1\section{Proofs of the general results}\label{app:proofs}23Throughout, $S_{j\ell,k}$ ($\ell = 1,\dots,s_j$) denotes the $\ell$-th sub-block of level $j$ of worker $k$, of size $m/s_j$; $U_S = \abs{S}^{-1}\sum_{i\in S}\psi(X_i)$; $w$ solves \eqref{eq:vandermonde}.45\begin{lemma}[Influence preservation]\label{lem:influence}6For every $k$ and $j$,7\[8\frac1{s_j}\sum_{\ell=1}^{s_j} U_{S_{j\ell,k}} = U_k, \qquad\text{and hence}\qquad \sum_{j=1}^r w_j\,\frac1{s_j}\sum_{\ell=1}^{s_j} U_{S_{j\ell,k}} = U_k .9\]10\end{lemma}11\begin{proof}12The sub-blocks of level $j$ partition $D_k$ into $s_j$ sets of equal size $m/s_j$, so13\[14\frac1{s_j}\sum_{\ell=1}^{s_j} U_{S_{j\ell,k}} = \frac1{s_j}\sum_{\ell=1}^{s_j}\frac{s_j}{m}\sum_{i\in S_{j\ell,k}}\psi(X_i) = \frac1m\sum_{i\in D_k}\psi(X_i) = U_k .15\]16The second claim uses $\sum_j w_j = 1$.17\end{proof}1819\begin{lemma}[Bias cancellation]\label{lem:bias}20Under \cref{ass:lin,ass:bias} with $m/s_r \ge n_0$, $\Jr{r,k} - \theta_0 = U_k + R^J_k$ with $R^J_k$ as in \eqref{eq:RJ}, and21\[22\E R^J_k = \sum_{j=1}^r w_j\,\rho_{m/s_j}, \qquad \norm{\E R^J_k} \le C_b\, m^{-r}\sum_{j=1}^r \abs{w_j}\, s_j^{\,r}.23\]24In particular $\E R^J_k$ is deterministic and does not depend on $k$.25\end{lemma}26\begin{proof}27Applying \eqref{eq:lin} to each sub-block and \cref{lem:influence},28\[29\Jr{r,k} - \theta_0 = \sum_{j=1}^r w_j\,\frac1{s_j}\sum_{\ell=1}^{s_j}\big(U_{S_{j\ell,k}} + R_{S_{j\ell,k}}\big) = U_k + R^J_k .30\]31Taking expectations with \eqref{eq:bias} at $n = m/s_j$,32\begin{align*}33\E R^J_k &= \sum_{j=1}^r w_j\Big[\sum_{a=1}^{r-1} b_a\Big(\frac{s_j}{m}\Big)^{a} + \rho_{m/s_j}\Big]34 = \sum_{a=1}^{r-1}\frac{b_a}{m^{a}}\underbrace{\sum_{j=1}^r w_j s_j^{\,a}}_{=\,0} + \sum_{j=1}^r w_j\,\rho_{m/s_j},35\end{align*}36and the inner sums vanish by \eqref{eq:vandermonde}. The bound follows from $\norm{\rho_{m/s_j}} \le C_b(s_j/m)^r$.37\end{proof}3839\begin{lemma}[Local remainder]\label{lem:remainder}40Under \cref{ass:lin}, $(\E\norm{R^J_k}^2)^{1/2} \le \sqrt C\,\Amp/m$ with $\Amp = \sum_j\abs{w_j}s_j$.41\end{lemma}42\begin{proof}43By Minkowski's inequality in $L_2$,44\[45\big(\E\norm{R^J_k}^2\big)^{1/2} \le \sum_{j=1}^r\frac{\abs{w_j}}{s_j}\sum_{\ell=1}^{s_j}\big(\E\norm{R_{S_{j\ell,k}}}^2\big)^{1/2}46\le \sum_{j=1}^r\frac{\abs{w_j}}{s_j}\cdot s_j\cdot\sqrt C\,\frac{s_j}{m} = \frac{\sqrt C\,\Amp}{m} .47\]48No independence between the nested sub-samples of a worker is used.49\end{proof}5051\begin{lemma}[Averaged centered remainder]\label{lem:average}52Under \cref{ass:lin}, $\E\big\|K^{-1}\sum_{k=1}^K(R^J_k - \E R^J_k)\big\|^2 \le C\Amp^2/(Km^2)$.53\end{lemma}54\begin{proof}55The blocks are disjoint, so the vectors $R^J_k - \E R^J_k$ are independent and centered; the second moment of the average is $K^{-2}\sum_k\E\norm{R^J_k - \E R^J_k}^2 \le K^{-1}\sup_k\E\norm{R^J_k}^2 \le C\Amp^2/(Km^2)$ by \cref{lem:remainder}.56\end{proof}5758\begin{lemma}[Pilot versus centralized MLE]\label{lem:pilot}59Under \cref{ass:lin,ass:bias}, $\pilot{r} - \thetaN = \E R^J_1 + K^{-1}\sum_k(R^J_k - \E R^J_k) - R_N = \op(m^{-r} + (m\sqrt K)^{-1} + N^{-1})$.60\end{lemma}61\begin{proof}62Averaging \cref{lem:bias} over $k$ and using $K^{-1}\sum_k U_k = N^{-1}\sum_{i=1}^N\psi(X_i) = U_N$ (equal block sizes),63\[64\pilot{r} - \theta_0 = U_N + \E R^J_1 + \frac1K\sum_{k=1}^K\big(R^J_k - \E R^J_k\big), \qquad \thetaN - \theta_0 = U_N + R_N,65\]66the second identity being \eqref{eq:lin} for the full sample. Subtracting, the influence terms cancel exactly. The orders follow from \cref{lem:bias,lem:average} and $\E\norm{R_N}^2 \le C N^{-2}$ via Markov's inequality.67\end{proof}6869\begin{lemma}[Quadratic contraction of one Newton step]\label{lem:newton}70Under \cref{ass:newton}, let $E_N$ be the event (of probability tending to one) on which $\thetaN \in U$ is an interior zero of $g_N$ and the bounds $M_N, \Lambda_N$ hold. On $E_N \cap \{\vartheta \in U\}$, $\norm{\mathcal N_N(\vartheta) - \thetaN} \le \tfrac12 M_N\Lambda_N\norm{\vartheta - \thetaN}^2$.71\end{lemma}72\begin{proof}73Let $e = \vartheta - \thetaN$. Since $U$ is convex, $\thetaN + te \in U$ for $t\in[0,1]$, and because $g_N(\thetaN) = 0$,74\[75g_N(\vartheta) = \int_0^1 H_N(\thetaN + te)\,e\,\dd t .76\]77Hence78\begin{align*}79\mathcal N_N(\vartheta) - \thetaN &= e - H_N(\vartheta)^{-1}g_N(\vartheta) \\80&= H_N(\vartheta)^{-1}\int_0^1\big[H_N(\vartheta) - H_N(\thetaN + te)\big]\,e\,\dd t .81\end{align*}82The Lipschitz bound gives $\norm{H_N(\vartheta) - H_N(\thetaN + te)} \le \Lambda_N(1-t)\norm{e}$, and $\int_0^1(1-t)\,\dd t = 1/2$.83\end{proof}8485\begin{lemma}[$q$ steps]\label{lem:qsteps}86Let $\iter{}{0} \pto \theta_0$ and define $\iter{}{j+1} = \mathcal N_N(\iter{}{j})$. Under \cref{ass:newton}, with probability tending to one all iterates $\iter{}{0},\dots,\iter{}{q}$ lie in $U$ and $\norm{\iter{}{q} - \thetaN} \le (\tfrac12 M_N\Lambda_N)^{2^q - 1}\norm{\iter{}{0} - \thetaN}^{2^q}$.87\end{lemma}88\begin{proof}89Let $\kappa_N = \tfrac12 M_N\Lambda_N = \op(1)$ and $e_j = \norm{\iter{}{j} - \thetaN}$. Since $e_0 \pto 0$ and $\thetaN \pto\theta_0$, there is a ball $B \subset U$ around $\theta_0$ such that $P(\iter{}{0}\in B, \thetaN\in B) \to 1$. On $E_N$, induction with \cref{lem:newton} gives $e_{j+1} \le \kappa_N e_j^2$ and, because $\kappa_N e_0 \pto 0$, $e_{j+1} \le e_j$ eventually, so that each iterate remains in the ball and the recursion applies $q$ times: $e_q \le \kappa_N^{1+2+\dots+2^{q-1}}e_0^{2^q} = \kappa_N^{2^q-1}e_0^{2^q}$.90\end{proof}9192\begin{proof}[Proof of \cref{thm:frontier}]93By \cref{lem:pilot} and $m \to \infty$ (which forces $K = o(N)$), $\pilot{r} - \theta_0 = U_N + \op(1) \pto 0$. \Cref{lem:qsteps} applies and, with $\kappa_N = \op(1)$,94\[95\sqrt N\norm{\iter{r}{q} - \thetaN} = \op\Big(\sqrt N\big[m^{-r} + (m\sqrt K)^{-1} + N^{-1}\big]^{2^q}\Big) \le \op\Big(3^{2^q}\sqrt N\big[m^{-r2^q} + (m\sqrt K)^{-2^q} + N^{-2^q}\big]\Big),96\]97using $(a+b+c)^{2^q} \le 3^{2^q}\max(a,b,c)^{2^q}$. With $m = N/K$ the three terms are98\begin{align*}99\sqrt N\, m^{-r2^q} &= N^{1/2}\Big(\frac KN\Big)^{r2^q} \to 0 \quad\Longleftrightarrow\quad K^{r2^q} = o\big(N^{r2^q - 1/2}\big) \quad\Longleftrightarrow\quad K = o\big(N^{1 - 1/(2^{q+1}r)}\big);\\100\sqrt N\,(m\sqrt K)^{-2^q} &= N^{1/2}K^{2^{q-1}}N^{-2^q} =101\begin{cases} (K/N)^{1/2} \to 0, & q = 0 \ (\text{since } K = o(N)),\\ \le N^{1/2 - 2^{q-1}} \le N^{-1/2} \to 0, & q \ge 1,\ K \le N; \end{cases}\\102\sqrt N\, N^{-2^q} &\to 0 .103\end{align*} The limit law follows from $\sqrt N(\thetaN - \theta_0) \dto \mathcal N(0,\Sigma_\psi)$ and Slutsky's lemma.104\end{proof}105106\begin{proof}[Proof of \cref{cor:inference}]107$\sqrt N(\iter{r}{q} - \theta_0) \dto \mathcal N(0,\Sigma_\psi)$ and $\widehat\Sigma_N \pto\Sigma_\psi$ give $\sqrt N\,\widehat\Sigma_{N,jj}^{-1/2}(\iter{r}{q,j} - \theta_{0,j}) \dto \mathcal N(0,1)$ by Slutsky's lemma. Consistency of $-H_N(\iter{r}{q})^{-1}$ under correct specification follows from \cref{ass:newton} (Lipschitz continuity of $H_N$ on $U$, $\iter{r}{q} \pto \theta_0$) and $-H_N(\theta_0) \pto I(\theta_0)$; the sandwich case is analogous with a law of large numbers for the aggregated score outer products.108\end{proof}109110\begin{proof}[Proof of \cref{thm:inexact}]111Let $e = \vartheta - \thetaN$ and $\bar H := \int_0^1 H_N(\thetaN + te)\,\dd t$, so that $g_N(\vartheta) = \bar H e$ on $E_N\cap\{\vartheta\in U\}$. Then112\[113\tilde\vartheta - \thetaN = e - \widehat H^{-1}(\bar H e + \delta_g) = \widehat H^{-1}\big(\widehat H - \bar H\big)\,e - \widehat H^{-1}\delta_g,114\]115and116\[117\norm{\widehat H - \bar H} \le \norm{\widehat H - H_N(\thetaN)} + \int_0^1\norm{H_N(\thetaN) - H_N(\thetaN + te)}\,\dd t \le \norm{\widehat H - H_N(\thetaN)} + \tfrac12\Lambda_N\norm{e},118\]119which gives \eqref{eq:inexact}. Iterating with a fixed $\widehat H$ (or with a fresh $\widehat H$ of the same accuracy at each step) yields $e_{j+1} = \op(\delta_N e_j + e_j^2)$ and, by induction, $e_q = \op(e_0\delta_N^q + \sum_{\text{mixed}} \cdots + e_0^{2^q})$ in which every mixed term is dominated by $e_0\delta_N^{q}$ or $e_0^{2^q}$ once $e_0 \to 0$ and $\delta_N\to0$; the iterates remain in $U$ as in \cref{lem:qsteps} because $\delta_N + \kappa_N e_0 \pto 0$. With $e_0 = \op(m^{-r} + \sqrt K/N + N^{-1})$ the binding condition is $\sqrt N\,m^{-r}\delta_N^q \to 0$ (the terms $\sqrt N(\sqrt K/N)\delta_N^q$ and $\sqrt N N^{-1}\delta_N^q$ vanish for any $\delta_N \le 1$ and $K \le N$), together with the exact-curvature condition $\sqrt N e_0^{2^q} \to 0$ of \cref{thm:frontier}.120121(a) Write122\[123\widehat H - H_N(\thetaN) = \frac1K\sum_{k=1}^K\big[H_{D_k}(\thetahat_{D_k}) - H_{D_k}(\theta_0)\big] + \big[H_N(\theta_0) - H_N(\thetaN)\big].124\] The second bracket is $\op(\norm{\thetaN - \theta_0}) = \op(N^{-1/2})$ by \cref{ass:newton}. For the first, \cref{ass:third} gives, for each $k$ and $\Delta_k := \thetahat_{D_k} - \theta_0$,125\[126H_{D_k}(\thetahat_{D_k}) - H_{D_k}(\theta_0) = \nabla H_{D_k}(\theta_0)[\Delta_k] + r_k, \qquad \norm{r_k} \le \tfrac12\bar L_k\norm{\Delta_k}^2,127\]128with $\bar L_k$ the block average of $L(X_i)$. By Cauchy--Schwarz and \cref{ass:third},129\[130\E\norm{r_k} \le \tfrac12\big(\E\bar L_k^2\big)^{1/2}\big(\E\norm{\Delta_k}^4\big)^{1/2} = O(m^{-1}).131\] Write $\nabla H_{D_k}(\theta_0) = T + V_k$ with $T = \E\nabla h(X;\theta_0)$ and $V_k$ a centered block average with $\E\norm{V_k}^2 = O(m^{-1})$, and $\Delta_k = U_k + R_{D_k}$. Then $K^{-1}\sum_k T[U_k] = T[U_N] = \op(N^{-1/2})$; $K^{-1}\sum_k T[R_{D_k}] = O_p(m^{-1})$ by \cref{ass:lin}; and $\E\norm{V_k[U_k]} \le (\E\norm{V_k}^2)^{1/2}(\E\norm{U_k}^2)^{1/2} = O(m^{-1})$. Collecting terms, $\delta_N = m^{-1} + N^{-1/2}$, and the condition $\sqrt N m^{-r}m^{-q} \to 0$ reads $K = o(N^{1 - 1/(2(r+q))})$; the $N^{-1/2}$ part of $\delta_N$ never binds. Since $r + q \le 2^q r$ for $q \ge 1$, this condition implies that of \cref{thm:frontier}.132133(b) Decompose134\begin{align*}135H_{D_1}(\vartheta) - H_N(\thetaN) &= \big[H_{D_1}(\vartheta) - H_{D_1}(\theta_0)\big] + \big[H_{D_1}(\theta_0) - \bar H(\theta_0)\big] \\136&\quad + \big[\bar H(\theta_0) - H_N(\theta_0)\big] + \big[H_N(\theta_0) - H_N(\thetaN)\big],137\end{align*}138where $\bar H(\theta_0) = \E h(X;\theta_0)$. The first and last brackets are $\op(\norm{\vartheta - \theta_0} + N^{-1/2})$ by the Lipschitz property (applied to $H_{D_1}$, for which the same bound is assumed as for $H_N$ on $U$, with probability tending to one); the second is $\op(m^{-1/2})$ and the third $\op(N^{-1/2})$ by the central limit theorem for block averages of $h(X_i;\theta_0)$. Hence $\delta_N = m^{-1/2} + \norm{\vartheta - \theta_0}$, and along the iteration $\norm{\iter{}{j} - \theta_0} = \op(N^{-1/2} + e_j)$, so the binding condition is $\sqrt N m^{-r}m^{-q/2} \to 0$, i.e.\ $K = o(N^{1 - 1/(2r+q)})$.139140(c) Rounding each entry of $H_N(\vartheta)$ to $b$ significant bits yields $\widehat H$ with $\abs{\widehat H_{ab} - H_{N,ab}} \le 2^{-b}\abs{H_{N,ab}}$, hence $\norm{\widehat H - H_N(\vartheta)} \le p\,2^{-b}\norm{H_N(\vartheta)}_{\max}$, and $\norm{H_N(\vartheta) - H_N(\thetaN)} \le \Lambda_N\norm{e}$; so $\delta_N = O_p(\varepsilon) + \op(\norm{e})$, the second part being absorbed in the $\norm{e}^2$ term. Likewise $\norm{\delta_g} \le p\,\varepsilon\,\norm{g_N(\vartheta)}_{\max} \le p\,\varepsilon\,\norm{\bar H}\norm{e} = \op(\varepsilon\norm{e})$. Inserting into \eqref{eq:inexact} with $\norm{e_0} = \op(m^{-r} + \sqrt K/N + N^{-1})$, first-order equivalence requires $\sqrt N\varepsilon m^{-r} \to 0$ (the other two terms vanish for $\varepsilon \le 1$). With $K \sim N^\alpha$, $m^{-r} = N^{-r(1-\alpha)}$ up to constants, so $\sqrt N\varepsilon m^{-r} \to 0$ iff $\log_2(1/\varepsilon) = b \ge (\tfrac12 - r(1-\alpha))\log_2 N + \omega(1)$ when $\tfrac12 - r(1-\alpha) \ge 0$, i.e.\ $\alpha \ge 1 - 1/(2r)$; otherwise $\sqrt N m^{-r} \to 0$ already and any $b \ge 0$ works.141\end{proof}142143\begin{proof}[Proof of \cref{prop:robust}]144(i) With weights $m_k/N$, $\sum_k (m_k/N)U_k = N^{-1}\sum_i\psi(X_i) = U_N$, so \cref{lem:pilot} holds with $\E R^J$ replaced by $\sum_k(m_k/N)\E R^J_k$, whose norm and whose centered second moment satisfy145\[146\Big\|\sum_{k}\frac{m_k}{N}\E R^J_k\Big\| \le C_b\Big(\sum_j\abs{w_j}s_j^{\,r}\Big)\frac1N\sum_{k=1}^K m_k^{\,1-r} = C_b\Big(\sum_j\abs{w_j}s_j^{\,r}\Big)m_{\mathrm{eff}}^{-r}, \qquad147\sum_{k=1}^K\Big(\frac{m_k}{N}\Big)^2\frac{C\Amp^2}{m_k^2} = \frac{C\Amp^2 K}{N^2}.148\] The Newton steps use $g_N = \sum_k(m_k/N)g_{D_k}$ and $H_N = \sum_k(m_k/N)H_{D_k}$, which are the full-sample derivatives, so \cref{lem:newton,lem:qsteps} are unchanged. With equal weights, $K^{-1}\sum_k U_k - U_N = \sum_i c_i\psi(X_i)$ with $c_i = (Km_{k(i)})^{-1} - N^{-1}$; its covariance is $(\sum_i c_i^2)\Sigma_\psi$ and $\sum_i c_i^2 = \sum_k m_k[(Km_k)^{-1} - N^{-1}]^2 = K^{-2}\sum_k m_k^{-1} - N^{-1}$, which is $\ge 0$ with equality iff all $m_k$ are equal (Jensen); when the sizes are bounded away from equality in the sense $K^{-1}\sum_k (N/(Km_k)) \ge 1 + \eta$, this is $\ge \eta/N$, of exact order $N^{-1}$, so the discrepancy is of exact order $N^{-1/2}$ and $\sqrt N(\pilot{} - \thetaN)$ does not vanish. One Newton step squares it to $\op(N^{-1})$ by \cref{lem:newton}. (ii) The only properties of $\psi$ used in \cref{lem:influence,lem:bias,lem:remainder,lem:average,lem:pilot} are $\E\psi = 0$ and that the same $\psi$ appears in \eqref{eq:lin} for every sub-sample; \cref{ass:newton} is stated for the pseudo-MLE with the sandwich limit. The proofs go through verbatim.149\end{proof}150