Power and Exponential Moments of the Number of Visits and Related Quantities for Perturbed Random Walks

Abstract

Let $(\xi _1,\eta _1),(\xi _2,\eta _2),\ldots$ be a sequence of i.i.d. copies of a random vector $(\xi ,\eta )$ taking values in $\mathbb{R }^2$, and let $S_n:= \xi _1+\cdots +\xi _n$. The sequence $(S_{n-1} + \eta _n)_{n \ge 1}$ is then called perturbed random walk. We study random quantities defined in terms of the perturbed random walk: $\tau (x)$, the first time the perturbed random walk exits the interval $(-\infty ,x]; \,N(x)$, the number of visits to the interval $(-\infty ,x]$; and $\rho (x)$, the last time the perturbed random walk visits the interval $(-\infty ,x]$. We provide criteria for the almost sure finiteness and for the finiteness of exponential moments of these quantities. Further, we provide criteria for the finiteness of power moments of $N(x)$ and $\rho (x)$. In the course of the proofs of our main results, we investigate the finiteness of power and exponential moments of shot-noise processes and provide complete criteria for both, power and exponential moments.

Appendix: Auxiliary Results

1.1 Auxiliary Results from Classical Random Walk Theory

This section contains some facts from classical random walk theory that are either reformulations or slight extensions of known results. The first result is a combination of Theorems 2.1 and 2.2 in [27].

Proposition 5.1

For $a>0$, let $V^*_a(I) := \sum _{n \ge 0} e^{an} \mathbb{P }\{S_n \in I\},\,I \subseteq \mathbb{R }$ Borel and $V^*_a(x) := V^*_a((-\infty ,x]),\,x \in \mathbb{R }$. Further, let $R:=-\log \inf _{t \ge 0} \mathbb{E }e^{-t \xi }$.

• (a)

  • (i) Assume that $\mathbb{P }\{\xi \ge 0\}=1$ and let $\beta :=\mathbb{P }\{\xi =0\}\in [0,1)$. Then for $a>0$ the following conditions are equivalent:

    $$\begin{aligned}&V^*_a(x) < \infty \text{ for} \text{ some/all} x\ge 0;\end{aligned}$$

    (5.1)
    $$\begin{aligned}&0 < V^*_a(I) < \infty \text{ for} \text{ some} \text{ bounded} \text{ interval} I \subseteq \mathbb{R };\end{aligned}$$

    (5.2)
    $$\begin{aligned}&a<-\log \beta \end{aligned}$$

    (5.3)

    where $-\log \beta := \infty$ if $\beta = 0$.

  • (ii) Assume that $\mathbb{P }\{\xi <0\}>0$. Then for $a>0$ condition (5.1) (with $x\in \mathbb{R }$) is equivalent to

    $$\begin{aligned} a < R \quad \text{ or} \quad a=R \ \text{ and} \ \mathbb{E }\xi e^{-\gamma _0\xi }>0, \end{aligned}$$

    (5.4)

    where $\gamma _0$ is the unique positive value defined by $\mathbb{E }e^{-\gamma _0\xi }= e^{-R}$.

• (b) Whenever $V^*_a(x)$ is finite,

  $$\begin{aligned} 0 < \liminf _{x \rightarrow \infty } e^{-\gamma x} V^*_a(x)\le \limsup _{x\rightarrow \infty } e^{-\gamma x}V^*_a(x) < \infty . \end{aligned}$$

Part (a) of the Proposition contains more equivalent criteria for the finiteness of the exponential renewal function of a random walk than Theorem 2.1 in [3]. For this reason, we decided to include a proof.

Proof

We begin with part (a)(i) and assume that $\mathbb{P }\{\xi \ge 0\} = 1$. Then the equivalence between (5.1) and (5.3) follows from [27, Theorem 2.1(b) and (c)]. Moreover, the implication “(5.1)$\Rightarrow$(5.2)” is trivial. It remains to prove that $0 < V^*_a(I) < \infty$ implies that $a<-\log \beta$. We will use contraposition and assume that $a \ge -\log \beta$, in particular, $\beta > 0$. Then let $I \subseteq [0,\infty )$ denote an arbitrary bounded interval with $V^*_a(I) > 0$. We have to show that $V^*_a(I)=\infty$. To this end, notice that $V^*_a(I) > 0$ implies that $\mathbb{P }\{S_n \in I\} > 0$ for some $n \in \mathbb{N }$. Then, for any $k \ge 0$, we infer

$$\begin{aligned} \mathbb{P }\{S_{n+k} \in I\} \ge \mathbb{P }\{S_n \in I, \xi _{n+1} = \ldots = \xi _{n+k} = 0\} = \mathbb{P }\{S_n \in I\} \beta ^k. \end{aligned}$$

In conclusion,

$$\begin{aligned} V^*_a(I)&= \sum _{k \ge 0} e^{ak} \mathbb{P }\{S_k \in I\} \ge \sum _{k \ge 0} e^{a(n+k)} \mathbb{P }\{S_{n+k} \in I\} \\&\ge e^{an} \mathbb{P }\{S_n \in I\} \sum _{k \ge 0} (e^a\beta )^k = \infty . \end{aligned}$$

Part (a)(ii) follows from Theorem 2.1(a) in [27].

Part (b) follows from Theorem 2.2 in [27]. $\square$

Lemma 5.2 will be used in the proof of Theorem 2.9.

Lemma 5.2

Let $p>0$ and $I \subseteq \mathbb{R }$ be an open interval such that $0 < \mathbb{E }\left( \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in I\}}\right)^p < \infty$. Then $\mathbb{E }\left( \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in J\}}\right)^p < \infty$ for any bounded interval $J\subseteq \mathbb{R }$. In particular, $\mathbb{E }N^*(x)^p<\infty$ for some $x \in \mathbb{R }$ entails $\mathbb{E }N^*(y)^p<\infty$ for every $y \in \mathbb{R }$.

Remark 5.3

In the case that $x \ge 0$ the second assertion was known from [31].

Proof

Let $I = (a,b)$ such that $0 < \mathbb{E }\left( \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in I\}}\right)^p < \infty$. We assume w.l.o.g. that $-\infty < a < b < \infty$. We first show that

$$\begin{aligned} \mathbb{E }\bigg (\sum _{n \ge 0} \mathbb{1 }_{\{|S_n| < \varepsilon \}}\bigg )^p < \infty \quad \text{ for} \text{ some} \varepsilon > 0. \end{aligned}$$

(5.5)

Pick $\varepsilon > 0$ so small that $I_{\varepsilon } := (a+\varepsilon ,b-\varepsilon )$ satisfies $\mathbb{E }\left( \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in I_{\varepsilon }\}}\right)^p > 0$. Then $\mathbb{P }\{S_n \in I_{\varepsilon }\} > 0$ for some $n \in \mathbb{N }$. In particular, $\mathbb{P }\{\tau ^{*}(I_{\varepsilon }) < \infty \} > 0$, where $\tau ^{*}(I_{\varepsilon }) = \inf \{n \ge 0: S_n \in I_{\varepsilon }\}$. Using the strong Markov property at $\tau ^{*}(I_{\varepsilon })$, we get

$$\begin{aligned} \infty&> \mathbb{E }\bigg ( \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in I\}}\bigg )^p \\&\ge \mathbb{E }\bigg ( \mathbb{1 }_{\{\tau ^{*}(I_{\varepsilon })<\infty \}} \sum _{n \ge \tau ^{*}(I_{\varepsilon })} \mathbb{1 }_{\{|S_n - S_{\tau ^{*}(I_{\varepsilon })}| < \varepsilon \}}\bigg )^p \\&= \mathbb{P }\{\tau ^{*}(I_{\varepsilon }) < \infty \} \, \mathbb{E }\bigg (\sum _{n \ge 0} \mathbb{1 }_{\{|S_n| < \varepsilon \}}\bigg )^p. \end{aligned}$$

Hence, (5.5) holds. Now let $J$ be a non-empty bounded interval $\subseteq \mathbb{R }$, and $J_1, \ldots , J_m$ open intervals of length at most $\varepsilon$ such that $J \subseteq J_1 \cup \ldots \cup J_m$. Using the inequality $(x_1+\ldots +x_m)^p \le (m^{p-1}\vee 1)(x_1^p+\ldots +x_m^p),\,x_j \ge 0$ for $j=1,\ldots ,m$, leads to

$$\begin{aligned} \mathbb{E }\bigg ( \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in J\}}\bigg )^p&\le \mathbb{E }\bigg ( \sum _{k=1}^m \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in J_k\}}\bigg )^p \\&\le (m^{p-1}\vee 1) \sum _{k=1}^m \mathbb{E }\bigg ( \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in J_k\}}\bigg )^p. \end{aligned}$$

Therefore, it suffices to prove the result under the additional assumption that the length of $J$ is at most $\varepsilon$. Using the strong Markov property at $\tau ^{*}(J) := \inf \{n \ge 0: S_n \in J\}$ gives

$$\begin{aligned} \mathbb{E }\bigg ( \sum _{n \ge 0} \mathbb{1 }_{\{S_n \in J\}}\bigg )^p&\le \mathbb{E }\bigg ( \mathbb{1 }_{\{\tau ^{*}(J)<\infty \}} \sum _{n \ge \tau ^{*}(J)} \mathbb{1 }_{\{|S_n -S_{\tau ^{*}(J)}|<\varepsilon \}}\bigg )^p \\&= \mathbb{P }{\{\tau ^{*}(J)<\infty \}} \mathbb{E }\bigg ( \sum _{n \ge 0} \mathbb{1 }_{\{|S_n|<\varepsilon \}}\bigg )^p < \infty . \end{aligned}$$

This proves the first assertion of the lemma. Concerning the second, assume that $\mathbb{E }N^*(x)^p<\infty$ for some $x \in \mathbb{R }$. Then, for any $y > x$,

$$\begin{aligned} \mathbb{E }N^*(y)^p ~\le ~ (2^{p-1} \vee 1) \bigg (\mathbb{E }N^*(x)^p+\mathbb{E }\bigg [\sum _{n \ge 0}\mathbb{1 }_{\{x<S_n \le y\}}\bigg ]^p \bigg ) ~<~ \infty , \end{aligned}$$

where the last term is finite by the first part of the lemma. $\square$

The following lemma summarizes properties of the functions $A_+$ and $J_+$ that are frequently used throughout the proofs. These properties were known before and are stated here only for the reader’s convenience. Recall from () that

$$\begin{aligned} A_+(x) := \int \limits _0^x \mathbb{P }\{\xi >y\} \, \mathrm{d}y = \mathbb{E }\min (\xi ^+, x) \quad \text{ and} \quad J_+(x):= \frac{x}{A_+(x)} \end{aligned}$$

whenever $A_+(x) > 0$. Further, recall that $\mathbb{U }_{p-1}(x) = \sum _{n \ge 1} n^{p-1} \mathbb{P }\{S_n \le x\}$ and, analogously, $\mathbb{U }_{p-1}^>(x) = \sum _{n \ge 1} n^{p-1} \mathbb{P }\{S_{\tau _n^*} \le x\}$ where $\tau _n^*$ is the $n$th strictly ascending ladder index of the random walk $(S_n)_{n \ge 0}$.

Lemma 5.4

Assume that $S_n \rightarrow \infty$ a.s. Then the following assertions are true:

• (a) $A_+(x) > 0$ for all $x > 0$; $A_+$ and $J_+$ are non-decreasing.

• (b) $\lim _{x \rightarrow \infty } J_+(x) = \infty$.

• (c) $J_+$ is subadditive, i.e., $J_+(x+y) \le J_+(x)+J_+(y)$ for all $x, y \ge 0$. In particular, $J_+(x+y) \sim J_+(x)$ as $x \rightarrow \infty$ for any $y \in \mathbb{R }$.

• (d) For any $p > 0$ if $\mathbb{U }_{p-1}(0)<\infty$, then

  $$\begin{aligned} \mathbb{U }_{p-1}(x) \asymp \mathbb{U }_{p-1}^>(x) \asymp J_+(x)^p \end{aligned}$$

  (5.6)

  as $x \rightarrow \infty$. Moreover, with $W(\cdot )$ denoting either $\tau ^*(\cdot ),\,N^*(\cdot )$ or $\rho ^*(\cdot )$, then $\mathbb{E }W(x)^p \asymp J_+(x)^p$ whenever $\mathbb{E }W(0)^p < \infty$.

Proof

(a) Since $S_n \rightarrow \infty$ a.s. is assumed, $\mathbb{P }\{\xi ^+ > 0\}=\mathbb{P }\{\xi > 0\} > 0$ and therefore $\mathbb{P }\{\xi >y\}>0$ in a right neighborhood of $0$. The monotonicity of $A_+$ follows from its definition. The monotonicity of $J_+$ and assertion (b) follow from the following representation

$$\begin{aligned} J_+(x) = \left(\int \limits _0^1 \mathbb{P }\{\xi >xy\} \, \mathrm{d}y \right)^{-1} \!\!\!, \quad x > 0. \end{aligned}$$

Regarding (c) notice that the subadditivity of $J_+$ follows from the monotonicity of $A_+$. $J_+(x+y) \sim J_+(x)$ as $x \rightarrow \infty$ immediately follows from the subadditivity of $J_+$ together with (b).

(d) follows from equations [31, Theorems 2.1 and 2.2, Eq. (4.5)] and one of the displayed formulas on p. 28 of the cited reference. $\square$

Lemma 5.5

Let $p>0$ and assume $\lim _{k \rightarrow \infty } S_k=+\infty$ a.s. Then the following assertions are equivalent:

$$\begin{aligned} \mathbb{E }J_+ \Big (\Big | \min _{0 \le k \le \tau ^*-1} S_k \Big |\Big )^{p+1} ~&<~ \infty ; \end{aligned}$$

(5.7)

$$\begin{aligned} \mathbb{E }J_+ \Big (\Big | \inf _{k \ge 0} S_k \Big | \Big )^p ~&<~ \infty ; \end{aligned}$$

(5.8)

$$\begin{aligned} \mathbb{E }J_+(\xi ^-)^{p+1} ~&<~ \infty \end{aligned}$$

(5.9)

where $\tau ^*:=\inf \{k \in \mathbb{N }: S_k>0\}$.

Lemma 5.5 has several predecessors, e.g., [30, Theorem 1], [2, Theorem 3], [31, Proposition 4.1], [3, Lemma 3.5]. Even though Lemma 5.5 does not follow directly from either of these results, the proofs given in [30] and [31] can be adopted to treat the present case after the observation that the function $x \mapsto J_+(x)$ is nondecreasing and subadditive. Therefore, we omit a proof.

1.2 Elementary Facts

Lemma 5.6

Let $1 \le p = n+ \delta$ with $n \in \mathbb{N }_0$ and $\delta \in (0,1]$. Then, for any $x,y \ge 0$,

$$\begin{aligned} (x+y)^p ~\le ~ x^p+y^p+p2^{p-1}(xy^{p-1} + x^ny^{\delta }). \end{aligned}$$

(5.10)

This estimate is a variant of an estimate we have learned from [18]. For the reader’s convenience, we include a brief proof which is a slight modification of the argument given in the cited reference.

Proof

For any $0 \le r \le 1$, we have $(1+r)^p = 1+p \int \limits _0^r (1+t)^{p-1} \mathrm{d}t$. By the mean value theorem for integrals, for some $\gamma \in (0,r)$,

$$\begin{aligned} (1\!+\!r)^p ~=~ 1+p r (1\!+\!\gamma )^{p-1} ~\le ~ 1+ p2^{p-1}r ~\le ~ 1+p2^{p-1}r^{\delta }, \end{aligned}$$

(5.11)

where in the last step we have used that $0 \le r \le 1$. Now let $x, y \ge 0$. When $x \le y$, use the first estimate in (5.11) to get $(x+y)^p \le y^p +p2^{p-1}xy^{p-1}$. When $y \le x$ use the second estimate in (5.11) to infer $(x+y)^p \le x^p + p2^{p-1} x^n y^{\delta }$. Thus, in any case, (5.10) holds. $\square$

The next auxiliary result is an elementary consequence of a version of the summation by parts formula.

Lemma 5.7

Let $b_n \ge 0$ for all $n \in \mathbb{N }$ and $p>1$. Then

$$\begin{aligned} \sum _{n \ge 1} n^{p-1} \sum _{k=n}^\infty b_k ~<~ \infty \quad \text{ iff} \quad \sum _{n \ge 1} n^p b_n ~<~ \infty . \end{aligned}$$

Proof

For arbitrary $m \in \mathbb{N }$,

$$\begin{aligned} \sum _{n=1}^m n^{p-1}\sum _{k=n}^\infty b_k ~=~ \sum _{n=1}^m n^{p-1} \sum _{k=m}^\infty b_k + \sum _{k=1}^{m-1} b_k \sum _{n=1}^k n^{p-1}. \end{aligned}$$

(5.12)

In particular,

$$\begin{aligned} \sum _{n=1}^m n^{p-1} \sum _{k=n}^\infty b_k ~\ge ~ \sum _{n=1}^{m-1} b_n\sum _{k=1}^n k^{p-1}. \end{aligned}$$

Consequently, if $\sum _{n \ge 1} n^{p-1}\sum _{k=n}^\infty b_k$ converges, then so does $\sum _{n \ge 1} b_n\sum _{k=1}^n k^{p-1}$ and thus also $\sum _{n \ge 1} b_n n^p$. Conversely, if the series $\sum _{n \ge 1} n^p b_n$ converges, then $\sum _{n \ge 1} b_n\sum _{k=1}^n k^{p-1}$ converges and, in particular,

$$\begin{aligned} \lim _{m\rightarrow \infty } \sum _{n=m}^{\infty } b_n \sum _{k=1}^n k^{p-1} ~=~ 0. \end{aligned}$$

Further,

$$\begin{aligned} 0 ~\le ~ \sum _{k=1}^m k^{p-1} \sum _{n=m}^\infty b_n ~\le ~ \sum _{n=m}^{\infty } b_n\sum _{k=1}^n k^{p-1} ~\rightarrow ~ 0 \quad \text{ as} m \rightarrow \infty . \end{aligned}$$

Letting $m$ tend to $\infty$ in (5.12), we conclude that $\sum _{n \ge 1} n^{p-1}\sum _{k=n}^\infty b_k$ converges. $\square$