\section{Exact analysis in the exponential model and sharpness}\label{sec:exact}

Let $X_i \sim \mathrm{Exp}(\theta_0)$ in the rate parametrization. The average log-likelihood, score and Hessian on $n$ observations are
\begin{equation}\label{eq:exp-model}
L_n(\theta) = \log\theta - \theta\,\bar X_n, \qquad g_n(\theta) = \frac1\theta - \bar X_n, \qquad H_n(\theta) = -\frac{1}{\theta^2}, \qquad \thetahat_n = \frac{1}{\bar X_n}.
\end{equation}
The model is trivial for distributed inference, since $\sum_{i\in D_k}X_i$ is a sufficient statistic that any worker can upload, and the Hessian does not depend on the data, so that single-machine curvature is exact. It is used here because every quantity in \cref{sec:construction,sec:results} has a closed form, which makes it possible to check the assumptions, the constants and the boundary behaviour rather than assume them. The statements below are proved in \cref{app:exp} from Gamma and Beta moments and verified symbolically in the replication package. Write $Y = \theta_0\bar X_n$, so that $nY \sim \mathrm{Gamma}(n,1)$, and
\begin{equation}\label{eq:exp-psi}
\psi(x) = \theta_0 - \theta_0^2\,x, \qquad R_n = \thetahat_n - \theta_0 - \frac1n\sum_{i=1}^n\psi(X_i) = \frac{(1 - \theta_0\bar X_n)^2}{\bar X_n} .
\end{equation}

\begin{theorem}[Exact exponential identities and boundary behaviour]\label{thm:exponential}
\begin{enumerate}[label=(\roman*)]
\item \emph{Moments.} For $n > 2$,
\begin{align}
\E\thetahat_n &= \theta_0\,\frac{n}{n-1}, &
\Var\thetahat_n &= \theta_0^2\,\frac{n^2}{(n-1)^2(n-2)}, \label{eq:exp-mom1}\\
\E R_n &= \frac{\theta_0}{n-1} = \theta_0\sum_{a\ge1} n^{-a}, &
\E R_n^2 &= \theta_0^2\,\frac{3n+2}{n(n-1)(n-2)}, \label{eq:exp-mom2}
\end{align}
and $\sup_{n\ge3} n^2\,\E R_n^2/\theta_0^2 = 33/2$. Hence \cref{ass:lin,ass:bias} hold for every $r$ with $n_0 = 3$, $C = 33\theta_0^2/2$ and $b_a = \theta_0$ for all $a$.
\item \emph{Exact bias of the corrected estimators.} For any pattern with $m/s_r > 1$,
\begin{equation}\label{eq:exp-bias}
\E\Jr{r,k} - \theta_0 = \theta_0\sum_{j=1}^r \frac{w_j\, s_j}{m - s_j} = \betar\,\theta_0\, m^{-r}\big(1 + O(m^{-1})\big),
\end{equation}
in particular
\begin{equation}\label{eq:exp-bias23}
\E\Jr{2,k} - \theta_0 = -\frac{2\theta_0}{(m-1)(m-2)}, \qquad \E\Jr{3,k} - \theta_0 = \frac{8\theta_0}{(m-1)(m-2)(m-4)} .
\end{equation}
\item \emph{Exact variance of the half-block correction.} For $m > 4$,
\begin{equation}\label{eq:exp-varJ2}
\Var\Jr{2,k} = \theta_0^2\,\frac{m^2\,(m^2 - 6m + 17)}{(m-4)(m-2)^2(m-1)^2} = \theta_0^2\Big(\frac1m + \frac{4}{m^2} + O(m^{-3})\Big),
\end{equation}
the same first two terms as $\Var\thetahat_m$: the variance cost of the half-block correction is $O(m^{-3})$ in this model.
\item \emph{Newton identities and basin.} For every $x > 0$,
\begin{equation}\label{eq:exp-newton}
\mathcal N_N(x) = 2x - x^2\bar X_N, \qquad \mathcal N_N(x) - \thetaN = -\frac{(x - \thetaN)^2}{\thetaN}, \qquad \mathcal N_N^{\,q}(x) - \thetaN = -\frac{(x-\thetaN)^{2^q}}{\thetaN^{\,2^q-1}} .
\end{equation}
The iteration converges to $\thetaN$ if and only if $0 < x < 2\thetaN$; it diverges to $-\infty$ when $x < 0$ or $x > 2\thetaN$, and $x = 2\thetaN$ is mapped to $0$.
\item \emph{Boundary.} If $m \to \infty$ and $K/N^{\alphafr{r}{q}} \to c \in (0,\infty)$,
\begin{equation}\label{eq:exp-boundary}
\sqrt N\,\big(\iter{r}{q} - \thetaN\big) \;\pto\; \begin{cases} -\theta_0\,(\betar\, c^{\,r})^{2^q}, & q \ge 1, \\[2pt] \betar\,\theta_0\, c^{\,r}, & q = 0, \end{cases}
\end{equation}
so that for $(r,q) = (2,1)$ the limit is $-4\theta_0c^4$ and $\sqrt N(\iter{2}{1} - \theta_0) \dto \mathcal N(-4\theta_0c^4,\, \theta_0^2)$.
\item \emph{Necessity.} If $m \to \infty$ and $K/N^{\alphafr{r}{q}} \to \infty$, then $\sqrt N\,\abs{\iter{r}{q} - \thetaN} \pto \infty$.
\end{enumerate}
\end{theorem}

Items (v)--(vi) show that the exponent of \cref{thm:frontier} is sharp for the construction in this model: the small-$o$ cannot be weakened to $O$, and beyond the frontier the estimator is not $\sqrt N$-consistent for $\thetaN$. Three qualifications matter. The boundary regime $m \asymp N^{1/(2^{q+1}r)}$ is extremely slow to reach---for $(2,1)$, $m \asymp N^{1/8}$, so that $m \le 16$ for all $N \le 2^{24}$---and the finite-$m$ bias in \eqref{eq:exp-bias23} exceeds its leading term $-2\theta_0m^{-2}$ by the factor $m^2/((m-1)(m-2))$, which equals $1.5$ at $m = 6$; \cref{sec:mc} therefore reports both the limit and an exact finite-$m$ prediction. The sharpness is for this estimator family: it is not a lower bound over all distributed procedures, as pooling $\sum_i X_i$ shows. And the signs alternate, $\beta_2 = -2 < 0 < \beta_3 = 8$, so the higher-order pilot over-corrects where the half-block pilot under-corrects, one reason why finite-sample accuracy is not monotone in $r$.

Item (iv) also gives an explicit failure gallery. Because the weights have negative entries, $\Jr{r,k}$ can be negative when a sub-block average is small; the pilot then lies outside the basin $(0, 2\thetaN)$ with positive probability for every finite $m$, and the Newton iteration diverges rather than converges. The probability is negligible for $m \ge 32$ and visible for $m \le 8$ in \cref{sec:mc}, where such replications are reported, not discarded. A production implementation would damp or project the step; the theory above describes the undamped map.