c\cdot X)-\mathbf{1}_{\mathcal{H}}(c\cdot M)\bigr)^{2}$$

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

$$\mathbb{E}\bigl[(T-\hat{T})^{2}\bigr]=\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}(c\cdot\varepsilon)^{2}\bigr].$$

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

$$\lVert T-g(\mathsf{D})\rVert_{2}^{2}=\mathbb{E}\bigl[(T-g(\mathsf{D}))^{2}\bigr]=\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}(c\cdot\varepsilon)^{2}\bigr].$$

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 hypotheses of Directional Form of the Multivariate van Trees Inequality hold for the certificate data, with $l$ there replaced by $d$, the data space $(\mathsf{Y},\mathcal{Y})$ with the measure $\varrho_{0}$, the random variables $\Theta_{1},\dots,\Theta_{d}$, the map $\mathsf{D}$, the information matrix $\mathcal{I}$, and the vectors $\alpha$ and $z$ in place of $a$ and $z$ there. Applying that corollary to the function $h$ gives

$$\bigl\lVert h(\mathsf{D})-\alpha\cdot\Theta\bigr\rVert_{2}^{2}=\mathbb{E}\bigl[(h(\mathsf{D})-\alpha\cdot\Theta)^{2}\bigr]\ \ge\ \frac{(\alpha\cdot z)^{2}}{z\cdot(\mathcal{I}z)}\ \ge\ \varkappa,$$

the last step by (C4), division by $z\cdot(\mathcal{I}z)$ being legitimate because $z\cdot(\mathcal{I}z)>0$ by claim 1 of Rank-One Lower Bound for the Inverse of a Positive Definite Matrix. Since the mean-square norm is nonnegative, $\lVert h(\mathsf{D})-\alpha\cdot\Theta\rVert_{2}\ge\sqrt{\varkappa}$.

At every point of $\Omega$,

$$h(\mathsf{D})-\alpha\cdot\Theta=\bigl(g(\mathsf{D})-T\bigr)+\bigl(T-G(\mathsf{D})-\alpha\cdot\Theta\bigr),$$

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

$$\sqrt{\varkappa}\ \le\ \bigl\lVert h(\mathsf{D})-\alpha\cdot\Theta\bigr\rVert_{2}\ \le\ \bigl\lVert g(\mathsf{D})-T\bigr\rVert_{2}+\bigl\lVert T-G(\mathsf{D})-\alpha\cdot\Theta\bigr\rVert_{2}\ \le\ \bigl\lVert T-g(\mathsf{D})\bigr\rVert_{2}+\epsilon,$$

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

$$\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}(c\cdot\varepsilon)^{2}\bigr]=\lVert T-g(\mathsf{D})\rVert_{2}^{2}\ \ge\ \bigl(\sqrt{\varkappa}-\epsilon\bigr)^{2},$$

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.