1\section{Proofs for the exponential model}\label{app:exp}23Let $Y = \theta_0\bar X_n$, so that $nY \sim \mathrm{Gamma}(n,1)$, $\thetahat_n = \theta_0/Y$ and $R_n = \theta_0(1 - Y)^2/Y = \theta_0(Y^{-1} - 2 + Y)$. The Gamma moments $\E(nY)^{-k} = \Gamma(n-k)/\Gamma(n)$ ($k < n$) give $\E Y = 1$, $\E Y^2 = 1 + 1/n$, $\E Y^{-1} = n/(n-1)$ and $\E Y^{-2} = n^2/((n-1)(n-2))$.45\begin{proof}[Proof of \cref{thm:exponential}(i)]6From the Gamma moments,7\begin{align*}8\E\thetahat_n &= \theta_0\,\E Y^{-1} = \theta_0\,\frac{n}{n-1}, \\9\Var\thetahat_n &= \theta_0^2\big(\E Y^{-2} - (\E Y^{-1})^2\big) = \theta_0^2\,\frac{n^2}{(n-1)^2(n-2)},\\10\E R_n &= \theta_0\Big(\frac{n}{n-1} - 2 + 1\Big) = \frac{\theta_0}{n-1} = \theta_0\sum_{a\ge1}n^{-a},11\end{align*}12so $b_a = \theta_0$ for all $a$ and the remainder $\rho_n = \theta_0\sum_{a\ge r}n^{-a}$ satisfies $\norm{\rho_n} \le 2\theta_0 n^{-r}$ for $n \ge 2$. Expanding $(Y^{-1} - 2 + Y)^2 = Y^{-2} - 4Y^{-1} + 6 - 4Y + Y^2$,13\[14\E R_n^2 = \theta_0^2\Big[\frac{n^2}{(n-1)(n-2)} - \frac{4n}{n-1} + 3 + \frac1n\Big] = \theta_0^2\,\frac{3n+2}{n(n-1)(n-2)},15\]16and $n^2\E R_n^2/\theta_0^2 = n(3n+2)/((n-1)(n-2))$ decreases from $33/2$ at $n = 3$ to $3$, which gives $C = 33\theta_0^2/2$ with $n_0 = 3$. The influence function $\psi(x) = \theta_0 - \theta_0^2x$ is the same for all sub-samples and $\Sigma_\psi = \theta_0^2 = I(\theta_0)^{-1}$.17\end{proof}1819\begin{proof}[Proof of (ii)]20A sub-block of level $j$ has $n = m/s_j$ observations, so its MLE has mean $\theta_0 n/(n-1) = \theta_0 m/(m - s_j)$. Using $\sum_j w_j = 1$,21\[22\E\Jr{r,k} - \theta_0 = \theta_0\sum_{j=1}^r w_j\Big[\frac{m}{m-s_j} - 1\Big] = \theta_0\sum_{j=1}^r\frac{w_j\,s_j}{m-s_j}23= \theta_0\sum_{a\ge1}\frac{1}{m^{a}}\sum_{j=1}^r w_j s_j^{\,a},24\]25where the terms $a < r$ vanish by \eqref{eq:vandermonde} and the leading term is $\betar\theta_0 m^{-r}$. For the two default patterns,26\begin{align*}27s = (1,2):\qquad & \theta_0\Big[\frac{2}{m-1} - \frac{2}{m-2}\Big] = -\frac{2\theta_0}{(m-1)(m-2)},\\28s = (1,2,4):\qquad & \theta_0\Big[\frac{8/3}{m-1} - \frac{4}{m-2} + \frac{4/3}{m-4}\Big] = \frac{8\theta_0}{(m-1)(m-2)(m-4)} .29\end{align*} Both identities, and the constants $\beta_2 = -2$, $\beta_3 = 8$ and $\beta_4 = -64$, are verified in rational arithmetic in \texttt{theory/symbolic\_checks.py}.30\end{proof}3132\begin{proof}[Proof of (iii)]33Let $h = m/2$ and let $S_A, S_B$ be the sums over the two halves, $S = S_A + S_B \sim \mathrm{Gamma}(m,\theta_0)$ and $W = S_A/S \sim \mathrm{Beta}(h,h)$, independent of $S$. Then34\[35\thetahat_{D_k} = \frac mS, \qquad \thetahat_A = \frac{h}{SW}, \qquad \thetahat_B = \frac{h}{S(1-W)}, \qquad36\Jr{2,k} = \frac{G(W)}{S}, \quad G(W) = 2m - \frac h2\Big[\frac1W + \frac{1}{1-W}\Big].37\] Independence gives $\E\Jr{2,k} = \E S^{-1}\,\E G(W)$ and $\E\Jr{2,k}^2 = \E S^{-2}\,\E G(W)^2$, with38\[39\E S^{-k} = \theta_0^k\,\frac{\Gamma(m-k)}{\Gamma(m)}, \qquad40\E\big[W^{-a}(1-W)^{-c}\big] = \frac{\Gamma(h-a)\,\Gamma(h-c)\,\Gamma(2h)}{\Gamma(h)^2\,\Gamma(2h-a-c)}, \qquad a, c < h .41\] Simplifying the resulting rational functions (SymPy, check S6) gives the stated $\Var\Jr{2,k}$ and the expansion $\theta_0^2(m^{-1} + 4m^{-2} + O(m^{-3}))$; the same expansion holds for $\Var\thetahat_m = \theta_0^2m^2/((m-1)^2(m-2))$. The condition $m > 4$ ensures $h > 2$ so that all inverse moments exist.42\end{proof}4344\begin{proof}[Proof of (iv)]45With $g_N(x) = 1/x - \bar X_N$ and $H_N(x) = -1/x^2$,46\[47\mathcal N_N(x) = x + x^2\Big(\frac1x - \bar X_N\Big) = 2x - x^2\bar X_N,48\qquad49\mathcal N_N(x) - \thetaN = -\bar X_N x^2 + 2x - \frac{1}{\bar X_N} = -\bar X_N\Big(x - \frac{1}{\bar X_N}\Big)^2 = -\frac{(x-\thetaN)^2}{\thetaN},50\]51since $\thetaN = 1/\bar X_N$. Writing $e_j = \iter{}{j} - \thetaN$ gives $e_{j+1} = -e_j^2/\thetaN$ and, by induction, $e_q = -e_0^{2^q}/\thetaN^{2^q-1}$. In units of $\thetaN$, $u_{j+1} = -u_j^2$ with $u_j = e_j/\thetaN$, so $\abs{u_j} \to 0$ iff $\abs{u_0} < 1$, i.e.\ $0 < x < 2\thetaN$; $\abs{u_0} > 1$ gives $\abs{u_j} \to \infty$ with $u_j < 0$ for $j \ge 1$; $u_0 = \pm1$ gives $u_1 = -1$ and then $u_j = -1$, i.e.\ the iterate $0$, where the log-likelihood is undefined.52\end{proof}5354\begin{proof}[Proof of (v)--(vi)]55Let $\alpha = \alphafr{r}{q}$ and $K/N^\alpha \to c \in (0,\infty]$, with $m \to \infty$. By \cref{lem:pilot} and (ii), $\pilot{r} - \thetaN = \E R^J_1 + \bar\xi_N - R_N$ with $\E R^J_1 = \betar\theta_0m^{-r}(1 + O(m^{-1}))$, $\bar\xi_N = \op(\sqrt K/N)$ and $R_N = \op(N^{-1})$; part (i) supplies the constants. Since $m^{-r} = (K/N)^r$ and $r(1-\alpha) = 2^{-(q+1)}$,56\[57N^{1/2^{q+1}}\,m^{-r} = N^{1/2^{q+1}}\,(K/N^{\alpha})^r\,N^{-r(1-\alpha)} = (K/N^\alpha)^r \to c^r,58\]59while $N^{1/2^{q+1}}\sqrt K/N \le N^{1/2^{q+1} + \alpha/2 - 1} \to 0$ because $\alpha < 1$ and $1/2^{q+1} \le 1/2$ give $1/2^{q+1} + \alpha/2 - 1 < 0$; likewise $N^{1/2^{q+1}}N^{-1} \to 0$. Hence $N^{1/2^{q+1}}(\pilot{r} - \thetaN) \pto \betar\theta_0c^r$ when $c < \infty$ and $\pto \pm\infty$ when $c = \infty$. For $q = 0$, $N^{1/2^{q+1}} = \sqrt N$ and the claim follows. For $q \ge 1$, (iv) gives60\[61\sqrt N\big(\iter{r}{q} - \thetaN\big) = -\frac{\sqrt N\,(\pilot{r} - \thetaN)^{2^q}}{\thetaN^{\,2^q-1}} = -\frac{\big[N^{1/2^{q+1}}(\pilot{r} - \thetaN)\big]^{2^q}}{\thetaN^{\,2^q - 1}},62\]63and $\thetaN \pto \theta_0$ with the continuous mapping theorem yields $-\theta_0(\betar c^r)^{2^q}$ when $c < \infty$ and divergence to $-\infty$ in probability when $c = \infty$. Combining with $\sqrt N(\thetaN - \theta_0)\dto\mathcal N(0,\theta_0^2)$ proves the displayed limit law for $(2,1)$. Strictly, (iv) requires the pilot to be positive for the iteration to be meaningful; $P(\pilot{r} \le 0) \to 0$ since $\pilot{r} \pto \theta_0 > 0$.64\end{proof}65