A max-type recursive model: some properties and open questions

TL;DR

This paper investigates a max-type recursive model related to depinning transitions, analyzing its critical behavior including extinction probability and moments, and presents several conjectures on its properties.

Contribution

It introduces a detailed analysis of the critical regime of a max-type recursive model, highlighting new properties and open questions.

Findings

Analysis of extinction probability in the critical regime

Results on the first moment and moment generating function

Identification of open conjectures for future research

Abstract

We consider a simple max-type recursive model which was introduced in the study of depinning transition in presence of strong disorder, by Derrida and Retaux. Our interest is focused on the critical regime, for which we study the extinction probability, the first moment and the moment generating function. Several stronger assertions are stated as conjectures.

Equations

$$X_{n+1}=(X_n^{(1)}+\cdots +X_n^{(m)}-1{)}^{+},n\ge 0,$$

$$m\mathbf{E}(X_n)-1\le \mathbf{E}(X_{n+1})\le m\mathbf{E}(X_n),$$

$$F_{\infty }:=\underset{n\to \infty }{\lim }\downarrow \frac{\mathbf{E}(X_n)}{m^n}=\underset{n\to \infty }{\lim }\uparrow \frac{\mathbf{E}(X_n)-\frac{1}{m-1}}{m^n},$$

$$P_{X_0}=(1-p){\delta }_0+p{P}_{Y_0},$$

$$F_{\infty}(p)=\exp\Big{(}-\frac{K+o(1)}{(p-p_{c})^{1/2}}\Big{)},\qquad p\downarrow p_{c}\,,$$

$$(m-1)\mathbf{E}(X_0{m}^{X_0})=\mathbf{E}(m^{X_0})<\infty ;$$

$$(m-1)\mathbf{E}(X_0{m}^{X_0})=\mathbf{E}(m^{X_0})<\infty .$$

$$\mathbf{P}(X_n\ne 0)\sim \frac{4}{(m-1{)}^2}\frac{1}{n^2},n\to \infty .$$

$$\mathbf{E}(X_n)\sim \frac{4m}{(m-1{)}^3}\frac{1}{n^2},n\to \infty ,$$

$$\mathbf{E}(X_n^r)\sim \frac{c(r)}{n^2},n\to \infty ,$$

$$\mathbf{E}(s^{X_n})-1\sim \frac{4m}{(m-1{)}^2}\frac{s-1}{m-s}\frac{1}{n^2}.$$

$$\mathbf{E}(m^{X_n})-1\sim \frac{2}{m-1}\frac{1}{n}.$$

$$c_1\frac{[\mathbf{P}(X_0=0){]}^{1/2}}{{\Lambda }_0^{1/2}}\frac{1}{n}\le \mathbf{E}(m^{X_n})-1\le c_2\frac{{\Lambda }_0^{1/2}}{[\mathbf{P}(X_0=0){]}^{1/2}}\frac{1}{n},n\ge 1.$$

$$G_n(s):=\mathbf{E}(s^{X_n}),$$

$$G_{n+1}(s)=\frac{1}{s}{G}_n(s{)}^m+(1-\frac{1}{s})G_n(0{)}^m.$$

$$G_{n+1}^{\prime }(s)=\frac{m}{s}{G}_n^{\prime }(s)G_n(s{)}^{m-1}-\frac{1}{s^2}{G}_n(s{)}^m+\frac{1}{s^2}{G}_n(0{)}^m.$$

$$(s-1)s{G}_{n+1}^{\prime }(s)-G_{n+1}(s)=[m(s-1)G_n^{\prime }(s)-G_n(s)]G_n(s{)}^{m-1}.$$

$$(m-1)m{G}_{n+1}^{\prime }(m)-G_{n+1}(m)=[(m-1)m{G}_n^{\prime }(m)-G_n(m)]G_n(m{)}^{m-1}.$$

$$G_{n+1}^{\prime \prime }(s)$$

$$G_{n+1}^{\prime \prime \prime }(s)=\frac{6m}{s^3}{G}_n^{\prime }(s)G_n(s{)}^{m-1}-\frac{3m}{s^2}{G}_n^{\prime \prime }(s)G_n(s{)}^{m-1}$$

$$-\frac{3m(m-1)}{s^2}{G}_n^{\prime }(s{)}^2{G}_n(s{)}^{m-2}-\frac{6}{s^4}{G}_n(s{)}^m+\frac{6}{s^4}{G}_n(0{)}^m+\frac{m}{s}{G}_n^{\prime \prime \prime }(s)G_n(s{)}^{m-1}$$

$$+\frac{3m(m-1)}{s}{G}_n^{\prime }(s)G_n^{\prime \prime }(s)G_n(s{)}^{m-2}+\frac{m(m-1)(m-2)}{s}{G}_n^{\prime }(s{)}^3{G}_n(s{)}^{m-3}.$$

$$(m-1)m{G}_0^{\prime }(m)=G_0(m).$$

$$(m-1)m{G}_n^{\prime }(m)=G_n(m).$$

$$G_{n+1}^{\prime \prime }(m)=G_n^{\prime \prime }(m)G_n(m{)}^{m-1}+\frac{m-2}{m^3(m-1)}{G}_n(m{)}^m-\frac{2}{m^3}{G}_n(0{)}^m,$$

$$G_{n+1}^{\prime\prime}(m)+\frac{2}{m^{2}(m-1)}\,G_{n+1}(m)=\big{[}G_{n}^{\prime\prime}(m)+\frac{1}{m^{2}(m-1)}\,G_{n}(m)\big{]}\,G_{n}(m)^{m-1}\,.$$

$$G_{n+1}^{\prime \prime \prime }(m)=\frac{-2m^2+7m-6}{m^4(m-1{)}^2}{G}_n(m{)}^m+\frac{6}{m^4}{G}_n(0{)}^m+G_n^{\prime \prime \prime }(m)G_n(m{)}^{m-1}.$$

$$m{G}_{n+1}^{\prime \prime \prime }(m)+3G_{n+1}^{\prime \prime }(m)$$

$${\mathcal{D}}_n(m):=m(m-1)G_n^{\prime \prime \prime }(m)+(4m-5)G_n^{\prime \prime }(m)+\frac{2(m-2)}{m^2(m-1)}{G}_n(m)$$

$${\mathcal{D}}_{n+1}(m)={\mathcal{D}}_n(m)G_n(m{)}^{m-1}.$$

$$\prod _{i=0}^n{G}_i(m{)}^{m-1}\le \frac{c_3}{\mathbf{P}(X_0=0)}{n}^2,n\ge 1.$$

Full text

A max-type recursive model: some properties

and open questions

Xinxing Chen, Bernard Derrida, Yueyun Hu, Mikhail Lifshits and Zhan Shi

In honor of Professor Charles M. Newman

on the occasion of his 70th birthday

Summary. We consider a simple max-type recursive model which was introduced in the study of depinning transition in presence of strong disorder, by Derrida and Retaux [6]. Our interest is focused on the critical regime, for which we study the extinction probability, the first moment and the moment generating function. Several stronger assertions are stated as conjectures.

Keywords. Max-type recursive model, critical regime, free energy, survival probability.

2010 Mathematics Subject Classification. 60G50, 82B20, 82B27.

1 Introduction

Fix an integer $m\geq 2$. Let $X_{0}\geq 0$ be a non-negative random variable (but very soon, taking values in ${\mathbb{Z}}_{+}:=\{0,\,1,\,2,\ldots\}$). To avoid trivial discussions, we assume that $X_{0}$ is not almost surely a constant. Consider the following recurrence relation:

[TABLE]

where $X_{n}^{(1)}$, $\ldots$, $X_{n}^{(m)}$ are independent copies of $X_{n}$, and for all $x\in{\mathbb{R}}$, $x^{+}:=\max\{x,\,0\}$ is the positive part of $x$.

The model with recursion defined in (1) was introduced by Derrida and Retaux [6] as a simple hierarchical renormalization model to understand depinning transition of a line in presence of strong disorder. The study of depinning transition has an important literature both in mathematics and in physics. Of particular interest are problems about the relevance of the disorder, and if it is, the precise description of the transition. Similar problems are raised when the line has hierarchical constraints. We refer to Derrida, Hakim and Vannimenus [5], Giacomin, Lacoin and Toninelli [7], Lacoin [10], Berger and Toninelli [3], Derrida and Retaux [6], and Hu and Shi [9] for more details and references. Let us mention that the recursion (1) appears as a special case in the survey paper of Aldous and Bandyopadhyay [2], in a spin glass toy-model in Collet et al. [4], and is also connected to a parking scheme recently studied by Goldschmidt and Przykucki [8].

Since $x-1\leq(x-1)^{+}\leq x$ for all $x\geq 0$, we have, by (1),

[TABLE]

so the free energy

[TABLE]

is well-defined.

In a sense, the free energy is positive for “large” random variables and vanishes for “small” ones. The two regimes are separated by a “surface” (called “critical manifold” in [6]) in the space of distributions that exhibits a critical behavior. Many interesting questions are related to the behavior of $X_{n}$ at the critical regime or near it.

As an example, let us recall a key conjecture, due to Derrida and Retaux [6], which handles the following parametric setting.

For any random variable $X$, we write $P_{X}$ for its law. Assume

[TABLE]

where $\delta_{0}$ is the Dirac measure at [math], $Y_{0}$ is a positive random variable, and $p\in[0,\,1]$ a parameter. Since $F_{\infty}=:F_{\infty}(p)$ is non-decreasing in $p$, there exists $p_{c}\in[0,\,1]$ such that $F_{\infty}>0$ for $p>p_{c}$ and that $F_{\infty}(p)=0$ for $p<p_{c}$. A conjecture of Derrida and Retaux [6] says that if $p_{c}>0$ (and possibly under some additional integrability conditions on $Y_{0}$), then

[TABLE]

for some constant $K\in(0,\,\infty)$.

When $p_{c}=0$, it is possible to have other exponents than $\frac{1}{2}$ in (3), see [9], which also contains several open problems in the regime $p\downarrow p_{c}\,$.

We have not been able to prove or disprove the conjecture. In this paper, we are interested in the critical regime, i.e., $p=p_{c}$ in the Derrida–Retaux conjecture setting. However, we do not formulate the model in a parametric way. When $X_{0}$ is integer valued, the critical regime is characterized by the following theorem.

Theorem 1**.**

(Collet et al. [4]).* If $X_{0}$ takes values in ${\mathbb{Z}}_{+}:=\{0,\,1,\,2,\ldots\}$, the critical regime is given by*

[TABLE]

more precisely, $F_{\infty}>0$ if either ${\bf E}(m^{X_{0}})<(m-1){\bf E}(X_{0}\,m^{X_{0}})<\infty$, or ${\bf E}(X_{0}\,m^{X_{0}})=\infty$ (a fortiori, if ${\bf E}(m^{X_{0}})=\infty$), and $F_{\infty}=0$ otherwise.

We assume from now on that $X_{0}$ is ${\mathbb{Z}}_{+}$-valued, and we work in the critical regime, i.e., assuming

[TABLE]

A natural question is whether ${\bf E}(X_{n})\to 0$ in the critical regime. The answer is positive.

Theorem 2**.**

Assume $\eqref{variete-critique}$. Then $\lim_{n\to\infty}{\bf E}(m^{X_{n}})=1$. A fortiori, $\lim_{n\to\infty}{\bf E}(X_{n})=0$.

It is natural to study the asymptotic behavior of $X_{n}$ quantitatively. Although we have not succeeded in making many of our arguments rigorous, we are led by a general asymptotic picture described by the following two conjectures. The first of them (Conjecture 1) describes how ${\bf P}(X_{n}\not=0)$ tends to [math], while the second one (Conjecture 2) describes the conditional asymptotic behavior of $X_{n}$ provided that $X_{n}\not=0$.

We use the notation $a_{n}\sim b_{n}$, $n\to\infty$, to denote $\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=1$.

Conjecture 1**.**

Assume $\eqref{variete-critique}$. Then we have

[TABLE]

When $m=2$, Conjecture 1 was already given by Collet et al. [4] (see their equation (AIII.10)) and by Derrida and Retaux [6].

Our next conjecture concerns weak convergence of $X_{n}$ given $X_{n}>0$.

Conjecture 2**.**

Assume $\eqref{variete-critique}$. Then, conditionally on $X_{n}\not=0$, the random variable $X_{n}$ converges weakly to a limit $Y_{\infty}$ with geometric law: ${\bf P}(Y_{\infty}=k)=\frac{m-1}{m^{k}}$ for integers $k\geq 1$.

The two conjectures above immediately lead to more specific quantitative assertions. In view of Theorem 2, it is natural to study how ${\bf E}(X_{n})$ goes to zero in the critical regime.

Conjecture 3**.**

Assume $\eqref{variete-critique}$. Then we have

[TABLE]

and more generally, for all real numbers $r\in(0,\,\infty)$,

[TABLE]

where $c(r)=c(r,\,m):=\frac{4}{m-1}\sum_{k=1}^{\infty}\frac{k^{r}}{m^{k}}$.

In view of Conjectures 1 and 2, we may also guess how fast the moment generating function converges.

Conjecture 4**.**

Assume $\eqref{variete-critique}$. Then

a)* We have, for $n\to\infty$ and $s\in(0,\,m),\;s\not=1$,*

[TABLE]

b)* We have, for $n\to\infty$,*

[TABLE]

Possibly some additional integrability conditions, such as ${\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$ in Theorem 3 below, are necessary for our conjectures to hold.

The following weaker version of (6) can be rigorously proved.

Theorem 3**.**

Assume $\eqref{variete-critique}$. If $\Lambda_{0}:={\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$, then there exist constants $c_{2}\geq c_{1}>0$, depending only on $m$, such that

[TABLE]

Remark. (i) The assumption (4) guarantees that ${\bf P}(X_{0}=0)>0$.

(ii) It will be seen (in Lemma 3) that in case $m\geq 3$, we have ${\bf P}(X_{0}=0)\geq\frac{m-2}{m-1}$.∎

The content of the rest of the paper is as follows:

$\bullet$ Section 2: A few key facts about the moment generating functions of $X_{n}$ and their interrelations, with some technical proofs postponed to Section 7;

$\bullet$ Section 3: Proof of Theorem 1;

$\bullet$ Section 4: Proof of Theorem 2;

$\bullet$ Section 5: Heuristics for Conjectures 1 and 4b;

$\bullet$ Section 6: Heuristics for Conjectures 2, 3, and 4a;

$\bullet$ Section 8: Proof of Theorem 3;

$\bullet$ Section 9: Proofs of weaker versions of some of our conjectures.

Throughout the paper, $c_{i}=c_{i}(m)$, for $1\leq i\leq 17$, denote finite and positive constants whose values depend only on $m$.

2 Moment generating functions and their evolution

2.1 Derivatives and evolution

All the techniques used in this article are based on the evaluation of the moment generating functions and on their evolution during the recursive process (1). Write, for $n\geq 0$,

[TABLE]

the moment generating function of $X_{n}$.

In terms of generating functions, the recursion (1) writes as

[TABLE]

Moreover, if $G_{n}^{\prime}(s)$ is well defined, then so is $G_{n+1}^{\prime}(s)$ and differentiation yields

[TABLE]

Eliminating $G_{n}(0)$ from the two identities, it follows that

[TABLE]

Taking $s=m$ yields a formula particularly convenient for iterations, namely,

[TABLE]

Further differentiation of (8) yields (if the involved derivatives are well defined for $G_{0}$)

[TABLE]

and

[TABLE]

Notice that assumption (4) can be rewritten in the language of generating functions as

[TABLE]

It follows immediately from (10) that we have in this case for all $n\geq 0$,

[TABLE]

Assuming ${\bf E}(X_{0}^{2}\,m^{X_{0}})<\infty$ and plugging (13) into (2.1) with $s=m$, we obtain, for all $n\geq 0$,

[TABLE]

which, in combination with (7), yields

[TABLE]

This is still not a perfectly iterative relation because of the difference of the nominators on both sides.

However, let us continue the same operation with the third derivative. Assuming that ${\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$ and plugging (13) into (12) with $s=m$, we obtain for all $n\geq 0$,

[TABLE]

By excluding $G_{n}(0)$ from (16) and (14), it follows that

[TABLE]

Now we may get a completely iterative aggregate

[TABLE]

because combining (15) and (17) yields, for all $n\geq 0$,

[TABLE]

This is another perfectly iterative relation along with (10). Recall that it is proved under $\eqref{variete-critique}$ and assuming that ${\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$.

The approach based on analysis of the moment generating function is already adopted by Collet et al. [4] and Derrida and Retaux [6], where the main attention is focused on the case $m=2$. As seen from (18), the case $m=2$ only involves the first half of ${\mathcal{D}}_{n}(m)$. Our work reveals the importance of the second half; together with the first half, they serve as a useful tool in the study of the moment generating function, as we will see in Section 7.2.

2.2 Products

The main technical properties of generating functions are contained in the following two complementary propositions.

Proposition 1**.**

Assume $\eqref{variete-critique}$. There exists a constant $c_{3}\in(0,\,\infty)$ such that

[TABLE]

Proposition 2**.**

Assume $\eqref{variete-critique}$. If $\Lambda_{0}:={\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$, then there exists a positive constant $c_{4}$, depending only on $m$, such that

[TABLE]

The proofs of these propositions, rather technical, are postponed to Section 7.

3 Proof of Theorem 1

The proof of Theorem 1 essentially follows Collet et al. [4]. It is presented here, not only for the sake of self-containedness, but also for some simplification, which we consider as interesting, in both the upper and the lower bounds.

Assume for a while that ${\bf E}(X_{0}\,m^{X_{0}})<\infty$, which means, in the language of generating functions, that $G_{0}^{\prime}(s)$ is well defined for all $0\leq s\leq m$. Then, as we know from (8), the derivative $G_{n}^{\prime}(s)$ is well defined for all $n\geq 0$ and $0\leq s\leq m$.

Proof of Theorem 1: Upper bound. Suppose $(m-1){\bf E}(X_{0}\,m^{X_{0}})\leq{\bf E}(m^{X_{0}})$ and ${\bf E}(X_{0}\,m^{X_{0}})<\infty$. Let us prove that $F_{\infty}=0$. In the language of generating functions our assumption simply means $(m-1)m\,G_{0}^{\prime}(m)-G_{0}(m)\leq 0$ and $G_{0}^{\prime}(m)<\infty$. Then iterative identity (10) yields that we have the same relations $(m-1)m\,G_{n}^{\prime}(m)-G_{n}(m)\leq 0$ and $G_{n}^{\prime}(m)<\infty$ for all $n\geq 0$. Back to the moments’ language, we obtain ${\bf E}(X_{n}\,m^{X_{n}})<\infty$ and ${\bf E}(m^{X_{n}})\geq(m-1){\bf E}(X_{n}\,m^{X_{n}})$. The latter expression, by the FKG inequality, is greater than or equal to $(m-1){\bf E}(X_{n})\,{\bf E}(m^{X_{n}})$. Therefore, ${\bf E}(X_{n})\leq\frac{1}{m-1}$, for all $n\geq 0$. By definition, we get $F_{\infty}=0$.∎

We now turn to the lower bound for the free energy.

Lemma 1**.**

If ${\bf E}(m^{X_{0}})<(m-1){\bf E}(X_{0}\,m^{X_{0}})<\infty$, then there exists $s\in(1,\,m)$ such that

[TABLE]

Proof. Taking $s\in(1,\,m)$ sufficiently close to $m$ we may assure that

[TABLE]

In the language of generating functions, this means

[TABLE]

By (9),

[TABLE]

where at the last step we used $G_{n}(s)\geq 1$ for all $s>1$. By induction, we obtain

[TABLE]

which is precisely equivalent to the claim of our lemma.∎

Lemma 2**.**

If $F_{\infty}=0$, then $\sup_{n\geq 0}{\bf E}(X_{n}\,s^{X_{n}})<\infty$ for all $s\in(0,\,m)$.

Proof. Let $k\geq 1$ and $n\geq 0$. Clearly,

[TABLE]

where, as before, $X_{n}^{(i)}$, $i\geq 1$, are independent copies of $X_{n}$. It follows that

[TABLE]

Suppose $F_{\infty}=0$. Then by (2), ${\bf E}(X_{n+k})\leq\frac{1}{m-1}$ for all $n\geq 0$ and $k\geq 1$; hence

[TABLE]

for all $n\geq 0$ and $k\geq 1$, implying the assertion of our lemma.∎

Proof of Theorem 1: Lower bound. Assume first ${\bf E}(m^{X_{0}})<(m-1){\bf E}(X_{0}\,m^{X_{0}})<\infty$. Let us prove that $F_{\infty}>0$.

By Lemma 1, there exists $s\in(1,\,m)$ such that

[TABLE]

If $F_{\infty}=0$, then by Lemma 2 we would have $\sup_{n\geq 0}{\bf E}(X_{n}\,s^{X_{n}})<\infty$, contradicting (20).

Consider now the remaining case ${\bf E}(X_{0}\,m^{X_{0}})=\infty$. For any $k\geq 1$ let $\widetilde{X}_{0,k}:=\min\{X_{0},\,k\}$ be the trimmed version of $X_{0}$. By choosing $k$ sufficiently large, one obtains

[TABLE]

The just proved part of our theorem asserts that the free energy associated with $\widetilde{X}_{0,k}$ is positive, and a fortiori $F_{\infty}>0$ in this case.∎

4 Proof of Theorem 2

We prove Theorem 2 in this section by means of Proposition 1. Write as before $G_{n}(s):={\bf E}(s^{X_{n}})$.

Lemma 3**.**

Assume $\eqref{variete-critique}$.

(i)* For any $n\geq 0$, $s\mapsto\frac{G_{n}(s)^{m-1}}{s}$ is non-increasing on $[1,\,m]$. In particular, we have $\frac{G_{n}(s)^{m-1}}{s}\geq\frac{G_{n}(m)^{m-1}}{m}$ for $n\geq 0$ and $s\in[1,\,m]$.*

(ii)* We have $\sup_{n\geq 0}G_{n}(m)\leq m^{1/(m-1)}$.*

(iii)* We have $\inf_{n\geq 0}{\bf P}(X_{n}=0)\geq\frac{m-2}{m-1}$.*

Proof. (i) Since $s\mapsto\frac{s\,G_{n}^{\prime}(s)}{G_{n}(s)}$ is non-decreasing on $[1,\,\infty)$ (this is a general property of moment generating functions, and has nothing to do with assumption (4)), we have, for $s\in[1,\,m]$,

[TABLE]

by (13). This implies that

[TABLE]

for $s\in[1,\,m]$; hence $s\mapsto\frac{G_{n}(s)^{m-1}}{s}$ is non-increasing on $[1,\,m]$.

(ii) By (i), $\frac{G_{n}(m)^{m-1}}{m}\leq G_{n}(1)^{m-1}=1$, so $G_{n}(m)^{m-1}\leq m$.

(iii) From (4), we get

[TABLE]

This implies $(m-1){\bf P}(X_{n}=0)\geq(m-2){\bf E}(m^{X_{n}})\geq m-2$, as desired.∎

Proof of Theorem 2. Assume (4). Let, for $n\geq 0$,

[TABLE]

The proof of Theorem 2 consists of showing that $\varepsilon_{n}\to 0$.

By (7), $G_{n+1}(s)\leq\frac{1}{s}\,G_{n}(s)^{m}+(1-\frac{1}{s})$. In particular, taking $s=m$ gives that

[TABLE]

We will now use the following elementary inequality

[TABLE]

Indeed, the function $h(x):=\tfrac{x}{\log(1+x)}$ satisfies $h^{\prime}(x)\leq\tfrac{1}{2}$ for all $x>0$. Therefore,

[TABLE]

Let now $n>k\geq 0$. Since $\varepsilon_{n-k}\leq\frac{(1+\varepsilon_{n-k-1})^{m}-1}{m}$ (see (21)), we have

[TABLE]

the last inequality being a consequence of (22). Iterating the inequality yields that for all integers $0\leq k\leq n$,

[TABLE]

the second inequality following from the inequality $\frac{1}{\log(1+x)}<\frac{1}{x}+1$ (for $x>0$). As a consequence, for integers $n>j\geq 0$,

[TABLE]

The sum on the left-hand side is $\sum_{i=j+1}^{n}\log(1+\varepsilon_{i})$, whereas the integral on the right-hand side is equal to

[TABLE]

for some constant $c_{5}>0$ whose value depends only on $m$ (recalling from Lemma 3 (ii) that $\varepsilon_{n}\leq m^{1/(m-1)}-1$). This yields that for $n>j\geq 0$,

[TABLE]

Replacing the pair $(n,\,j)$ by $(j,\,0)$, we also have

[TABLE]

for $j\geq 1$. Therefore,

[TABLE]

which is bounded by $\frac{c_{3}^{1/2}}{[{\bf P}(X_{0}=0)]^{1/2}}\,n$ (see Proposition 1). Consequently, with $c_{6}:=\frac{c_{3}^{1/2}}{c_{5}^{2}}$, we have $\varepsilon_{j}\,\varepsilon_{n}\leq\frac{c_{6}}{[{\bf P}(X_{0}=0)]^{1/2}}\,\frac{n}{j(n-j)}$ for $n>j\geq 1$. In particular,

[TABLE]

This yields

[TABLE]

i.e., $\varepsilon_{j}\to 0$.∎

5 Around Conjectures 1 and 4b

Let

[TABLE]

By Theorem 2, $\varepsilon_{n}\to 0$, $n\to\infty$. Propositions 1 and 2 together say that $\prod_{i=0}^{n}(1+\varepsilon_{i})$ is of order of magnitude $n^{2/(m-1)}$ when $n$ is large. So if $n\mapsto\varepsilon_{n}$ were sufficiently regular, we would have

[TABLE]

This is (6) in Conjecture 4b.

From the relation (7) we obtain, with $s=m$,

[TABLE]

Since $G_{n}(0)=1-{\bf P}(X_{n}\not=0)$, whereas $\varepsilon_{n}\to 0$, this implies that

[TABLE]

Let us look back at (25). If we were able to show that $n^{2}(\varepsilon_{n}-\frac{2}{m-1}\,\frac{1}{n})$ admits a finite limit when $n\to\infty$, (26) would give an affirmative answer to Conjecture 1.

6 About Conjectures 2, 3 and 4a

Let us first look at Conjecture 4a. Theorem 2 yields ${\bf P}(X_{n}\not=0)\to 0$. Therefore,

[TABLE]

Using this fact and the identity (7), we obtain

[TABLE]

Suppose we were able to prove that ${\bf P}(X_{n+1}\not=0)\sim{\bf P}(X_{n}\not=0)$ (which would be a consequence of Conjecture 1), and that

[TABLE]

for $s\in(0,m)$, with some real-valued measurable function $H(\,\cdot\,)$. Then

[TABLE]

which would lead to

[TABLE]

for $s\in(0,\,m)$. This, in view of Conjecture 1, is what have led us to Conjecture 4a.

To see why Conjecture 2 would be true, let us note from (27) and $H(s)=\frac{m(s-1)}{m-s}$ that

[TABLE]

On the right-hand side, $\frac{(m-1)s}{m-s}={\bf E}(s^{Y_{\infty}})$ if $Y_{\infty}$ is such that ${\bf P}(Y_{\infty}=k)=\frac{m-1}{m^{k}}$ for integers $k\geq 1$. In words, $Y_{\infty}$ has a geometric distribution with parameter $\frac{1}{m}$. This would give a proof of Conjecture 2.

Conjecture 3 would be an immediate consequence of Conjecture 1 and (28).

We end this section with an elementary argument showing that our prediction for ${\bf E}(X_{n})$ in Conjecture 3 would be a consequence of Conjecture 1 (without using the conjectured convergence in (28)).

Lemma 4**.**

Assume ${\bf E}(X_{0})<\infty$. For all $n\geq 0$, we have ${\bf E}(X_{n})<\infty$, and

[TABLE]

Proof. Taking $s=1$ in (8) gives that for $n\geq 0$, ${\bf E}(X_{n})<\infty$ and

[TABLE]

which implies the lemma by noting that $G_{n}(0)={\bf P}(X_{n}=0)$.∎

If Conjecture 1 were true, then it would follow from Lemma 4 (and Theorem 2 which guarantees ${\bf E}(X_{n})\to 0$) that

[TABLE]

Applying Lemma 5 below to $a_{n}:={\bf E}(X_{n})$, $\lambda:=\frac{1}{m}$ and $b:=2$, this would yield the prediction for ${\bf E}(X_{n})$ in Conjecture 3.

Lemma 5**.**

Let $0<\lambda<1$ and $b>0$. Let $(a_{n},\,n\geq 1)$ be a bounded sequence such that

[TABLE]

for some real $x$. Then

[TABLE]

Proof. Let $p_{n}:=a_{n}-\lambda a_{n+1}$. Then $\lim_{n\to\infty}n^{b}p_{n}=x$ and

[TABLE]

In particular, the sequence $(p_{n})$ is bounded.

By iterating the identity $a_{n}=p_{n}+\lambda a_{n+1}$, we get a series representation

[TABLE]

where the series converges because $(p_{n})$ is bounded. We may apply the dominated convergence theorem to the identity

[TABLE]

because for every $k\geq 0$ we have $n^{b}p_{n+k}\to x$ by assumption, and the dominating summable majorant $(M_{k},k\geq 0)$ is given by $M_{k}:=M\lambda^{k}$. We arrive at

[TABLE]

as required. ∎

7 Proofs of Propositions 1 and 2

7.1 Proof of Proposition 1

Let us define, for $n\geq 0$ and $s\in[0,\,m)$,

[TABLE]

Then by (9), (8) and (2.1),

[TABLE]

Lemma 6**.**

Assume $\eqref{variete-critique}$. Then $\Delta_{n}(s)\in[0,\,1]$ for $n\geq 0$ and $s\in[0,\,m)$.

Proof. By the definition of $\Delta_{n}$ , for $s\in[0,\,m)$,

[TABLE]

which is non-positive. Hence $\Delta_{n}$ is non-increasing on $(0,\,m)$. Since $\Delta_{n}(0)=G_{n}(0)={\bf P}(X_{n}=0)\leq 1$, whereas under assumption (4), it is easily checked that $\lim_{s\to m-}\Delta_{n}(s)=0$, the lemma follows.∎

Lemma 7**.**

Assume $\eqref{variete-critique}$. For all $n\geq 0$,

[TABLE]

Proof. By using (13) and writing $x:=\frac{s}{m}\in[0,\,1)$ for brevity, we have

[TABLE]

[In particular, $G_{n}(0)=\sum_{k=1}^{\infty}m^{k}(km-k-1)\,{\bf P}(X_{n}=k)$.] Furthermore,

[TABLE]

Consequently,

[TABLE]

By the Cauchy–Schwarz inequality, $(\sum_{k=1}^{\infty}a_{k}q_{k})^{2}\leq(\sum_{k=1}^{\infty}q_{k})(\sum_{k=1}^{\infty}a_{k}^{2}q_{k})$, with $a_{k}:=1-x^{k}$ and $q_{k}:=m^{k}(km-k-1){\bf P}(X_{n}=k)$, the proof of the lemma is reduced to showing the following: for $x\in[0,1]$ and $k\geq 1$,

[TABLE]

This can be rewritten as $2k\,x^{k}(1-x)\leq 1-x^{2k}$, which is proved by

[TABLE]

(using that $2x^{k}\leq 2x^{k-1/2}\leq x^{k+\ell}+x^{k-\ell-1}$ for $0\leq\ell<k$).∎

Proof of Proposition 1. By (29) and Lemma 7, for $s\in[0,\,m)$,

[TABLE]

Hence for any $s\in(\frac{m}{2},\,m)$,

[TABLE]

where we used Lemma 3 (i) in the last inequality. Therefore, for $s\in(\frac{m}{2},\,m)$,

[TABLE]

the second inequality being a consequence of Lemma 6. Taking $s=m-\frac{m}{n+2}$ gives that with $c_{7}:=\sup_{n\geq 1}(1+\frac{2}{n})^{n+1}\in(1,\,\infty)$ and all $n\geq 1$,

[TABLE]

By (30) and (31), with $x:=1-\frac{1}{n+2}$,

[TABLE]

On the right-hand side, $(1-x)^{2}=\frac{1}{(n+2)^{2}}$, whereas

[TABLE]

which is ${\bf P}(X_{0}=0)$ because under (4), we have ${\bf E}\{[(m-1)X_{0}-1]m^{X_{0}}\}=0$. Consequently,

[TABLE]

for all $n\geq 1$. This yields the proposition.∎

7.2 Proof of Proposition 2

Now we start to prepare the proof of Proposition 2. In the case $m=2$, it is proved in Appendix III of [4].

Lemma 8**.**

Assume $\eqref{variete-critique}$ and let $n\geq 0$. If $G_{n}^{\prime\prime\prime}(m)<\infty$, then

[TABLE]

where $c_{8}=c_{8}(m)\in(0,\,\infty)$ is a constant whose value depends only on $m$.

Proof. According to Lemma 3 (ii), $G_{n}(m)\leq m^{1/(m-1)}$. By plugging this into (4), we have $G_{n}^{\prime}(m)\leq\frac{m^{1/(m-1)}}{m(m-1)}=:c_{9}$. The function $G_{n}^{\prime}(\,\cdot\,)$ being convex, we have

[TABLE]

By the convexity of $G_{n}^{\prime\prime}(\,\cdot\,)$, this implies that, for $s\in[0,\,m)$,

[TABLE]

The lemma follows by taking $s:=m-\frac{1}{\max\{G_{n}^{\prime\prime\prime}(m)^{1/2},1\}}$.∎

Lemma 9**.**

Assume $\eqref{variete-critique}$ and let $n\geq 0$. If $G_{n}^{\prime\prime\prime}(m)<\infty$, then

[TABLE]

where $c_{10}=c_{10}(m)\in(0,\,\infty)$ is a constant that does not depend on $n$.

Proof. If $m=2$ and ${\bf P}(X_{n}=0)\leq\tfrac{1}{2}$, then $G_{n}(2)-1\geq{\bf P}(X_{n}\geq 1)\geq\tfrac{1}{2}$. If $m=2$ and ${\bf P}(X_{n}=0)>\tfrac{1}{2}$, the equation (AIII.5) of [4] says that $G_{n}^{\prime\prime}(2)(G_{n}(2)-1)>\tfrac{1}{4}\,[{\bf P}(X_{n}=0)]^{2}>\tfrac{1}{16}$. So the lemma (in case $m=2$) follows from Lemma 8.

In the rest of the proof, we assume $m\geq 3$.

Write $\eqref{variete-critique}$ as

[TABLE]

It follows that

[TABLE]

Writing

[TABLE]

it follows from the Cauchy–Schwarz inequality that

[TABLE]

Write $\varepsilon_{n}:=G_{n}(m)-1$ as before. We have $G_{n}(m)\geq 1+(m-1){\bf P}(X_{n}=1)$, i.e.,

[TABLE]

On the other hand, ${\bf P}(X_{n}=0)\geq\frac{m-2}{m-1}$ by Lemma 3 (iii). Therefore,

[TABLE]

which yields

[TABLE]

for some constant $c_{11}=c_{11}(m)\in(0,\,\infty)$ and all $n$. The lemma (in case $m\geq 3$) follows from Lemma 8.∎

Proof of Proposition 2. Write $b_{j}=\prod_{i=0}^{j}G_{i}(m)^{m-1}$. By (19),

[TABLE]

By definition (18), we have ${\mathcal{D}}_{n}(m)\geq m(m-1)\,G_{n}^{\prime\prime\prime}(m)$ and $\frac{{\mathcal{D}}_{0}(m)}{m(m-1)}\leq c_{12}\Lambda_{0}$ for some $c_{12}=c_{12}(m)\in(0,\,\infty)$, where $\Lambda_{0}:={\bf E}(X_{0}^{3}m^{X_{0}})$. Therefore,

[TABLE]

We know that $b_{n-1}\geq 1$, and that $\Lambda_{0}\geq{\bf E}(X_{0}m^{X_{0}})=\frac{1}{m-1}{\bf E}(m^{X_{0}})\geq\frac{1}{m-1}$. So, with $c_{13}:=\max\{c_{12},\,m-1\}$, we have

[TABLE]

On the other hand, by Lemma 9, $\max\{G_{n}^{\prime\prime\prime}(m),\,1\}\geq(\frac{c_{10}}{G_{n}(m)-1})^{2}$. Therefore, with $c_{14}:=\frac{c_{10}}{c_{13}^{1/2}}$, we have $G_{n}(m)\geq 1+\frac{c_{14}}{\Lambda_{0}^{1/2}b_{n-1}^{1/2}}$, i.e.,

[TABLE]

Since $\frac{c_{14}}{\Lambda_{0}^{1/2}b_{n-1}^{1/2}}\leq\frac{c_{14}}{\Lambda_{0}^{1/2}}\leq(m-1)^{1/2}c_{14}$, this yields that for some $c_{15}=c_{15}(m)\in(0,\,\infty)$ depending only on $m$,

[TABLE]

Hence $b_{n}^{1/2}-b_{n-1}^{1/2}\geq\frac{c_{15}}{\Lambda_{0}^{1/2}}$. Thus $b_{n}^{1/2}\geq\frac{c_{15}}{\Lambda_{0}^{1/2}}n$, as desired. ∎

8 Proof of Theorem 3

Assume $\eqref{variete-critique}$. Write as before $G_{n}(m):={\bf E}(m^{X_{n}})$ and $\varepsilon_{n}:=G_{n}(m)-1$. Recall from (23) that for integers $n>j\geq 0$,

[TABLE]

Taking $j:=\lfloor\frac{n}{2}\rfloor$ gives that for $n\geq 1$,

[TABLE]

By Proposition 1,

[TABLE]

whereas by Proposition 2, under the assumption $\Lambda_{0}:={\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$,

[TABLE]

(for some constant $c_{16}=c_{16}(m)>0$ depending only on $m$; noting that the case $n=1$ is trivial because $\Lambda_{0}\geq\frac{1}{m-1}$). Hence

[TABLE]

Consequently, $\varepsilon_{n}\leq c_{17}\frac{\Lambda_{0}^{1/2}}{{\bf P}(X_{0}=0)^{1/2}}\,\frac{1}{n}$ with $c_{17}:=\frac{2c_{3}^{1/2}}{c_{5}c_{16}}$.

To prove the lower bound, we note that under the assumption $\Lambda_{0}:={\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$, we have, by (33), $\varepsilon_{n}\geq\frac{c_{10}}{\max\{G_{n}^{\prime\prime\prime}(m)^{1/2},1\}}$; since

[TABLE]

(see (35)), which is bounded by $c_{12}\Lambda_{0}\,\frac{c_{3}}{{\bf P}(X_{0}=0)}\,n^{2}$ according to Proposition 1, this yields that

[TABLE]

as desired.∎

9 Some additional rigorous results

Let $(b_{i})$ be a numerical sequence. We define its harmonic limit as

[TABLE]

if the latter exists. The $h\hbox{-}\!\limsup$ and $h\hbox{-}\!\liminf$ are defined in the same way.

Proposition 3**.**

Assume $\eqref{variete-critique}$. If $\,{\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$, then

[TABLE]

This statement is a (much) weaker version of Conjecture 4b and a complement to Theorem 3. However, it has the advantage that the limiting constant $\tfrac{2}{m-1}$ appears explicitly in the result.

Proof. We denote as usual $\varepsilon_{i}:=G_{i}(m)-1={\bf E}(m^{X_{i}})-1$. Rewrite the assertions of Propositions 1 and 2 as

[TABLE]

Using $\log(1+x)\leq x\leq\log(1+x)+x^{2}/2$ for all $x>0$ we obtain

[TABLE]

where $S:=\sum_{i=1}^{\infty}\varepsilon_{i}^{2}<\infty$ by Theorem 3. It follows immediately that

[TABLE]

proving the proposition.∎

Proposition 4**.**

Assume $\eqref{variete-critique}$. If $\,{\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$, then

[TABLE]

This result is a weaker version of Conjecture 1, but at least we are able to say something rigorous about the order $i^{-2}$ for ${\bf P}(X_{i}\not=0)$.

Proof. It is more convenient to use in the calculations the quantity

[TABLE]

Recall that (7) with $s=m$ may be written as

[TABLE]

whereas

[TABLE]

or equivalently, as in (26),

[TABLE]

It follows that, for $n\to\infty$,

[TABLE]

By using (36) and Theorem 3, we immediately obtain

[TABLE]

and

[TABLE]

where $c_{1}$ and $c_{2}$ are the constants from Theorem 3.∎

Proposition 5**.**

Assume $\eqref{variete-critique}$. If $\,{\bf E}(X_{0}^{3}\,m^{X_{0}})<\infty$, then

[TABLE]

This result is a weaker version of Conjecture 3, but gives a rigorous statement about the order $i^{-2}$ for ${\bf E}(X_{i})$.

Proof. As in the previous proof, we will use $p_{i}:=m^{-1}(1-G_{i}(0)^{m})$. We already know that

[TABLE]

whereas

[TABLE]

By summing up, we obtain

[TABLE]

Notice that the last term is bounded because by Jensen’s inequality,

[TABLE]

using Theorem 3 at the last step. It follows that

[TABLE]

The proof is completed by an application of (37) and (38).∎

Acknowledgments. X. C. was supported by NSFC grants Nos. 11771286 and 11531001. M. L. was supported by RFBR grant 16-01-00258. Part of the work was carried out when M. L. and Z. S. were visiting, respectively, LPMA Université Pierre et Marie Curie in June and July 2016, and New York University Shanghai in spring 2016; we are grateful to LPMA and NYUSH for their hospitality.