Proof. Expand \(\sum_{\alpha\in I_n}(X_{\alpha,n}+Y_{\alpha,n})^2\) into the three terms \[\sum_{\alpha\in I_n} X_{\alpha,n}^2 + 2\sum_{\alpha\in I_n} X_{\alpha,n}Y_{\alpha,n} + \sum_{\alpha\in I_n} Y_{\alpha,n}^2.\] The first term converges in probability to \(1\) by assumption. For the last term, \(\Ex*{\sum_{\alpha\in I_n} Y_{\alpha,n}^2} \to 0\) by assumption, so \(\sum_{\alpha\in I_n} Y_{\alpha,n}^2 \xrightarrow{\mathbb{P}}0\) by Markov’s inequality. For the cross term, Cauchy–Schwarz gives \[\abs*{2\sum_{\alpha\in I_n} X_{\alpha,n}Y_{\alpha,n}} \leq 2 \left(\sum_{\alpha\in I_n} X_{\alpha,n}^2\right)^{1/2} \left(\sum_{\alpha\in I_n} Y_{\alpha,n}^2\right)^{1/2} \xrightarrow{\mathbb{P}}0.\] For expectations, the same inequality and Cauchy–Schwarz give \[\Ex*{\abs*{2\sum_{\alpha\in I_n} X_{\alpha,n}Y_{\alpha,n}}} \leq 2\left(\sum_{\alpha\in I_n}\Ex*{X_{\alpha,n}^2}\right)^{1/2} \left(\sum_{\alpha\in I_n}\Ex*{Y_{\alpha,n}^2}\right)^{1/2} \longrightarrow 0.\] Together with \(\sum_{\alpha\in I_n}\Ex*{X_{\alpha,n}^2}\to1\) and \(\sum_{\alpha\in I_n}\Ex*{Y_{\alpha,n}^2}\to0\), this proves the expectation claim. ◻

Proof. This is exactly the row-wise i.i.d.triangular-array weak law highlighted in Section 2; see [25]. In the present setting, the proof is a direct application of the von Bahr–Esseen inequality [26]. Write \[W_{n,i} \mathrel{\vcenter{:}}= \frac{h_s^{(1)}(Z_i)^2}{\zeta_{s,\omega}^{1}}, \qquad X_{n,i} \mathrel{\vcenter{:}}= W_{n,i} - 1, \qquad i = 1,\dotsc,n.\] Because the first projection is centered, \(\Ex{W_{n,1}} = 1\), so the variables \(X_{n,1},\dotsc,X_{n,n}\) are row-wise i.i.d.and mean zero. Let \(r \in (1,2]\) be as in 2. Then the von Bahr–Esseen inequality gives \[\mathbb{E}\left[ \abs*{ \frac{1}{n}\sum_{i=1}^{n} X_{n,i} }^{r} \right] \leq \frac{2}{n^{r}} \sum_{i=1}^{n} \Ex*{\abs*{X_{n,i}}^{r}} = \frac{2\Ex*{\abs*{X_{n,1}}^{r}}}{n^{r-1}}.\] Since \(r \geq 1\), \[\abs*{x-1}^{r} \leq 2^{r-1}\left(\abs*{x}^{r} + 1\right),\] so \[\frac{\Ex*{\abs*{X_{n,1}}^{r}}}{n^{r-1}} \lesssim \frac{\Ex*{W_{n,1}^{r}} + 1}{n^{r-1}} \longrightarrow 0\] by 2. Therefore, \[\frac{1}{n\zeta_{s,\omega}^{1}} \sum_{i=1}^{n} h_s^{(1)}(Z_i)^2 = \frac{1}{n}\sum_{i=1}^{n} W_{n,i} \xrightarrow{\mathbb{P}}1.\] ◻

Proof of 1. \[\begin{align} 1 & \leq \frac{n}{s^2}\frac{\Varb*{\mathrm{U}_{n, s, \omega}\left(D_{[n]}\right)}}{\zeta_{s, \omega}^{1}} = \left(\frac{s^2}{n} \zeta_{s, \omega}^{1}\right)^{-1} \left(\sum_{j=1}^{s-1}\binom{s}{j}^2\binom{n}{j}^{-1} V_{s, \omega}^{j} + \binom{n}{s}^{-1} V_{s}^{s}\right) \\ & \leq 1 + \left(\frac{s^2}{n} \zeta_{s, \omega}^{1}\right)^{-1} \frac{s^2}{n^2} \left(\sum_{j=2}^{s-1}\binom{s}{j} V_{s, \omega}^{j} + V_{s}^{s}\right) \\ & \leq 1 + \frac{s}{n}\left(\frac{\zeta_{s}^{s}}{s\zeta_{s, \omega}^{1}} - 1\right) \longrightarrow 1. \end{align}\] ◻

Proof of 1.
* For each delete set \(\ell \in L_{n,d}\), write \[\Delta_{\ell} \mathrel{\vcenter{:}}= \Delta_{\ell}[U_{n,s,\omega}].\] Then the delete-\(d\) jackknife variance estimator takes the form \[\widehat{\sigma}_{JKD}^2\left(D_{[n]}; d\right) = \frac{n-d}{d}\binom{n}{d}^{-1}\sum_{\ell \in L_{n,d}} \Delta_{\ell}^{2}.\] We first expand each delete-\(d\) difference by Hoeffding order and isolate its first-projection, or Hájek, component. We then show that the remaining higher-order terms are negligible on the \(\zeta_{s,\omega}^{1}\) scale, identify the averaged square of the leading term with the empirical average of \(h_i^2\) up to negligible off-diagonal corrections, and finally invoke 3 2. The last step is to compare the resulting scale back to the actual variance target via 1. By the generalized Hoeffding decomposition, \[\Delta_{\ell} = \sum_{j=1}^{s-1}\binom{s}{j}\delta_{j,\ell} + \delta_{s,\ell},\] where the projection-level differences \(\delta_{j,\ell}\), \(1 \leq j \leq s\), are defined in 7. To isolate the leading Hájek term, write \[h_{i} \mathrel{\vcenter{:}}= h_{s,\omega}^{(1)}(Z_{i}), \qquad i = 1, \dotsc, n.\] Since \[G_{1,1}^{(\ell)} = \sum_{i \in \ell} h_{i}, \qquad G_{1,0}^{(\ell)} = \sum_{i \in [n]\backslash \ell} h_{i},\] and \[q_{1}^{(n,d)} = \frac{1}{n}, \qquad r_{1}^{(n,d)} = \frac{1}{n} - \frac{1}{n-d} = -\frac{d}{n(n-d)},\] the first projection contributes \[s \delta_{1,\ell} = \frac{s}{n}\sum_{i \in \ell} h_{i} - \frac{sd}{n(n-d)}\sum_{i \in [n]\backslash \ell} h_{i}.\] We therefore write \[\Delta_{\ell} = \frac{s\sqrt{d}}{n} \left[ \frac{1}{\sqrt{d}}\sum_{i \in \ell} h_{i} + T_{\ell} \right],\] where \[T_{\ell} \mathrel{\vcenter{:}}= -\frac{\sqrt{d}}{n-d}\sum_{i \in [n]\backslash \ell} h_{i} + \frac{n}{s\sqrt{d}} \left\{ \sum_{j=2}^{s-1}\binom{s}{j}\delta_{j,\ell} + \delta_{s,\ell} \right\}.\] Consequently, \[\widehat{\sigma}_{JKD}^2\left(D_{[n]}; d\right) = \frac{(n-d)s^2}{n^2}\binom{n}{d}^{-1} \sum_{\ell \in L_{n,d}} \left[ \frac{1}{\sqrt{d}}\sum_{i \in \ell} h_{i} + T_{\ell} \right]^2.\] Write \[A_{\ell} \mathrel{\vcenter{:}}= \frac{1}{\sqrt{d}}\sum_{i \in \ell} h_{i}.\] We first show that \(T_{\ell}\) is negligible relative to the variance scale \(\zeta_{s,\omega}^{1}\). Since distinct Hoeffding orders are orthogonal, we have \[\Ex*{T_{\ell}^{2}} = \frac{d}{n-d}\zeta_{s,\omega}^{1} + \frac{n^2}{s^2 d} \left\{ \sum_{j=2}^{s-1}\binom{s}{j}^{2}\Ex*{\delta_{j,\ell}^{2}} + \Ex*{\delta_{s,\ell}^{2}} \right\}.\] The key simplification is that, once the delete-\(d\) difference has been organized by Hoeffding order and overlap strata, the second-moment calculation is mostly combinatorial. Distinct Hoeffding orders are orthogonal, and within a fixed canonical order the degeneracy of the projection kills the mixed overlap terms. What remains is to count how many tuples land in each overlap stratum and multiply by the variance attached to that order. The useful partition is therefore not really the pair of cases \(d<j\) and \(d \geq j\), but the simpler split between \(j\)-tuples that hit the delete set and \(j\)-tuples that avoid it. The exact overlap level \(a=\abs*{\iota\cap\ell}\) is only a device for counting the first class: all nonzero-overlap strata carry the same coefficient \(q_j^{(n,d)}\), while the zero-overlap stratum carries \(r_j^{(n,d)}\). With the standard convention that impossible binomial coefficients are zero, Vandermonde’s identity gives \[\sum_{a=1}^{\min\{j,d\}}\binom{d}{a}\binom{n-d}{j-a} = \binom{n}{j}-\binom{n-d}{j},\] while \(\abs*{\mathcal{O}_{j,0}(\ell)}=\binom{n-d}{j}\). This is why the two apparent regimes collapse to the same coefficient after squaring and taking expectations. For each \(2 \leq j \leq s-1\), the overlap-stratum representation and degeneracy of the canonical \(j\)th projection imply \[\begin{align} \Ex*{\delta_{j,\ell}^{2}} & = \left[ \left(q_{j}^{(n,d)}\right)^{2} \sum_{a=1}^{\min\{j,d\}} \abs*{\mathcal{O}_{j,a}(\ell)} + \left(r_{j}^{(n,d)}\right)^{2}\abs*{\mathcal{O}_{j,0}(\ell)} \right] V_{s,\omega}^{j} \\ & = \left[ \binom{n-d}{j}^{-1} - \binom{n}{j}^{-1} \right] V_{s,\omega}^{j}, \end{align}\] and likewise \[\Ex*{\delta_{s,\ell}^{2}} = \left[ \binom{n-d}{s}^{-1} - \binom{n}{s}^{-1} \right] V_{s}^{s}.\] Substituting these identities gives \[\begin{align} \Ex*{T_{\ell}^{2}} & = \frac{d}{n-d}\zeta_{s,\omega}^{1} \\ & \quad + \frac{n}{s d}\Bigg\{ \sum_{j=2}^{s-1} \frac{\binom{s-1}{j-1}}{\binom{n-1}{j-1}} \left( \binom{n}{j}\binom{n-d}{j}^{-1} - 1 \right)\binom{s}{j}V_{s,\omega}^{j} \\ & \qquad\qquad + \binom{n-1}{s-1}^{-1} \left( \binom{n}{s}\binom{n-d}{s}^{-1} - 1 \right)V_{s}^{s} \Bigg\}. \end{align}\] For \(2 \leq j \leq s\), write \(B_{j,n,d}\) for the amount by which the \(j\)-tuple normalization changes after deleting \(d\) observations. The theorem’s row-wise delete-domain condition \(s+d\leq n\) ensures that the denominator \(\binom{n-d}{j}\) is nonzero for every \(2\leq j\leq s\) considered here. Since each deleted position affects a \(j\)-tuple only at order \(j/n\), the product below formalizes the heuristic that the cumulative perturbation is \(O(jd/n)\). \[B_{j,n,d} \mathrel{\vcenter{:}}= \binom{n}{j}\binom{n-d}{j}^{-1} = \prod_{m=0}^{d-1}\frac{n-m}{n-j-m} = \prod_{m=0}^{d-1}\left(1 + \frac{j}{n-j-m}\right).\] Since \(sd = o(n)\), we have \(jd/(n-j-d+1) \leq 1/2\) uniformly for \(2 \leq j \leq s\) and \(n\) large enough. A Weierstrass-product bound then yields \[B_{j,n,d} - 1 \lesssim \frac{2jd}{n-j-d+1} \lesssim \frac{3jd}{n}, \qquad 2 \leq j \leq s.\] Using also \[\frac{\binom{s-1}{j-1}}{\binom{n-1}{j-1}} \leq \left(\frac{e(s-1)}{n-1}\right)^{j-1},\] we obtain \[\begin{align} \Ex*{T_{\ell}^{2}} & \lesssim \frac{d}{n-d}\zeta_{s,\omega}^{1} + \frac{3}{s}\Bigg\{ \sum_{j=2}^{s-1} j \left(\frac{e(s-1)}{n-1}\right)^{j-1} \binom{s}{j}V_{s,\omega}^{j} + s\left(\frac{e(s-1)}{n-1}\right)^{s-1}V_{s}^{s} \Bigg\}. \end{align}\] Splitting \(j = 1 + (j-1)\), bounding the first part by \(\frac{e(s-1)}{n-1}\), and using \[\zeta_{s}^{s} = \sum_{j=1}^{s-1}\binom{s}{j}V_{s,\omega}^{j} + V_{s}^{s},\] we further obtain \[\begin{align} \Ex*{T_{\ell}^{2}} & \lesssim \frac{d}{n-d}\zeta_{s,\omega}^{1} + \frac{3e(s-1)}{s(n-1)} \left(\zeta_{s}^{s} - s\zeta_{s,\omega}^{1}\right) \\ & \quad + \frac{3}{s}\zeta_{s}^{s} \sum_{j=1}^{\infty} j\left(\frac{e(s-1)}{n-1}\right)^{j}. \end{align}\] Evaluating the geometric derivative series gives \[\begin{align} \Ex*{T_{\ell}^{2}} & \lesssim \Bigg[ \frac{d}{n-d} + \frac{3e(s-1)(n-1)}{\left(n-1-e(s-1)\right)^{2}} \Bigg]\zeta_{s,\omega}^{1} \\ & \quad + \frac{3e(s-1)}{s} \Bigg[ \frac{1}{n-1} + \frac{n-1}{\left(n-1-e(s-1)\right)^{2}} \Bigg] \left(\zeta_{s}^{s} - s\zeta_{s,\omega}^{1}\right). \end{align}\] Therefore, \[\frac{\Ex*{T_{\ell}^{2}}}{\zeta_{s,\omega}^{1}} \longrightarrow 0\] by \(sd = o(n)\), \(s = o(n)\), and 1. Now define \[X_{\ell} \mathrel{\vcenter{:}}= \sqrt{ \frac{n-d}{n\binom{n}{d}\zeta_{s,\omega}^{1}} }\,A_{\ell}, \qquad Y_{\ell} \mathrel{\vcenter{:}}= \sqrt{ \frac{n-d}{n\binom{n}{d}\zeta_{s,\omega}^{1}} }\,T_{\ell}.\] Then \[\sum_{\ell \in L_{n,d}} \left(X_{\ell} + Y_{\ell}\right)^{2} = \frac{n\widehat{\sigma}_{JKD}^{2}\left(D_{[n]}; d\right)}{s^2\zeta_{s,\omega}^{1}}.\] Moreover, by exchangeability, \[\sum_{\ell \in L_{n,d}} \Ex*{Y_{\ell}^{2}} = \frac{n-d}{n\zeta_{s,\omega}^{1}}\Ex*{T_{\ell}^{2}} \longrightarrow 0,\] and \[\sum_{\ell \in L_{n,d}} \Ex*{X_{\ell}^{2}} = \frac{n-d}{n\zeta_{s,\omega}^{1}}\Ex*{A_{\ell}^{2}} = \frac{n-d}{n} \longrightarrow 1.\] It remains to verify \(\sum_{\ell \in L_{n,d}} X_{\ell}^{2} \xrightarrow{\mathbb{P}}1\). This is the diagonal-square step. Averaging \(A_{\ell}^{2}\) over all delete sets extracts the empirical average of the squared first-projection values, while the cross-products are washed out by the combinatorial averaging over deletion patterns. In that sense, the jackknife behaves here as though it were applied to a linear statistic with influence values \(h_1,\dotsc,h_n\). Here the exact averaged-square identity is \[\begin{align} \binom{n}{d}^{-1}\sum_{\ell \in L_{n,d}} A_{\ell}^{2} & = \frac{1}{d}\binom{n}{d}^{-1} \sum_{\ell \in L_{n,d}}\sum_{i \in \ell}\sum_{j \in \ell} h_{i}h_{j} \\ & = \frac{1}{n}\sum_{i=1}^{n} h_{i}^{2} + \frac{d-1}{n(n-1)} \sum_{i \neq j} h_{i}h_{j}. \end{align}\] Note that despite \(d > 1\), the averaged-square identity reduces the diagonal to \(n^{-1}\sum_i h_i^2\)—the same empirical average of squared first-projection values that appears in the delete-\(1\) case and is controlled by 2. The off-diagonal term is negligible because \[\begin{align} \Ex*{\abs*{ \frac{d-1}{n(n-1)} \sum_{i \neq j} h_{i}h_{j} }} & \leq \frac{d-1}{n(n-1)} \Ex*{\abs*{ \left(\sum_{i=1}^{n} h_{i}\right)^{2} - \sum_{i=1}^{n} h_{i}^{2} }} \\ & \leq \frac{2(d-1)}{n-1}\zeta_{s,\omega}^{1} = o\left(\zeta_{s,\omega}^{1}\right). \end{align}\] Hence \[\sum_{\ell \in L_{n,d}} X_{\ell}^{2} = \frac{n-d}{n\zeta_{s,\omega}^{1}} \left[ \frac{1}{n}\sum_{i=1}^{n} h_{i}^{2} + o_{p}\left(\zeta_{s,\omega}^{1}\right) \right].\] By 3, \[\frac{1}{n\zeta_{s,\omega}^{1}}\sum_{i=1}^{n} h_{i}^{2} \xrightarrow{\mathbb{P}}1.\] Granting this input, \(\sum_{\ell \in L_{n,d}} X_{\ell}^{2} \xrightarrow{\mathbb{P}}1\), so 2 yields \[\frac{n\widehat{\sigma}_{JKD}^{2}\left(D_{[n]}; d\right)}{s^2\zeta_{s,\omega}^{1}} \xrightarrow{\mathbb{P}} 1.\] The desired ratio consistency then follows from 1. ◻

Proof. Chernoff’s inequality gives \(\Pb*{G^c}\leq C\exp(-c\mu)\), after reducing \(c\) to absorb the deterministic shift from \(B\) to \(K=1+B\). The same Chernoff bound and the standard identity \[\Ex*{K\1*{G^c}} = \int_0^\infty \Pb*{K\1*{G^c}>t} \,\;\mathrm{d}t\] give \(\Ex*{K\1*{G^c}}\leq C\mu\exp(-c\mu)\), again with a possibly smaller \(c\). Applying the same integration argument to \(K^2\1*{G^c}\) gives \(\Ex*{K^2\1*{G^c}}\leq C\mu^2\exp(-c\mu)\). On \(G\), \(K^{-3}\leq C\mu^{-3}\) and \(K^{-5}\leq C\mu^{-5}\). On \(G^c\), the trivial bound \(K^{-3}\leq 1\) and the tail bound show that the contribution is exponentially small, hence \(O(\mu^{-3})\). Similarly, \[\Ex*{(K-\mu)^2K^{-5}\1*{G}} \leq C\mu^{-5}\Ex*{(K-\mu)^2} \leq C\mu^{-4},\] while on \(G^c\), \((K-\mu)^2K^{-5}\leq (K+\mu)^2\) and the preceding tail estimates make the contribution exponentially small. ◻

Proof. Write \(A=\set*{\widehat N>0}\) and \(B_\ell=\set*{\widehat N_\ell^\circ>0}\). Since \(B_\ell\subset A\), the zero-count convention gives \[U_h=U_{h^0}+\theta_s\1*{A}, \qquad U_{h,\ell}=U_{h^0,\ell}+\theta_s\1*{B_\ell},\] and hence \[\Delta_{\ell,h} = \Delta_{\ell,h^0} + C_\ell, \qquad C_\ell \mathrel{\vcenter{:}}= \theta_s\1*{A\cap B_\ell^c}.\] If \(H_1\) is the number of selected subsamples intersecting the delete set, then \[A\cap B_\ell^c = \set*{\widehat N_\ell^\circ=0,\;H_1>0}.\] The avoid and hit Bernoulli layers are independent, so \[\Pb*{A\cap B_\ell^c} = (1-p)^{\binom{n-d}{s}} \left\{ 1-(1-p)^{\binom{n}{s}-\binom{n-d}{s}} \right\} \leq \exp(-N_d^\circ).\] Define \[B_n \mathrel{\vcenter{:}}= \frac{n}{s^2\zeta_{s,\omega}^{1}} \frac{n-d}{d}\binom{n}{d}^{-1} \sum_{\ell\in L_{n,d}} C_\ell^2.\] By exchangeability and the preceding bound, \[\Ex*{B_n} \leq \theta_s^2 \frac{n(n-d)}{d\,s^2\zeta_{s,\omega}^{1}} \exp(-N_d^\circ).\] The assumptions imply this upper bound tends to zero: indeed \(1/\zeta_{s,\omega}^{1}=o(Ns/n)\), \(d\geq1\), \(s\geq1\), and \(N_d^\circ\asymp N\) reduce the polynomial factor to \(o(Nn)\), which is dominated by \(\exp(-N_d^\circ)\) because \(N/n\to\infty\). Hence \(B_n\xrightarrow{\mathbb{P}}0\). With \[A_n \mathrel{\vcenter{:}}= \frac{n\widehat\sigma_{JKD,h^0}^{2}}{s^2\zeta_{s,\omega}^{1}},\] Cauchy–Schwarz gives \[\frac{n}{s^2\zeta_{s,\omega}^{1}} \abs*{\widehat\sigma_{JKD,h}^{2}-\widehat\sigma_{JKD,h^0}^{2}} \leq 2A_n^{1/2}B_n^{1/2}+B_n \xrightarrow{\mathbb{P}} 0.\]

For the variance target, \(U_h=U_{h^0}+\theta_s\1*{A}\) and \(\Varb*{\theta_s\1*{A}}\leq\theta_s^2\Pb*{A^c}\leq\theta_s^2\exp(-N)\). The inequality \[\abs*{\Varb*{X+Y}-\Varb*{X}} \leq 2\sqrt{\Varb*{X}\Varb*{Y}}+\Varb*{Y}\] with \(X=U_{h^0}\) and \(Y=\theta_s\1*{A}\), together with the centered variance tightness and the same exponential domination argument, proves the variance-transfer claim. ◻

Proof of 2.
* We first prove the result for the centered kernel, so \(\theta_{s}=0\) throughout the main normalization argument. The zero-count centering-transfer step at the end removes this temporary restriction. The reciprocal-count expressions below use the zero-count conventions from 1 3. Whenever a selected kernel term appears inside the expectation, the corresponding realized count is automatically positive, so the reciprocal manipulations are on positive-count events. Let \[\hbar_{s}\left(D_{[s]}; \rho, \omega\right) \mathrel{\vcenter{:}}= \frac{\rho}{p} h_{s}\left(D_{[s]}; \omega\right), \qquad p \mathrel{\vcenter{:}}= \frac{N}{\binom{n}{s}},\] and define the Horvitz-Thompson normalized auxiliary complete generalized U-statistic \[\overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right) \mathrel{\vcenter{:}}= \binom{n}{s}^{-1} \sum_{\iota \in L_{n,s}} \hbar_{s}\left(D_{\iota}; \rho_{\iota}, \omega\right) = \frac{1}{N} \sum_{\iota \in L_{n,s}} \rho_{\iota} h_{s}\left(D_{\iota}; \omega\right).\] This differs from \(U_{n,s,N,\omega}\) only through the denominator: the random count \(\widehat N\) is replaced by its expectation \(N\). Working first with \(\overline{U}_{n,s,\rho,\omega}\) separates the random-normalization effect from the delete-\(d\) Hoeffding expansion. The overlap decomposition can then be carried out for a kernel with the same first projection as the original incomplete statistic, and the comparison between \(N\) and \(\widehat N\) is handled afterward as a separate normalization step.

By construction, for every \(1 \leq j \leq s-1\), \[\hbar_{s}^{(j)}\left(D_{[j]}\right) = h_{s}^{(j)}\left(D_{[j]}\right).\] In particular, \[\hbar_{s}^{(1)}(Z_1) = h_{s}^{(1)}(Z_1),\] so the normalized first-projection square is unchanged: \[\frac{\hbar_{s}^{(1)}(Z_1)^2}{\zeta_{s,\omega}^{1}} = \frac{h_{s}^{(1)}(Z_1)^2}{\zeta_{s,\omega}^{1}}.\] Consequently, the Horvitz-Thompson kernel has the same first-projection scale and satisfies the same row-wise square law. It remains to show that Bernoulli sampling noise is a negligible remainder.

Write \(\overline{V}_{s}^{s}\) and \(\overline{\zeta}_{s}^{s}\) for the final-order variance terms corresponding to \(\hbar_{s}\). Since the lower Hoeffding orders are unchanged and \(\theta_{s}=0\), \[\overline{\zeta}_{s}^{s} = \Varb*{\hbar_{s}\left(D_{[s]}; \rho, \omega\right)} = \Varb*{\frac{\rho}{p} h_{s}\left(D_{[s]}; \omega\right)} = \frac{1}{p}\zeta_{s}^{s},\] and therefore \[\overline{V}_{s}^{s} = V_{s}^{s} + \frac{1-p}{p}\zeta_{s}^{s}.\] Thus Horvitz-Thompson normalization leaves every lower Hoeffding order untouched and only inflates the final, purely \(s\)-fold component.

Let \(\overline{\sigma}_{n}^{2} \mathrel{\vcenter{:}}=\Varb*{\overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right)}\). The variance decomposition from the proof of 1 yields \[\begin{align} 1 & \leq \frac{n}{s^{2}} \frac{\overline{\sigma}_{n}^{2}}{\zeta_{s,\omega}^{1}} \\ & = \left(\frac{s^{2}}{n}\zeta_{s,\omega}^{1}\right)^{-1} \left( \sum_{j=1}^{s-1} \binom{s}{j}^{2}\binom{n}{j}^{-1}V_{s,\omega}^{j} + \binom{n}{s}^{-1}\overline{V}_{s}^{s} \right) \\ & \leq 1 + \frac{s}{n} \left( \frac{\zeta_{s}^{s}}{s\zeta_{s,\omega}^{1}} - 1 \right) + \frac{n}{s^{2}\zeta_{s,\omega}^{1}} \binom{n}{s}^{-1}\frac{1-p}{p}\zeta_{s}^{s} \\ & \leq 1 + \frac{s}{n} \left( \frac{\zeta_{s}^{s}}{s\zeta_{s,\omega}^{1}} - 1 \right) + \frac{n\zeta_{s}^{s}}{N s^{2}\zeta_{s,\omega}^{1}} \longrightarrow 1, \end{align}\] where the last step uses 1 3 and the boundedness of \(\zeta_{s}^{s}\). Thus, \[\label{eq:var95scale95u95bar} \frac{n}{s^{2}} \frac{\overline{\sigma}_{n}^{2}}{\zeta_{s,\omega}^{1}} \longrightarrow 1.\tag{11}\] So the Horvitz-Thompson normalized statistic already lives on the same variance scale as the complete statistic. The asymptotically-sufficient sampling condition is used here in its most transparent role: it says that the extra top-order variance created by Bernoulli thinning is too small to compete with the first projection.

We now analyze the delete-\(d\) jackknife built from \(\overline{U}_{n,s,\rho,\omega}\). For each delete set \(\ell \in L_{n,d}\), define \[\overline{\Delta}_{\ell} \mathrel{\vcenter{:}}= \Delta_{\ell}[\overline{U}_{n,s,\rho,\omega}] = \overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n],-\ell}\right).\] By the generalized Hoeffding decomposition, \[\overline{\Delta}_{\ell} = \sum_{j=1}^{s-1}\binom{s}{j}\delta_{j,\ell} + \overline{\delta}_{s,\ell},\] where the lower-order terms \(\delta_{j,\ell}\), \(1 \leq j \leq s-1\), are exactly the complete-case projection-level differences from 7, while the final-order term becomes \[\overline{\delta}_{s,\ell} = q_{s}^{(n,d)} \sum_{a=1}^{\min\{s,d\}} \overline{G}_{s,a}^{(\ell)} + r_{s}^{(n,d)} \overline{G}_{s,0}^{(\ell)},\] with \(\overline{G}_{s,a}^{(\ell)}\) denoting the overlap-stratum sums built from \(\hbar_{s}^{(s)}\). Thus the delete-\(d\) expansion differs from the complete-case expansion only through the final-order term. All lower-order overlap combinatorics and leading-term algebra are unchanged, so it remains to bound this modified top-order contribution.

Write \[h_{i} \mathrel{\vcenter{:}}= h_{s,\omega}^{(1)}(Z_i) = \hbar_{s}^{(1)}(Z_i), \qquad i = 1,\dotsc,n.\] Since the first projection is unchanged, the same algebra as in the proof of 1 gives \[\overline{\Delta}_{\ell} = \frac{s\sqrt d}{n} \left[ A_{\ell} + \overline{T}_{\ell} \right],\] where \[A_{\ell} \mathrel{\vcenter{:}}= \frac{1}{\sqrt d} \sum_{i \in \ell} h_{i}\] and \[\overline{T}_{\ell} \mathrel{\vcenter{:}}= -\frac{\sqrt d}{n-d} \sum_{i \in [n]\backslash \ell} h_{i} + \frac{n}{s\sqrt d} \left\{ \sum_{j=2}^{s-1}\binom{s}{j}\delta_{j,\ell} + \overline{\delta}_{s,\ell} \right\}.\]

Orthogonality of distinct Hoeffding orders yields \[\Ex*{\overline{T}_{\ell}^{2}} = \frac{d}{n-d}\zeta_{s,\omega}^{1} + \frac{n^{2}}{s^{2}d} \left\{ \sum_{j=2}^{s-1}\binom{s}{j}^{2}\Ex*{\delta_{j,\ell}^{2}} + \Ex*{\overline{\delta}_{s,\ell}^{2}} \right\}.\] For \(2 \leq j \leq s-1\), the lower-order terms are unchanged, so \[\Ex*{\delta_{j,\ell}^{2}} = \left[ \binom{n-d}{j}^{-1} - \binom{n}{j}^{-1} \right] V_{s,\omega}^{j},\] while the final-order term becomes \[\Ex*{\overline{\delta}_{s,\ell}^{2}} = \left[ \binom{n-d}{s}^{-1} - \binom{n}{s}^{-1} \right]\overline{V}_{s}^{s}.\] Using the same product and geometric-series bounds as in the complete-case analysis gives \[\begin{align} \Ex*{\overline{T}_{\ell}^{2}} & \lesssim \Bigg[ \frac{d}{n-d} + \frac{3e(s-1)(n-1)}{\left(n-1-e(s-1)\right)^{2}} \Bigg]\zeta_{s,\omega}^{1} \\ & \quad + \frac{3e(s-1)}{s} \Bigg[ \frac{1}{n-1} + \frac{n-1}{\left(n-1-e(s-1)\right)^{2}} \Bigg] \left( \zeta_{s}^{s} - s\zeta_{s,\omega}^{1} \right) \\ & \quad + 3\binom{n-1}{s-1}^{-1}\frac{1-p}{p}\zeta_{s}^{s}. \end{align}\] The last term is the only new Bernoulli-sampling contribution, and \[\binom{n-1}{s-1}^{-1}\frac{1-p}{p} = \binom{n-1}{s-1}^{-1} \left( \frac{\binom{n}{s}}{N} - 1 \right) = \frac{n}{Ns} - \binom{n-1}{s-1}^{-1}.\] Therefore, \[\frac{1}{\zeta_{s,\omega}^{1}} \binom{n-1}{s-1}^{-1}\frac{1-p}{p}\zeta_{s}^{s} \longrightarrow 0\] by 3 and the boundedness of \(\zeta_{s}^{s}\). The remaining terms satisfy the corresponding complete-case bounds, so \[\label{eq:overline95T95negligible} \frac{\Ex*{\overline{T}_{\ell}^{2}}}{\zeta_{s,\omega}^{1}} \longrightarrow 0.\tag{12}\] Thus, after subtracting off the linear Hájek piece, the remainder is negligible on the \(\zeta_{s,\omega}^{1}\) scale. The additional Bernoulli-sampling contribution is absorbed by 3.

Define \[\overline{X}_{\ell} \mathrel{\vcenter{:}}= \sqrt{ \frac{n-d}{n\binom{n}{d}\zeta_{s,\omega}^{1}} }\,A_{\ell}, \qquad \overline{Y}_{\ell} \mathrel{\vcenter{:}}= \sqrt{ \frac{n-d}{n\binom{n}{d}\zeta_{s,\omega}^{1}} }\,\overline{T}_{\ell}.\] Then \[\sum_{\ell \in L_{n,d}} \left( \overline{X}_{\ell} + \overline{Y}_{\ell} \right)^{2} = \frac{n\widehat{\overline{\sigma}}_{JKD}^{2}\left(D_{[n]}; d, \rho, \omega\right)}{s^{2}\zeta_{s,\omega}^{1}},\] where \[\widehat{\overline{\sigma}}_{JKD}^{2}\left(D_{[n]}; d, \rho, \omega\right) \mathrel{\vcenter{:}}= \frac{n-d}{d}\binom{n}{d}^{-1} \sum_{\ell \in L_{n,d}} \overline{\Delta}_{\ell}^{2}.\] Moreover, 12 implies \[\sum_{\ell \in L_{n,d}} \Ex*{\overline{Y}_{\ell}^{2}} = \frac{n-d}{n\zeta_{s,\omega}^{1}} \Ex*{\overline{T}_{\ell}^{2}} \longrightarrow 0.\] The corresponding statements for \(\overline{X}_{\ell}\) are exactly the same as in the complete-case proof because \(A_{\ell}\) is unchanged: \[\sum_{\ell \in L_{n,d}} \Ex*{\overline{X}_{\ell}^{2}} \longrightarrow 1, \qquad \sum_{\ell \in L_{n,d}} \overline{X}_{\ell}^{2} \xrightarrow{\mathbb{P}}1,\] where the second convergence again follows from the exact averaged-square identity and 3. Hence 2 gives \[\label{eq:scaled95ht95jkd} \frac{n\widehat{\overline{\sigma}}_{JKD}^{2}\left(D_{[n]}; d, \rho, \omega\right)}{s^{2}\zeta_{s,\omega}^{1}} \xrightarrow{\mathbb{P}}1.\tag{13}\] The limit in 13 is the Horvitz-Thompson target that remains to transfer to realized-count normalization. At this point the delete-\(d\) jackknife has already been shown to work for the Horvitz-Thompson normalized statistic. What remains is no longer a Hoeffding-decomposition problem. It is purely a question of whether replacing the deterministic counts \(N\) and \(N_{d}^{\circ}\) by their realized versions \(\widehat N\) and \(\widehat N_{\ell}^{\circ}\) can change the jackknife differences at first order.

It remains to replace the Horvitz-Thompson normalization by the empirical counts. For each delete set \(\ell\), define \[\widehat N_{\ell}^{\circ} \mathrel{\vcenter{:}}= \sum_{\iota \in \mathcal{O}_{s,0}(\ell)} \rho_{\iota}, \qquad N_{d}^{\circ} \mathrel{\vcenter{:}}= \Ex*{\widehat N_{\ell}^{\circ}} = p\binom{n-d}{s} = N\binom{n-d}{s}\binom{n}{s}^{-1}.\] Then \[U_{n,s,N,\omega}\left(D_{[n]}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right) = \left( \widehat N^{-1} - N^{-1} \right) \sum_{\iota \in L_{n,s}} \rho_{\iota} h_{s}\left(D_{\iota}; \omega\right),\] and \[U_{n,s,N,\omega}\left(D_{[n],-\ell}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n],-\ell}\right) = \left( \left(\widehat N_{\ell}^{\circ}\right)^{-1} - \left(N_{d}^{\circ}\right)^{-1} \right) \sum_{\iota \in \mathcal{O}_{s,0}(\ell)} \rho_{\iota} h_{s}\left(D_{\iota}; \omega\right).\] The actual incomplete statistic and its Horvitz-Thompson analogue differ only through the reciprocal-count factors. It remains to show that count fluctuations are negligible after jackknife rescaling. We start with the full-sample normalization error. Write \[S \mathrel{\vcenter{:}}= \sum_{\iota \in L_{n,s}} \rho_{\iota} h_{s}\left(D_{\iota}; \omega\right).\] By Cauchy-Schwarz, \[S^{2} \leq \widehat N \sum_{\iota \in L_{n,s}} \rho_{\iota} h_{s}\left(D_{\iota}; \omega\right)^{2}.\] Therefore, \[\begin{align} \Ex*{ \left( U_{n,s,N,\omega}\left(D_{[n]}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right) \right)^{2} } & = \Ex*{ \left( \frac{N-\widehat N}{N\widehat N} \right)^{2} S^{2} } \\ \qquad\leq & \frac{1}{N^{2}} \Ex*{ \frac{(N-\widehat N)^{2}}{\widehat N} \sum_{\iota \in L_{n,s}} \rho_{\iota} h_{s}\left(D_{\iota}; \omega\right)^{2} }. \end{align}\] By exchangeability and independence of \(\rho_{\iota}\) from the kernel, \[\begin{align} & \Ex*{ \left( U_{n,s,N,\omega}\left(D_{[n]}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right) \right)^{2} } \\ & \quad \leq \frac{\binom{n}{s}\zeta_{s}^{s}}{N^{2}} \Ex*{ \frac{(N-\widehat N)^{2}}{\widehat N} \rho_{[s]} }. \end{align}\] Conditioning on \(\rho_{[s]}=1\), we have \(\widehat N = 1 + B\) with \(B \sim \mathrm{Binomial}\left(\binom{n}{s}-1,p\right)\), so \[\begin{align} & \Ex*{ \frac{(N-\widehat N)^{2}}{\widehat N} \;\middle|\; \rho_{[s]} = 1 } \\ & \quad = \Ex*{\widehat N \;\middle|\;\rho_{[s]} = 1} - 2N + N^{2}\Ex*{\widehat N^{-1} \;\middle|\;\rho_{[s]} = 1}. \end{align}\] The binomial identity \[\Ex*{\frac{1}{1+B}} = \frac{1-(1-p)^{\binom{n}{s}}}{\binom{n}{s}p} \leq \frac{1}{N}\] gives \[\Ex*{ \frac{(N-\widehat N)^{2}}{\widehat N} \;\middle|\; \rho_{[s]} = 1 } \leq 1-p.\] Consequently, \[\label{eq:full95norm95error95bound} \Ex*{ \left( U_{n,s,N,\omega}\left(D_{[n]}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right) \right)^{2} } \leq \frac{\zeta_{s}^{s}}{N}.\tag{14}\]

The deleted-sample normalization error is handled in exactly the same way. For fixed \(\ell\), write \[S_{0}^{(\ell)} \mathrel{\vcenter{:}}= \sum_{\iota \in \mathcal{O}_{s,0}(\ell)} \rho_{\iota} h_{s}\left(D_{\iota}; \omega\right).\] Repeating the preceding argument with \(\widehat N_{\ell}^{\circ}\) in place of \(\widehat N\) and \(N_{d}^{\circ}\) in place of \(N\) yields \[\label{eq:deleted95norm95error95bound} \Ex*{ \left( U_{n,s,N,\omega}\left(D_{[n],-\ell}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n],-\ell}\right) \right)^{2} } \leq \frac{\zeta_{s}^{s}}{N_{d}^{\circ}}.\tag{15}\] Moreover, \[\frac{N_{d}^{\circ}}{N} = \frac{\binom{n-d}{s}}{\binom{n}{s}} = \prod_{m=0}^{s-1} \frac{n-d-m}{n-m} \longrightarrow 1\] because \(sd = o(n)\). Thus the expected retained count changes only by a \(1+o(1)\) factor, and 15 gives the same normalization bound for deleted samples.

For the normalization transfer, the relevant cancellation comes from separating the avoided and hit subsamples. For fixed \(\ell\), define \[\mathcal{O}_{s,1}(\ell) \mathrel{\vcenter{:}}= \set*{\iota \in L_{n,s}: \iota \cap \ell \neq \emptyset},\] and write \[H_{1} \mathrel{\vcenter{:}}= \sum_{\iota \in \mathcal{O}_{s,1}(\ell)} \rho_{\iota}, \qquad S_{1} \mathrel{\vcenter{:}}= \sum_{\iota \in \mathcal{O}_{s,1}(\ell)} \rho_{\iota} h_{s}\left(D_{\iota}; \omega\right), \qquad N_{1} \mathrel{\vcenter{:}}= \Ex*{H_{1}} = N - N_{d}^{\circ}.\] Thus \(\widehat N = \widehat N_{\ell}^{\circ} + H_{1}\) and \[\label{eq:hit95fraction95bound} \frac{N_{1}}{N} = 1 - \frac{\binom{n-d}{s}}{\binom{n}{s}} = 1 - \prod_{m=0}^{s-1} \left( 1 - \frac{d}{n-m} \right) \lesssim \frac{sd}{n},\tag{16}\] where the last step uses \(sd=o(n)\) and the elementary bound \(1-\prod_{m}(1-a_m)\leq \sum_m a_m\) for \(a_m \in [0,1]\). In particular, 16 gives \(N_{1}=o(N)\), while \(N_{d}^{\circ}\asymp N\).

Now let \[\Delta_{\ell} \mathrel{\vcenter{:}}= \Delta_{\ell}[U_{n,s,N,\omega}] = U_{n,s,N,\omega}\left(D_{[n]}\right) - U_{n,s,N,\omega}\left(D_{[n],-\ell}\right), \qquad R_{\ell} \mathrel{\vcenter{:}}= \Delta_{\ell} - \overline{\Delta}_{\ell}.\] Using \(S=S_{0}^{(\ell)}+S_{1}\) and \(\widehat N=\widehat N_{\ell}^{\circ}+H_{1}\), \[R_{\ell} = \left( \frac{1}{\widehat N} - \frac{1}{N} \right)S_{1} + \left( \frac{N_{1}}{N N_{d}^{\circ}} - \frac{H_{1}}{\widehat N \widehat N_{\ell}^{\circ}} \right)S_{0}^{(\ell)}.\] We bound the two terms on the right separately. First, use \[S_{1}^{2} \leq H_{1}\sum_{\iota \in \mathcal{O}_{s,1}(\ell)} \rho_{\iota}h_{s}(D_{\iota};\omega)^{2} \leq \widehat N\sum_{\iota \in \mathcal{O}_{s,1}(\ell)} \rho_{\iota}h_{s}(D_{\iota};\omega)^{2}.\] Then \[\left( \left( \frac{1}{\widehat N} - \frac{1}{N} \right)S_{1} \right)^{2} \leq \frac{(N-\widehat N)^{2}}{N^{2}\widehat N} \sum_{\iota \in \mathcal{O}_{s,1}(\ell)} \rho_{\iota}h_{s}(D_{\iota};\omega)^{2}.\] By exchangeability within \(\mathcal{O}_{s,1}(\ell)\), conditioning on any fixed hit subset being selected gives the same reciprocal-count factor as in the proof of 14 . Hence \[\Ex*{ \left( \left( \frac{1}{\widehat N} - \frac{1}{N} \right)S_{1} \right)^{2} } \lesssim \frac{N_{1}}{N^{2}}\zeta_{s}^{s} \lesssim \frac{sd}{n}\frac{\zeta_{s}^{s}}{N}.\] For the second term, on the event \(\widehat N_{\ell}^{\circ}>0\), write \[g_{k}(u) \mathrel{\vcenter{:}}= \frac{u}{k(k+u)}, \qquad k,u>0,\] so that \[\frac{N_{1}}{N N_{d}^{\circ}} - \frac{H_{1}}{\widehat N \widehat N_{\ell}^{\circ}} = g_{N_{d}^{\circ}}(N_{1}) - g_{\widehat N_{\ell}^{\circ}}(H_{1}).\] On the complementary zero-count event, \(S_{0}^{(\ell)}=0\) by convention, so the same product is identically zero and there is nothing to bound. Let \[G_\ell \mathrel{\vcenter{:}}= \set*{ \frac{1}{2} N_d^\circ \leq \widehat N_\ell^\circ \leq 2N_d^\circ }.\] On \(G_\ell\), \(N_d^\circ\asymp\widehat N_\ell^\circ\). Since \(N_1=o(N_d^\circ)\), the derivatives \[\partial_u g_k(u)=(k+u)^{-2}, \qquad \partial_k g_k(u)=-\frac{u(2k+u)}{k^2(k+u)^2}\] are uniformly controlled along the two line segments joining \((N_d^\circ,N_1)\), \((\widehat N_\ell^\circ,N_1)\), and \((\widehat N_\ell^\circ,H_1)\). Therefore \[\abs*{ g_{N_d^\circ}(N_1) - g_{\widehat N_\ell^\circ}(H_1) } \1*{G_\ell} \lesssim \left( \frac{\abs*{H_1-N_1}}{(N_d^\circ)^2} + \frac{N_1\abs*{\widehat N_\ell^\circ-N_d^\circ}}{(N_d^\circ)^3} \right) \1*{G_\ell}.\] Using \(\left(S_{0}^{(\ell)}\right)^2\leq\widehat N_\ell^\circ\sum_{\iota\in\mathcal{O}_{s,0}(\ell)}\rho_\iota h_s(D_\iota;\omega)^2\) and \(\widehat N_\ell^\circ\leq2N_d^\circ\) on \(G_\ell\), we obtain \[\begin{align} & \Ex*{ \left[ \left( \frac{N_{1}}{N N_{d}^{\circ}} - \frac{H_{1}}{\widehat N \widehat N_{\ell}^{\circ}} \right)S_{0}^{(\ell)} \right]^2 \1*{G_\ell} } \\ & \quad \lesssim \frac{1}{(N_d^\circ)^3} \Ex*{ (H_1-N_1)^2 \sum_{\iota\in\mathcal{O}_{s,0}(\ell)} \rho_\iota h_s(D_\iota;\omega)^2 } \\ & \qquad + \frac{N_1^2}{(N_d^\circ)^5} \Ex*{ (\widehat N_\ell^\circ-N_d^\circ)^2 \sum_{\iota\in\mathcal{O}_{s,0}(\ell)} \rho_\iota h_s(D_\iota;\omega)^2 }. \end{align}\] Now \(H_1\) depends only on the Bernoulli variables indexed by \(\mathcal{O}_{s,1}(\ell)\), while \(\widehat N_\ell^\circ\) and \(S_0^{(\ell)}\) depend only on those indexed by \(\mathcal{O}_{s,0}(\ell)\), so the hit count is independent of the avoid-layer quantities. Also, \[\Ex*{ \sum_{\iota\in\mathcal{O}_{s,0}(\ell)} \rho_\iota h_s(D_\iota;\omega)^2 } = N_d^\circ\zeta_s^s\] because the centered kernel has second moment \(\zeta_s^s\). For the second expectation, exchangeability within the avoid layer and conditioning on a fixed avoid subset being selected gives \(\widehat N_\ell^\circ=1+B_0\), with \(B_0\sim\mathrm{Binomial}(\abs*{\mathcal{O}_{s,0}(\ell)}-1,p)\); hence \[\Ex*{ (\widehat N_\ell^\circ-N_d^\circ)^2 \sum_{\iota\in\mathcal{O}_{s,0}(\ell)} \rho_\iota h_s(D_\iota;\omega)^2 } \lesssim (N_d^\circ)^2\zeta_s^s.\] Since \(\Varb*{H_1}=N_1(1-p)\leq N_1\), this gives \[\Ex*{ \left[ \left( \frac{N_{1}}{N N_{d}^{\circ}} - \frac{H_{1}}{\widehat N \widehat N_{\ell}^{\circ}} \right)S_{0}^{(\ell)} \right]^2 \1*{G_\ell} } \lesssim \frac{N_1}{(N_d^\circ)^2}\zeta_s^s + \frac{N_1^2}{(N_d^\circ)^3}\zeta_s^s.\] On \(G_\ell^c\), use the crude coefficient bound \[\abs*{ \frac{N_1}{NN_d^\circ} - \frac{H_1}{\widehat N\widehat N_\ell^\circ} } \leq \frac{N_1}{NN_d^\circ} + \frac{1}{\widehat N_\ell^\circ}\] on the positive-count event. Together with \(S_0^{(\ell)2}\leq \widehat N_\ell^\circ\sum_{\mathcal{O}_{s,0}(\ell)}\rho_\iota h_s(D_\iota;\omega)^2\), \(\ref{lem:reciprocal95binomial95good95count}\), and the same exchangeability argument, this yields, after reducing the exponential constant if necessary, \[\Ex*{ \left[ \left( \frac{N_1}{NN_d^\circ} - \frac{H_1}{\widehat N\widehat N_\ell^\circ} \right)S_0^{(\ell)} \right]^2 \1*{G_\ell^c} } \lesssim \zeta_s^s\exp(-cN_d^\circ).\] Combining the good-count and complement bounds, and using \(N_d^\circ\asymp N\), we conclude \[\Ex*{ \left[ \left( \frac{N_{1}}{N N_{d}^{\circ}} - \frac{H_{1}}{\widehat N \widehat N_{\ell}^{\circ}} \right)S_{0}^{(\ell)} \right]^{2} } \lesssim \frac{N_{1}}{\left(N_{d}^{\circ}\right)^{2}}\zeta_{s}^{s} + \frac{N_{1}^{2}}{\left(N_{d}^{\circ}\right)^{3}}\zeta_{s}^{s} + \zeta_s^s\exp(-cN_d^\circ)\] and therefore, by 16 , \[\Ex*{ \left[ \left( \frac{N_{1}}{N N_{d}^{\circ}} - \frac{H_{1}}{\widehat N \widehat N_{\ell}^{\circ}} \right)S_{0}^{(\ell)} \right]^{2} } \lesssim \frac{sd}{n}\frac{\zeta_{s}^{s}}{N} + \zeta_s^s\exp(-cN_d^\circ).\] Combining the last two displays and \(\abs*{a+b}^{2}\leq 2(a^{2}+b^{2})\), we obtain the sharpened coupled bound \[\label{eq:coupled95R95bound} \Ex*{R_{\ell}^{2}} \lesssim \frac{sd}{n}\frac{\zeta_{s}^{s}}{N} + \zeta_s^s\exp(-cN_d^\circ).\tag{17}\] Define \[\overline{Z}_{\ell} \mathrel{\vcenter{:}}= \sqrt{ \frac{n(n-d)}{s^{2}d\binom{n}{d}\zeta_{s,\omega}^{1}} }\,R_{\ell}.\] Using 17 , exchangeability gives \[\sum_{\ell \in L_{n,d}} \Ex*{\overline{Z}_{\ell}^{2}} = \frac{n(n-d)}{s^{2}d\zeta_{s,\omega}^{1}} \Ex*{R_{\ell}^{2}} \lesssim \frac{n-d}{N s\zeta_{s,\omega}^{1}} \zeta_{s}^{s} + \frac{n(n-d)}{s^2d\zeta_{s,\omega}^{1}} \zeta_s^s\exp(-cN_d^\circ) \longrightarrow 0\] by 3, the boundedness of \(\zeta_{s}^{s}\), \((n-d)/n \leq 1\), and the exponential domination used in 5.

Since \[\frac{n\widehat{\sigma}_{JKD}^{2}\left(D_{[n]}; d, \omega\right)}{s^{2}\zeta_{s,\omega}^{1}} = \sum_{\ell \in L_{n,d}} \left( \overline{X}_{\ell} + \overline{Y}_{\ell} + \overline{Z}_{\ell} \right)^{2},\] while \(\sum \Ex*{\overline{X}_{\ell}^{2}} \to 1\), \(\sum \overline{X}_{\ell}^{2} \xrightarrow{\mathbb{P}}1\), and \[\sum_{\ell \in L_{n,d}} \Ex*{ \left( \overline{Y}_{\ell} + \overline{Z}_{\ell} \right)^{2} } \leq 2 \sum_{\ell \in L_{n,d}} \Ex*{\overline{Y}_{\ell}^{2}} + 2 \sum_{\ell \in L_{n,d}} \Ex*{\overline{Z}_{\ell}^{2}} \longrightarrow 0,\] 2 yields \[\label{eq:scaled95incomplete95jkd} \frac{n\widehat{\sigma}_{JKD}^{2}\left(D_{[n]}; d, \omega\right)}{s^{2}\zeta_{s,\omega}^{1}} \xrightarrow{\mathbb{P}}1.\tag{18}\] Thus realized counts add only a second-order perturbation to the Horvitz-Thompson jackknife argument.

Finally, 14 implies \[\Varb*{ U_{n,s,N,\omega}\left(D_{[n]}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right) } \leq \frac{\zeta_{s}^{s}}{N} = o\left( \frac{s^{2}}{n}\zeta_{s,\omega}^{1} \right),\] again by 3 and bounded \(\zeta_{s}^{s}\). Hence \[\abs*{ \sigma_{n} - \overline{\sigma}_{n} } \leq \sqrt{ \Varb*{ U_{n,s,N,\omega}\left(D_{[n]}\right) - \overline{U}_{n,s,\rho,\omega}\left(D_{[n]}\right) } } = o\left( \sqrt{ \frac{s^{2}}{n}\zeta_{s,\omega}^{1} } \right),\] so 11 implies \[\frac{n\sigma_{n}^{2}}{s^{2}\zeta_{s,\omega}^{1}} \longrightarrow 1.\] The same comparison that transfers the jackknife estimator from \(\overline{U}_{n,s,\rho,\omega}\) to \(U_{n,s,N,\omega}\) also transfers the variance target itself. So both the estimator and the target are asymptotically governed by the same first-projection scale. Combining this with 18 gives, for centered kernels, \[\frac{\widehat{\sigma}_{JKD}^{2}\left(D_{[n]}; d, \omega\right)}{\sigma_{n}^{2}} \xrightarrow{\mathbb{P}}1.\] For a general kernel with bounded \(\theta_s\), the lower Hoeffding projections and \(\zeta_{s,\omega}^1\) are unchanged after replacing \(h_s\) by \(h_s-\theta_s\). Moreover, 3 and the boundedness of \(\zeta_s^s\) imply \(N/n\to\infty\) and \(1/\zeta_{s,\omega}^1=o(Ns/n)\). Since \(N_d^\circ/N\to1\) by \(sd=o(n)\), 5 transfers both the jackknife estimator and the variance target from the centered kernel back to the original kernel. This proves the stated ratio consistency under the zero-count convention. ◻

Proof of 8.
* By symmetry, it suffices to consider \(i=1\) and \(j=c+1\) for the first two equations. \[\begin{align} & \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D^{\prime}_{[s]}\right)}\\ & \quad = \Ex*{\kappa\left(x; Z_{1}, D_{[c]}\right)\kappa\left(x; Z_{1}, D_{(c+1):s}\right)\kappa\left(x; Z_{1}, D^{\prime}_{(c+1):s}\right)} \\ & \quad = \Ex*{\kappa\left(x; Z_{1}, D_{[2s - c]}\right)} = (2s - c)^{-1} \end{align}\] For the remaining two cases we use the reciprocal-binomial identity \[\label{eq:reciprocal95binomial95identity} \sum_{r = 0}^{a}\binom{a}{r}\binom{N}{r}^{-1} = \frac{N+1}{N+1-a}, \qquad 0 \leq a \leq N.\tag{19}\] This follows by writing \(\binom{a}{r}\binom{N}{r}^{-1} = \binom{N-r}{a-r}\binom{N}{a}^{-1}\) and applying the hockey-stick identity. Considering the second case, we find the following. \[\begin{align} & \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)} \\ & = \frac{1}{(2s - c)!} \sum_{r = 0}^{s - c - 1}\binom{s - c - 1}{r} r! \left((s - 1) + (s - c - 1 - r)\right)! \\ & = \frac{1}{(2s - c)!}\sum_{r = 0}^{s - c - 1}\binom{s - c - 1}{r} r! \left(2s - c - 2 - r\right)! \\ & = \frac{1}{(2s - c)(2s - c - 1)} \sum_{r = 0}^{s - c - 1}\binom{s - c - 1}{r}\binom{2s - c - 2}{r}^{-1} \\ & \overset{\text{\ref{eq:reciprocal95binomial95identity}}}{=} \frac{1}{s(2s-c)}. \end{align}\] While unintuitive at first, the terms in this expression have intuitive meaning when we consider this as a combinatorial problem. Consider lining up the observations in order of their distance to the point of interest and counting the cases for which the expression in the expectation is equal to one. First, there are \((2s-c)!\) possible orderings of the observations with probability one, leading to the denominator. Next, notice that only those orderings where \(\norm*{X_{c+1}^{\prime} - x} \leq \norm*{X_{1}-x}\) and \(\norm*{X_{1} - x} \leq \norm*{X_{i} - x}\) for any \(i = 2, \dotsc, c\) can possibly lead to a non-zero realization of the kernel term. Furthermore, out of the \((s-c-1)\) observations in \(D^{\prime}_{(c+2):s}\), it is possible for \(i = 0, \dotsc, s-c-1\) observations to lie at a distance to the point of interest that is smaller than \(\norm*{X_{1}-x}\) but larger than \(\norm*{X_{c+1}^{\prime} - x}\) in any permutation. The sum adjusts for those possible configurations. Because \(c \leq s-1\), the exact value is bounded from below by \((2s^2)^{-1}\) and is therefore \(\omega(e^{-s})\).

Considering the third case, without loss of generality, we consider the case of \(i = j = c+1\). We find the following. \[\begin{align} & \Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)} \\ & = \mathbb{E}\left[ \kappa\left(x; Z_{c+1}, D_{[c]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[c]}\right) \kappa\left(x; Z_{c+1}, D_{(c+1):s}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{(c+1):s}^{\prime}\right) \right] \\ & = \frac{2}{(2s - c)!} \sum_{r = 0}^{s-c-1} \binom{s - c - 1}{r}(s - 1 + r)!(s-c-1-r)! \\ & = \frac{2(2s - c - 2)!}{(2s-c)!} \sum_{r = 0}^{s-c-1} \binom{s - c - 1}{r}\binom{2s - c - 2}{s-1+r}^{-1} \\ & = \frac{2}{(2s-c)(2s-c-1)}\sum_{r = 0}^{s-c-1} \binom{s - c - 1}{r}\binom{2s - c - 2}{s-1+r}^{-1} \\ & = \frac{2}{(2s-c)(2s-c-1)}\sum_{r = 0}^{s-c-1} \binom{s - c - 1}{r}\binom{2s - c - 2}{s-c-1-r}^{-1} \\ & \overset{\text{\ref{eq:reciprocal95binomial95identity}}}{=} \frac{2}{s(2s-c)}. \end{align}\] The third case follows from a similar combinatorial logic as the second. By symmetry, take \(\norm*{X_{c+1}^{\prime} - x} \leq \norm*{X_{c+1}-x}\) and multiply by two. Any number \(i = 0, \dotsc, s-c-1\) of observations in \(D^{\prime}_{(c+2):s}\) can be farther from \(x\) than \(X_{c+1}\) or lie between \(\norm*{X_{c+1}^{\prime} - x}\) and \(\norm*{X_{c+1}-x}\). The summation counts these configurations. The exact value is again bounded from below by \((s^2)^{-1}\), and hence it is \(\omega(e^{-s})\). ◻

Proof of 10.
* We will largely argue along the same lines as the original proof in [4]. Thus, consider first the following inequalities. \[\begin{align} & \abs*{(A) - f^{2}(x)} = \left|\mathbb{E}\left[f^{2}(X_{1}) (2s - c) \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}} \right] - f^{2}(x)\right| \\ & \leq \mathbb{E}\left[\abs*{f^{2}(X_{1}) - f^{2}(x)} (2s - c) \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}} \right] \end{align}\] \[\begin{align} & \abs*{(B) - f^{2}(x)} \\ & = \left|\mathbb{E}\left[ f(X_{1}) f(X_{c+1}^{\prime}) \frac{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \right] - f^{2}(x)\right| \\ & \leq \mathbb{E}\left[ \abs*{f(X_{1}) f(X_{c+1}^{\prime}) - f^{2}(x)} \frac{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \right] \end{align}\] \[\begin{align} & \abs*{(C) - f^{2}(x)} \\ & = \left|\mathbb{E}\left[ f(X_{c+1}) f(X_{c+1}^{\prime}) \frac{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{c+1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \right] - f^{2}(x)\right| \\ & \leq \mathbb{E}\left[ \abs*{f(X_{c+1}) f(X_{c+1}^{\prime}) - f^{2}(x)} \frac{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{c+1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \right] \end{align}\] Now, fix an arbitrary \(\epsilon > 0\). By continuity of \(f\) at \(x\), there exists a \(\delta > 0\), such that the following holds. \[\forall X, X^{\prime} \in B(x, \delta): \quad \abs*{f(X) f(X^{\prime}) - f^{2}(x)} < \epsilon\] We can consider decompositions of these terms in analogy to [4], i.e., by considering cases with observations lying within this sphere or outside of it, and observe the following. \[\begin{align} & \mathbb{E}\Big[\abs*{f^{2}(X_{1}) - f^{2}(x)} (2s - c) \\ & \quad \times \mathbb{E}\left[\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D^{\prime}_{[s]}\right) \1*{X_1 \in B(x, \delta)} \middle| X_{1} \right] \Big] \\ & \leq \epsilon \mathbb{E}\left[(2s - c) \mathbb{E}\left[\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D^{\prime}_{[s]}\right) \1*{X_1 \in B(x, \delta)} \middle| X_{1} \right] \right] \\ & \leq \epsilon \mathbb{E}\left[(2s - c) \mathbb{E}\left[\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D^{\prime}_{[s]}\right) \middle| X_{1} \right] \right] = \epsilon \end{align}\] \[\begin{align} & \mathbb{E}\Bigg[ \abs*{f(X_{1}) f(X_{c+1}^{\prime}) - f^{2}(x)} \1*{X_1, X_{c+1}^{\prime} \in B(x, \delta)} \\ & \qquad \times \left. \frac{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \right] \\ & \leq \epsilon \mathbb{E}\left[ \frac{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \1*{X_1, X_{c+1}^{\prime} \in B(x, \delta)} \right] \\ & \leq \epsilon \mathbb{E}\left[ \frac{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \right] = \epsilon \end{align}\] \[\begin{align} & \mathbb{E}\Bigg[ \abs*{f(X_{c+1}) f(X_{c+1}^{\prime}) - f^{2}(x)} \frac{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{c+1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \\ & \quad \times \1*{X_{c+1}, X_{c+1}^{\prime} \in B(x, \delta)} \Bigg] \\ & \leq \epsilon \mathbb{E}\left[ \frac{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{c+1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \1*{X_{c+1}, X_{c+1}^{\prime} \in B(x, \delta)} \right] \\ & \leq \epsilon \mathbb{E}\left[ \frac{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{c+1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \right] = \epsilon \end{align}\] For the complementary events, use the fact that if \(X\) or \(X^{\prime}\) does not lie in \(B(x,\delta)\), then \[B(x, \delta) \subseteq B(x, \max\left(\norm*{X - x}, \norm*{X^{\prime} - x}\right)).\] Therefore, \[\begin{align} & \mathbb{E}\Bigg[\abs*{f^{2}(X_{1}) - f^{2}(x)} (2s - c) \\ & \qquad \times \mathbb{E}\left[\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D^{\prime}_{[s]}\right) \left(1 - \1*{X_1 \in B(x, \delta)}\right) \middle| X_{1} \right] \Bigg] \\ & \leq \mathbb{E}\left[\abs*{f^{2}(X_{1}) - f^{2}(x)} (2s - c) \left\{1-\varphi\left(B\left(x,\delta\right)\right)\right\}^{2s-c-1} \left(1 - \1*{X_1 \in B(x, \delta)}\right) \right] \\ & \leq (2s - c) 1 \left\{1-\varphi\left(B\left(x,\delta\right)\right)\right\}^{2s-c-1} \Ex*{\abs*{f^{2}(X_{1}) - f^{2}(x)}} \end{align}\] In the second case, first recall the form of the conditional expectation from 9. \[\begin{align} & \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_1, X_{c+1}^{\prime}} \\ & = \1*{\norm*{X_{c+1}^{\prime} - x} \leq \norm*{X_{1} - x}} \\ & \qquad \times \left\{1-\varphi\left(B\left(x,\norm*{X_1-x}\right)\right)\right\}^{s-1} \left\{1-\varphi\left(B\left(x,\norm*{X_{c+1}^{\prime}-x}\right)\right)\right\}^{s-c-1} \end{align}\] The indicator is nonzero only if \(\max\left(\norm*{X_{1} - x}, \norm*{X_{c+1}^{\prime} - x}\right) = \norm*{X_{1} - x}\), in which case \[B(x, \delta) \subseteq B(x, \norm*{X_{1} - x})\] Hence \[\begin{align} & \mathbb{E}\Bigg[ \abs*{f(X_{1}) f(X_{c+1}^{\prime}) - f^{2}(x)} \frac{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right)}} \\ & \quad \left(1 - \1*{X_1, X_{c+1}^{\prime} \in B(x, \delta)}\right) \Bigg] \\ & \overset{(\text{\ref{lem:expec95kernel95prod}})}{\leq} (2s-c)^{2} \mathbb{E}\Big[ \abs*{f(X_{1}) f(X_{c+1}^{\prime}) - f^{2}(x)} \\ & \qquad \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D^{\prime}_{[s]}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}} \left(1 - \1*{X_1, X_{c+1}^{\prime} \in B(x, \delta)}\right) \Big] \\ & \overset{(\text{\ref{lem:cond95expec95kernel95prod}})}{=} (2s-c)^{2} \mathbb{E}\Big[ \abs*{f(X_{1}) f(X_{c+1}^{\prime}) - f^{2}(x)} \1*{\norm*{X_{c+1}^{\prime} - x} \leq \norm*{X_{1} - x}} \\ & \qquad \times \left\{1-\varphi\left(B\left(x,\norm*{X_1-x}\right)\right)\right\}^{s-1} \left\{1-\varphi\left(B\left(x,\norm*{X_{c+1}^{\prime}-x}\right)\right)\right\}^{s-c-1} \\ & \qquad \times \left(1 - \1*{X_1, X_{c+1}^{\prime} \in B(x, \delta)}\right) \Big] \\ & \leq (2s-c)^{2} \left\{1-\varphi\left(B\left(x, \delta \right)\right)\right\}^{s-1} \\ & \qquad \times \mathbb{E}\Big[ \abs*{f(X_{1}) f(X_{c+1}^{\prime}) - f^{2}(x)} \1*{\delta < \norm*{X_{c+1}^{\prime} - x} \leq \norm*{X_{1} - x}} \Big] \\ & \leq (2s-c)^{2} \left\{1-\varphi\left(B\left(x, \delta \right)\right)\right\}^{s-1} \Ex*{\abs*{f(X_{1}) f(X_{c+1}^{\prime}) - f^{2}(x)}} \end{align}\]

 ◻

Proof of 11.
* Both inequalities are immediate from the exact identities in 8. ◻

Proof of 12. This result follows immediately from 7 and the following observation. \[\begin{align} & \Ex*{Y_1 s \Ex*{\kappa\left(x; Z_1, D_{[s]}\right) \;\middle|\;X_{1}}} = \Ex*{\left(\mu\left(X_1\right) + \varepsilon_1\right) s \Ex*{\kappa\left(x; Z_1, D_{[s]}\right) \;\middle|\;X_{1}}} \\ & = \Ex*{\left(\mu\left(X_1\right) + \Ex*{\varepsilon_1 \;\middle|\;X_1}\right) s \Ex*{\kappa\left(x; Z_1, D_{[s]}\right) \;\middle|\;X_{1}}} \\ & = \Ex*{\mu\left(X_1\right) s \Ex*{\kappa\left(x; Z_1, D_{[s]}\right) \;\middle|\;X_{1}}} \\ & \overset{\text{(\ref{lem:dem13})}}{\longrightarrow} \mu\left(x\right) \quad \text{as} \quad s \longrightarrow \infty \end{align}\] ◻

Proof of 13. \[\begin{align} & \psi_{s}^{1}\left(x; z_1\right) = \Ex*{h_{s}\left(x; D_{[s]}\right) \;\middle|\;Z_1 = z_1} = \Ex*{\sum_{i = 1}^{s} \kappa\left(x; Z_{i}, D_{[s]}\right) Y_{i} \;\middle|\;Z_1 = z_1} \\ & = \mathbb{E}\Bigg[\left(\mu(x_1) + \varepsilon_1\right)\kappa\left(x; Z_1, D_{[s]}\right) \\ & \qquad\qquad\quad + \sum_{i = 2}^{s} \kappa\left(x; Z_{i}, D_{[s]}\right) \mu(X_{i})\;\Bigg|\; Z_1 = z_1 \Bigg] \\ & = \mu(x_1)\Ex*{\kappa\left(x; Z_1, D_{[s]}\right)\;\middle|\;X_1 = x_1} \\ & \qquad + \varepsilon_1 \Ex*{\kappa\left(x; Z_1, D_{[s]}\right)\;\middle|\;X_1 = x_1} \\ & \qquad + \Ex*{\sum_{i = 2}^{s} \kappa\left(x; Z_{i}, D_{[s]}\right) \mu(X_{i})\;\middle|\;X_1 = x_1} \end{align}\] ◻

Proof of 14.
* This result follows immediately from 7 and the following observation. \[\begin{align} & \Omega_{s}\left(x\right) = \Ex*{h_{s}^{2}\left(x; D_{[s]}\right)} = \Ex*{\left(\sum_{i = 1}^{s}\kappa\left(x; Z_{i}, D_{[s]}\right)Y_{i}\right)^2} \\ & = \Ex*{\sum_{i = 1}^{s}\sum_{j = 1}^{s}\left(\kappa\left(x; Z_{i}, D_{[s]}\right)\kappa\left(x; Z_{j}, D_{[s]}\right)Y_{i}Y_{j}\right)} = \Ex*{s \kappa\left(x; Z_{1}, D_{[s]}\right)Y_{1}^2} \\ & = \Ex*{Y_{1}^2 s \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right) \;\middle|\;X_{1}}} = \Ex*{\left(\mu\left(X_1\right) + \varepsilon_1\right)^2 s \Ex*{\kappa\left(x; Z_1, D_{[s]}\right) \;\middle|\;X_{1}}} \\ & = \Ex*{\left(\mu^2\left(X_1\right) + 2\mu\left(X_1\right)\varepsilon_1 + \varepsilon_1^2\right)s \Ex*{\kappa\left(x; Z_1, D_{[s]}\right) \;\middle|\;X_{1}}} \\ & = \mathbb{E}\left[\left(\mu^2\left(X_1\right) + 2\mu\left(X_1\right) \Ex*{\varepsilon_1 \;\middle|\;X_1} + \Ex*{\varepsilon_1^2 \;\middle|\;X_1}\right) s \Ex*{\kappa\left(x; Z_1, D_{[s]}\right) \;\middle|\;X_{1}}\right] \\ & = \Ex*{\left(\mu^2\left(X_1\right) +\sigma_{\varepsilon}^{2}(X_1)\right) s \Ex*{\kappa\left(x; Z_1, D_{[s]}\right) \;\middle|\;X_{1}}} \\ & \overset{\text{(\ref{lem:dem13})}}{\longrightarrow} \mu^2\left(x\right) +\sigma_{\varepsilon}^{2}(x) \quad \text{as} \quad s \longrightarrow \infty \end{align}\] Furthermore, we have the following inequality. \[\mu^2(x) + \sigma_{\varepsilon}^2(x) \leq \mu^2\left(x\right) + \overline{\sigma}_{\varepsilon}^{2}\] Thus, we obtain the desired result. ◻

Proof of 15. \[\begin{align} & \Omega_{s}^{c}\left(x\right) = \mathbb{E}\left[h_{s}\left(x; D_{[s]}\right) h_{s}\left(x; D_{[s]}^{\prime}\right)\right] \\ & = \mathbb{E}\left[ \left(\sum_{i = 1}^{s}\kappa\left(x; Z_{i}, D_{[s]}\right)Y_{i}\right) \left(\sum_{j = 1}^{c}\kappa\left(x; Z_{j}, D_{[s]}^{\prime}\right)Y_{j} + \sum_{j = c+1}^{s}\kappa\left(x; Z_{j}^{\prime}, D_{[s]}^{\prime}\right)Y_{j}^{\prime}\right) \right] \\ & = \underbrace{\Ex*{c \kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D_{[s]}^{\prime}\right)Y_{1}^{2}}}_{(A)} \\ & \quad + 2 \underbrace{\Ex*{c(s-c) \kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right)Y_{1}Y_{c+1}^{\prime}}}_{(B)} \\ & \quad + \underbrace{\Ex*{(s-c)^2 \kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right)Y_{c+1}Y_{c+1}^{\prime}}}_{(C)} \end{align}\] Starting from this decomposition, we will analyze the terms one by one using 10. \[\begin{align} (A) & = \Ex*{c\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D_{[s]}^{\prime}\right)Y_{1}^{2}} \\ & = \frac{c}{2s-c} \Ex*{\left(\mu(X_1) + \varepsilon_1\right)^2 (2s-c) \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D_{[s]}^{\prime}\right) \;\middle|\;X_{1}}} \\ & = \frac{c}{2s-c} \Ex*{\left(\mu^{2}(X_1) + \sigma_{\varepsilon}^{2}(X_1)\right) (2s-c) \Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{1}, D_{[s]}^{\prime}\right) \;\middle|\;X_{1}}} \\ & \overset{\text{(\ref{lem:kernel95prod95dirac95convergence})}}{\lesssim} \frac{c}{2s-c} \left(\mu^2(x) + \sigma_{\varepsilon}^{2}(x)\right) + o(1) \end{align}\] Similarly, we can find the following. \[\begin{align} (B) & = \Ex*{c(s-c) \kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right)Y_{1}Y_{c+1}^{\prime}} \\ & \overset{(\text{\ref{lem:expec95kernel95prod}})}{=} \frac{c(s-c)}{s(2s-c)} \mathbb{E}\Bigg[ \left(\mu(X_1) + \varepsilon_{1}\right)\left(\mu(X_{c+1}^{\prime}) + \varepsilon_{c+1}^{\prime}\right) \\ & \qquad \times \left. \frac{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right)}} \right] \\ & = \frac{c(s-c)}{s(2s-c)} \mathbb{E}\left[ \mu(X_1) \mu(X_{c+1}^{\prime}) \frac{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right) \;\middle|\;X_{1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right)}} \right] \\ & \overset{(\text{\ref{lem:kernel95prod95dirac95convergence}})}{\lesssim} \frac{c(s-c)}{s(2s-c)} \mu^2(x) + o(1) \end{align}\] The third term can be asymptotically bounded in the following way. \[\begin{align} (C) & = \Ex*{(s-c)^2 \kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right)Y_{c+1}Y_{c+1}^{\prime}} \\ & \overset{(\text{\ref{lem:expec95kernel95prod}})}{=} \frac{2(s-c)^2}{s(2s-c)} \mathbb{E}\Bigg[ \mu(X_{c+1}) \mu(X_{c+1}^{\prime}) \\ & \qquad\qquad\qquad \times \frac{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right) \;\middle|\;X_{c+1}, X_{c+1}^{\prime}}}{\Ex*{\kappa\left(x; Z_{c+1}, D_{[s]}\right)\kappa\left(x; Z_{c+1}^{\prime}, D_{[s]}^{\prime}\right)}} \Bigg] \\ & \overset{(\text{\ref{lem:kernel95prod95dirac95convergence}})}{\lesssim} \frac{2(s-c)^2}{s(2s-c)} \mu^2(x) + o(1) \end{align}\] The coefficients \(c/(2s-c)\), \(c(s-c)/(s(2s-c))\), and \(2(s-c)^2/(s(2s-c))\) are uniformly bounded over \(1 \leq c \leq s-1\). The result of 15 follows immediately by summing up the asymptotic bounds for the individual terms. ◻

Proof. Observation \(1\) is the nearest neighbor among \(s\) draws exactly when the remaining \(s-1\) observations all lie outside \(B(x,R_1)\). Hence 20 follows. The density condition implies that \(U_1=F_x(R_1)\) is uniform on \((0,1)\), so the displayed moment identities follow by integrating powers of \(1-U_1\) over the unit interval. ◻

Proof. By 14, \[\zeta_s^s(x) \leq \Omega_s(x) \lesssim 1.\] For the lower bound on the first-projection variance, use the selector profile \(q_s\) from 16. The decomposition in 13 gives \(h_s^{(1)}(x; Z_1) = q_s(X_1)\varepsilon_1 + b_s(X_1)\) for an \(X_1\)-measurable remainder \(b_s\). Since \(\Ex{\varepsilon_1 \;\middle|\;X_1} = 0\), the law of total variance gives \[\zeta_s^1(x) \geq \Ex*{q_s(X_1)^2\sigma_\varepsilon^2(X_1)} \geq \underline{\sigma}_{\varepsilon}^{2}\, \Ex*{q_s(X_1)^2}.\] 21 with \(a=2\) gives \(\Ex*{q_s(X_1)^2} = (2(s-1)+1)^{-1}\), and hence \(\zeta_s^1(x) \gtrsim s^{-1}\). The cross-scale identity in 22 is used later in the TDNN non-cancellation argument. The reverse bound \(\zeta_s^1(x) \lesssim s^{-1}\) is the first-projection variance rate from [4], adapted to the present heteroskedastic setup by using the uniform variance bound in 5 [asm:tdnn95design95variance]. Consequently, \[\zeta_s^1(x) \asymp s^{-1}.\] Therefore, \[\frac{s}{n} \left( \frac{\zeta_s^s(x)}{s\zeta_s^1(x)} - 1 \right) = O\left(\frac{s}{n}\right) \longrightarrow 0\] as \(s=o(n)\), which is exactly 1 for the single-scale DNN estimator. ◻

Proof. Write \(r=r_\eta\) for the exponent fixed in the statement. The \(b_s\) bound below uses the upper rate \(\zeta_s^1(x) \lesssim s^{-1}\) from 17, which rests on the first-projection variance calculation of [4]. Selector moment. For \(q_s\) from 16, 21 with \(a=2r\) gives \[\Ex*{q_s(X_1)^{2r}} = \frac{1}{2r(s-1)+1} \lesssim s^{-1}.\] By 13, the first projection decomposes as \(h_s^{(1)}(x; Z_1) = q_s(X_1)\,\varepsilon_1 + b_s(X_1)\), where \[b_s(u) \coloneq \mu(u)\,q_s(u) + \mathbb{E}\!\left[ \sum_{i=2}^{s}\kappa\!\left(x; Z_i, D_{[s]}\right)\mu(X_i) \,\middle|\, X_1 = u \right] - \theta_s(x).\] Bias remainder. Since \(\mathcal{X}\) is compact and \(\mu\) is continuous under 5 [asm:tdnn95design95compact] [asm:tdnn95design95regression], write \(M_\mu \mathrel{\vcenter{:}}=\sup_{v\in\mathcal{X}}\abs*{\mu(v)} < \infty\). For the \(b_s\) remainder: since the selector weights \(\kappa(x; Z_i, D_{[s]})\) sum to one almost surely (6), \[\mu(u)\,q_s(u) + \Ex*{\sum_{i=2}^{s}\kappa\mu(X_i) \;\middle|\;X_1=u} = \Ex*{\sum_{i=1}^{s}\kappa(x;Z_i,D_{[s]})\mu(X_i) \;\middle|\;X_1=u}.\] The right-hand side is bounded by \(M_\mu\) in absolute value. Together with \(\abs*{\theta_s(x)} \leq M_\mu\) (since \(\theta_s(x)\) is itself an expectation of the same convex combination of \(\mu\) values, bounded by \(M_\mu\) via 6), we have \(\abs*{b_s(u)} \leq 2M_\mu\) uniformly in \(u\). Because \(\Ex{\varepsilon_1 \;\middle|\;X_1} = 0\) kills the cross term, \[\zeta_s^1(x) = \Varb*{h_s^{(1)}(x; Z_1)} = \Ex*{q_s(X_1)^2\,\sigma_\varepsilon^2(X_1)} + \Ex*{b_s(X_1)^2} \geq \Ex*{b_s(X_1)^2},\] so \(\Ex{b_s(X_1)^2} \leq \zeta_s^1(x) \lesssim s^{-1}\) by 17. The \(L^\infty\) bound then gives \[\Ex*{\abs*{b_s(X_1)}^{2r}} \leq (2M_\mu)^{2(r-1)}\,\Ex*{b_s(X_1)^2} \lesssim s^{-1}.\] Conditional error moment. By 8, the choice of \(r=r_\eta\), and the boundedness of \(\mu\) on \(\mathcal{X}\), there is a constant \(C_{\varepsilon,r}<\infty\) such that \[\label{eq:dnn95conditional95error95moment} \sup_{u\in\mathcal{X}} \Ex*{\abs*{\varepsilon}^{2r}\;\middle|\;X=u} \leq C_{\varepsilon,r}.\tag{24}\] Since \(q_s(X_1)\) is \(X_1\)-measurable, iterated expectation and 21 24 give \[\label{eq:dnn95selector95error95moment} \begin{align} \Ex*{ q_s(X_1)^{2r}\abs*{\varepsilon_1}^{2r} } & = \Ex*{ q_s(X_1)^{2r} \Ex*{\abs*{\varepsilon_1}^{2r}\;\middle|\;X_1} } \\ & \leq C_{\varepsilon,r} \Ex*{q_s(X_1)^{2r}} \lesssim s^{-1}. \end{align}\tag{25}\] Combining 25 with the \(b_s\) bound and \(\abs*{a+b}^{2r}\leq 2^{2r-1}\left(\abs*{a}^{2r}+\abs*{b}^{2r}\right)\) yields \[\label{eq:dnn95first95projection95moment} \Ex*{\abs*{h_s^{(1)}(x; Z_1)}^{2r}} \lesssim s^{-1}.\tag{26}\] By 17, \(\zeta_s^1(x) \asymp s^{-1}\). Consequently, \[\Ex*{ \left( \frac{h_s^{(1)}(x; Z_1)^2}{\zeta_s^1(x)} \right)^r } = \frac{\Ex*{\abs*{h_s^{(1)}(x; Z_1)}^{2r}}}{\left(\zeta_s^1(x)\right)^r} \lesssim s^{r-1}.\] Since \(r>1\) and \(s=o(n)\), \[\frac{1}{n^{r-1}} \Ex*{ \left( \frac{h_s^{(1)}(x; Z_1)^2}{\zeta_s^1(x)} \right)^r } \lesssim \left(\frac{s}{n}\right)^{r-1} \longrightarrow 0.\] This proves 2. ◻

Proof. Write \(\rho \coloneq s_1 / s_2\). Since \[w_1^* = \frac{1}{1-\rho^{-2/k}} = -\frac{\rho^{2/k}}{1-\rho^{2/k}}, \qquad w_2^* = 1-w_1^* = \frac{1}{1-\rho^{2/k}},\] we have \[a_{\mathfrak S} = -\frac{\rho^{1+2/k}}{1-\rho^{2/k}}, \qquad b_{\mathfrak S} = \frac{1}{1-\rho^{2/k}}.\] Under 7, both coefficients are uniformly bounded and \(b_{\mathfrak S} \geq 1\).

The selector-profile moments from 16 give \[\begin{align} \Ex*{q_{s_1}(X_1)^2} & = \frac{1}{2s_1-1}, \\ \Ex*{q_{s_2}(X_1)^2} & = \frac{1}{2s_2-1}, \\ \Ex*{q_{s_1}(X_1) q_{s_2}(X_1)} & = \frac{1}{s_1+s_2-1}. \end{align}\] The following matrix records the \(L^2\) geometry of the two selector profiles \(q_{s_1}\) and \(q_{s_2}\). Its determinant lower bound says that, under the scale-separation condition, these profiles do not become asymptotically collinear, so the signed TDNN coefficients cannot cancel the stochastic first-projection signal. Let \[\Gamma_{\mathfrak S} \coloneq s_2 \begin{pmatrix} \Ex*{q_{s_1}(X_1)^2} & \Ex*{q_{s_1}(X_1) q_{s_2}(X_1)} \\ \Ex*{q_{s_1}(X_1) q_{s_2}(X_1)} & \Ex*{q_{s_2}(X_1)^2} \end{pmatrix}.\] Then \[s_2 \Ex*{\tilde{q}_{\mathfrak S}(X_1)^2} = \begin{pmatrix} a_{\mathfrak S} & b_{\mathfrak S} \end{pmatrix} \Gamma_{\mathfrak S} \begin{pmatrix} a_{\mathfrak S} \\ b_{\mathfrak S} \end{pmatrix}.\] Moreover, \[\det\left(\Gamma_{\mathfrak S}\right) = \frac{s_2^2 (s_2-s_1)^2}{(2s_1-1)(2s_2-1)(s_1+s_2-1)^2}.\] Using 7, \[s_1 \geq \mathfrak c s_2, \qquad s_2-s_1 \geq \mathfrak c s_2, \qquad s_1 \leq (1-\mathfrak c)s_2,\] so \[\det\left(\Gamma_{\mathfrak S}\right) \geq \frac{\mathfrak c^2}{16(1-\mathfrak c)}.\] Since \[\operatorname{tr}\left(\Gamma_{\mathfrak S}\right) = \frac{s_2}{2s_1-1} + \frac{s_2}{2s_2-1} \leq \frac{1}{\mathfrak c} + 1,\] the eigenvalues of \(\Gamma_{\mathfrak S}\) are uniformly bounded above and away from zero. In particular, there exist constants \(0 < \underline\lambda \leq \overline{\lambda} < \infty\) such that \[\underline\lambda I_2 \preceq \Gamma_{\mathfrak S} \preceq \overline{\lambda} I_2.\] Because \(a_{\mathfrak S}\) and \(b_{\mathfrak S}\) are uniformly bounded and \(b_{\mathfrak S} \geq 1\), there exists \(C_v < \infty\) such that \[1 \leq a_{\mathfrak S}^2 + b_{\mathfrak S}^2 \leq C_v.\] Combining the last two displays gives \[\underline\lambda \leq s_2 \Ex*{\tilde{q}_{\mathfrak S}(X_1)^2} \leq \overline{\lambda} C_v,\] which is exactly the claimed bound. ◻

Proof. Write \(r=r_\eta\) for the exponent fixed in the statement. By the exact identity 34 and Minkowski’s inequality in \(L^{2r}\), \[\begin{align} \left( \Ex*{\abs*{h_{\mathfrak S}^{(1)}(x; Z_1)}^{2r}} \right)^{1/(2r)} & \leq \abs*{w_1^*}\frac{s_1}{s_2} \left( \Ex*{\abs*{h_{s_1}^{(1)}(x; Z_1)}^{2r}} \right)^{1/(2r)} \\ & \quad + \abs*{w_2^*} \left( \Ex*{\abs*{h_{s_2}^{(1)}(x; Z_1)}^{2r}} \right)^{1/(2r)}. \end{align}\] The bounded-ratio condition keeps \(w_1^*\) and \(w_2^*\) uniformly bounded, and 26 gives \[\Ex*{\abs*{h_{s_k}^{(1)}(x; Z_1)}^{2r}} \lesssim s_k^{-1}, \qquad k \in \{1,2\}.\] Since \(s_1 \leq s_2\), \[\left( \frac{s_1}{s_2} \right) s_1^{-1/(2r)} \leq s_2^{-1/(2r)}.\] Therefore, \[\Ex*{\abs*{h_{\mathfrak S}^{(1)}(x; Z_1)}^{2r}} \lesssim s_2^{-1}.\] Dividing by 39 yields \[\Ex*{ \left( \frac{h_{\mathfrak S}^{(1)}(x; Z_1)^2}{\zeta_{\mathfrak S}^{1}(x)} \right)^r } = \frac{\Ex*{\abs*{h_{\mathfrak S}^{(1)}(x; Z_1)}^{2r}}}{\left(\zeta_{\mathfrak S}^{1}(x)\right)^r} \lesssim s_2^{r-1}.\] Since \(r>1\) and \(s_2=o(n)\), \[\frac{1}{n^{r-1}} \Ex*{ \left( \frac{h_{\mathfrak S}^{(1)}(x; Z_1)^2}{\zeta_{\mathfrak S}^{1}(x)} \right)^r } \lesssim \left(\frac{s_2}{n}\right)^{r-1} \longrightarrow 0.\] This proves 2. ◻