The left tail of renewal measure

TL;DR

This paper derives precise asymptotics for the left tail of the renewal measure in two-sided random walks, using a change of measure approach under broad conditions.

Contribution

It provides a new, nearly equivalent formulation to Blackwell's Theorem for the left tail asymptotics of renewal measures with finite exponential moments.

Findings

Exact asymptotics for the left tail of renewal measure

Applicable to broad class of two-sided random walks

Uses a simple change of measure approach

Abstract

In the paper, we find exact asymptotics of the left tail of renewal measure for a broad class of two-sided random walks. We only require that an exponential moment of the left tail is finite. Through a simple change of measure approach, our result turns out to be almost equivalent to Blackwell's Theorem.

Equations

$$H(B):=\sum _{n=0}^{\infty }\mathbb{P}(S_n\in B),B\in \mathcal{B}(\mathbb{R})$$

$$H\left((x,x+h]\right)⟶\frac{h}{\mathbb{E}X}\text{mbox}asx\to \infty .$$

$$H\left((dn,dn+h]\right)⟶\frac{d\lfloor h/d\rfloor }{\mathbb{E}X}\text{mbox}asn\to \infty .$$

$$m(x)H\left((x,x+h]\right)⟶\frac{h}{\Gamma (\alpha )\Gamma (2-\alpha )}\text{mbox}asx\to \infty ,$$

$$\underset{\delta \to 0}{\lim }\underset{x\to \infty }{\mathrm{limsup}}{\int }_1^{\delta x}\frac{F(x)-F(x-z)}{\overline{F}(z)z^2}\mathrm{d}z=0,$$

$$\mathbb{P}(X\le -x)=o(e^{-rx})\text{mbox}asx\to \infty$$

$$H((-\infty ,-x))=o(e^{-r_{1}x})\text{mbox}asx\to \infty .$$

$$H_{\rho }(B):=\sum _{n=0}^{\infty }{\rho }^n\mathbb{P}(S_n\in B),B\in \mathcal{B}(\mathbb{R}),$$

$$\frac{\mu (x+s+\Delta )}{\mu (x+\Delta )}\to 1\text{mbox}asx\to \infty ,$$

$${\mu }^{\ast 2}(x+\Delta )\sim 2\mu (x+\Delta ).$$

$${\int }_{\mathbb{R}}{e}^{-\epsilon x}\mu (\mathrm{d}x)<\infty \text{mbox}forsome\epsilon >0.$$

$$\eta =\sum _{n=0}^{\infty }{\rho }^n{\mu }^{\ast n}.$$

$$\eta (x+\Delta )\sim \frac{\rho }{(1-\rho {)}^2}\mu (x+\Delta ).$$

$$g(\theta ):=\mathbb{E}{e}^{-\theta X}.$$

$$\mathbb{P}(X<0)>0\text{mbox}andg(\theta )<\infty \text{mbox}forsome\theta >0.$$

$$\kappa :=\sup \{\theta >0:g(\theta )<1\}\text{mbox}and\rho :=g(\kappa ).$$

$$\underset{x\to \infty }{\lim }{e}^{\kappa x}H((-\infty ,-x))=\frac{1}{\kappa g^{\prime }(\kappa )}\in [0,\infty ).$$

$$\mathbb{E}{e}^{-\kappa X}{1}_{\{-X>t\}}\sim \frac{L(t)}{t^{\alpha }}$$

$$e^{\kappa x}H((-\infty ,-x))\sim \frac{1}{\Gamma (\alpha )\Gamma (2-\alpha )}\frac{1}{\kappa m(x)},$$

$$\underset{n\to \infty }{\lim }{e}^{\kappa dn}H((-\infty ,-nd))=\frac{d}{(e^{\kappa d}-1)g^{\prime }(\kappa )}.$$

$$\underset{x\to \infty }{\lim }{e}^{\kappa x}H((-\infty ,-x))=0.$$

$$e^{\kappa x}H((-\infty ,-x))\sim \frac{\mathbb{E}{e}^{-\kappa X}{1}_{x<-X\le x+1}}{\kappa (1-\rho {)}^2}.$$

$$\mathbb{P}(-X>t)=\frac{\alpha }{\kappa }\frac{L(t)}{t^{\alpha +1}}{e}^{-\kappa t},t>0.$$

$${\int }_x^{\infty }{t}^{\beta }L(t)\mathrm{d}t\sim x^{\beta +1}L(x)/(-\beta -1).$$

$$\underset{x\to \infty }{\lim }{e}^{\kappa x}H((-x-\delta ,-x))=\frac{1-e^{-\delta \kappa }}{\kappa g^{\prime }(\kappa )}.$$

$$\mathbb{Q}((X_1,\dots ,X_n)\in B)={\rho }^{-n}\mathbb{E}{e}^{-\kappa S_n}{1}_{\{-(X_1,\dots ,X_n)\in B\}},B\in \mathcal{B}({\mathbb{R}}^n),$$

$$\mathbb{E}f(S_n)={\rho }^n{\mathbb{E}}_{\mathbb{Q}}{e}^{-\kappa S_n}f(-S_n).$$

$$\mathbb{P}(S_n<-x)={\rho }^n{\mathbb{E}}_{\mathbb{Q}}{e}^{-\kappa S_n}{1}_{S_n>x}={\rho }^n{\int }_{(x,\infty )}{e}^{-\kappa t}{\mathbb{Q}}_X^{\ast n}(\mathrm{d}t).$$

$$H((-\infty ,-x))=H_{\mathbb{P}}((-\infty ,-x))=\sum _{n=1}^{\infty }\mathbb{P}(S_n<-x)={\int }_{(x,\infty )}{e}^{-\kappa t}{H}_{\mathbb{Q}}(\mathrm{d}t),$$

$$H_{\mathbb{P}}((-\infty ,-x))=\kappa {\int }_x^{\infty }{e}^{-\kappa s}{H}_{\mathbb{Q}}((x,s])\mathrm{d}s=\kappa e^{-\kappa x}{\int }_0^{\infty }{e}^{-\kappa h}{H}_{\mathbb{Q}}((x,x+h])\mathrm{d}h.$$

Full text

1. Introduction

Let $(X_{k})_{k\geq 1}$ be a sequence of independent copies of a random variable $X$ with ${\mathbb{E}}X>0$ (we allow ${\mathbb{E}}X=\infty$). Further, define $S_{n}=X_{1}+\ldots+X_{n}$, $n\geq 1$ and $S_{0}=0$. The measure defined by

[TABLE]

is called the renewal measure of $(S_{n})_{n\geq 1}$.

We say that the distribution of a random variable $X$ is $d$-arithmetic ($d>0$) if it is concentrated on $d{\mathbb{Z}}$ and not concentrated on $d^{\prime}{\mathbb{Z}}$ for any $d^{\prime}>d$. A distribution is said to be non-arithmetic if it is not $d$-arithmetic for any $d>0$.

A fundamental result of renewal theory is the Blackwell Theorem (Blackwell (1953)): if the distribution of $X$ is non-arithmetic, then for any $h>0$,

[TABLE]

If the distribution of $X$ is $d$-arithmetic, then for any $h>0$,

[TABLE]

The above results remain true if ${\mathbb{E}}X=\infty$ with the usual convention that $c/\infty=0$ for any finite $c$. In the infinite-mean case the exact asymptotics of $H((x,x+h])$ are also known. Assume that $X$ is a non-negative random variable with a non-arithmetic law such that ${\mathbb{P}}(X>x)=L(x)x^{-\alpha}$ with $\alpha\in(0,1)$, where $L$ is a slowly varying function. Then ${\mathbb{E}}X=\infty$. If $\alpha\in(1/2,1)$, then without additional assumptions the so called Strong Renewal Theorem holds, for $h>0$,

[TABLE]

where $m(x)=\int_{0}^{x}{\mathbb{P}}(X>t)\mathrm{d}t\sim L(x)x^{1-\alpha}/(1-\alpha)\to\infty$. Here and later on $f(x)\sim g(x)$ means that $f(x)/g(x)\to 1$ as $x\to\infty$.

The case of $\alpha\in(0,1/2]$ is much harder and was completely solved just recently by Caravenna and Doney (2016). It was shown that if $\alpha\in(0,1/2]$ and $X$ is a non-negative random variable with regularly varying tail, then (3) holds if and only if ((Caravenna and Doney, 2016, Proposition 1.11))

[TABLE]

where $F$ is the cumulative distribution function of $X$ and $\overline{F}=1-F$. It was already observed by (Kevei, 2016, Theorem 3.1) that this result generalizes to $X$ attaining negative values as well if additionally

[TABLE]

for some $r>0$. This will be our setup. Full picture of SRT for random walks is also known ((Caravenna and Doney, 2016, Theorem 1.12)).

It is clear that $\lim_{x\to\infty}H((-\infty,-x))=0$. There are considerably fewer papers dedicated to analysis of exact asymptotics of such object than of $H((x,x+h])$ as in Blackwell’s Theorem. Under some additional assumptions we know more about the asymptotic behaviour of the left tail. Stone (1965) proved that if for some $r>0$ (5) holds, then for some $r_{1}>0$,

[TABLE]

Stone’s result was strengthened by van der Genugten (1969), where exact asymptotics as well the speed of convergence of the remainder term are given for $d$-arithmetic and spread-out laws (i.e. laws, whose $n$th convolution has a nontrivial absolutely continuous part for some $n\in\mathbb{N}$). An important contribution regarding the asymptotics of the left tail of renewal measure was made by Carlsson (1983), who concerned with the case when ${\mathbb{E}}|X|^{m}<\infty$ for some $m\geq 2$, but this does not fit well into our setup. We allow ${\mathbb{E}}X_{+}=\infty$, but on the other hand we require that some exponential moments of $X_{-}$ exist. The results mentioned above were obtained using some analytical methods, whereas we will use a simple probabilistic argument, which boils down the asymptotics of $H((-\infty,-x))$ to the asymptotics of $\widetilde{H}((x,x+h])$, where $\widetilde{H}$ is some new (possibly defective, see below) renewal measure.

1.1. Defective renewal measure

For $\rho\in(0,1)$ consider

[TABLE]

where $(S_{n})_{n\geq 1}$ is, as in the previous section, a random walk starting from [math]. $H_{\rho}$ is called a defective renewal measure of $(S_{n})_{n\geq 1}$. In contrast to the renewal measure, $H_{\rho}$ is a finite measure. Let $\tau$ be independent of $(S_{n})_{n\geq 1}$ and ${\mathbb{P}}(\tau=n)=(1-\rho)\rho^{n}$, $n=0,1,\ldots$. Then $H_{\rho}(B)={\mathbb{P}}(S_{\tau}\in B)/(1-\rho)$. It is well known that if the distribution of $S_{1}$ is subexponential, then ${\mathbb{P}}(S_{\tau}>x)\sim{\mathbb{E}}\tau{\mathbb{P}}(S_{1}>x)$. Here, we are interested in exact asymptotics of $H_{\rho}(B)$ when $B=(x,x+T]$ for any $T>0$. In this context, local subexponentiality is the key concept (Asmussen et al. (2003)).

Let $\mu$ be a probability measure on ${\mathbb{R}}$. For $T>0$ we write $\Delta=(0,T]$ and $x+\Delta=(x,x+T]$. We say that $\mu$ belongs to the class $\mathcal{L}_{\Delta}$ if $\mu(x+\Delta)>0$ for sufficiently large $x$ and

[TABLE]

uniformly in $s\in[0,1]$.

We say that $\mu$ is $\Delta$-subexponential if $F\in\mathcal{L}_{\Delta}$ and

[TABLE]

Then we write $\mu\in\mathcal{S}_{\Delta}$. Finally, $\mu$ is called locally subexponential if $\mu\in\mathcal{S}_{\Delta}$ for any $T>0$. We denote this class by $\mathcal{S}_{loc}$.

The following Theorem is an obvious conclusion from (Watanabe and Yamamuro, 2009, Theorem 1.1).

Theorem 1.1**.**

Assume that $\mu$ is a probability measure on ${\mathbb{R}}$ such that

[TABLE]

For $0<\rho<1$ define

[TABLE]

Then $\mu\in\mathcal{S}_{\Delta}$ if and only if $\eta/(1-\rho)\in\mathcal{S}_{\Delta}$ if and only if

[TABLE]

Some examples of measures from $\mathcal{S}_{loc}$ may be found in (Asmussen et al., 2003, Section 4).

2. Main result

Assume that $X$ is a random variable with ${\mathbb{E}}X\in(0,\infty]$. We define the Laplace transform of the distribution of $X$ by

[TABLE]

Function $g$ is convex and lower-semicontinuous. We are interested in a situation of an exponentially decaying left tail, that is,

[TABLE]

Under (8) we define

[TABLE]

Since $g^{\prime}(0)=-{\mathbb{E}}X<0$, $\kappa$ is strictly positive. Moreover, we have $g(\theta)\to\infty$ as $\theta\to\infty$ and thus $\kappa$ is finite. In general we have $0<\rho\leq 1$ and a sufficient condition for $\rho=1$ is that $g(\theta)<\infty$ for all $\theta>0$.

Theorem 2.1**.**

Assume $X$ is a random variable with a positive (possibly infinite) expectation such that (8) holds. Let $H$ be the renewal measure of $(S_{n})_{n\geq 0}$, where $S_{n}=\sum_{k=1}^{n}X_{k}$ for $n\in\mathbb{N}$, $S_{0}=0$ and $X_{k}$ are independent copies of $X$. Define $\kappa$ and $\rho$ as in (9).

• (a)

Assume that $\rho=1$.

• (a-i)

Assume that $X$ has a non-arithmetic distribution. Then

[TABLE]

Moreover, if

[TABLE]

for some $\alpha\in(0,1)$ and a slowly varying function $L$, then $g^{\prime}(\kappa)=\infty$. For $\alpha\in(0,1/2]$, assume additionally that $F(t)={\mathbb{E}}e^{-\kappa X}\mathbf{1}_{\{-X\leq t\}}$ satisfies (4). In such case,

[TABLE]

where $m(x)\sim L(x)x^{1-\alpha}/(1-\alpha)$.

• (a-ii)

Assume that $X$ has a $d$-arithmetic distribution. Then

[TABLE]

• (b)

If $\rho<1$, then

[TABLE]

Moreover, if $\rho^{-1}{\mathbb{E}}e^{-\kappa X}\mathbf{1}_{\{-X\in\cdot\}}\in\mathcal{S}_{loc}$, then

[TABLE]

Remark 2.2**.**

Condition (10) is implied by

[TABLE]

Indeed, for any slowly varying function $L$ and $\beta<-1$, (Bingham et al., 1989, Proposition 1.5.10) asserts that

[TABLE]

Remark 2.3**.**

Under the same assumptions, a stronger result concerning (a-ii) is proved in (van der Genugten, 1969, Theorem 2) (the remainder term is also exponential).

Remark 2.4**.**

If $X$ has a non-arithmetic distribution, for any $\delta>0$, we obtain “more local” behaviour:

[TABLE]

Proof of Theorem 2.1.

Note that $g^{\prime}(\kappa)=-{\mathbb{E}}Xe^{-\theta X}$ is positive ($1=g(0)=g(\kappa)$ and $g$ is convex), but may infinite.

Define $\mathcal{F}_{n}=\sigma(X_{1},\ldots,X_{n})$ and let $\mathcal{F}_{\infty}$ be the smallest $\sigma$-field containing all $\mathcal{F}_{n}$. On $(\Omega,\mathcal{F}_{\infty})$ we define a new measure ${\mathbb{Q}}$ via projections

[TABLE]

where $S_{n}=X_{1}+\ldots+X_{n}$, $n\in{\mathbb{N}}$ and $S_{0}=0$. By the definition of $\rho$, ${\mathbb{Q}}$ is a probability measure. Moreover, $(X_{n})_{n\geq 1}$ is an iid sequence under ${\mathbb{Q}}$ as well. Let ${\mathbb{E}}_{{\mathbb{Q}}}$ denote the corresponding expectation. For any Borel function $f\colon{\mathbb{R}}\to{\mathbb{R}}_{+}$ one has

[TABLE]

Thus,

[TABLE]

Moreover, observe that ${\mathbb{E}}_{{\mathbb{Q}}}X=-{\mathbb{E}}Xe^{-\kappa X}=g^{\prime}(\kappa){\in(0,\infty]}$, thus $(S_{n})_{n}$ has a positive drift under ${\mathbb{Q}}$ as well. Hence, for $x>0$,

[TABLE]

where $H_{{\mathbb{Q}}}=\sum_{n=0}^{\infty}{\rho^{n}}{\mathbb{Q}}_{X}^{\ast n}$ is the (defective if $\rho<1$) renewal measure of $(S_{n})_{n\geq 0}$ under ${\mathbb{Q}}$.

Writing $e^{-\kappa t}=\kappa\int_{t}^{\infty}e^{-\kappa s}\mathrm{d}s$, through Tonelli’s Theorem, we arrive at key identity:

[TABLE]

Consider first the case of $\rho=1$. For any renewal measure $H$ we have $H((x,x+h])\leq\alpha h+\beta$ for some $\alpha,\beta>0$ and all $x$, thus by Lebesgue’s Dominated Convergence Theorem and (1),

[TABLE]

which gives the first part of (a-i). For (a-ii) use (2), instead of (1).

For the second part of (a-i), observe that

[TABLE]

thus the result follows by the Strong Renewal Theorem.

If $\rho<1$, then $H_{\mathbb{Q}}$ is a finite measure and (11) follows again by Lebesgue’s Dominated Convergence Theorem.

Consider now the case, when ${\mathbb{Q}}_{X}\in\mathcal{S}_{loc}$. Since ${\mathbb{E}}_{\mathbb{Q}}e^{-\kappa X}=1<\infty$, by Theorem 1.1, we have $(1-\rho)H_{{\mathbb{Q}}}\in\mathcal{S}_{loc}$ and

[TABLE]

Define $L(y):=H_{{\mathbb{Q}}}((\log y,\log y+1])$. By (7), $L$ is a slowly varying function. Moreover, for any $h>0$

[TABLE]

By Potter bounds ((Bingham et al., 1989, Theorem 1.5.6)) for any $\varepsilon>0$ and $C>1$ there exists $x_{0}$ such that for $x>x_{0}$,

[TABLE]

Observe that (7) implies for $h>0$,

[TABLE]

Indeed, for $h=k/n\in{\mathbb{Q}}_{+}$ one gets

[TABLE]

and

[TABLE]

By monotonicity, $f(h):=\lim_{x\to\infty}H_{{\mathbb{Q}}}((x,x+h])/H_{{\mathbb{Q}}}((x,x+1])$ exists for all $h>0$ and $f(h)=h$. Thus, by (12) and Lebesgue’s Dominated Convergence Theorem we conclude that

[TABLE]

The use of (13) completes the proof. ∎

Acknowledgements

The author was partially supported by NCN Grant No. 2015/19/D/ST1/03107. I am grateful to the referee for a number of helpful suggestions for improvement in the article.