1 Introduction

Fix a probability space \((\Omega ,{\mathcal {A}},{\mathbb {P}})\) and a metric space X. Let \({\mathcal {B}}(X)\) denote the \(\sigma \)-algebra of all Borel subsets of X.

We say that \(f:X\times \Omega \rightarrow X\) is a random-valued function (shortly: an rv-function) if it is measurable for the product \(\sigma \)-algebra \({\mathcal {B}}(X)\otimes {\mathcal {A}}\). The iterates of such an rv-function are given by

$$\begin{aligned} f^{0}(x,\omega _1,\omega _2,\ldots )=x,\quad f^{n}(x,\omega _1,\omega _2,\ldots )=f(f^{n-1}(x,\omega _1,\omega _2,\ldots ),\omega _{n}) \end{aligned}$$

for \(n\in {\mathbb {N}}\), \(x\in X\) and \((\omega _1,\omega _2,\ldots )\) from \(\Omega ^{\infty }\) defined as \(\Omega ^{{\mathbb {N}}}\). Note that \(f^n:X\times \Omega ^\infty \rightarrow X\) is an rv-function on the product probability space \((\Omega ^\infty ,{\mathcal {A}}^\infty , \mathbb P^\infty )\). More exactly, one can show that for \(n\in {\mathbb {N}}\) the n-th iterate \(f^n\) is measurable for \({\mathcal {B}}(X)\otimes \mathcal A_n\), where \({\mathcal {A}}_n\) denotes the \(\sigma \)-algebra of all sets of the form

$$\begin{aligned} \{(\omega _1,\omega _2,\ldots )\in \Omega ^{\infty }:(\omega _1,\ldots ,\omega _n)\in A\} \end{aligned}$$

with A from the product \(\sigma \)-algebra \({\mathcal {A}}^n\).

The iterates so defined were introduced independently in [4] and [9] with reference to functional equations (see e.g. [3, 7, 18]). In a broader context they form random forward iterations (see e.g. [8, 13]), also known as outer iterations (see e.g. [11]). These iterates are prototypes of random dynamical systems (see [1, Section 1.1]; cf. [22]) and they have Markov property. The family \(\{f(\cdot ,\omega ): \omega \in \Omega \}\) forms an iterated function system (IFS for abbreviation) in which functions \(f(\cdot ,\omega )\) are choosing independently with probability \({\mathbb {P}}\). A generalization of this concept, devoted to random iteration with place-dependent probabilities, can be found, e.g., in [16, Section 3] and [25]. As in the mentioned papers we will express asymptotic behaviour of our iterates by the convergence in law. In fact, this type of convergence of iterations is closely related to the asymptotic stability of Markov operators with the kernel of the form \((x,B)\mapsto \int _\Omega \mathbbm {1}_B(f(x,\omega ))\mathbb P(d\omega )\), determined by a fixed rv-function f. For details see [14] and for a more complete point of view we refer the reader to [13, 17, 20] and the references therein; we only mention here that Markovian operator P is asymptotically stable, if it has an invariant measure \(\mu ^*\), i.e., \(P\mu ^*=\mu ^*\), which attracts any probability Borel measure. There are many papers in which the convergence in law of iterates of rv-functions and the stability of Markov operators with the kernel determined by an rv-function are investigated; however usually a kind of Lipschitz contraction on the rv-function considered is assumed (see e.g. [2, 6, 8, 15]).

This paper aims to extend the results on convergence in law of iterates of random-valued functions that are mean contractive in the conventional sense to the case where only a weak (non-linear) form of the mean contractivity is assumed and examine how fast the sequence of iterates converges. The speed of convergence obtained does not have to be geometric as in the Lipschitz case. Let us mention that results [21, Theorem 9.2] and [23, Theorem 6.3.2] involve the same form of the contractivity property as that employed in the manuscript pertain to IFS with place-dependent probabilities of choosing them, but for finite many transformations and bring only asymptotic stability of IFS.

2 Preliminaries

Let \((X,\rho )\) be a complete and separable metric space. By \({\mathcal {M}}_1(X)\) we denote the set of all probability measures defined on \({\mathcal {B}}(X)\). \(\textrm{Lip}_{\alpha }(X)\) denotes the set of all real functions defined on X that meet a Lipschitz condition with a constant \(\alpha \in [0,\infty )\), and

$$\begin{aligned} \mathrm Lip_1^b({ X})=\{\varphi \in \mathrm Lip_1({ X}):\varphi \hbox { is bounded}\}. \end{aligned}$$

It is well known (see [10, Theorem 11.3.3]) that the weak convergence of probability Borel measures on X is metrizable by the Fortet-Mourier metric \(d_{FM}:{\mathcal {M}}_1(X)\times \mathcal M_1(X)\rightarrow [0,\infty )\) given by

$$\begin{aligned} d_{FM}(\mu ,\nu )=\sup \left\{ \left| \int _X\varphi d\mu -\int _X\varphi d\nu \right| :\varphi \in \mathrm Lip_1({ X}),\,{|\varphi (x)|\le 1}\hbox { for }{x\in X}\right\} . \end{aligned}$$

Moreover we will use the Hutchinson metric (see [12, 19]), also known as Wasserstein or Kantorovich-Rubinstein distance [24], defined by

$$\begin{aligned} d_H(\mu ,\nu )=\sup \left\{ \left| \int _X\varphi d\mu -\int _X\varphi d\nu \right| :\varphi \in \mathrm Lip_1^b({ X})\right\} . \end{aligned}$$

According to [15, Lemma 3.1(i)],

$$\begin{aligned} \begin{aligned} d_H(\mu ,\nu )=\sup \left\{ \left| \int _X\varphi d\mu -\int _X\varphi d\nu \right| :\varphi \in \mathrm Lip_1({ X})\right\} \\ \hbox {for }\mu ,\nu \in {\mathcal {M}}_1^1(X), \end{aligned} \end{aligned}$$
(1)

where

$$\begin{aligned} {\mathcal {M}}_1^1(X)=\left\{ \mu \in \mathcal M_1(X):\int _X\rho (x,x_0)\mu (dx)<\infty \right\} \end{aligned}$$

with an arbitrarily fixed \(x_0\in X\); the definition does not depend on the choice of \(x_0\).

Remark 2.1

[see [24, Theorem 6.18], cf. [15, Theorem 3.3 and Remark 3.2]]. The metric space \(({\mathcal {M}}_1^1(X),d_H|_{\mathcal M_1^1(X)\times {\mathcal {M}}_1^1(X)})\) is complete.

Remark 2.2

[see [15, Theorem 3.3]]. The set \({\mathcal {M}}_1^1(X)\) is dense in \(({\mathcal {M}}_1(X),d_{FM})\).

3 Main result

We employ the following hypothesis.

  1. (H)

    \((X,\rho )\) is a separable and complete metric space and \(f:X\times \Omega \rightarrow X\) is an rv-function such that

    $$\begin{aligned} \int _\Omega \rho (f(x,\omega ),f(z,\omega ))\mathbb P(d\omega )\le \psi (\rho (x,z))\quad \hbox {for }x,z\in X \end{aligned}$$

    with a concave function \(\psi :[0,\infty )\rightarrow [0,\infty )\).

Remark 3.1

If \(\psi :[0,\infty )\rightarrow [0,\infty )\) is concave, then it is non-decreasing.

Proof

Suppose, towards a contradiction, that there are \(t_1,t_2\in [0,\infty )\) such that \(t_1<t_2\) and \(\psi (t_2)<\psi (t_1)\). Fix \(\alpha \in (\frac{\psi (t_2)}{\psi (t_1)},1)\) and put \(t=\frac{t_2-\alpha t_1}{1-\alpha }\). Then

$$\begin{aligned} \psi (t_2)=\psi (\alpha t_1+(1-\alpha )t)\ge \alpha \psi (t_1)+(1-\alpha )\psi (t), \end{aligned}$$

and hence \(\psi (t)\le \frac{\psi (t_2)-\alpha \psi (t_1)}{1-\alpha }<0\), a contradiction. \(\square \)

Proposition 3.1

Assume (H) and define \(P:\mathcal M_1(X)\rightarrow {\mathcal {M}}_1(X)\) by

$$\begin{aligned} (P\mu )(B)=\int _X\left( \int _\Omega \mathbbm {1}_B(f(x,\omega ))\mathbb P(d\omega )\right) \mu (dx) \quad \hbox {for }B\in {\mathcal {B}}(X). \end{aligned}$$
(2)

Then

$$\begin{aligned} d_H(P\mu ,P\nu )\le \psi (d_H(\mu ,\nu ))\quad \hbox {for }\mu ,\nu \in {\mathcal {M}}_1^1(X). \end{aligned}$$

Proof

Observe first that for any Borel \(\varphi :X\rightarrow {\mathbb {R}}\), which is non-negative or bounded, we have

$$\begin{aligned} \int _X\varphi d(P\mu )=\int _X\left( \int _\Omega \varphi (f(x,\omega ))\mathbb P(d\omega )\right) \mu (dx) \quad \hbox {for }\mu \in {\mathcal {M}}_1(X). \end{aligned}$$
(3)

Fix \(\mu ,\nu \in {\mathcal {M}}_1^1(X)\) and denote by \(\Lambda (\mu ,\nu )\) the collection of all probability Borel measures \(\lambda \) on \(X\times X\) such that

$$\begin{aligned} \lambda (B\times X)=\mu (B)\quad \hbox {and}\quad \lambda (X\times B)=\nu (B)\quad \hbox {for }B\in {\mathcal {B}}(X). \end{aligned}$$

If \(\lambda \in \Lambda (\mu ,\nu )\), then

$$\begin{aligned} \int _{X\times X}\rho d\lambda&\le \int _{X\times X}\left( \rho (x,x_0)+\rho (x_0,z)\right) \lambda (d(x,z))\\&=\int _{X}\rho (x,x_0)\mu (dx)+\int _{X}\rho (x_0,z)\nu (dz)<\infty . \end{aligned}$$

Therefore, the formula

$$\begin{aligned} T(\lambda )=\int _{X\times X}\rho d\lambda \quad \hbox {for }\lambda \in \Lambda (\mu ,\nu ) \end{aligned}$$

defines a functional \(T:\Lambda (\mu ,\nu )\rightarrow [0,\infty )\).

If \(\lambda \in \Lambda (\mu ,\nu )\) and \(\varphi \in \mathrm Lip_1^b({ X})\), then by (3), (H) and Jensen’s inequality (see [10, 10.2.6]), we have

$$\begin{aligned} \left| \int _X \varphi d(P\mu )\right.&\left. -\int _X\varphi d(P\nu )\right| \\&=\left| \int _{X\times X} \left( \int _{\Omega } (\varphi (f(x,\omega ))-\varphi (f(z,\omega )))\mathbb P(d\omega )\right) \lambda (d(x,z))\right| \\&\le \int _{X\times X} \left( \int _{\Omega } \big |\varphi (f(x,\omega ))-\varphi (f(z,\omega ))\big |\mathbb P(d\omega )\right) \lambda (d(x,z))\\&\le \int _{X\times X}\left( \int _{\Omega } \rho (f(x,\omega ),f(z,\omega )){\mathbb {P}}(d\omega )\right) \lambda (d(x,z))\\&\le \int _{X\times X}\psi \big (\rho (x,z)\big )\lambda (d(x,z)) \le \psi \left( \int _{X\times X}\rho (x,z)\lambda (d(x,z))\right) \\&=\psi \big (T(\lambda )\big ), \end{aligned}$$

and hence

$$\begin{aligned} d_H(P\mu ,P\nu )\le \psi \big (T(\lambda )\big )\quad \hbox {for }\lambda \in \Lambda (\mu ,\nu ). \end{aligned}$$
(4)

Applying (1) and the Kantorovich-Rubinstein Theorem (see [10, Theorem 11.8.2]) we conclude that there exists \(\lambda _0\in \Lambda (\mu ,\nu )\) such that

$$\begin{aligned} d_H(\mu ,\nu )=\inf \left\{ \int _{X\times X}\rho d\lambda :\lambda \in \Lambda (\mu ,\nu )\right\} = \int _{X\times X}\rho d\lambda _0. \end{aligned}$$

This jointly with (4) implies

$$\begin{aligned} d_H(P\mu ,P\nu )\le \psi (T(\lambda _0))=\psi \left( \int _{X\times X}\rho d\lambda _0\right) =\psi (d_H(\mu ,\nu )), \end{aligned}$$

and the proof is complete. \(\square \)

Corollary 3.1

Assume (H) and let \(P:\mathcal M_1(X)\rightarrow {\mathcal {M}}_1(X)\) be the operator given by (2). If there exists \(x_0\in X\) such that

$$\begin{aligned} \int _\Omega \rho (f(x_0,\omega ),x_0){\mathbb {P}}(d\omega )<\infty , \end{aligned}$$
(5)

then \(P({\mathcal {M}}_1^1(X))\subset {\mathcal {M}}_1^1(X)\) and for every \(n\in {\mathbb {N}}\) we have

$$\begin{aligned} d_H(P^n\mu ,P^n\nu )\leqslant \psi ^n (d_H(\mu ,\nu ))\quad \hbox {for }\mu ,\nu \in {\mathcal {M}}_1^1(X). \end{aligned}$$
(6)

Proof

If \(\mu \in {\mathcal {M}}_1^1(X)\), then by (3) with \(\varphi =\rho (\cdot ,x_0)\) we obtain

$$\begin{aligned} \int _X\rho (x,x_0)(P\mu )(dx)&=\int _X\left( \int _\Omega \rho (f(x,\omega ),x_0)\mathbb P(d\omega )\right) \mu (dx)\\&\le \int _X\left( \int _\Omega \rho (f(x,\omega ),f(x_0,\omega ))\mathbb P(d\omega )\right) \mu (dx)\\&\quad +\int _\Omega \rho (f(x_0,\omega ),x_0){\mathbb {P}}(d\omega )\\&\le \int _X\psi (\rho (x,x_0))\mu (dx)+\int _\Omega \rho (f(x_0,\omega ),x_0)\mathbb P(d\omega )\\&\le \psi \left( \int _X\rho (x,x_0)\mu (dx)\right) +\int _\Omega \rho (f(x_0,\omega ),x_0)\mathbb P(d\omega ). \end{aligned}$$

Thus (5) implies \(P\mu \in {\mathcal {M}}_1^1(X)\).

By Proposition 3.1 we see that (6) holds for \(n=1\). If (6) holds for some \(n\in {\mathbb {N}}\), then Proposition 3.1 and Remark 3.1 for \(\mu ,\nu \in {\mathcal {M}}_1^1(X)\) imply

$$\begin{aligned} d_H(P^{n+1}\mu ,P^{n+1}\nu )&\le \psi (d_H(P^n\mu ,P^n\nu ))\le \psi (\psi ^n (d_H(\mu ,\nu )))\\&=\psi ^{n+1}(d_H(\mu ,\nu )), \end{aligned}$$

which completes the proof. \(\square \)

Given an rv-function \(f:X\times \Omega \rightarrow X\) we denote by \(\pi ^f_n(x,\cdot )\) the distribution of \(f^n(x,\cdot )\), i.e.,

$$\begin{aligned} \pi ^f_n(x,B)={\mathbb {P}}^\infty (f^n(x,\cdot )\in B)\quad \hbox {for }n\in {\mathbb {N}}\cup \{0\},\, x\in X\hbox { and }B\in {\mathcal {B}}(X). \end{aligned}$$

Clearly, for every \(x\in X\), \(\pi ^f_0(x,\cdot )=\delta _x\), the Dirac measure concentrated at x, and \(\pi ^f_1(x,\cdot )\) is the distribution of \(f(x,\cdot )\). One can check that \(\pi ^f_{n+1}(x,\cdot )= P\pi _f^n(x,\cdot )\) holds for \(x\in X\) and \(\mu \in {\mathcal {M}}_1(X)\) (see [14]), which implies \(\pi ^f_{n+1}(\mu ,\cdot ) = P\pi ^f_n(\mu ,\cdot )\) for any \(\mu \in {\mathcal {M}}_1(X)\), with \(\pi ^f_n(\mu ,\cdot )=\int _X \pi ^f_n(x,\cdot )\mu (dx)\), and shows that operator P given by (2) is the transition operator of the sequence of iterates under consideration. It turns out that this operator is asymptotically stable. In fact, the following theorem gives what follows.

Theorem 3.1

Assume (H) with \(\psi \) satisfying also

$$\begin{aligned} \psi (t)<t\quad \hbox {for }t\in (0,\infty ). \end{aligned}$$
(7)

If (5) holds with some \(x_0\in X\), then the operator \(P:{\mathcal {M}}_1(X)\rightarrow {\mathcal {M}}_1(X)\) given by (2) admits an invariant measure \(\pi ^f\in {\mathcal {M}}_1^1(X)\) and

$$\begin{aligned}{} & {} d_H(P^n\mu ,\pi ^f)\le \psi ^n(d_H(\mu ,\pi ^f))\quad \hbox {for }\mu \in {\mathcal {M}}_1^1(X)\hbox { and }n\in {\mathbb {N}}, \end{aligned}$$
(8)
$$\begin{aligned}{} & {} \begin{aligned} d_H(\pi ^f_n(x,\cdot ),\pi ^f)\le \psi ^n(d_H(\delta _x,\pi ^f)) \le \psi ^n\left( \int _X\rho (x,z)\pi ^f(dz)\right) \\ \hbox {for }x\in X\hbox { and }n\in {\mathbb {N}}. \end{aligned} \end{aligned}$$
(9)

Moreover,

$$\begin{aligned} \lim _{n\rightarrow \infty }d_{FM}(P^n\mu ,\pi ^f)=0\quad \hbox {for }\mu \in {\mathcal {M}}_1(X). \end{aligned}$$

Proof

From Corollary 3.1, Remark 2.1 and the Boyd-Wong Theorem (see [5, Theorem 1]) we conclude that there exists a measure \(\pi ^f\in {\mathcal {M}}_1^1(X)\) such that

$$\begin{aligned} P\pi ^f=\pi ^f \end{aligned}$$

and

$$\begin{aligned} \lim _{n\rightarrow \infty }d_H(P^n\mu ,\pi ^f)=0\quad \hbox {for }\mu \in \mathcal M_1^1(X). \end{aligned}$$
(10)

By Corollary 3.1 we have (8).

To get the first inequality in (9), observe that by a simple induction for all \(x\in X\) and \(n\in {\mathbb {N}}\cup \{0\}\) we have

$$\begin{aligned} P^n\delta _x=\pi ^f_n(x,\cdot ); \end{aligned}$$
(11)

indeed, \(P^0\delta _x=\delta _x=\pi _0^f(x,\cdot )\) and if (11) holds for some \(n\in {\mathbb {N}}\cup \{0\}\), then

$$\begin{aligned} (P^{n+1}\delta _x)(B)&=(P(\pi ^f_n(x,\cdot )))(B)=\int _X\left( \int _\Omega \mathbbm {1}_B(f(z,\omega )){\mathbb {P}}(d\omega )\right) \pi ^f_n(x,dz)\\&=\int _{\Omega ^\infty }\left( \int _\Omega \mathbbm {1}_B(f(f^n(x,\omega ),\omega ')){\mathbb {P}}(d\omega ')\right) \mathbb P^\infty (d\omega )\\&=\int _{\Omega ^\infty }\mathbbm {1}_B(f^{n+1}(x,\omega ))\mathbb P^\infty (d\omega )=\pi ^f_{n+1}(x,B) \end{aligned}$$

for every \(B\in {\mathcal {B}}\)(X) (cf. [14, Proposition 2.1]). Since \(\delta _x\in {\mathcal {M}}_1^1(X)\) for every \(x\in X\), Corollary 3.1 and (11) imply \(\pi ^f_n(x,\cdot )\in \mathcal M_1^1(X)\) for all \(x\in X\) and \(n\in {\mathbb {N}}\). Therefore, the first inequality in (9) follows from (8) and (11).

For the prove of the second inequality in (9), note that if \(x\in X\) and \(\varphi \in \mathrm Lip_1^b({ X})\), then

$$\begin{aligned} \left| \int _X \varphi d\delta _x-\int _X\varphi d\pi ^f\right|&=\left| \varphi (x)-\int _X\varphi (z)\pi ^f(dz)\right| \\&\le \int _X|\varphi (x)-\varphi (z)|\pi ^f(dz)\le \int _X\rho (x,z)\pi ^f(dz). \end{aligned}$$

Hence

$$\begin{aligned} d_H(\delta _x,\pi ^f)\le \int _X\rho (x,z)\pi ^f(dz)\quad \hbox {for }x\in X, \end{aligned}$$

which jointly with Remark 3.1 gives the second inequality in (9).

It remains to prove the moreover part. For this purpose, we fix \(\mu \in {\mathcal {M}}_1(X)\).

If \(\varphi \in \mathrm Lip_1({ X})\) and \(|\varphi (x)|\le 1\) for every \(x\in X\), then due to (H) and (7) the function \(\phi :X\rightarrow {\mathbb {R}}\) given by

$$\begin{aligned} \phi (x)=\int _\Omega \varphi (f(x,\omega )){\mathbb {P}}(d\omega ) \end{aligned}$$

belongs to \(\mathrm Lip_1({ X})\) and \(|\phi (x)|\le 1\) for every \(x\in X\), and moreover, by (3), for every \(\nu \in {\mathcal {M}}_1(X)\) we have

$$\begin{aligned} \left| \int _X\varphi d(P\mu )-\int _X\varphi d(P\nu )\right| =\left| \int _X\phi d\mu -\int _X\phi d\nu \right| \le d_{FM}(\mu ,\nu ), \end{aligned}$$

whence

$$\begin{aligned} d_{FM}(P\mu ,P\nu )\le d_{FM}(\mu ,\nu )\quad \hbox {for }\mu ,\nu \in {\mathcal {M}}_1(X). \end{aligned}$$

Fix \(\varepsilon >0\). By Remark 2.2 there exists \(\nu \in \mathcal M_1^1(X)\) such that \(d_{FM}(\mu ,\nu )\le \frac{\varepsilon }{2}\) and by (10) there exists \(n_0\in {\mathbb {N}}\) such that \(d_H(P^n\nu ,\pi ^f)\le \frac{\varepsilon }{2}\) for every \(n\ge n_0\). Hence

$$\begin{aligned} d_{FM}(P^n\mu ,\pi ^f)&\le d_{FM}(P^n\mu ,P^n\nu )+d_{FM}(P^n\nu ,\pi ^f)\\&\le d_{FM}(\mu ,\nu )+d_H(P^n\nu ,\pi ^f)\le \frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon \end{aligned}$$

for every \(n\ge n_0\). \(\square \)

Remark 3.2

Assume \(f:X\times \Omega \rightarrow X\) is an rv-function. If for some \(x \in X\) the sequence \((\pi _n^f(x,\cdot ))_{n\in {\mathbb {N}}}\) converges weakly to a \(\pi \), then \(\textrm{supp}\pi \subset \textrm{cl}f(X \times \Omega )\).

Proof

Suppose there exists \(x_0 \in \textrm{supp}\pi \cap (X{\setminus }\textrm{cl}f(X \times \Omega ))\) and let B be a closed ball in X with center at \(x_0\) and contained in \(X{\setminus }\textrm{cl}f(X\times \Omega )\). Since \(x_0\in \textrm{supp}\pi \), it follows that \(\pi (B)>0\), and by Urysohn’s lemma there is a continuous \(\varphi :X \rightarrow [0,1]\) such that

$$\begin{aligned} \varphi (z)=1\hbox { for }z\in B\quad \hbox { and }\quad \varphi (z)=0\hbox { for }z\in \textrm{cl}f(X\times \Omega ). \end{aligned}$$

Then \(\varphi \big (f^n(x,\omega )\big )=0\) for \(n \in {\mathbb {N}}\) and \(\omega \in \Omega ^\infty \), and

$$\begin{aligned} \int _X\varphi (z)\pi (dz)=\lim \limits _{n \rightarrow \infty }\int _X\varphi (z) \pi _n^f(x,dz)=\lim \limits _{n \rightarrow \infty }\int _{\Omega ^\infty }\varphi (f^n(x,\omega )){\mathbb {P}}^\infty (d\omega )=0. \end{aligned}$$

Hence

$$\begin{aligned} \pi (B)= \int _B\varphi d\pi =0, \end{aligned}$$

a contradiction. \(\square \)

By Remark 3.1 we can set

$$\begin{aligned} \psi (\infty )=\lim _{x\rightarrow \infty }\psi (x) \end{aligned}$$

and consider \(\psi \) as a mapping of \([0,\infty ]\) into itself.

Remark 3.3

Since by the Remark 3.2 the support of the invariant measure \(\pi ^f\) from Theorem 3.1 is included in the closure of \(f(X\times \Omega )\), it follows that for all \(x\in X\) and \(y\in f(X\times \Omega )\), we have

$$\begin{aligned} \int _{X}\rho (x,z)\pi ^f(dz)\le \rho (x,y)+\int _{\textrm{supp}\pi ^f}\rho (y,z)\pi ^f(dz) \le \rho (x,y)+\textrm{diam}f(X\times \Omega ). \end{aligned}$$

Therefore, (9) yields

$$\begin{aligned} d_H\left( \pi ^f_n(x,\cdot ),\pi ^f\right) \le \psi ^n\big (\textrm{dist}(x,f(X\times \Omega )) +\textrm{diam}f(X\times \Omega )\big ) \end{aligned}$$

for all \(x\in X\) and \(n\in {\mathbb {N}}\).

4 Examples

Fix \(\xi \in L^1(\Omega ,{\mathcal {A}},{\mathbb {P}})\), a non-zero \(\eta \in L^1(\Omega ,{\mathcal {A}},{\mathbb {P}})\), and put \(\alpha =\frac{1}{\Vert \eta \Vert _1}\). Let \(\psi :[0,\infty )\rightarrow [0,\infty )\) be a concave function such that \(\eta \psi (x)+\xi \) is non-negative for every \(x\in [0,\infty )\) and

$$\begin{aligned} \frac{|\psi (\alpha x)-\psi (\alpha z)|}{\alpha }\le \psi (|x-z|)<|x-z|\quad \hbox {for }x,z\in [0,\infty )\hbox { with }x\ne z. \end{aligned}$$
(12)

Note that \(\psi \) is non-expansive and \(\psi (0)=0\). In particular, the formula

$$\begin{aligned} f(x,\omega )=\eta (\omega )\psi (\alpha x)+\xi (\omega )\quad \hbox {for }x\in [0,\infty )\hbox { and }\omega \in \Omega \end{aligned}$$

defines a random affine map \(f:[0,\infty )\times \Omega \rightarrow [0,\infty )\). It is clear that (5) holds with any \(x_0\in [0,\infty )\) and

$$\begin{aligned} \int _{\Omega }|f(x,\omega )-f(z,\omega )|\mathbb P(d\omega )\le \psi (|x-z|) \quad \hbox {for }x,z\in [0,\infty ). \end{aligned}$$

In consequence (H) holds, and one can apply Theorem 3.1. Note that neither [2, Theorem 3.1] nor [8, Theorem 1.1] do not apply, whenever

$$\begin{aligned} \psi \not \in \bigcup _{\alpha \in (0,1)}\textrm{Lip}_\alpha ([0,\infty )). \end{aligned}$$
(13)

Let \(\pi ^f\) be the measure resulting from Theorem 3.1. Then (9) yields

$$\begin{aligned} d_H\left( \pi ^f_{n}(x,\cdot ),\pi ^f\right) \le \psi ^{n-1}(\psi (\infty )) \quad \hbox {for }x\in [0,\infty )\hbox { and }n\in {\mathbb {N}}. \end{aligned}$$
(14)

Example 4.1

Consider \(\psi :[0,\infty )\rightarrow [0,\infty )\) given by

$$\begin{aligned} \psi (t)=\frac{t}{1+t}. \end{aligned}$$

It is easy to see that \(\psi \) is concave, satisfies (12) with \(\alpha =1\), (13) holds, and (14) leads to

$$\begin{aligned} d_H\left( \pi ^f_{n}(x,\cdot ),\pi ^f\right) \le \psi ^{n-1}(1)=\frac{1}{n} \quad \hbox {for }x\in [0,\infty )\hbox { and }n\in {\mathbb {N}}. \end{aligned}$$

Example 4.2

Consider now \(\psi :[0,\infty )\rightarrow [0,\infty )\) given by

$$\begin{aligned} \psi (t)=\arctan t. \end{aligned}$$

It is easy to check that \(\psi \) is concave, satisfies (12) with \(\alpha =1\), and (13) holds. Now, (14) gives

$$\begin{aligned} d_H\left( \pi ^f_{n}(x,\cdot ),\pi ^f\right) \le \psi ^{n-1}\left( \frac{\pi }{2}\right) \quad \hbox {for }x\in [0,\infty )\hbox { and }n\in {\mathbb {N}}. \end{aligned}$$

Applying [18, Theorem 1.3.6] we conclude that for every \(c\in \left( \sqrt{\frac{3}{2}},\infty \right) \) there exists \(n_0\in {\mathbb {N}}\) such that

$$\begin{aligned} d_H\left( \pi ^f_{n}(x,\cdot ),\pi ^f\right) \le \frac{c}{\sqrt{n}}\quad \hbox {for }x\in [0,\infty )\hbox { and } n\ge n_0. \end{aligned}$$

Example 4.3

Fix \(\psi _0:[0,\infty )\rightarrow [0,\infty )\) of the form \(\psi _0(t)=\frac{t}{1+t}\) or \(\psi _0(t)=\arctan t\), and consider \(\psi :[0,\infty )\rightarrow [0,\infty )\) given by

$$\begin{aligned} \psi (t)=\sin \psi _0(t). \end{aligned}$$

Note that \(\psi \) is concave and (13) holds. Observe also that \(\psi \) satisfies (12) with \(\alpha =2\); indeed, for all \(x\ne z\) we have

$$\begin{aligned} |\sin \psi _0(2x)-\sin \psi _0(2z)|&\le 2\sin \frac{|\psi _0(2x)-\psi _0(2z)|}{2} \le 2\sin \frac{\psi _0(2|x-z|)}{2}\\&\le 2\sin \psi _0(|x-z|)\le 2|x-z|. \end{aligned}$$

Since \(\psi \le \psi _0\), we conclude that (14) implies

$$\begin{aligned} d_H\left( \pi ^f_{n}(x,\cdot ),\pi ^f\right) \le \psi _0^{n-1}(\psi _0(\infty )) \quad \hbox {for }x\in [0,\infty )\hbox { and }n\in {\mathbb {N}}. \end{aligned}$$