Ergodicity and central limit theorem for random interval homeomorphisms



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i

|fn(I)|≤ qn for i Ω and n ∈ N.
Set b = inf I. We claim that there exists r N and a sequence (i1,..., ir) Σr such that f(i1 ,...,ir )([a, 1 a]) I. Suppose, contrary to our claim, that for every r N and (i1,..., ir) Σr, if f(i1 ,...,ir )(1 a) I, then f(i1,...,ir )(a) b. This implies that
Prδa([0, b]) Prδ1a([0, b] I) = Prδ1a([0, b]) + Prδ1a(I)
for every r N. Since the iterated function system (f1,... , fN ; p1,... , pN ) is asymptotically stable and μ is atomless, we have Prδa([0, b]) μ([0, b]), Prδ1a([0, b]) → μ([0, b]), Prδ1a(I) → μ(I) as r → ∞, by Alexandrov’s theorem. This is a contradiction with the above inequality, for μ(I) > 0.
Now choose (i1,..., ir) Σr such that f(i1 ,...,ir )([a, 1 a]) I and define

Ω = {(i1,... , ir)i : i Ω}∈ Σ.
We see at once that Ω satisfies the desired condition. The proof is complete.

Lemma 4: Let (f1,..., fN ; p1,..., pN ) be an admissible iterated function sys- tem and let α, δ, ε, M be the positive constants given in Lemma 1. Assume that


J = [a, 1 a] for some a (0, 1 ) such that M = εα < a−α . Then there exists
2 6
n0 N such that for every integer n n0

i

n
P({i Σ : f √4 n([ε , 1 − ε ]) ⊂ J}) 1 .
n 5



9
Proof. Let α, δ, ε, M be the positive constants given in Lemma 1. From Lemma 1 it follows that there exists n0 N such that for n ≥ n0 we have
k
P({i Σ : k 4 nfi (εn) ε}) 10 .

6
From (3.1) and the fact that M = εα < a−α we obtain

4 n
k α

P({i Ω : fi (εn) a}) P({i Σ : k 4 nfi (εn) ε}) · (1 2Ma )
9 2 3
10 · 3 = 5 .
In the same way we may prove that for n0 sufficiently large we have


i

n
P({i Ω : f √4 n(1 − ε
3
) 1 a}) 5
for n n0.

Combining these two estimates we obtain that there exists n0 N such that
P({i Σ : f √4 n([ε , 1 ε ]) J}) 1 2 + 2 = 1 for n n .

i n n

The proof is complete.


5 5 5 0



Definition 3: Let A Σ Σ. We say that a sequence i Σ is dominated by A if there exists j ∈A such that i j. Additionally, we shall assume that the empty sequence is dominated by any A.
Lemma 5: Let A⊂ Σ be 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 Pnn \ 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 ... , σk1ij is dominated by A.
Proof. Let k, n N, k < n and let A⊂ Σ with P(A) > 0 be given. Let β > 0 be such that P(A) ≥ β. Define
B1 := {i Σn : i is dominated by A}

and observe that

Pnn \ B1) 1 β.



By C2 denote the set of all i Σ such that i is not dominated by B1 but i||i|−1 is dominated. If C2 is empty, then putting im to be the empty sequences for


m = 2,...,k will finish the proof. In the other case define
B2 := {ij|n : i C2, j is dominated by A},
where ij|n denotes the natural projection on Σn of the concatenation of i and j. Observe that Pn(B2|Σn \ B1) > β. This gives
Pnn \ (B1 B2)) =Pnn \ B1) · Pnn \ B2|Σn \ B1)
=Pnn \ B1)(1 Pn(B2|Σn \ B1))

2
≤(1 − β) .
Obviously, if i B2, then i = i1i2 and for j = 1, 2 either ij or σij is dominated by A.
Further, if B1,..., Bl1 are given for some l k, then
Pnn \ (B1 ··· Bl1)) (1 β)l1
and if i ∈ Bm for m = 1,... ,l 1, then i = i1 ··· im and at least one of the sequences ij, σij,..., σl2ij is dominated by A for j = 1,... , m. Define now the set Cl of all i Σ such that for some m = 1,... ,l the se- quences i, i||i|−1,... , i||i|−m+1 are not dominated by Bl1, but i||i|−m is domi- nated. If Cl is empty, then putting im to be the empty sequence for m = l,...,k
will finish the proof. In the other case define
Bl := {ij|n : i Cl, j is dominated byA}.
Observe that
Pnn\(B1 ··· Bl))
= Pnn \ (B1 ··· Bl1))Pnn \ Bl | Σn \ (B1 ··· Bl1))

l
≤ (1 − β) .
Moreover, from the definition of Bl it follows that if i Bl, then i = i1 ··· il and at least one of the sequences im, σim,..., σl1im for m = 1,... ,l is dominated by A.
Taking l = k and setting A = B1 ··· Bk we finish the proof. In- deed, if i ∈ A there exist i1, i2,..., ikΣ, some of them possibly empty sequences, such that i = i1i2 ··· ik and for j = 1, 2,...,k at least one of the sequences ij, σij,..., σk1ij is dominated by A. The proof is complete.



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