Stochastic recursions: between Kesten's and Grincevi\v{c}ius-Grey's assumptions

TL;DR

This paper analyzes the tail behavior of solutions to stochastic recursions with Lipschitz functions near affine transformations, focusing on cases where the multiplicative component's moments are critical and the additive noise has heavy tails.

Contribution

It characterizes the tail asymptotics of stationary solutions under critical moment conditions and extends second order asymptotics for affine recursions.

Findings

Tail of stationary solution follows a power law with index -α.

Second order asymptotics are derived for affine recursions.

Results connect Kesten's and Grincevičius-Grey's assumptions.

Abstract

We study the stochastic recursion $X_n=\Psi_n(X_{n-1})$, where $(\Psi_n)_{n\geq 1}$ is a sequence of i.i.d. random Lipschitz mappings close to the random affine transformation $x\mapsto Ax+B$. We describe the tail behaviour of the stationary solution $X$ under the assumption that there exists $\alpha>0$ such that $\mathbb{E} |A|^{\alpha}=1$ and the tail of $B$ is regularly varying with index $-\alpha<0$. We also find the second order asymptotics of the tail of $X$ when $\Psi(x)=Ax+B$.

Equations

$$X_n={\Psi }_n(X_{n-1}),n\ge 1$$

$$X\stackrel{d}{=}\Psi (X),X\text{mbox}and\Psi \text{mbox}areindependent,$$

$$Ax+B_1\le \Psi (x)\le Ax+B,\text{mbox}a.s.$$

$$\max \{Ax,B_1\}\le \Psi (x)\le Ax+B,\text{mbox}a.s.$$

$$x^{\alpha }\mathbb{P}(X_{\max }>x)\sim \frac{1}{\rho }{\int }_0^x\frac{L(t)}{t}\mathrm{d}t\text{mbox}asx\to \infty ,$$

$$x^{\alpha }\mathbb{P}(X>x)\sim \frac{1}{\rho }{\int }_0^x\frac{L(t)}{t}\mathrm{d}t$$

$$\tilde{L}(x)={\int }_0^x\frac{L(t)}{t}\mathrm{d}t$$

$$f(x)\sim g(x)\text{mbox}if\underset{x\to \infty }{\lim }f(x)/g(x)=1.$$

$$x^{\alpha }\mathbb{P}(R>x)\sim \frac{1}{\rho }{\int }_0^x\frac{L(t)}{t}\mathrm{d}t$$

$$R_1\stackrel{d}{=}A{R}_1+B_1,R_1\text{mbox}and(A,B_1)\text{mbox}areindependent.$$

$$x^{\alpha }\mathbb{P}(R<-x)\sim \frac{1}{\rho }{\int }_0^x\frac{L_1(t)}{t}\mathrm{d}t.$$

$$\rho =\mathbb{E}{A}^{\alpha }\log A$$

$${\int }_0^x\frac{L(t)}{t}\mathrm{d}t\sim \frac{1}{\alpha }\mathbb{E}{B}_{+}^{\alpha }{1}_{B\le x}.$$

$$\Big{|}x^{\alpha}{\mathbb{P}}(R>x)-\frac{1}{\rho}\int_{0}^{x}\frac{L(t)}{t}\mathrm{d}t-C\Big{|}=O(L(x))+o(1),\quad\mbox{as}\ x\to\infty,$$

$$\mathbf{H}((x,x+h])\le c\max \{h^{\beta },h\}$$

$$\mathbb{P}(R>x)\sim c\mathbb{P}(B>x),$$

$$\mathbb{P}(R>x)\sim c\mathbb{P}(A>x)$$

$$\mathbb{P}(R>x)\sim c{x}^{-\alpha }$$

$$\mathbb{P}(R>x)\sim cl(x)x^{-\alpha }.$$

$$\underset{n\to \infty }{\lim }\frac{1}{n}\log L({\Psi }_{n,1})<0a.s.$$

$${\tilde{L}}_r(x)={\int }_0^x\frac{L_r(t)}{t}\mathrm{d}t\text{mbox}and{\tilde{L}}_{1,r}(x)={\int }_0^x\frac{L_{1,r}(t)}{t}\mathrm{d}t.$$

$$\max \{Ax,B_1\}\le \Psi (x)\le Ax+B,\text{mbox}a.s.,x\ge 0.$$

$$\frac{1-\epsilon }{\rho }{\tilde{L}}_{1,r}(x)\le x^{\alpha }\mathbb{P}(X>x)\le \frac{1+\epsilon }{\rho }{\tilde{L}}_r(x).$$

$$x^{\alpha }\mathbb{P}(X>x)\sim \frac{1}{\rho }{\tilde{L}}_r(x).$$

$$Ax+B_1\le \Psi (x)\le Ax+B,\text{mbox}a.s.,x\in \mathbb{R},$$

$$R_n=A_n{R}_{n-1}+B_n,R_{n,1}=A_n{R}_{n-1,1}+B_{n,1}$$

$$R_{n,1}\le X_n\le R_n\text{mbox}a.s.$$

$$R_1\le X\le R,\text{mbox}a.s$$

$$x^{\alpha }\mathbb{P}(X>x)/{\tilde{L}}_r(x)\le x^{\alpha }\mathbb{P}(R>x)/{\tilde{L}}_r(x).$$

$$\underset{x\to \infty }{\mathrm{limsup}}{x}^{\alpha }\mathbb{P}(X>x)/{\tilde{L}}_r(x)\le \underset{x\to \infty }{\lim }{x}^{\alpha }\mathbb{P}(R>x)/{\tilde{L}}_r(x)=\frac{1}{\rho },$$

Full text

1. Introduction

1.1. Results and motivation

Let $(\Psi_{n})_{n\geq 1}$ be a sequence of i.i.d (independent identically distributed) random Lipschitz real mappings. Given $X_{0}$ independent of $(\Psi_{n})_{n\geq 1}$ we study stochastic recursions

[TABLE]

known also as iterated function systems (IFS). Beginning from the early nineties IFS modeled on Lipschitz functions attracted a lot of attention [2, 3, 7, 15, 16, 18, 23, 32]. Under mild contractivity hypotheses, $X_{n}$ converges in law to a random variable $X$ satisfying (in distribution)

[TABLE]

where $\Psi$ is a generic element of the sequence $(\Psi_{n})_{n\geq 1}$ [15, 18]. However, to describe the tail of $X$ some further assumptions are needed. Usually one assumes that $\Psi$ is close to an affine mapping or, more precisely, that for every $x\in{\mathbb{R}}$

[TABLE]

with $A$, $B$ and $B_{1}$ nice enough. The reason is that if $\Psi(x)=Ax+B$, then the tail of stationary distribution is thoroughly described under various assumptions on $A$ and $B$, see Section 1.3.

In the present paper we consider two kinds of approximations: (3) and the case when $\Psi\colon{\mathbb{R}}\to{\mathbb{R}}$ is a random Lipschitz mapping satisfying for all $x\geq 0$

[TABLE]

Under suitable conditions on $A$, $B$ and $B_{1}$ we obtain asymptotics of ${\mathbb{P}}(X>x)$ as $x\to\infty$ in both cases, see Theorems 1.3 and 1.4.

There is a number of the papers on the subject [2, 8, 13, 17, 32], where the IFS are modeled on the assumptions needed to handle the tail in the affine recursion. Typical conditions exhibit existence of certain moments of $A$ and $B$ or regular behaviour of their tails and in all the settings considered up to now either $A$ or $B$ has basically the ultimate influence on the tail, not both. A short overview is given in Section 1.3.

We study an opposite situation. For the time being, we assume $A\geq 0$ a.s., ${\mathbb{E}}A^{\alpha}=1$ and $B,B_{1}$ have right tails regularly varying with index $-\alpha$ for some $\alpha>0$ such that ${\mathbb{E}}|B|^{\alpha},{\mathbb{E}}|B_{1}|^{\alpha}$ are infinite111If ${\mathbb{P}}(B>x)\sim x^{-\alpha}L(x)$ then ${\mathbb{E}}|B|^{\alpha}$ may be finite or infinite depending on the slowly varying function $L$. Our starting point is the tail behaviour of $X_{\max}$ being the stationary solution to “so called” extremal recursion, corresponding to $\Psi(x)=\max\{Ax,B\}$. Then

[TABLE]

see [14], where $L(x)=x^{\alpha}{\mathbb{P}}(B>x)$ is assumed to be slowly varying function, $\sim$ is defined in (7) and $\rho$ is as in (A-2)222Note that there is no issue with integrability of $L(t)/t$ near $0+$ because $L(t)\leq t^{\alpha}$ for $t>0$.. More precisely, if conditions (A-1), (A-2), (B-1), (AB-1) defined in Theorem 1.1 hold then (5) follows and the right hand side of (5) is due to both the behaviour of $B$ and of an appropriate renewal measure determined by $A$. Moreover, $\int_{0}^{x}L(t)t^{-1}\mathrm{d}t/L(x)$ tends to infinity as $x\to\infty$ and $x\mapsto\int_{0}^{x}L(t)t^{-1}\mathrm{d}t$ is again slowly varying.

The next step is to prove a result in the spirit of (5) for $\Psi(x)=Ax+B$, see Theorem 1.1 below. While the behaviour of the right tails of stationary distribution of the extremal and the affine recursion turn out to be the same, the asymptotics

[TABLE]

of $X$ corresponding to (4) is a straight forward conclusion, Theorem 1.3 in Section 1.4. Neither the affine recursion nor iterated function systems have been considered under these assumptions and the appearance of the function

[TABLE]

is probably the most interesting phenomenon here. For the IFS satisfying (3) we prove that both ${\mathbb{P}}(X>x)$ and ${\mathbb{P}}(X<-x)$ have similar behaviour for large $x$, Theorem 1.4.

1.2. Perpetuities

Before we formulate precisely the results for Lipschitz iterations let us discuss solutions to the affine recursion with $\Psi(x)=Ax+B$. Such solutions, if exist, are called perpetuities and throughout the paper they will be denoted by $R$. It exists and it is unique if ${\mathbb{E}}\log|A|<0$ and ${\mathbb{E}}\log^{+}|B|<\infty$ which is guaranteed by assumptions of Theorem 1.1. For two functions $f,g$ we write

[TABLE]

Recall that $L$ is slowly varying if $L(x)\sim L(\lambda x)$ for any $\lambda>0$. Let $B_{+}=\max\{B,0\}$ and $B_{-}=\max\{-B,0\}$. We have the following theorem

Theorem 1.1**.**

Suppose that

• (A-1)

$A\geq 0$* a.s. and the law of $\log A$ given $A>0$ is non-arithmetic,*

• (A-2)

there exists $\alpha>0$ such that ${\mathbb{E}}A^{\alpha}=1$, $\rho:={\mathbb{E}}A^{\alpha}\log A<\infty$,

• (B-1)

$L(x):=x^{\alpha}{\mathbb{P}}(B>x)$* is slowly varying, ${\mathbb{E}}B_{+}^{\alpha}=\infty$ and ${\mathbb{E}}B_{-}^{\alpha-\varepsilon}<\infty$ for all $\varepsilon\in(0,\alpha)$,*

• (AB-1 )

${\mathbb{E}}A^{\eta}B_{+}^{\alpha-\eta}<\infty$* for some $\eta\in(0,\alpha)\cap(0,1]$.*

Then

[TABLE]

We see that the behaviour of ${\mathbb{P}}(R>x)$ as $x\to\infty$ is described in terms of the behaviour of the tail of $B_{+}$. Accordingly, the behaviour of $P(R<-x)$ depends on the tail of $B_{-}$. To see this, let us denote $B_{1}=-B$. Then, $R_{1}=-R$ satisfies

[TABLE]

and the right tail of $R_{1}$ is the same as the left tail of $R$. We thus obtain the following result.

Corollary 1.2**.**

Assume (A-1) and (A-2) and

• (B-2)

$L_{1}(x):=x^{\alpha}{\mathbb{P}}(B<-x)$* is slowly varying, ${\mathbb{E}}B_{-}^{\alpha}=\infty$ and ${\mathbb{E}}B_{+}^{\alpha-\varepsilon}<\infty$ for all $\varepsilon\in(0,\alpha)$,*

• (AB-2)

${\mathbb{E}}A^{\eta}B_{-}^{\alpha-\eta}<\infty$* for some $\eta\in(0,\alpha)\cap(0,1]$,*

then

[TABLE]

Finally, if all the above assumptions and additionally (B-1), (AB-1) are satisfied, then we have both (8) and (9) with possibly different slowly varying functions $L$ and $L_{1}$.

To obtain tail asymptotics one usually applies an appropriate renewal theorem and so do we. However, what we need goes beyond existing results and we prove a new one, Theorem 3.1. Note that under (A-1) and (A-2)

[TABLE]

is strictly positive. Indeed, consider $f(\beta):={\mathbb{E}}A^{\beta}$. Since $f(0)=1=f(\alpha)$, $f$ is convex, we have $f^{\prime}(\alpha)=\rho>0$. Secondly, observe that, under ${\mathbb{E}}|B|^{\alpha-\eta}<\infty$, (8) depends only on the regular behaviour of the right tail of $B_{+}$ and so we may obtain different asymptotics for ${\mathbb{P}}(R>x)$ and ${\mathbb{P}}(R<-x)$ if it is so for $B$. It follows from (36) that

[TABLE]

and so the right hand sides of (8) and (9) tend to $\infty$ when $x\to\infty$. Finally, conditions (AB-1) and (AB-2) require a comment. If ${\mathbb{E}}A^{\alpha+\varepsilon}<\infty$ for some $\varepsilon>0$ then they both are satisfied by Hölder inequality. But much less is needed. Namely, if ${\mathbb{E}}A^{\alpha}W(A)<\infty$, where $W(x)=\max\{\widetilde{L}(x),\log x\}$ then (AB-1) and (AB-2) hold, see the Appendix.

Next we study the second order asymptotics of the right tail of $R$. Assuming more regularity of $\log A$, we prove that

[TABLE]

for some constant $C$; see Theorem 4.4. Notice that either $L(x)$ or $1$ may dominate the right hand side of (10). (10) holds when the renewal measure determined by $\log A$ satisfies

[TABLE]

for some $\beta\in(0,1)$ and for all $x,h\geq 0$ - see Lemma 2.1. In view of [19] and [11] it is not much of a surprise that stronger assumptions on ${\mathbf{H}}$ are needed to describe the second order asymptotics of the tail of a perpetuity.

Finally, we develop a new approach to deal with signed $A$. We show how to reduce “signed $A$” to “non-negative $A$” (see Theorem 5.4 (i)) and we apply our result to the case when ${\mathbb{E}}|A|^{\alpha}=1$. The method is quite general and it is applicable beyond our particular assumptions.

1.3. Previous results on perpetuities

${\mathbb{P}}(R>x)$ converges to zero when $x$ tends to infinity and a natural problem consists of describing the rate at which this happens. Depending on the assumptions on $(A,B)$ we may obtain light-tailed $R$ (all the moments exist) or a heavy tailed $R$ (certain moments of $|R|$ are infinite). The first case occurs when ${\mathbb{P}}(|A|\leq 1)=1$ and $B$ has the moment generating function in some neighbourhood of the origin, see [33, 20, 24, 30, 12, 31].

The second case is when ${\mathbb{P}}(|A|>1)>0$ with ${\mathbb{E}}\log|A|<0$ and $|A|,|B|$ have some positive moments. Then the tail behaviour of $R$ may be determined by $A$ or $B$ alone, or by both of them. The first case happens when the tail of $B$ is regularly varying with index $-\alpha<0$, ${\mathbb{E}}|A|^{\alpha}<1$ and ${\mathbb{E}}|A|^{\alpha+\varepsilon}<\infty$ for some $\varepsilon>0$. Then

[TABLE]

see [22, 21]. Also it may happen that

[TABLE]

when ${\mathbb{E}}|A|^{\alpha}<1$ but ${\mathbb{P}}(|B|>x)=O({\mathbb{P}}(A>x))$, see [13]. When ${\mathbb{E}}|A|^{\alpha}=1$, ${\mathbb{E}}|B|^{\alpha}<\infty$, ${\mathbb{E}}|A|^{a}\log^{+}|A|<\infty$ and the law of $\log|A|$ given $\{A\neq 0\}$ is non-arithmetic, then [26, 22, 19]

[TABLE]

and it is $A$ that plays the main role. When ${\mathbb{E}}|A|^{a}\log^{+}|A|=\infty$ an extra slowly varying function $l$ appears in (12), i.e.

[TABLE]

(13) was proved by [28] for $A\geq 0$ but applying our approach to signed $A$ (see Section 5) we may conclude (13) also there 333For the results in the case when $\max\{|A|,|B|\}$ does not have positive moments we refer to [17]..

In view of all that it is natural to go a step further and to ask what happens when at the same time $A$ and $B$ contribute significantly to the tail in the sense of (A-2) and (B-1).

1.4. Lipschitz iterations

In this section we state the results for IFS and we show how do they follow from (5) and Theorem 1.1. We assume that $\Psi$ satisfies conditions sufficient for existence of stationary solution. Let $L(\Psi)$, $L(\Psi_{n,1})$ be the Lipschitz constants of $\Psi$, $\Psi_{n,1}=\Psi_{n}\circ\dots\circ\Psi_{1}$ respectively. If ${\mathbb{E}}\log^{+}L(\Psi)<\infty$, ${\mathbb{E}}\log^{+}|\Psi(0)|<\infty$ and

[TABLE]

then $X_{n}$ converges in distribution to a random variable $X$, which does not depend on $X_{0}$ and satisfies (2).

For slowly varying functions $L_{r}$ and $L_{1,r}$ let us denote

[TABLE]

Theorem 1.3**.**

Suppose that (A-1), (A-2), (B-1) and (AB-1) are satisfied both for $B$ and $B_{1}$ with $L=L_{r}$ and $L=L_{1,r}$ respectively. Let $\Psi$ be such that

[TABLE]

Then for every $\varepsilon>0$ and $x$ sufficiently large

[TABLE]

Particularly, if $\widetilde{L}_{r}(x)\sim\widetilde{L}_{1,r}(x)$ then

[TABLE]

Theorem 1.4**.**

If a function $\Psi$ satisfies

[TABLE]

then under the assumptions of Theorem (1.3), assertions (16) and (17) hold true.

If (A-1), (A-2), (B-2) and (AB-2) hold both for $B$ and $B_{1}$ then we have analogous conclusions for ${\mathbb{P}}(X<-x)$.

Theorems 1.3 and 1.4 follow quickly from Theorem 1.1 and (5) (i.e. Theorem 4.2 of [14]). To see this let us consider Theorem 1.4. Let

[TABLE]

with $R_{0}=R_{0,1}=X_{0}$. Then for every $n$,

[TABLE]

and so

[TABLE]

where $R\stackrel{{\scriptstyle d}}{{=}}AR+B$ with $(A,B)$ independent of $R$ and similarly for $(A_{1},B_{1},R_{1})$. Hence

[TABLE]

Letting $x\to\infty$, we obtain

[TABLE]

which implies the right hand side of (16). The left hand side is obtained analogously. Clearly $\widetilde{L}_{r}(x)\sim\widetilde{L}_{1,r}(x)$ implies (17). In the same way we proceed for the proof of Theorem 1.3.

Let us comment on stochastic iterations that fall under the assumption of Theorem (1.3). Subtracting $Ax$ from (15) we arrive at

[TABLE]

where we have defined $\Theta(x):=\Psi(x)-Ax$. Analyzing this condition geometrically, we see that $\Psi(x)=Ax+\Theta(x)$ satisfies (15) if for each $x\geq 0$, the value of $\Theta(x)$ belongs (a.s.) to patterned part of the figure below. If $\Theta(x)\in[B_{1},B]$ a.s. for each $x$, then $\Psi$ satisfies (18). Moreover, $\Theta$ may be chosen in the way that ${\mathbb{E}}\log L(\Theta)<0$ which implies (14).

0$-B_{1}/A$0B_{1}$$B\,$$x

1.5. Structure of the paper

Theorem 1.1 is proved in Section 4.1. Section 4.2. is devoted to the second order asymptotics. Before, we need some preliminaries on the renewal theory. A renewal theorem which is the basic tool is formulated in Section 3 and proved in the last section. Subsection 2.3 contains material needed only for the second order asymptotics. We deal with general $A$ in Section 5.

2. Preliminaries

2.1. Regular variation

A measurable function $L\colon(0,\infty)\to(0,\infty)$ is called slowly varying, (denoted $L\in R(0)$), if for all $\lambda>0$,

[TABLE]

For $\rho\in{\mathbb{R}}$ we write $R(\rho)$ for the class of regularly varying functions with index $\rho$, which consists of functions $f$ of the form $f(x)=x^{\rho}L(x)$ for some $L\in R(0)$.

If $L\in R(0)$ is bounded away from [math] and $\infty$ on every compact subset of $[0,\infty)$, then for any $\delta>0$ there exists $A=A(\delta)>1$ such that (Potter’s Theorem, see e.g [10], Appendix B)

[TABLE]

Assume that $L\in R(0)$ is locally bounded on $(x_{0},\infty)$ for some $x_{0}>0$. Then, for $\alpha>0$ one has

[TABLE]

[5, Proposition 1.5.9a]. Define $\widetilde{L}_{x_{0}}(x):=\int_{x_{0}}^{x}t^{-1}L(t)\mathrm{d}t$. Function $\widetilde{L}_{x_{0}}$ is sometimes called de Haan function. It is again slowly varying and has the property that for any $\lambda>0$,

[TABLE]

To prove it, use the fact that convergence in (19) is locally uniform [5, Theorem 1.5.2].

2.2. Renewal theory

Let $(Z_{k})_{k\geq 1}$ be the sequence of independent copies of random variable $Z$ with ${\mathbb{E}}Z>0$. We write $S_{n}=Z_{1}+\ldots+Z_{n}$ for $n\in\mathbb{N}$ and $S_{0}=0$. The measure ${\mathbf{H}}$ defined on Borel sets $\mathcal{B}({\mathbb{R}})$ by

[TABLE]

is called the renewal measure of $(S_{n})_{n\geq 1}$, $H(x):={\mathbf{H}}((-\infty,x])$ is called the renewal function. If ${\mathbb{E}}Z>0$, then $H(x)$ is finite for all $x\in{\mathbb{R}}$ if and only if ${\mathbb{E}}Z_{-}^{2}<\infty$ ([27]).

We say that the distribution of $Z$ is arithmetic if its support is contained in $d{\mathbb{Z}}$ for some $d>0$; otherwise it is non-arithmetic. Equivalently, the distribution of $Z$ is arithmetic if and only if there exists $0\neq t\in{\mathbb{R}}$ such that $f_{Z}(t)=1$, where $f_{Z}$ is the characteristic function of the distribution of $Z$. The law of $Z$ is strongly non-lattice if the Cramer’s condition is satisfied, that is, $\limsup_{|t|\to\infty}|f_{Z}(t)|<1$.

A fundamental result of renewal theory is the Blackwell theorem (see [6]): if the distribution of $Z$ is non-arithmetic, then for any $h>0$,

[TABLE]

Note that in the non-arithmetic case, since ${\mathbf{H}}((x,x+h])$ is convergent as $x\to\infty$ we have $C=\sup_{x}{\mathbf{H}}((x,x+1])<\infty$ and so

[TABLE]

for some positive $\alpha$, $\beta$ and any $h>0$.

Under additional assumptions we know more about the asymptotic behaviour of ${\mathbf{H}}$ and $H$ (see [34]). If for some $r>0$ one has ${\mathbb{P}}(Z\leq x)=o(e^{rx})$ as $x\to-\infty$, then there is some $r_{1}>0$ such that

[TABLE]

More accurate asymptotics of $H(x)$ as $x\to-\infty$ is given in [29]. If $Z$ has finite second moment, for some $r>0$, ${\mathbb{P}}(Z>x)=o(e^{-rx})$ as $x\to\infty$ and the distribution of $Z$ is strongly non-lattice, then there is $r_{1}>0$ such that (see [34])

[TABLE]

2.3. Renewal measure with extra regularity

For the second order asymptotics we need a better control of ${\mathbf{H}}\left((x,x+h]\right)$ in terms of $h$ than (24); something in the spirit of

[TABLE]

for some $\beta>0$. Observe that with $C_{n}=\sup_{x}{\mathbf{H}}((x,x+1/n])<\infty$ we have

[TABLE]

thus (27) holds for all $x$ and $h>1/n$ with $\beta=1$. Hence, we have to investigate the case of small $h$ only. We have the following statement.

Lemma 2.1**.**

Assume that ${\mathbb{P}}(Z>x)=o(e^{-rx})$ as $x\to\infty$ for some $r>0$, ${\mathbb{E}}Z_{-}^{2}<\infty$ and that the law of $Z$ is strongly non-lattice. If there exists $\beta>0$ such that

[TABLE]

then there exists $\tilde{\beta}>0$ and $c>0$ such that for $x\geq 0$ and $h\geq 0$,

[TABLE]

Remark 2.2**.**

Notice that (28) is satisfied when the law of $Z$ has density in $L^{p}$ for some $1<p\leq\infty$.

Before we write the proof let us describe a certain factorization of ${\mathbf{H}}$ that will be used in it. In renewal theory it is usually easier to consider first a non-negative $Z$, and then to extend some argument to arbitrary $Z$ using the following approach (see e.g. proof of Lemma 2.1). Let $N=\inf\{n\in{\mathbb{N}}\colon S_{n}>0\}$ be the first ladder epoch of $(S_{n})_{n\geq 1}$. We define a measure by

[TABLE]

The support of ${\mathbf{V}}$ is contained in $(-\infty,0]$ and ${\mathbf{V}}({\mathbb{R}})={\mathbb{E}}N$. Since $(S_{n})_{n\geq 1}$ has a positive drift, ${\mathbb{E}}N$ is finite. Let $Z_{1}^{>}\stackrel{{\scriptstyle d}}{{=}}S_{N}$ be the first ladder height of $(S_{n})_{n\geq 1}$ and consider an i.i.d. sequence $(Z_{n}^{>})_{n\geq 1}$. Then

[TABLE]

where ${\mathbf{H}}^{>}$ is the renewal measure of $(S_{n}^{>})_{n\geq 1}$ and $S_{n}^{>}=\sum_{k=1}^{n}Z_{k}^{>}$ ([6, Theorem 2], see also [1, Lemma 2.63] for more general formulation).

Proof of Lemma 2.1.

We will first consider the case when $Z\geq 0$ a.s. Let $F$ be the cumulative distribution function of $Z$. From condition (28) we infer that there exists $\beta,c,\varepsilon>0$ such that for any $a\geq 0$ and any $h\in(0,\varepsilon]$ one has $F(a+h)-F(a)={\mathbb{P}}(a<Z\leq a+h)\leq ch^{\beta}$. Decreasing $\varepsilon$, if needed, we can and do assume that $F(\varepsilon)<1$. Since $H(x)=\mathbf{1}_{x\geq 0}+H\ast F(x)$ we have for any $x\geq 0$ and $h\in(0,\varepsilon]$,

[TABLE]

and thus

[TABLE]

Let now $Z$ be arbitrary and let $S_{N}$ be the first ladder height of $(S_{n})_{n\geq 1}$. Since ${\mathbb{E}}N<\infty$ and, for $a\geq 0$ and small enough $h$,

[TABLE]

by (28) and it follows that

[TABLE]

Thus, using factorization ${\mathbf{H}}={\mathbf{V}}\ast{\mathbf{H}}^{>}$ we obtain for $x\geq 0$ and $h\in(0,\varepsilon]$,

[TABLE]

For $0\leq x\leq h^{-\delta}$ with $\delta<\beta$ this implies that

[TABLE]

On the other hand, for $x>h^{-\delta}$ and $r>0$ we have

[TABLE]

where we have used the fact that $x\exp(-x)<1$ for $x>0$. The conclusion follows by (26), since then

[TABLE]

∎

3. Renewal Theorem

A function $f\colon{\mathbb{R}}\to{\mathbb{R}}_{+}$ is called directly Riemann integrable on ${\mathbb{R}}$ (dRi) if for any $h>0$,

[TABLE]

and

[TABLE]

If $f$ is locally bounded and a.e. continuous on ${\mathbb{R}}$, then an elementary calculation shows that (30) with $h=1$ implies direct Riemann integrability of $f$. For directly Riemann integrable function $f$, we have the following Key Renewal Theorem ([4]):

[TABLE]

There are many variants of this theorem, when $f$ is not necessarily $L_{1}$ - see [25, Section 6.2.3]. Such results are usually obtained under the additional requirement that $f$ is (ultimately) monotone or $f$ is asymptotically equivalent to a monotone function.

Neither of them is sufficient for us. To prove Theorem 1.1 we need to integrate the function $e^{\alpha x}{\mathbb{E}}g(e^{-x}B)$ with respect to ${\mathbf{H}}$, where $g$ is $C^{1}$ function “approximating” $\mathbf{1}_{(1,\infty)}$. Therefore, we prove the following result.

Theorem 3.1**.**

Assume that $0<{\mathbb{E}}Z<\infty$, the law of $Z$ is non-arithmetic and ${\mathbb{P}}(Z\leq x)=o(e^{rx})$ as $x\to-\infty$. Assume further that there is a random variable $B$ and a slowly varying function $L$ such that ${\mathbb{P}}(B>x)=x^{-\alpha}L(x)$. Let $g$ be a bounded function, $\mathrm{supp}\,g\subset[1,\infty)$ and there exists a constant $c$ such that

[TABLE]

Then

[TABLE]

Assume additionally that ${\mathbb{E}}\exp(\varepsilon Z)<\infty$ for some $\varepsilon>0$ and that the law of $Z$ is strongly non-lattice. Then as $x\to\infty$,

[TABLE]

where $C$ depends on $\|g\|_{sup}\sup_{x\in{\mathbb{R}}}|g(x)|$ and the constant $c$ in (31). $C\to\infty$ if either $\|g\|_{sup}\to\infty$ or $c\to\infty$.

To obtain asymptotics of ${\mathbb{P}}(R>e^{x})$ one may integrate $e^{\alpha x}{\mathbb{P}}(B>e^{x})$ with respect to ${\mathbf{H}}$ and control other components as it is explained in the proof of Theorem 4.1. However, using $C^{1}$ functions $g$ instead of $\mathbf{1}_{(1,\infty)}(x)$ allows us to avoid many technical obstacles, without requiring stronger regularity of ${\mathbf{H}}$. Basically we need $g$ as defined in (38) i.e. approximating $\mathbf{1}_{(1,\infty)}(x)$. Observe that when $g(x)$ is replaced by $\mathbf{1}_{(1,\infty)}(x)$ we obtain $e^{\alpha(x-z)}{\mathbb{E}}\mathbf{1}_{(1,\infty)}(e^{z-x}B)=L(e^{x-z})$ and so Theorem 3.1 is in analogy to Theorems 3.1 and 3.3 in [14] which say that

[TABLE]

or with more regularity on $Z$,

[TABLE]

The proof of Theorem 3.1 is postponed to the last section.

4. Perpetuities

4.1. First order asymptotics

In this section we prove Theorem 1.1. The assumptions are the same as in [14, Theorem 4.2 (i)], where the extremal recursion was considered. The proof, however, is not that simple. Therefore, we use a different approach, introduced in [9]. Instead of proving directly the asymptotics of ${\mathbb{P}}(R>x)$ we look for the asymptotics of ${\mathbb{E}}g(R/x)$, where $g$ is a $C^{1}$ function and $\mathrm{supp}\,g\subset[1,\infty)$. The advantage of such approach is that certain function is easily shown to be dRi (see Proposition 4.3). Moreover, the asymptotics of ${\mathbb{P}}(R>x)$ follows straightforward from the asymptotics of ${\mathbb{E}}g(R/x)$ and the whole proof is quite simple.

Theorem 1.1 is an immediate consequence of (34) below.

Theorem 4.1**.**

Suppose that conditions (A-1), (A-2), (B-1), (AB-1) are satisfied. Let $g$ be a bounded function supported in $[1,\infty)$. Suppose that (31) holds. Then

[TABLE]

Moreover, as $x\to\infty$

[TABLE]

where $C$ depends on $\|g\|_{sup}$ and $c$ in (31).

Using Hölder continuous or $C^{1}$ functions approximating indicators instead of indicators themselves is a standard procedure which usually allows to reduce regularity requirement for the probability distribution in question. By regularity we mean here assumptions similar to (42) or even existence of density. They seem to be needed, if indicators are used, but with Hölder continuous functions one can handle calculations differently. In various problems this approach is very successful.

Although we use regularity of functions in intermediate steps, what we obtain at the end allows us to take the limit and to eliminate the dependence on Hölder constants or derivatives, see e.g Sections 3.1, 3.2 or Appendix D in [10]. This can be done in (34) because the right hand side depends only on the integral of $g$. However this is not the case in (35) because $C\to\infty$ if $\|g^{\prime}\|_{sup}\to\infty$ which takes place when indicators are approximated by $C^{1}$ functions. Therefore, for the second order asymptotics we have to proceed differently. The problem is treated in the next Section.

Finally, $\alpha\widetilde{L}(x)$ in (34) and (35) may be replaced by ${\mathbb{E}}B_{+}^{\alpha}\mathbf{1}_{B\leq x}$. As an easy consequence of (21) we obtain

Proposition 4.2**.**

Assume that the first condition in (B-1) holds. Then, we have

[TABLE]

and for $r>0$,

[TABLE]

Assuming additionally that the second condition in (B-1) holds we have $\widetilde{L}(x)\uparrow\infty$ as $x\to\infty$.

Proof of Theorem 1.1.

It is enough to prove that for a $\xi>1$

[TABLE]

Let $\xi>1$ and $\eta>0$ be such that $\xi-\eta>1$. Let $g_{1}$ be a $C^{1}$ function such that $0\leq g_{1}\leq 1$ and

[TABLE]

Let $g_{2}(x)=g_{1}(x-\eta)$.

Then $g_{1},g_{2}$ satisfy the assumptions of Theorem 4.1, because $g^{\prime}_{1}(x)=g^{\prime}_{2}(x)=0$ for $x\leq\xi-\eta$ and $x\geq\xi+\eta$. We have

[TABLE]

Moreover,

[TABLE]

Letting $\eta\to 0$ we infer that

[TABLE]

Finally,

[TABLE]

Hence the conclusion follows. ∎

Proof of Theorem 4.1.

The proof presented here follows very closely the proof of Theorem 4.2 in [14]. Let us denote

[TABLE]

and

[TABLE]

where $A$ and $R$ are independent, $\mathrm{supp}\,g\subset[1,\infty)$. Let us define the distribution of $Z$ by

[TABLE]

Then, we have for any $x>0$,

[TABLE]

Iterating the above equation (see page 8 in [14]), one arrives at

[TABLE]

where ${\mathbf{H}}$ is the renewal measure of $(S_{n})_{n\geq 1}$ and $S_{n}=Z_{1}+\ldots+Z_{n}$, where $(Z_{i})_{i}$ are independent copies of $Z$ and $S_{0}=0$. Let us define

[TABLE]

Let us note that (31) is equivalent to condition $|xg^{\prime}(x)-\alpha g(x)|\leq c$ for $x\geq 1$. Thus, (31) along with boundedness of $g$ imply that $g^{\prime}$ is also bounded and so $g$ is Lipschitz continuous.

By Proposition 4.3, $\psi_{0}$ is directly Riemann integrable and so

[TABLE]

The main contribution to the asymptotics of $f$ comes from $\int_{{\mathbb{R}}}\psi_{B}(x-z){\mathbf{H}}(\mathrm{d}z)$. Observe that

[TABLE]

and that since the law of $Z$ have the same supports as $\log A$ given $A>0$, it is also non-arithmetic. Moreover

[TABLE]

as $x\to\infty$. By Theorem 3.1 we obtain the assertion. ∎

In the next proposition we do not need to assume that $A\geq 0$ with probability $1$ nor that the law of $R$ is the solution of the equation $R\stackrel{{\scriptstyle d}}{{=}}AR+B$. We require only that the moments of $|R|$ of order strictly smaller then $\alpha$ are finite, which is satisfied in our framework; see [10, Lemma 2.3.1]. For $0<\varepsilon\leq 1$, we define $H^{\varepsilon}$ to be the set of bounded functions $g$ satisfying

[TABLE]

Clearly, due to boundedness of $g$, $H^{\varepsilon_{1}}\subset H^{\varepsilon}$ if $\varepsilon_{1}\leq\varepsilon$.

Proposition 4.3**.**

Suppose that $A,B,R$ are real valued random variables and $(A,B)$ is independent of $R$. Fix $0<\varepsilon<\alpha$, $\varepsilon\leq 1$ and assume further that ${\mathbb{E}}|A|^{\alpha}<\infty$, ${\mathbb{E}}\left[|A|^{\varepsilon}|B|^{\alpha-\varepsilon}\right]<\infty$, ${\mathbb{E}}|R|^{\beta}<\infty$ for every $\beta<\alpha$. Then for every $g\in H^{\varepsilon}$ such that $0\notin\mathrm{supp}\ g$ the function

[TABLE]

is directly Riemann integrable.

Proof.

Since $\psi_{0}$ is continuous it is enough to prove that

[TABLE]

For $x,y\in{\mathbb{R}}$ we have

[TABLE]

Interchanging the roles of $x$ and $y$, we arrive at

[TABLE]

where $\mathrm{supp}\,g\subset\{x:|x|>\eta\}$. Thus, for any $x\in{\mathbb{R}}$

[TABLE]

Since $\alpha-\varepsilon>0$, we have

[TABLE]

and

[TABLE]

where

[TABLE]

Hence, there is a constant $C=C(\eta,\alpha,\varepsilon,\|g\|_{\varepsilon})$ such that

[TABLE]

Let us first consider the case when $\alpha-2\varepsilon>0$. We have

[TABLE]

Since $R$ and $(A,B)$ are independent, the first term above is finite by assumption. For the second term, we have

[TABLE]

where we have used $|A|^{\alpha-2\varepsilon}\mathbf{1}_{|B|>|A|}\leq|B|^{\alpha-2\varepsilon}$. On the other hand, if $\alpha-2\varepsilon\leq 0$, then we have $0<\alpha-\varepsilon\leq\varepsilon$ and the right hand side of (41) up to a multiplicative constant $C$ is equal to

[TABLE]

It is clear that both terms are finite; for the second use $|B|^{2\varepsilon-\alpha}\mathbf{1}_{|B|\leq|AR|}\leq|AR|^{2\varepsilon-\alpha}$. ∎

4.2. Perpetuity - second order asymptotics

In this section we study the second order asymptotics i.e the size of

[TABLE]

when $x\to\infty$. For that we need more stringent assumptions on the distribution of $A$. Recall that $\widetilde{L}(x)/L(x)\to\infty$ as $x\to\infty$.

Theorem 4.4**.**

Assume (A-1), (A-2), (B-1). Suppose further that there exists $\beta>0$ such that

[TABLE]

and ${\mathbb{E}}A^{\gamma}<\infty$ for some $\gamma>\alpha+\alpha^{2}/\beta$. If the distribution of $Z$ defined by (39) is strongly non-lattice, then as $x\to\infty$,

[TABLE]

where $a_{+}=\max\{a,0\}$.

Remark 4.5**.**

Depending on $L$ either the constant or $O(L(x))$ may dominate in (43). If $L(x)$ is asymptotically bounded away from zero, then (43) says that

[TABLE]

when $x\to\infty$.

If $L(x)\to 0$ then (43) is more precise and it implies

[TABLE]

Remark 4.6**.**

In Theorem 4.4 it is required that the law of $Z$ is strongly non-lattice, but it is somehow desirable to have a sufficient condition in terms of the distribution of $\log A$. It is enough to assume that the law of $\log A$ is spread-out, i.e. for some $n$ its $n$-th convolution has a non-zero absolutely continuous part for some $n\in\mathbb{N}$. If the law of $\log A$ is spread-out then the law of $Z$ is spread-out as well. This in turn implies that the distribution of $Z$ is strongly non-lattice.

We begin with the following technical Lemma.

Lemma 4.7**.**

Under assumptions of Theorem 4.4, both functions

[TABLE]

and

[TABLE]

are $O(L(e^{x}))$ as $x\to\infty$.

Proof.

By assumption we have $\gamma>\alpha+\alpha^{2}/\beta$. Take $\delta$ such that

[TABLE]

and $\gamma^{\prime}\in(\alpha+\alpha^{2}/\beta,\gamma)$ such that

[TABLE]

Then, we have

[TABLE]

It is clear that $K_{1}=O(L(e^{x}))$. Furthermore, taking $\eta$ such that

[TABLE]

we obtain

[TABLE]

for some $s>0$. Indeed, ${\mathbb{E}}|ARB|^{\alpha-\eta}=({\mathbb{E}}|AB|^{\alpha-\eta})({\mathbb{E}}|R|^{\alpha-\eta})$ and applying Hölder inequality with $p=\gamma/(\alpha-\eta)$ and $q=\gamma/(\gamma-\alpha+\eta)$ we obtain

[TABLE]

and $(\alpha-\eta)q<\alpha$ in view of (46). Moreover, since $1-\gamma/\gamma^{\prime}<0$ we have

[TABLE]

for some $s>0$ and so $K_{2}$ and $K_{3}$ are $O(L(e^{x}))$ as well. For $K_{4}$ define $\lambda(x)=1-e^{-(1-\delta)x}\to 1$ and recall that $\alpha/\gamma^{\prime}<1$. Then, by (42),

[TABLE]

which is $O(e^{-sx})$ for some $s>0$ in view of (45).

We proceed similarly for $I_{2}$ writing

[TABLE]

Then one can show that there exists $\delta>0$ small enough to ensure that $I_{2}(x)=O(L(e^{x}))$.

∎

Proof of Theorem 4.4.

We begin the proof in the same way as in the proof of Theorem 4.1 (see also proof of [14, Theorem 4.2]) but with $f(x)=e^{\alpha x}{\mathbb{P}}(R>e^{x})$, $\psi(x)=e^{\alpha x}\left({\mathbb{P}}(AR+B>e^{x})-{\mathbb{P}}(AR>e^{x})\right)$, $\psi_{B}(x)=e^{\alpha x}{\mathbb{P}}(B>e^{x})$. Then

[TABLE]

where

[TABLE]

In view of Theorem 3.3 in [14] we know that

[TABLE]

Hence it remains to show that

[TABLE]

as $x\to\infty$. Let us denote

[TABLE]

so that

[TABLE]

In the proof of Theorem 4.2 in [14] we have already shown (under weaker assumptions) that

[TABLE]

and that ${\mathbb{E}}\min\{AR,B\}_{+}^{\alpha}<\infty$. By the preceding Lemma we know that $I_{i}(x)=O(L(e^{x}))$ for $i=1,2$ and this implies that as $x\to\infty$,

[TABLE]

Indeed, consider $\int_{(-\infty,0]}\frac{L(e^{x-z})}{L(e^{x})}{\mathbf{H}}(\mathrm{d}z)$. For any $\delta>0$, the integrand is bounded by $ce^{-\delta z}$ for some $c>1$ by Potter’s bound (20). Combining this with (25) and Lebesgue’s Dominated Convergence Theorem we conclude that

[TABLE]

Observe that there exists $\beta^{\ast}>0$ such that

[TABLE]

Indeed, let $p=\frac{\alpha+\varepsilon}{\alpha}$, $q=\frac{\alpha+\varepsilon}{\varepsilon}$ with $\alpha+\varepsilon\leq\gamma$. Then

[TABLE]

Hence

[TABLE]

and (49) follows by (42). In view of (28) we have the following easy result for $x>u$ and $d>u$,

[TABLE]

for some $>0$, where, the first inequality follows from monotonicity of the integrand and the second one by Lemma 2.1.

Moreover, notice that for $0<\lambda\leq 1$ and all $x>0$ one has $\log(1+x)\leq\lambda^{-1}x^{\lambda}$. Let us note that on the event $\{AR+B>\max\{AR,B\}\}$, both $AR$ and $B$ are positive and, on the space restricted to this event, random variables $U=\log\max\{AR,B\}$ and $D=\log(AR+B)$ are well defined. Then, by (50)

[TABLE]

For the first term above we have

[TABLE]

provided $\lambda<\alpha$. An analogous calculation shows that ${\mathbb{E}}(AR+B)^{\alpha}(D-U)\mathbf{1}_{D>U}<\infty$ and so $\int_{(x-D,x-U]\cap(0,\infty)}e^{\alpha(x-z)}{\mathbf{H}}(\mathrm{d}z)\mathbf{1}_{D>U}$ is dominated by an integrable random variable which does not depend on $x$. Thus, by Lebesgue’s Dominated Convergence Theorem we have

[TABLE]

and for $d>u$ as $x\to\infty$,

[TABLE]

where we have used the Key Renewal Theorem since $x\mapsto e^{\alpha x}\mathbf{1}_{[0,d-u)}$ is dRi (it has compact support, is bounded and a.e. continuous). Thus

[TABLE]

We proceed similarly with $I_{2}$. With $D=\log\max\{AR,B\}$ and $U=\log(AR+B)$ (analogously as before, $D$ and $U$ are well defined on the events $\{\max\{AR,B\}>0\}$ and $\{AR+B>0\}$, respectively), we have

[TABLE]

and by (50)

[TABLE]

Again, as before we do calculations for the term with . It is bounded by

[TABLE]

as before. The second term equals

[TABLE]

Now, since $\min\{AR,B\}\leq 0$ and

[TABLE]

we have

[TABLE]

and

[TABLE]

Similarly as before, Lebesgue’s Dominated Convergence Theorem implies that as $x\to\infty$,

[TABLE]

and so as $x\to\infty$, after straightforward simplification,

[TABLE]

∎

5. Perpetuities with general $A$

Now we are going to consider perpetuities with $A$ attaining negative values as well. More precisely, we assume that ${\mathbb{P}}(A<0)>0$, possibly with ${\mathbb{P}}(A\leq 0)=1$. Our aim is to reduce the general case to the one already solved: non-negative $A$. We propose a unified approach to perpetuities, which applies beyond our particular assumptions.

Assume that ${\mathbb{E}}\log|A|<0$ and ${\mathbb{E}}\log^{+}|B|<\infty$. Then the stochastic equation $R\stackrel{{\scriptstyle d}}{{=}}AR+B$ with $(A,B)$ and $R$ independent has a unique solution, or equivalently, that $R_{n}=A_{n}R_{n-1}+B_{n}$, $n\geq 1$, converges in distribution to $R$ for any $R_{0}$ independent of $(A_{n},B_{n})_{n\geq 1}$, where $(A_{n},B_{n})_{n\geq 1}$ is a sequence of independent copies of the pair $(A,B)$.

Define the filtration $\mathbb{F}=\{\mathcal{F}_{n}\colon n\geq 1\}$, where $\mathcal{F}_{n}=\sigma((A_{k},B_{k})_{k=1}^{n})$. Following [35, Lemma 1.2], for any stopping time $N$ (with respect to $\mathbb{F}$) which is finite with probability one, $R$ satisfies

[TABLE]

where $R_{n}^{\ast}=B_{1}+A_{1}B_{2}+\ldots+A_{1}\ldots A_{n-1}B_{n}$ for $n\geq 1$. For $n\geq 1$ we write $\Pi_{n}=A_{1}\cdot\ldots\cdot A_{n}$ and $\Pi_{0}=1$. Let $N:=\inf\{n\in\mathbb{N}\colon\Pi_{n}\geq 0\}$. Then, $N$ is a stopping time with respect to $\mathbb{F}$ and $N$ is finite with probability $1$. Indeed, if ${\mathbb{P}}(A\leq 0)=1$ then $N=2$. If ${\mathbb{P}}(A>0)>0$ then $N=\infty$ if and only if $A_{1}<0$ and for every $n\geq 2$, $A_{n}>0$ which means that for every $n$

[TABLE]

Let now ${\mathbb{P}}(A<0)>0$ and $A_{+}=A\mathbf{1}_{A\geq 0}$, $A_{-}=-A\mathbf{1}_{A<0}$.

Since $\{N\geq k\}=\{A_{1}<0,A_{2}>0,\ldots,A_{k-1}>0\}$ for $k\geq 2$ we have

[TABLE]

Let us denote the expression in brackets by $S$. Then, $S$ is independent of $((A_{1})_{-},B_{1})$ and it is the unique solution to

[TABLE]

Summing up, we obtain

Lemma 5.1**.**

Assume that ${\mathbb{P}}(A<0)>0$ with ${\mathbb{E}}\log|A|<0$ and ${\mathbb{E}}\log^{+}|B|<\infty$. Let $R$ be the solution to

[TABLE]

Then $R$ is also a solution to (51), where

[TABLE]

and $S$ satisfies (53).

Thanks to the above lemma, we can reduce the case of signed $A$ to the case on non-negative $A$. The properties of $\Pi_{N}$ and $R_{N}^{\ast}$ will be inherited by the properties of the original $(A,B)$.

The main result of this section is

Theorem 5.2**.**

Suppose that

• (sA-1)

${\mathbb{P}}(A<0)>0$, ${\mathbb{E}}\log|A|<0$,

• (sA-2)

there exists $\alpha>0$ such that ${\mathbb{E}}|A|^{\alpha}=1$, $\rho={\mathbb{E}}|A|^{\alpha}\log|A|<\infty$,

• (sA-3)

the distribution of $\log|A|$ given $|A|>0$ is non-arithmetic,

• (sA-4)

there exists $\varepsilon>0$ such that ${\mathbb{E}}|A|^{\alpha+\varepsilon}<\infty$,

• (sB-1)

[TABLE]

• (sB-2)

${\mathbb{E}}|B|^{\alpha}=\infty$.

Then

[TABLE]

The proof relies on Lemma 5.1. The tail asymptotics of $S$ follows from [21] as it is explained below in the proof of Theorem 5.4. In view of (51) to conclude Theorem 5.2 it remains to prove that $\Pi_{N}$ and $R_{N}^{\ast}$ satisfy assumptions of Theorem 1.1. First we will prove that $\Pi_{N}$ inherits its properties from $A$. The following result is strongly inspired by [19, (9.11)-(9.13)] (see also [1, Lemma 4.12]). For completeness, the proof is included below.

Theorem 5.3**.**

• (i)

If the law of $\log|A|$ given $A\neq 0$ is non-arithmetic (spread-out), then the law of $\log\Pi_{N}$ given $\Pi_{N}>0$ is non-arithmetic (spread-out),

• (ii)

If ${\mathbb{E}}|A|^{\alpha}=1$ and ${\mathbb{E}}|A|^{\alpha+\varepsilon}<\infty$ for some $\varepsilon>0$ then there exists $\bar{\varepsilon}>0$ such that ${\mathbb{E}}\Pi_{N}^{\alpha+\bar{\varepsilon}}<\infty$,

• (iii)

If ${\mathbb{E}}|A|^{\alpha}=1$ then ${\mathbb{E}}\Pi_{N}^{\alpha}=1$ and ${\mathbb{E}}\Pi_{N}^{\alpha}\log\Pi_{N}=2{\mathbb{E}}|A|\log|A|$.

Proof.

If $P(A\leq 0)=1$ then $\Pi_{N}=A_{1}A_{2}$ and the law of $\log\Pi_{N}$ given $\log\Pi_{N}>0$ is ${\mathbb{P}}_{<}\ast{\mathbb{P}}_{<}$, where ${\mathbb{P}}_{<}$ is the law of {\mathbb{P}}_{\log|A|\big{|}A<0}. ${\mathbb{P}}_{<}\ast{\mathbb{P}}_{<}$ is non-arithmetic or spread out respectively if so is ${\mathbb{P}}_{<}$. Also the remaining of the above statements are clear in this case so for the rest of the proof we assume that $P(A>0)>0$.

• (i)

Denote by ${\mathbb{P}}_{>}$ and ${\mathbb{P}}_{<}$ the laws of {\mathbb{P}}_{\log A\big{|}A>0} and {\mathbb{P}}_{\log|A|\big{|}A<0}, respectively. Set $p={\mathbb{P}}(A>0)$ and $q={\mathbb{P}}(A<0)$. By [19, (9.11)], we have

[TABLE]

If $p\,{\mathbb{P}}_{>}+q\,{\mathbb{P}}_{<}$ is spread out then there are $k,l\geq 0$ such that ${\mathbb{P}}_{>}^{\ast k}\ast{\mathbb{P}}_{<}^{\ast l}$ has a non zero absolutely continuous component. Hence ${\mathbb{P}}_{>}\ast{\mathbb{P}}_{<}^{\ast 2}$ is spread out and the mixture of measures, one of which is spread-out is spread-out as well.

If $p\,{\mathbb{P}}_{>}+q\,{\mathbb{P}}_{<}$ is non-arithmetic then the supports of ${\mathbb{P}}_{>}$ and ${\mathbb{P}}_{<}^{\ast 2}$ generate a dense subgroup of ${\mathbb{R}}$ (see the argument below [19, (9.13)]). Thus, we conclude that ${\mathbb{P}}_{\log\Pi_{N}|\Pi_{N}>0}$ is non-arithmetic.

• (ii)

Let $\mu_{+}^{(\varepsilon)}:={\mathbb{E}}A^{\alpha+\varepsilon}\mathbf{1}_{A\geq 0}$. Since the function $\varepsilon\mapsto\mu_{+}^{(\varepsilon)}$ is continuous and $\mu_{+}^{(0)}<1$, then there exists $\varepsilon_{1}>0$ such that $\mu_{+}^{(\varepsilon_{1})}<1$.

Then, we have

[TABLE]

• (iii)

Define a measure $\mathbb{Q}_{n}$ on $(\Omega,\mathcal{F}_{n})$ by

[TABLE]

Let $\mathcal{F}_{\infty}$ be the smallest $\sigma-$field containing all $\mathcal{F}_{n}$. The sequence of measures $\mathbb{Q}_{n}$ is consistent, thus by Kolmogorov theorem there exists a unique measure $\mathbb{Q}$ on $\mathcal{F}_{\infty}$ such that $\mathbb{Q}(S)=\mathbb{Q}_{n}(S)$ for $S\in\mathcal{F}_{n}$. Note that $(A_{n})_{n\geq 1}$ are i.i.d. also under ${\mathbb{Q}}$. We have

[TABLE]

and for any $k>1$,

[TABLE]

Hence ${\mathbb{E}}_{{\mathbb{Q}}}N=2$, where ${\mathbb{E}}_{{\mathbb{Q}}}$ is the expectation with respect to ${\mathbb{Q}}$.

Since $\mathcal{F}_{N}\subset\mathcal{F}_{\infty}$, for any $S\in\mathcal{F}_{N}$ we have

[TABLE]

Putting $S=\Omega$ we obtain that ${\mathbb{E}}\Pi_{N}^{\alpha}=1$. Further, since $\Pi_{N}$ is $\mathcal{F}_{N}$ measurable, we have

[TABLE]

where the Wald’s identity was used.

∎

Secondly we show that the tails of $R_{N}^{\ast}$ behave like ${\mathbb{P}}(|B|>x)$. Let now ${\mathbb{P}}(A>0)>0$ and $A_{+}=A\,\mathbf{1}_{A\geq 0}$, $A_{-}=-A\,\mathbf{1}_{A<0}$.

Theorem 5.4**.**

Assume additionally that

[TABLE]

and ${\mathbb{E}}|A|^{\alpha+\varepsilon}<\infty$ for some $\varepsilon>0$. If $\mu_{+}={\mathbb{E}}A^{\alpha}\mathbf{1}_{A>0}<1$, then

[TABLE]

and

[TABLE]

Proof.

Tail asymptotic of $S$ follow from the application of [21, Theorem 3] to $(M,Q,R)=(A_{+},B,S)$. We have ${\mathbb{E}}|M|^{\alpha}={\mathbb{E}}A^{\alpha}\mathbf{1}_{A>0}<1$ and ${\mathbb{E}}|M|^{\alpha+\varepsilon}\leq{\mathbb{E}}|A|^{\alpha+\varepsilon}<\infty$ by the assumption.

Tail asymptotics of $R_{N}^{\ast}$ then follow from [21, Lemma 4], since $R_{N}^{\ast}\stackrel{{\scriptstyle d}}{{=}}B+A^{-}S$. Here $(M,Q,Y)=(A_{-},B,S)$ and ${\mathbb{E}}|M|^{\alpha}={\mathbb{E}}A^{\alpha}\mathbf{1}_{A<0}<1$ and ${\mathbb{E}}|M|^{\alpha+\varepsilon}$ is finite as above. One easily checks that ${\mathbb{P}}(R_{N}^{\ast}>t)\sim{\mathbb{P}}(|B|>t)$. To obtain ${\mathbb{P}}(R_{N}^{\ast}<-t)\sim{\mathbb{P}}(|B|>t)$ we apply the above argument to $-R\stackrel{{\scriptstyle d}}{{=}}A(-R)-B$. ∎

6. Proof of Theorem 3.1

First we prove that

[TABLE]

Since $g$ is bounded and its support is contained in $[1,\infty)$, there exists a constant $c$ such that $g(x)\leq c\,\mathbf{1}_{x>1-\varepsilon}$ for any $\varepsilon>0$. Thus, with $c_{\varepsilon}=c(1-\varepsilon)^{-1}$ we have $e^{\alpha(x-z)}{\mathbb{E}}g(e^{z-x}B)\leq c\,e^{\alpha(x-z)}{\mathbb{P}}(B>(1-\varepsilon)e^{x-z})=c_{\varepsilon}L((1-\varepsilon)e^{x-z})$ and therefore

[TABLE]

by (48).

For the main part we have

[TABLE]

The first term is easily seen to be $O(L(e^{x}))$. Indeed, observe that the integral $\int_{(0,\infty)}e^{-\alpha z}{\mathbf{H}}(\mathrm{d}z)=\int_{0}^{\infty}\alpha e^{-\alpha x}{\mathbf{H}}\left((0,x]\right)\mathrm{d}x$ is finite by (24). Bounding $g$ by an indicator as before, we have

[TABLE]

Let us decompose $I_{2}(x)$ in the following way

[TABLE]

The first term above is

[TABLE]

and it constitutes the main ingredient in (32). To see this, change the variable $r=e^{-(x-z)}B$, to obtain

[TABLE]

by the fact that $\mathrm{supp}\,g\subset[1,\infty)$. But (36) gives us that ${\mathbb{E}}B_{+}^{\alpha}\mathbf{1}_{B\leq e^{x}}=\alpha\widetilde{L}(e^{x})-L(e^{x}).$ It remains to prove that the second term in (58) is $o(\widetilde{L}(e^{x}))$. Let us denote

[TABLE]

The equality $g(1)=0$ is a consequence of $\mathrm{supp}\,g\subset[1,\infty)$ and differentiability of $g$. Since $\lim_{z\to\infty}e^{-\alpha z}R(z)=0$, after integrating by parts we see that

[TABLE]

where we have substituted $t=z-x+\log B$. Moreover, notice that

[TABLE]

and so

[TABLE]

By the assumption (31), there exists a constant $C$ such that for all $t>0$,

[TABLE]

so it amounts to estimate

[TABLE]

Define

[TABLE]

We will show that $J(x)\to 0$, and since the denominator equals $\alpha\widetilde{L}(e^{x})$ this will be the end of the proof.

Define the law of $C_{x}$ by

[TABLE]

Note that ${\mathbb{P}}(0<C_{x}\leq e^{x})=1$. Thus, $J(x)$ may be rewritten as

[TABLE]

Since for any positive $x$ and $t$, $|R(t+x)-R(x)|=|{\mathbf{H}}((x,x+t])-\frac{t}{{\mathbb{E}}Z}|\leq c\,t+b$ for some $c,b>0$, we have

[TABLE]

Moreover, $x-C_{x}$ converges to infinity in probability, as $x\to\infty$. Indeed, for any $N>0$ we have

[TABLE]

because $\widetilde{L}$ is slowly varying. Since, $\left|R(t+x)-R(x)\right|\to 0$ as $x\to\infty$, we infer that

[TABLE]

converges to [math] in probability, as $x\to\infty$. But (60) is bounded in $x$, thus the convergence holds also in $L_{1}$ and we may finally conclude that

[TABLE]

which completes the proof of (32).

If additionally ${\mathbb{E}}\exp(\varepsilon Z)<\infty$ for some $\varepsilon>0$ and the law of $Z$ is strongly non-lattice then (59) is bounded by $CL(\exp(x))$ which was proved in [14] - see the end of the proof of Theorem 3.3 there just before the references.

7. Appendix

Suppose that ${\mathbb{E}}|B|^{\beta}<\infty$ for any $\beta<\alpha$ and that there is $\varepsilon>0$ such that ${\mathbb{E}}|A|^{\alpha+\varepsilon}<\infty$. Then by Hölder inequality we may conclude that for every $\eta<\alpha$, ${\mathbb{E}}|B|^{\alpha-\eta}|A|^{\eta}<\infty$. However, if the tail of $B$ exhibits some more regularity, a weaker condition implies the same conclusion.

Suppose that $x^{\alpha}{\mathbb{P}}(|B|>x)\leq L(x)$, where $L$ is a slowly varying function bounded away from [math] and $\infty$ on any compact subset of $(0,\infty)$. Let $W$ be a non-decreasing function such that

[TABLE]

For instance $W(x)=\max\{\sup_{0<w\leq x}L(w),\log(x)\}$ or $W(x)=\max\{\widetilde{L}(x),\log(x)\}$ will do.

Lemma 7.1**.**

Assume that $W$ is as above, $\eta<\alpha$, $D>\frac{2\alpha}{\eta}-1$ and

[TABLE]

Then

[TABLE]

Proof.

Since ${\mathbb{E}}|A|^{\alpha}<\infty$ and ${\mathbb{E}}|B|^{\alpha-\eta}<\infty$, it is enough to prove that for a fixed $C_{0}$

[TABLE]

We choose $\beta>0$ and $\gamma$ such that

[TABLE]

For $m\geq k$ consider the sets

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Let $C_{0}=e^{k_{0}}W(e^{k_{0}})^{\beta}$, where $k_{0}$ is such that $W(k_{0})\geq 1$. Then

[TABLE]

[TABLE]

and

[TABLE]

Let $\delta>0$. By the Potter bounds (20), since $W(e^{m})\geq 1$ for $m\geq k_{0}$

[TABLE]

Hence

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Further,

[TABLE]

because $W$ is non-decreasing and $W(e^{k})\geq 1$. Therefore, we have

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Notice that in view of (61)

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In order to sum up over $\beta$, we choose $\beta$ such that

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Finally, we take $\delta$ sufficiently small to ensure

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Then

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Hence

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Finally, we need to guarantee that

[TABLE]

Suppose that $\beta(\alpha-\eta-\alpha\gamma)=-\gamma-\xi$ for some $\xi>0$. Then (62) becomes

[TABLE]

and we may minimize $D$ by an appropriate choice of $\gamma$. Notice that if $\xi=0$ and $\gamma=1-\eta/\alpha$ the right hand side of (63) becomes $2\alpha/\eta-1$. Since $\gamma$ may be arbitrary close to $1-\eta/\alpha$ and $\xi$ arbitrary close to [math], $D>2\alpha/\eta-1$ will do. ∎

Acknowledgements

Ewa Damek was partially supported by the NCN Grant UMO-2014/15/B/ST1/00060. Bartosz Kołodziejek was partially supported by the NCN Grant UMO-2015/19/D/ST1/03107.