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Central Limit Theorem for Random “Contractive” Functions on Interval Cover

Central Limit Theorem for Random “Contractive” Functions on Interval

Open Access
|Apr 2025

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1. Introduction

We are given a family of continuous functions {fi : [0, 1] → [0, 1] : i ∈ {1, . . . , N}} that are increasing and injective, possessing the following “contractive” properties:

  • (1) ∀x∈(0,1)∃i,j∈{1,...,N} fi(x) < x < fj (x),

  • (2) ∃i∈{1,...,N} fi(0) > 0,

  • (3) ∃i∈ {1,...,N} fi(1) < 1.

We are also given a probability vector (p1, . . . , pN). The family (f1, . . . , fN; p1, . . . , pN) is called an iterated function system with probabilities.

This system generates a Markov chain, the distribution of which we will investigate. First, we are interested in whether this chain possesses a unique, non-atomic invariant measure. We will show that every measure will converge weakly to the unique invariant measure. Finally, we will prove the central limit theorem for this chain.

We use the approach from [1] as arguments for proving the properties turn out to be similar and easier to grasp. However, we took a different approach to proving the existence of a unique invariant measure as we use e-property introduced in [2] which made the proof quicker.

2. Theoretical base

Let (S, d) be a metric space. By (C(S), ‖ · ‖), we denote the family of all continuous and bounded functions f : S → ℝ, equipped with the supremum norm ‖ · ‖.

Let ℳ(S) denote the set of all finite measures on the σ-algebra ℬ(S) of Borel subsets of the set S. Let ℳ1(S) ⊂ ℳ(S) denote the subset of all probability measures on S.

An operator P : ℳ(S) → ℳ(S) is called a Markov operator if it satisfies the following conditions:

  • (1) P(λ1μ1 + λ2μ2) = λ1Pμ1 + λ2Pμ2 for λ1, λ2 ≥ 0, μ1, μ2 ∈ ℳ(S),

  • (2) Pμ(S) = μ(S) for μ ∈ ℳ(S).

A Markov operator P is called a Feller operator if there exists a linear operator U : C(S) → C(S) such that ∫S Ufdμ = ∫S fdPμ for every f ∈ C(S) and μ ∈ ℳ(S).

A measure μ∗ is called P-invariant for the Markov operator P if Pμ∗ = μ∗.

A Markov operator P is called asymptotically stable if it has a unique invariant measure μ∗ ∈ ℳ1(S) and, moreover, for every measure μ ∈ ℳ1(S), the sequence (Pnμ)n∈ℕ converges weakly to μ∗, i.e.,

∀f∈ CSlimn→∞∫SfxPnμdx=∫Sfxμ*dx.

Theorem 2.1 (Krylov-Bogolyubov [3]).

Let (S, d) be a compact metric space, and let P : ℳ(S) → ℳ(S) be a Feller operator defined on finite Borel measures on this space.

Then P has at least one invariant probability measure, i.e., there exists a measure μ ∈ ℳ1(S) such that for any A ∈ ℬ(S),

PμA=μA.

Let (Ω, ℱ, μ) be a measure space. A set A ∈ ℱ is called an atom if, for any B ∈ ℱ, B ⊂ A, and μ(B) < μ(A), we have μ(B) = 0. A measure is called non-atomic if the measure space has no atoms, i.e., if A, B ∈ ℱ and B ⊂ A, then μ(A) > μ(B) > 0.

We will consider the following Feller operator. Assume that fi : [0, 1] → [0, 1] for i ∈ {1, . . . ,N} are continuous functions and let (p1, . . . , pN) be a probability vector. The family (f1, . . . , fN; p1, . . . , pN) generates a Markov operator P : ℳ(S) → ℳ(S) of the form

PμA=∑i=1Npiμfi−1A   for A∈ℬ0,1.
This operator is a Feller operator, whose predual operator is the operator U : C([0, 1]) → C([0, 1]) given by the formula
Uϕx=∑i=1Npiϕfix   for ϕ∈C0,1,x∈0,1.

Let H be the space of continuous functions f : [0, 1] → [0, 1] that are increasing and injective. Let {f1, . . . , fN} ⊂ H be a finite set of functions satisfying the following properties:

  • (1) ∀x∈(0,1) ∃i,j∈{1,...,N} fi(x) < x < fj(x),

  • (2) ∃i∈{1,...,N} fi(0) > 0,

  • (3) ∃i∈{1,...,N} fi(1) < 1

and (p1, . . . , pN) is a probability vector. fd denotes a function satisfying condition (2), and fg denotes a function satisfying condition (3). The corresponding probabilities are denoted by pd and pg.

The system (f1, . . . , fN; p1, . . . , pN) has a corresponding Markov operator P defined as before. For each measure ν ∈ ℳ1(S), we describe the Markov chain (Xn) with the transition probability π(x, A) = Pδx(A) for x ∈ [0, 1], A ∈ ℬ([0, 1]), and initial distribution ν using the probabilistic measure ℙν on the space ([0, 1]ℕ, ℬ([0, 1])⊗ℕ) such that:

ℙν[Xn+1∈A|Xn=x]=πx,A   and   ℙνX0∈A=νA,
where x ∈ [0, 1], A ∈ ℬ([0, 1]).

Since the interval [0, 1] is a compact set, the Markov operator P has at least one invariant measure μ∗, which is a consequence of Krylov-Bogoliubov’s theorem. We now need to show that there is only one invariant measure. We also need to show that it is nonatomic.

3. The invariant measure

We introduce a few theorems and lemmas needed to prove atomlessness and existence of a unique measure.

Theorem 3.1 ([4], Corollary 2.13).

Let ω→gωnn≥0 be a non-degenerate random walk generated by a subgroup G ⊂ Homeo([0, 1]), such that:

  • (1) There is no non-trivial interval I ⊂ [0, 1] invariant under G.

  • (2) There exists at least one probability measure μ on (0, 1) that is stationary for the random walk.

Then there exists a q < 1 such that for every x ∈ (0, 1), there exists a neighborhood Ix of x, such that for ℙ almost all i ∈ Σ, the following holds:
ginIx≤qn   for every n∈ℕ.
Moreover, if the first condition is violated, the theorem holds for
x∈infsupp μ,supsupp μ.

Remark 3.1.

Note that the theorem is also true on the interval [a, b].

Theorem 3.2.

There exists a q < 1 such that for every x ∈ (0, 1), there exists a neighborhood Ix of x, such that for ℙ almost all i ∈ Σ, the following holds:

finIx≤qn   for every n∈ℕ.

Proof

Let gi ∈ Homeo([−ε, 1 + ε]), such that for x ∈ [0, 1] we have gi(x) = fi(x). This family does not satisfy the first assumption of the previous theorem, because [0, 1] is invariant under G. We will prove that there exists a stationary measure for the random walk generated by G.

By Krylov–Bogoliubov’s theorem, we know that the Markov operator generated by the family (f1, . . . , fn, p1, . . . , pn) has an invariant measure μ. Let μ∗ ∈ M([−ε, 1 + ε]) be defined by μ∗(A) = μ(A ∩ [0, 1]). It is easy to verify that μ∗ is an invariant measure for the operator PG generated by G. We will show that μ∗ is a stationary measure for the random walk generated by G. We need to show that for every measurable set A, the following holds:

∫−ε,1+επx,Adμ*x=μ*A.
However,
∫−ε,1+επx,Adμ*x=∫−ε,1+εPGδxAdμ*x=∑i=1npi∫−ε,1+εδxgi−1Adμ*x.
But the last integral is μ*gi−1A , so μ∗ is a stationary measure. From the definition of μ∗, it follows that supp(μ∗) = supp(μ), and later in the work, we will prove that {0, 1} ∈ supp(μ), so by the previous theorem, the statement holds.

Before we proceed to prove the uniqueness of the invariant measure, we will prove that invariant measures must be atomless, and their supports contain the endpoints of the interval.

Lemma 3.1.

Every invariant measure of the operator P is atomless.

Proof

Suppose there exists an invariant measure μ∗ for the operator P that has an atom. Let a ∈ (0, 1) be a point such that

μ*a=  supx∈0,1 μ*x.
Then a is an atom of the measure μ∗, which means that μ∗({a}) > 0. Since μ∗ is invariant under the operator P, we know that
μ*a=∑i=1npiμ*fi−1a.
For each function fi, there is a point fi−1a such that the measure μ∗ assigns the same value μ({a}), because fi−1a for i ∈ Σ∗ forms an infinite set of points mapped to a. This means that for each i, we have μ*fi−1a=μ*a , which suggests that μ({a}) should be split between infinitely many different points fi−1a . However, this leads to a contradiction with the property that μ∗ is a finite measure, as it would imply that μ∗((0, 1)) = ∞, which is impossible. Thus, we arrive at a contradiction, and therefore the measure μ∗ cannot have atoms. Hence, every invariant measure of the operator P is atomless.

Lemma 3.2.

Every invariant measure of the operator P contains the points 0 and 1 in its support.

Proof

Suppose that 0 ∉ supp(μ). Then, by the atomlessness of μ, there exists the largest a > 0 such that μ([0, a]) = 0. However, from the invariance of μ, we obtain:

0=μ0,a=∑i=1npiμfi−10,a.
This implies that for every i, we have μfi−10,a=0 . However, there exists some i for which fi(a) < a, which implies that 0,a⊂fi−10,a , contradicting the maximality of a. A similar proof can be conducted for 1.

Definition 3.1 ([2]).

Let S be a compact metric space. We say that a Feller operator P has the e-property at the point x ∈ S if for every Lipschitz function ϕ: S → ℝ, the following holds:

limy→x supn∈ℕ Unϕx−Unϕy=0.
If P has the e-property at every point, we say that P has the e-property.

Theorem 3.3 ([3, Lemma 2.8]).

Let P be a Feller operator that possesses the e-property. Then for any two distinct ergodic measures μ, ν ∈ M1(S), the following holds:

supp μ∩supp ν=∅.
In particular, from the proof of the lemma, it follows that if P has the e-property at x, then x ∉ supp μ ∩ supp ν.

We now proceed with the proof of the uniqueness of the invariant measure.

Theorem 3.4.

There exists exactly one invariant measure for the Feller operator P.

Proof

We know that if μ is an invariant measure, then 0 ∈ supp μ. Therefore, we only need to check that the operator P has the e-property at 0. Let ε > 0 be given. We need to show that there exists a δ > 0 such that if h < δ, then for every n ≥ N0, the following holds:

∑i∈Σnpifih−∑i∈Σnpifi0=∑i∈Σnpifi0,h<ε.
Let N1 be large enough such that (1 − pd)N1 < ε/3. Let Σn1=i∈Σn:fik≠fd for 1≤k≤N1 . Note that ℙnΣn1=(1−pd)N1 . Let Σn2=i∈Σn:i|N1∉ΣN11 . For i∈Σn2 , let xi = fi(0). Since 0 < xi < 1, there exists a neighborhood Ji such that for almost all j ∈ Σ, we have fjkJi<qk . Let Σ3=j∈Σ:for every i∈Σn2,fjkJi<qk . Clearly, ℙ(Σ3) = 1. Choose δ small enough such that for every i∈Σn2 , we have fi([0, δ]) ⊂ Ji. Let N2 > N1 be sufficiently large such that qN2−N1 < ε/3. Let Σn4=j*=ij|n:i∈Σn2 and j∈Σ3 and Σn5=j*=(ij)|n:i∈Σn2 and j∉Σ3 . Let N3 > N2 be sufficiently large such that ℙN3Σn5<ε/3 . For n > N3, the following holds:
∑i∈Σnpifi0,h=∑i∈Σn1pifi0,h+∑i∈Σn4pifi0,h+∑i∈Σn5pifi0,h≤∑i∈Σn1pi+∑i∈Σn4piqn−N1+∑i∈Σn5pi≤(1−pd)N1+qN2−N1+∑i∈Σn5pi≤ε.
Consequently, P has the e-property at 0, so if μ and ν are two distinct ergodic measures, then 0 ∉ supp μ∩supp ν and on the other hand, 0 ∈ supp μ∩supp ν – a contradiction, which completes the proof.

We will now proceed with the proof of the asymptotic stability of the operator P.

Theorem 3.5 ([1]).

Let μ∗ be the unique invariant measure of the operator P. Then every probabilistic measure μ converges weakly to μ∗. For continuous functions ϕ, we have:

limn→∞Pnμ,ϕ=μ*,ϕ.

Proof

For ψ ∈ C([0, 1]), define the sequence of random variables on (Σ, ℝ) by

ξnψi=μ*,ψ∘fin,in−1,…,i1.
We will show that ξnψ is a bounded martingale. The boundedness is obvious, so it is sufficient to show that
𝔼(ξn+1ψ|ξ1ψ,…,ξnψ)=ξnψ.
Notice that U∈σξ1ψ,…,ξnψ has the form Un × Σ1 × Σ for some Un ∈ Σn. We have:
∫U𝔼(ξn+1ψ|ξ1ψ,…,ξnψ)dℙ=∫Uξn+1ψdℙn+1=∑i∈Un+1piμ*,ψ∘fi=∑i∈Σ1piμ*,∑j∈Unpjψ∘fjfix=μ*,U∑j∈Unpjψ∘fj=μ*,∑j∈Unpjψ∘fj=∫Unξnψdℙn=∫Uξnψdℙ.
This completes this part of the proof. By the Martingale Convergence Theorem, we know that ξnψ converges for almost every i ∈ Σ. The space of continuous functions on the interval is central, so there exists a set Σ0 ⊂ Σ of full measure such that for every i ∈ Σ0, the sequence ξnψ converges. This follows from the fact that we can specify such a set for the center. From the Riesz Representation Theorem, we know that for i ∈ Σ0, there exists a probabilistic measure μi such that
(3.1)
limn→∞ξnψi=μi,ψ.
We will show that for almost every i, the measure μi = δv(i) for some v(i). It is sufficient to show that for ε > 0, there exists a set Σε ⊂ Σ of full measure such that for each i ∈ Σε, there exists an interval I of length less than ε such that
μiI≥1−ε.
Then for i∈Σ1=∩n=1∞Σ1/n , we have μi = δv(i). The set Σ1 is a set of full measure. For a fixed ε > 0, choose a natural number l such that 1/l < ε. Let the interval [a, b] satisfy
μ*a,b>1−ε
and contain fd(fg(0)) and fd(fg(1)). We can specify j ∈ Σ such that fjn(b) → 0. Notice that there exists an n1 such that fjn1 (b) < a. Let i1 = jn1. In general, for k ≤ l, there exists nk such that
fjnkb<fjnk−1a.
Let ik = jnk. Notice that the intervals Jn = fin[a, b] are disjoint. It follows that for some n∗, we have
Jn*<1/l<ε/2.
Notice that
Σ1=i∈Σ:for infinitely many n,fina,b<ε/2.
is a set of full measure. This is true because if i ∉ Σ1, there exists N such that for n > N |fin[a, b]| ≥ ε/2. However, with probability 1, the sequence σN i contains the fragment (id, ig)in∗. So with probability 1, there exists n′ > N such that |fin′ [a, b]| < ε/2. By the compactness of the interval, it follows that for infinitely many n, the image fin′ [a, b] ∈ I, where I is an interval of length ε. Notice that Σ1 ⊂ Σε, which completes this part of the proof.

To show the asymptotic stability, it is sufficient to show that for a Lipschitz function ψ and any points x, y ∈ (0, 1), the following holds:

limn→∞Pnδx,ψ−Pnδy,ψ=0.
This is true because for any measure μ ∈ M1(0, 1), we have
Pnμ,ψ−μ*,ψ≤∫0,1∫0,1Pnδx,ψ−Pnδy,ψμdxμ*dy.
Let us fix points x and y belonging to the interval (0, 1), with x < y. Choose any ε > 0. Since the measure μ∗ is invariant, according to the proof of uniqueness from Theorem 4.4, its support contains the points 0 and 1, which means that μ∗((0, x)) > 0 and μ∗((y, 1)) > 0. We know that for almost every sequence i = (i1, i2, . . . ) ∈ Σ with respect to the measure ℙ, the sequence of measures
μ*∘fi1,i2,…,in−1vi−ε2,vi+ε2∩0,1→1
as n → ∞. Since μ∗((0, x)) > 0 and μ∗((y, 1)) > 0, we can find points un ∈ (0, x) and vn ∈ (y, 1) such that for sufficiently large n, we have un,vn∈fi1,i2,…,in−1vi−ε2,vi+ε2∩0,1 . Then it follows that f(i1,...,in) (un) and f(i1,...,in) (vn) lie in the interval vi−ε2,vi+ε2 , and in particular, for sufficiently large n, f(i1,...,in) (x) and f(i1,...,in) (y) lie in this interval. Since ε > 0 was arbitrary, we conclude that for almost every i = (i1, i2, . . . ) ∈ Σ with respect to ℙ, the following convergence holds:
limn→∞fi1,…,inx−fi1,…,iny=0.
By equation (3.1) and the fact that ⟨Pnδz, ϕ⟩ = Unϕ(z) for z ∈ [0, 1], we have:
(3.2)
Pnδx,ϕ−Pnδy,ϕ=Unϕx−Unϕy≤L∑i1,…,in∈Σnfi1,…,inx−fi1,…,inypi1⋯pin,
where L is the Lipschitz constant for the function ϕ. We will analyze why the right-hand side of the above inequality tends to zero as n → ∞. For i = (i1, . . . , in, . . . ), define gn(i) := |fin (x) − fin (y)|, where in = (i1, . . . , in). Then, gn(i) → 0 as n → ∞ for almost every i ∈ Σ with respect to ℙ. From the construction of the probability measures ℙ and ℙn for n ∈ ℕ, it follows that the following relationship holds
ℙBn×Σ1×Σ1×…=ℙnBn   for Bn∈Σn.
Since gn(i) depends solely on the first n coordinates, it follows that
∫Σgnidℙi=∫Σngnidℙni=∑i1,…,in∈Σnfi1,…,inx−fi1,…,inypi1⋯⋅⋅pin.
As a result, according to Lebesgue’s theorem, we have
limn→∞∑i1,…,in∈Σnfi1,…,inx−fi1,…,inypi1⋯⋅⋅pin=0,
which proves inequality (3.2) and thus completes the proof.

4. Central Limit Theorem

Lemma 4.1 ([1]).

Let the family (f1, . . . , fN; p1, . . . , pN) be a contracting iterated function system, and let a∈0,12 . Then there exists r ∈ ℕ and Ω ⊆ Σ with probability ℙ(Ω) > 0 such that J = [a, 1 − a] and we get

∑n=1∞finJ≤r+q1−q
for every i ∈ Ω, where q is a constant given by Theorem 3.2.

Lemma 4.2.

Let J = [a, 1 − a] be such that fd(fg(0)), fd(fg(1)) ∈ J. Let En=i∈Σ:fin1/40,1∈J . Then, for n ≥ 16, there exists β = pdpg > 0 such that ℙ(En) ≥ β.

Lemma 4.3.

Let 𝒜 ⊂ Σ be a set such that ℙ(𝒜) ≥ β for some β > 0 and let k, n ∈ ℕ with k < n. Then, there exists a set A ⊂ Σn such that ℙn(Σn\A) ≤ (1 − β)k and for any i ∈ A there exist i1, i2, . . . , ik ∈ Σ∗ such that i = i1i2 . . . ik and for j = 1, . . . , k at least one of the sequences ij, σij , . . . , σk−1ij is dominated by A.

Theorem 4.1 (Central Limit Theorem).

Let Xn be a stationary Markov chain generated by a random walk with the initial distribution given by μ∗. If ϕ: [0, 1] → ℝ is a Lipschitz function such that ∫[0,1] ϕdμ∗ = 0, then the process ϕ(Xn) satisfies the central limit theorem. That is,

σ2=limn→∞𝔼ϕX0+⋯+ϕXnn2
exists, and
ϕX0+⋯+ϕXnn⇒𝒩0,σ   as   n→∞.

Proof

From the uniqueness of the ergodic measure μ∗, we know that the chain is ergodic. Therefore, by Theorem 1 from [5], it is sufficient to show that

∑n=1∞n−3/2∑j=1nUjϕL2<∞.
Let En be as in Lemma 4.2. Clearly, ℙ(En) ≥ β > 0. Let Ω be as in Lemma 4.1. Let
A=ij|n:i∈Enn1/4,j∈Ω.
Clearly, ℙ(𝒜) ≥ α > 0 for some α (independent of n). Then, by Lemma 4.3, for k = ⌊n1/8⌋, we get a sequence An ∈ Σn satisfying ℙ(Bn = Σn/An) ≤ (1 − α)⌊n1/8⌋. Moreover, if i ∈ An, then i = i1 . . . ik, where at least one of im, σim, . . . , σk−1im is dominated by 𝒜. Therefore, for any x, y, we have:
∑j=1imfijx−fijy≤n1/8+n1/4+r+q1−q≤2n1/4+c.
Notice that
∑j=1nfijx−fijy≤2kn1/4+c≤Cn3/4.
Let L be the Lipschitz constant of the function ϕ. Since
∫0,1ϕdμ*=∫0,1Ujϕdμ*=0,
we have:
∫0,1∑j=1nUjϕx2dμ*x≤∫0,1∫0,1∑j=1nUjϕx−Ujϕy2dμ*xμ*y.
We know that
∑j=1nUjϕx−Ujϕy≤∫An∑j=1nfijϕx−fijϕydℙi   + ∫Bn∑j=1nfijϕx−fijϕydℙi≤LCn3/8+2nLℙBn≤C*n3/8.
This implies that
∫0,1∫0,1∑j=1nUjϕx−Ujϕy2dμ*xμ*y≤C*2n3/4.
Therefore, ∑j=1nUjϕL2<C*n3/8 . By Theorem 1 from [5], we obtain the result.

DOI: https://doi.org/10.2478/amsil-2025-0005 | Journal eISSN: 2391-4238 (formerly 0860-2107) | Journal ISSN: 0860-2107
Language: English
Page range: 1 - 12
Submitted on: Dec 26, 2024
Accepted on: Mar 12, 2025
Published on: Apr 26, 2025
Published by: University of Silesia in Katowice, Institute of Mathematics
In partnership with: Paradigm Publishing Services

© 2025 Maciej Block, Wiktoria Kacprzak, Dorian Falęcki, published by University of Silesia in Katowice, Institute of Mathematics
This work is licensed under the Creative Commons Attribution 4.0 License.