On large deviation probabilities for empirical distribution of supercritical branching random walks with unbounded displacements

Given a branching random walk on \({\mathbb {R}}\) started from the origin, where the tail of the branching law decays at least exponentially fast and the offspring number is at least one, let \(Z_n(\cdot )\) be the counting measure which counts the number of individuals at the n-th generation located in a given set. Under some mild conditions, it is known (Biggins in Stoch. Process. Appl. 34:255–274, 1990) that for any interval \(A\subset {\mathbb {R}}\), \(\frac{Z_n(\sqrt{n}A)}{Z_n({\mathbb {R}})}\) converges a.s. to \(\nu (A)\), where \(\nu \) is the standard Gaussian measure. In this work, we investigate the convergence rates of

for \(\Delta \in (0, 1-\nu (A))\). We consider both the Schröder case, where the offspring number could be one, and the Böttcher case, where the offspring number is at least two.

We are grateful to Prof. Zhan Shi for help with Lemma 3.3. We also would like to thank the referee for carefully reading our paper, spotting several inaccuracies and for making valuable suggestions to improve the presentation. Hui He is supported by NSFC (Nos. 11671041, 11531001, 11371061).

Authors and Affiliations

1. Institut Camille Jordan, CNRS UMR 5208, Universite Claude Bernard Lyon 1, 69622, Villeurbanne Cedex, France

   Xinxin Chen

2. School of Mathematical Sciences, Beijing Normal University, Beijing, 100875, People’s Republic of China

   Hui He

Corresponding author

Correspondence to Xinxin Chen.

Appendix A

Lemma A.1

Proof

Let \(x_0:=\sup \{x\in {\mathbb {R}}: F(x)<1\}\) and \(\theta _0:=\sup \{t\in {\mathbb {R}}_+: \Lambda (t)<\infty \}\). According to Sect. 2.6 of [6], there are three cases:

1. (A)

   if \(x_0<\infty \), then \(\gamma (a)\uparrow -\log {\mathbb {P}}(X=x_0)\) as \(a\uparrow x_0\);

2. (B)

   if \(x_0=\infty \) and \(\Lambda '(t)\uparrow \infty \) as \(t\uparrow \theta _0\), then for any \(a>0\) there exists \(t\in D_\lambda ^o\) such that \(\gamma (a)=t\Lambda '(t)-\Lambda (t)\);

3. (C)

   if \(x_0=\infty \), \(\theta _0<\infty \) and \(\Lambda '(t)\uparrow a_0<\infty \) as \(t\uparrow \theta _0\), then \(\Lambda (\theta _0)<\infty \) and \(\gamma (a)=a\theta _0-\Lambda (\theta _0)\) for \(a\ge a_0\).

Recall (2.8). Let \(\Lambda _{\infty }'=\sup _{t\in \mathcal{D}_{\Lambda }^o}\Lambda '(t)\). In all cases above, we have

A simple calculation shows that the above infimum is taken when \(\Lambda (t)=-\log p_1.\) Then (A.1) follows readily. \(\square \)

Lemma A.2

For any \(\alpha >0\), let \(\{L_k: k\ge 0\}\) be a sequence such that \(0<L_0< {\bar{\Lambda }}(p_1)\) and

If \(0<\alpha \Lambda '({\bar{\Lambda }}(p_1))\le 1\), then the sequence \((L_k)_{k\ge 0}\) is non-decreasing and

Proof

Notice that

Thus if \(0<L_0\le {\bar{\Lambda }}(p_1)=:\theta \), then using the facts that \(\log \frac{1}{p_1}=\Lambda (\theta )\) yields

which implies \(L_1\ge L_0\). Furthermore, since \(0<\alpha \Lambda '({\bar{\Lambda }}(p_1))<1\), then by convexity of \(\Lambda \),

Therefore, \(\theta \ge L_1\ge L_0\). Inductively, we have \(\theta \ge L_{k+1}\ge L_k>0\). So the sequence \((L_k)_{k\ge 1}\) is non-decreasing and hence its limit exists with \(L:=\lim _{k\rightarrow \infty }L_k\le \theta \). On the other hand, as \(L_0\le L_k\le \theta \),

This shows that \(\theta -L_k\rightarrow 0\) as \(k\rightarrow \infty \). \(\square \)

Appendix B

The following lemma concerns large deviation probabilities of sums of i.i.d. random variables. The results are possibly well-known to some experts or implicitly contained in some articles.

Lemma B.1

Suppose that \(\{X_i\}_{i\ge 1}\) is a sequence of i.i.d. random variables, having the same distribution as X.

1. (1)

   If \({\mathbb {P}}(X\ge x)=\Theta (1) e^{-\lambda x^{\alpha }}\) as \(x\rightarrow \infty \) with some \(\lambda >0\) and \(\alpha >1\), then for \(a>0\), and for a sequence of integers \((t_n)\) such that \(t_n=o(n^{\frac{1}{3}})\) and \(t_n\rightarrow \infty \), we have

   $$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{t_n^{\alpha -1}}{n^{\alpha /2}}\log {\mathbb {P}}\left( \sum _{i=1}^{t_n}X_i \ge a\sqrt{n}\right) \le - \lambda a^\alpha . \end{aligned}$$
2. (2)

   If \({\mathbb {P}}(X\ge x)=\Theta (1) e^{-e^{x^{\alpha }}}\) as \(x\rightarrow \infty \) with some \(\alpha >0\), then for any \(a>0\) and any sequence \(t_n\uparrow \infty \) such that \(t_n=o\left( \frac{\sqrt{n}}{(\log n)^{2/\alpha +1}}\right) \),

   $$\begin{aligned} \liminf _{n\rightarrow \infty }\frac{t_n^{\alpha }}{n^{{\alpha }/{2}}}\log \left[ -\log {\mathbb {P}}\left( \sum _{i=1}^{t_n}X_i \ge a\sqrt{n}\right) \right] \ge a^{\alpha }. \end{aligned}$$

Proof

Notice that \({\mathbb {P}}\left( \sum _{i=1}^{t_n}X_i \ge a\sqrt{n}\right) \le {\mathbb {P}}\left( \sum _{i=1}^{t_n}X_i^+ \ge a\sqrt{n}\right) \), where \(X_i^+=X_i\vee 0.\) Thus it suffices to prove the desired results when X is non-negative. In the sequel of this proof, we always assume \(X\ge 0 \) a.s. Observe that

Proof of (1): Note that there exists \(c_{16}>0\) such that for any \(x\ge 0\),

Therefore,

Meanwhile,

where in the third inequality we use the fact that the convexity of mapping \(x\mapsto x^\alpha \) for \(\alpha >1\) implies

Plugging (B.2) and (B.3) into (B.1), we obtain that

which suffices to conclude (1) of Lemma B.1.

Proof of (2): Similarly, for any \(\varepsilon >0\) there exists some constant \(c_{17}\ge 1\) such that

where we use the fact

Consequently,

We have completed the proof. \(\square \)

Appendix C

We shall show how the arguments in Sect. 2.3 in [13] can be extended to our settings in Theorems 1.5 and 1.6.

Proof of Theorem 1.5

Lower bound. Since \(\text {ess sup }X=L\), then for any \(0<\eta <L/2\),

According to the property of \(I_A(p)\), for any \(\varepsilon >\frac{2\eta }{L}\), there exists \(x\in {{\mathbb {R}}}\), \(\delta >0\) such that for all \(y\in \text {sgn}(x)\left( x\sqrt{n}, x\sqrt{n}+\frac{2\eta x\sqrt{n}}{L}\right] =:B\),

Set

and

By the Markov property,

Note that

Meanwhile, for any \(\zeta \in {{\mathcal {M}}}_{\eta }\), as \({\bar{Z}}_m^{\zeta }(\sqrt{n}A)\ge \min _{y\in \zeta }{\bar{Z}}^{\delta _y}_m(\sqrt{n}A)\) because of (3.18),

Then by exactly the same arguments for bounding (2.24) in [13], we have for any \(\zeta \in {{\mathcal {M}}}_{\eta }\),

which, together with (C.1) and (C.2), gives

Letting \(\eta \downarrow 0\) yields the lower bound.

Upper bound. For any \(\varepsilon >0\), set

Then we only need to copy the arguments from (2.29) to (2.34) in [13] to conclude that

\(\square \)

Proof of Theorem 1.6

Lower bound We first notice that there exist \(L>L_0>0\) such that \({\mathbb {P}}(X\in (L_0, L))>\frac{1}{4}(1-{\mathbb {P}}(X=0))>0.\) Then for any n large enough, there exists \(z_n\in (L_0, L)\) such that

By property of \(J_A(p)\), we may find \(r\in (0,1),\, x\in {\mathbb R}\) and \(\delta >0\), such that for all \(y\in \text {sgn}(x)(|x|\sqrt{n}-L-2, |x|\sqrt{n}+L+2)=:B\),

Set

and

By the Markov property,

Notice that

by having all particles give birth to b children in the first s generations, move with step size only in \([z_n, z_n+\frac{1}{n})\) or only in \((-z_n, -\frac{1}{n}-z_n]\) in the first w generations (depending on the sign of x), and then alternate between \([z_n, z_n+\frac{1}{n})\) and \((-z_n-\frac{1}{n}, -z_n]\) steps in the succeeding q generations. So in the s-th generation, all the \(b^s\) particles are located in B if \(\frac{w}{n}<1.\) Again,like (C.3), for any \(\zeta \in {{\mathcal {M}}}_{n}\),

Then by exactly the same arguments for bounding (2.38) in [13], we have for any \(\zeta \in {{\mathcal {M}}}_{\eta }\),

which, together with (C.5) and (C.6), implies the lower bound.

Upper bound The proof for upper bound is also exactly the same as the one in [13]; see pages 12–13 there. We omit it here. \(\square \)

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