5 Finite trees and the total progeny
Two fields agreeing at the vertices of one of their samples cut out the same tree. Hence the probability that the sample is a prescribed finite tree \(T\) is \(\prod _{v\in T}\theta _{c(v)}\), and it is at least \(p^{\lvert T \rvert }\) whenever every offspring weight along \(T\) is at least \(p\).
Choose a field \(e\) with sample \(T\). We first show, by induction on the length of a word, that any field \(c\) satisfying \(c(v)=e(v)\) for every \(v\in T\) has the same sample as \(e\). Indeed, membership of a child \(vj\) is determined by membership of \(v\) and the single inequality \(j{\lt}c(v)\), so the induction never uses a coordinate outside \(T\).
Since \(T\) is finite, agreement with \(e\) on \(T\) is a finite cylinder event. Independence of the coordinate field gives it mass \(\prod _{v\in T}\theta _{e(v)}\). This cylinder is contained in the event that the sample is \(T\). If every factor is at least \(p\), its mass is at least \(p^{\lvert T \rvert }\), which proves the point-mass bound.
The truncated progeny \(Z_k\) is the number of vertices of the sample within the first \(k\) generations.
Writing \(F_k(s)=\mathbb {E}\bigl[s^{Z_k}\bigr]\), one has \(F_{k+1}(s)=s\, f(F_k(s))\), and any \(y{\gt}1\) with \(s f(y)=y\) bounds every \(F_k(s)\).
The truncations satisfy the deterministic recursion
where \(c_i(w)=c(iw)\) is the field below the \(i\)-th child. On the event \(c(\emptyset )=j\), the \(j\) fields \(c_i\) are independent copies of the original field. Therefore
Summing over \(j\) gives \(F_{k+1}(s)=s f(F_k(s))\).
Since \(Z_0=0\), we have \(F_0(s)=1\leq y\). If \(F_k(s)\leq y\), monotonicity of \(f\) on the nonnegative half-line gives
Induction proves the uniform bound.
For a subcritical law there are \(s,y{\gt}1\) with \(s f(y)=y\).
Use the factorisation from the proof of theorem 2.7,
Continuity of \(g\) gives \(y{\gt}1\) sufficiently close to \(1\) that \(g(y){\lt}1\). Since \(1-y{\lt}0\), the factorisation implies \(f(y){\lt}y\). On the other hand, monotonicity and \(y{\gt}1\) give \(f(y)\geq f(1)=1\). Thus \(s=y/f(y)\) is well defined and satisfies \(s{\gt}1\) and \(s f(y)=y\).
For a subcritical offspring law there is \(\sigma {\gt}1\) with \(\mathbb {E}\bigl[\sigma ^{\lvert \mathcal{T} \rvert }\bigr]{\lt}\infty \). The same holds for the size of a bush once the conjugate law is subcritical.
By lemma 5.4, choose \(\sigma ,y{\gt}1\) such that \(\sigma f(y)=y\). Then lemma 5.3 gives
The variables \(Z_k\) increase with \(k\). If the sample is finite, then \(Z_k=\lvert \mathcal{T} \rvert \) for all sufficiently large \(k\), and hence \(\sigma ^{\lvert \mathcal{T} \rvert }\leq \sup _k\sigma ^{Z_k}\). If the sample is infinite, the same inequality holds with Mathlib’s convention that the finite cardinality of an infinite set is zero. The monotone convergence theorem therefore gives
For the bush law, apply the same result to the subcritical conjugate law and use theorem 4.5 to identify the resulting Galton–Watson tree law with the law of a bush.