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5.4 Central Limit Theorems 363

where A(u):={y ∈ S : h(y) ≤ u} belongs to the splitting class A (see

Chapter 3 (5.23)). Thus, by Corollary 5.2 (Chapter 3),

∞



n=m+1

h(x)|≤(d − c)

∞



n=m+1

(1 − ˜χ )

[n/N]

→ 0asm →∞,

(4.11)

implying that the series (3.9) converges uniformly to a solution g of

the Poisson equation (3.3). Note also, that X

(x):= α

···α

x is con-

tinuous and monotone nondecreasing in x for every n ≥ 0, which im-

plies that T

h(x) is continuous and monotone nondecreasing in x for

every n ≥ 0. Hence g is bounded, continuous, and monotone non-

decreasing. Suppose σ

> 0. Then, as in the proof of Theorem 4.1,

k,n

:= n

−1/2

[g(X

) − Tg(X

k−1

)] for 1 ≤ k ≤ n (n > 1) is a bounded

martingale difference sequence for which L

(ε) = 0 for all sufﬁciently

large n. Since g is bounded, there exists a constant c

such that g − c

nonnegative, bounded, continuous, and monotone nondecreasing, as are

T (g − c

), (g − c

)

, and T (g − c

)

. Hence, with any of these functions

as h, (4.10) and (4.11) hold. Therefore, as in the proof of Theorem 4.1,

the relations (4.5)–(4.7) hold with g replaced by g − c

. By expanding

(g − c

)

as g

− 2 c

g + c

, it now follows that (4.5)–(4.7) hold (for

g), verifying the hypothesis of the CLT. This proves the theorem for all

bounded continuous monotone nondecreasing h.Ifh = f

− f

, where

both f

and f

are bounded, continuous, and monotone nondecreas-

ing, then



∞

n=m+1

h(x)|≤



∞

n=m+1

(x)|+



∞

n=m+1

(x)|

→ 0asm →∞, uniformly for all x, by the same argument lead-

ing to (4.11). The limits g



∞

n=0

are bounded, continuous,

monotone nondecreasing (i = 1, 2), and the same is true for T (g

)

and (Tg

)(Tg

)(i, j = 1, 2). Hence with g = g

− g

as the solution

of the Poisson equation (3.3) corresponding to

h =

−

, one has

(see (4.4)–(4.8))

= n

−1



k=1



k−1

) + Tg

k−1

) − 2(Tg

)(X

k−1

)

− (Tg

)

k−1

) − (Tg

)

k−1

) + 2Tg

k−1

) · Tg

k−1

)



→ σ

in probability. (4.12)

The hypothesis of the martingale CLT is thus satisﬁed by the martingale

difference sequence X

k,n

= [g(X

) − Tg(X

k−1

)]/n

1/2

364 Invariant Distributions: Estimation and Computation

Remark 4.1 In Section 5.3 and in the present section, we have focused

on estimating λ

hdπ for bounded h. While this is ﬁne for estimat-

ing probabilities π (A) =

dπ for appropriate A, one is not able to

estimate interesting quantities such as moments

dπ (r = 1, 2,...)

on unbounded state spaces. For the latter purpose, one ﬁrst shows that if,

for an h ∈ L

(π), the Poisson equation (3.3) has a solution g ∈ L

(π),

then the CLT (4.1) holds under P

, i.e., under the initial distribution π

(see Complements and Details). Although one may then derive the CLT

under certain other initial distributions µ (e.g., if µ is absolutely con-

tinuous w.r.t. π ), one cannot, in general, derive (4.1) under every initial

distribution. In addition, one typically does not know enough about π to

decide which h ∈ L

(π) (except for bounded h) and, even if one has a

speciﬁc h ∈ L

(π), for which such h the Poisson equation has a solution

in L

(π).

Remark 4.2 The hypothesis of continuity of the i.i.d. nondecreasing

maps α

(n ≥ 1) in Theorem 4.2 may be relaxed to measurability,in

view of Theorem C5.1 in Chapter 3, Complements and Details. It may

be shown that the Poisson equation (3.3) in this case has a solution

g ∈ L

(π), if h ∈ L

(π) is such that it may be expressed as the difference

between two nondecreasing measurable functions in L

(π). It may be also

shown that for such h, the CLT (4.1) holds, no matter what the initial

distribution µ may be (see Complements and Details).

Remark 4.3 In Section 2.10, the CLT for positive recurrent Markov

chains was derived using the so-called renewal method, in which the

chain {X

: n ≥ 0} is divided up into disjoint independent blocks over the

random time intervals [η

(r−1)

,η

(r)

), r ≥ 1, where η

(r)

is the rth return

time to a speciﬁed state, say, i(r ≥ 1), η

(0)

= 0. One may then apply the

classical CLT for independent summands, together with a general result

(Proposition 10.1 (Chapter 2)) on the classical CLT, but with a random

number of summands. Such a method may be extended to irreducible

processes satisfying (a) the Doeblin minorization (Theorem 9.1 (Chap-

ter 2) or Corollary 5.3 (Chapter 3)) and, more generally, to irreducible

Harris recurrent processes satisfying (b) a local Doeblin minorization

(Theorem 9.2 (Chapter 2)). In case (a), following the proof of Corol-

lary 5.3 (Chapter 3), if one decides the nature of β

(n ≥ 1) by i.i.d.

coin tosses θ

(n ≥ 1), where P(θ

= 1) = χ, P(θ

= 0) = 1 −χ , and

whenever θ

= 1, one lets β

= Z

(n ≥ 1), being i.i.d. with common

5.5 The Nature of the Invariant Distribution 365

distribution λ

, as deﬁned in Chapter 3 (5.32)), then η

(r)

is replaced by

the rth return time with θ

= 1. For the general case (b), the succes-

sive renewal, or regeneration, times are described in Complements and

Details in Section 2.9.

5.5 The Nature of the Invariant Distribution

It is in general quite difﬁcult to compute invariant distributions of

Markov processes. A notable exception is the class of (positive recurrent)

birth–death chains, for which the invariant probabilities are calculated in

Propositions 8.3 and 8.4 of Chapter 2. The problem of explicit analytic

computation is particuarly hard on state spaces S, which are uncount-

able as, e.g., in the case S is an interval or a rectangle (ﬁnite or inﬁ-

nite). Two particularly interesting (classes of) examples of analytic com-

putations of invariant probabilities on S = (0, ∞) are given in Section

4.6 on random continued fractions, which also illustrate the delicacy of

the general problem. Given such difﬁculties, one may turn to somewhat

qualitative descriptions or broad properties of invariant probabilities on

continuum state spaces S. As an example, consider Markov processes

= α

···α

on an interval S, where α

(n ≥ 1) are i.i.d. monotone

and the splitting condition (H) holds. The following result is proved in

Remark 5.4, Section 3.5.

Proposition 5.1 In addition to the hypothesis of Theorem 5.1 (Chapter

3), assume α

are strictly monotone. Then the unique invariant proba-

bility is nonatomic.

A simple general result on absolute continuity of π is the following.

Proposition 5.2 Suppose the transition probabilities p(x, ·) of a Markov

process on a state space (S, S) are, for each x ∈ S, absolutely continuous

with respect to a sigmaﬁnite measure ν on (S, S). Then, if π is invariant

under p, π is absolutely continuous with respect to ν.

Proof. Let B ∈ S such that ν(B) = 0. Then p(x, B) = 0 for all x,so

that

π(B) =

p(x, B)π(dx) =

0π(dx) = 0. (5.1)

366 Invariant Distributions: Estimation and Computation

Beyond such simple results on absolute continuity with respect to

Lebesgue measure ν = m, say, proving absolute continuity or singularity

turns out to be a formidable mathematical task in general. To illustrate

the difﬁculty, we will consider S = [0, 1] and  consisting of two afﬁne

linear maps f

, f

given by

(x) = θ x, f

(x) = θ x + (1 − θ), x ∈ [0, 1], (5.2)

picked with probabilities p and 1 − p: P(α

= f

) = p, P(α

= f

) =

1 − p for some p, 0 < p < 1. Here the parameter θ lies in (0, 1). One

may also represent this process as

n+1

= θ X

+ ε

n+1

(n ≥ 0), (5.3)

where ε

(n ≥ 1) are i.i.d. Bernoulli: P(ε

= 0) = p, P(ε

= 1 − θ) =

1 − p (0 < p < 1), θ ∈ (0, 1). If 0 <θ <1/2, then f

maps [0, 1] onto

[0,θ] = I

,say,and f

maps [0, 1] onto [1 − θ,1] = I

, with I

and

disjoint, each of length θ. With two iterations, the possible maps

are f

◦ f

, f

◦ f

, f

◦ f

, f

◦ f

; the ﬁrst two maps [0, 1] onto two

disjoint subintervals of I

, namely, I

= [0,θ

] and I

= [θ (1 − θ ),θ]

with a gap in the middle of I

. Similarly, f

◦ f

and f

◦ f

map [0, 1]

onto two disjoint subintervals of I

, namely, I

= [1 − θ,θ

+ 1 −θ]

and I

= [1 − θ

, 1] with a gap in the middle of I

. All these four

intervals are of the same length θ

. In this way the 2

possible n-stage

iterations map [0, 1] onto 2

disjoint subintervals, dividing each of the

n−1

intervals at the (n − 1)th stage into two disjoint intervals with a

gap in the middle. Each of the 2

n-stage intervals is of length θ

. Thus

the Lebesgue measure of the union D

, say, of all these 2

intervals is

, converging to 0 as n →∞,if0<θ<1/2. Thus D ≡



∞

n=1

is a Cantor set of Lebesgue measure 0, containing the support of the

invariant probability π of the Markov process (5.3) since π (D

) = 1

∀n. For the case p = 1 − p = 1/2, and θ = 1/2, one may directly check

from (5.3) that the uniform distribution is invariant. Thus we have proved

the following result (use Proposition 5.1 to show π is nonatomic).

Proposition 5.3 Consider the Markov process (5.3) on S = [0, 1] with

(n ≥ 1) i.i.d.: P(ε

= 0) = p, P(ε

= 1 − θ) = 1 − p, where 0 <

θ<1, 0 < p < 1. (a) If 0 <θ <1/2, then the unique invariant proba-

bility π is nonatomic and the support of π is a Cantor set of Lebesgue

measure zero. In particular, π is singular w.r.t. Lebesgue measure.

5.5 The Nature of the Invariant Distribution 367

(b) If θ = 1/2 and p = 1/2, then the unique invariant probability π

is the uniform distribution, i.e., the Lebesgue measure on [0, 1].

What happens if 1/2 <θ <1? It is simple to check that for each n, the

union D

of the 2

n-stage intervals (namely, ranges of f

◦ f

◦···◦

,(i

= 0or1,∀j )) is [0, 1] if θ ≥ 1/2. It has been known for many

years that there are values of θ ∈ (1/2, 1) for which π is absolutely con-

tinuous with respect to Lebesgue measure, and other θ>1/2 for which

π is singular with respect to Lebesgue measure. A famous conjecture

that, for the symmetric case p = 1/2, π is absolutely continuous for all

θ ∈ [1/2, 1) outside a set of Lebesgue measure zero stood unresolved for

nearly 60 years, until Solomyak (1995) proved it. Peres and Solomyak

(1996) extended this result to arbitrary p ∈ (0, 1). Further, it is known

that, if p = 1/2 then the set of θ ∈ (1/2, 1) for which π is singular is

inﬁnite; but it is still not known if this set is countable. For comprehensive

surveys of the known results in this ﬁeld, along with some new results

and open questions, we refer to Peres, Schlag, and Solomyak (1999) and

Mitra, Montrucchio, and Privileggi (2004). The latter article is of partic-

ular interest to economists as it relates (5.2), (5.3) to a one-sector optimal

growth model.

5.5.1 Random Iterations of Two Quadratic Maps

In this subsection we consider the Markov process X

:= α

n−1

(n ≥

1) as deﬁned in Section 4.4, with P(α

= F

) = η and P(α

= F

) =

1 − η for appropriate pairs µ<λand η ∈ (0, 1). If 0 ≤ µ, λ ≤ 1, then

it follows from Proposition 7.4(a) of Chapter 1 that the backward iterates

(x) → 0 a.s. for all x ∈ [0, 1], so that the Dirac measure δ

is the

unique invariant probability. Now take 1 <µ<λ≤ 2, and let a = p

≡

1 − 1/µ and b = p

. Then F

, F

are both strictly increasing on [ p

, p

]

and leave this interval invariant. It follows from Example 4.1 of Chapter

4 that there exists a unique invariant probability π on the state space

(0, 1) and that π is nonatomic by Proposition 5.1. Since p

and p

are

attractive ﬁxed points of F

and F

, respectively, they belong to the

support S(π )ofπ as do the set of all points of the form (see Lemma 6.2,

Chapter 4)

···ε

≡ f

··· f

( f

: = F

, f

:= F

···ε

(k ≥ 1), (5.4)

368 Invariant Distributions: Estimation and Computation

for all k-tuples (ε

,ε

,...,ε

) of 0’s and 1’s and for all k ≥ 1. Write

Orb(x; µ, λ) ={F

···ε

x: k ≥ 0, ε

= 0or1∀i} (k = 0 corresponds to

x). By Lemma 6.2 (Chapter 4), if x ∈ S(π) then S(π ) =

Orb(x; µ, λ).

This support, however, need not be [ p

, p

]. Indeed, if F

) > F

i.e.,

−

(1 <µ<λ≤ 2), (5.5)

then S(π)isaCantor subset of [p

, p

]. Before proving this asser-

tion we identify pairs µ, λ satisfying (5.5). On the interval [1, 2] the

function g(x):= x

−2

− x

−3

is strictly increasing on [1, 3/2] and strictly

decreasing on [3/2, 2], and g(1) = 0, g(3/2) = 4/27, g(2) = 1/8.

Therefore, (5.5) holds iff

λ ∈ (3/2, 2] and µ ∈ [

λ, λ), (5.6)

where

λ ≤ 3/2 is uniquely deﬁned for a given λ ∈ (3/2, 2] by g(

λ) =

g(λ). Since the smallest value of

λ as λ varies over (3/2, 2] is

√

5 − 1,

which occurs when λ = 2(g(2) = 1/8), it follows that µ can not be

smaller than

√

5 − 1 if (5.5) or (5.6) holds.

To show that S(π ) is a Cantor set (i.e., a closed, nowhere dense

set having no isolated point), for µ, λ satisfying (5.5), or (5.6), write

I = [p

, p

], I

= F

(I ), I

= F

(I ), I

···ε

= F

···ε

(I ) for k ≥ 1

and k-tuples (ε

,ε

, ···,ε

) of 0’s and 1’s. Here F

···ε

= f

··· f

as deﬁned in (5.4). Under the present hypothesis F

) > F

) (or

(5.5)), the 2

intervals I

···ε

are disjoint, as may be easily shown by in-

duction, using the fact that F

, F

are strictly increasing on [ p

, p

]. Let

=∪I

···ε

where the union is over the 2

k-tuples (ε

,ε

,...,ε

Since X

(x) ∈ J

for all x ∈ I , and J

↓ as k ↑, S(π) ⊂ J

for all k,so

that S(π ) ⊂ J :=∩

∞

k=1

. Further, F

is a strict contraction on [p

, p

]

while F

(I ) ↓{p

} as n ↑∞. Hence the lengths of I

...ε

go to zero

as k →∞, for a sequence (ε

,ε

,...) which has only ﬁnitely many

0’s or ﬁnitely many 1’s. If there are inﬁnitely many 0’s and inﬁnitely

many 1’s in (ε

,ε

,...), then for large k with ε

= 0, one may express

··· f

as a large number of compositions of functions of the type

or F

(n ≥ 1). Since for all µ, λ satisfying (5.5) the derivatives of

these functions on [ p

, p

] are bounded by F



) < 1 (use in-

duction on n and the estimate F



) < 1), it follows that the lengths

of the nested intervals I

···ε

go to zero as k →∞(for every sequence

(ε

,...)). Thus, J does not contain any (nonempty) open interval.

5.6 Complements and Details 369

Also, J ⊂ Orb (p

; µ, λ) = Orb ( p

; µ, λ) by the same reasoning, so

that J = S(π ). Since π is nonatomic, S(π) does not include any isolated

point, completing the proof that S(π) is a Cantor set.

Write |A| for the Lebesgue measure of A. If, in addition to (5.5),

λ<



µ − 1

2 − µ



µ, (5.7)

then |J |=0. Indeed, for any subinterval I



of I , one has |F



)|≤



)|I



|, |F



)|≤F



)|I



|, from which it follows that |J

k+1

|≤

(2 − µ)(1 + λ/µ)|J

|. If (5.7) holds then c ≡ (2 − µ)(1 + λ/µ) < 1, so

that |J |=0 if (5.5), (5.7) hold, i.e., S(π) is a Cantor set of Lebesgue

measure zero.

Note that the proof |I

···ε

|→0 depends only on the facts that on

, p

], (i) F

is a contraction, (ii) F

(I )↓{p

}, and (iii) F



)< 1,



) < 1. The last condition (iii) may be expressed as

λ − 2λ

(µ − 1)/µ

< 1, λ<µ/(2 − µ)

(1 <µ<λ≤ 2). (5.8)

If (5.8) holds, but (5.5) does not, then the 2

intervals I

···ε

cover

I = [p

, p

]. Since the endpoints of I

···ε

are in S(π), it follows that

S(π) = [ p

, p

There are many open problems in the topic of this subsection. For

example, if 1 <µ<λ≤ 2, but (5.5) (or (5.6)) does not hold, and (5.8)

does not hold, then S(π) = [ p

, p

]. However, we do not know when π is

singular or absolutely continuous in this case. Similarly, we do not know

for which µ, λ satisfying 2 <µ<λ≤ 3, π is absolutely continuous

(see Example 4.2 in Chapter 4).

An additional point of complexity, as well as interest, here is that

there are µ, λ satisfying 1 <µ<λ<4 for which the process X

(n ≥ 0)

deﬁned above does not have a unique invariant probability on S = (0, 1),

as has been shown by Athreya and Dai (2002) by an example. One would

like to know precisely for which pairs (µ, λ) does the process X

(n ≥ 0)

have a unique invariant probability or, equivalently, for which pairs it

does not.

5.6 Complements and Details

Section 5.3. This section is mainly based on Bhattacharya and

Majumdar (2001). Also see Athreya and Majumdar (2003).

370 Invariant Distributions: Estimation and Computation

Section 5.4.

Proof of Proposition 4.1 (Martingale CLT). Assume that, as n →∞,

(i) s

→ 1 in probability and (ii) L

(ε) → 0 in probability, for every

ε>0. Then S

converges in distribution to N (0, 1).

Proof. Consider the conditional characteristic functions

k,n

(ξ):= E(exp{iξ X

k,n

}|F

k−1,n

), (ξ ∈ R

). (C4.1)

Provided |

(ϕ

k,n

(ξ))

−1

|  δ(ξ ), a constant, it would be enough to

show that

(a) E(exp{iξ S

k,n

(ξ)) = 1, and

(b)

k,n

(ξ) → exp{−ξ

/2} in probability.

Indeed, if |(

k,n

(ξ))

−1

)|  δ(ξ ), then (a), (b) imply

)

E exp{iξ S

}−exp{−ξ

/2}

)

= exp{−ξ

/2}

)

E exp{iξ S

}

exp{−ξ

/2}

− E



exp{i ξ S

}

k,n

(ξ)



)

 exp{−ξ

/2}E

)

exp{−ξ

/2}

−

k,n

(ξ)

)

→ 0. (C4.2)

Now part (a) follows by taking successive conditional expectations

given F

k−1,n

(k = k

, k

− 1,...,1), if (ϕ

k,n

(ξ))

−1

is integrable. Note

that the martingale difference is not needed for this. It turns out, how-

ever, that in general (ϕ

k,n

(ξ)) cannot be bounded away from zero. Our

ﬁrst task is then to replace X

k,n

by new martingale differences Y

k,n

for

which this integrability does hold, and whose sum has the same asymp-

totic distribution as S

. To construct Y

k,n

, ﬁrst use assumption (ii) to

check that M

≡ max{σ

k,n

:1 k

 k

}→0 in probability. Therefore,

there exists a nonrandom sequence δ

↓ 0 such that

P(M

 δ

) → 0asn →∞. (C4.3)

Similarly, there exists, for each ε>0, a nonrandom sequence 

(ε) ↓ 0

such that

P(L

(ε)  

(ε)) → 0asn →∞. (C4.4)

5.6 Complements and Details 371

Write L

k,n

in place of L

(ε) for the sum over the indices j =

1, 2,...,k(1

 k  k

). Consider the events

k,n

(ε):={σ

k,n

<δ

, L

k,n

<

(ε), s

k,n

< 2}, (1  k  k

(C4.5)

Then A

k,n

(ε)isF

k−1,n

-measurable. Therefore, Y

k,n

deﬁned by

k,n

:= X

k,n

(ε)

(C4.6)

has zero-conditional expectation, given F

k−1,n

. Although Y

k,n

depends

on ε, we will suppress this dependence for notational convenience. Note

also that

P(Y

k,n

= X

k,n

for 1  k  k

)  P





k=1

k,n

(ε)



= P(M

<δ

, L

(ε) <

(ε), s

< 2) → 1. (C4.7)

We will use the notation (4.3) and (C4.1) with a prime (



) to denote

the corresponding quantities for {Y

k,n

}. For example, using the fact

E(Y

k,n

k−1,n

) = 0 and a Taylor expansion,

)



k,n

(ξ) −



1 −

2

k,n



)



exp(iξ Y

k,n

}−



1 + iξ Y

k,n

(iξ )

k,n



| F

k−1,n



)



−ξ

k,n

(1 − u)(exp{iuξ Y

k,n

}−1) du |F

k−1,n



)

 ε

|ξ|

2

k,n

+ ξ

E(Y

k,n

{|Y

k,n

|>ε}

k−1,n

)

 ε

|ξ|

2

k,n

+ ξ

E(X

k,n

{|Y

k,n

|>ε}

k−1,n

). (C4.8)

Fix ξ ∈

. Since M



<δ

,0 1 −(ξ

/2)σ

2

k,n

 1(1 k  k

) for

all large n. Therefore, using (C4.8),

)



k,n

(ξ) −



1 −

2

k,n



)





k=1

)



k,n

(ξ) −



1 −

2

k,n



)

 |ξ |

ε + ξ



(ε), (C4.9)

372 Invariant Distributions: Estimation and Computation

and

)



1 −

2

k,n



− exp



−

2



)



1 −

2

k,n



−

exp



−

2

k,n



)





4

k,n



2



. (C4.10)

Therefore,

)



k,n

(ξ) − exp



−

2



)

 |ξ |

ε + ξ



(ε) +

. (C4.11)

Moreover, (C4.11) implies

)



k,n

(ξ)

)



exp



−

2



−|ξ |

ε + ξ



(ε) +

 exp{−ξ

}−|ξ |

ε −





(ε) +



. (C4.12)

By choosing ε sufﬁciently small, one has for all sufﬁciently large n

(depending on ε), |ϕ



k,n

(ξ)| bounded away from zero. Therefore, (a)

holds for {Y

k,n

} for all sufﬁciently small ε (and all sufﬁciently large n,

depending on ε). By using relations as in (C4.2) and the inequalities

(C4.11) and (C4.12) and the fact that s

2

→ 1 in probability, we get (for

ε sufﬁciently small)

lim

n→∞

)



exp{i ξ S



}−exp



−



)

 exp



−



lim

n→∞

)



exp



−



−1

−



k,n

(ξ)

−1

)

 exp



−



exp







exp



−



−|ξ |



−1

lim

n→∞

)



k,n

(ξ) − e

−ξ

)





exp



−



−|ξ |



−1

|ξ|

ε. (C4.13)