Heavy-Tailed Random Walks on Complexes of Half-Lines

Abstract

We study a random walk on a complex of finitely many half-lines joined at a common origin; jumps are heavy-tailed and of two types, either one-sided (towards the origin) or two-sided (symmetric). Transmission between half-lines via the origin is governed by an irreducible Markov transition matrix, with associated stationary distribution $\mu _k$. If $\chi _k$ is 1 for one-sided half-lines k and 1 / 2 for two-sided half-lines, and $\alpha _k$ is the tail exponent of the jumps on half-line k, we show that the recurrence classification for the case where all $\alpha _k \chi _k \in (0,1)$ is determined by the sign of $\sum _k \mu _k \cot ( \chi _k \pi \alpha _k )$. In the case of two half-lines, the model fits naturally on ${{\mathbb {R}}}$ and is a version of the oscillating random walk of Kemperman. In that case, the cotangent criterion for recurrence becomes linear in $\alpha _1$ and $\alpha _2$; our general setting exhibits the essential nonlinearity in the cotangent criterion. For the general model, we also show existence and non-existence of polynomial moments of return times. Our moments results are sharp (and new) for several cases of the oscillating random walk; they are apparently even new for the case of a homogeneous random walk on ${{\mathbb {R}}}$ with symmetric increments of tail exponent $\alpha \in (1,2)$.

1 Introduction

We study Markov processes on a complex of half-lines ${{\mathbb {R}}}_+\times {{\mathcal {S}}}$, where ${{\mathcal {S}}}$ is finite, and all half-lines are connected at a common origin. On a given half-line, a particle performs a random walk with a heavy-tailed increment distribution, until it would exit the half-line, when it switches (in general, at random) to another half-line to complete its jump.

To motivate the development of the general model, we first discuss informally some examples; we give formal statements later.

The one-sided oscillating random walk takes place on two half-lines, which we may map onto ${{\mathbb {R}}}$. From the positive half-line, the increments are negative with density proportional to $y^{-1-\alpha }$, and from the negative half-line, the increments are positive with density proportional to $y^{-1-\beta }$, where $\alpha , \beta \in (0,1)$. The walk is transient if and only if $\alpha + \beta < 1$; this is essentially a result of Kemperman [15].

The oscillating random walk has several variations and has been well studied over the years (see, e.g. [17,18,19, 21]). This previous work, as we describe in more detail below, is restricted to the case of two half-lines. We generalize this model to an arbitrary number of half-lines, labelled by a finite set ${{\mathcal {S}}}$, by assigning a rule for travelling from half-line to half-line.

First, we describe a deterministic rule. Let T be a positive integer. Define a routing schedule of length T to be a sequence $\sigma = (i_1, \ldots , i_T)$ of T elements of ${{\mathcal {S}}}$, dictating the sequence in which half-lines are visited, as follows. The walk starts from line $i_1$, and, on departure from line $i_1$ jumps over the origin to $i_2$, and so on, until departing $i_T$ it returns to $i_1$; on line $k \in {{\mathcal {S}}}$, the walk jumps towards the origin with density proportional to $y^{-1-\alpha _k}$ where $\alpha _k \in (0,1)$. One simple example takes a cyclic schedule in which $\sigma$ is a permutation of the elements of ${{\mathcal {S}}}$. In any case, a consequence of our results is that now the walk is transient if and only if

$$\begin{aligned} \sum _{k \in {{\mathcal {S}}}} \mu _k \cot ( \pi \alpha _k ) > 0, \end{aligned}$$

(1.1)

where $\mu _k$ is the number of times k appears in the sequence $\sigma$. In particular, in the cyclic case the transience criterion is $\sum _{k \in {{\mathcal {S}}}} \cot ( \pi \alpha _k ) > 0$.

It is easy to see that, if ${{\mathcal {S}}}$ contains two elements, the cotangent criterion (1.1) is equivalent to the previous one for the one-sided oscillating walk ($\alpha _1 + \alpha _2 < 1$). For more than two half-lines, the criterion is nonlinear, and it was necessary to extend the model to more than two lines in order to see the essence of the behaviour.

More generally, we may choose a random routing rule between lines: on departure from half-line $i \in {{\mathcal {S}}}$, the walk jumps to half-line $j \in {{\mathcal {S}}}$ with probability p(i, j). The deterministic cyclic routing schedule is a special case in which $p(i, i') =1$ for $i'$ the successor to i in the cycle. In fact, this set-up generalizes the arbitrary deterministic routing schedule described above, as follows. Given the schedule sequence $\sigma$ of length T, we may convert this to a cyclic schedule on an extended state space consisting of $\mu _k$ copies of line k and then reading $\sigma$ as a permutation. So the deterministic routing model is a special case of the model with Markov routing, which will be the focus of the rest of the paper.

Our result again will say that (1.1) is the criterion for transience, where $\mu _k$ is now the stationary distribution associated with the stochastic matrix p(i, j). Our general model also permits two-sided increments for the walk from some of the lines, which contribute terms involving $\cot ( \pi \alpha _k /2 )$ to the cotangent criterion (1.1). These two-sided models also generalize previously studied classical models (see, e.g. [15, 17, 23]). Again, it is only in our general setting that the essential nature of the cotangent criterion (1.1) becomes apparent.

Rather than ${{\mathbb {R}}}_+\times {{\mathcal {S}}}$, one could work on ${{\mathbb {Z}}}_+\times {{\mathcal {S}}}$ instead, with mass functions replacing probability densities; the results would be unchanged.

The paper is organized as follows. In Sect. 2, we formally define our model and describe our main results, which as well as a recurrence classification include results on existence of moments of return times in the recurrent cases. In Sect. 3, we explain how our general model relates to the special case of the oscillating random walk when ${{\mathcal {S}}}$ has two elements, and state our results for that model; in this context, the recurrence classification results are already known, but the existence-of-moments results are new even here, and are in several important cases sharp. The present work was also motivated by some problems concerning many-dimensional, partially homogeneous random walks similar to models studied in [5, 6, 13]: we describe this connection in Sect. 4. The main proofs are presented in Sects. 5, 6, and 7, the latter dealing with the critical boundary case which is more delicate and requires additional work. We collect various technical results in the Appendix.

2 Model and Results

Consider $(X_n, \xi _n ; n \in {{\mathbb {Z}}}_+)$, a discrete-time, time-homogeneous Markov process with state space ${{\mathbb {R}}}_+\times {{\mathcal {S}}}$, where ${{\mathcal {S}}}$ is a finite non-empty set. The state space is equipped with the appropriate Borel sets, namely sets of the form $B \times A$ where $B \in {{\mathcal {B}}}({{\mathbb {R}}}_+)$ is a Borel set in ${{\mathbb {R}}}_+$, and $A \subseteq {{\mathcal {S}}}$. The process will be described by:

• an irreducible stochastic matrix labelled by ${{\mathcal {S}}}$, $P = ( p(i,j) ; i,j \in {{\mathcal {S}}})$; and

• a collection $(w_i ; i \in {{\mathcal {S}}})$ of probability density functions, so $w_i : {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}_+$ is a Borel function with $\int _{{\mathbb {R}}}w_i (y) {\mathrm d}y =1$.

We view ${{\mathbb {R}}}_+\times {{\mathcal {S}}}$ as a complex of half-lines ${{\mathbb {R}}}_+\times \{k\}$, or branches, connected at a central origin ${{\mathcal {O}}}:= \{0 \} \times {{\mathcal {S}}}$; at time n, the coordinate $\xi _n$ describes which branch the process is on, and $X_n$ describes the distance along that branch at which the process sits. We will call $X_n$ a random walk on this complex of branches.

To simplify notation, throughout we write ${{\mathbb {P}}}_{x,i} [ \, \cdot \, ]$ for ${{\mathbb {P}}}[ \, \cdot \mid (X_0 , \xi _0 ) = (x,i) ]$, the conditional probability starting from $(x,i) \in {{\mathbb {R}}}_+\times {{\mathcal {S}}}$; similarly, we use ${{\mathrm{{\mathbb {E}}}}}_{x,i}$ for the corresponding expectation. The transition kernel of the process is given for $(x,i) \in {{\mathbb {R}}}_+\times {{\mathcal {S}}}$, for all Borel sets $B \subseteq {{\mathbb {R}}}_+$ and all $j \in {{\mathcal {S}}}$, by

$$\begin{aligned}&{{\mathbb {P}}}\left[ ( X_{n+1}, \xi _{n+1} ) \in B \times \{ j \} \mid (X_n, \xi _n) = (x , i) \right] \nonumber \\&\quad = {{\mathbb {P}}}_{x,i} \left[ ( X_{1}, \xi _{1} ) \in B \times \{ j \} \right] \nonumber \\&\quad = p (i,j) \int _B w_i (-z-x) {\mathrm d}z +{{\mathbf {1}}}{\{ i = j \}} \int _B w_i (z-x) {\mathrm d}z. \end{aligned}$$

(2.1)

The dynamics of the process represented by (2.1) can be described algorithmically as follows. Given $(X_n, \xi _n ) = (x,i) \in {{\mathbb {R}}}_+\times {{\mathcal {S}}}$, generate (independently) a spatial increment $\varphi _{n+1}$ from the distribution given by $w_i$ and a random index $\eta _{n+1} \in {{\mathcal {S}}}$ according to the distribution $p(i, \, \cdot \,)$. Then,

• if $x + \varphi _{n+1} \ge 0$, set $(X_{n+1} , \xi _{n+1}) = (x + \varphi _{n+1} , i)$; or

• if $x + \varphi _{n+1} < 0$, set $(X_{n+1}, \xi _{n+1} ) = (| x + \varphi _{n+1} | , \eta _{n+1} )$.

In words, the walk takes a $w_{\xi _n}$-distributed step. If this step would bring the walk beyond the origin, it passes through the origin and switches onto branch $\eta _{n+1}$ (or, if $\eta _{n+1}$ happens to be equal to $\xi _n$, it reflects back along the same branch).

The finite irreducible stochastic matrix P is associated with a (unique) positive invariant probability distribution $(\mu _k ; k \in {{\mathcal {S}}})$ satisfying

$$\begin{aligned} \sum _{j \in {{\mathcal {S}}}} \mu _j p (j,k) - \mu _k =0 , \quad \text {for all } k \in {{\mathcal {S}}}. \end{aligned}$$

(2.2)

For future reference, we state the following.

• (A0) Let $P = ( p(i,j) ; i,j \in {{\mathcal {S}}})$ be an irreducible stochastic matrix, and let $(\mu _k ; k \in {{\mathcal {S}}})$ denote the corresponding invariant distribution.

Our interest here is when the $w_i$ are heavy-tailed. We allow two classes of distribution for the $w_i$: one-sided or symmetric. It is convenient, then, to partition ${{\mathcal {S}}}$ as ${{\mathcal {S}}}= {{\mathcal {S}}}^\mathrm{one}\cup {{\mathcal {S}}}^\mathrm{sym}$ where ${{\mathcal {S}}}^\mathrm{one}$ and ${{\mathcal {S}}}^\mathrm{sym}$ are disjoint sets, representing those branches on which the walk takes, respectively, one-sided and symmetric jumps. The $w_k$ are then described by a collection of positive parameters $(\alpha _k ; k \in {{\mathcal {S}}})$.

For a probability density function $v: {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}_+$, an exponent $\alpha \in (0,\infty )$, and a constant $c \in (0,\infty )$, we write $v \in {{\mathfrak {D}}}_{\alpha ,c}$ to mean that there exists $c : {{\mathbb {R}}}_+\rightarrow (0,\infty )$ with $\sup _{y} c(y) < \infty$ and $\lim _{y \rightarrow \infty } c (y) = c$ for which

$$\begin{aligned} v (y) = {\left\{ \begin{array}{ll} c (y) y^{-1-\alpha } &{} \text { if } y > 0 \\ 0 &{} \text {if } y \le 0. \end{array}\right. } \end{aligned}$$

(2.3)

If $v \in {{\mathfrak {D}}}_{\alpha ,c}$ is such that (2.3) holds and c(y) satisfies the stronger condition $c(y) = c + O (y^{-\delta })$ for some $\delta >0$, then we write $v \in {{\mathfrak {D}}}_{\alpha ,c}^+$.

Our assumption on the increment distributions $w_i$ is as follows.

• (A1) Suppose that, for each $k \in {{\mathcal {S}}}$, we have an exponent $\alpha _k \in (0,\infty )$, a constant $c_k \in (0,\infty )$, and a density function $v_k \in {{\mathfrak {D}}}_{\alpha _k,c_k}$. Then suppose that, for all $y \in {{\mathbb {R}}}$, $w_k$ is given by

  $$\begin{aligned} w_k (y) = {\left\{ \begin{array}{ll} v_k (-y) &{} \text { if } k \in {{\mathcal {S}}}^\mathrm{one}\\ \frac{1}{2} v_k (|y|) &{} \text { if } k \in {{\mathcal {S}}}^\mathrm{sym}. \end{array}\right. } \end{aligned}$$

  (2.4)

We say that $X_n$ is recurrent if $\liminf _{n \rightarrow \infty } X_n = 0$, a.s., and transient if $\lim _{n \rightarrow \infty } X_n = \infty$, a.s. An irreducibility argument shows that our Markov chain $(X_n,\xi _n)$ displays the usual recurrence/transience dichotomy and exactly one of these two situations holds; however, our proofs establish this behaviour directly using semimartingale arguments, and so we may avoid discussion of irreducibility here.

Throughout we define, for $k \in {{\mathcal {S}}}$,

$$\begin{aligned} \chi _k := \frac{ 1 + {{\mathbf {1}}}{\{ k \in {{\mathcal {S}}}^\mathrm{one}\}} }{2} = {\left\{ \begin{array}{ll} \frac{1}{2} &{} \text { if } k \in {{\mathcal {S}}}^\mathrm{sym};\\ 1 &{} \text { if } k \in {{\mathcal {S}}}^\mathrm{one}. \end{array}\right. } \end{aligned}$$

Our first main result gives a recurrence classification for the process.

Theorem 1

Suppose that (A0) and (A1) hold.

• (a) Suppose that $\max _{k \in {{\mathcal {S}}}} \chi _k \alpha _k \ge 1$. Then, $X_n$ is recurrent.

• (b) Suppose that $\max _{k \in {{\mathcal {S}}}} \chi _k \alpha _k < 1$.

  • (i) If $\sum _{k \in {{\mathcal {S}}}} \mu _k \cot ( \chi _k \pi \alpha _k ) < 0$, then $X_n$ is recurrent.

  • (ii) If $\sum _{k \in {{\mathcal {S}}}} \mu _k \cot ( \chi _k \pi \alpha _k ) > 0$, then $X_n$ is transient.

  • (iii) Suppose in addition that the densities $v_{k}$ of (A1) belong to ${{\mathfrak {D}}}_{\alpha _k,c_k}^+$ for each k. Then, $X_n$ is recurrent if $\sum _{k \in {{\mathcal {S}}}} \mu _k \cot ( \chi _k \pi \alpha _k ) = 0$.

In the recurrent cases, it is of interest to quantify recurrence via existence or non-existence of passage-time moments. For $a > 0$, let $\tau _a := \min \{ n \ge 0 : X_n \le a \}$, where throughout the paper we adopt the usual convention that $\min \emptyset := +\infty$. The next result shows that in all the recurrent cases, excluding the boundary case in Theorem 1(b)(iii), the tails of $\tau _a$ are polynomial.

Theorem 2

Suppose that (A0) and (A1) hold. In cases (a) and (b)(i) of Theorem 1, there exist $0< q_0 \le q_1 < \infty$ such that for all $x>a$ and all $k\in {{\mathcal {S}}}$,

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}_{x,k}\left[ \tau _a^{q} \right]< \infty , \quad \text {for } q<q_0 \quad \text { and } {{\mathrm{{\mathbb {E}}}}}_{x,k}\left[ \tau _a^{q} \right] = \infty , \quad \text {for } q>q_1. \end{aligned}$$

Remark 1

Our general results have $q_0 < q_1$ so that Theorem 2 does not give sharp estimates; these remain an open problem.

We do have sharp results in several particular cases for two half-lines, in which case our model reduces to the oscillating random walk considered by Kemperman [15] and others. We present these sharp moments results (Theorems 4 and 6) in the next section, which discusses in detail the case of the oscillating random walk, and also describes how our recurrence results relate to the known results for this classical model.

3 Oscillating Random Walks and Related Examples

3.1 Two Half-Lines Become One Line

In the case of our general model in which ${{\mathcal {S}}}$ consists of two elements, ${{\mathcal {S}}}= \{ -1, +1\}$, say, it is natural and convenient to represent our random walk on the whole real line ${{\mathbb {R}}}$. Namely, if $\omega (x, k) := k x$ for $x \in {{\mathbb {R}}}_+$ and $k = \pm 1$, we let $Z_n = \omega (X_n, \xi _n)$.

The simplest case has no reflection at the origin, only transmission, i.e. $p(i,j) = {{\mathbf {1}}}{\{ i \ne j \}}$, so that $\mu = (\frac{1}{2}, \frac{1}{2})$. Then, for $B \subseteq {{\mathbb {R}}}$ a Borel set,

$$\begin{aligned} {{\mathbb {P}}}\left[ Z_{n+1} \in B \mid (X_n, \xi _n ) = (x, i) \right]&= {{\mathbb {P}}}_{x,i} \left[ (X_{1}, \xi _{1} ) \in B^+ \times \{ +1 \} \right] \\&\quad + {{\mathbb {P}}}_{x,i} \left[ (X_{1}, \xi _{1} ) \in B^- \times \{ -1 \} \right] , \end{aligned}$$

where $B^+ = B \cap {{\mathbb {R}}}_+$ and $B^- = \{ - x : x \in B, \, x <0 \}$. In particular, by (2.1), writing $w_+$ for $w_{+1}$, for $x \in {{\mathbb {R}}}_+$,

$$\begin{aligned} {{\mathbb {P}}}\left[ Z_{n+1} \in B \mid (X_n, \xi _n ) = (x, +1) \right]&= \int _{B^+} w_{+} (z-x ) {\mathrm d}z + \int _{B^-} w_{+} (-z-x) {\mathrm d}z \\&= \int _B w_{+} (z-x) {\mathrm d}z , \end{aligned}$$

and, similarly, writing $w_-(y)$ for $w_{-1}(-y)$, for $x \in {{\mathbb {R}}}_+$,

$$\begin{aligned} {{\mathbb {P}}}\left[ Z_{n+1} \in B \mid (X_n, \xi _n ) = (x, -1) \right] = \int _B w_{-} (z+x) {\mathrm d}z. \end{aligned}$$

For $x \ne 0$, $\omega$ is invertible with

$$\begin{aligned} \omega ^{-1} (x) = {\left\{ \begin{array}{ll} \left( |x| , +1 \right) &{} \text { if } x > 0 \\ \left( |x| , -1\right) &{} \text { if } x < 0 ,\end{array}\right. } \end{aligned}$$

and hence we have for $x \in {{\mathbb {R}}}{\setminus } \{ 0 \}$ and Borel $B \subseteq {{\mathbb {R}}}$,

$$\begin{aligned} {{\mathbb {P}}}\left[ Z_{n+1} \in B \mid Z_n = x \right] = {\left\{ \begin{array}{ll} \int _B w_{+} (z-x) {\mathrm d}z &{} \text { if } x > 0 \\ \int _B w_{-} (z-x) {\mathrm d}z &{} \text { if } x < 0 . \end{array}\right. } \end{aligned}$$

(3.1)

We may make an arbitrary non-trivial choice for the transition law at $Z_n = 0$ without affecting the behaviour of the process, and then, (3.1) shows that $Z_n$ is a time-homogeneous Markov process on ${{\mathbb {R}}}$. Now, $Z_n$ is recurrent if $\liminf _{n \rightarrow \infty } | Z_n | = 0$, a.s., or transient if $\lim _{n \rightarrow \infty } | Z_n | = \infty$, a.s. The one-dimensional case described at (3.1) has received significant attention over the years. We describe several of the classical models that have been considered.

3.2 Examples and Further Results

3.2.1 Homogeneous Symmetric Random Walk

The most classical case is as follows.

• (Sym) Let $\alpha \in (0,\infty )$. For $v \in {{\mathfrak {D}}}_{\alpha ,c}$, suppose that $w_+(y) = w_-(y) = \frac{1}{2} v ( |y| )$ for $y\in {{\mathbb {R}}}$.

In this case, $Z_n$ describes a random walk with i.i.d. symmetric increments.

Theorem 3

Suppose that (Sym) holds. Then, the symmetric random walk is transient if $\alpha < 1$ and recurrent if $\alpha > 1$. If, in addition, $v \in {{\mathfrak {D}}}^+_{\alpha ,c}$, then the case $\alpha =1$ is recurrent.

Theorem 3 follows from our Theorem 1, since in this case

$$\begin{aligned} \sum _{k \in {{\mathcal {S}}}} \mu _k \cot \left( \chi _k \pi \alpha _k \right) = \cot ( \pi \alpha / 2) . \end{aligned}$$

Since it deals with a sum of i.i.d. random variables, Theorem 3 may be deduced from the classical theorem of Chung and Fuchs [7], via, e.g. the formulation of Shepp [23]. The method of the present paper provides an alternative to the classical (Fourier analytic) approach that generalizes beyond the i.i.d. setting (note that Theorem 3 is not, formally, a consequence of Shepp’s most accessible result, Theorem 5 of [23], since v does not necessarily correspond to a unimodal distribution in Shepp’s sense).

With $\tau _a$ as defined previously, in the setting of the present section we have $\tau _a= \min \{ n \ge 0 : | Z_n | \le a \}$. Use ${{\mathrm{{\mathbb {E}}}}}_x [ \, \cdot \, ]$ as shorthand for ${{\mathrm{{\mathbb {E}}}}}[ \, \cdot \, \mid Z_0 = x]$. We have the following result on existence of passage-time moments, whose proof is in Sect. 6; while part (i) is well known, we could find no reference for part (ii).

Theorem 4

Suppose that (Sym) holds, and that $\alpha > 1$. Let $x \notin [ -a, a]$.

• (i) If $\alpha \ge 2$, then ${{\mathrm{{\mathbb {E}}}}}_{x} \left[ \tau _a^q \right] < \infty$ if $q < 1/2$ and ${{\mathrm{{\mathbb {E}}}}}_{x} \left[ \tau _a^q \right] = \infty$ if $q > 1/2$.

• (ii) If $\alpha \in (1,2)$, then ${{\mathrm{{\mathbb {E}}}}}_x \left[ \tau _a^q \right] < \infty$ if $q < 1 - \frac{1}{\alpha }$ and ${{\mathrm{{\mathbb {E}}}}}_{x} \left[ \tau _a^q \right] = \infty$ if $q > 1 - \frac{1}{\alpha }$.

Our main interest concerns spatially inhomogeneous models, i.e. in which $w_x$ depends on x, typically only through ${{\mathrm{sgn}}}x$, the sign of x. Such models are known as oscillating random walks and were studied by Kemperman [15], to whom the model was suggested in 1960 by Anatole Joffe and Peter Ney (see [15, p. 29]).

3.2.2 One-Sided Oscillating Random Walk

The next example, following [15], is a one-sided oscillating random walk:

• (Osc1) Let $\alpha , \beta \in (0,\infty )$. For $v_+ \in {{\mathfrak {D}}}_{\alpha ,c_+}$ and $v_- \in {{\mathfrak {D}}}_{\beta ,c_-}$, suppose that

  $$\begin{aligned} w_+ ( y ) = v_+ (-y) , \quad \text {and}\quad w_- (y ) = v_- (y) . \end{aligned}$$

In other words, the walk always jumps in the direction of (and possibly over) the origin, with tail exponent $\alpha$ from the positive half-line and exponent $\beta$ from the negative half-line. The following recurrence classification applies.

Theorem 5

Suppose that (Osc1) holds. Then, the one-sided oscillating random walk is transient if $\alpha + \beta < 1$ and recurrent if $\alpha + \beta > 1$. If, in addition, $v_+ \in {{\mathfrak {D}}}^+_{\alpha ,c_+}$ and $v_- \in {{\mathfrak {D}}}^+_{\beta ,c_-}$, then the case $\alpha + \beta = 1$ is recurrent.

Theorem 5 was obtained in the discrete-space case by Kemperman [15, p. 21]; it follows from our Theorem 1, since in this case

$$\begin{aligned} \sum _{k \in {{\mathcal {S}}}} \mu _k \cot \left( \chi _k \pi \alpha _k \right) = \frac{1}{2} \cot \left( \pi \alpha \right) + \frac{1}{2} \cot \left( \pi \beta \right) = \frac{\sin \left( \pi \left( \alpha + \beta \right) \right) }{ 2 \sin \left( \pi \alpha \right) \sin \left( \pi \beta \right) } . \end{aligned}$$

The special case of (Osc1) in which $\alpha = \beta$ was called antisymmetric by Kemperman; here, Theorem 5 shows that the walk is transient for $\alpha < 1/2$ and recurrent for $\alpha > 1/2$. We have the following moments result, proved in Sect. 6.

Theorem 6

Suppose that (Osc1) holds, and that $\alpha = \beta > 1/2$. Let $x \notin [ -a, a]$.

• (i) If $\alpha \ge 1$, then ${{\mathrm{{\mathbb {E}}}}}_x [ \tau _a^q ] < \infty$ if $q < \alpha$ and ${{\mathrm{{\mathbb {E}}}}}_x [ \tau _a^q ] = \infty$ if $q > \alpha$.

• (ii) If $\alpha \in (1/2,1)$, then ${{\mathrm{{\mathbb {E}}}}}_x [ \tau _a^q ] < \infty$ if $q < 2 - \frac{1}{\alpha }$ and ${{\mathrm{{\mathbb {E}}}}}_x [ \tau _a^q ] = \infty$ if $q > 2 - \frac{1}{\alpha }$.

Problem 1

Obtain sharp moments results for general $\alpha , \beta \in (0,1)$.

3.2.3 Two-Sided Oscillating Random Walk

Another model in the vein of [15] is a two-sided oscillating random walk:

• (Osc2) Let $\alpha , \beta \in (0,\infty )$. For $v_+ \in {{\mathfrak {D}}}_{\alpha ,c_+}$ and $v_- \in {{\mathfrak {D}}}_{\beta ,c_-}$, suppose that

  $$\begin{aligned} w_+ ( y ) = \frac{1}{2} v_+ (|y|) , ~~ \text {and}~~ w_- (y ) = \frac{1}{2} v_- (|y|) . \end{aligned}$$

Now, the jumps of the walk are symmetric, as under (Sym), but with a tail exponent depending upon which side of the origin the walk is currently on, as under (Osc1).

The most general recurrence classification result for the model (Osc2) is due to Sandrić [21]. A somewhat less general, discrete-space version was obtained by Rogozin and Foss (Theorem 2 of [17, p. 159]), building on [15]. Analogous results in continuous time were given in [4, 11]. Here is the result.

Theorem 7

Suppose that (Osc2) holds. Then, the two-sided oscillating random walk is transient if $\alpha + \beta < 2$ and recurrent if $\alpha + \beta > 2$. If, in addition, $v_+ \in {{\mathfrak {D}}}^+_{\alpha ,c_+}$ and $v_- \in {{\mathfrak {D}}}^+_{\beta ,c_-}$, then the case $\alpha + \beta = 2$ is recurrent.

Theorem 7 also follows from our Theorem 1, since in this case

$$\begin{aligned} \sum _{k \in {{\mathcal {S}}}} \mu _k \cot ( \chi _k \pi \alpha _k ) = \frac{1}{2} \cot ( \pi \alpha /2 ) + \frac{1}{2} \cot (\pi \beta /2 ) = \frac{\sin (\pi ( \alpha + \beta ) /2 )}{ 2 \sin (\pi \alpha /2 ) \sin (\pi \beta /2 ) } . \end{aligned}$$

Problem 2

Obtain sharp moments results for $\alpha , \beta \in (0,2)$.

3.2.4 Mixed Oscillating Random Walk

A final model is another oscillating walk that mixes the one- and two-sided models:

• (Osc3) Let $\alpha , \beta \in (0,\infty )$. For $v_+ \in {{\mathfrak {D}}}_{\alpha ,c_+}$ and $v_- \in {{\mathfrak {D}}}_{\beta ,c_-}$, suppose that

  $$\begin{aligned} w_+ ( y ) = \frac{1}{2} v_+ (|y|) , ~~ \text {and}~~ w_- (y ) = v_- ( y ) . \end{aligned}$$

In the discrete-space case, Theorem 2 of Rogozin and Foss [17, p. 159] gives the recurrence classification.

Theorem 8

Suppose that (Osc3) holds. Then, the mixed oscillating random walk is transient if $\alpha + 2 \beta < 2$ and recurrent if $\alpha + 2 \beta > 2$. If, in addition, $v_+ \in {{\mathfrak {D}}}^+_{\alpha ,c_+}$ and $v_- \in {{\mathfrak {D}}}^+_{\beta ,c_-}$, then the case $\alpha + 2 \beta = 2$ is recurrent.

Theorem 8 also follows from our Theorem 1, since in this case

$$\begin{aligned} \sum _{k \in {{\mathcal {S}}}} \mu _k \cot ( \chi _k \pi \alpha _k ) = \frac{1}{2} \cot ( \pi \alpha /2 ) + \frac{1}{2} \cot (\pi \beta ) = \frac{\sin (\pi ( \alpha + 2 \beta ) /2 )}{ 2 \sin (\pi \alpha /2 ) \sin (\pi \beta ) } . \end{aligned}$$

3.3 Additional Remarks

It is possible to generalize the model further by permitting the local transition density to vary within each half-line. Then, we have the transition kernel

$$\begin{aligned} {{\mathbb {P}}}[ Z_{n+1} \in B \mid Z_n = x ] = \int _B w_x (z -x ) {\mathrm d}z, \end{aligned}$$

(3.2)

for all Borel sets $B \subseteq {{\mathbb {R}}}$. Here, the local transition densities $w_x : {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}_+$ are Borel functions. Variations of the oscillating random walk, within the general setting of (3.2), have also been studied in the literature. Sandrić [19, 21] supposes that the $w_x$ satisfy, for each $x \in {{\mathbb {R}}}$, $w_x (y) \sim c(x) | y |^{-1-\alpha (x)}$ as $|y| \rightarrow \infty$ for some measurable functions c and $\alpha$; he refers to this as a stable-like Markov chain. Under a uniformity condition on the $w_x$, and other mild technical conditions, Sandrić [19] obtained, via Foster–Lyapunov methods similar in spirit to those of the present paper, sufficient conditions for recurrence and transience: essentially $\liminf _{x \rightarrow \infty } \alpha (x) >1$ is sufficient for recurrence and $\limsup _{x \rightarrow \infty } \alpha (x) <1$ is sufficient for transience. These results can be seen as a generalization of Theorem 3. Some related results for models in continuous-time (Lévy processes) are given in [20, 22, 24]. Further results and an overview of the literature are provided in Sandrić’s Ph.D. thesis [18].

4 Many-Dimensional Random Walks

The next two examples show how versions of the oscillating random walk of Sect. 3 arise as embedded Markov chains in certain two-dimensional random walks.

Example 1

Consider $\xi _n =(\xi _n^{(1)}, \xi _n^{(2)})$, $n \in {{\mathbb {Z}}}_+$, a nearest-neighbour random walk on ${{\mathbb {Z}}}^2$ with transition probabilities

$$\begin{aligned} {{\mathbb {P}}}\left[ \xi _{n+1} = (y_1, y_2 ) \mid \xi _n = (x_1, x_2 ) \right] = p \left( x_1, x_2 ; y_1, y_2 \right) . \end{aligned}$$

Suppose that the probabilities are given for $x_2 \ne 0$ by,

$$\begin{aligned} p \left( x_1, x_2 ; x_1 , x_2 + 1 \right)&= p \left( x_1, x_2 ; x_1 , x_2 - 1 \right) = \frac{1}{3} ; \nonumber \\ p \left( x_1, x_2 ; x_1 + 1, x_2 \right)&= \frac{1}{3} {{\mathbf {1}}}{\{ x_2 < 0 \}}; \nonumber \\ p \left( x_1, x_2 ; x_1 - 1, x_2 \right)&= \frac{1}{3} {{\mathbf {1}}}{\{ x_2 > 0 \}}; \end{aligned}$$

(4.1)

(the rest being zero) and for $x_2 = 0$ by $p (x_1, 0 ; x_1 , 1 ) = 1$ for all $x_1 > 0$, $p (x_1, 0 ; x_1 , -1 ) = 1$ for all $x_1 < 0$, and $p (0, 0 ; 0 , 1 ) = p (0, 0 ; 0 , -1 ) = 1/2$. See Fig. 1 for an illustration.

Fig. 1

Pictorial representation of the non-homogeneous nearest-neighbour random walk on ${{\mathbb {Z}}}^2$ of Example 1, plus a simulated trajectory of 5000 steps of the walk. We conjecture that the walk is recurrent

Set $\tau _0 :=0$ and define recursively $\tau _{k+1} = \min \{ n > \tau _k : \xi ^{(2)}_n = 0 \}$ for $k \ge 0$; consider the embedded Markov chain $X_n = \xi ^{(1)}_{\tau _n}$. We show that $X_n$ is a discrete version of the oscillating random walk described in Sect. 3. Indeed, $| \xi ^{(2)}_n |$ is a reflecting random walk on ${{\mathbb {Z}}}_+$ with increments taking values $-1, 0, +1$ each with probability 1 / 3. We then (see, e.g. [9, p. 415]) have that for some constant $c \in (0,\infty )$,

$$\begin{aligned} {{\mathbb {P}}}\left[ \tau _1 > r \right] = \left( c + o(1) \right) r^{-1/2}, \quad \text {as}\quad r \rightarrow \infty . \end{aligned}$$

Suppose that $\xi ^{(1)}_0 = x >0$. Since between times $\tau _0$ and $\tau _1$, $\xi ^{(1)}_n$ is monotone, we have

$$\begin{aligned} {{\mathbb {P}}}\left[ \tau _1> 3r + r^{3/4} \right] = \,&{{\mathbb {P}}}\left[ \tau _1> 3r + r^{3/4} , \xi _{\tau _1} - \xi _{\tau _0}< -r \right] \\&+ {{\mathbb {P}}}\left[ \tau _1 > 3r + r^{3/4} , \xi _{\tau _1} - \xi _{\tau _0} \ge -r \right] \\ \le \,&{{\mathbb {P}}}\left[ \xi _{\tau _1} - \xi _{\tau _0} < -r \right] + {{\mathbb {P}}}\left[ \xi _{{\lceil 3r + r^{3/4}\rceil }} - \xi _{\tau _0} \ge -r \right] . \end{aligned}$$

In other words,

$$\begin{aligned} {{\mathbb {P}}}\left[ \xi _{\tau _1}^{(1)} - \xi _{\tau _0}^{(1)} < -r \right] \ge {{\mathbb {P}}}\left[ \tau _1 > 3r + r^{3/4} \right] - {{\mathbb {P}}}\left[ \xi ^{(1)}_{\lceil 3r + r^{3/4}\rceil } - \xi ^{(1)}_0 \ge -r \right] . \end{aligned}$$

Here, $\xi ^{(1)}_n + \frac{n}{3}$ is a martingale with bounded increments, so, by the Azuma–Hoeffding inequality, for some $\varepsilon >0$ and all $r \ge 1$,

$$\begin{aligned} {{\mathbb {P}}}\left[ \xi ^{(1)}_{\lceil 3r + r^{3/4} \rceil } - \xi ^{(1)}_0 \ge -r \right]&\le {{\mathbb {P}}}\left[ \xi ^{(1)}_{\lceil 3r + r^{3/4} \rceil } + \tfrac{1}{3} { \lceil 3r + r^{3/4} \rceil } - \xi ^{(1)}_0 \ge \tfrac{1}{3} r^{3/4} \right] \\&\le \exp \left\{ - \varepsilon r^{1/2} \right\} . \end{aligned}$$

Similarly,

$$\begin{aligned} {{\mathbb {P}}}\left[ \xi _{\tau _1}^{(1)} - \xi _{\tau _0}^{(1)} < -r \right] \le {{\mathbb {P}}}\left[ \tau _1 > 3r - r^{3/4} \right] + {{\mathbb {P}}}\left[ \xi ^{(1)}_{\lceil 3r- r^{3/4} \rceil } - \xi ^{(1)}_0 \le -r \right] , \end{aligned}$$

where, by Azuma–Hoeffding once more,

$$\begin{aligned} {{\mathbb {P}}}\left[ \xi ^{(1)}_{\lceil 3r- r^{3/4} \rceil } - \xi ^{(1)}_0 \le -r \right]&\le {{\mathbb {P}}}\left[ \xi ^{(1)}_{\lceil 3r- r^{3/4} \rceil } + \tfrac{1}{3} {\lceil 3r- r^{3/4} \rceil } - \xi ^{(1)}_0 \le -\tfrac{1}{3} r^{3/4} + 1 \right] \\&\le \exp \left\{ - \varepsilon r^{1/2} \right\} . \end{aligned}$$

Combining these bounds, and using the symmetric argument for $\{ \xi _{\tau _1}^{(1)} > r\}$ when $\xi ^{(1)}_0 = x <0$, we see that for $r > 0$,

$$\begin{aligned} {{\mathbb {P}}}\left[ X_{n+1} - X_n< -r \mid X_n = x \right]&= u(r), \quad \text {if } x> 0, \quad \text { and } \nonumber \\ {{\mathbb {P}}}\left[ X_{n+1} - X_n > r \mid X_n = x \right]&= u(r), \quad \text {if } x < 0 , \end{aligned}$$

(4.2)

where $u(r) = ( c + o(1) ) r^{-1/2}$. Thus, $X_n$ satisfies a discrete-space analogue of (Osc1) with $\alpha = \beta = 1/2$. This is the critical case identified in Theorem 5, but that result does not cover this case due to the rate of convergence estimate for u; a finer analysis is required. We conjecture that the walk is recurrent.

Example 2

We present two variations on the previous example, which are superficially similar but turn out to be less delicate. First, modify the random walk of the previous example by supposing that (4.1) holds but replacing the behaviour at $x_2 = 0$ by $p (x_1, 0 ; x_1 , 1 ) = p (x_1, 0 ; x_1, -1 ) = 1/2$ for all $x_1 \in {{\mathbb {Z}}}$. See the left-hand part of Fig. 2 for an illustration.

The embedded process $X_n$ now has, for all $x \in {{\mathbb {Z}}}$ and for $r \ge 0$,

$$\begin{aligned} {{\mathbb {P}}}\left[ X_{n+1} - X_n < -r \mid X_n = x \right] = {{\mathbb {P}}}\left[ X_{n+1} - X_n > r \mid X_n = x \right] = u(r),\qquad \end{aligned}$$

(4.3)

where $u(r) = (c/2) ( 1 + o(1) ) r^{-1/2}$. Thus, $X_n$ is a random walk with symmetric increments, and the discrete version of our Theorem 3 (and also a result of [23]) implies that the walk is transient. This walk was studied by Campanino and Petritis [5, 6], who proved transience via different methods.

Fig. 2

Pictorial representation of the two non-homogeneous nearest-neighbour random walks on ${{\mathbb {Z}}}^2$ of Example 2. Each of these walks is transient

Next, modify the random walk of Example 1 by supposing that (4.1) holds but replacing the behaviour at $x_2 = 0$ by $p (x_1, 0 ; x_1 , 1 ) = p (x_1, 0 ; x_1, -1 ) = 1/2$ if $x_1 \ge 0$, and $p(x_1, 0 ; x_1 , -1) = 1$ for $x_1 < 0$. See the right-hand part of Fig. 2 for an illustration. This time the walk takes a symmetric increment as at (4.3) when $x \ge 0$ but a one-sided increment as at (4.2) when $x < 0$. In this case, the discrete version of our Theorem 8 (and also a result of [17]) shows that the walk is transient.

One may obtain the general model on ${{\mathbb {Z}}}_+\times {{\mathcal {S}}}$ as an embedded process for a random walk on complexes of half-spaces, generalizing the examples considered here; we leave this to the interested reader.

5 Recurrence Classification in the Non-critical Cases

5.1 Lyapunov Functions

Our proofs are based on demonstrating appropriate Lyapunov functions; that is, for suitable $\varphi : {{\mathbb {R}}}_+\times {{\mathcal {S}}}\rightarrow {{\mathbb {R}}}_+$ we study $Y_n = \varphi ( X_n , \xi _n )$ such that $Y_n$ has appropriate local supermartingale or submartingale properties for the one-step mean increments

$$\begin{aligned} D\varphi (x,i)&:= {{\mathrm{{\mathbb {E}}}}}\left[ \varphi \left( X_{n+1}, \xi _{n+1} \right) - \varphi \left( X_n, \xi _n \right) \mid \left( X_n, \xi _n \right) = (x, i) \right] \\&= {{\mathrm{{\mathbb {E}}}}}_{x, i} \left[ \varphi (X_{1}, \xi _{1} ) - \varphi \left( X_0, \xi _0 \right) \right] . \end{aligned}$$

First, we note some consequences of the transition law (2.1). Let $\varphi : {{\mathbb {R}}}_+\times {{\mathcal {S}}}\rightarrow {{\mathbb {R}}}_+$ be measurable. Then, we have from (2.1) that, for $(x, i) \in {{\mathbb {R}}}_+\times {{\mathcal {S}}}$,

$$\begin{aligned} D\varphi (x,i) =&\sum _{j \in {{\mathcal {S}}}} p (i,j) \int _{-\infty }^{-x} \left( \varphi ( -x-y , j) - \varphi (x,i) \right) w_i (y) {\mathrm d}y \nonumber \\&+ \int _{-x}^\infty \left( \varphi (x+y, i) - \varphi (x,i) \right) w_i (y) {\mathrm d}y. \end{aligned}$$

(5.1)

Our primary Lyapunov function is roughly of the form $x \mapsto | x| ^\nu$, $\nu \in {{\mathbb {R}}}$, but weighted according to an ${{\mathcal {S}}}$-dependent component (realized by a collection of multiplicative weights $\lambda _k$); these weights provide a crucial technical tool.

For $\nu \in {{\mathbb {R}}}$ and $x \in {{\mathbb {R}}}$, we write

$$\begin{aligned} f_\nu (x) := \left( 1 + |x| \right) ^\nu . \end{aligned}$$

Then, for parameters $\lambda _k > 0$ for each $k \in {{\mathcal {S}}}$, define for $x \in {{\mathbb {R}}}_+$ and $k \in {{\mathcal {S}}}$,

$$\begin{aligned} f_\nu (x , k ) := \lambda _k f_\nu (x) = \lambda _k (1+x)^\nu . \end{aligned}$$

(5.2)

Now, for this Lyapunov function, (5.1) gives

$$\begin{aligned} Df_\nu (x, i) =&\sum _{j \in {{\mathcal {S}}}} p (i,j) \int _{-\infty }^{-x} \left( \lambda _j f_\nu (x+y) - \lambda _i f_\nu (x) \right) w_i (y) {\mathrm d}y \nonumber \\&+ \lambda _i \int _{-x}^\infty \left( f_\nu ( x+ y) - f_\nu (x) \right) w_i (y) {\mathrm d}y. \end{aligned}$$

(5.3)

Depending on whether $i \in {{\mathcal {S}}}^\mathrm{sym}$ or $i \in {{\mathcal {S}}}^\mathrm{one}$, the above integrals can be expressed in terms of $v_i$ as follows. For $i\in {{\mathcal {S}}}^\mathrm{sym}$,

$$\begin{aligned} Df_\nu (x, i) =&\sum _{j \in {{\mathcal {S}}}} p (i,j) \frac{\lambda _j}{2} \int _{x}^\infty f_\nu (y-x) v_i (y) {\mathrm d}y - \frac{\lambda _i}{2} \int _x^\infty f_\nu (x) v_i (y) {\mathrm d}y \nonumber \\&+ \frac{\lambda _i}{2} \int _{0}^x \left( f_\nu ( x+ y) + f_\nu (x-y) - 2f_\nu (x) \right) v_i ( y ) {\mathrm d}y \nonumber \\&+ \frac{\lambda _i}{2} \int _x^\infty \left( f_\nu ( x+ y) - f_\nu (x) \right) v_i (y) {\mathrm d}y . \end{aligned}$$

(5.4)

For $i\in {{\mathcal {S}}}^\mathrm{one}$,

$$\begin{aligned} Df_\nu (x, i) =&\sum _{j \in {{\mathcal {S}}}} p (i,j) \lambda _j \int _x^\infty f_\nu (y-x) v_i (y) {\mathrm d}y - \lambda _i \int _x^\infty f_\nu (x) v_i (y) {\mathrm d}y \nonumber \\&+\lambda _i \int _{0}^x \left( f_\nu ( x - y) - f_\nu (x) \right) v_i (y) {\mathrm d}y. \end{aligned}$$

(5.5)

5.2 Estimates of Functional Increments

In the course of our proofs, we need various integral estimates that can be expressed in terms of classical transcendental functions. For the convenience of the reader, we gather all necessary integrals in Lemmas 1 and 2; the proofs of these results are deferred until the Appendix. Recall that the Euler gamma function ${\varGamma }$ satisfies the functional equation $z {\varGamma }(z) = {\varGamma }(z+1)$, and the hypergeometric function ${}_m F_n$ is defined via a power series (see [1]).

Lemma 1

Suppose that $\alpha > 0$ and $-1<\nu <\alpha$. Then,

$$\begin{aligned} i_0^{\nu ,\alpha }&:=\int _1^\infty \frac{(1+u)^\nu -1}{u^{1+\alpha }} {\mathrm d}u= \frac{1}{\alpha -\nu } \ {}_2 F_1 \left( -\nu , \alpha -\nu ; \alpha -\nu +1 ; -1\right) -\frac{1}{\alpha };\\ i_{2,0}^\alpha&:=\int _1^\infty \frac{1}{u^{1+\alpha }} {\mathrm d}u= \frac{1}{\alpha }; \\ i_{2,1}^{\nu ,\alpha }&:= \int _1^\infty \frac{(u-1)^\nu }{u^{1+\alpha }} {\mathrm d}u = \frac{{\varGamma }(1+\nu ){\varGamma }(\alpha -\nu )}{{\varGamma }(1+\alpha )}. \end{aligned}$$

Suppose that $\alpha \in (0,2)$ and $\nu > -1$. Then,

$$\begin{aligned} i_1^{\nu ,\alpha } :=&\int _0^1 \frac{(1+u)^\nu +(1-u)^\nu -2 }{u^{1+\alpha }} {\mathrm d}u \\ =&\,\frac{\nu (\nu -1)}{2-\alpha } \ {}_4 F_3 \left( 1,1-\tfrac{\nu }{2},1-\tfrac{\alpha }{2}, \tfrac{3-\nu }{2} ; \tfrac{3}{2},2, 2-\tfrac{\alpha }{2} ; 1\right) . \end{aligned}$$

Suppose that $\alpha \in (0,1)$ and $\nu > -1$. Then,

$$\begin{aligned} {\tilde{i}}_1^{\nu ,\alpha }:= \int _0^1 \frac{(1-u)^\nu -1 }{u^{1+\alpha }} {\mathrm d}u= \frac{1}{\alpha } \left( 1 - \frac{{\varGamma }(1+\nu ){\varGamma }(1-\alpha )}{{\varGamma }(1-\alpha +\nu )}\right) . \end{aligned}$$

Recall that the digamma function is $\psi (z) = \frac{{\mathrm d}}{{\mathrm d}z} \log {\varGamma }(z) = {\varGamma }' (z) / {\varGamma }(z)$, which has $\psi (1) = - \gamma$ where $\gamma \approx 0.5772$ is Euler’s constant.

Lemma 2

Suppose that $\alpha >0$. Then,

$$\begin{aligned} j_0^\alpha&:=\int _1^\infty \frac{\log (1+u)}{u^{1+\alpha }} {\mathrm d}u= \frac{1}{\alpha }\left( \psi (\alpha )-\psi \left( \tfrac{\alpha }{2}\right) \right) ; \\ j_2^\alpha&:=\int _1^\infty \frac{\log (u-1)}{u^{1+\alpha }} {\mathrm d}u= -\frac{1}{\alpha }(\gamma +\psi (\alpha )). \end{aligned}$$

Suppose that $\alpha \in (0,2)$. Then,

$$\begin{aligned} j_1^\alpha :=\int _0^1 \frac{\log (1-u^2)}{u^{1+\alpha }} {\mathrm d}u= \frac{1}{\alpha }\left( \gamma +\psi \left( 1-\tfrac{\alpha }{2} \right) \right) . \end{aligned}$$

Suppose that $\alpha \in (0,1)$. Then,

$$\begin{aligned} {\tilde{j}}_1^\alpha :=\int _0^1 \frac{\log (1-u)}{u^{1+\alpha }} {\mathrm d}u= \frac{1}{\alpha }\left( \gamma +\psi (1-\alpha )\right) . \end{aligned}$$

Remark 2

The j integrals can be obtained as derivatives with respect to $\nu$ of the i integrals, evaluated at $\nu =0$.

The next result collects estimates for our integrals in the expected functional increments (5.4) and (5.5) in terms of the integrals in Lemma 1.

Lemma 3

Suppose that $v \in {{\mathfrak {D}}}_{\alpha ,c}$. For $\alpha >0$ and $-1< \nu < \alpha$ we have

$$\begin{aligned} \int _x^\infty f_\nu ( y -x ) v (y) {\mathrm d}y&= c x^{\nu - \alpha } i^{\nu , \alpha }_{2,1} + o \left( x^{\nu - \alpha } \right) ; \end{aligned}$$

(5.6)

$$\begin{aligned} \int _x^\infty f_\nu (x) v (y) {\mathrm d}y&= c x^{\nu -\alpha } i_{2,0}^{\nu ,\alpha } + o \left( x^{\nu - \alpha } \right) ; \end{aligned}$$

(5.7)

$$\begin{aligned} \int _x^\infty \left( f_\nu (x+y) - f_\nu (x) \right) v (y) {\mathrm d}y&= c x^{\nu -\alpha } i_{0}^{\nu ,\alpha }+ o \left( x^{\nu - \alpha } \right) . \end{aligned}$$

(5.8)

For $\alpha \in (0,2)$ and $\nu > -1$ we have

$$\begin{aligned} \int _{0}^x \left( f_\nu (x+y) + f_\nu (x-y) - 2 f_\nu (x) \right) v ( y ) {\mathrm d}y = c x^{\nu - \alpha } i_1^{\nu ,\alpha } + o \left( x^{\nu - \alpha } \right) . \end{aligned}$$

(5.9)

For $\alpha \in (0,1)$ and $\nu > -1$ we have

$$\begin{aligned} \int _{0}^x \left( f_\nu (x-y) - f_\nu (x) \right) v (y) {\mathrm d}y = c x^{\nu - \alpha } {{\tilde{i}}}_1^{\nu ,\alpha } + o \left( x^{\nu - \alpha } \right) . \end{aligned}$$

(5.10)

Moreover, if $v \in {{\mathfrak {D}}}_{\alpha ,c}^+$ then stronger versions of all of the above estimates hold with $o (x^{\nu - \alpha } )$ replaced by $O (x^{\nu - \alpha -\delta } )$ for some $\delta >0$.

Proof

These estimates are mostly quite straightforward, so we do not give all the details. We spell out the estimate in (5.6); the others are similar. We have

$$\begin{aligned} \int _x^\infty f_\nu ( y -x ) v (y) {\mathrm d}y&= x^\nu \int _x^\infty \left( \tfrac{1+ y}{x} - 1 \right) ^\nu c(y) y^{-1-\alpha } {\mathrm d}y . \end{aligned}$$

With the substitution $u = \frac{1+y}{x}$, this last expression becomes

$$\begin{aligned} x^{1+\nu } \int _{\frac{1+x}{x}}^\infty ( u -1 )^\nu u^{-1-\alpha } (x - u^{-1} )^{-1-\alpha } c ( u x -1 ) {\mathrm d}u . \end{aligned}$$

Let $\varepsilon \in (0,c)$. Then, there exists $y_0 \in {{\mathbb {R}}}_+$ such that $| c(y) - c | < \varepsilon$ for all $y \ge y_0$, so that $| c ( ux -1 ) - c | < \varepsilon$ for all u in the range of integration, provided $x \ge y_0$. Writing

$$\begin{aligned} f(u) = (u-1)^\nu u^{-1-\alpha }, \quad \text {and}\quad g(u) = \left( x - u^{-1} \right) ^{-1-\alpha } c ( ux -1 ) , \end{aligned}$$

for the duration of the proof, we have that

$$\begin{aligned} \int _x^\infty f_\nu ( y -x ) v (y) {\mathrm d}y = x^{1+\nu } \int _{\frac{1+x}{x}}^\infty f(u) g(u) {\mathrm d}u . \end{aligned}$$

(5.11)

For $u \ge \frac{1+x}{x}$ and $x \ge y_0$, we have

$$\begin{aligned} g_- := (c-\varepsilon ) x^{-1-\alpha } \le g(u) \le (c+\varepsilon ) (x-1)^{-1-\alpha } =: g_+ , \end{aligned}$$

so that $g_+ - g_- \le 2 \varepsilon (x-1)^{-1-\alpha } + C_1 x^{-2-\alpha }$ for a constant $C_1 < \infty$ not depending on $x \ge y_0$ or $\varepsilon$. Moreover, it is easy to see that $\int _1^\infty | f(u) | {\mathrm d}u \le C_2$ for a constant $C_2$ depending only on $\nu$ and $\alpha$, provided $\nu \in (-1,\alpha )$. Hence, Lemma 18 shows that

$$\begin{aligned} \left| \int _{\frac{1+x}{x}}^\infty f(u) g(u) {\mathrm d}u - (c-\varepsilon ) x^{-1-\alpha } \int _{\frac{1+x}{x}}^\infty f(u) {\mathrm d}u \right| \le 2 C_2 \varepsilon (x-1)^{-1-\alpha } + C_1 C_2 x^{-2-\alpha } , \end{aligned}$$

for all $x \ge y_0$. Since also

$$\begin{aligned} \left| \int _{\frac{1+x}{x}}^\infty f(u) {\mathrm d}u - i^{\nu ,\alpha }_{2,1} \right| \le \int _1^{\frac{1+x}{x}} | f(u) | {\mathrm d}u \rightarrow 0 , \end{aligned}$$

as $x \rightarrow \infty$, it follows that for any $\varepsilon >0$ we may choose x sufficiently large so that

$$\begin{aligned} \left| \int _{\frac{1+x}{x}}^\infty f(u) g(u) {\mathrm d}u - c x^{-1-\alpha } i^{\nu ,\alpha }_{2,1} \right| \le \varepsilon x^{-1-\alpha } , \end{aligned}$$

and since $\varepsilon >0$ was arbitrary, we obtain (5.6) from (5.11). $\square$

We also need the following simple estimates for ranges of $\alpha$ when the asymptotics for the final two integrals in Lemma 3 are not valid.

Lemma 4

Suppose that $v \in {{\mathfrak {D}}}_{\alpha ,c}$.

• (i) For $\alpha \ge 2$ and any $\nu \in (0,1)$, there exist $\varepsilon >0$ and $x_0 \in {{\mathbb {R}}}_+$ such that, for all $x \ge x_0$,

  $$\begin{aligned} \int _{0}^x \left( f_\nu (x+y) + f_\nu (x-y) - 2 f_\nu (x) \right) v ( y ) {\mathrm d}y \le {\left\{ \begin{array}{ll} - \varepsilon x^{\nu -2} \log x &{} \text { if } \alpha =2, \\ - \varepsilon x^{\nu - 2} &{} \text { if } \alpha > 2. \end{array}\right. } \end{aligned}$$

• (ii) For $\alpha \ge 1$ and any $\nu >0$, there exist $\varepsilon >0$ and $x_0 \in {{\mathbb {R}}}_+$ such that, for all $x \ge x_0$,

  $$\begin{aligned} \int _{0}^x \left( f_\nu (x-y) - f_\nu (x) \right) v (y) {\mathrm d}y \le {\left\{ \begin{array}{ll} - \varepsilon x^{\nu -1} \log x &{} \text { if } \alpha =1, \\ - \varepsilon x^{\nu - 1} &{} \text { if } \alpha > 1. \end{array}\right. } \end{aligned}$$

Proof

For part (i), set $a_\nu (z) = (1 + z)^\nu + (1-z)^\nu - 2$, so that

$$\begin{aligned}&\int _{0}^x \left( f_\nu (x+y) + f_\nu (x-y) - 2 f_\nu (x) \right) v ( y ) {\mathrm d}y \\&\quad = (1+x)^\nu \int _0^x a_\nu \left( \tfrac{y}{1+x} \right) c( y) y^{-1-\alpha } {\mathrm d}y. \end{aligned}$$

Suppose that $\alpha \ge 2$ and $\nu \in (0,1)$. For $\nu \in (0,1)$, calculus shows that $a_\nu (z)$ has a single local maximum at $z=0$, so that $a_\nu (z) \le 0$ for all z. Moreover, Taylor’s theorem shows that for any $\nu \in (0,1)$ there exists $\delta _\nu \in (0,1)$ such that $a_\nu (z) \le - (\nu /2) (1-\nu ) z^2$ for all $z \in [0, \delta _\nu ]$. Also, $c(y) \ge c/2 > 0$ for all $y \ge y_0$ sufficiently large. Hence, for all $x \ge y_0 / \delta _\nu$,

$$\begin{aligned} \int _{0}^x a_\nu \left( \tfrac{y}{1+x} \right) c (y) y^{-1-\alpha } {\mathrm d}y&\le \int _{y_0}^{\delta _\nu x} a_\nu \left( \tfrac{y}{1+x} \right) c (y) y^{-1-\alpha } {\mathrm d}y \\&\le - \frac{c \nu (1-\nu )}{4(1+x)^2} \int _{y_0}^{\delta _\nu x} y^{1-\alpha } {\mathrm d}y , \end{aligned}$$

which yields part (i) of the lemma.

For part (ii), suppose that $\alpha \ge 1$ and $\nu >0$. For any $\nu >0$, there exists $\delta _\nu \in (0,1)$ such that $(1-z)^\nu -1 \le -(\nu /2) z$ for all $z \in [0,\delta _\nu ]$. Moreover, $c(y) \ge c/2 >0$ for all $y \ge y_0$ sufficiently large. Hence, since the integrand is non-positive, for $x > y_0 / \delta _\nu$,

$$\begin{aligned}&(1+x)^\nu \int _{0}^x \left( \left( 1 -\tfrac{y}{1+x}\right) ^\nu - 1 \right) c (y) y^{-1-\alpha } {\mathrm d}y \\&\quad \le \frac{c}{2} (1+x)^\nu \int _{y_0}^{\delta _\nu x} \left( \left( 1 -\tfrac{y}{1+x} \right) ^\nu - 1 \right) y^{-1-\alpha } {\mathrm d}y\\&\quad \le - \frac{c \nu (1+x)^\nu }{4( 1+x)} \int _{y_0}^{\delta _\nu x} y^{-\alpha } {\mathrm d}y , \end{aligned}$$

and part (ii) follows. $\square$

Lemma 5

Suppose that (A1) holds and $\chi _i\alpha _i<1$. Then for $\nu \in (-1,1 \wedge \alpha _i)$, $\flat \in \{\mathrm{one}, \mathrm{sym} \}$, and $i \in {{\mathcal {S}}}^\flat$, as $x\rightarrow \infty$,

$$\begin{aligned} Df_\nu (x,i)= \chi _i\lambda _i c_i x^{\nu -\alpha _i} i_{2,1}^{\nu ,\alpha _i} \left( \frac{(P{\varvec{\lambda }})_i}{\lambda _i}+R^\flat (\alpha _i,\nu )\right) + o (x^{\nu - \alpha _i} ), \end{aligned}$$

where ${\varvec{\lambda }}=(\lambda _k; k\in {{\mathcal {S}}})$,

$$\begin{aligned} R^\mathrm{sym}(\alpha ,\nu ) = \frac{i_0^{\nu ,\alpha }+i_1^{\nu ,\alpha } -i_{2,0}^{\alpha }}{i_{2,1}^{\nu ,\alpha }}, \quad \text {and}\quad R^\mathrm{one}(\alpha ,\nu ) = \frac{{\tilde{i}}_1^{\nu ,\alpha } -i_{2,0}^{\alpha }}{i_{2,1}^{\nu ,\alpha }}. \end{aligned}$$

Moreover, if $v_i \in {{\mathfrak {D}}}_{\alpha _i,c_i}^+$ then, for some $\delta >0$,

$$\begin{aligned} Df_\nu (x,i)= \chi _i\lambda _i c_i x^{\nu -\alpha _i} i_{2,1}^{\nu ,\alpha _i} \left( \frac{(P{\varvec{\lambda }})_i}{\lambda _i}+R^\flat (\alpha _i,\nu )\right) + O (x^{\nu - \alpha _i-\delta } ). \end{aligned}$$

Finally, as $\nu \rightarrow 0$,

$$\begin{aligned} R^\flat (\alpha ,\nu )= {\left\{ \begin{array}{ll} R^\mathrm{sym}(\alpha , \nu ) = -1 + \nu \pi \cot (\pi \alpha _i / 2) + o(\nu ) &{} \text { if } \flat =\mathrm{sym}\\ R^\mathrm{one}(\alpha , \nu ) = -1 + \nu \pi \cot (\pi \alpha _i ) + o(\nu ) &{} \text { if } \flat =\mathrm{one}. \end{array}\right. } \end{aligned}$$

(5.12)

Proof

The above expressions for $Df_\nu (x,i)$ follow from (5.4) and (5.5) with Lemma 3.

Additionally, we compute that $R^\mathrm{sym}(\alpha ,0)=R^\mathrm{one}(\alpha ,0)=-1$. For $\nu$ in a neighbourhood of 0, uniformity of convergence of the integrals over $(1,\infty )$ enables us to differentiate with respect to $\nu$ under the integral sign to get

$$\begin{aligned} \frac{\partial }{\partial \nu } R^\mathrm{sym}(\alpha , \nu )\vert _{\nu =0}&= \frac{j_0^\alpha +j_1^\alpha }{i_{2,1}^{0,\alpha }} -R^\mathrm{sym}(\alpha ,0) \frac{j_2^\alpha }{i_{2,1}^{0,\alpha }}\\&= \psi \left( 1- \tfrac{\alpha }{2} \right) - \psi \left( \tfrac{\alpha }{2} \right) \\&= \pi \cot \left( \tfrac{\pi \alpha }{2} \right) , \end{aligned}$$

using Lemma 2 and the digamma reflection formula (equation 6.3.7 from [1, p. 259]), and then, the first formula in (5.12) follows by Taylor’s theorem. Similarly, for the second formula in (5.12),

$$\begin{aligned} \frac{\partial }{\partial \nu } R^\mathrm{one}(\alpha , \nu )\vert _{\nu =0}&= \frac{{{\tilde{j}}}_1^\alpha }{i_{2,1}^{0,\alpha }} -R^\mathrm{one}(\alpha ,0) \frac{j_2^\alpha }{i_{2,1}^{0,\alpha }}\\&= \psi (1-\alpha )- \psi (\alpha ) \\&= \pi \cot (\pi \alpha ). \end{aligned}$$

This completes the proof. $\square$

We conclude this subsection with two algebraic results.

Lemma 6

Suppose that (A0) holds. Given $(b_k ; k \in {{\mathcal {S}}})$ with $b_k \in {{\mathbb {R}}}$ for all k, there exists a solution $( \theta _k ; k \in {{\mathcal {S}}})$ with $\theta _k \in {{\mathbb {R}}}$ for all k to the system of equations

$$\begin{aligned} \sum _{j \in {{\mathcal {S}}}} p (k,j) \theta _j - \theta _k = b_k, \quad ( k \in {{\mathcal {S}}}) , \end{aligned}$$

(5.13)

if and only if $\sum _{k \in {{\mathcal {S}}}} \mu _k b_k = 0$. Moreover, if a solution to (5.13) exists, we may take $\theta _k >0$ for all $k \in {{\mathcal {S}}}$.

Proof

As column vectors, we write ${\varvec{\mu }}= (\mu _k ; k \in {{\mathcal {S}}})$ for the stationary probabilities as given in (A0), ${{\mathbf {b}}}= (b_k ; k \in {{\mathcal {S}}})$, and ${\varvec{\theta }}= (\theta _k ; k \in {{\mathcal {S}}})$. Then, in matrix-vector form, (5.13) reads $( P - I ) {\varvec{\theta }}= {{\mathbf {b}}}$, while ${\varvec{\mu }}$ satisfies (2.2), which reads $( P - I)^\top {\varvec{\mu }}= {{\mathbf {0}}}$, the homogeneous system adjoint to (5.13) (here I is the identity matrix and ${{\mathbf {0}}}$ is the vector of all 0s).

A standard result from linear algebra (a version of the Fredholm alternative) says that $( P - I ) {\varvec{\theta }}= {{\mathbf {b}}}$ admits a solution ${\varvec{\theta }}$ if and only if the vector ${{\mathbf {b}}}$ is orthogonal to any solution ${{\mathbf {x}}}$ to $(P - I)^\top {{\mathbf {x}}}= {{\mathbf {0}}}$; but, by (A0), any such ${{\mathbf {x}}}$ is a scalar multiple of ${\varvec{\mu }}$. In other words, a solution ${\varvec{\theta }}$ to (5.13) exists if and only if ${\varvec{\mu }}^\top {{\mathbf {b}}}= 0$, as claimed.

Finally, since P is a stochastic matrix, $(P - I ) {\mathbf {1}} = {{\mathbf {0}}}$, where ${\mathbf {1}}$ is the column vector of all 1s; hence, if ${\varvec{\theta }}$ solves $( P - I ) {\varvec{\theta }}= {{\mathbf {b}}}$, then so does ${\varvec{\theta }}+ \gamma {\mathbf {1}}$ for any $\gamma \in {{\mathbb {R}}}$. This implies the final statement in the lemma. $\square$

Lemma 7

Let $U=(U_{k,\ell } ; k,\ell =0,\ldots , M )$ be a given upper triangular matrix having all its upper triangular elements non-negative ($U_{k,\ell }\ge 0$ for $0\le k<\ell \le M$ and vanishing all other elements) and $A =(A_k ; k=1, \ldots , M )$ a vector with positive components. Then, there exists a unique lower triangular matrix $L=(L_{k,\ell } ; k, \ell = 0,\ldots , M)$ (so diagonal and upper triangular elements vanish) satisfying

• (i) $L_{m,m-1} =(UL)_{m,m} +A_m$ for $m=1, \ldots , M$;

• (ii) $L_{k, \ell }=\sum _{r=0}^{k-\ell -1} L_{\ell +r+1, \ell +r}= L_{\ell +1,\ell }+ \cdots + L_{k, k-1}$ for $0\le \ell < k \le M$.

Also, all lower triangular elements of L are positive, i.e. $L_{k,\ell }>0$ for $0\le \ell <k\le M$.

Proof

We construct L inductively. Item (i) demands

$$\begin{aligned} L_{m,m-1} =\sum _{\ell =0}^M U_{m,\ell }L_{\ell ,m} +A_m= \sum _{\ell =m+1}^M U_{m,\ell }L_{\ell ,m} +A_m. \end{aligned}$$

(5.14)

In the case $m=M$, with the usual convention that an empty sum is 0, the demand (5.14) is simply $L_{M,M-1}= A_M$. So we can start our construction taking $L_{M, M-1} = A_M$, which is positive by assumption [item (ii) makes no demands in the case $k = M$, $\ell = M-1$].

Suppose now that all matrix elements $L_{k,\ell }$ have been computed in the lower-right corner ${\varLambda }_m$ ($1 \le m \le M$):

$$\begin{aligned} {\varLambda }_m = \begin{matrix} L_{m,m-1} &{} &{} &{} &{}\\ L_{m+1, m-1} &{} L_{m+1, m} &{} &{}\\ \vdots &{} \vdots &{} \ddots &{} \\ L_{M,m-1} &{} L_{M, m} &{} \ldots &{} L_{M,M-1} \end{matrix} \end{aligned}$$

The elements of L involved in statement (i) (for given m) and in statement (ii) for $\ell = m-1$ are all in ${\varLambda }_m$; thus, as part of our inductive hypothesis we may suppose that the elements of ${\varLambda }_m$ are such that (i) holds for the given m, and (ii) holds with $\ell = m-1$ and all $m \le k \le M$. We have shown that we can achieve this for ${\varLambda }_M$.

The inductive step is to construct from ${\varLambda }_m$ ($2 \le m \le M$) elements $L_{k, m-2}$ for $m-1 \le k \le M$ and hence complete the array ${\varLambda }_{m-1}$ in such a way that (i) holds for $m-1$ replacing m, that (ii) holds for $\ell = m-2$, and that all elements are positive. Now, (5.14) reveals the demand of item (i) as

$$\begin{aligned} L_{m-1,m-2} = \sum _{\ell =m}^M U_{m-1,\ell } L_{\ell , m-1} +A_{m-1} , \end{aligned}$$

which we can achieve since the elements of L on the right-hand side are all in ${\varLambda }_m$, and since $A_{m-1} > 0$ we get $L_{m-1, m-2} > 0$.

The $\ell = m-2$ case of (ii) demands that for $m \le k \le M$ we have

$$\begin{aligned} L_{k,m-2} =\sum _{r=0}^{k-m+1} L_{m-1+r, m-2+r}= L_{m-1, m-2} +\cdots +L_{k,k-1} \end{aligned}$$

which involves only elements of ${\varLambda }_m$ in addition to $L_{m-1, m-2}$, which we have already defined, and positivity of all the $L_{k, m-2}$ follows by hypothesis. This gives us the construction of ${\varLambda }_{m-1}$ and establishes the inductive step.

This algorithm can be continued down to ${\varLambda }_1$. But then the lower triangular matrix L is totally determined. The diagonal and upper triangular elements of L do not influence the construction, and may be set to zero. $\square$

Corollary 1

Let the matrix U and the vector A be as in Lemma 7. Let ${{\mathcal {L}}}$ be the set of lower triangular matrices ${\tilde{L}}$ satisfying

• (i) ${\tilde{L}}_{m,m-1} >(U{\tilde{L}})_{m,m} +A_m$ for $m=1, \ldots , M$;

• (ii) ${\tilde{L}}_{k,\ell }=\sum _{r=0}^{k-\ell -1}{\tilde{L}}_{\ell +r+1, \ell +r}= {\tilde{L}}_{\ell +1,\ell }+ \cdots + {\tilde{L}}_{k, k-1}$ for $0\le \ell < k \le M$,

viewed as subset of the positive cone ${\mathcal {V}}= (0,\infty )^{\frac{M(M-1)}{2}}$. Then, ${{\mathcal {L}}}$ is a non-empty, open subset of ${\mathcal {V}}$.

5.3 Supermartingale Conditions and Recurrence

We use the notation

We start with the case $\max _{k \in {{\mathcal {S}}}} \chi _k \alpha _k < 1$. We will obtain a local supermartingale by choosing the $\lambda _k$ carefully. Lemma 6, which shows how the stationary probabilities $\mu _k$ enter, is crucial; a similar idea was used for random walks on strips in Section 3.1 of [10]. Next is our key local supermartingale result in this case.

Proposition 1

Suppose that (A0) and (A1) hold, and that $\max _{k \in {{\mathcal {S}}}} \chi _k \alpha _k < 1$.

• (i) If $\sum _{k \in {{\mathcal {S}}}} \mu _k \cot ( \chi _k \pi \alpha _k ) < 0$, then there exist $\nu \in (0, \underline{\alpha })$, $\lambda _k >0$ ($k \in {{\mathcal {S}}}$), $\varepsilon >0$, and $x_0 \in {{\mathbb {R}}}_+$ such that for all $x \ge x_0$ and all $i \in \mathcal {S}$,

  $$\begin{aligned} Df_\nu (x,i)= & {} {{\mathrm{{\mathbb {E}}}}}_{x,i} \left[ f_\nu ( X_1, \xi _1 ) - f_\nu ( X_0, \xi _0 ) \right] \le - \varepsilon x^{\nu - {{\bar{\alpha }}}} , \end{aligned}$$

• (ii) If $\sum _{k \in {{\mathcal {S}}}} \mu _k \cot ( \chi _k \pi \alpha _k ) > 0$, then there exist $\nu \in (-1,0)$, $\lambda _k >0$ ($k \in {{\mathcal {S}}}$), $\varepsilon >0$, and $x_0 \in {{\mathbb {R}}}_+$ such that for all $x \ge x_0$ and all $i \in \mathcal {S}$,

  $$\begin{aligned} Df_\nu (x,i)= & {} {{\mathrm{{\mathbb {E}}}}}_{x,i} [ f_\nu ( X_1, \xi _1 ) - f_\nu ( X_0, \xi _0 ) ] \le - \varepsilon x^{\nu - {{\bar{\alpha }}}}, \end{aligned}$$

Proof

Write $a_k = \pi \cot ( \chi _k \pi \alpha _k )$, $k \in {{\mathcal {S}}}$. First, we prove part (i). By hypothesis, $\sum _{k \in {{\mathcal {S}}}} \mu _k a_k = - \delta$ for some $\delta > 0$. Set $b_k = -a_k - \delta$ for all $k \in {{\mathcal {S}}}$, so that $\sum _{k \in {{\mathcal {S}}}} \mu _k b_k = 0$. Lemma 6 shows that for these $b_k$, we can find a collection of $\theta _k \in (0,\infty )$ so that

$$\begin{aligned} \sum _{j \in {{\mathcal {S}}}} p (i,j) \theta _j - \theta _i + a_i = - \delta , \quad \text {for all}~ i \in {{\mathcal {S}}}; \end{aligned}$$

(5.15)

fix these $\theta _k$ for the rest of the proof. We then choose ${\varvec{\lambda }}=(\lambda _k ; k\in {{\mathcal {S}}})$ of the form $\lambda _k:=\lambda _k(\nu ) = 1 + \theta _k \nu$, for some $\nu \in (0, 1 \wedge \underline{\alpha })$.

Since $i_{2,1}^{\nu ,\alpha }>0$ for all $\alpha >0$ and all $\nu \in (0,\alpha )$, Lemma 5 shows that $Df_\nu (x,i)$ will be negative for all i and all x sufficiently large provided that we can find $\nu$ such that $\frac{(P{\varvec{\lambda }})_i}{\lambda _i}+R^\flat (\alpha _i,\nu ) < 0$ for all i. By (5.12), writing ${\varvec{\theta }}= (\theta _k ; k \in {{\mathcal {S}}})$,

$$\begin{aligned} \frac{(P{\varvec{\lambda }})_i}{\lambda _i}+R^\flat (\alpha _i,\nu )&= \frac{(P{\varvec{\lambda }})_i}{\lambda _i} -1 + \nu a_i +o(\nu ) \\&= \frac{1+ \nu (P{\varvec{\theta }})_i}{1+\nu \theta _i} -1 + \nu a_i +o(\nu )\\&= \frac{\nu \left( ( P {\varvec{\theta }})_i - \theta _i \right) }{1+\nu \theta _i} + \nu a_i + o (\nu ) \\&= \nu \left( (P{\varvec{\theta }})_i- \theta _i\right) + \nu a_i +o(\nu ) \\&= -\nu \delta +o(\nu ), \end{aligned}$$

by (5.15). Therefore, by Lemma 5, we can always find sufficiently small $\nu > 0$ and a vector ${\varvec{\lambda }}={\varvec{\lambda }}(\nu )$ with strictly positive elements for which $Df_\nu (x,i) \le -\varepsilon x^{\nu - \alpha _i}$ for some $\varepsilon >0$, all i, and all x sufficiently large. Maximizing over i gives part (i).

The argument for part (ii) is similar. Suppose $\nu \in (-1,0)$. This time, $\sum _{k \in {{\mathcal {S}}}} \mu _k a_k = \delta$ for some $\delta > 0$, and we set $b_k = -a_k + \delta$, so that $\sum _{k \in {{\mathcal {S}}}} \mu _k b_k = 0$ once more. Lemma 6 now shows that we can find $\theta _k$ so that

$$\begin{aligned} \sum _{j \in {{\mathcal {S}}}} p (i,j) \theta _j - \theta _i + a_i = \delta , \quad \text {for all}\quad i \in {{\mathcal {S}}}. \end{aligned}$$

(5.16)

With this choice of $\theta _k$, we again set $\lambda _k = 1 + \theta _k\nu$; note we may assume $\lambda _k >0$ for all k for $\nu$ sufficiently small.

Again, $Df_\nu (x,i)$ will be non-positive for all i and all x sufficiently large provided that we can find ${\varvec{\lambda }}$ and $\nu$ such that $\frac{(P{\varvec{\lambda }})_i}{\lambda _i}+R^\flat (\alpha _i,\nu ) < 0$ for all i. Following a similar argument to before, we obtain with (5.16) that

$$\begin{aligned} \frac{(P{\varvec{\lambda }})_i}{\lambda _i}+R^\flat (\alpha _i,\nu )&= \nu \left( (P{\varvec{\theta }})_i- \theta _i\right) + \nu a_i +o(\nu ) = \nu \delta +o(\nu ). \end{aligned}$$

Thus, we can find for $\nu < 0$ close enough to 0 a vector ${\varvec{\lambda }}={\varvec{\lambda }}(\nu )$ with strictly positive elements for which $Df_\nu (x,i)\le -\varepsilon x^{\nu - \alpha _i}$ for all i and all x sufficiently large. $\square$

Now we examine the case $\max _{k \in {{\mathcal {S}}}} \chi _k\alpha _k \ge 1$.

Proposition 2

Suppose that (A0) and (A1) hold, and $\max _{k \in {{\mathcal {S}}}} \chi _k \alpha _k \ge 1$. Then, there exist $\nu \in (0, \alpha _\star )$, $\lambda _k >0$ $(k \in {{\mathcal {S}}})$, $\varepsilon >0$, and $x_0 \in {{\mathbb {R}}}_+$ such that for all $i\in {{\mathcal {S}}}$,

$$\begin{aligned} Df_\nu (x,i)={{\mathrm{{\mathbb {E}}}}}_{x,i} \left[ f_\nu ( X_1, \xi _1 ) - f_\nu ( X_0, \xi _0 ) \right] \le - \varepsilon x^{\nu - \alpha ^\star } , \quad \text {for all}\quad x \ge x_0.\qquad \quad \end{aligned}$$

(5.17)

Before starting the proof of this proposition, we introduce the following notation. For $k\in {{\mathcal {S}}}$, denote by $a_k=\pi \cot (\pi \chi _k \alpha _k)$ and define the vector ${{\mathbf {a}}}=(a_k ; k\in {{\mathcal {S}}})$. For $i\in {{\mathcal {S}}}$ and $A \subseteq {{\mathcal {S}}}$, write $P(i,A) =\sum _{j\in A} p(i,j)$. Define ${{\mathcal {S}}}_0=\{i\in {{\mathcal {S}}}: \chi _i \alpha _i \ge 1\}$ and recursively, for $m \ge 1$,

$$\begin{aligned} {{\mathcal {S}}}_m:= \left\{ i \in {{\mathcal {S}}}{\setminus } \cup _{\ell =0}^{m-1} {{\mathcal {S}}}_\ell : P(i,{{\mathcal {S}}}_{m-1})>0 \right\} . \end{aligned}$$

Denote by $M:=\max \{m\ge 0: {{\mathcal {S}}}_m\ne \emptyset \}$. Since P is irreducible, the collection $({{\mathcal {S}}}_m ; m=0,\ldots , M )$ is a partition of ${{\mathcal {S}}}$.

Proof of Proposition 2

It suffices to find $x_0\in {{\mathbb {R}}}_+$, $\varepsilon >0$, $\nu \in (0, \alpha _\star )$, and an open, non-empty subset G of the positive cone ${\mathcal {C}}:= (0,\infty )^{|{{\mathcal {S}}}|}$ such that

$$\begin{aligned} G\subseteq \cap _{i\in {{\mathcal {S}}}} \{{\varvec{\lambda }}\in {\mathcal {C}}: Df_\nu (x,i) \text { satisfies condition } (5.17)\}. \end{aligned}$$

Now, for $i\in {{\mathcal {S}}}_0$, inequality (5.17) is satisfied thanks to (5.4), (5.5), and Lemmas 3 and 4 for every choice of ${\varvec{\lambda }}$ (with positive components) and $\nu \in (0, \underline{\alpha })$. Hence, the previous condition reduces to the requirement

$$\begin{aligned} G\subseteq \cap _{i\in {{\mathcal {S}}}{\setminus } {{\mathcal {S}}}_0} \{{\varvec{\lambda }}\in {\mathcal {C}}: Df_\nu (x,i) \text { satisfies condition } (5.17)\}. \end{aligned}$$

(5.18)

The rest of the proof is devoted into establishing this fact.

Suppose that $i\in {{\mathcal {S}}}_m$ with $m=1,\ldots , M$. Then, $\chi _i \alpha _i < 1$ by construction, and Lemma 5 shows that condition (5.17) will be satisfied if the system of inequalities

$$\begin{aligned} \frac{(P{\varvec{\lambda }})_i}{\lambda _i} +R^\flat (\alpha _i, \nu )<0, ~ i=1, \ldots , M \end{aligned}$$

(5.19)

has non-trivial solutions ${\varvec{\lambda }}$ for sufficiently small $\nu$. Thanks to Lemma 5, we have $R^\flat (\alpha _i,\nu )= -1+\nu a_i +o(\nu )$. We will obtain (5.17) if, for $\nu$ sufficiently small,

$$\begin{aligned} {{\mathcal {R}}}={{\mathcal {R}}}(P, \nu ,{{\mathbf {a}}}):=\bigcap _{m=1}^M \left\{ {\varvec{\lambda }}\in {\mathcal {C}}: \frac{(P{\varvec{\lambda }})_i}{\lambda _i} < 1 - \nu a_i, i\in {{\mathcal {S}}}_m\right\} \ne \emptyset . \end{aligned}$$

(5.20)

We seek a solution ${\varvec{\lambda }}\in {{\mathcal {R}}}$ (for sufficiently small $\nu$), under the Ansatz that the $\lambda _j$ are constant on every ${{\mathcal {S}}}_\ell$, i.e. the vector ${\varvec{\lambda }}$ has the form ${\hat{{\varvec{\lambda }}}}$ with ${\hat{\lambda }}_j=\lambda ^{(\ell )}$ for all $j\in {{\mathcal {S}}}_\ell$. Suppose that $i \in {{\mathcal {S}}}_m$. Then, $p(i,j) = 0$ for $j \in {{\mathcal {S}}}_\ell$ with $\ell < m-1$, so that

$$\begin{aligned} \frac{(P{\hat{{\varvec{\lambda }}}})_i}{{\hat{\lambda }}_i}= \sum _{j\in {{\mathcal {S}}}} p(i,j) \frac{{\hat{\lambda }}_j}{{\hat{\lambda }}_i} = \sum _{\ell =0}^M P(i,{{\mathcal {S}}}_\ell ) \frac{\lambda ^{(\ell )}}{\lambda ^{(m)}} = \sum _{\ell =m-1}^MP(i,{{\mathcal {S}}}_\ell ) \frac{\lambda ^{(\ell )}}{\lambda ^{(m)}}. \end{aligned}$$

(5.21)

We introduce the auxiliary matrix $\rho =(\rho _{k,\ell } ; k,\ell \in \{0, \ldots , M\} )$ defined by $\rho _{k,\ell }:= \lambda ^{(k)}/ \lambda ^{(\ell )}$. By construction, $\rho _{k,k}=1$ and $\rho _{k,\ell }= 1/\rho _{\ell ,k}$. Let

$$\begin{aligned} L_{k, \ell } = \frac{1}{\nu } \log \rho _{k,\ell } = - \frac{1}{\nu } \log \rho _{\ell , k} = \frac{1}{\nu } \left( \log \lambda ^{(k)} - \log \lambda ^{(\ell )} \right) . \end{aligned}$$

(5.22)

It suffices to determine the upper triangular part of $\rho$, or, equivalently, the lower triangular array $(L_{k, \ell } ; 0 \le \ell < k \le M )$. We do so recursively, starting with $L_{M, M-1}$. In the case $i\in {{\mathcal {S}}}_M$, the condition in (5.20) reads, by (5.21),

$$\begin{aligned} \rho _{M-1, M}< & {} \frac{1-P(i,{{\mathcal {S}}}_M) -\nu a_i}{P(i,{{\mathcal {S}}}_{M-1})}= \frac{P\left( i, {{\mathcal {S}}}_M^{\mathrm {c}}\right) }{P(i,{{\mathcal {S}}}_{M-1})} -\nu \frac{a_i}{P(i,{{\mathcal {S}}}_{M-1})} \\= & {} 1-\nu \frac{a_i}{P(i,{{\mathcal {S}}}_{M-1})}. \end{aligned}$$

On introducing the constant $A_M=\max _{i\in {{\mathcal {S}}}_M} \frac{a_i}{P(i,{{\mathcal {S}}}_M)}$, it is enough to choose $\rho _{M-1,M} < 1-\nu A_M= \exp (-\nu A_M) +o(\nu )$. In other words, with $\rho _{\ell ,k}= \exp (-\nu L_{k,\ell })$, we see that the choice $L_{M, M-1} >A_M$ satisfies the condition in (5.20) for $i\in {{\mathcal {S}}}_M$.

Suppose now that we have determined the condition in (5.20) for $i\in \cup _{\ell = m}^M {{\mathcal {S}}}_\ell$. Then, for $i \in {{\mathcal {S}}}_{m-1}$ the condition amounts, via (5.21), to

$$\begin{aligned} \rho _{m-1, m} P(i, {{\mathcal {S}}}_{m-1}) < 1- \nu a_i -P(i, {{\mathcal {S}}}_m) - \sum _{\ell =m+1}^M \rho _ {\ell ,m} P(i, {{\mathcal {S}}}_\ell ). \end{aligned}$$

Using the fact that for $\ell < m$ we have $\rho _{\ell ,m} = \exp (-\nu L_{m, \ell })$ and for $\ell > m$ we have $\rho _{\ell , m} = \exp (\nu L_{\ell , m})$, the above expression becomes, up to $o(\nu )$ terms

$$\begin{aligned} 1-\nu L_{m,m-1}&< \frac{P\left( i, \left( \cup _{\ell =m}^M {{\mathcal {S}}}_\ell \right) ^c\right) }{P(i,{{\mathcal {S}}}_{m-1})} -\nu \sum _{\ell =m+1}^M \frac{P(i, {{\mathcal {S}}}_l)}{P(i, {{\mathcal {S}}}_{m-1})} L_{\ell ,m} - \nu \frac{a_i}{P(i, {{\mathcal {S}}}_{m-1})}\\&= 1- \nu \sum _{\ell =m+1}^M \frac{P(i, {{\mathcal {S}}}_\ell )}{P(i, {{\mathcal {S}}}_{m-1})} L_{\ell ,m} - \nu \frac{a_i}{P(i, {{\mathcal {S}}}_{m-1})}. \end{aligned}$$

Introducing the upper triangular matrix $U=(U_{m,n} ; 0 \le m<n\le M )$ defined by $U_{m,n}= \max _{i\in {{\mathcal {S}}}_m} \frac{P(i, {{\mathcal {S}}}_n)}{P(i, {{\mathcal {S}}}_{m-1})}$ for $m\ge 1$, and the vector $A_m=\max _{i\in {{\mathcal {S}}}_m} \frac{a_i}{P(i, {{\mathcal {S}}}_{m-1})}$ for $m=1, \ldots , M$, the condition in (5.20) is satisfied if we solve the recursion

$$\begin{aligned} L_{m,m-1} >(UL)_{m, m} +A_m, \quad \text { for}\quad m=M, M-1, \ldots , 1, \end{aligned}$$

with initial condition $L_{M,M-1}>A_M$ and condition $L_{\ell ,m} >0$ for $0 \le m < \ell \le M$. Additionally, we have from (5.22) that

$$\begin{aligned} L_{k, \ell } = L_{k, k-1} + L_{k-1, k-2} + \cdots + L_{\ell +1, \ell }, ~~ 0 \le \ell < k \le M . \end{aligned}$$

Hence, by Corollary 1, there exist non-trivial solutions for the lower triangular matrix L within an algorithmically determined region ${\mathcal {L}}$. The positivity of the lower triangular part of L implies that the components of ${\varvec{\lambda }}$ are ordered: $\lambda ^{(m)}<\lambda ^{(m+1)}$ for $0\le m<M-1$. Given L, the ratios of the $\lambda ^{(k)}$ are determined, and by construction, the $\lambda ^{(k)}$ are positive. $\square$

We are almost ready to complete the proof of Theorem 1, excluding part (b)(iii); first, we need one more technical result concerning non-confinement.

Lemma 8

Suppose that (A0) and (A1) hold. Then, $\limsup _{n \rightarrow \infty } X_n = \infty$, a.s.

Proof

We claim that for each $x \in {{\mathbb {R}}}_+$, there exists $\varepsilon _x >0$ such that

$$\begin{aligned} {{\mathbb {P}}}\left[ X_{n+1} - X_n \ge 1 \mid ( X_n , \xi _n) = (y,i) \right] \ge \varepsilon _x, ~\text {for all}~ y \in [0,x] \text { and all } i\in {{\mathcal {S}}}.\nonumber \\ \end{aligned}$$

(5.23)

Indeed, given $(x,i) \in {{\mathbb {R}}}_+\times {{\mathcal {S}}}$, we may choose $j \in {{\mathcal {S}}}$ so that $p(i,j) >0$ and we may choose $z_0 \ge x$ sufficiently large so that, for some $\varepsilon >0$, $v_i (z) \ge \varepsilon z^{-1-\alpha _i}$ for all $z \ge z_0$. Then if $y \in [0,x]$,

$$\begin{aligned} {{\mathbb {P}}}\left[ X_{n+1} \ge y + 1 \mid (X_n , \xi _n) = (y,i) \right]&\ge p(i,j) \int _{2z_0+1}^\infty \varepsilon z^{-1-\alpha _i} {\mathrm d}z = \varepsilon _{x, i} >0 , \end{aligned}$$

which gives (5.23). The local escape property (5.23) implies the $\limsup$ result by a standard argument: see, e.g. [16, Proposition 3.3.4]. $\square$

Proof of Theorem 1

We are not yet ready to prove part (b)(iii): we defer that part of the proof until Sect. 7.

The other parts of the theorem follow from the supermartingale estimates in this section together with the technical results from the Appendix. Indeed, under the conditions of part (a) or (b)(i) of the theorem, we have from Propositions 2 or 1(i), respectively, that for suitable $\nu >0$ and $\lambda _k$,

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ f_\nu ( X_{n+1}, \xi _n ) - f_\nu (X_n , \xi _n ) \mid X_n , \xi _n \right] \le 0 , ~~\text {on}~ \{ X_n \ge x_0\} . \end{aligned}$$

Thus, we may apply Lemma 16, which together with Lemma 8 shows that $\liminf _{n \rightarrow \infty } X_n \le x_0$, a.s. Thus, there exists an interval $I \subseteq [0,x_0+1]$ such that $(X_n,\eta _n) \in I \times \{ i \}$ i.o., where i is some fixed element of ${{\mathcal {S}}}$. Let $\tau _0 := 0$ and for $k \in {{\mathbb {N}}}$ define $\tau _k = \min \{ n > \tau _{k-1}+1 : (X_n, \eta _n) \in I \times \{ i \} \}$. Given $i \in {{\mathcal {S}}}$, we may choose $j, k \in {{\mathcal {S}}}$ such that $p(i,j) > \delta _1$ and $p(j,k) > \delta _1$ for some $\delta _1 >0$; let $\gamma = \alpha _i \vee \alpha _j$. Then, we may choose $\delta _2 \in (0,1)$ and $z_0 \in {{\mathbb {R}}}_+$ such that $v_i (z) > \delta _2 z^{-1-\gamma }$ and $v_j (z ) > \delta _2 z^{-1-\gamma }$ for all $z \ge z_0$. Then, for any $\varepsilon \in (0,1)$,

$$\begin{aligned} {{\mathbb {P}}}\left[ X_{\tau _k +2 }< \varepsilon \mid X_{\tau _k} \right]&\ge {{\mathbb {P}}}\left[ X_{\tau _k +2 } < \varepsilon , X_{\tau _k + 1} \in \left[ z_0 +1, z_0 +2 \right] , \eta _{\tau _k+1} = j \mid X_{\tau _k} \right] \\&\ge \delta _1^2 \delta _2^2 \varepsilon (z_0 +3)^{-1-\gamma } (x_0 + z_0 + 3)^{-1-\gamma } , \end{aligned}$$

uniformly in k. Thus, Lévy’s extension of the Borel–Cantelli lemma shows $X_n < \varepsilon$ infinitely often. Thus, since $\varepsilon \in (0,1)$ was arbitrary, $\liminf _{n \rightarrow \infty } X_n = 0$, a.s.

On the other hand, under the conditions of part (b)(ii) of the theorem, we have from Proposition 1(ii) that for suitable $\nu <0$ and $\lambda _k$,

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ f_\nu ( X_{n+1}, \xi _n ) - f_\nu (X_n , \xi _n ) \mid X_n , \xi _n \right] \le 0 , \quad \text {on}\quad \{ X_n \ge x_1\} , \end{aligned}$$

for any $x_1$ sufficiently large. Thus, we may apply Lemma 17, which shows that for any $\varepsilon >0$ there exists $x \in (x_1, \infty )$ for which, for all $n \ge 0$,

$$\begin{aligned} {{\mathbb {P}}}\Bigl [ \inf _{m \ge n} X_m \ge x_1 \; \Bigl | \;{{\mathcal {F}}}_n \Bigr ] \ge 1 - \varepsilon , \quad \text {on}\quad \{ X_n \ge x \} . \end{aligned}$$

Set $\sigma _x = \min \{ n \ge 0 : X_n \ge x \}$. Then, on $\{ \sigma _x < \infty \}$,

$$\begin{aligned} {{\mathbb {P}}}\Bigl [ \inf _{m \ge \sigma _x} X_m > x_1 \; \Bigl | \;{{\mathcal {F}}}_{\sigma _x} \Bigr ] \ge 1 - \varepsilon , \ \text {a.s.}\end{aligned}$$

But on $\{ \sigma _x < \infty \} \cap \{ \inf _{m \ge \sigma _x} X_m > x_1 \}$, we have $\liminf _{m \rightarrow \infty } X_m \ge x_1$, so

$$\begin{aligned} {{\mathbb {P}}}\Bigl [ \liminf _{m \rightarrow \infty } X_m \ge x_1 \Bigr ]&\ge {{\mathrm{{\mathbb {E}}}}}\left[ {{\mathbb {P}}}\Bigl [ \inf _{m \ge \sigma _x} X_m > x_1 \; \Bigl | \;{{\mathcal {F}}}_{\sigma _x} \Bigr ] {{\mathbf {1}}}{\{ \sigma _x< \infty \}} \right] \\&\ge (1 -\varepsilon ) {{\mathbb {P}}}[ \sigma _x < \infty ] = (1-\varepsilon ), \end{aligned}$$

by Lemma 8. Since $\varepsilon >0$ was arbitrary, we get $\liminf _{m \rightarrow \infty } X_m \ge x_1$, a.s., and since $x_1$ was arbitrary we get $\lim _{m \rightarrow \infty } X_m = \infty$, a.s. $\square$

6 Existence or Non-existence of Moments

6.1 Technical Tools

The following result is a straightforward reformulation of Theorem 1 of [2].

Lemma 9

Let $Y_n$ be an integrable ${{\mathcal {F}}}_n$-adapted stochastic process, taking values in an unbounded subset of ${{\mathbb {R}}}_+$, with $Y_0 = x_0$ fixed. For $x >0$, let $\sigma _x := \inf \{ n \ge 0 : Y_n \le x \}$. Suppose that there exist $\delta >0$, $x >0$, and $\gamma < 1$ such that for any $n \ge 0$,

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ Y_{n+1}-Y_n \mid {{\mathcal {F}}}_n \right] \le -\delta Y_n^{\gamma }, \text { on } \{n < \sigma _x \}. \end{aligned}$$

(6.1)

Then, for any $p \in [0,1/(1-\gamma ))$, ${{\mathrm{{\mathbb {E}}}}}[ \sigma _x^p ] < \infty$.

The following companion result on non-existence of moments is a reformulation of Corollary 1 of [2].

Lemma 10

Let $Y_n$ be an integrable ${{\mathcal {F}}}_n$-adapted stochastic process, taking values in an unbounded subset of ${{\mathbb {R}}}_+$, with $Y_0 = x_0$ fixed. For $x >0$, let $\sigma _x := \inf \{ n \ge 0 : Y_n \le x \}$. Suppose that there exist $C_1, C_2 >0$, $x >0$, $p >0$, and $r > 1$ such that for any $n \ge 0$, on $\{ n < \sigma _x \}$ the following hold:

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ Y_{n+1}-Y_n \mid {{\mathcal {F}}}_n \right]&\ge - C_1 ; \end{aligned}$$

(6.2)

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ Y^r_{n+1}-Y^r_n \mid {{\mathcal {F}}}_n \right]&\le C_2 Y_n^{r-1} ; \end{aligned}$$

(6.3)

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ Y_{n+1}^p - Y_n^p \mid {{\mathcal {F}}}_n \right]&\ge 0 . \end{aligned}$$

(6.4)

Then, for any $q >p$, ${{\mathrm{{\mathbb {E}}}}}[ \sigma _x^q ] = \infty$ for $x_0 > x$.

6.2 Proof of Theorem 2

Proof of Theorem 2

Under conditions (a) or (b)(i) of Theorem 1, we have from Propositions 2 or 1, respectively, that there exist positive $\lambda _k$ and constants $\varepsilon >0$, $\beta >0$ and $\nu \in (0,\beta )$ such that,

$$\begin{aligned} D f_\nu (x,i) \le - \varepsilon x^{\nu - \beta } , \quad \text { for all }\quad x \ge x_0 \quad \text { and all }\quad i. \end{aligned}$$

Let $Y_n = f_\nu ( X_n, \xi _n)$. Then, $Y_n$ is bounded above and below by positive constants times $(1+X_n)^\nu$, so we have that (6.1) holds for x sufficiently large with $\gamma = 1 - (\beta / \nu )$. It follows from Lemma 9 that ${{\mathrm{{\mathbb {E}}}}}[ \sigma _x^p ] < \infty$ for $p \in (0, \nu / \beta )$, which gives the claimed existence of moments result.

It is not hard to see that some moments of the return time fail to exist, due to the heavy-tailed nature of the model, and an argument is easily constructed using the ‘one big jump’ idea: a similar idea is used in [14]. We sketch the argument. For any x, i, for all y sufficiently large we have ${{\mathbb {P}}}_{x,i} [ X_1 \ge y-x ] \ge \varepsilon y^{-{{\bar{\alpha }}}}$. Given such a first jump, with uniformly positive probability the process takes time at least of order $y^\beta$ to return to a neighbourhood of zero (where $\beta$ can be bounded in terms of $\underline{\alpha }$); this can be proved using a suitable maximal inequality as in the proof of Theorem 2.10 of [14]. Combining these two facts shows that with probability of order $y^{-{{\bar{\alpha }}}}$ the return time to a neighbourhood of the origin exceeds order $y^\beta$. This polynomial tail bound yields non-existence of sufficiently high moments. $\square$

6.3 Explicit Cases: Theorems 4 and 6

We now restrict attention to the case ${{\mathcal {S}}}= \{ 1,2 \}$ with $\alpha _1 = \alpha _2 = \alpha$ and $\chi _1 = \chi _2 = \chi$, so both half-lines are of the same type. Take $\lambda _1 = \lambda _2 = 1$ and $\nu \in (0, \alpha )$, so that $f_\nu (x , i ) = (1 +x)^\nu$. Then, Lemma 5 shows that, for $i \in {{\mathcal {S}}}^\flat$, $\flat \in \{ \mathrm{sym}, \mathrm{one} \}$,

$$\begin{aligned} D f_\nu (x,i)&= \chi c_i x^{\nu - \alpha } C^\flat (\nu , \alpha ) + o (x^{\nu -\alpha } ) , \end{aligned}$$

(6.5)

where

$$\begin{aligned} C^\mathrm{sym} (\nu , \alpha )&= i^{\nu , \alpha }_{2,1} + i^{\nu ,\alpha }_0 + i^{\nu ,\alpha }_1 - i^\alpha _{2,0} ; \\ C^\mathrm{one} (\nu , \alpha )&= i^{\nu , \alpha }_{2,1} + \tilde{i}^{\nu ,\alpha }_1 - i^\alpha _{2,0} . \end{aligned}$$

The two cases we are interested in are the recurrent two-sided symmetric case, where $\chi =\frac{1}{2}$ (i.e. ${{\mathcal {S}}}= {{\mathcal {S}}}^\mathrm{sym}$) with $\alpha > 1$, and the recurrent one-sided antisymmetric case, where $\chi =1$ (i.e. ${{\mathcal {S}}}= {{\mathcal {S}}}^\mathrm{one}$) with $\alpha > \frac{1}{2}$.

Lemma 11

Let $\flat \in \{ \mathrm{sym}, \mathrm{one} \}$ and $\chi \alpha \in (\frac{1}{2},1)$. The function $\nu \mapsto C^\flat (\nu ,\alpha )$ is continuous for $\nu \in [0, \alpha )$ with $C (0,\alpha ) =0$ and $\lim _{\nu \uparrow \alpha } C^\flat (\nu , \alpha ) = \infty$. There exists $\nu ^\flat _0 = \nu ^\flat _0 (\alpha ) \in (0,\alpha )$ such that $C^\flat (\nu ,\alpha ) < 0$ for $\alpha \in (0, \nu ^\flat _0)$, $C^\flat (\nu ^\flat _0, \alpha ) = 0$, and $C^\flat (\nu ,\alpha ) > 0$ for $\alpha \in (\nu ^\flat _0 , \alpha )$

Proof

We give the proof only in the case $\flat = \mathrm{sym}$; the other case is very similar. Thus, $\chi = \frac{1}{2}$, and for ease of notation, we write just C instead of $C^\flat$.

Clearly $C ( 0 ,\alpha ) = 0$. For $\nu \ge 1$, convexity of the function $z \mapsto z^\nu$ on ${{\mathbb {R}}}_+$ shows that $(1+u)^\nu + (1-u)^\nu - 2 \ge 0$ for all $u \in [0,1]$, so that $i_1^{\nu , \alpha } \ge 0$; clearly, $i^{\nu , \alpha }_{2,1}$ and $i_0^{\nu , \alpha }$ are also non-negative. Hence, by the expression for $i_{2,1}^{\nu ,\alpha }$ in Lemma 1,

$$\begin{aligned} \liminf _{\nu \uparrow \alpha } C (\nu ,\alpha ) \ge - \frac{1}{\alpha } + \liminf _{\nu \uparrow \alpha } \frac{{\varGamma }(\nu +1) {\varGamma }(\alpha - \nu )}{{\varGamma }(1+\alpha )} , \end{aligned}$$

which is $+\infty$. Moreover, by Lemma 2 and the subsequent remark,

$$\begin{aligned} \left. \frac{\partial }{\partial \nu } C (\nu , \alpha ) \right| _{\nu =0} = \frac{1}{\alpha } \left( \psi \left( 1 -\tfrac{\alpha }{2} \right) - \psi \left( \tfrac{\alpha }{2} \right) \right) = \frac{\pi }{\alpha } \cot \left( \tfrac{\pi \alpha }{2} \right) , \end{aligned}$$

which is negative for $\alpha \in (1,2)$. Hence, $C (\nu , \alpha ) <0$ for $\nu > 0$ small enough.

Since $\nu \mapsto C (\nu , \alpha )$ is a non-constant analytic function on $[0,\alpha )$, its zeros can accumulate only at $\alpha$, but this is ruled out by the fact that $C (\nu ,\alpha ) \rightarrow \infty$ as $\nu \rightarrow \alpha$. Hence, $C ( \, \cdot \, ,\alpha )$ has only finitely many zeros in $[0,\alpha )$; one is at 0, and there must be at least one zero in $(0 , \alpha )$, by Rolle’s theorem. Define $\nu _- := \nu _- (\alpha )$ and $\nu _+ := \nu _+ (\alpha )$ to be the smallest and largest such zeros, respectively.

Suppose $0< \nu _1 \le \nu _2 < \alpha$. By Jensen’s inequality, ${{\mathrm{{\mathbb {E}}}}}_{x,i} [ (1 + X_{1} )^{\nu _2} ] \ge \left( {{\mathrm{{\mathbb {E}}}}}_{x,i} [ (1+ X_1 )^{\nu _1} ] \right) ^{\nu _2/\nu _1}$. Hence

$$\begin{aligned} (1+ x)^{\nu _2} + D f_{\nu _2} ( x, i)&\ge \left( (1+x)^{\nu _1} + D f_{\nu _1} (x, i) \right) ^{\nu _2/\nu _1} \\&= (1+x)^{\nu _2} \left( 1 + (1+x)^{-\nu _1} D f_{\nu _1} (x, i) \right) ^{\nu _2/\nu _1} \\&= (1+x)^{\nu _2} + \frac{\nu _2}{\nu _1} x^{\nu _2-\nu _1} D f_{\nu _1} (x, i) + o (x^{\nu _2 - \alpha }), \end{aligned}$$

using Taylor’s theorem and the fact that $D f_{\nu _1} (x, i) = O (x^{\nu _1 -\alpha } )$, by (6.5). By another application of (6.5), it follows that

$$\begin{aligned} \chi c_i x^{\nu _2 - \alpha } C (\nu _2, \alpha ) + o (x^{\nu _2 - \alpha }) \ge \frac{\nu _2}{\nu _1} \chi c_i x^{\nu _2 - \alpha } C (\nu _1, \alpha ) + o (x^{\nu _2 - \alpha }) . \end{aligned}$$

Multiplying by $x^{\alpha -\nu _2}$ and taking $x \rightarrow \infty$, we obtain

$$\begin{aligned} C (\nu _2, \alpha ) \ge \frac{\nu _2}{\nu _1} C (\nu _1, \alpha ),\quad \text {for}\quad 0< \nu _1 \le \nu _2 < \alpha . \end{aligned}$$

In particular, (i) if $C ( \nu _1 ,\alpha ) \ge 0$ then $C ( \nu , \alpha ) \ge 0$ for all $\nu \ge \nu _1 >0$; and (ii) if $C( \nu _1, \alpha ) > 0$ and $\nu _2 > \nu _1$, we have $C (\nu _2, \alpha ) > C (\nu _1, \alpha )$. It follows from these two observations that $C ( \nu , \alpha ) = 0$ for $\nu \in [ \nu _- , \nu _+ ]$, which is not possible unless $\nu _- = \nu _+$. Hence, there is exactly one zero of $C ( \, \cdot \, ,\alpha )$ in $(0,\alpha )$; call it $\nu _0 (\alpha )$. $\square$

Lemma 12

The positive zero of $C^\flat ( \, \cdot \, , \alpha )$ described in Lemma 11 is given by $v^\mathrm{one}_0 ( \alpha ) = 2 \alpha -1$ or $v^\mathrm{sym}_0 (\alpha ) = \alpha -1$.

Proof

First suppose $\flat = \mathrm{one}$. Then, from Lemma 1 we verify that for $\alpha \in ( \frac{1}{2}, 1 )$,

$$\begin{aligned} C^\mathrm{one} ( 2 \alpha -1 , \alpha ) = \frac{ {\varGamma }(2\alpha ) {\varGamma }( 1 - \alpha )}{{\varGamma }(1 + \alpha ) } - \frac{{\varGamma }(2\alpha ) {\varGamma }( 1-\alpha )}{\alpha {\varGamma }(\alpha ) } = 0 , \end{aligned}$$

since $\alpha {\varGamma }(\alpha ) = {\varGamma }(1 +\alpha )$.

Now, suppose that $\flat = \mathrm{sym}$ and $\alpha \in (1,2)$. To verify $C^\mathrm{sym} (\alpha -1 ,\alpha ) = 0$, it is simpler to work with the integral representations directly, rather than the hypergeometric functions. After the substitution $z = 1/u$, we have

$$\begin{aligned} i^{\alpha -1,\alpha }_0 = \int _0^1 \left( ( 1 + z)^{\alpha -1} - z^{\alpha -1} \right) {\mathrm d}z = \frac{2^\alpha -2}{\alpha } . \end{aligned}$$

Similarly, after the same substitution,

$$\begin{aligned} i_1^{\alpha -1 , \alpha } = \int _1^\infty \left( (z+1)^{\alpha -1} + (z-1)^{\alpha -1} - 2 z^{\alpha -1} \right) {\mathrm d}z, \end{aligned}$$

which we may evaluate as

$$\begin{aligned} i_1^{\alpha -1, \alpha }&= \lim _{y\rightarrow \infty } \int _1^y\left( (z+1)^{\alpha -1} + (z-1)^{\alpha -1} - 2 z^{\alpha -1} \right) {\mathrm d}z \\&= \frac{2-2^\alpha }{\alpha } + \frac{1}{\alpha } \lim _{y \rightarrow \infty } y^\alpha \left( (1+y^{-1} )^\alpha + (1 -y^{-1} )^\alpha - 2 \right) \\&= \frac{2-2^\alpha }{\alpha } . \end{aligned}$$

Finally, we have that $i^{\alpha -1, \alpha }_{2,1} - i^\alpha _{2,0} = \frac{ {\varGamma }(\alpha )}{{\varGamma }(1+\alpha )} - \frac{1}{\alpha } =0$, so altogether we verify that $C^\mathrm{sym} (\alpha -1 ,\alpha ) = 0$. $\square$

We can now complete the proofs of Theorems 4 and 6.

Proof of Theorem 4

Let $Y_n = f_\nu (X_n, \xi _n )$. First suppose that $\alpha \in (1,2)$. Then, we have from (6.5) together with Lemmas 11 and 12 that, for any $\nu \in (0, \alpha - 1)$,

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}[ Y_{n+1} - Y_n \mid {{\mathcal {F}}}_n ] \le - \varepsilon Y_n^{1-(\alpha /\nu )} , \quad \text { on }\quad \{ Y_n \ge y_0 \} , \end{aligned}$$

for some $\varepsilon >0$ and $y_0 \in {{\mathbb {R}}}_+$. It follows from Lemma 9 that ${{\mathrm{{\mathbb {E}}}}}[ \sigma ^p ] < \infty$ for $p < \nu / \alpha$ and since $\nu < \alpha - 1$ was arbitrary we get ${{\mathrm{{\mathbb {E}}}}}[ \sigma ^p ] < \infty$ for $p < 1 - (1/\alpha )$.

For the non-existence of moments when $\alpha \in (1,2)$, we will apply Lemma 10 with $Y_n = f_\nu (X_n , \xi _n) = (1+X_n)^\nu$ for some $\nu \in (0,\alpha )$. Then, condition (6.2) follows from (6.5), which also shows that for $r \in (1, \alpha / \nu )$,

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ Y_{n+1}^r - Y_n^r \mid (X_n, \xi _n) = (x,i) \right] = \frac{c_i}{2} x^{r \nu - \alpha } C^\mathrm{sym} (r \nu , \alpha ) + o (x^{r \nu -\alpha } ) , \end{aligned}$$

so that ${{\mathrm{{\mathbb {E}}}}}[ Y_{n+1}^r - Y_n^r \mid {{\mathcal {F}}}_n ] \le c_i C^\mathrm{sym} (r \nu , \alpha ) Y_n^{r - (\alpha / \nu )}$, for all $Y_n$ sufficiently large. Since $\alpha / \nu > 1$ condition (6.3) follows. Finally, we may choose $\nu < \alpha$ close enough to $\alpha$ and then take $\gamma \in (\alpha -1, \nu )$ so that from (6.5), with Lemmas 11 and 12, ${{\mathrm{{\mathbb {E}}}}}[ Y_{n+1}^{\gamma /\nu } - Y_n^{\gamma /\nu } \mid {{\mathcal {F}}}_n ] \ge 0$, for all $Y_n$ sufficiently large. Thus, we may apply Lemma 10 to obtain ${{\mathrm{{\mathbb {E}}}}}[ \sigma ^p ] = \infty$ for $p > \gamma / \nu$, and taking $\gamma$ close to $\alpha -1$ and $\nu$ close to $\alpha$ we can achieve any $p > 1 - (1/\alpha )$, as claimed.

Next, suppose that $\alpha \ge 2$. A similar argument to before but this time using Lemmas 3 and 4 shows that for any $\nu \in (0,1)$,

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}[ Y_{n+1} - Y_n \mid {{\mathcal {F}}}_n ] \le - \varepsilon Y_n^{1-(2/\nu )} , \quad \text { on }\quad \{ Y_n \ge y_0 \} , \end{aligned}$$

for some $\varepsilon >0$ and $y_0 \in {{\mathbb {R}}}_+$. Lemma 9 then shows that ${{\mathrm{{\mathbb {E}}}}}[ \sigma ^p ] < \infty$ for $p < \nu / 2$ and since $\nu \in (0,1)$ was arbitrary we get ${{\mathrm{{\mathbb {E}}}}}[ \sigma ^p ] < \infty$ for $p < 1/2$.

We sketch the argument for ${{\mathrm{{\mathbb {E}}}}}[ \sigma ^{p} ] = \infty$ when $p > 1/2$. For $\nu \in (1,\alpha )$, it is not hard to show that $D f_\nu (x,i) \ge 0$ for all x sufficiently large, and we may verify the other conditions of Lemma 10 to show that ${{\mathrm{{\mathbb {E}}}}}[ \sigma ^p ] = \infty$ for $p > 1/2$. $\square$

Proof of Theorem 6

Most of the proof is similar to that of Theorem 4, so we omit the details. The case where a different argument is required is the non-existence part of the case $\alpha \ge 1$. We have that for some $\varepsilon >0$ and all y sufficiently large, ${{\mathbb {P}}}_{x,i} [ X_1 \ge y ] \ge \varepsilon (x + y)^{-\alpha }$. A similar argument to Lemma 4 shows that for any $\nu \in (0,1)$, for some $C \in {{\mathbb {R}}}_+$, ${{\mathrm{{\mathbb {E}}}}}[ X_{n+1}^\nu - X_n^\nu \mid X_n = x ] \ge - C$. Then, a suitable maximal inequality implies that with probability at least 1 / 2 started from $X_1 \ge y$ it takes at least $c y^{\nu }$ steps for $X_n$ to return to a neighbourhood of 0, for some $c>0$. Combining the two estimates gives

$$\begin{aligned} {{\mathbb {P}}}_{x,i} \left[ \sigma \ge c y^{\nu } \right] \ge \frac{1}{2}(x + y)^{-\alpha } , \end{aligned}$$

which implies ${{\mathrm{{\mathbb {E}}}}}[ \sigma ^p ] =\infty$ for $p \ge \alpha / \nu$, and since $\nu \in (0,1)$ was arbitrary, we can achieve any $p > \alpha$.$\square$

7 Recurrence Classification in the Critical Cases

7.1 Logarithmic Lyapunov Functions

In this section, we prove Theorem 1(b)(iii). Throughout this section, we write $a_k := \cot ( \chi _k \pi \alpha _k )$ and suppose that $\max _{k \in {{\mathcal {S}}}} \chi _k \alpha _k < 1$, that $\sum _{k \in {{\mathcal {S}}}} \mu _k a_k = 0$, and that $v_i \in {{\mathfrak {D}}}^+_{\alpha _i,c_i}$ for all $i \in {{\mathcal {S}}}$, that is, for $y >0$, $v_i ( y ) = c_i (y) y^{-1-\alpha _i}$, with $\alpha _i \in (0,\infty )$ and $c_i (y) = c_i + O (y ^{-\delta } )$, where $\delta >0$ may be chosen so as not to depend upon i.

To prove recurrence in the critical cases, we need a function that grows more slowly than any power; now, the weights $\lambda _k$ are additive rather than multiplicative. For $x \in {{\mathbb {R}}}$, write $g(x) := \log ( 1 + |x| )$. Then, for $x \in {{\mathbb {R}}}_+$ and $k \in {{\mathcal {S}}}$, define

$$\begin{aligned} g (x, k) := g (x) + \lambda _k = \log \left( 1 + |x| \right) + \lambda _k , \end{aligned}$$

(7.1)

where $\lambda _k >0$ for all $k \in {{\mathcal {S}}}$. Also write

$$\begin{aligned} h (x , k ) := \left( g (x , k) \right) ^{1/2} = \left( \log ( 1 +|x| ) + \lambda _k \right) ^{1/2} . \end{aligned}$$

Lemma 6 shows that there exist $\lambda _k > 0$ ($k \in {{\mathcal {S}}}$) such that

$$\begin{aligned} a_k + \sum _{j \in {{\mathcal {S}}}} p (i, j) ( \lambda _j - \lambda _i ) = 0 ; \end{aligned}$$

(7.2)

we fix such a choice of the $\lambda _k$ from now on.

We prove recurrence by establishing the following result.

Lemma 13

Suppose that the conditions of Theorem 1(b)(iii) hold, and that $(\lambda _k ; k \in {{\mathcal {S}}})$ are such that (7.2) holds. Then, there exists $x_0 \in {{\mathbb {R}}}_+$ such that

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ h \left( X_{n+1}, \xi _{n+1} \right) - h \left( X_n, \xi _n \right) \mid X_n = x, \xi _n = i \right] \le 0 , ~\text {for} ~ x \ge x_0 \quad \text { and all }\quad i \in {{\mathcal {S}}}. \end{aligned}$$

It is not easy to perform the integrals required to estimate Dh(x, i) (and hence establish Lemma 13) directly. However, the integrals for Dg(x, i) are computable (they appear in Lemma 2), and we can use some analysis to relate Dh(x, i) to Dg(x, i). Thus, the first step in our proof of Lemma 13 is to estimate Dg(x, i).

For the Lyapunov function g, (5.1) gives for $x \in {{\mathbb {R}}}_+$ and $i \in {{\mathcal {S}}}$,

$$\begin{aligned} Dg(x,i)&= {{\mathrm{{\mathbb {E}}}}}\left[ g (X_{n+1}, \xi _{n+1}) - g (X_n,\xi _n) \mid (X_n, \xi _n ) = (x, i) \right] \nonumber \\&= \sum _{j \in {{\mathcal {S}}}} p (i,j) (\lambda _j - \lambda _i ) \int _{-\infty }^{-x} w_i (y) {\mathrm d}y + \int _{-\infty }^\infty \left( g(x+y) - g(x) \right) w_i (y) {\mathrm d}y \nonumber \\&= \chi _i \sum _{j \in {{\mathcal {S}}}} p (i,j) (\lambda _j - \lambda _i )T_{i} (x) + G^\flat _{i} (x), \end{aligned}$$

(7.3)

where we have used the fact that g(x) is defined for all $x \in {{\mathbb {R}}}$ and symmetric about 0, and we have introduced the notation

$$\begin{aligned} T_i (x) := \int _x^\infty v_i (y) {\mathrm d}y = \frac{c_i}{\alpha _i} x^{-\alpha _i} + O (x ^{-\alpha _i - \delta } ), \end{aligned}$$

and

$$\begin{aligned} G^\flat _{i} (x) := {\left\{ \begin{array}{ll} \frac{1}{2} \int _{-\infty }^\infty \left( g (x+y) - g (x) \right) v_i (|y|) {\mathrm d}y &{} \text { if } \flat =\mathrm{sym}\\ \int _{0}^\infty \left( g(x-y) - g(x) \right) v_i (y) {\mathrm d}y &{} \text { if } \flat =\mathrm{one.} \end{array}\right. } \end{aligned}$$

(7.4)

The next lemma, proved in the next subsection, estimates the integrals in (7.4).

Lemma 14

Suppose that $v_i \in {{\mathfrak {D}}}^+_{\alpha _i,c_i}$. Then, for $\flat \in \{ one , sym \}$, for $\alpha \in (0, 1/\chi _i)$, for some $\eta >0$, as $x \rightarrow \infty$,

$$\begin{aligned} G^\flat _{i} ( x ) = \chi _i c_{i} x^{-{\alpha _i}} \frac{\pi }{{\alpha _i}} \cot ( \chi _i \pi \alpha _i ) + O ( x^{-{\alpha _i}-\eta } ) . \end{aligned}$$

Note that Lemma 14 together with (7.3) shows that

$$\begin{aligned} D g (x,i)&= \frac{\chi _i c_i}{\alpha _i} \left( a_i + \sum _{j \in {{\mathcal {S}}}} p (i,j) (\lambda _j - \lambda _i ) \right) x^{-{\alpha _i}} + O ( x^{-{\alpha _i}-\eta } ) = O ( x^{-\alpha _i -\eta } ) , \end{aligned}$$

(7.5)

by (7.2). This is not enough by itself to establish recurrence, since the sign of the $O ( x^{-\alpha _i -\eta } )$ term is unknown. This is why we need the function h(x, i).

7.2 Proof of Recurrence in the Critical Case

Proof of Lemma 14

To ease notation, we drop the subscripts i everywhere for the duration of this proof. From (7.4) we obtain, for $x >0$,

$$\begin{aligned} G^\mathrm{sym}(x ) =&\frac{1}{2} \int _x^\infty \log \left( \frac{1+y -x}{1+x} \right) v (y) {\mathrm d}y \nonumber \\&+ \frac{1}{2} \int _x^\infty \log \left( \frac{1+x+y}{1+x} \right) v (y) {\mathrm d}y \nonumber \\&+ \frac{1}{2} \int _0^x \left[ \log \left( \frac{1+x+y}{1+x} \right) + \log \left( \frac{1+x-y}{1+x} \right) \right] v (y) {\mathrm d}y . \end{aligned}$$

(7.6)

Let $\alpha \in (0,2)$. The claim in the lemma for $\flat = sym$ will follow from the estimates

$$\begin{aligned} \int _x^\infty \log \left( \frac{1+x+y}{1+x} \right) v (y) {\mathrm d}y&= c x^{-\alpha } j_0^\alpha + O (x^{-\alpha -\eta } ) ; \nonumber \\ \int _x^\infty \log \left( \frac{1+y -x}{1+x} \right) v (y) {\mathrm d}y&= c x^{-\alpha } j_2^\alpha + O (x^{-\alpha -\eta } ) ; \nonumber \\ \int _0^x \left[ \log \left( \frac{1+x+y}{1+x} \right) + \log \left( \frac{1+x-y}{1+x} \right) \right] v (y) {\mathrm d}y&= c x^{-\alpha } j_1^\alpha + O (x^{-\alpha -\eta } ); \end{aligned}$$

(7.7)

since then we obtain from (7.6) with (7.7) and Lemma 2 that

$$\begin{aligned} G^\mathrm{sym}(x ) = \frac{c}{2} x^{-\alpha } \frac{1}{\alpha } \left( \psi \left( 1 - \tfrac{\alpha }{2} \right) - \psi \left( \tfrac{\alpha }{2} \right) \right) + O (x^{-\alpha -\eta } ), \end{aligned}$$

which yields the stated result via the digamma reflection formula (equation 6.3.7 from [1, p. 259]). We present here in detail the proof of only the final estimate in (7.7); the others are similar. Some algebra followed by the substitution $u = y/(1+x)$ shows that the third integral in (7.7) is

$$\begin{aligned}&\int _0^x \log \left( 1 - \frac{y^2}{(1+x)^2} \right) c (y) y^{-1-\alpha } {\mathrm d}y \\&\quad = (1+x)^{-\alpha } \int _0^{\frac{x}{1+x}} \log ( 1 -u^2 ) c ( u (1+x) ) u^{-1-\alpha } {\mathrm d}u . \end{aligned}$$

There is a constant $C \in {{\mathbb {R}}}_+$ such that for all x sufficiently large,

$$\begin{aligned} \int _0^{\frac{\sqrt{x}}{1+x}} \left| \log ( 1 -u^2 ) c ( u (1+x) ) u^{-1-\alpha } \right| {\mathrm d}u \le C \int _0^{\frac{\sqrt{x}}{1+x}} u^{1-\alpha } {\mathrm d}u = O \left( x^{(\alpha /2)-1} \right) , \end{aligned}$$

using Taylor’s theorem for $\log$ and the fact that c(y) is uniformly bounded. On the other hand, for $u > \frac{\sqrt{x}}{1+x}$ we have $c ( u (1+x) ) = c + O (x^{-\delta /2} )$, so that

$$\begin{aligned} \int _{\frac{\sqrt{x}}{1+x}}^{\frac{x}{1+x}} \log ( 1 -u^2 ) c ( u (1+x) ) u^{-1-\alpha } {\mathrm d}u = c \int _{\frac{\sqrt{x}}{1+x}}^{\frac{x}{1+x}} \log ( 1 -u^2 ) u^{-1-\alpha } {\mathrm d}u + O \left( x^{-\delta /2} \right) . \end{aligned}$$

Here we have that

$$\begin{aligned} \int _{\frac{\sqrt{x}}{1+x}}^{\frac{x}{1+x}} \log ( 1 -u^2 ) u^{-1-\alpha } {\mathrm d}u = j_1^\alpha + O \left( x^{(\alpha /2)-1 } \right) + O \left( x^{-1} \log x \right) . \end{aligned}$$

Combining these estimates and using the fact that $\alpha \in (0,2)$, we obtain the final estimate in (7.7). The claim in the lemma for $\flat = one$ follows after some analogous computations, which we omit. $\square$

Now, we relate Dh(x, i) to Dg(x, i), by comparing the individual integral terms.

Lemma 15

Suppose that $v_i \in {{\mathfrak {D}}}^+_{\alpha _i,c_i}$. Then, for all x and all i,

$$\begin{aligned}&\int _x^\infty \bigl ( h (x+y, i) -h(x,i) \bigr ) v_i (y) {\mathrm d}y \le \frac{1}{2 h(x,i)} \int _x^\infty \left( g (x+y ) -g(x ) \right) v_i (y) {\mathrm d}y ; \\&\int _0^x \bigl ( h (x-y ,i) - h(x,i) \bigr ) v_i (y) {\mathrm d}y \le \frac{1}{2 h (x,i)} \int _0^x \left( g (x-y) - g(x) \right) v_i (y) {\mathrm d}y ; \\&\qquad \text {and } \int _0^x \bigl ( h (x+y,i) + h(x-y,i) - 2 h (x, i ) \bigr ) v_i (y) {\mathrm d}y \\&\quad \le \frac{1}{2 h (x,i)} \int _0^x \bigl ( g(x+y) + g (x-y) - 2 g(x) \bigr ) v_i (y) {\mathrm d}y . \end{aligned}$$

Finally, there exists $\varepsilon >0$ such that, for all $i, j \in {{\mathcal {S}}}$ and all x sufficiently large,

$$\begin{aligned}&\int _x^\infty \left( h (y-x, j) - h (x, i) \right) v_i (y) {\mathrm d}y \\&\quad \le \frac{1}{2 h (x,i)} \int _x^\infty \left( g (y-x,j) - g(x,i) \right) v_i (y) {\mathrm d}y - \varepsilon \frac{x^{-\alpha _i}}{\log ^{3/2} x} . \end{aligned}$$

Proof

The proof is based on the observation that, since $( h(x,i) )^2 = g(x,i)$,

$$\begin{aligned} h(z,j) - h(x,i) = \frac{g(z,j) - g (x,i)}{h(z,j) + h (x, i)} . \end{aligned}$$

Thus, for $y \ge 0$,

$$\begin{aligned} h (x+y, i) -h(x,i) = \frac{g(x+y) - g (x)}{h(x+y,i) + h (x, i)} \le \frac{g (x+y) - g(x)}{2 h (x, i)} , \end{aligned}$$

(7.8)

since $g(x+y) - g (x) \ge 0$ and $h(x+y,i) \ge h(x,i)$. This gives the first inequality in the lemma. Similarly, for $y \in [0,x]$,

$$\begin{aligned} h (x-y ,i) - h(x,i) = \frac{g (x -y) - g(x)}{h(x-y,i)+h(x,i)} \le \frac{g (x -y) - g(x)}{2h(x,i)} , \end{aligned}$$

since $g (x -y) - g(x) \le 0$ and $h(x-y,i) \le h(x,i)$. This gives the second inequality and also yields the third inequality once combined with the $y \in [0,x]$ case of (7.8).

Finally, for $y \ge x$ note that

$$\begin{aligned} h (y-x ,j ) -h (x ,i )&= \frac{g (y-x ,j ) -g (x ,i ) }{ h ( y-x , j ) + h (x , i )} . \end{aligned}$$

(7.9)

Also note that, for $y \ge x > 0$, $g (x , i ) = g(x) + \lambda _i$ and $g (y-x , j ) = g( y-x ) +\lambda _j$, so

$$\begin{aligned} g (y -x , j ) -g (x , i ) = \log \left( {\mathrm {e}}^{\lambda _j-\lambda _i} \frac{1+y-x}{1+x} \right) . \end{aligned}$$

So the sign of the expression in (7.9) is non-positive for $y \le \psi (x) := x -1 + (1+x) {\mathrm {e}}^{\lambda _i - \lambda _j}$ and non-negative for $y \ge \psi (x)$, and

$$\begin{aligned} g (\psi (x) - x, j ) -g (x , i ) = 0 . \end{aligned}$$

(7.10)

By the monotonicity in y of the denominator, the expression in (7.9) satisfies

$$\begin{aligned} \frac{g (y-x ,j ) -g (x ,i ) }{ h ( y-x , j ) + h (x , i )} \le \frac{g (y-x ,j ) -g (x ,i ) }{ h ( \psi (x) -x , j ) + h (x , i )} \end{aligned}$$

both for $y \in [0,\psi (x)]$ and for $y \in [\psi (x), \infty )$. Here $h ( \psi (x) - x,j) = h (x , i )$, by (7.10). Hence, we obtain the bound

$$\begin{aligned} \int _x^\infty \left( h (y-x , j ) -h (x , i ) \right) v_i (y) {\mathrm d}y \le \int _x^\infty \left( \frac{g (y-x,j ) -g (x,i ) }{ 2 h ( x , i ) } \right) v_i (y) {\mathrm d}y. \end{aligned}$$

To improve on this estimate, suppose that $y \ge K x$ where $K \in {{\mathbb {N}}}$ is such that $K x > \psi (x)$. Then, using the fact that the numerator in (7.9) is positive, we may choose $K \in {{\mathbb {N}}}$ such that for all j and all $y \ge Kx$,

$$\begin{aligned} h (y-x ,j ) -h (x ,i )\le & {} \frac{g (y-x ,j ) -g (x ,i ) }{ h ( (K-1) x , j ) + h (x , i )} \\\le & {} \frac{g (y-x ,j ) -g (x ,i ) }{ \left( \log ( 1 + |x| ) + \lambda _i + 1 \right) ^{1/2} + h (x,i) } , \end{aligned}$$

for K sufficiently large (depending on $\max _j \lambda _j$) and and all x sufficiently large. Here

$$\begin{aligned} \left( \log ( 1 + |x| ) + \lambda _i + 1 \right) ^{1/2} = h (x,i) \left( 1 + \frac{1}{g(x,i)} \right) ^{1/2} = h (x,i) + \frac{1 + o(1)}{2 \log ^{1/2} x } . \end{aligned}$$

It follows that, for some $\varepsilon >0$, for all x sufficiently large, for $y \ge Kx$,

$$\begin{aligned} h (y-x ,j ) -h (x ,i ) \le \frac{g (y-x ,j ) -g (x ,i ) }{ 2 h(x,i) } - \varepsilon \left( \frac{g (y-x ,j ) -g (x ,i ) }{ \log ^{3/2} x } \right) . \end{aligned}$$

The final inequality in the lemma now follows since, for all x sufficiently large,

$$\begin{aligned}&\int _{Kx} ^\infty \left( g (y-x ,j ) -g (x ,i ) \right) v_i (y) {\mathrm d}y \\&\quad \ge \frac{c_i}{2} \int _{K x} ^\infty \log \left( {\mathrm {e}}^{\lambda _j - \lambda _i }\left( \frac{2 + y }{1+x} - 1 \right) \right) y^{-1-\alpha _i} {\mathrm d}y \\&\quad \ge \frac{c_i}{2} x^{-\alpha _i} \int _{K/2}^\infty \log \left( {\mathrm {e}}^{\lambda _j - \lambda _i } ( u -1 ) \right) u^{-1-\alpha _i} {\mathrm d}u , \end{aligned}$$

where we used the substitution $u = \frac{2+y}{1+x}$. For K sufficiently large, the term inside the logarithm is uniformly positive, and the claimed bound follows. $\square$

Now, we may complete the proofs of Lemma 13 and then Theorem 1(b)(iii).

Proof of Lemma 13

Lemma 15 together with (7.5) shows that,

$$\begin{aligned} D h ( x, i) \le \frac{ D g (x,i)}{ 2 h (x, i) } - \varepsilon \frac{ x^{-\alpha _i}}{\log ^{3/2} x} \le - \varepsilon \frac{ x^{-\alpha _i}}{\log ^{3/2} x} + O ( x^{-\alpha _i - \eta } ) \le 0 , \end{aligned}$$

for all x sufficiently large. $\square$

Proof of Theorem 1(b)(iii)

Lemma 13 with Lemma 16 shows that $\liminf _{n \rightarrow \infty } X_n \le x_0$, a.s., and then, a similar argument to that in the proof of parts (a) and (b)(i) of Theorem 1 shows that $\liminf _{n \rightarrow \infty } X_n =0$, a.s. $\square$

Appendix: Technical Results

1.1 Semimartingale Results

Lemma 16

Let $(X_n, \xi _n)$ be an ${{\mathcal {F}}}_n$-adapted process taking values in ${{\mathbb {R}}}_+\times {{\mathcal {S}}}$. Let $f: {{\mathbb {R}}}_+\times {{\mathcal {S}}}\rightarrow {{\mathbb {R}}}_+$ be such that $\lim _{x \rightarrow \infty } f(x,i) = \infty$ for all $i \in {{\mathcal {S}}}$, and ${{\mathrm{{\mathbb {E}}}}}f (X_0, \xi _0 ) < \infty$. Suppose that there exist $x_0 \in {{\mathbb {R}}}_+$ and $C < \infty$ for which, for all $n \ge 0$,

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}\left[ f\left( X_{n+1} , \xi _{n+1} \right) - f\left( X_n, \xi _n \right) \mid {{\mathcal {F}}}_n \right]&\le 0, \quad \text {on}\quad \{ X_n > x_0 \}, \ \text {a.s.}; \\ {{\mathrm{{\mathbb {E}}}}}\left[ f\left( X_{n+1} , \xi _{n+1} \right) - f\left( X_n , \xi _n \right) \mid {{\mathcal {F}}}_n \right]&\le C, \quad \text {on}\quad \{ X_n \le x_0 \}, \ \text {a.s.}\end{aligned}$$

Then,

$$\begin{aligned} {{\mathbb {P}}}\Bigl [ \bigl \{ \limsup _{n \rightarrow \infty } X_n < \infty \bigr \} \cup \bigl \{ \liminf _{n \rightarrow \infty } X_n \le x_0 \bigr \} \Bigr ] = 1 . \end{aligned}$$

Proof

First note that, by hypothesis, ${{\mathrm{{\mathbb {E}}}}}f(X_{1}, \xi _1 ) \le {{\mathrm{{\mathbb {E}}}}}f (X_0 ,\xi _0 ) + C < \infty$, and iterating this argument, it follows that ${{\mathrm{{\mathbb {E}}}}}f (X_n, \xi _n) < \infty$ for all $n \ge 0$.

Fix $n \in {{\mathbb {Z}}}_+$. For $x_0 \in {{\mathbb {R}}}_+$ in the hypothesis of the lemma, write $\lambda = \min \{ m \ge n : X_m \le x_0 \}$. Let $Y_m = f(X_{m\wedge \lambda }, \xi _{m \wedge \lambda } )$. Then, $(Y_m, m \ge n)$ is an $({{\mathcal {F}}}_m , m \ge n )$-adapted non-negative supermartingale. Hence, by the supermartingale convergence theorem, there exists $Y_\infty \in {{\mathbb {R}}}_+$ such that $\lim _{m \rightarrow \infty } Y_m = Y_\infty$, a.s. In particular,

$$\begin{aligned} \limsup _{m \rightarrow \infty } f(X_m, \xi _m) \le Y_\infty ,\quad \text {on}\quad \{ \lambda = \infty \} . \end{aligned}$$

Set $\zeta _i = \sup \{ x \ge 0 : f(x,i) \le 1+ Y_\infty \}$, which has $\zeta _i < \infty$ a.s. since $\lim _{x \rightarrow \infty } f(x,i) = \infty$. Then, $\limsup _{m \rightarrow \infty } X_m \le \max _i \zeta _i < \infty$ on $\{ \lambda = \infty \}$. Hence

$$\begin{aligned} {{\mathbb {P}}}\Bigl [ \bigl \{ \limsup _{n \rightarrow \infty } X_n < \infty \bigr \} \cup \bigl \{ \inf _{m \ge n } X_m \le x_0 \bigr \} \Bigr ] = 1 . \end{aligned}$$

Since $n \in {{\mathbb {Z}}}_+$ was arbitrary, the result follows:

$$\begin{aligned} {{\mathbb {P}}}\Bigl [ \bigl \{ \limsup _{n \rightarrow \infty } X_n < \infty \bigr \} \cup \bigcap _{n \ge 0} \bigl \{ \inf _{m \ge n } X_m \le x_0 \bigr \} \Bigr ] = 1 . \end{aligned}$$

This completes the proof. $\square$

Lemma 17

Let $(X_n, \xi _n)$ be an ${{\mathcal {F}}}_n$-adapted process taking values in ${{\mathbb {R}}}_+\times {{\mathcal {S}}}$. Let $f: {{\mathbb {R}}}_+\times {{\mathcal {S}}}\rightarrow {{\mathbb {R}}}_+$ be such that $\sup _{x,i} f(x,i) < \infty$ and $\lim _{x \rightarrow \infty } f(x,i) = 0$ for all $i \in {{\mathcal {S}}}$. Suppose that there exists $x_1 \in {{\mathbb {R}}}_+$ for which $\inf _{y \le x_1} f(y,i) >0$ for all i and

$$\begin{aligned} {{\mathrm{{\mathbb {E}}}}}[ f(X_{n+1}, \xi _{n+1} ) - f(X_n, \xi _n) \mid {{\mathcal {F}}}_n ] \le 0, \text { on } \{X_n \ge x_1 \}, \quad \text {for all}\quad n \ge 0. \end{aligned}$$

Then, for any $\varepsilon >0$ there exists $x \in (x_1, \infty )$ for which, for all $n \ge 0$,

$$\begin{aligned} {{\mathbb {P}}}\Bigl [ \inf _{m \ge n} X_m \ge x_1 \; \Bigl | \;{{\mathcal {F}}}_n \Bigr ] \ge 1 - \varepsilon , \quad \text {on}\quad \{ X_n \ge x \} . \end{aligned}$$

Proof

Fix $n \in {{\mathbb {Z}}}_+$. For $x_1 \in {{\mathbb {R}}}_+$ in the hypothesis of the lemma, write $\lambda = \min \{ m \ge n : X_m \le x_1 \}$ and set $Y_m = f(X_{m \wedge \lambda }, \xi _{m \wedge \lambda } )$. Then, $(Y_m , m \ge n )$ is an $({{\mathcal {F}}}_m , m \ge n )$-adapted non-negative supermartingale, and so converges a.s. as $m \rightarrow \infty$ to some $Y_\infty \in {{\mathbb {R}}}_+$. Moreover, by the optional stopping theorem for supermartingales,

$$\begin{aligned} Y_{n} \ge {{\mathrm{{\mathbb {E}}}}}\left[ Y_\infty \mid {{\mathcal {F}}}_{n} \right] \ge {{\mathrm{{\mathbb {E}}}}}\left[ Y_\infty {{\mathbf {1}}}{\{ \lambda < \infty \}} \mid {{\mathcal {F}}}_{n} \right] , \ \text {a.s.}\end{aligned}$$

Here, we have that, a.s.,

$$\begin{aligned} Y_\infty {{\mathbf {1}}}{\{ \lambda< \infty \}}&= \lim _{m \rightarrow \infty } Y_m {{\mathbf {1}}}{\{ \lambda< \infty \}} = f \left( X_{ \lambda } , \xi _\lambda \right) {{\mathbf {1}}}{\{ \lambda< \infty \}} \\&\ge \min _i \inf _{y \le x_1} f(y,i) {{\mathbf {1}}}{\{ \lambda < \infty \}} . \end{aligned}$$

Combining these inequalities we obtain

$$\begin{aligned} \min _i \inf _{y \le x_1} f(y,i) {{\mathbb {P}}}\left[ \lambda < \infty \mid {{\mathcal {F}}}_{n} \right] \le Y_{n}, \ \text {a.s.}\end{aligned}$$

In particular, on $\{ X_{n} \ge x > x_1 \}$, we have $Y_{n} = f( X_{n}, \xi _n )$ and so

$$\begin{aligned} \min _i \inf _{y \le x_1} f(y) {{\mathbb {P}}}\left[ \lambda < \infty \mid {{\mathcal {F}}}_n \right] \le f \left( X_n, \xi _n\right) \le \max _i \sup _{y \ge x } f(y, i) . \end{aligned}$$

Since $\lim _{y \rightarrow \infty } f(y,i) = 0$ and $\inf _{y \le x_1} f(y,i) >0$, given $\varepsilon >0$ we can choose $x > x_1$ large enough so that

$$\begin{aligned} \frac{ \max _i \sup _{y \ge x } f(y,i)}{\min _i \inf _{y \le x_1} f(y,i)} < \varepsilon ; \end{aligned}$$

the choice of x depends only on f, $x_1$, and $\varepsilon$, and, in particular, does not depend on n. Then, on $\{ X_n \ge x \}$, ${{\mathbb {P}}}[ \lambda< \infty \mid {{\mathcal {F}}}_n ] < \varepsilon$, as claimed.

1.2 Proofs of Integral Computations

Proof of Lemma 1

Consider $i_0^{\nu ,\alpha }$. With the change of variable $v = 1/u$, we get

$$\begin{aligned} \int _0^1 u^{-1-\alpha } (1+u)^\nu {\mathrm d}u&= \int _0^1 v^{\alpha - \nu - 1} (1+ v)^\nu {\mathrm d}v \\&= \frac{{\varGamma }(\alpha -\nu ) {\varGamma }(1)}{{\varGamma }(\alpha - \nu +1 )} {}_2 F_1 \left( - \nu , \alpha -\nu ; \alpha -\nu +1 ; -1 \right) , \end{aligned}$$

by the integral representation for the Gauss hypergeometric function (see equation 15.3.1 of [1, p. 558]). The given expression for $i_0^{\nu ,\alpha }$ follows.

The integral $i_{2,0}^{\alpha }$ is trivial. Consider $i_{2,1}^{\nu ,\alpha }$. The substitution $v = 1/u$ gives

$$\begin{aligned} \int _1^\infty u^{-1-\alpha } ( u -1)^{\nu } {\mathrm d}u = \int _0^1 (1- v)^{\nu } v^{\alpha -\nu -1} {\mathrm d}v = \frac{{\varGamma }(1 +\nu ) {\varGamma }(\alpha -\nu )}{{\varGamma }(1+\alpha )} , \end{aligned}$$

by the integral formula for the Beta function (see equation 6.2.1 of [1, p. 258]), provided $\nu > - 1$ and $\alpha - \nu >0$. Hence, we obtain the given expression for $i_{2,1}^{\nu ,\alpha }$.

Next, consider $i_1^{\nu ,\alpha }$. By considering separately the asymptotics of the integrand as $u \downarrow 0$ and $u \uparrow 1$, we see that $i_1^{\nu ,\alpha }$ is finite provided $\alpha \in (0,2)$ and $\nu > -1$. For $u\in [-1,1]$, we have the Taylor series expansion $(1+u)^\nu = \sum _{n\ge 0} \frac{u^n}{n!} \frac{{\varGamma }(\nu +1)}{{\varGamma }(\nu +1-n)}$. Hence,

$$\begin{aligned} \left( (1+u)^\nu +(1-u)^\nu -2 \right) u^{-1-\alpha }&= 2 \sum _{n\ge 1} \frac{u^{2n-1-\alpha }}{(2n)!} \frac{{\varGamma }(\nu +1)}{{\varGamma }(\nu +1-2n)} . \end{aligned}$$

Here, the power series for $n \ge 2$ converges normally (hence uniformly) over $|u| \le 1$. This remark allows interchanging summation and integration to obtain $i_1^{\nu , \alpha } = W^{\nu ,\alpha } (1)$, where for $|z| \le 1$ we define

$$\begin{aligned} W^{\nu ,\alpha }(z)&:= 2\sum _{n\ge 1} \frac{z^{n}}{2n-\alpha } \frac{1}{(2n)!} \frac{{\varGamma }(\nu +1)}{{\varGamma }(\nu +1-2n)}\\&=\frac{1}{2-\alpha } \nu (\nu -1)z\left[ 1 + \frac{2-\alpha }{4-\alpha } \frac{1}{4!} (\nu -2)(\nu -3) z \right. \nonumber \\&\quad \left. + \frac{2-\alpha }{6-\alpha } \frac{1}{6!}(\nu -2)(\nu -3)(\nu -4)(\nu -5) z^2+\cdots \right] \nonumber \\&= \frac{1}{2-\alpha } \nu (\nu -1)z\left[ \sum _{n\ge 0} c_n z^n\right] , \nonumber \end{aligned}$$

(7.11)

where $c_n= \frac{2-\alpha }{2n+2-\alpha } \frac{1}{(2(n+1))!} (\nu -2)(\nu -3)\cdots (\nu -2n-1)$, for $n\ge 1$ and $c_0=1$. An elementary computation yields

$$\begin{aligned} \frac{c_{n+1}}{c_n}= \frac{(n+1-\alpha /2) (n+1-\nu /2) (n+3/2-\nu /2) (n+1)}{ (n+2 -\alpha /2) (n+2) (n+1/2)}\frac{1}{n+1}. \end{aligned}$$

Therefore (see [3, p. 10] or [12, equation 5.81, p. 207] for a more easily accessible reference), $\sum _{n\ge 0} c_n z^n= {}_4 F_3 ( 1, 1-\frac{\nu }{2}, 1-\frac{\alpha }{2}, \frac{3-\nu }{2}, 1 ; 2, \frac{3}{2}, 2-\frac{\alpha }{2} ; z )$. Now, the series defining the generalized hypergeometric function $_pF_q ( \beta _1, \ldots , \beta _p ; \gamma _1, \ldots , \gamma _q ; z)$ for $p=q+1$ converges for all z with $|z|<1$; for $z=1$, the series converges for $\sum _{i=1}^q\gamma _i > \sum _{j=1}^p \beta _j$, a condition that reduces to $\nu > -1$ in the present case. Hence,

$$\begin{aligned} i_1^{\nu ,\alpha }= \frac{ \nu (\nu -1) }{2-\alpha } \ {}_4 F_3 \left( 1, 1-\tfrac{\nu }{2}, 1-\tfrac{\alpha }{2}, \tfrac{3-\nu }{2}, 1 ; 2, \tfrac{3}{2}, 2-\tfrac{\alpha }{2} ; 1 \right) . \end{aligned}$$

Finally, consider ${{\tilde{i}}}^{\nu ,\alpha }_1$. Due to the singularity at $u=0$, we compute

$$\begin{aligned} \int _0^1\left( (1-u)^{\nu } - 1 \right) u^{-1-\alpha } {\mathrm d}u = \lim _{t \downarrow 0} \int _t^1 \left( (1-u)^{\nu } - 1 \right) u^{-1-\alpha } {\mathrm d}u . \end{aligned}$$

For the last integral, we get, for $\alpha > 0$ and $t >0$,

$$\begin{aligned} \int _t^1 \left( (1-u)^{\nu } - 1 \right) u^{-1-\alpha } {\mathrm d}u = \int _t^1 (1-u)^{\nu } u^{-1-\alpha } {\mathrm d}u + \frac{1 -t^{-\alpha }}{\alpha } . \end{aligned}$$

With the substitution $v = \frac{1-u}{1-t}$, the last integral becomes

$$\begin{aligned} \int _t^1 (1-u)^{\nu } u^{-1-\alpha } {\mathrm d}u&= (1-t)^{1+\nu } \int _0^1 v^{\nu } (1 - (1-t)v )^{-\alpha -1} {\mathrm d}v \\&= \frac{(1-t)^{1+ \nu } }{1+\nu } {}_2 F_1 \left( 1+\alpha , 1+\nu ; 2+\nu ; 1- t\right) , \end{aligned}$$

provided $\nu > -1$, by the integral representation for the Gauss hypergeometric function (equation 15.3.1 of [1, p. 558]). Now, by equation 15.3.6 from [1, p. 559],

$$\begin{aligned} {}_2 F_1 ( 1+\alpha , 1+\nu ; 2+\nu ; 1- t) =&\frac{{\varGamma }(2+\nu ) {\varGamma }(-\alpha )}{{\varGamma }(1-\alpha +\nu )} {}_2 F_1 \left( 1+\alpha , 1+\nu ; 1+\alpha ; t\right) \\&+ t^{-\alpha } \frac{1+\nu }{\alpha } {}_2 F_1 \left( 1-\alpha + \nu , 1 ; 1-\alpha ; t \right) \\ =&\frac{{\varGamma }(2+\nu ) {\varGamma }(-\alpha )}{{\varGamma }(1-\alpha + \nu )} + \frac{1+\nu }{\alpha } t^{-\alpha } + O \left( t^{1-\alpha } \right) , \end{aligned}$$

as $t \downarrow 0$, provided $\alpha \in (0,1)$. Combining these results, we get

$$\begin{aligned} \int _t^1 ( (1-u)^{\nu } - 1 ) u^{-1-\alpha } {\mathrm d}u = \frac{1}{\alpha } + \frac{ {\varGamma }(1+\nu ) {\varGamma }(-\alpha )}{{\varGamma }(1-\alpha + \nu )} + O \left( t^{1-\alpha } \right) . \end{aligned}$$

Letting $t \downarrow 0$ and using the fact that $- \alpha {\varGamma }(-\alpha ) = {\varGamma }(1-\alpha )$ we obtain the given expression for ${{\tilde{i}}}_{1}^{\nu ,\alpha }$. $\square$

Proof of Lemma 2

We appeal to tables of standard integrals (Mellin transforms) from Section 6.4 of [8]. In particular, the given formulae for $j_0^\alpha$, $j_2^\alpha$, and ${{\tilde{j}}}_1^\alpha$ follow from, respectively, equations 6.4.17, 6.4.20, and 6.4.19 of [8, pp. 315–316]; also used are formulas 6.3.7 and 6.3.8 from [1, p. 259] and the fact that $\psi (1) = -\gamma$. Lastly, for $j_1^\alpha$ we use the substitution $u^2 = s$ to obtain

$$\begin{aligned} j_1^\alpha = \frac{1}{2} \int _0^1 \frac{\log (1-s)}{s^{1+(\alpha /2 )}} {\mathrm d}s = \frac{1}{2} {{\tilde{j}}}_1^{\alpha /2} . \end{aligned}$$

This completes the proof. $\square$

Finally, we need the following elementary fact.

Lemma 18

Let $A \subseteq {{\mathbb {R}}}_+$ be a Borel set, and let f and g be measurable functions from A to ${{\mathbb {R}}}$. Suppose that there exist constants $g_-$ and $g_+$ with $0< g_-< g_+ < \infty$ such that $g_- \le g(u) \le g_+$ for all $u \in A$. Then,

$$\begin{aligned} \left| \int _A f(u) g(u) {\mathrm d}u - g_- \int _A f(u) {\mathrm d}u \right| \le (g^+ - g^-) \int _A | f(u) | {\mathrm d}u . \end{aligned}$$

Proof

It suffices to suppose that $\int _A | f(u) | {\mathrm d}u < \infty$. Then, $\int _A | f(u) g(u) | {\mathrm d}u \le g_+ \int _A | f(u) | {\mathrm d}u < \infty$, and

$$\begin{aligned} \left| \int _A f(u) g(u) {\mathrm d}u - g_- \int _A f(u) {\mathrm d}u \right|&= \left| \int _A f(u) ( g(u) - g_- ) {\mathrm d}u \right| \\&\le \int _A | f(u) | ( g (u) - g_- ) {\mathrm d}u . \end{aligned}$$

This completes the proof. $\square$