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Compactness and fractal dimensions of inhomogeneous continuum random trees

Abstract

We introduce a new stick-breaking construction for inhomogeneous continuum random trees. This new construction allows us to prove the necessary and sufficient condition for compactness conjectured by Aldous et al. (Probab Theory Relat Fields 129(2):182–218, 2004) by comparison with Lévy trees. We also compute the fractal dimensions (Minkowski, Packing, Hausdorff).

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Acknowledgements

Thanks are due to Nicolas Broutin for interesting conversations and numerous advice on earlier versions of this paper.

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Correspondence to Arthur Blanc-Renaudie.

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Appendix

Appendix

We first prove an exponential concentration inequality for general Pólya urns.

Lemma A.1

Let \(\{m_n\}_{n\ge 0}\) be a positive real-valued sequence. Let \((A_n)_{n\ge 0}\) be a sequence of positive real-valued random variables such that \(A_0\le m_0\) and such that for every \(n\ge 0\),

$$\begin{aligned} \mathbb {P}\left( \left. A_{n+1}=A_n+m_{n+1} \right| A_n \right) =\frac{A_n}{M_n}; \quad \mathbb {P}\left( \left. A_{n+1}=A_n \right| A_n \right) =\frac{M_n-A_n}{M_n}, \end{aligned}$$

where for every \(n\ge 0\), \(M_n=\sum _{i=0}^n m_n\). We say that in this case \((A_n)_{n\ge 0}\) is a \((A_0,\{m_i\}_{i\ge 0} )\) Pólya urn.

  1. (a)

    If \(\sum _{n=0}^\infty \frac{m_n^2}{M_n^2}<\infty \), then almost surely for every \(a\ge 0\) and \(t\in \mathbb {R}^+\),

    $$\begin{aligned}&\mathbb {P}\left( \left. \sup _{i\ge a} \left| \frac{A_i}{M_i} - \frac{A_a}{M_a} \right|> t\frac{A_a}{M_a} \right| A_a \right) \\&\le 2 \exp \left( -\frac{\frac{t^2}{4}\frac{A_a}{M_a}}{\sum _{n> a} \frac{m_n^2}{M_n^2}+t\max \left( \sum _{n> a} \frac{m_n^2}{M_n^2}, \max _{n>a} \frac{m_n}{M_n} \right) }\right) . \end{aligned}$$
  2. (b)

    If \(\{m_n\}_{n\in \mathbb {N}}\) is bounded, then almost surely for every \(a\ge 0\) and \(t\in \mathbb {R}^+\),

    $$\begin{aligned} \mathbb {P}\left( \left. \sup _{i\ge a} \left| \frac{A_i}{M_i} - \frac{A_a}{M_a} \right|> t\frac{A_a}{M_a} \right| A_a \right) \le 2 \exp \left( -\frac{t^2}{4(1+t)}\frac{A_a}{\max _{n>a} m_n} \right) . \end{aligned}$$

Remark

Note that Lemma A.1 implies that almost surely \(\{\frac{A_i}{M_i}\}_{i\in \mathbb {N}}\) is a Cauchy sequence and so converges. The statement should then be seen as an estimate on the speed of convergence.

Proof

First let us explain why (b) follows from (a). We have for every \(a\in \mathbb {N}\),

$$\begin{aligned} \sum _{n>a}\frac{m_n^2}{M_n^2} \le \max _{n>a} m_n \sum _{n>a}\frac{m_n}{M_n^2} \le \max _{n>a} m_n \int _{M_a}^{+\infty } \frac{dx}{x^2} =\frac{\max _{n>a} m_n}{M_a}, \end{aligned}$$

and (b) follows by replacing \(\max _{n>a} \frac{m_n}{M_n}\) and \(\sum _{n>a}\frac{m_n^2}{M_n^2}\) by the upper bound \(\frac{\max _{n>a} m_n}{M_a}\) in (a).

We focus henceforth on (a). To simplify the notation set for every \(n\in \mathbb {N}\), \(X_n:=\frac{A_n}{M_n}\) and \(\delta _n:=\frac{m_n}{M_n}\). Also we write for every \(a\in \mathbb {N}\), \(\mathbb {E}^{(a)}[\dots ]=\mathbb {E}[\dots \vert X_a]\). We first prove by induction that for every \(a,b,c\in \mathbb {N}\) and \(\lambda \in \mathbb {R}\) satisfying

$$\begin{aligned} a\le b\le c\quad ; \quad \vert \lambda \vert \le \Lambda _a:=\frac{1}{4\max \left\{ \sum _{n> a} \delta _n^2, \max _{n>a} \delta _n \right\} } \end{aligned}$$
(33)

we have

figure a

Note that when \(b=c\), \(P(a,b,c,\lambda )\) is trivial. Therefore it suffices to prove that for every \(a,b,c, \lambda \) such that \(a\le b< c\) and \(\lambda \le \Lambda _a\) that \(f(a,b+1,c,\lambda )\le f(a,b,c,\lambda )\). Fix \(a\le b<c\), \(\lambda \le \Lambda _a\) and let \(\gamma := \lambda \left( 1+\lambda \sum _{n=b+2}^{c}\delta _n^2\right) \). We have,

$$\begin{aligned} f(a,b+1,c,\lambda )&= \mathbb {E}^{(a)}\left[ e^{\gamma X_{b+1}} \right] \nonumber \\&= \mathbb {E}^{(a)}\left[ e^{\gamma X_b}\mathbb {E}^{(b)} \left[ e^{\gamma \left( X_{b+1}-X_b \right) } \right] \right] \nonumber \\&= \mathbb {E}^{(a)}\left[ e^{\gamma X_b} \left( X_b e^{\gamma \left( \frac{A_b+m_{b+1}}{M_{b+1}} -X_b\right) }+\left( 1- X_b\right) e^{ \gamma \left( \frac{A_b}{M_{b+1}}- X_b \right) } \right) \right] \nonumber \\&= \mathbb {E}^{(a)}\left[ e^{\gamma X_b} \left( X_b e^{\gamma \delta _{b+1}\left( 1- X_b\right) }+\left( 1- X_b\right) e^{-\gamma \delta _{b+1} X_b} \right) \right] . \end{aligned}$$
(34)

Furthermore by (33), \(\vert \gamma \vert \le \frac{5}{4}\vert \lambda \vert \le \frac{5}{16\delta _{b+1}}\), hence since for every \(0\le x\le 1\) and \(\vert c\vert \le \frac{5}{16}\), \(xe^{c(1-x)}+(1-x)e^{-cx}\le e^{\frac{16}{25}c^2x}\) , we have

$$\begin{aligned} X_b e^{\gamma \delta _{b+1}\left( 1- X_b\right) }+\left( 1- X_b\right) e^{-\gamma \delta _{b+1} X_b} \le e^{\frac{16}{25}\gamma ^2\delta _{b+1}^2 X_b}. \end{aligned}$$
(35)

Finally by (34), (35), and \(\vert \gamma \vert \le \frac{5}{4}\vert \lambda \vert \),

$$\begin{aligned}f(a,b+1,c,\lambda ) \le \mathbb {E}^{(a)}\left[ e^{\left( \gamma +\frac{16}{25}\gamma ^2\delta _{b+1}^2 \right) X_b} \right] \le \mathbb {E}^{(a)}\left[ e^{\left( \gamma +\lambda ^2\delta _{b+1}^2 \right) X_b} \right] =f(a,b,c,\lambda ) . \end{aligned}$$

This concludes our proof by induction of \(P(a,b,c,\lambda )\).

We now fix \(a\in \mathbb {N}\). For every \(n\in \mathbb {N}\) and \(0\le \lambda \le \Lambda _a\), by \(P(a,a,n,\lambda )\) and \(P(a,a,n,-\lambda )\), we have the sub-Gaussian bound, \(\mathbb {E}^{(a)}\left[ e^{\lambda \vert X_n-X_a\vert } \right] \le 2e^{\lambda ^2V_a/2}\) where \(V_a:=2X_a\sum _{i>a} \delta _i^2\). Furthermore note that \(\{X_n\}_{n\ge a}\) is a martingale, and hence that for every \(\lambda \in \mathbb {R}^+\), \(\{e^{\lambda \vert X_n-X_a\vert }\}_{n\ge a}\) is a sub-martingale. It follows by Doob’s inequality that for every \(t \in \mathbb {R}^+\) and \(0\le \lambda \le \Lambda _a\),

$$\begin{aligned} \mathbb {P}^{(a)} \left( \sup _{n\ge a} \vert X_n-X_a\vert \ge t \right)&= \mathbb {P}^{(a)} \left( \sup _{n\ge a} e^{\lambda \vert X_n-X_a\vert } \ge e^{\lambda t} \right) \nonumber \\&\le \sup _{n\ge a} \frac{\mathbb {E}^{(a)}\left[ e^{\lambda \vert X_n-X_a\vert }\right] }{e^{\lambda t}} \nonumber \\&\le 2 e^{\lambda ^2\frac{V_a}{2} -\lambda t} \end{aligned}$$
(36)

On the one hand, for every \(0\le t\le V_a \Lambda _a\), taking \(\lambda := t/V_a\) in (36) gives,

$$\begin{aligned} \mathbb {P}^{(a)} \left( \sup _{n\ge a} \vert X_n-X_a\vert \ge t \right) \le 2e^{-\frac{t^2}{2V_a}}. \end{aligned}$$
(37)

On the other hand, for every \(t > V_a\Lambda _a\), taking \( \lambda := \Lambda _a\) in (36) gives,

$$\begin{aligned} \mathbb {P}^{(a)} \left( \sup _{n\ge a} \vert X_n-X_a\vert \ge t \right) \le 2e^{\Lambda _a^2\frac{V_a}{2}-\Lambda _at} = 2e^{\frac{t^2}{V_a} \left( \frac{1}{2}\left( \frac{t}{\Lambda _a V_a} \right) ^{-2}- \left( \frac{t}{\Lambda _a V_a} \right) ^{-1}\right) }, \end{aligned}$$

hence since for every \(x\ge 1\), \(\frac{1}{2x^2}-\frac{1}{x}\le -\frac{1}{2+x}\),

$$\begin{aligned} \mathbb {P}^{(a)} \left( \sup _{n\ge a} \vert X_n-X_a\vert \ge t \right) \le 2e^{-\frac{t^2}{2V_a+t/\Lambda _a}}. \end{aligned}$$
(38)

By (37) the last inequality is also true for \(0\le t\le V_a \Lambda _a\). The desired inequality then directly follows from a reorganization of the different terms in (38). We omit the straightforward details. \(\square \)

The following lemma is a version of the strong law of large number.

Lemma A.2

Let \(p>0\), \((X_i)_{i\in \mathbb {N}}\) be a family of independent Bernoulli random variables with mean p. Let \((a_i)_{i\in \mathbb {N}}\) be a positive real-valued sequence and let for every \(n\in \mathbb {N}\), \(A_n:=\sum _{i=1}^n a_i\). Suppose that \(A_n\longrightarrow \infty \) and \(\sum _{n=1}^{\infty } \frac{a_n^2}{ A_n^2}<\infty \), then almost surely

$$\begin{aligned} \sum _{i=1}^{n} a_iX_i\sim pA_n.\end{aligned}$$

Proof

Since \(\sum _{n=1}^{\infty } \frac{a_n^2}{ A_n^2}<\infty \) and \(\{X_i\}_{i\in \mathbb {N}}\) are independent random variables, the classical three series theorem implies that \(S_n:=\sum _{i=1}^n \frac{a_i (X_i-p)}{A_i}\) almost surely converges as \(n\rightarrow \infty \). Therefore,

$$\begin{aligned} \sum _{i=1}^n \frac{a_i (X_i-p)}{A_n}=\sum _{i=1}^n \frac{A_i}{A_n}(S_i-S_{i-1})=S_n-\sum _{i=1}^{n-1}S_i \frac{A_{i+1}-A_{i}}{A_n} \longrightarrow _{n\rightarrow \infty } 0. \end{aligned}$$

\(\square \)

Remark

If there exists \(\varepsilon >0\) such that \(a_i=O(A_i^{1-\varepsilon })\), then as n goes to infinity,

$$\begin{aligned} \sum _{n=1}^{n} \frac{a_i^2}{A_i^2} =O\left( \sum _{i=1}^{n} \frac{a_i}{A_i^{1+\varepsilon }} \right) = O \left( \int _1^{A_n} \frac{dx}{x^{1+\varepsilon }} \right) =O(1).\end{aligned}$$

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Blanc-Renaudie, A. Compactness and fractal dimensions of inhomogeneous continuum random trees. Probab. Theory Relat. Fields 185, 961–991 (2023). https://doi.org/10.1007/s00440-022-01138-9

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Keywords

  • ICRT
  • Inhomogeneous continuum random trees
  • Compactness
  • Fractal dimensions
  • Stick breaking
  • Pólya urn

Mathematics Subject Classification

  • 60D05
  • 60F10