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Global regime for general additive functionals of conditioned Bienaymé-Galton-Watson trees

Abstract

We give an invariance principle for very general additive functionals of conditioned Bienaymé-Galton-Watson trees in the global regime when the offspring distribution lies in the domain of attraction of a stable distribution, the limit being an additive functional of a stable Lévy tree. This includes the case when the offspring distribution has finite variance (the Lévy tree being then the Brownian tree). We also describe, using an integral test, a phase transition for toll functions depending on the size and height.

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Appendices

Appendix A. A space of measures

Let \((S,\rho )\) be a Polish metric space, \(S_0 \subset S\) be a closed set in S and \(0 \in S_0\) be a distinguished point. Denote by \(\mathcal {K}\) the class of compact sets \(K\subset S\). For any \(x \in S\) and \(A\subset S\), the distance from x to A is defined by \(\rho (x,A) = \inf \{\rho (x,y) :\, y \in A\}\). Let \(\mathfrak {F}\) be a countable set of measurable \([0,+\infty ]\)-valued functions on S satisfying the following assumptions:

  1. (H1)

    The constant function \(\mathbf {1}\) belongs to \(\mathfrak {F}\).

  2. (H2)

    All \(f\in \mathfrak {F}\) are continuous on \(S_0^c\).

  3. (H3)

    All \(f\in \mathfrak {F}\) are bounded away from zero and infinity on \(\{x \in S:\, \rho (x,S_0) \ge \varepsilon , \, \rho (x,0) \le M\}\) for every \(0<\varepsilon< M<+\infty \).

  4. (H4)

    For all \(f\in \mathfrak {F}\), the set \(\mathfrak {F}^\star (f)\subset \mathfrak {F}\) of functions \(f^\star \in \mathfrak {F}\) such that \(f/f^\star \) is bounded on \(S_0^c\) and \(\lim _{\rho (x,S_0) \rightarrow 0+}f(x)/f^\star (x) = 0\) is non-empty.

Note that assumption (H3) is automatically satisfied when S is compact and every \(f \in \mathfrak {F}\) is positive on \(S_0^c\). Notice that  (H4) implies that \(\mathfrak {F}^\star (f)\) is infinitely countable for any \(f\in \mathfrak {F}\). We shall write \(f^\star \) for any element of \(\mathfrak {F}^\star (f)\). By  (H1) and  (H4), we have \(\lim _{\rho (x,S_0) \rightarrow 0+} \mathbf {1}^\star (x) =+ \infty \). By convention, we take \(\mathbf {1}^\star \equiv + \infty \) on \(S_0\) and \(f/f^\star \equiv 0\) on \(S_0\) for every \(f\in \mathfrak {F}\). We will occasionally need the following additional assumption:

  1. (H5)

    S is compact or \(\inf _{K \in \mathcal {K}} \sup _{x \in K^c} f(x)/f^\star (x) = 0\) for every \(f\in \mathfrak {F}\) (and some \(f^\star \in \mathfrak {F}^\star (f)\)).

Denote by \(\mathcal {M}= \mathcal {M}(S)\) the space of nonnegative finite measures on S endowed with the weak topology. Recall that \((\mathcal {M}, d_{\mathrm {BL}})\), with \(d_{\mathrm {BL}}\) the bounded Lipschitz distance is a Polish metric space. Recall that, for \(\mu \in \mathcal {M}\) and \(f\in {{\mathcal {B}}}_+(S)\), the notation \(f\mu \) stands for the measure \(f(x) \mu (\mathrm {d}x)\). Set

$$\begin{aligned} \mathcal {M}_\mathfrak {F}= \mathcal {M}_\mathfrak {F}(S) :=\left\{ \mu \in \mathcal {M}:\, \mu (f) < \infty \, \text { for all } f \in \mathfrak {F}\right\} . \end{aligned}$$
(A.1)

For \(\mu \in \mathcal {M}_\mathfrak {F}\), we have \(\mu (S_0) = 0\) (as \(\mathbf {1}^\star \equiv + \infty \) on \(S_0\)) and \(f\mu \in \mathcal {M}\) for every \(f\in \mathfrak {F}\). In particular, since \((f/f^\star )f^\star =f\) on \(S_0^c\), we have \((f/f^\star )f^\star \mu = f \mu \) for every \(f\in \mathfrak {F}\) (and \(f^\star \in \mathfrak {F}^\star (f)\)). We say a sequence \((\mu _n, \, n \in \mathbb {N})\) of elements of \(\mathcal {M}_\mathfrak {F}\) converges to \(\mu \in \mathcal {M}_\mathfrak {F}\) if and only if \((f \mu _n, \, n \in \mathbb {N})\) converges to \( f \mu \) in \(\mathcal {M}\) for every \(f\in \mathfrak {F}\). We consider the following distance \(d_\mathfrak {F}\) on \(\mathcal {M}_\mathfrak {F}\) which defines the same topology:

$$\begin{aligned} d_\mathfrak {F}(\mu ,\nu ) = \sum _{k\in {{\mathbb {N}}}} \frac{1}{2^k}\left( 1\wedge d_{\mathrm {BL}}\left( f_k \mu ,f_k \nu \right) \right) \quad \text {for} \quad \mu , \nu \in \mathcal {M}_\mathfrak {F}, \end{aligned}$$
(A.2)

where \(\{f_k:\, k\in {{\mathbb {N}}}\}\) is an enumeration of \(\mathfrak {F}\). (The choice of the enumeration is unimportant, as the corresponding distances all define the same topology on \(\mathcal {M}_\mathfrak {F}\).) Notice that the mapping \(\mu \mapsto f\mu \) is continuous from \(\mathcal {M}_\mathfrak {F}\) to \(\mathcal {M}\). In particular, taking \(f=\mathbf {1}\) gives that every sequence which converges in \(\mathcal {M}_\mathfrak {F}\) also converges in \(\mathcal {M}\) to the same limit.

We shall see that the space \((\mathcal {M}_\mathfrak {F},d_\mathfrak {F})\) is complete and separable (Proposition A.1) and give a complete description of its compact subsets (Proposition A.2). The main goal of this section is to give conditions which allow to strengthen a convergence in \(\mathcal {M}\) to a convergence in \(\mathcal {M}_\mathfrak {F}\) for deterministic measures (Corollary A.3) and then to extend this result to random measures (Proposition 7.6 and Corollary A.11).

Proposition A.1

The space \((\mathcal {M}_\mathfrak {F},d_\mathfrak {F})\) is complete and separable.

Proof

Let \((\mu _n, \, n \in \mathbb {N})\) be a Cauchy sequence in \(\mathcal {M}_\mathfrak {F}\). Then, by definition of \(d_\mathfrak {F}\), the sequence \((f \mu _n, \, n \in \mathbb {N})\) is Cauchy in \(\mathcal {M}\) for every \(f \in \mathfrak {F}\). By completeness of \(\mathcal {M}\), for every \(f\in \mathfrak {F}\), there exists a measure \(\nu _f \in \mathcal {M}\) such that \(\lim _{n\rightarrow \infty }f\mu _n= \nu _f\) in \(\mathcal {M}\). We claim that \(\nu _f(S_0) = 0\) for every \(f\in \mathfrak {F}\). Indeed, fix \(f\in \mathfrak {F}\) and \(f^\star \in \mathfrak {F}^\star (f)\). As \(f^\star \in \mathfrak {F}\), we have \(\lim _{n\rightarrow \infty }f^\star \mu _n= \nu _{f^\star }\) in \(\mathcal {M}\). By (H4), the function \(f/f^\star \) is continuous and bounded on S, so that the mapping \(\pi \mapsto (f/f^\star ) \pi \) is continuous on \(\mathcal {M}\). In particular, we have \(\lim _{n\rightarrow \infty }f \mu _n = (f/f^\star ) \nu _{f^\star }\) in \(\mathcal {M}\). On the other hand, we have \(\lim _{n\rightarrow \infty }f \mu _n =\nu _f\) in \(\mathcal {M}\). We deduce that \(\nu _f = (f/f^\star )\nu _{f^\star }\). It follows that \(\nu _f(S_0) = 0\) since \(f/f^\star = 0\) on \(S_0\).

We set \(\mu = \nu _\mathbf {1}\) so that \(\lim _{n\rightarrow \infty }\mu _n = \mu \) in \(\mathcal {M}\). Let \(f\in \mathfrak {F}\). We shall prove that \(f\mu =\nu _f\). Consider the closed set \(F_k=\{f\ge 1/k\}\) for \(k\in {{\mathbb {N}}}^*\). Notice that \(F_k \subset {\text {int}}(F _ {k+1})\). Therefore, by Urysohn’s lemma, there exists, for \(k\in {{\mathbb {N}}}^*\), a continuous function \(\chi _k :S \rightarrow [0,1]\) such that \(\chi _k = 1\) on \(F_ k\) and \({\text {supp}}(\chi _k) \subset {\text {int}}(F_{k+1})\). Notice that \((\chi _k f/f)\mu _n = \chi _k \mu _n\) since \((f/f)=1\) on \( S_0^c\) and \(\mu _n(S_0) = 0\). Since \(\chi _k\) and \(\chi _k/f\) are continuous and bounded, the mappings \(\nu \mapsto \chi _k \nu \) and \(\nu \mapsto (\chi _k/f) \nu \) are continuous from \(\mathcal {M}\) to itself. We deduce that \(\chi _k \mu = \lim _{n\rightarrow \infty } \chi _k \mu _n= \lim _{n\rightarrow \infty } (\chi _k/f) f \mu _n=(\chi _k/f) \nu _f\) in \(\mathcal {M}\). Letting k go to infinity, as \(\chi _k \uparrow \mathbf {1}\) on \(S_0^c\) since f is positive on \(S_0^c\), and \(\mu (S_0)=\nu _f(S_0)=0\), we deduce (using the monotone convergence theorem) that \(\mu =(1/f) \nu _f\) and thus \(f\mu = \nu _f\). Since this holds for all \(f\in \mathfrak {F}\), this proves that \(\mu \in \mathcal {M}_\mathfrak {F}\) and that \(\lim _{n\rightarrow \infty }f\mu _n = f \mu \) in \(\mathcal {M}\) for every \(f\in \mathfrak {F}\). Thus \(\mathcal {M}_\mathfrak {F}\) is complete.

Next, define \(F'_n = \{x \in S :\, \rho (x,S_0)\ge 1/n, \, \rho (x,0) \le n\}\). We will identify the space \(\mathcal {M}(F'_n)\) with the subset of \(\mathcal {M}\) consisting of the measures whose support lies in \(F'_n\). Notice that \(F'_n\) is a Polish space (when endowed with the topology induced by \(\rho \)) as a closed subset of the Polish space S. In particular, the set \(\mathcal {M}(F'_n)\) endowed with the bounded Lipschitz distance is a Polish space. Let \(f\in \mathfrak {F}\). By (H3), the functions f and 1/f are both continuous and bounded on \(F'_n\), so it is easy to check that the topology induced by \(d_\mathfrak {F}\) on \(\mathcal {M}(F'_n)\) coincides with the topology of weak convergence, i.e. the one induced by \(d_{\mathrm {BL}}\). Therefore, the space \((\mathcal {M}(F'_n), d_\mathfrak {F})\) is separable. To prove that \(\mathcal {M}_\mathfrak {F}\) is separable, it suffices to show that \(\mathcal {M}_\mathfrak {F}\) is equal to the completion of \(\bigcup _{n\ge 1} \mathcal {M}(F'_n)\) with respect to \(d_{\mathrm {BL}}\). Notice that \(F'_n \subset {\text {int}}(F'_ {n+1})\). Therefore, by Urysohn’s lemma, there exists a continuous function \(\chi '_n :S \rightarrow [0,1]\) such that \(\chi '_n = 1\) on \(F'_ n\) and \({\text {supp}}(\chi '_n) \subset {\text {int}}(F'_{n+1})\). Let \(\mu \in \mathcal {M}_\mathfrak {F}\) and set \(\mu _n = \chi '_n\mu \). Then it is clear that \(\mu _n\) has support in \(F'_{n+1}\) and thus \(\mu _n\in \mathcal {M}(F'_{n+1})\). Moreover, for every \(f\in \mathfrak {F}\) and every nonnegative \(h\in {{\mathcal {C}}}_b(S)\), we have

$$\begin{aligned} \mu _n(hf) =\mu (hf \chi '_n)\xrightarrow [n\rightarrow \infty ]{} \mu (hf ) \end{aligned}$$

by the monotone convergence theorem, since \(\chi '_n \uparrow \mathbf {1}_{S_0^c}\) and \(\mu (S_0) = 0\). This proves that \((f\mu _n, n \in \mathbb {N})\) converges to \(f\mu \) in \(\mathcal {M}\) for every \(f\in \mathfrak {F}\), thus \(d_\mathfrak {F}(\mu _n,\mu ) \rightarrow 0\). This concludes the proof. \(\square \)

A set of measures \(A \subset \mathcal {M}\) is said to be bounded if \(\sup _{\mu \in A} \mu (\mathbf {1}) < \infty \). We now give a characterization of compactness in \(\mathcal {M}_\mathfrak {F}\).

Proposition A.2

Let \(A \subset \mathcal {M}_\mathfrak {F}\).

  1. (i)

    A is relatively compact if and only if for every \(f\in \mathfrak {F}\), the family \(\{f\mu :\, \mu \in A\}\) of finite measures is bounded and tight.

  2. (ii)

    If (H5) holds, then A is relatively compact if and only if for every \(f\in \mathfrak {F}\), the family \(\{f \mu :\, \mu \in A\}\) is bounded.

Proof

To prove (i), start by assuming that A is relatively compact. For every \(\mu \in \mathcal {M}_\mathfrak {F}\) and every \(f\in \mathfrak {F}\), set \(F_f(\mu ) = f \mu \). This defines a continuous mapping \(F _f:\mathcal {M}_\mathfrak {F}\rightarrow \mathcal {M}\). It follows that the set

$$\begin{aligned} F_f(A) = \{ f \mu :\mu \in A\} \end{aligned}$$

is relatively compact in \(\mathcal {M}\), i.e. it is bounded and tight by Prokhorov’s theorem.

Conversely, let us assume that \(\{ f \mu :\mu \in A\}\) is bounded and tight in \(\mathcal {M}\) for all \(f\in \mathfrak {F}\). Let \((\mu _n, \, n \in \mathbb {N})\) be a sequence in A. Since the sequence of measures \((f \mu _n, \, n \in \mathbb {N})\) is bounded and tight, it is relatively compact in \(\mathcal {M}\) for every \(f\in \mathfrak {F}\). Therefore, by diagonal extraction, there exists a subsequence still denoted by \((f \mu _n, \, n \in \mathbb {N})\) which converges in \(\mathcal {M}\) for every \(f\in \mathfrak {F}\). By the same argument as in the proof of Proposition A.1, it follows that \((\mu _n, \, n \in \mathbb {N})\) converges in \(\mathcal {M}_\mathfrak {F}\). This proves that A is relatively compact.

To prove (ii), assume that  (H5) holds. The statement for a compact S follows immediately since a family of finite measures on a compact space is always tight. Now assume that S is not compact and let \(A \subset \mathcal {M}_\mathfrak {F}\) such that the family \(\{f \mu :\mu \in A\}\) is bounded for every \(f\in \mathfrak {F}\). To prove that \(A \subset \mathcal {M}_\mathfrak {F}\) is relatively compact, it is enough to show that \(\{f \mu :\mu \in A\}\) is tight and to apply the first point. Let \(f^\star \in \mathfrak {F}^\star (f)\) be the one appearing in (H5) and \(K \subset S\) be a compact subset. For every \(\mu \in A\), since \(\mu (S_0) = 0\), we have

$$\begin{aligned} \int _{K^c} f(x)\,\mu (\mathrm {d}x)&= \int _{K^c} f(x)\mathbf {1}_{S_0 ^c}(x)\,\mu (\mathrm {d}x)\\&=\int _{K^c} \frac{f(x)}{f^\star (x)} \mathbf {1}_{S_0^c}(x)f^\star (x)\, \mu (\mathrm {d}x) \\&\le \mu (f^\star )\, \sup _{ K^c}\frac{f}{f^\star }\cdot \end{aligned}$$

It follows that

$$\begin{aligned} \sup _{\mu \in A} \int _{K^c} f(x)\,\mu (\mathrm {d}x) \le \sup _{\mu \in A}\mu (f^\star )\,\, \sup _{ K^c}\frac{f}{f^\star }, \end{aligned}$$

and taking the infimum over all compact subsets \(K \in \mathcal {K}\) yields, thanks to (H5)

$$\begin{aligned} \inf _{K\in \mathcal {K}} \, \sup _{\mu \in A} \int _{K^c} f(x)\,\mu (\mathrm {d}x) = 0, \end{aligned}$$

i.e. the family \(\{f \mu :\mu \in A\}\) is tight. This completes the proof. \(\square \)

The next result gives sufficient conditions allowing to strengthen convergence in \(\mathcal {M}\) to convergence in \(\mathcal {M}_\mathfrak {F}\).

Corollary A.3

Let \((\mu _n, \, n \in \mathbb {N})\) be a sequence of elements of \(\mathcal {M}_\mathfrak {F}\) converging in \(\mathcal {M}\) to some \(\mu \in \mathcal {M}\). Then \(\mu \in \mathcal {M}_\mathfrak {F}\) and \(\lim _{n\rightarrow \infty }\mu _n = \mu \) in \(\mathcal {M}_\mathfrak {F}\) under either of the following conditions:

  1. (i)

    \((f\mu _n, \, n \in \mathbb {N})\) is bounded and tight for every \(f\in \mathfrak {F}\).

  2. (ii)

     (H5) holds and \((f\mu _n, \, n \in \mathbb {N})\) is bounded for every \(f\in \mathfrak {F}\).

Proof

Either condition guarantees that the sequence \((\mu _n, \, n \in \mathbb {N})\) is relatively compact in \(\mathcal {M}_\mathfrak {F}\) by Proposition A.2. Let \(\hat{\mu }\in \mathcal {M}_\mathfrak {F}\) be a limit point of \((\mu _n, \, n \in \mathbb {N})\). Then there exists a subsequence, still denoted by \((\mu _n, \, n \in \mathbb {N})\) such that \(\lim _{n\rightarrow \infty }\mu _n = \hat{\mu }\) in \(\mathcal {M}_\mathfrak {F}\). In particular, we have \(\lim _{n\rightarrow \infty }\mu _n = \hat{\mu }\) in \(\mathcal {M}\). Since \(\lim _{n\rightarrow \infty }\mu _n = \mu \) in \(\mathcal {M}\) by assumption, it follows that \(\hat{\mu } = \mu \). This proves that \(\mu \in \mathcal {M}_\mathfrak {F}\) and that \(\lim _{n\rightarrow \infty }\mu _n = \mu \) in \(\mathcal {M}_\mathfrak {F}\) since the sequence \((\mu _n, \, n \in \mathbb {N})\) is relatively compact in \(\mathcal {M}_\mathfrak {F}\) and has only one limit point \(\mu \). \(\square \)

The compactness criterion of Proposition A.2 yields a tightness criterion for random measures in \(\mathcal {M}_\mathfrak {F}\).

Proposition A.4

Let \(\Xi \) be a family of \(\mathcal {M}_\mathfrak {F}\)-valued random variables.

  1. (i)

    The family \(\Xi \) is tight (in distribution) in \(\mathcal {M}_\mathfrak {F}\) if and only if for every \(f\in \mathfrak {F}\), the family \(\{f\xi :\, \xi \in \Xi \}\) is tight (in distribution) in \(\mathcal {M}\), i.e. if and only if

    $$\begin{aligned} \lim _{r\rightarrow \infty } \sup _{\xi \in \Xi } {\mathbb {P}}\left( \xi (f) > r\right) = 0 \end{aligned}$$
    (A.3)

    and

    $$\begin{aligned} \inf _{K \in \mathcal {K}} \sup _{\xi \in \Xi } {\mathbb {E}}\left[ 1\wedge \int _{K^c}f(x) \xi (\mathrm {d}x)\right] = 0. \end{aligned}$$
    (A.4)
  2. (ii)

    If (H5) holds, then \(\Xi \) is tight (in distribution) in \(\mathcal {M}_\mathfrak {F}\) if and only if (A.3) holds for every \(f\in \mathfrak {F}\).

Proof

To prove (i), assume that \(\Xi \) is tight in \(\mathcal {M}_\mathfrak {F}\). Since the mapping \(F_f :\mu \mapsto f \mu \) is continuous from \(\mathcal {M}_\mathfrak {F}\) to \(\mathcal {M}\) for every \(f\in \mathfrak {F}\) and since tightness is preserved by continuous mappings, it follows that the family \(F_f(\Xi ) = \{f\xi :\, \xi \in \Xi \}\) is tight in \(\mathcal {M}\) for every \(f\in \mathfrak {F}\). The result now follows from Theorem 4.10 in [31].

Conversely, assume that (A.3) and (A.4) hold for all \(f\in \mathfrak {F}\) and let \(\varepsilon >0\). Let \(\{f_k:\, k\in {{\mathbb {N}}}^*\}\) be an enumeration of \(\mathfrak {F}\). We set for \(k\in {{\mathbb {N}}}^*\):

$$\begin{aligned} C_k= k \left( 1+ \sup _{j\le k,\, f_k\in \mathfrak {F}^\star (f_j)} \left\Vert f_j/f_k\right\Vert _\infty \right) , \end{aligned}$$

with the convention that \(\sup \emptyset =0\). For every \(k \in {{\mathbb {N}}}^*\), there exists \(r_k > 0\) and a compact set \(K_k \in \mathcal {K}\) such that

$$\begin{aligned} \sup _{\xi \in \Xi } {\mathbb {P}}\left( \xi (f_k) > r_k\right) \le \frac{\varepsilon }{2^{k}} \quad \text {and}\quad \sup _{\xi \in \Xi } {\mathbb {E}}\left[ 1\wedge \int _{K_k ^c} f_k(x) \xi (\mathrm {d}x)\right] \le \frac{\varepsilon }{C_k 2^{k}}\cdot \end{aligned}$$

Set

$$\begin{aligned} A_\varepsilon =\bigcap _{k\in {{\mathbb {N}}}^*} \left\{ \mu \in \mathcal {M}_\mathfrak {F}:\, \mu (f_k) \le r_k \text { and } \int _{K_k ^c} f_k(x) \mu (\mathrm {d}x) \le \frac{1}{C_k} \right\} . \end{aligned}$$

Then for every \(\xi \in \Xi \), we have

$$\begin{aligned} {\mathbb {P}}\left( \xi \in A_\varepsilon ^c\right)&= {\mathbb {P}}\left( \exists k \in {{\mathbb {N}}}^*, \, \xi (f_k)> r_k \text { or } \int _{K_k ^c} f_k(x) \xi (\mathrm {d}x)> \frac{1}{C_ k} \right) \\&\le \sum _{k\in {{\mathbb {N}}}^*} {\mathbb {P}}\left( \xi (f_k)> r_k\right) + \sum _{k\in {{\mathbb {N}}}^*} {\mathbb {P}}\left( \int _{K_k ^c} f_k(x) \xi (\mathrm {d}x) > \frac{1}{C_ k}\right) \le 2\varepsilon , \end{aligned}$$

where in the last inequality we used that

$$\begin{aligned} {\mathbb {P}}\left( \int _{K_k ^c} f(x) \xi (\mathrm {d}x)> \frac{1}{C_k}\right)= & {} {\mathbb {P}}\left( 1 \wedge \int _{K_k ^c} f_k(x) \xi (\mathrm {d}x) > \frac{1}{C_k}\right) \\\le & {} C_k {\mathbb {E}}\left[ 1\wedge \int _{K_k ^c} f_k(x) \xi (\mathrm {d}x)\right] \le \frac{\varepsilon }{2^{k}}\cdot \end{aligned}$$

Thus, to prove that \(\Xi \) is tight in \(\mathcal {M}_\mathfrak {F}\), it remains to show that \(A_\varepsilon \subset \mathcal {M}_\mathfrak {F}\) is relatively compact. We have \(\sup _{\mu \in A_\varepsilon } \mu (f_k) \le r_k < \infty \) so that the family \(\{f_k\mu :\, \mu \in A_\varepsilon \}\) is bounded for every \(k \in {{\mathbb {N}}}^*\). Moreover, for every \(i \ge k\) such that \(f_i\in \mathfrak {F}^\star (f_k)\), we have

$$\begin{aligned} \sup _{\mu \in A_\varepsilon }\int _{K_i ^c} f_k(x) \mu (\mathrm {d}x) \le \left\Vert f_k/f_i\right\Vert _{\infty } \, \sup _{\mu \in A_\varepsilon }\int _{K_i ^c} f_i(x) \mu (\mathrm {d}x) \le \frac{1}{i}\cdot \end{aligned}$$

This implies that \(\inf _{K \in \mathcal {K}} \sup _{\mu \in A_\varepsilon } \int _{K^c} f_k(x) \mu (\mathrm {d}x) \le 1/i\) for \(i \ge k\) such that \(f_i\in \mathfrak {F}^\star (f_k)\). Since there are infinitely many such i, we deduce that

$$\begin{aligned} \inf _{K \in \mathcal {K}} \sup _{\mu \in A_\varepsilon } \int _{K^c} f_k(x) \mu (\mathrm {d}x) = 0, \end{aligned}$$

i.e. the family \(\{f_k \mu :\, \mu \in A_\varepsilon \}\) is tight. As this holds for all \(k\in {{\mathbb {N}}}^*\), we get by Proposition A.2 that \(A_\varepsilon \) is relatively compact in \(\mathcal {M}_\mathfrak {F}\) (in fact, \(A_\varepsilon \) is compact as it is closed). This proves (i). The proof of (ii) is similar. \(\square \)

We now give a sufficient condition for tightness in the space \(\mathcal {M}_\mathfrak {F}\).

Corollary A.5

Assume that (H5) holds. Let \(\Xi \) be a family of \(\mathcal {M}_\mathfrak {F}\)-valued random variables such that for every \(f\in \mathfrak {F}\),

$$\begin{aligned} \sup _{\xi \in \Xi } {\mathbb {E}}\left[ \xi (f)\right] < \infty . \end{aligned}$$
(A.5)

Then \(\Xi \) is tight (in distribution) in \(\mathcal {M}_\mathfrak {F}\).

Proof

By the Markov inequality, we have for every \(f\in \mathfrak {F}\),

$$\begin{aligned} \sup _{\xi \in \Xi }{\mathbb {P}}\left( \xi (f) > r\right) \le \frac{1}{r} \sup _{\xi \in \Xi } {\mathbb {E}}\left[ \xi (f)\right] \xrightarrow [r \rightarrow \infty ]{} 0. \end{aligned}$$

This proves that \(\Xi \) is tight in \(\mathcal {M}_\mathfrak {F}\) by Proposition A.4-(ii). \(\square \)

We denote by \(\mathcal {B}\) (resp. \(\mathcal {B}_{\mathfrak {F}}\)) the Borel \(\sigma \)-field on \((\mathcal {M},d_{\mathrm {BL}})\) (resp. on \((\mathcal {M}_\mathfrak {F},d_\mathfrak {F})\)). We also denote by \(\mathcal {B}_{\mathrm {tr}}= \{A \cap \mathcal {M}_\mathfrak {F}:\, A \in \mathcal {B}\}\) the trace \(\sigma \)-field of \(\mathcal {B}\) on \(\mathcal {M}_\mathfrak {F}\).

Lemma A.6

We have \(\mathcal {B}_{\mathfrak {F}}= \mathcal {B}_{\mathrm {tr}}\).

Proof

Step 1. We first prove that \(\mathcal {M}_\mathfrak {F}\) is a Borel subset in \(\mathcal {M}\). For \(g\in {{\mathcal {B}}}_+(S)\), we consider the function \(\Theta _g\) defined on \(\mathcal {M}\) by \(\Theta _g(\mu ) = g\mu \). Denote \({{\mathcal {B}}}_{b+}={{\mathcal {B}}}_b(S) \cap {{\mathcal {B}}}_+(S)\) the set of bounded nonnegative measurable functions defined on S. We follow the proof of [10, Theorem 15.13] to prove that, for every \(g\in {{\mathcal {B}}}_{b+}\), \(\Theta _g\) is a measurable function from \(\mathcal {M}\) to \(\mathcal {M}\). Denote by \({{\mathcal {F}}}=\{g\in {{\mathcal {B}}}_{b+} :\, \Theta _g \text { is measurable}\}\). The function \(\Theta _g\) is continuous for g belonging to \({{\mathcal {C}}}_{b+}= {{\mathcal {C}}}_b(S) \cap {{\mathcal {C}}}_+(S)\). Furthermore, the set \({{\mathcal {F}}}\) is closed under bounded pointwise convergence: if \(g_n \rightarrow g\) pointwise, with \(g\in {{\mathcal {B}}}_{b+}\) and \((g_n, \,n\in {{\mathbb {N}}})\) a bounded sequence of elements of \({{\mathcal {F}}}\) (i.e. \(\sup _{n\in {{\mathbb {N}}}} \left\Vert g_n\right\Vert _\infty <\infty \)), then \(\Theta _g(\mu )=\lim _{n\rightarrow \infty } \Theta _{g_n}(\mu )\) by dominated convergence and thus g belongs to \({{\mathcal {F}}}\). An immediate extension of [10, Theorem 4.33] gives that \({{\mathcal {B}}}_{b+}\subset {{\mathcal {F}}}\).

We then deduce that the function \(\theta _g :\mathcal {M}\rightarrow [0, +\infty ]\) defined by \(\theta _g(\mu ) = g\mu (\mathbf {1})=\mu (g)\) is measurable for every \(g\in {{\mathcal {B}}}_{b+}\), and as \(g\in {{\mathcal {B}}}_+(S)\) is the limit of \(g\wedge n \in {{\mathcal {B}}}_{b+}\) as n goes to infinity, we deduce by monotone convergence that \(\theta _g=\lim _{n\rightarrow \infty } \theta _{g\wedge n}\), and thus \(\theta _g\) is measurable for every \(g\in {{\mathcal {B}}}_+(S)\). By definition of \(\mathcal {M}_\mathfrak {F}\), we have that \(\mathcal {M}_\mathfrak {F}= \bigcap _{f\in \mathfrak {F}}\theta _f ^{-1} ({{\mathbb {R}}}_+),\) and thus \(\mathcal {M}_\mathfrak {F}\) is a Borel subset in \(\mathcal {M}\).

Step 2. We prove that for every \(\mu \in \mathcal {M}_\mathfrak {F}\), the mapping \(\nu \mapsto d_\mathfrak {F}(\mu ,\nu )\) defined on \(\mathcal {M}_\mathfrak {F}\) is \(\mathcal {B}_{\mathrm {tr}}\)-measurable. Let \(g\in {{\mathcal {B}}}_{b+}\). Since the function \(\Theta _g\) is measurable from \(\mathcal {M}\) to itself by step 1, it is \(\mathcal {B}/\mathcal {B}\)-measurable. By definition of the trace \(\sigma \)-field, it follows that the mapping \(\Theta _g\) from \(\mathcal {M}_\mathfrak {F}\) to \(\mathcal {M}\) is \(\mathcal {B}_{\mathrm {tr}}/\mathcal {B}\)-measurable. Let \(f\in \mathfrak {F}\). By monotone convergence we get that \(\Theta _f=\lim _{n\rightarrow \infty } \Theta _{f\wedge n}\), and thus \(\Theta _{f}\) is \(\mathcal {B}_{\mathrm {tr}}/\mathcal {B}\)-measurable.

Since \(\mu \in \mathcal {M}_\mathfrak {F}\), we have \(f \mu \in \mathcal {M}\) and the mapping \(\pi \mapsto d_{\mathrm {BL}}(f\mu , \pi )\) from \(\mathcal {M}\) to \(\mathbb {R}\) is continuous hence \(\mathcal {B}\)-measurable. Thus, by composition we get that the mapping \(\nu \mapsto d_{\mathrm {BL}}(f\mu , f\nu )\) from \(\mathcal {M}_\mathfrak {F}\) to \(\mathbb {R}\) is \(\mathcal {B}_{\mathrm {tr}}\)-measurable. Finally, the mapping \(\nu \mapsto d_\mathfrak {F}(\mu ,\nu )\) from \(\mathcal {M}_\mathfrak {F}\) to \(\mathbb {R}\) is \(\mathcal {B}_{\mathrm {tr}}\)-measurable as a sum of \(\mathcal {B}_{\mathrm {tr}}\)-measurable mappings.

Step 3. We conclude the proof of the lemma. For every \(\mu \in \mathcal {M}_\mathfrak {F}\) and every \(\varepsilon > 0\), we have

$$\begin{aligned} B(\mu , \varepsilon ) = \{ \nu \in \mathcal {M}_\mathfrak {F}:\, d_\mathfrak {F}(\mu ,\nu ) < \varepsilon \} \in \mathcal {B}_{\mathrm {tr}}\end{aligned}$$

by Step 2. Since \(\mathcal {M}_\mathfrak {F}\) is a Polish space, every open set is the countable union of open balls and it follows that every open set lies in \(\mathcal {B}_{\mathrm {tr}}\). Hence we get \(\mathcal {B}_{\mathfrak {F}}\subset \mathcal {B}_{\mathrm {tr}}\).

Conversely, notice that the identity mapping from \( (\mathcal {M}_\mathfrak {F}, d_\mathfrak {F})\) to \((\mathcal {M}_\mathfrak {F}, d_{\mathrm {BL}})\) is continuous. Therefore, if \(V \subset \mathcal {M}\) is an open set, \(V \cap \mathcal {M}_\mathfrak {F}\) is open in \((\mathcal {M}_\mathfrak {F}, d_{\mathrm {BL}})\) hence also in \((\mathcal {M}_\mathfrak {F}, d_\mathfrak {F})\). In particular, we have \(V \cap \mathcal {M}_\mathfrak {F}\in \mathcal {B}_{\mathfrak {F}}\). Since this is true for every open set \(V \subset \mathcal {M}\), we deduce that \(\mathcal {B}_{\mathrm {tr}}\subset \mathcal {B}_{\mathfrak {F}}\). \(\square \)

The following two results are a direct consequence of Lemma A.6.

Corollary A.7

Let \(\xi \) be a \(\mathcal {M}\)-valued random variable such that a.s. \(\xi (f) < \infty \) for every \(f\in \mathfrak {F}\). Then \(\xi \) is a \(\mathcal {M}_\mathfrak {F}\)-valued random variable. Conversely, if \(\xi \) is a \(\mathcal {M}_\mathfrak {F}\)-valued random variable then \(\xi \) is also a \(\mathcal {M}\)-valued random variable.

Corollary A.8

Let \(\xi \) and \(\zeta \) be \(\mathcal {M}_\mathfrak {F}\)-valued random variables. Then the following conditions are equivalent:

  1. (i)

    \(\xi \overset{\scriptscriptstyle (d)}{=}\zeta \) when viewed as \(\mathcal {M}_\mathfrak {F}\)-valued random variables.

  2. (ii)

    \(\xi \overset{\scriptscriptstyle (d)}{=}\zeta \) when viewed as \(\mathcal {M}\)-valued random variables.

  3. (iii)

    \(\xi (h) \overset{\scriptscriptstyle (d)}{=}\zeta (h)\) for every \(h\in {{\mathcal {C}}}_b(S)\).

  4. (iv)

    \(\xi (fh) \overset{\scriptscriptstyle (d)}{=}\zeta (fh)\) for every \(h\in {{\mathcal {C}}}_b(S)\) and \(f\in \mathfrak {F}\).

We now characterize convergence in distribution of random measures in \(\mathcal {M}_\mathfrak {F}\). Recall that (H1)–(H4) are in force.

Proposition A.9

Let \(\xi _n\) and \(\xi \) be \(\mathcal {M}_\mathfrak {F}\)-valued random variables. Then \(\xi _n\) converges in distribution to \(\xi \) in \(\mathcal {M}_\mathfrak {F}\) if and only if \(\xi _n(f h) \xrightarrow [n \rightarrow \infty ]{(d)}\xi (f h)\) for every \(h\in {{\mathcal {C}}}_b(S) \) and every \(f\in \mathfrak {F}\).

Proof

Assume that \(\xi _n\) converges in distribution to \(\xi \) in \(\mathcal {M}_\mathfrak {F}\). Let \(f\in \mathfrak {F}\). Since \(F :\mu \mapsto f \mu \) is continuous from \(\mathcal {M}_\mathfrak {F}\) to \(\mathcal {M}\) and \(\nu \mapsto \nu (h)\) is continuous from \(\mathcal {M}\) to \(\mathbb {R}\) for every \(h\in {{\mathcal {C}}}_b(S)\), it follows that the mapping \(\mu \mapsto \mu (fh)\) is continuous from \(\mathcal {M}_\mathfrak {F}\) to \(\mathbb {R}\). By the continuous mapping theorem, we get \(\xi _n(fh) \xrightarrow []{\scriptscriptstyle (d)}\xi (fh)\).

Conversely, for every \(f\in \mathfrak {F}\), \(f\xi _n\) and \(f\xi \) are \(\mathcal {M}\)-valued random variables, and we have \(\xi _n(fh) \xrightarrow []{\scriptscriptstyle (d)}\xi (fh)\) for every \(h\in {{\mathcal {C}}}_b(S)\). By [31, Theorem 4.11], this implies that \(f\xi _n \xrightarrow [n \rightarrow \infty ]{(d)}f \xi \) in the space \(\mathcal {M}\). In particular, \((f \xi _n, \, n \in \mathbb {N})\) is tight (in distribution) in \(\mathcal {M}\) for every \(f\in \mathfrak {F}\). By Proposition  A.4, it follows that \((\xi _n, \, n \in \mathbb {N})\) is tight in \(\mathcal {M}_\mathfrak {F}\). Since \(\mathcal {M}_\mathfrak {F}\) is Polish, Prokhorov’s theorem ensures that \((\xi _n, \, n \in \mathbb {N})\) is relatively compact (in distribution) in \(\mathcal {M}_\mathfrak {F}\). Let \(\hat{\xi }\) be a limit point (in distribution) of \((\xi _n, \, n \in \mathbb {N})\). There exists a subsequence, still denoted by \(\xi _n\), such that \(\xi _n \xrightarrow []{\scriptscriptstyle (d)}\hat{\xi }\) in \(\mathcal {M}_\mathfrak {F}\). Let \(h\in {{\mathcal {C}}}_b(S)\). Applying the first part of the proof, we get that \(\xi _n(fh) \xrightarrow [n \rightarrow \infty ]{(d)}\hat{\xi }(fh)\) for every \(f\in \mathfrak {F}\). Therefore, we have \(\hat{\xi }(fh) \overset{\scriptscriptstyle (d)}{=}\xi (fh)\) for every \(h\in {{\mathcal {C}}}_b(S)\). It follows from Corollary  A.8 that \(\hat{\xi } \overset{\scriptscriptstyle (d)}{=}\xi \) in \(\mathcal {M}_\mathfrak {F}\). Thus the sequence \((\xi _n, \, n \in \mathbb {N})\) is relatively compact and has only one limit point \(\xi \) in \(\mathcal {M}_\mathfrak {F}\). This proves the result. \(\square \)

We state now the main result of this section. Recall that (H1)–(H4) are in force.

Proposition 7.6

Let \((\xi _n, \, n \in \mathbb {N})\) be a sequence of \(\mathcal {M}_\mathfrak {F}\)-valued random variables and \(\xi \) be a \(\mathcal {M}\)-valued random variable such that \(\xi _n \xrightarrow []{\scriptscriptstyle (d)}\xi \) in \(\mathcal {M}\) and \((\xi _n, n \in \mathbb {N})\) is tight (in distribution) in \(\mathcal {M}_\mathfrak {F}\). Then \(\xi \) is a \(\mathcal {M}_\mathfrak {F}\)-valued random variable and we have the convergence in distribution \(\xi _n \xrightarrow []{\scriptscriptstyle (d)}\xi \) in \(\mathcal {M}_\mathfrak {F}\).

Proof

By assumption, the sequence \((\xi _n, \, n \in \mathbb {N})\) is relatively compact (in distribution) in the space \(\mathcal {M}_\mathfrak {F}\). Let \(\hat{\xi } \in \mathcal {M}_\mathfrak {F}\) be a limit point in distribution and let \(h\in {{\mathcal {C}}}_b(S)\). On the one hand, Proposition  A.9 applied with \(f=\mathbf {1}\) yields the convergence \(\xi _n(h) \xrightarrow []{\scriptscriptstyle (d)}\hat{\xi }(h)\). On the other hand, since \(\xi _n \xrightarrow []{\scriptscriptstyle (d)}\xi \) in \(\mathcal {M}\) it follows that \(\xi _n(h) \xrightarrow []{\scriptscriptstyle (d)}\xi (h)\). Therefore \(\hat{\xi }(h) \overset{\scriptscriptstyle (d)}{=}\xi (h)\) for every \(h\in {{\mathcal {C}}}_b(S)\), i.e. \(\hat{\xi } \overset{\scriptscriptstyle (d)}{=}\xi \) in \(\mathcal {M}\). Since the distribution of \(\hat{\xi }\) is concentrated on \(\mathcal {M}_\mathfrak {F}\), the same is true for \(\xi \). In other words \(\xi \in \mathcal {M}_\mathfrak {F}\) a.s., and so \(\xi \) is a \(\mathcal {M}_\mathfrak {F}\)-valued random variable by Corollary A.7. Now, applying Corollary A.8 we get \(\hat{\xi } \overset{\scriptscriptstyle (d)}{=}\xi \) in the space \(\mathcal {M}_\mathfrak {F}\). Thus the sequence \((\xi _n, \, n \in \mathbb {N})\) is relatively compact in \(\mathcal {M}_\mathfrak {F}\) and has only one limit point \(\xi \), so \(\xi _n \xrightarrow []{\scriptscriptstyle (d)}\xi \) in \(\mathcal {M}_\mathfrak {F}\). \(\square \)

The following special case is particularly useful. Recall that (H1)–(H4) are in force.

Corollary A.11

Assume that (H5) holds. Let \((\xi _n, \, n \in \mathbb {N})\) and \(\xi \) be \(\mathcal {M}\)-valued random variables such that \(\xi _n \xrightarrow []{\scriptscriptstyle (d)}\xi \) in \(\mathcal {M}\) and for every \(f\in \mathfrak {F}\),

$$\begin{aligned} \sup _{n} {\mathbb {E}}\left[ \xi _n(f)\right] < \infty . \end{aligned}$$
(A.6)

Then \((\xi _n, \, n \in \mathbb {N})\) and \(\xi \) are \(\mathcal {M}_\mathfrak {F}\)-valued random variables and we have the convergence in distribution \( \xi _n \xrightarrow []{\scriptscriptstyle (d)}\xi \) in \(\mathcal {M}_\mathfrak {F}\). Moreover, for every \(f\in \mathfrak {F}\), we have

$$\begin{aligned} {\mathbb {E}}\left[ \xi (f)\right] \le \liminf _{n \rightarrow \infty } {\mathbb {E}}\left[ \xi _n(f)\right] < \infty . \end{aligned}$$

Furthermore, if \(({\mathbb {E}}\left[ \xi _n(\bullet )\right] , \, n\in \mathbb {N})\) converges to \( {\mathbb {E}}\left[ \xi (\bullet )\right] \) in \(\mathcal {M}\) then the convergence actually holds in \(\mathcal {M}_\mathfrak {F}\).

Proof

The random variable \(\xi _n\) is \(\mathcal {M}\)-valued and satisfies \(\xi _n(f) < \infty \) a.s. since \({\mathbb {E}}\left[ \xi _n(f)\right] < \infty \) for every \(f\in \mathfrak {F}\), so by Corollary A.7, \(\xi _n\) is a \(\mathcal {M}_\mathfrak {F}\)-valued random variable. By Corollary A.5, the assumption (A.6) implies that \((\xi _n, \, n \in \mathbb {N})\) is tight (in distribution) in \(\mathcal {M}_\mathfrak {F}\). Therefore Proposition 7.6 applies and gives the convergence in distribution \(\xi _n \xrightarrow []{\scriptscriptstyle (d)}\xi \) in \(\mathcal {M}_\mathfrak {F}\). Moreover, Skorokhod’s representation theorem in conjunction with Fatou’s lemma implies that for every \(f \in \mathfrak {F}\),

$$\begin{aligned} {\mathbb {E}}\left[ \xi (f)\right] \le \liminf _{n \rightarrow \infty }{\mathbb {E}}\left[ \xi _n(f)\right] < \infty . \end{aligned}$$

Now set \(\mu _n = {\mathbb {E}}\left[ \xi _n(\bullet )\right] \) and \(\mu = {\mathbb {E}}\left[ \xi (\bullet )\right] \) and assume that \(\mu _n \rightarrow \mu \) in \(\mathcal {M}\). Notice that the assumption (A.6) implies that \(\mu _n \in \mathcal {M}_\mathfrak {F}\) for every \(n \in \mathbb {N}\) and that the sequence of measures \((f\mu _n, \, n \in \mathbb {N})\) is bounded for every \(f\in \mathfrak {F}\). Thus Corollary  A.3 gives the convergence \(\lim _{n\rightarrow \infty }\mu _n = \mu \) in \(\mathcal {M}_\mathfrak {F}\). \(\square \)

Appendix B. Sub-exponential tail bounds for the height of conditioned BGW trees

Assume that \(\xi \) satisfies  \((\xi 1)\) and  \((\xi 2)\) and denote by \(\tau ^n\) a BGW(\(\xi \)) tree conditioned to have n vertices. Then by [35, Theorem 1] which is stated for the aperiodic case but is trivially extended to the general case, for every \(\alpha \in (0,\gamma /(\gamma - 1))\), there exist two constants \(C_0, c_0 >0\) such that for every \(y \ge 0\) and every \(n \in \Delta \)

$$\begin{aligned} {\mathbb {P}}\left( \frac{b_n}{n}\mathfrak {h}(\tau ^n)\le y\right) \le C_0 \exp \left( -c_0 y^{-\alpha }\right) . \end{aligned}$$
(A.7)

We will show that under the stronger assumption \((\xi 3)\), the previous inequality holds with \(\alpha = \gamma /(\gamma -1)\). Since the finite variance case has already been treated in [5], we assume henceforth that \(\xi \) has infinite variance.

Recall that L is a slowly varying function such that \({\mathbb {E}}\left[ \xi ^2 \mathbf {1}_{\{\xi \le n\}}\right] = n^{2-\gamma }L(n)\). On the other hand, the slowly varying function appearing in the appendix of [35], which we denote by K, satisfies \({\text {Var}}\left( \xi \mathbf {1}_{\{\xi \le n\}}\right) = n^{2-\gamma }K(n)\). Since \({\text {Var}} (\xi ) = +\infty \), we have as n goes to infinity that

$$\begin{aligned} {\mathbb {E}}\left[ \xi ^2 \mathbf {1}_{\{\xi \le n\}}\right] \sim n^{2-\gamma } K(n)+1 \sim n^{2-\gamma } K(n), \end{aligned}$$

see the appendix in [35]. Therefore, we get \(K(n) \sim L(n)\) and K is bounded above.

Following the proof of [35, Theorem 1] to get (A.7) holds for \(\alpha =\gamma /(\gamma -1)\), it is enough to prove the analogue of Proposition 8 therein with \(\alpha = \gamma /(\gamma -1)\), that is Proposition B.1 below. Let \((W_n, \, n\in {{\mathbb {N}}})\) be a random walk with starting point \(W_0 = 0\) and jump distribution \(\xi -1\).

Proposition B.1

Assume that \(\xi \) satisfies \((\xi 1)\) and \((\xi 3)\) . There exist two constants \(C_0, c_0 >0\) such that for every \(u \ge 0\) and every \(n \ge 1\),

$$\begin{aligned} {\mathbb {P}}\left( \min _{1 \le i \le n} W_i \le -u\,\!b_n\right) \le C_0 \exp \left( -c_0 u^{\gamma /(\gamma -1)}\right) . \end{aligned}$$
(A.8)

Proof

Note that \({\mathbb {P}}\left( \min _{1 \le i \le n} W_i \le -u\, \!b_n\right) = 0\) if \(u\, \!b_n > n\), so that it is enough to prove (A.8) for \(1\le u \le n/b_n\). Write, for \(h >0\)

$$\begin{aligned} {\mathbb {P}}\left( \min _{1 \le i \le n}W_i \le -u\, \!b_n \right)= & {} {\mathbb {P}}\left( \max _{1\le i \le n} \mathrm {e}^{-hW_i} \ge \mathrm {e}^{hu\, \!b_n}\right) \nonumber \\\le & {} \mathrm {e}^{-hu\, \! b_n} {\mathbb {E}}\left[ \mathrm {e}^{-hW_n}\right] = \mathrm {e}^{-hu \, \! b_n} {\mathbb {E}}\left[ \mathrm {e}^{-hW_1}\right] ^n, \end{aligned}$$
(A.9)

where the inequality follows from Doob’s maximal inequality applied to the submartingale \((\mathrm {e}^{-hW_n}, \, n \in {{\mathbb {N}}})\). We shall apply (A.9) with \(h = \varepsilon u^{\eta }/ b_n\) where \(\eta = 1/(\gamma - 1)\) and \(\varepsilon >0\) is a constant to be chosen later. Note that \(\gamma /(\gamma - 1)=\eta \gamma = 1+\eta \). Observe that \(\varepsilon u^\eta / b_n\) is bounded uniformly in \(1 \le u \le n/b_n\) and \(n \ge 1\). Indeed, since \(b_n \ge \underline{b} \, \! n^{1/\gamma }\), we have

$$\begin{aligned} \frac{u^\eta }{b_n} \le \left( \frac{n}{b_n}\right) ^\eta \frac{1}{b_n}\le \frac{1}{\underline{b}^{1+\eta }}\cdot \end{aligned}$$

Therefore, using [35, Eq. (42)] together with the inequality \(1+x \le \mathrm {e}^x\), we have for every \(n \ge 1\) and every \(1\le u \le n/b_n\)

$$\begin{aligned} {\mathbb {E}}\left[ \mathrm {e}^{-\varepsilon \frac{u^\eta }{b_n}W_1}\right] ^n\le \exp \left\{ Cn \left( \varepsilon \frac{u^\eta }{b_n}\right) ^\gamma K\left( \frac{b_n}{\varepsilon u^\eta }\right) \right\} \le \exp \left( C'\varepsilon ^\gamma u^{\eta \gamma }\right) , \end{aligned}$$

as K is bounded from above and \(b_n \ge \underline{b} \, \! n^{1/\gamma }\). Thus, we deduce from (A.9) that for \(1\le u\le n/b_n\)

$$\begin{aligned} {\mathbb {P}}\left( \min _{1 \le i \le n}W_i \le -u\, \!b_n \right) \le \exp \left( -\left( \varepsilon - C' \varepsilon ^\gamma \right) u^{1+\eta }\right) . \end{aligned}$$

The conclusion readily follows by choosing \(\varepsilon >0\) small enough such that \(\varepsilon - C' \varepsilon ^\gamma >0\). \(\square \)

Remark B.2

In fact, this proof is valid if we only assume that the slowly varying function L of \((\xi 3)\) is bounded from above, in which case \(n^{-1/\gamma }b_n\) is bounded below.

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Abraham, R., Delmas, JF. & Nassif, M. Global regime for general additive functionals of conditioned Bienaymé-Galton-Watson trees. Probab. Theory Relat. Fields 182, 277–351 (2022). https://doi.org/10.1007/s00440-021-01095-9

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  • Issue Date:

  • DOI: https://doi.org/10.1007/s00440-021-01095-9

Keywords

  • Galton-Watson trees
  • Lévy trees
  • Additive functionals
  • Scaling limit
  • Phase transition

Mathematics Subject Classification

  • 60J80
  • 60F17
  • 05C05