Sub-exponential tail bounds for conditioned stable Bienaymé–Galton–Watson trees

Abstract

We establish uniform sub-exponential tail bounds for the width, height and maximal outdegree of critical Bienaymé–Galton–Watson trees conditioned on having a large fixed size, whose offspring distribution belongs to the domain of attraction of a stable law. This extends results obtained for the height and width by Addario-Berry, Devroye and Janson in the finite variance case.

Appendix

In this appendix, we prove several useful results concerning the asymptotic behavior of Laplace transforms of critical offspring distributions belonging to domains of attractions of stable laws and of their associated size-biased distributions. As before, assume that $\mu$ is a critical offspring distribution belonging to the domain of attraction of a stable law of index $\alpha \in (1,2]$. Let $\sigma ^{2} \in (0,\infty ]$ be the variance of $\mu$ and let X is a random variable with law $\mu$. Recall from the Introduction that L is a slowly varying function such that $\text {Var}(X \cdot {\mathbbm {1}}_{X \leqslant n})= n^{2-\alpha } L(n)$. Note that $L(n) \rightarrow \sigma ^{2}$ when $\sigma ^{2}<\infty$, and that $L(n) \rightarrow \infty$ when $\sigma ^{2}=\infty$ and $\alpha =2$. Hence, if X is a random variable with law $\mu$, we have

$$\begin{aligned} {\mathbb {E}}\left[ X^{2} {\mathbbm {1}}_{X \leqslant n}\right] \quad \mathop {\sim }_{n \rightarrow \infty } \quad n^{2-\alpha } L(n) +1. \end{aligned}$$

since ${\mathbb {E}}\left[ X {\mathbbm {1}}_{X \leqslant n}\right] \rightarrow 1$ as $n \rightarrow \infty$. The term “$+1$” is not negligible only when $\sigma ^{2}<\infty$ (in which case $\alpha =2$).

Offspring distributions. Set $G(s)= \sum _{i \geqslant 0} \mu (i) s^{i}$ for $0 \leqslant s \leqslant 1$. Then by e.g. [8, Lemma 4.7]

$$\begin{aligned} G(s)-s \quad \mathop {\sim }_{s \uparrow 1} \quad \frac{\Gamma (3-\alpha )}{\alpha (\alpha -1)} \cdot (1-s)^{\alpha } L( (1-s)^{-1}). \end{aligned}$$

(41)

We stress that this holds in the both cases $\sigma ^{2}<\infty$ and $\sigma ^{2}=\infty$.

Also, if W is a random variable with distribution ${\mathbb {P}}\left( W=i\right) = \mu (i+1)$ for $i \geqslant -1$, since ${\mathbb {E}}\left[ e^{-\lambda W}\right] = e^{\lambda }G(e^{-\lambda })$ for $\lambda >0$, we have

$$\begin{aligned} {\mathbb {E}}\left[ e^{-\lambda W}\right] -1 \quad \mathop {\sim }_{\lambda \downarrow 0} \quad \frac{\Gamma (3-\alpha )}{\alpha (\alpha -1)} \cdot \lambda ^{\alpha } L(1/\lambda ). \end{aligned}$$

(42)

Again, this holds in the both cases $\sigma ^{2}<\infty$ and $\sigma ^{2}=\infty$.

Size-biased offspring distributions. Let ${\mu }^{*}$ be the so-called size-biased probability distribution on $\mathbb {Z}_{+}$ defined by ${\mu }^{*}(i)=i\mu (i)$ for $i \geqslant 0$. Note that ${\mu }^{*}$ is indeed a probability distribution since $\mu$ is critical. Let $X^{*}$ be a random variable having law $\mu ^{*}$. When $\mu$ has finite variance, we claim that

$$\begin{aligned} 1-{\mathbb {E}}\left[ s^{X^{*}}\right] \quad \mathop {\sim }_{s \uparrow 1} \quad (1-s)(\sigma ^{2}+1), \end{aligned}$$

(43)

and when $\mu$ has infinite variance, we claim that

$$\begin{aligned}&1-{\mathbb {E}}\left[ s^{X^{*}}\right] \,\, \mathop {\sim }_{s \uparrow 1} \,\, \frac{\Gamma (3-\alpha )}{\alpha -1} \cdot (1-s)^{\alpha -1} L((1-s)^{-1}),\nonumber \\&\qquad 1- {\mathbb {E}}\left[ e^{-\lambda {X}^{*}}\right] \,\,\mathop {\sim }_{\lambda \downarrow 0} \,\, \frac{\Gamma (3-\alpha )}{\alpha -1} \cdot \lambda ^{\alpha -1} L(1/\lambda ). \end{aligned}$$

(44)

When $\mu$ has finite variance the claim (43) simply follows from the fact that ${\mathbb {E}}\left[ X^{*}\right] =\sigma ^{2}+1$.

Now assume that $\mu$ has infinite variance. Then there exists a slowly varying function $L_{1}$ such that ${\mathbb {P}}\left( X \geqslant n\right) = \mu ([n,\infty ))=L_{1}(n)/n^{\alpha }$ (see [20, Corollary XVII.5.2 and (5.16)]) when $\alpha <2$, we have $L_{1}(n)= \frac{2-\alpha }{\alpha } L(n)$, and $L_{1}(n)/L(n) \rightarrow 0$ as $n \rightarrow \infty$ when $\alpha =2$. As a consequence, $\mu ^{*}$ belongs to the domain of attraction of a stable law of index $\alpha -1$, because

$$\begin{aligned} {\mu }^{*}([n,\infty )) \quad \mathop {\sim }_{n \rightarrow \infty } \quad \frac{\alpha }{\alpha -1} \cdot \frac{L_{1}(n)}{n^{\alpha -1}} \end{aligned}$$

(45)

since we can write ${\mu }^{*}([n,\infty )) = (n-1) {\mu }([n,\infty )+ \sum _{j=n}^{\infty } \mu ([j,\infty ))$.

If $\alpha <2$, (45) and [20, Corollary XVII.5.2 and (5.16)] give that

$$\begin{aligned} {\mathbb {E}}\left[ (X^*)^{2} {\mathbbm {1}}_{X^{*} \leqslant n}\right] \quad \mathop {\sim }_{n \rightarrow \infty } \quad n^{3-\alpha } \cdot \frac{2-\alpha }{3-\alpha } L(n), \end{aligned}$$

and (44) result follows e.g. by [8, Lemma 4.6].

Now assume that $\alpha =2$ and set $q^{*}_{i}= {\mathbb {P}}\left( X^{*} > i\right)$ for $i \geqslant 0$. Then

$$\begin{aligned} \sum _{i=0}^{n}q^{*}_{i}= {\mathbb {E}}\left[ X^{2} {\mathbbm {1}}_{X \leqslant n}\right] + (n+1) \mu ^{*}([n+1,\infty )) \quad \mathop {\sim }_{n \rightarrow \infty } \quad L(n). \end{aligned}$$

Indeed, we know that $L_{1}(n)/L(n) \rightarrow 0$ as $n \rightarrow \infty$. Hence, by [20, Theorem XIII.5.5], we have $\sum _{i=0}^{\infty }q^{*}_{i} s^{i} \sim L((1-s)^{-1})$ as $s \uparrow 1$. Then

$$\begin{aligned} 1-{\mathbb {E}}\left[ s^{X^{*}}\right] =(1-s) \sum _{i=0}^{\infty }q^{*}_{i} s^{i} \quad \mathop {\sim }_{s \uparrow 1} \quad (1-s)L((1-s)^{-1}). \end{aligned}$$

Our claim (44) then follows by taking $s=e^{-\lambda }$.

Finally, from (43) and (44) it is a simple matter to see that the estimates

$$\begin{aligned}&1-{\mathbb {E}}\left[ s^{X^{*}-1}\right] \, \mathop {\sim }_{s \uparrow 1} \, \frac{\Gamma (3-\alpha )}{\alpha -1} \cdot (1-s)^{\alpha -1} L((1-s)^{-1}),\nonumber \\&\qquad 1- {\mathbb {E}}\left[ e^{-\lambda ({X}^{*}-1) } \right] \,\mathop {\sim }_{\lambda \downarrow 0} \, \frac{\Gamma (3-\alpha )}{\alpha -1} \cdot \lambda ^{\alpha -1} L(1/\lambda ) \end{aligned}$$

(46)

hold in both the cases $\sigma ^{2}<\infty$ and $\sigma ^{2}=\infty$ (when $\mu$ has infinite variance and $\alpha =2$ we use the fact that $L(n) \rightarrow \infty$ as $n \rightarrow \infty$).