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

Abstract

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

$$\begin{aligned} {\mathbb {P}}\left( \frac{Z_n(\sqrt{n}A)}{Z_n({\mathbb {R}})}-\nu (A)>\Delta \right) , \end{aligned}$$

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.

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Acknowledgements

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).

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Correspondence to Xinxin Chen.

Appendices

Appendix A

Lemma A.1

$$\begin{aligned} \inf _{a>0}\frac{\gamma (a)-\log p_1}{a}=\Lambda ^{-1} \left( \log \frac{1}{p_1}\right) \in (0,\infty ). \end{aligned}$$
(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

$$\begin{aligned} \inf _{a>0}\frac{\gamma (a)-\log p_1}{a}=\inf _{a<\Lambda _{\infty }'} \frac{\gamma (a)-\log p_1}{a}=\inf _{t\in \mathcal{D}_{\Lambda }^o}\frac{\log \frac{1}{p_1}+t\Lambda '(t)-\Lambda (t) }{\Lambda '(t)}. \end{aligned}$$

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

$$\begin{aligned} L_{k+1}=F(L_k)=\alpha \inf _{u\in {\mathbb {R}}}(-\log p_1+\gamma (u)-uL_k)+L_k, \,k\ge 0. \end{aligned}$$

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

$$\begin{aligned} \lim _{k\rightarrow \infty }L_k={\bar{\Lambda }}(p_1)=\inf _{a>0}\frac{\gamma (a)-\log p_1}{a}. \end{aligned}$$

Proof

Notice that

$$\begin{aligned} \inf _{u\in {{\mathbb {R}}}}(\gamma (u)-ux)=\inf _{t\in \mathcal{D}_{\Lambda }^o }(t\Lambda '(t)-\Lambda (t)-\Lambda '(t)x )=-\Lambda (x),\quad x\in {{\mathcal {D}}}_{\Lambda }. \end{aligned}$$

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

$$\begin{aligned} \inf _{u\in {\mathbb {R}}}(-\log p_1+\gamma (u)-uL_0)&=\inf _{u>0}(-\log p_1+\gamma (u)-uL_0) \\&=\Lambda (\theta )-\Lambda (L_0)\ge 0, \end{aligned}$$

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

$$\begin{aligned} L_{1}=\alpha (\Lambda (\theta )-\Lambda (L_0))+L_0\le \alpha \Lambda '(\theta )(\theta -L_0)+L_0\le \theta . \end{aligned}$$

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 \(L_0\le L_k\le \theta \),

$$\begin{aligned} L_{k+1}-L_k=\alpha (\Lambda (\theta )-\Lambda (L_k))\ge \alpha \Lambda '(L_k)(\theta -L_k)\ge \alpha \Lambda '(L_0)(\theta -L_k)\ge 0. \end{aligned}$$

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

$$\begin{aligned} {\mathbb {P}}\left( \sum _{i=1}^{t_n}{X_i}\ge a\sqrt{n}\right)\le & {} {\mathbb {P}}\left( \sup _{1\le i\le t_n}{X_i}\ge n\right) \nonumber \\&+\,{\mathbb {P}}\left( \sum _{i=1}^{t_n}{X_i} \ge a\sqrt{n}, \sup _{1\le i\le t_n}{X_i}< n\right) . \end{aligned}$$
(B.1)

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

$$\begin{aligned} {\mathbb {P}}({X}\ge x)\le c_{16} e^{-\lambda x^{\alpha }}. \end{aligned}$$

Therefore,

$$\begin{aligned} {\mathbb {P}}\left( \sup _{1\le i\le t_n}{X_i}\ge n\right) \le t_n {\mathbb {P}}({X}\ge n)\le c_{16} t_n e^{-\lambda n^{\alpha }}. \end{aligned}$$
(B.2)

Meanwhile,

$$\begin{aligned}&{\mathbb {P}}\left( \sum _{i=1}^{t_n}{X_i} \ge a\sqrt{n}, \sup _{1\le i\le t_n}{X_i}< n\right) \nonumber \\&\quad = \sum _{x_i\in [0,n)\cap {{\mathbb {N}}}, i=1,\ldots , t_n}{\mathbb {P}}\left( \sum _{i=1}^{t_n}{X_i} \ge a\sqrt{n}, \sup _{1\le i\le t_n}{X_i}< n, {X_i}\in [x_i, x_i+1)\right) \nonumber \\&\quad \le \sum _{\begin{array}{c} \sum _{i=1}^{t_n}x_i\ge a\sqrt{n}-t_n; \\ x_i\in [0,n)\cap {{\mathbb {N}}}, i=1,\ldots , t_n \end{array}} {\mathbb {P}}\left( {X_i}\in [x_i, x_i+1],\, \forall 1\le i\le t_n\right) \nonumber \\&\quad \le \sum _{\begin{array}{c} \sum _{i=1}^{t_n}x_i\ge a\sqrt{n}-t_n; \\ x_i\in [0,n)\cap {{\mathbb {N}}}, i=1,\ldots , t_n \end{array}} c_{16}^{t_n}\exp \left\{ -\lambda \sum _{i=1}^{t_n}x_i^{\alpha }\right\} \nonumber \\&\quad \le \sum _{\begin{array}{c} \sum _{i=1}^{t_n}x_i\ge a\sqrt{n}-t_n; \\ x_i\in [0,n)\cap {{\mathbb {N}}}, i=1,\ldots , t_n \end{array}} c_{16}^{t_n}\exp \left\{ -\lambda \frac{(a\sqrt{n}-t_n)^{\alpha }}{t_n^{\alpha -1}}\right\} \nonumber \\&\quad \le (nc_{16})^{t_n}\exp \left\{ -\lambda \frac{(a\sqrt{n}-t_n)^{\alpha }}{t_n^{\alpha -1}}\right\} , \end{aligned}$$
(B.3)

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

$$\begin{aligned} \sum _{i=1}^{t_n}x_i^{\alpha }\ge t_n \left( \frac{\sum _{i=1}^{t_n}x_i}{t_n}\right) ^\alpha \ge \frac{(a\sqrt{n}-t_n)^{\alpha }}{t_n^{\alpha -1}}. \end{aligned}$$

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

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

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

$$\begin{aligned} {\mathbb {P}}\left( \sum _{i=1}^{t_n}{X_i} \ge a\sqrt{n}\right)\le & {} c_{17} t_n e^{-e^{n^\alpha }}+{\mathbb {P}}\left( \sum _{i=1}^{t_n}{X_i} \ge a\sqrt{n}, \sup _{1\le i\le t_n}{X_i}< n\right) \nonumber \\\le & {} c_{17}t_n e^{-e^{n^\alpha }}+ (c_{17}n)^{t_n} \exp \left\{ - e^{\left( \frac{a\sqrt{n}}{t_n}\right) ^\alpha (1-\varepsilon )}\right\} , \end{aligned}$$
(B.4)

where we use the fact

$$\begin{aligned} \sum _{i=1}^{t_n}e^{x_i^\alpha }\ge \exp \left\{ \max _{1\le i\le t_n}x_i^\alpha \right\} \ge \exp \left\{ {\left( \frac{\sum _{i=1}^{t_n}x_i}{t_n}\right) ^\alpha }\right\} . \end{aligned}$$

Consequently,

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

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 \(0<\eta <L/2\),

$$\begin{aligned} q_{\eta }:={\mathbb {P}}\left( X\in (L-\eta , L]\right) ={\mathbb {P}}\left( X\in [-L, -L+\eta )\right) >0. \end{aligned}$$

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 \(y\in \text {sgn}(x)\left( x\sqrt{n}, x\sqrt{n}+\frac{2\eta x\sqrt{n}}{L}\right] =:B\),

$$\begin{aligned} \nu \left( A-\frac{y}{\sqrt{n}}\right) \ge p+\delta ,\quad \frac{|y|}{\sqrt{n}}<I_A(p)+\varepsilon . \end{aligned}$$

Set

$$\begin{aligned} w=\lceil |x|\sqrt{n}/(L-\eta )\rceil ;\quad m=n-|w| \end{aligned}$$

and

$$\begin{aligned} {{\mathcal {M}}}_{\eta }=\left\{ \zeta \in {{\mathcal {M}}}: |\zeta |=b^{w}~\hbox {and supp}~\zeta \subset B\right\} . \end{aligned}$$

By the Markov property,

$$\begin{aligned} {\mathbb {P}}({\bar{Z}}_n(\sqrt{n}A)\ge p)\ge \sum _{\zeta \in \mathcal{M}_{\eta }}{\mathbb {P}}(Z_{w}=\zeta ){\mathbb {P}}({\bar{Z}}_m^{\zeta }(\sqrt{n}A)\ge p). \end{aligned}$$
(C.1)

Note that

$$\begin{aligned} {\mathbb {P}}(Z_{w}\in {{\mathcal {M}}}_{\eta })\ge q_{\eta }^{\frac{b^{w+1}-b}{b-1}}p_b^{\frac{b^{w}-1}{b-1}}=e^{-c_{18}b^w}. \end{aligned}$$
(C.2)

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),

$$\begin{aligned} {\mathbb {P}}({\bar{Z}}_m^{\zeta }(\sqrt{n}A)\ge p)\ge \prod _{y\in \zeta } {\mathbb {P}}({\bar{Z}}_m(\sqrt{n}A-y)\ge p). \end{aligned}$$
(C.3)

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

$$\begin{aligned} {\mathbb {P}}\left( {\bar{Z}}_m^{\zeta }(\sqrt{n}A)\ge p\right) \ge e^{-c_{19}b^{w}}, \end{aligned}$$
(C.4)

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

$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{1}{\sqrt{n}}\log \left[ -\log {\mathbb {P}}({\bar{Z}}_n(\sqrt{n}A)\ge p) \right] \ge \frac{I_A(p)\log b}{L-\eta }. \end{aligned}$$

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

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

$$\begin{aligned} |w|=\lfloor (I_A(p)-\varepsilon )\sqrt{n}/L\rfloor ;\quad m=n-|w|. \end{aligned}$$

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

$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{1}{\sqrt{n}}\log \left[ -\log {\mathbb {P}}({\bar{Z}}_n(\sqrt{n}A)\ge p) \right] \le \frac{I_A(p)\log b}{L}. \end{aligned}$$

\(\square \)

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

$$\begin{aligned} \ell _{z_n}:={\mathbb {P}}\left( X\in [z_n, z_n+\frac{1}{n})\right) \ge \frac{1}{n^2}. \end{aligned}$$

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

$$\begin{aligned} r<J_A(p)+\varepsilon ,\quad \nu ((A-{y}/{\sqrt{n}})/\sqrt{1-r})\ge p+\delta . \end{aligned}$$

Set

$$\begin{aligned} q=2\lfloor rn/2\rfloor ;\quad w=\left\lfloor \frac{|x|\sqrt{n}}{z_n}\right\rfloor ;\quad s=q+w \end{aligned}$$

and

$$\begin{aligned} {{\mathcal {M}}}_n=\left\{ \zeta \in {{\mathcal {M}}}: |\zeta |=b^{s}\text { and }\text {supp}\zeta \subset B \right\} . \end{aligned}$$

By the Markov property,

$$\begin{aligned} {\mathbb {P}}({\bar{Z}}_n(\sqrt{n}A)\ge p)\ge \sum _{\zeta \in \mathcal{M}_{\eta }}{\mathbb {P}}(Z_{s}=\zeta ){\mathbb {P}}\left( {\bar{Z}}_{n-s}^{\zeta }(\sqrt{n}A)\ge p\right) . \end{aligned}$$
(C.5)

Notice that

$$\begin{aligned} {\mathbb {P}}(Z_{s}\in {{\mathcal {M}}}_n)\ge \ell _{z_n}^{\frac{b^{s+1}-b}{b-1}}p_b^{\frac{b^{s}-1}{b-1}}\ge e^{-c_{20}b^s\log n}, \end{aligned}$$
(C.6)

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 \(\zeta \in {{\mathcal {M}}}_{n}\),

$$\begin{aligned} {\mathbb {P}}\left( {\bar{Z}}_{n-s}^{\zeta }(\sqrt{n}A)\ge p\right) \ge \prod _{y\in \zeta } {\mathbb {P}}({\bar{Z}}_{n-s}(\sqrt{n}A-y)\ge p). \end{aligned}$$

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

$$\begin{aligned} {\mathbb {P}}\left( {\bar{Z}}_m^{\zeta }(\sqrt{n}A)\ge p\right) \ge e^{-c_{21}b^{s}}, \end{aligned}$$
(C.7)

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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Chen, X., He, H. On large deviation probabilities for empirical distribution of supercritical branching random walks with unbounded displacements. Probab. Theory Relat. Fields 175, 255–307 (2019). https://doi.org/10.1007/s00440-018-0891-4

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Keywords

  • Branching random walk
  • Large deviation
  • Schröder case
  • Böttcher case

Mathematics Subject Classification

  • 60J60
  • 60F10