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 $\Omega$,

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

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 $g$ gives

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, $\lVert g(\mathsf{D})-\alpha\cdot\Theta\rVert_{2}\ge\sqrt{\varkappa}$.

By claim 2 of the triangle inequality for the mean-square norm, applied to the decomposition $g(\mathsf{D})-\alpha\cdot\Theta=\bigl(g(\mathsf{D})-T\bigr)+\bigl(T-\alpha\cdot\Theta\bigr)$, and by (C3),

the last step also 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.