\chapter{A criterion for Poisson convergence}\label{s:poisson}

In this appendix, we prove the criterion used in the proof of \zcref{t:main}. It is a variant of \cite[Lemma~1]{FengWei2025}, where the conclusion is stated for half-lines.

\begin{lemma}\label{l:poisson}
Let $\Lambda > 0$, and for each $N$, let $\mathcal X_N = \sum_{i \in \mathcal I_N} \delta_{X_{i, N}}$ be a point process on $\R$ with an almost surely finite index set $\mathcal I_N$. Suppose that for every $k \ge 1$ and $\mathbf x = (x_1, \dots, x_k) \in \R^k$,
\begin{equation}\label{e:poissonhyp}
\lim_{N \to \infty} \E \sum_{\substack{i_1, \dots, i_k \in \mathcal I_N \\ \text{distinct}}} \prod_{j = 1}^k \big( X_{i_j, N} - x_j \big)_+ = \prod_{j = 1}^k \Lambda e^{-x_j} .
\end{equation}
Let $A_1, \dots, A_m \subset \R$ be disjoint intervals, each of the form $(a, b]$ with $a < b$ or $(a, \infty)$. Then $(\mathcal X_N(A_1), \dots, \mathcal X_N(A_m))$ converges in distribution to a vector of independent Poisson random variables with means $\Lambda \int_{A_l} e^{-x} \dd x$, $1 \le l \le m$. Consequently, $\mathcal X_N$ converges in distribution to $\PPP(\Lambda e^{-x} \dd x)$ with respect to the vague topology.
\end{lemma}

\begin{proof}
For $\mathbf x \in \R^k$, we define
\begin{equation}
\mathcal I_N(\mathbf x) = \E \sum_{\substack{i_1, \dots, i_k \in \mathcal I_N \\ \text{distinct}}} \prod_{j = 1}^k \1\{ X_{i_j, N} > x_j \} , \qquad f(x) = \Lambda e^{-x} .
\end{equation}
We first show that $\mathcal I_N(\mathbf x) \to \prod_j f(x_j)$. Given $h > 0$ and $x \in \R$, we set
\begin{equation}
u_{x, h}(t) = h^{-1} \big( (t - x)_+ - (t - x - h)_+ \big) ,
\end{equation}
so that
\begin{equation}\label{e:sandwich}
\1\{ t > x + h \} \le u_{x, h}(t) \le \1\{ t > x \} .
\end{equation}

Expanding the product $\prod_j u_{x_j, h}(X_{i_j, N})$ into $2^k$ terms and applying \eqref{e:poissonhyp} to each of them, we find
\begin{equation}
\lim_{N \to \infty} \E \sum_{\text{distinct}} \prod_{j = 1}^k u_{x_j, h}(X_{i_j, N}) = \prod_{j = 1}^k \frac{f(x_j) - f(x_j + h)}{h} .
\end{equation}
Combining this with \eqref{e:sandwich}, we find
\begin{equation}
\limsup_{N \to \infty} \mathcal I_N(\mathbf x) \le \prod_{j = 1}^k \frac{f(x_j - h) - f(x_j)}{h} , \qquad \liminf_{N \to \infty} \mathcal I_N(\mathbf x) \ge \prod_{j = 1}^k \frac{f(x_j) - f(x_j + h)}{h} ,
\end{equation}
where in the first bound we used \eqref{e:sandwich} with $x_j - h$ in place of $x_j$. Letting $h \downarrow 0$, we conclude that
\begin{equation}
\mathcal I_N(\mathbf x) \to \prod_j (- f'(x_j)) = \prod_j f(x_j) .
\end{equation}

Next, let $k_1, \dots, k_m \ge 0$. We define the falling factorial by
\begin{equation}
(n)_k = n (n - 1) \cdots (n - k + 1) .
\end{equation}
Since the intervals $A_l$ are disjoint, $\E \prod_l (\mathcal X_N(A_l))_{k_l}$ is the expected number of families of distinct indices $(i_{l, p})_{1 \le l \le m, 1 \le p \le k_l}$ with $X_{i_{l, p}, N} \in A_l$ for all $l$ and $p$. To compute this expectation, we write
\begin{equation}
\1\{ t \in (a, b] \} = \1\{ t > a \} - \1\{ t > b \}
\end{equation}
and
\begin{equation}
\1\{ t \in (a, \infty) \} = \1\{ t > a \} ,
\end{equation}
and we expand the product of these indicators. This expresses $\E \prod_l (\mathcal X_N(A_l))_{k_l}$ as a finite linear combination of the quantities $\mathcal I_N(\mathbf x)$, with $\mathbf x$ having coordinates among the endpoints of the $A_l$. Using the first part of the proof, we have
\begin{equation}
\lim_{N \to \infty} \E \prod_{l = 1}^m (\mathcal X_N(A_l))_{k_l} = \prod_{l = 1}^m \Big( \Lambda \int_{A_l} e^{-x} \dd x \Big)^{k_l} ,
\end{equation}
which is the corresponding factorial moment of the claimed Poisson vector. The method of moments then gives the convergence in distribution of $(\mathcal X_N(A_1), \dots, \mathcal X_N(A_m))$, since a vector of independent Poisson random variables is determined by its moments, and moments are finite linear combinations of factorial moments.

Finally, let $\mathcal X$ be a Poisson point process on $\R$ with intensity $\Lambda e^{-x} \dd x$. It is simple and has a nonatomic intensity measure. The convergence of count vectors proved above implies that $\mathcal X_N(B)$ converges in distribution to $\mathcal X(B)$ for every finite union $B$ of disjoint bounded intervals of the form $(a, b]$. This implies local tightness of the counts, convergence of the void probabilities, and convergence of the probabilities of at least two points. The one-dimensional version of Kallenberg's criterion \cite{Kallenberg1983}, as stated in \cite[Theorem~2.2]{BenArousBourgade2013} for point processes on $\R^2$, therefore gives the convergence in distribution of $\mathcal X_N$ to $\mathcal X$ with respect to the vague topology. Intervals of other types are handled by approximation using the expected-count limits above, since the limiting intensity has no atoms; see also \cite[Proposition~2.1]{BenArousBourgade2013} for the same deduction. This completes the proof.
\end{proof}