Solutions to complex smoothing equations

Abstract

We consider smoothing equations of the form

$$\begin{aligned} X ~\mathop {=}\limits ^{\text {law}}~ \sum _{j \ge 1} T_j X_j + C \end{aligned}$$

where $(C,T_1,T_2,\ldots )$ is a given sequence of random variables and $X_1,X_2,\ldots$ are independent copies of X and independent of the sequence $(C,T_1,T_2,\ldots )$. The focus is on complex smoothing equations, i.e., the case where the random variables $X, C,T_1,T_2,\ldots$ are complex-valued, but also more general multivariate smoothing equations are considered, in which the $T_j$ are similarity matrices. Under mild assumptions on $(C,T_1,T_2,\ldots )$, we describe the laws of all random variables X solving the above smoothing equation. These are the distributions of randomly shifted and stopped Lévy processes satisfying a certain invariance property called $(U,\alpha )$-stability, which is related to operator (semi)stability. The results are applied to various examples from applied probability and statistical physics.

Appendices

Appendix 1: The Choquet–Deny lemma

Given a probability measure $\mu$ on the similarity group ${\mathbb {S}}{} \textit{(d)}$, let U be the closed subgroup generated by the support of $\mu$—if $\mu$ is the step distribution of the associated multiplicative random walk $(L_n)_{n \in {\mathbb {N}}_0}$, see Sect. 3.4, then $U= {\mathbb {U}}$.

Lemma 5.1

Let $\psi :U \rightarrow {\mathbb {R}}$ be measurable and bounded. If

$$\begin{aligned} \int _{U} \psi (ug) \, \mu (\mathrm {d} u) = \psi (g) \end{aligned}$$

(5.1)

for all $u \in U$, then $\psi$ is constant $\mu$-a.e.

This is a consequence of [39, Theorem 3]. For the reader’s convenience, we state that theorem and show how the lemma can be derived from it.

In the following, let G be a locally compact, separable and unimodular group. A probability measure $\mu$ on G is called aperiodic, if the closed subgroup generated by the support of $\mu$ equals G. Write [G, G] for the commutator subgroup, i.e., the group generated by the commutators $[a,b] :=(ba)^{-1} ab$, $a,b \in G$, and $\overline{[G,G]}$ for its closure. Let $H \subsetneq G$ be a normal subgroup of G. Then G acts on H by conjugation (inner automorphisms), i.e.,

$$\begin{aligned} g.h ~:=~ g^{-1} h g, \quad g \in G, h \in H. \end{aligned}$$

For $A \subseteq H$ write

$$\begin{aligned} A^G ~:=~ \{g.a:\, g \in G, a \in A\}. \end{aligned}$$

The action of G on H is said to be compact if for each compact $A \subseteq H {\setminus } \{1_G\}$, $1_G$ the unit element of G, $A^G$ is relatively compact, i.e., has compact closure. Then [39, Theorem 3] reads as follows:

Theorem 5.2

Let $\mu$ be an aperiodic probability measure on G. If $\overline{[G,G]}$ is Abelian or compact and if the action of G on $\overline{[G,G]}$ is compact, then the only bounded, measurable functions $\psi$ satisfying

$$\begin{aligned} \int \psi (u g) \, \mu (\mathrm {d}u) = \psi (g) \quad \text {for all } g \in G \end{aligned}$$

are the $\mu$-almost everywhere constant functions.

Following the proof of [25, Theorem A.1], we show how Theorem 5.2 applies to the situation here, i.e., $G=U$ is the closed subgroup generated by the support of $\mu$. Then $\mu$ is aperiodic on U by the very definition of U. Referring to Proposition 4.1, there is a closed subgroup $\textit{A}_{U}$ of U, which is isomorphic to a closed subgroup of the multiplicative group ${\mathbb {R}}_>$, and a normal compact subgroup $\textit{C}_{U}=U \cap {\mathbb {O}}{} \textit{(d)}$, such that $U/\textit{C}_{U}\simeq \textit{A}_{U}$. The groups $\textit{A}_{U}$ and $\textit{C}_{U}$ (as a compact group, see [34, Theorem 1.4.1]) are unimodular and hence, by the Fubini formula for the Haar measure on $U=\textit{A}_{U}\textit{C}_{U}$, [34, Proposition 1.5.5], U is unimodular as well.

Clearly, the commutator subgroup of U is a subgroup of ${\mathbb {O}}{} \textit{(d)}$, hence its closure is compact. Moreover, for any compact $A \subseteq \overline{[U,U]} {\setminus } \{\textit{I}_{\textit{d}}\}$, $A^U$ is again a subset of ${\mathbb {O}}{} \textit{(d)}$ since

$$\begin{aligned} u.[a,b] = u^{-1} [a,b]\, u = o^{-1} [a,b]\, o, \end{aligned}$$

where $u = \left\| u \right\| o$ with $o \in \textit{C}_{U}\subseteq {\mathbb {O}}{} \textit{(d)}$. Hence, $A^U$ is relatively compact as a subset of a compact set.

Appendix 2: Evaluating the Lévy integrals

In this section, we compute

$$\begin{aligned} I(x) = \int _{{\mathbb {R}}^d {\setminus } \{0\}} \left( e^{\text {i}\left\langle x,y \right\rangle } - 1 - \text {i}\left\langle x,y \right\rangle \mathbbm {1}_{[0,1]}(\left| y \right| ) \right) \bar{\nu }(\text {d} \textit{y}), \quad x \in {\mathbb {R}}^d \end{aligned}$$

(6.1)

for a deterministic $(U,\alpha )$-invariant Lévy measure $\bar{\nu }$, i.e. satisfying (3.29).

Lemma 6.1

Let $\bar{\nu }$ be a deterministic Lévy measure satisfying (3.29) for some $1 \ne \alpha \in (0,2)$, and define I(x) via (6.1). Then, for $0 \not = x \in {\mathbb {R}}^d$,

$$\begin{aligned} I(x) = - |x|^\alpha \eta _1^\alpha (x) + \text {i}|x|^\alpha \eta _2^\alpha (x) + \text {i}\left\langle x,\gamma ^\alpha \right\rangle , \end{aligned}$$

(6.2)

with functions $\eta _1^\alpha ,\eta _2^\alpha$ defined in (1.19) and (1.20), respectively. $\eta _1^\alpha$ and $\eta _2^\alpha$ are bounded real functions satisfying $\eta _j^\alpha (u^\mathsf {T}x) = \eta _j^\alpha (x)$ for all $u \in {\mathbb {U}}$, $x \in {\mathbb {R}}^d{\setminus }\{0\}$, $j=1,2$, and $\eta _1^\alpha$ is nonnegative. The vector $\gamma ^\alpha$ satisfies $o \gamma ^\alpha = \gamma ^\alpha$ for all $o \in \textit{C}_{{\mathbb {U}}}$.

Proof

Fix $0 \not = x \in {\mathbb {R}}^d$ and notice that I(x) is finite since $\bar{\nu }$ is a Lévy measure. Further, according to Proposition 4.3, there is a $\textit{C}_{{\mathbb {U}}}$-invariant finite measure $\rho$ on $\textit{S}_{{\mathbb {U}}}$ such that (4.4) holds.

We set

$$\begin{aligned} \gamma ^\alpha :={\left\{ \begin{array}{ll} -\int y \mathbbm {1}_{[0,1]}(|y|) \bar{\nu }(\text {d} \textit{y})&{}\,\\ \quad = - \int _{\textit{A}_{{\mathbb {U}}}} \int _{\textit{S}_{{\mathbb {U}}}} ax \mathbbm {1}_{[0,1]}(|ax|) \left\| a \right\| ^{-\alpha } \rho (\text {d} \textit{x}) \, \textit{H}_{{} \textit{A}_{{\mathbb {U}}}}(\text {d} \textit{a}) &{} \text {if } \alpha < 1 \\ - \int _{\textit{A}_{{\mathbb {U}}}} \int _{\textit{S}_{{\mathbb {U}}}} ax \mathbbm {1}_{(1,\infty )}(|ax|) \left\| a \right\| ^{-\alpha } \rho (\text {d} \textit{x}) \, \textit{H}_{{} \textit{A}_{{\mathbb {U}}}}(\text {d} \textit{a}), &{} \text {if } \alpha \in (1,2). \end{array}\right. } \end{aligned}$$

The asserted $\textit{C}_{{\mathbb {U}}}$-invariance follows from the $\textit{C}_{{\mathbb {U}}}$-invariance of $\rho$.

Recalling the definitions

$$\begin{aligned} \eta _1^\alpha (x)&= \frac{1}{|x|^\alpha } \int \big (1 - \cos (\left\langle x,y \right\rangle ) \big ) \nu ^\alpha (\text {d} \textit{y}) \\ \eta _2^\alpha (x)&= \frac{1}{|x|^\alpha } \int \big (\sin (\left\langle x,y \right\rangle ) - \mathbbm {1}_{\{\alpha >1\}}\left\langle x,y \right\rangle \big ) \nu ^\alpha (\text {d} \textit{y}) , \end{aligned}$$

(6.2) holds, and it remains to prove boundedness and invariance properties of $\eta _i^\alpha$, $i=1,2$. Let $u \in {\mathbb {U}}$, then, using (3.29),

$$\begin{aligned} \eta _1^\alpha (u^\mathsf {T}x)&= \frac{1}{\left\| u \right\| ^\alpha |x|^\alpha } \int \big (1 - \cos (\left\langle x,uy \right\rangle ) \big ) \nu ^\alpha (\text {d} \textit{y}) \\&= \frac{\left\| u \right\| ^\alpha }{\left\| u \right\| ^\alpha |x|^\alpha } \int \big (1 - \cos (\left\langle x,y \right\rangle ) \big ) \nu ^\alpha (\text {d} \textit{y}) = \eta _1^\alpha (x), \end{aligned}$$

and the invariance of $\eta _2^\alpha$ is proved along the same lines. This implies in particular that the continuous functions $\eta _i^\alpha$ are determined by their respective values on the (relative) compact set $\textit{S}_{{\mathbb {U}}}$, hence the asserted boundedness follows. $\square$

For $\alpha =1$, we can compute a meaningful expression for $\eta ^1(x)$ only for $x \in E_1(Q^\mathsf {T})$. Notice that this implies $x \in E_1(t^Q)$ for all $t \ge 0$. Hence, using formula (4.6) for $\bar{\nu }$, we obtain

$$\begin{aligned} \eta ^1(x)&= \int _{{\mathbb {S}}^{d-1}} \int _{{\mathbb {R}}_>} \bigg ( e^{\text {i}\left\langle (t^{Q})^\mathsf {T}x,s \right\rangle } - 1 - \text {i}\left\langle (t^Q)^\mathsf {T}x,s \right\rangle \mathbbm {1}_{\{|(t^Q)^\mathsf {T}s| \le 1\}} \bigg ) t^{-2}\, \text {d} \textit{t}\, \rho (\text {d} \textit{s}) \nonumber \\&= \int _{{\mathbb {S}}^{d-1}} \int _{{\mathbb {R}}_>} \bigg ( e^{\text {i}\left\langle tx,s \right\rangle } - 1 - \text {i}\left\langle tx,s \right\rangle \mathbbm {1}_{\{t \le 1\}} \bigg ) t^{-2}\, \text {d} \textit{t}\, \rho (\text {d} \textit{s}) \nonumber \\&= -\int _{{\mathbb {S}}^{d-1}} \bigg ( \left| \left\langle x,s \right\rangle \right| + \text {i}\frac{2}{\pi }\left\langle x,s \right\rangle \log |\left\langle x,s \right\rangle | \bigg ) \rho (\text {d} \textit{s}) + \text {i}\left\langle \gamma ,x \right\rangle \end{aligned}$$

(6.3)

for a suitable $\gamma \in {\mathbb {R}}^d$, see [73, Theorem 14.10] for details.

Appendix 3: Auxiliary results from (Markov) renewal theory

Lemma 7.1

Let $(S_n)_{n \in {\mathbb {N}}_0}$ be a random walk with i.i.d. increments, $S_0=0$, ${\mathbb {E}}[S_1]>0$ and ${\mathbb {E}}[(S_1^+)^2]<\infty$. Let $\tau (t) :=\inf \{n \in {\mathbb {N}}_0: S_n > t\}$, $t \ge 0$. Then, for every $0<a<1$,

$$\begin{aligned} t {\mathbb {P}}(S_{\tau (t)-1} < at) ~\rightarrow ~ 0 \quad \text {as } t \rightarrow \infty . \end{aligned}$$

(7.1)

Proof

${\mathbb {E}}[(S_1^+)^2]<\infty$ implies $\lim _{t \rightarrow \infty } t^2{\mathbb {P}}(S_1 > t) = 0$. Further, it is known from standard random walk theory that $\lim _{t\rightarrow \infty } t^{-1} {\mathbb {E}}[\tau (t)] = {\mathbb {E}}[S_1]^{-1}$. Consequently, setting, for $n \in {\mathbb {N}}$, $A_n :=\{S_0 \le t, \ldots , S_{n-2} \le t, S_{n-1} < at\}$, we have

$$\begin{aligned} t {\mathbb {P}}(S_{\tau (t)-1} < at)= & {} t \sum _{n \ge 1} {\mathbb {P}}(A_n \cap \{S_{n+1}>t\}) \le t \sum _{n \ge 1} {\mathbb {P}}(A_n) {\mathbb {P}}(S_{1}>(1-a)t) \\\le & {} t^2 {\mathbb {P}}(S_{1}>(1-a)t) \cdot \frac{1}{t} \sum _{n \ge 1} {\mathbb {P}}(\tau (t) \ge n) \\= & {} t^2 {\mathbb {P}}(S_1 > (1-a)t) \cdot \frac{{\mathbb {E}}[\tau (t)]}{t} ~\rightarrow ~ 0 \qquad \text {as } t \rightarrow \infty . \end{aligned}$$

$\square$

1.1 A rate-of-convergence result in Markov renewal theory

Throughout this section, let $((O_n, S_n))_{n \in {\mathbb {N}}_0}$ be a random walk on ${\mathbb {O}}{} \textit{(d)}\times {\mathbb {R}}$ with increment law $\mu$, say. The components $O_n$ and $S_n$ may be dependent. We assume that $\mu$ satisfies the minorization condition (M) and that

$$\begin{aligned} {\mathbb {E}}[S_1] > 0 \text { and } {\mathbb {E}}\big [ |S_1|^{ \ell + 1 + \delta } \big ] < \infty \end{aligned}$$

(7.2)

for some $\ell > 0$. Let ${\mathbb {O}}$ be the closed subgroup of ${\mathbb {O}}{} \textit{(d)}$ generated by the support of $O_1$.

Proposition 7.2

Let $\ell >0$ and $g: {\mathbb {R}}\rightarrow [0,\infty )$ be a measurable function that is decreasing on $[0,\infty )$, with $g(t)=0$ for all $t<0$ and $\lim _{t \rightarrow \infty } t^{ \ell +1+\epsilon } g(t) =0$ for some $0<\epsilon < (\delta \wedge 1)$. Then

$$\begin{aligned} \lim _{t \rightarrow \infty } \sup _{\left| f \right| \le g} ~t^{ \ell +\epsilon } \left| {\mathbb {E}}\bigg [ \sum _{n=0}^\infty f(O_n, t-S_n) \bigg ] ~-~ \frac{1}{{\mathbb {E}}[S_1]} \int _0^\infty \int _{\mathbb {O}}f(o, r) \, H_{\mathbb {O}}(\text {d} \textit{o}) \, \text {d} \textit{r} \right| = 0. \end{aligned}$$

Here and below, $\sup _{|f| \le g}$ means the supremum over all measurable functions $f : {\mathbb {O}}\times {\mathbb {R}}\rightarrow {\mathbb {R}}$ satisfying $\sup _{o \in {\mathbb {O}}} |f(o,x)| \le g(x)$ for all $x \in {\mathbb {R}}$.

The main ingredient in the proof will be the use of regeneration techniques for general state space Markov chains as developed in [10, 68]. We sum up what is needed in the subsequent lemma.

Lemma 7.3

There is a measurable space $(\varOmega , {\mathcal {G}})$ together with a family of probability measures $({\mathbb {P}}_{o,r})_{o \in {\mathbb {O}}, r \in {\mathbb {R}}}$ and sequences of random variables $((M_n, R_n))_{n \ge 0}$ and $(\tau _n)_{n \ge 1}$ with

$$\begin{aligned} {\mathbb {P}}_{o,r}\big (((M_n, R_n))_{n \ge 0} \in \cdot \big ) = {\mathbb {P}}\big ( ((oO_n, r + S_n))_{n \ge 0} \in \cdot \big ) \end{aligned}$$

(7.3)

for all $o \in {\mathbb {O}}$, $r \in {\mathbb {R}}$. Further, the following properties hold:

• (i) There is a filtration $({\mathcal {G}}_n)_{n \in {\mathbb {N}}}$ such that $((M_n, R_n))_{n \ge 0}$ is Markov adapted to $({\mathcal {G}}_n)_{n \in {\mathbb {N}}}$, and $(\tau _n)_{n \ge 1}$ is a sequence of predictable $({\mathcal {G}}_n)_{n \in {\mathbb {N}}}$-stopping times, i.e., $\{\tau _n=k\} \in {\mathcal {G}}_{k-1}$.

• (ii) There are probability measures $\nu$ on ${\mathbb {O}}$ and $\eta$ on ${\mathbb {R}}$, $\eta$ having a bounded Lebesgue density, such that for all $n \ge 1$ and $o \in {\mathbb {O}}$, under ${\mathbb {P}}_o$, $((M_{\tau _n+k}, R_{\tau _n+k}-R_{\tau _n-1}))_{0 \le k \le \tau _{n+1}-\tau _n-1}$ is independent of $(M_0, R_0, \ldots , M_{\tau _n-1}, R_{\tau _n -1})$ and has law ${\mathbb {P}}_{\nu \otimes \eta }(((M_{k}, R_{k}))_{k=0}^{\tau _{1}-1} \in \cdot )$.

• (iii) For each bounded measurable function $f:{\mathbb {O}}\rightarrow {\mathbb {R}}$,

  $$\begin{aligned} {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{n=0}^{\tau _1-1} f(M_n) \bigg ] = {\mathbb {E}}_{\nu \otimes \eta }[\tau _1] \, \int _{\mathbb {O}}f(o) \, H_{\mathbb {O}}(\text {d} \textit{o}). \end{aligned}$$

  (7.4)

• (iv) There are $C, \lambda >0$ such that ${\mathbb {P}}_{o,r}(\tau _1 > n) \le C e^{-\lambda n}$ for all $o \in {\mathbb {O}}$, $r \in {\mathbb {R}}$ and $n \in {\mathbb {N}}$.

Here and below, we use the shorthand ${\mathbb {P}}_{\nu \otimes \eta } = \int _{{\mathbb {O}}} \int _{\mathbb {R}}{\mathbb {P}}_{o,r} \, \nu (\text {d} \textit{o}) \, \eta (\text {d} \textit{r})$, the same notation for expectations, and sometimes omit the initial value $R_0$ if it is irrelevant.

Now we turn to the proof of Proposition 7.2.

Proof (Proof of Proposition 7.2)

In order to prove this result, we will combine methods from [9] and [69]. Below, we describe the steps of the proof and defer the technicalities to several lemmata. By splitting f into its positive and negative part, it suffices to consider nonnegative functions which are bounded by g.

Let $((M_n, R_n))_{n \ge 0}$ and $(\tau _n)_{n \ge 0}$ be as in Lemma 7.3. Define $V_0 :=0$ and

$$\begin{aligned} V_n ~:=~ \sum _{k=1}^n (R_{\tau _{k+1}-1}-R_{\tau _k-1}) = R_{\tau _{n+1}-1} - R_{\tau _1-1}. \end{aligned}$$

Then, under each ${\mathbb {P}}_{o,r}$, $(V_n)_{n \ge 1}$ is a random walk with i.i.d. increments and increment law ${\mathbb {P}}_{\nu \otimes \eta }(R_{\tau _1-1} \in \cdot )$ which is absolutely continuous since the law $\eta$ of $R_0$ is absolutely continuous. Since ${\mathbb {E}}[S_1]>0$ and

$$\begin{aligned} 1 = {\mathbb {P}}(S_n/n \rightarrow {\mathbb {E}}[S_1] \text { as } n \rightarrow \infty ) = {\mathbb {P}}_{\textit{I}_{\textit{d}},0}(R_n/n \rightarrow {\mathbb {E}}[S_1] \text { as } n \rightarrow \infty ), \end{aligned}$$

we deduce that ${\mathbb {E}}_{\nu \otimes \eta }[V_1] = {\mathbb {E}}_{\nu \otimes \eta }[\tau _1]{\mathbb {E}}[S_1]$.

Let f be nonnegative and set

$$\begin{aligned} \hat{f}(t) ~:=~ {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{k=0}^{\tau _1-1} f(M_k, t- R_k) \bigg ]. \end{aligned}$$

Then we can proceed as in [9, Section 4] to obtain

$$\begin{aligned} {\mathbb {E}}\bigg [&\sum _{n=0}^\infty f(O_n, t-S_n) \bigg ] = {\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [ \sum _{n=0}^\infty f(M_n, t-R_n) \bigg ] \nonumber \\&= {\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [ \sum _{n=0}^{\tau _1 -1} f(M_n, t-R_n) \bigg ] \nonumber \\&\quad + {\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [ \sum _{n=1}^\infty \, {\mathbb {E}}\bigg [\sum _{k=\tau _n}^{\tau _{n+1}-1} f(M_k, t-(R_k-R_{\tau _{n}-1}) - R_{\tau _{n}-1}) \, \Big | \, {\mathcal {G}}_{\tau _n-1} \bigg ] \bigg ] \nonumber \\&= {\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [ \sum _{n=0}^{\tau _1 -1} f(M_n, t-R_n) \bigg ] + {\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [ \sum _{n=0}^\infty \hat{f}(t-R_{\tau _1-1}-V_n) \bigg ]\nonumber \\&=:E_1(t) + E_2(t). \end{aligned}$$

(7.5)

We show in Lemma 7.4 that $t^{ \ell +\epsilon } E_1(t)$ tends to zero, uniformly over all f with $\left| f \right| \le g$. We rewrite

$$\begin{aligned} E_2(t) = \int _{\mathbb {R}}{\mathbb {E}}_{\nu \otimes \eta } \bigg [\sum _{n=0}^\infty \hat{f}(t-s-V_n)\bigg ] \, {\mathbb {P}}_{\textit{I}_{\textit{d}},0}(R_{\tau _1-1} \in \text {d} \textit{s}) \end{aligned}$$

and use (7.4) to infer that (recall that $f \ge 0$)

$$\begin{aligned} \frac{1}{{\mathbb {E}}_{\nu \otimes \eta } [V_1]} \int _{\mathbb {R}}\hat{f}(r) \, \text {d} \textit{r}&= \frac{1}{{\mathbb {E}}_{\nu \otimes \eta } [V_1]} {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{k=0}^{\tau _1-1} \int _{\mathbb {R}}f(M_k, r-R_k) \, \text {d} \textit{r}\bigg ] \\&= \frac{1}{{\mathbb {E}}_{\nu \otimes \eta } [V_1]} {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{k=0}^{\tau _1-1} \int _{\mathbb {R}}f(M_k, r) \, \text {d} \textit{r}\bigg ] \\&= \frac{1}{{\mathbb {E}}[S_1]} \int _0^\infty \int _{{\mathbb {O}}} f(o,r) \, H_{\mathbb {O}}(\text {d} \textit{o}) \, \text {d} \textit{r}. \end{aligned}$$

Then the claimed convergence rate holds, if

$$\begin{aligned} \int _{\mathbb {R}}\, \sup _{\left| f \right| \le g} t^{ \ell +\epsilon } \left| {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{n=0}^\infty \hat{f}(t-s-V_n) \bigg ] - \frac{1}{{\mathbb {E}}_{\nu \otimes \eta }[V_1]}\int _0^\infty \hat{f}(r) \, \text {d} \textit{r} \right| \, {\mathbb {P}}_{\textit{I}_{\textit{d}},0}(R_{\tau _1-1} \in \text {d} \textit{s}) \end{aligned}$$

tends to 0 as $t \rightarrow \infty$. This result will be established in Lemma 7.5. $\square$

1.2 Lemmata needed in the proof of Proposition 7.2

Proof (Proof of Lemma 7.3)

Observe that $((oO_n, r+S_n))_{n \in {\mathbb {N}}_0}$ is indeed a Markov chain on ${\mathbb {O}}\times {\mathbb {R}}$ and that the increments of $S_n-S_{n-1}$ are independent of the past. Since $\mu$ satisfies the minorization condition (M), we have that for all $o \in \text {SO}{} \textit{(d)}$, $B \in \text {SO}{} \textit{(d)}$,

$$\begin{aligned} {\mathbb {P}}(oO_{1} \in B) = {\mathbb {P}}(O_{1} \in o^{-1}B) \ge \gamma H_{{\mathbb {O}}}((o^{-1}B) \cap \text {SO}{} \textit{(d)}) \ge \frac{\gamma }{2} H_{\text {SO}{} \textit{(d)}}(B). \end{aligned}$$

(7.6)

If ${\mathbb {O}}=\text {SO}{} \textit{(d)}$, this shows that $(oO_n)_{n \in {\mathbb {N}}}$ is a Doeblin chain on ${\mathbb {O}}$ (see [66, Section 16.2] for the definition).

If ${\mathbb {O}}$ contains elements with determinant $-1$ as well, then readily ${\mathbb {O}}={\mathbb {O}}{} \textit{(d)}$, for $\text {SO}{} \textit{(d)}\subseteq {\mathbb {O}}$, and the product of two matrices with negative determinant has a positive determinant. Moreover, this necessitates ${\mathbb {P}}(\det (O_1)=-1) >0$. Then, for all $o \in {\mathbb {O}}{\setminus } \text {SO}{} \textit{(d)}$ and $B \subseteq \text {SO}{} \textit{(d)}$,

$$\begin{aligned} {\mathbb {P}}(oO_{2} \in B) \ge&\int \mathbbm {1}_{\{\det (o')=-1 \}} {\mathbb {P}}(O_{1} \in (oo')^{-1}B) \, {\mathbb {P}}(O_1 \in \text {d} \textit{o}') \nonumber \\ \ge&{\mathbb {P}}(\det (O_1)=-1) \, \frac{\gamma }{2} H_{\text {SO}{} \textit{(d)}}(B). \end{aligned}$$

(7.7)

Thus, $(oO_n)_{n \in {\mathbb {N}}}$ is a Doeblin chain in the case ${\mathbb {O}}={\mathbb {O}}{} \textit{(d)}$, too. Its unique invariant probability measure is given by the normalized Haar measure $H_{\mathbb {O}}$ on ${\mathbb {O}}{} \textit{(d)}$.

Again by the minorization condition for $\mu$, it follows that there is an absolutely continuous measure $\eta (\text {d} \textit{x}):=\mathbbm {1}_I(x) dx$ such that

$$\begin{aligned} {\mathbb {P}}(oO_1 \in A, S_1 \in B) ~\ge ~ \frac{\gamma }{2} H_{\text {SO}{} \textit{(d)}}(A) \eta (B) \end{aligned}$$

(7.8)

for all $o \in \text {SO}{} \textit{(d)}$ and all measurable A, B.

(7.6) and (7.7) yield that there is some $q <1$ such that, for all $o \in {\mathbb {O}}$,

$$\begin{aligned} {\mathbb {P}}(oO_k \notin \text {SO}{} \textit{(d)}\text { for } k=1,2) \le q. \end{aligned}$$

(7.9)

It follows that $(oO_n)_{n \in {\mathbb {N}}}$ is $(\text {SO}{} \textit{(d)},\gamma /2,\nu ,1)$-recurrent in the sense of [9], with $\nu :=H_{\text {SO}{} \textit{(d)}}$. Then, $((M_n, R_n))_{n \in {\mathbb {N}}_0}$ can be constructed along the same lines as in [9, Section 3]: Under ${\mathbb {P}}_{o,r}$, let $((M_n, R_n))_{n \in {\mathbb {N}}}$ have the same transitions as $(oO_n,r+ S_n)_{n \in {\mathbb {N}}}$, but whenever $O_n$ enters $\text {SO}{} \textit{(d)}$, an independent $B(1,\gamma /2)$-distributed coin is flipped. If 1 shows up, then $(M_{n+1}, R_{n+1}-R_n)$ is generated according to $\nu \otimes \eta$, this event we call a regeneration; if 0 shows up, then $(M_{n+1}, R_{n+1}-R_n)$ is generated according to $(1-\frac{\gamma }{2})^{-1}\big ({\mathbb {P}}((M_nO_1, S_1) \in \cdot ) - \frac{\gamma }{2} \nu \otimes \eta \big )$. Thus, the total transition probabilities are still equal to that of $(oO_n, r+S_n)_{n \in {\mathbb {N}}}$. Let $\tau _0=0$ and $\tau _n$ be the nth regeneration time, see [10, 68] for details. This gives Assertions 1 and 2, while Assertion 3 is proved in [10, Theorem 6.1]. The construction together with (7.9) show that at least every third step, there is a uniform positive chance for regeneration, this yields Assertion 4. $\square$

Lemma 7.4

Let g be as in Proposition 7.2 and $((M_n,R_n))_{n \ge 0}$ as in Lemma 7.3. Then

$$\begin{aligned} \lim _{t \rightarrow \infty } \sup _{\left| f \right| \le g} t^{ \ell +\epsilon +1} \bigg |{\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [\sum _{n=0}^{\tau _1-1} f(M_n, t-R_n) \bigg ] \bigg | = 0. \end{aligned}$$

(7.10)

In particular, $\lim _{t \rightarrow \infty } t^{ \ell +\epsilon +1} \hat{g}(t)=0$. Moreover,

$$\begin{aligned} \lim _{t \rightarrow \infty } t^{ \ell +\epsilon +1} \, {\mathbb {P}}_{\textit{I}_{\textit{d}},0}(R_{\tau _1-1}>t/2) = 0. \end{aligned}$$

(7.11)

Proof

In order to prove (7.10), we assume that $g(0) \le 1$ and fix some $|f| \le g$. Recall that, by Lemma 7.3(iv), ${\mathbb {P}}_{\textit{I}_{\textit{d}},0}(\tau _1 >n) \le C e^{-\lambda n}$ for some $C,\lambda > 0$. Define $n_t= (\log t) (\ell +2)/\lambda$, $t>0$. Then

$$\begin{aligned}&t^{ \ell +\epsilon +1} \bigg | {\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [\sum _{n=0}^{\tau _1-1} f(M_n, t-R_n) \bigg ] \bigg | \le t^{ \ell +\epsilon +1} {\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [\sum _{n=0}^{\tau _1-1} g(t-R_n) \bigg ] \\&\quad \le t^{ \ell +\epsilon +1}{\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [\mathbbm {1}_{\{\tau _1 \le n_t\}}\sum _{n=0}^{\lfloor n_t \rfloor } g(t-R_n)\big (\mathbbm {1}_{\{R_n \le t/2\}} + \mathbbm {1}_{\{R_n > t/2\}} \big ) \bigg ] \\&\qquad + t^{ \ell +\epsilon +1}{\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [ \mathbbm {1}_{\{\tau _1 > n_t\}} \sum _{n=0}^{\tau _1-1} g(t-R_n) \bigg ] =:I_1(t) + I_2(t) + I_3(t). \end{aligned}$$

Recall that g is decreasing on $[0, \infty )$ and $\lim _{t \rightarrow \infty } t^{\ell +1+\delta } g(t)=0$ for some $\delta >0$. This gives

$$\begin{aligned} I_1(t)= & {} t^{ \ell +\epsilon +1}{\mathbb {E}}_{\textit{I}_{\textit{d}},0} \bigg [\mathbbm {1}_{\{\tau _1 \le n_t\}} \sum _{n=0}^{\lfloor n_t \rfloor } g(t-R_n) \mathbbm {1}_{\{R_n \le t/2\}} \bigg ]\\\le & {} t^{ \ell +\epsilon +1} (\lfloor n_t \rfloor + 1) g(t/2) ~\underset{t \rightarrow \infty }{\rightarrow }~0. \end{aligned}$$

By Jensen’s inequality, ${\mathbb {E}}[\left| S_n \right| ^\kappa ] \le n^{k-1} {\mathbb {E}}[\left| S_1 \right| ^\kappa ]$ for all $\kappa \ge 1$. Thus, applying Markov’s inequality,

$$\begin{aligned} I_2(t)&\le t^{ \ell +\epsilon +1} \sum _{n=0}^{\lfloor n_t \rfloor } {\mathbb {P}}(S_n > t/2) \le t^{ \ell +\epsilon +1} \sum _{n=0}^{\lfloor n_t \rfloor } \frac{{\mathbb {E}}[ \left| S_n \right| ^{\ell +1+\delta }]}{(t/2)^{\ell +1+\delta }} \\&\le \frac{2^{\ell +1+\delta }}{t^{ \delta -\epsilon }} \sum _{n=0}^{\lfloor n_t \rfloor } n^{\ell +\delta } {\mathbb {E}}[\left| S_1 \right| ^{\ell +1+\delta }] \le 2^{\ell +1+\delta } \frac{n_t^{\ell +1+\delta }}{t^{ \delta -\epsilon }} {\mathbb {E}}[\left| S_1 \right| ^{\ell +1+\delta }], \end{aligned}$$

which tends to zero as $t \rightarrow \infty$. Finally,

$$\begin{aligned} I_3(t)&\le t^{ \ell +\epsilon +1} {\mathbb {E}}_{\textit{I}_{\textit{d}},0}\big [ \tau _1 \mathbbm {1}_{\{\tau _1 > n_t\}} \big ] \\&\le t^{ \ell +\epsilon +1} \bigg ( n_t {\mathbb {P}}_{\textit{I}_{\textit{d}},0}(\tau _1 > n_t) + \int _{n_t}^\infty {\mathbb {P}}_{\textit{I}_{\textit{d}},0}(\tau _1 > r) \, \text {d} \textit{r}\bigg ) \\&\le t^{ \ell +\epsilon +1} \bigg ( n_t C e^{-\lambda n_t} + \frac{C}{\lambda } e^{-\lambda n_t} \bigg ) \\&\le C \frac{t^{ \ell +\epsilon +1} (n_t +1/\lambda )}{t^{\ell +2}} \rightarrow 0 \end{aligned}$$

as $t \rightarrow \infty$. Thus we have proved the first and second assertion. (7.11) follows from (7.10) with $g(s)=\mathbbm {1}_{[0,t/2)}(s)$. $\square$

Lemma 7.5

It holds that

$$\begin{aligned} \lim _{t \rightarrow \infty } \int _{\mathbb {R}}\, \sup _{\left| f \right| \le g} t^{ \ell + \epsilon } \left| {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{n=0}^\infty \hat{f}(t-s-V_n) \bigg ] - \frac{\int _0^\infty \hat{f}(r) \, \text {d} \textit{r}}{{\mathbb {E}}_{\nu \otimes \eta }[V_1]} \right| \, {\mathbb {P}}_{\textit{I}_{\textit{d}},0}(R_{\tau _1-1} \in \text {d} \textit{s}) = 0. \end{aligned}$$

(7.12)

Proof

We start by proving that the functions $\hat{f}$ are uniformly directly Riemann-integrable over $\left| f \right| \le g$. Write

$$\begin{aligned} \hat{f}(t) = \int {\mathbb {E}}_{\nu \otimes \delta _0} \bigg [\sum _{k=0}^{\tau _1-1} f(M_k,t-s-R_k) \bigg ] \, \eta (\text {d} \textit{s}), \end{aligned}$$

then observe that

$$\begin{aligned} \int \bigg | {\mathbb {E}}_{\nu \otimes \delta _0} \bigg [\sum _{k=0}^{\tau _1-1} f(M_k,t-R_k) \bigg ] \bigg | \text {d} \textit{t}&\le {\mathbb {E}}_{\nu \otimes \delta _0} \bigg [\sum _{k=0}^{\tau _1-1} \int g(t-R_k) \bigg ] \text {d} \textit{t}\\&= {\mathbb {E}}_{\nu \otimes \delta _0}[\tau _1] \int g(t) \text {d} \textit{t}\end{aligned}$$

and recall that $\eta$ has a bounded Lebesgue density. Consequently, $\hat{f}$ is the convolution of a Lebesgue integrable function with a bounded function and hence continuous, see [8, Lemma VII.1.2]. Further, arguing as in [9, Eq. (4.4)],

$$\begin{aligned} \sum _{l\in {\mathbb {Z}}} \sup _{t \in [l h,(l+1)h]} |\hat{f}(t)|&\le {\mathbb {E}}_{\nu \otimes \eta }[\tau _1] \int _{\mathbb {O}}\sum _{l \in {\mathbb {Z}}} \sup _{t \in [l 2h,(l+1)2h]} \left| f(o,t) \right| \, H_{\mathbb {O}}(\text {d} \textit{o}) \\&\le {\mathbb {E}}_{\nu \otimes \eta }[\tau _1] \sum _{l \in {\mathbb {Z}}} \sup _{t \in [l 2h,(l+1)2h]} g(t) =:C_g. \end{aligned}$$

Hence, it suffices to show that g is directly Riemann-integrable. The latter is clear since g is monotone and Lebesgue-integrable (see [8, Proposition V.4.1(v)]). We have the uniform bound (cf. [8, TheoremV.2.4(iii)])

$$\begin{aligned} \sup _{t \in {\mathbb {R}}} \sup _{\left| f \right| \le g} \, {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{n=0}^\infty |\hat{f}(t-V_n)| \bigg ] \le C_g {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{n=0}^\infty \mathbbm {1}_{[-4h,4h]}(V_n) \bigg ] =:D_g < \infty . \end{aligned}$$

Now we decompose the integral inside the limit in (7.12) according to the set $\{R_{\tau _1-1}> t/2\}$ to obtain the following upper bound

$$\begin{aligned} \sup _{s \le t/2} \sup _{\left| f \right| \le g} t^{ \ell + \epsilon } \bigg |{\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{n=0}^\infty \hat{f}(t-s-V_n) \bigg ] - \frac{1}{{\mathbb {E}}_{\nu \otimes \eta } [V_1]}\int _0^\infty \hat{f}(r) \text {d} \textit{r}\bigg |&\\ +\, 2 D_g \, t^{ \ell + \epsilon } \, {\mathbb {P}}_{\textit{I}_{\textit{d}},0}(R_{\tau _1-1} > t/2)&. \end{aligned}$$

The second term tends to zero by (7.11). For the first term, we invoke [69, Theorem 4.2(ii)] (with $G=\delta _0$), which gives (note that $\left| f \right| \le g$ implies $|\hat{f}| \le \hat{g}$)

$$\begin{aligned} \lim _{t \rightarrow \infty } t^{ \ell + \epsilon } \sup _{\left| \hat{f} \right| \le \hat{g}} \left| {\mathbb {E}}_{\nu \otimes \eta } \bigg [ \sum _{n=0}^\infty \hat{f}(t-s-V_n) \bigg ] -\frac{1}{{\mathbb {E}}_{\nu \otimes \eta } [V_1]}\int _0^\infty \hat{f}(r) dr \right| = 0 \end{aligned}$$

as soon as $V_1$ has positive drift (here, ${\mathbb {E}}_{\nu \otimes \eta }[V_1] = {\mathbb {E}}_{\nu \otimes \eta }[\tau _1]{\mathbb {E}}[S_1] > 0$ by the proof of Proposition 7.2), a spread-out law (here, the law of $V_1$ is even absolutely continuous) and ${\mathbb {E}}_{\nu \otimes \eta }[|V_1|^{ \ell +\epsilon +1}]<\infty$ (which is true by (7.2) and Lemma 7.3(iv)) and $\hat{g}$ is bounded, Lebesgue-integrable and satisfies

$$\begin{aligned} t^{ \ell + \epsilon } \int _t^{2t} \hat{g}(r) \text {d} \textit{r}\rightarrow 0 \quad \text { and } \quad t^{ \ell + \epsilon } \sup _{r \ge t} \hat{g}(r) \rightarrow 0 \quad \text { as } t\rightarrow \infty . \end{aligned}$$

(7.13)

Lemma 7.4 gives $\lim _{t \rightarrow \infty } t^{ \ell +\epsilon +1} \hat{g}(t) = 0$, which is sufficient for (7.13) to hold. $\square$