Cover Page The handle https://hdl.handle.net/1887/3135034

(1)

The handle https://hdl.handle.net/1887/3135034 holds various files of this Leiden

University dissertation.

Author: Ziemla_{ńska, M.A.}

Title: Approach to Markov Operators on spaces of measures by means of equicontinuity Issue Date: 2021-02-10

(2)

Chapter 5 Central Limit Theorem for some

non-stationary Markov chains

This chapter is based on:

Jacek Gulgowski, Sander C. Hille, Tomasz Szarek, Maria A. Ziemla´nska. Central Limit Theorem for some non-stationary Markov chains. Studia Mathematica, Number 246 (2019), Pages 109-131.

Abstract:

Using the classical Central Limit Theorem for stationary Markov chains proved by M. I. Gordin and B. A. Lif˘sic in [GL78] we show that it also holds for non–stationary Markov chains provided the transition probabilities satisfy the spectral gap property in the Kantorovich– Rubinstein norm.

(3)

5.1 Introduction

In this chapter we aim at showing that the Central Limit Theorem (CLT) obtained for stationary Markov chains by Gordin and Lif˘sic in [GL78] (see also its extension due to M. Maxwell and M. Woodroofe in [MW00]) may be extended to non–stationary ones provided that their transition probabilities satisfy the Spectral Gap Property in the Kantorovich– Rubinstein norm (see condition (A2) below). This condition implies that the iteration of the Markov operator associated to these transition probabilities constitute an equicon-tinuous family of conequicon-tinuous maps on probability measures, equipped with the Dudley metric, see Chapter 1, Theorem 2.8.4. In this way we obtain the result stronger than the almost sure result known in the literature as a quenched Central Limit Theorems (see [Pel15, BI95]).

Recently the CLT was proved for various non–stationary Markov processes (see [KW12, Kuk02]). For more details we refer readers to the book by T. Komorowski et al. [KLO12], where a more detailed description of recent results on central limit theorems is provided. Our result is in the same spirit as the main theorem in [KW12]. However some delicate approximation allow us to obtain the CLT for initial distributions with 2-nd moment finite instead of 2 δ.

In this chapter we introduce notation that deviates from previous chapters, mainly be-cause those used here are more common in the field of probability theory, while the topic considered is specially targeted to an audience from this field.

Suppose that X, ρ is a Polish space. By BX we denote the family of all Borel sets in X. Denote by BbX the set of all bounded Borel measurable functions equipped with

the supremum norm and let CbX be its subset consisting of all bounded continuous

functions.

By M1 and M we denote the spaces of all probability Borel measures and of all Borel

measures on X, respectively. Let π X BX 0, 1 be a transition probability on X and let U BbX BbX be defined by Ufx R_Xfyπx, dy for every f > BbX.

The operator U is dual to the Markov operator P defined onM and given by the formula P µ R_Xπx, µdx for µ > M, i.e.

S_XfxP µdx S

X

U fxµdx for any f > BbX and µ > M.

In particular, we have P δx πx, for x > X.

(4)

5.2. Assumptions

whose transition probability is π. If the distribution of X0 is µ0, then the distribution of

Xn equals Pnµ0, n C 1. By Fn we denote the natural filtration of the chain, i.e. the

increasing family of σ–algebrasFn σXi i B n.

For any x> X and n C 1 we define the measure Pn

x on Xn,BXn, where BXn_σA 1 A2 An Ai > BX, i 1, . . . , n by the formula Pn_xA1 An S_XnχA1Anx1, . . . , xnπxn1, dxnπx1, dx2δxdx1 S_XnχA1Anx1, . . . , xnP δxn1dxnP δx1dx2P δxdx1.

Here χA1Andenotes the characteristic function of A1An. Obviously, the distribution of the random vector X1, . . . , Xn if X0 x is given by Pnx. On the other hand, if the

distribution of X0 equals µ0, then the distribution of the random vector X1, . . . , Xn is

equal to Pn

µ0 RXPnx µ0dx.

By Px we will denote the probability measure on the Borel σ-algebra of the trajectory

space Xª associated with Xn with X0 x. Further, Ex denotes the expected value with

respect to Px. Analogously, if X0 is distributed with µ0, then Pµ0 RXPx µ0dx.

Similarly, we have then Eµ RXEx µdx.

5.2 Assumptions

Set Mp 1 µ > M1 S X ρx, x0pµdx @ ª for pC 1.

We equip the space M1

1 with the Kantorovich-Rubinstein distance

Yµ νY sup VS_Xf dµ S

X

f dνV for µ, ν> M1 1,

where the supremum is taken over all Lipschitz functions f X _{R with the Lipschitz} constant bounded by 1. If µ and ν are two Borel probability measures on some Polish spaces W and Z respectively, then byCµ, ν we denote the set of all joint Borel probability measures on W Z whose marginals are µ and ν (the so–called coupling). If µ, ν > M1

1,

(5)

Theorem 5.2.1. [Rubinstein Theorem [Bog07a], Theorem 8.10.45] The Kantorovich-Rubinstein distanceYµνY between Radon probability measures µ, ν > M1

1 can be represented

in the form

Yµ νY inf

γ>Cµ,νSXX

ρx, y γdx, dy.

Moreover, there exists a measure λ0> Cµ, ν at which the value Yµ νY is attained.

This immediately implies

Yδx δyY ρx, y for x, y> X.

Assumptions:

(A1) We assume that P leaves M1

1 invariant;

(A2) there exist C A 0 and 0 @ q @ 1 such that

YPn_µ_Pn_ν_{Y B Cq}n_{Yµ νY} _(5.1)

for all µ, ν> M1

1 and n> N;

(A3) we assume that there exists µ> M2

1 such that

sup

nC1 SX

ρx, x02Pnµdx @ ª for some (thus all) x0> X. (5.2)

The following theorem is standard but we provide its proof for completeness of our pre-sentation.

Proposition 5.2.1. If (A1) and (A2) hold, then P has a unique invariant measure µ> M1

1 and for any µ> M11 we have YPnµ µY 0 as n ª. Moreover, if (A3) holds,

then µ> M2 1.

Proof. For any µ> M1

1X and any m, n > N, m C n one obtains from (5.1) that

YPn_µ_Pm_µ_{Y B}mn1_Q k 0

YPnk_µ_Pnk1_µ_{Y B} Cqn

1 qYµ P µY. So Pn_µ is a Cauchy sequence in the Y Y-complete space M1

1 ([Vil08], Theorem 6.18).

Hence it converges to some invariant measure µ > M1

(6)

5.3. Gordin–Lif˘sic results for stationary case

µ> M1 1X,

YPn_µ_µ

Y YPnµ PnµY B CqnYµ µY 0 as n ª.

So µ must be unique. Now suppose that (A3) holds and let µ> M2

1 be such that (5.2) is

satisfied. Let x0> X. For any K A 0, one has

sup

nC1 SX

ρx, x0 , K2Pnµdx B sup nC1 SX

ρx, x02Pnµdx M0 @ ª.

Since x( ρx, x0 , K2 is a Lipschitz function on X andYPnµ µY 0, we have

S_X ρx, x0 , K2Pnµdx S X

ρx, x0 , K2µdx.

So _R_X ρx, x0 , K2µdx B M0 for every K. By the Monotone Convergence Theorem

we obtain that _R_X ρx, x02µdx B M0 too.

5.3 Gordin–Lif˘sic results for stationary case

We start with the following simple consequences of the Gordin and Lif˘sic result on the CLT for stationary Markov chains:

Proposition 5.3.1. Let P be a Markov operator that satisfies (A1) - (A3) and let Xn

be a stationary Markov chain corresponding to P . Then for any bounded Lipschitz function g X _{R such that} _R_Xgdµ 0, where µ is the unique invariant distribution for P , the limit σ2 lim n ªEµ gX0 gX_º1 gXn n 2

exists and is finite. Moreover, if σA 0, then

lim n ªP gX0 gX_º1 gXn n @ a 1 º 2πσ2 S a ªe y2 2σ2dy for all a> R.

Otherwise, if σ 0, then the sequence

gX0 gX1 gXn

º

n for nC 1

(7)

Proof. Let a bounded Lipschitz function g X _{R such that} _R_Xgdµ 0 be given. With no loss of generality we may assume that the Lipschitz constant of g equals 1. From Gordin and Lif˘sic [GL78] it follows that to finish the proof it is enough to show that

ª

Q

n 0

YUn_g_Y

L2_X,µ@ ª, where U is dual to P . In fact, since _R_XUi_gxµ

dx RXgxµdx 0 for all i C 0, we have ª Q n 1 YUn_g_Y L2_X,µ ª Q n 1SX Un_g_x2_µ dx 1~2 ª Q n 1SX Un_g_{x S} XU n_g_yµ dy2µdx 1~2 B Qª n 1SXSX SUn_g_{x U}n_g_ySµ dy 2 µdx 1~2 B Qª n 1SXSX Cqnρx, yµdy 2 µdx 1~2 B C Qª n 1 qnS XSX ρx, yµdy 2 µdx 1~2 B C Qª n 1 qnS Xρx, x0 SX ρx0, yµdy 2 µdx 1~2 B C1 q1_{2 S} X ρx, x02µdx 2 S X ρx0, yµdy 2 1~2 @ ª, by the fact that µ> M2

1. The proof is complete.

5.4 Auxiliary lemmas

We start with a theorem on the existence of a suitable coupling for the trajectories of a given Markov chain.

Theorem 5.4.1. Assume that a Markov operator P satisfies (A1) and (A2). Let l0 be a

positive integer such that Cql0 @ 1, where the constants C, q are given by (5.1). Then there exists a constant κA 0 such that for every integers l C l0 and mC 1 and every two points

x, y> X we have a measure Pml x,y > CPmlx , Pmly satisfying m Q i 1SXlmXlm ρxil, yilPmlx,ydx1, . . . , dxml, dy1, . . . , dyml B κρx, y. (5.3)

(8)

5.4. Auxiliary lemmas

Proof. Choose ˜q> Cql0_{, 1}. Fix l C l

0. By induction on m for every two points x, y> X we

construct Pml

x,y> CPmlx , Pmly and prove that

S_Xlm_Xlm m Q i 1 ρxil, yil Pmlx,ydx1, . . . , dxml, dy1, . . . , dyml B m Q j 1 ˜ qjρx, y. (5.4) Then the hypothesis will hold with κ q˜1 ˜q1. Fix x, y> X and let m 1. Since

YPl_δ

x PlδyY B Cql0ρx, y,

by the Kantorovich–Rubinstein theorem there exists µ1

x,y > CPlδx, Plδy such that

YPl_δ

x PlδyY B S XX

ρu, vµ1_x,ydu, dv B ˜qρx, y.

From the proof of the Kantorovich–Rubinstein theorem it follows that the function XX ? x, y µ1

x,y > CPlδx, Plδy is measurable if the space CPlδx, Plδy is endowed with

some metric of weak topology (see Theorem 11.8.2 in [Dud02]). We may assume that the measure µ1

x,y is absolutely continuous with respect to the product measure PlδxaPlδy. Let

gx,y X X 0,ª be the Radon–Nikodem density, i.e.

gx,yu, v dµ1 x,y dPl_δ xa Plδy u, v u, v> X. Define Pl_x,yA S_Xl S_XlχAx1, . . . , xl, y1, . . . , ylgx,yxl, ylπxl1, dxlπyl1, dyl πx1, dx2πy1, dy2δxdx1δydy1.

Let A A1 Al Xl for some Borel sets A1, . . . , Al` X. We have

Pl_x,yA

S_XlχA1Alx1, . . . , xl S_Xlgx,yxl, ylπyl1, dylπy1, dy2δydy1 πxl1, dxlπx1, dx2δxdx1.

Moreover, the measure

(9)

is absolutely continuous with respect to Pl_δ

x. Obviously,

S_Xlgx,yxl, ylπyl1, dylπy1, dy2δydy1 S_Xgx,yxl, ylP

l_δ

ydyl.

If we show that _R_Xgx,y , ylPlδydyl 1, Plδx-a.s., then

Pl_x,yA

S_XlχA1Alx1, . . . , xlπxl1, dxlπx1, dx2δxdx1 Pl

xA1 Al.

To do this set hxl R_Xgx,yxl, ylPlδydyl. From the definition of the coupling measure

µ1

x,y for any Borel set B` X we have

Pl_δ

xB µ1x,yB X S BSX

gx,yxl, ylPlδydyl Plδxdxl

S_BhxlPlδxdxl.

Since the above equality holds for an arbitrary Borel set B` X, we obtain that hxl 1,

Pl_δ

x-a.s., and consequently Plx,y A1AlXl PlxA1Al. In the same way we

show that Pl

x,yXl B1 Bl PylB1 Bl for Borel sets B1, . . . , Bl` X. Hence

Pl

x,y> CPlx, Ply. Finally, we have

S_Xl_Xlρxl, ylP

l

x,ydx1, . . . dxl, dy1, . . . , dyl

S_Xl_Xlρxl, ylgx,yxl, ylπxl1, dxlπyl1, dylπx1, dx2πy1, dy2δxdx1δydy1 S_X_Xρxl, ylgx,yxl, ylPlδxdxlPlδydyl S

XX

ρxl, ylµ1dxl, dyl B ˜qρx, y

and the first step of the proof is finished.

Assume now that for j 1, . . . , m and arbitrary x, y> X there exists Pjlx,y > CPjlx, Pjly such

that condition (5.4) holds with m replaced by j.

Define the measure Pm1lx,y on Xm1l Xm1l by the formula

Pm1lx,y A S Xm1lXm1l χAx1, . . . , xm1l, y1, . . . , ym1l Pml_x l,yldxl1, . . . , dxm1l, dyl1, . . . , dym1lP l x,ydx1, . . . , dxl, dy1, . . . , dyl

(10)

for any Borel set A` Xm1l Xm1l. From the Markov property it easily follows that Pm1lx,y > CPm1lx , Pm1ly . We also have

S_X_m1l_X_m1l m1 Q j 1 ρxjl, yjlPm1lx,y dx1, . . . , dxm1l, dy1, . . . , dym1l B S_Xl_XlS_Xml_Xml m1 Q j 2 ρxjl, yjlPmlxl,yldxl1, . . . , dxm1l, dyl1, . . . , dym1l Pl x,ydx1, . . . , dxl, dy1, . . . , dyl S Xl_Xlρxl, ylP l x,ydx1, . . . dxl, dy1, . . . , dyl BQm j 1 ˜ qj_S Xl_Xlρxl, ylP l x,ydx1, . . . dxl, dy1, . . . , dyl ˜qρx, y BQm j 1 ˜ qjqρ˜ x, y ˜qρx, y m1 Q j 1 ˜ qjρx, y

by the inductive hypothesis for 1 and m. This completes the proof.

As a consequence of Proposition 5.3.1 and Theorem 5.4.1 we have the following:

be a Markov chain corresponding to P . Let l0 be a positive integer such that Cql0 @ 1, where

the constants C, q are given by (5.1) and let l C l0 be given. Set Yn Xnl for n C 0 and

assume that X0 Y0 x for x > X. Then for any bounded Lipschitz function g X R

such that _R_Xgdµ 0, where µ is the unique invariant distribution for P , the limit ˜

σ2 _lim n ªEµ

gY0 gY1 . . . gYn

º

n

2

exists and is finite. Moreover, if ˜σ2A 0, then the sequence of random vectors Y

n satisfies

lim

n ªP

gY0 gY1_º . . . gYn

n @ a 1 º 2πσ2S a ªe y2 2˜σ2dy for all a> R.

Proof. Without loss of generality we may assume that g X _{R is bounded by 1 and} its Lipschitz constant is also bounded by 1. By ϕx

n, x > X, we denote the characteristic

function of the random variable

(11)

where Y0 x. When the distribution of Y0 is equal to µ the characteristic function is denoted by ϕµn. Obviously, ϕµ n t S Xϕ y ntµdy for t> R.

Moreover, from Proposition 5.3.1 it follows that lim n ªϕ µ n t e ˜ σ2t2 2 for t> R.

Theorem 5.4.1 allows us to evaluate the difference of characteristic functions ϕx

nt and

ϕynt for x, y > X and t > R. Fix x, y > X. We have

Sϕx nt ϕyntS S S Xlnexpit gx0 gxl . . . gxnl º n P nl xdx1, dx2, . . . , dxnl S_Xlnexpit

gy0 gy_ºl . . . gynl

n P nl y dy1, dy2,, dynlS B ºStS n SXln_Xln ρx, y ρxl, xl ρxnl, ynl P nl x,ydx1, . . . , dxnl, dy1, . . . , dynl B ºStS n1 κρx, y for t> R, where Pnl

x,y> CPnlx, Pnly is given by Theorem 5.4.1. Consequently, for t > R we obtain

Sϕx nt ϕµntS Sϕxnt S X ϕy_ntµdyS B lim n ª StS1 κ_º n SX ρx, yµdy 0 as n ª. Since limn ªϕµnt et 2_˜_σ2_~2

, we obtain that limn ªϕxnt et

2_˜_σ2_~2

and the proof is complete.

For any Markov chainZn with values in the space X and an arbitrary Lipschitz function

g X _{R we shall denote} S_mngZi n Q i m gZi for nC m C 0.

The key point in proving the main theorem is the following lemma:

(12)

Markov chain corresponding to P . Assume that g X _{R is a bounded Lipschitz function.} Then there exists a constant M A 0 such that for any n > N the function

x _Ex

Sn

0gX_º i

n

2

has Lipschitz constant B M.

Proof. Again assume that g X _{R is a 1–Lipschitz function bounded by 1. Let l}0 be a

positive integer such that Cql0 @ 1 with the constants C, q given by (5.1). We have Ex Sn 0gXi º n 2 1 n n Q i 0 ExgXi2 2 n n1 Q i 0 Q j 0@ji@l0,jBn ExgXigXj 2 n n Q i 0 Q j 0Bj@l0, l0jiBn Ex gXi ki,j Q m 1 gXml0ji,

where ki,j are maximal positive integers such that ki,jl0 j i B n. We show that all three

terms are Lipschitzean with a Lipschitz constant independent of n. To do this fix x, y> X. Since the function g 2 _{is a} B 2–Lipschitz function, by (5.1) we have

SExgXi2 EygXi2S B 2YPiδx PiδyY B 2Cρx, y for i 1, . . . , n

and the first term

1 n n Q i 1 ExgXi2 is a B 2C–Lipschitz function.

Further, for jA i we have

EzgXigXj EzgXiESXigXj for all z> X,

where ESXigXj E gXjSFi. Since ESXigXj hXi, where h is a bounded by 1 a B C–Lipschitz function, we obtain that

SExgXigXj EygXigXjS SExgXihXi EygXihXiS

(13)

for i 1, . . . , n and consequently the Lipschitz constant of the second term 2 n n1 Q i 1 Q j 0@ji@l0,jBn ExgXigXj

is bounded from above by 4l0C.

Finally, we have Ex gXi ki,j Q m 1 gXml0ji Ex gXiESXi ki,j Q m 1 gXml0j . Using Theorem 5.4.1 we show that the function

X ? z _Ez ki,j Q

m 1

gXml0j for all i, j C 0

has a Lipschitz constantB Cκqj_{, where κ is given by Theorem 5.4.1. Indeed, for any i, j} C 0

we have Ez ki,j Q m 1 gXml0j EzESXj ki,j Q m 1 gXml0j. In turn, from Theorem 5.4.1 we obtain that the function

X ? u ru Eu ki,j Q

m 1

gXml0 has a Lipschitz constant B κ. Indeed, fix u, v > X. We have

Sru rvS SEu ki,j Q m 1 gxml0 Ev ki,j Q m 1 gxml0S B S S_Xki,j l02 ki,j Q m 1 gxml0 gyml0P ki,jl0

u,v dx1, . . . , dxki,jl0; dy1, . . . , dyki,jl0 B S_Xki,j l02 ki,j Q m 1 ρxml0, yml0P ki,jl0

u,v dx1, . . . , dxki,jl0; dy1, . . . , dyki,jl0 B κρu, v, by Theorem 5.4.1. Since Ez ki,j Q m 1 gXml0j S X ruPjδzdu,

(14)

5.4. Auxiliary lemmas the function X? z _Ez ki,j Q m 1 gXml0j

has a Lipschitz constant B Cκqj_{, by (5.1). Moreover its supremum norm is bounded by}

n. Since the Lipschitz constant of the function z gzrz is bounded by n Cκqj _we

obtain that the function

X? x Ex gXi ki,j Q m 1 gXml0ji has a Lipschitz constant B Cqin Cκqj. Thus the third term

2 n n Q i 0 Q j 0Bj@l0, l0jiBn Ex gXi ki,j Q m 1 gXml0ji has Lipschitz constant bounded by

2 n n Q i 0 l01 Q j 0 Cqin Cκqj B 2C n n Q i 0 l0n Cκqi B 2l0C1 Cκ 1 q . Thus the function

x _Ex

Sn

0gX_º i

n

2

has Lipschitz constant B M, where

M 2C 4l0C

2l0C1 Cκ

1 q . This completes the proof.

Lemma 5.4.3. Let P be a Markov operator that satisfies (A1) - (A3) and let Xn be a

Markov chain corresponding to P . Assume that g X _{R is a bounded Lipschitz function.} Let x> X and ε A 0. Then there exists K A 0 and N0> N such that

EPn_δ x <@ @@ @> Sn 0gX_º i m 2 χK,ªW Sn 0gX_º i m W =A AA A?B ε (5.5)

for all m> N and n C N0.

(15)

such that Cql0 @ 1, where the constants C, q are given by (5.1) and let l C l

0 be given. Set

Yn Xnl for nC 0. We first show that the hypothesis holds if we replace Xn with Yn. Set

Sm

gY0 gY1_º . . . gYm

m for mC 0.

The Markov chainYn corresponds to the operator Pland the assumptions of Lemma 5.4.2

are satisfied with P and Xn replaced by Pl and Yn, respectively. Thus the functions

y _EySm2 are Lipschitzean with the Lipschitz constant bounded by, say ˜M , independent

of m. The Markov operator P satisfies (A2), so that we may find N0 > N such that

SEPn_δ xS

2

m EµSm2S B ˜MYPnδx µY B ε~9l3 for n C N0 and m> N. (5.6)

Moreover, from Theorem 5.4.1 it follows that for any Lipschitz function Θ R _{R with} the Lipschitz constant ˆM A 0 such that we have for all m C 1

SEyΘSm EzΘSmS B

κ ˆM º

mρy, z for y, z> X. Consequently, for some N1 > N we have

SEPn_δ

xΘSm EµΘSmS B ε~9l3 for n C N1 and m> N. (5.7)

Let ϕK R R, K C 1 be a Lipschitz function such that

χ0,K1y B ϕKy B χ0,Ky for all x> R.

We are going to show that there exists KC 1 such that

EµSm21 ϕKSSmS @ ε~9l3 for all mC 1. By Proposition 5.3.1 we have EµSm2 σ˜2 as m ª and EµSm2ϕKSSmS 1 º 2πSR x2ϕKSySey 2_~2˜σ2 dy as m ª.

(16)

Since the integral on the right hand side converges to σ2 _{as K} ª, we obtain that there

exist constants m0 and k0 such that

EµSm21 ϕK0SSmS EµSm2 EµSm2ϕK0SSmS @ ε~9l

3_{for all m C m} 0.

Enlarging K0 if necessary we obtain

EµSm21 ϕK0SSmS @ ε~9l

3_{for all m C 1.} _(5.8)

Observe that the function y Θy given by the formula Θy y2_ϕ

K0SyS is a Lipschitz function. Combining (5.6) - (5.8), we obtain that for nC maxN0, N1 we have

EPn_δ x S 2 mχK0,ªSSmS B EPnδxS 2 m1 ϕK0SSmS B SEPn_δ xS 2 m EµSm2S SEPn_δ xΘSm EµΘSmS EµSm21 ϕK0SSmS B ε~9l3_ε~9l3_ε~9l3_ε~3l3_.

Having this we may show that (5.5) holds with K lK0. Set mi maxj i jl B m for

i 0, . . . , l 1. We have EPn_δ x gX0 . . . gX_º m m 2 χK,ªW gX0 . . . gX_º m m W B l2 EPn_δ x max 0BiBl1 gXi gXilº . . . gXimil mi 2 χK,ªW gX0 . . . gX_º m m W B l2 EPn_δ x max 0BiBl1 gXi gXilº . . . gXimil mi 2 χK0,ªW gXi . . . gXº imil mi W B l2 EPn_δ x gX0 gXl . . . gXm0l º m1 2 χK0,ªW gX0 gXl . . . gXm0l º m0 W l2 EPn_δ x gX1 gX1l . . . gX1m1l º m1 2 χK0,ªW gX1 gX1l . . . gX1m1l º m1 W l2 EPn_δ x gXl1 gX2l1 . . . gXl1ml1l º ml1 2 χK0,ªW gXl1 gX2l1 . . . gXl1ml1l º ml1 W B l 2_lε_~l3 _ε.

Since εA 0 is arbitrary, the proof is complete.

(17)

be a Markov chain corresponding to P . Let l0 be a positive integer such that Cql0 @ 1,

where the constants C, q are given by (5.1) and let l C l0 be given. Assume that Yn Xnl

for nC 0. If g X _{R is a bounded Lipschitz function, then}

lim

m ªEx

gY0 gY1 . . . gYm

º m 2 ˜ σ2, where ˜ σ2 lim m ªEµ gX0 gX_º1 . . . gXn n 2 .

Proof. First observe that for any kC 1 we have RRRRR

RRRRR REx

n

2

Ex

gYk gYk1_º . . . gYnk

n 2 RRRRR RRRRR R 0 as n ª. Further we have Ex

gYk gYk1_º . . . gYnk

n

2

EPk_δ x

n

2

,

by the Markov property. On the other hand, the Markov chain Yn corresponds to the

operator Pl _{and the assumptions of Lemma 5.4.2 are satisfied with P and} X

n replaced

with Pl _and Y

n, respectively. Thus the functions

X ? x Ex

gY0 gY1 . . . gYn

º

n

2

for nC 0

are Lipschitzean with the Lipschitz constant bounded by, say ˜M , independent of n. Con-sequently, we obtain RRRRR RRRRR REP k_δ x

n

2

Eµ

n 2 RRRRR RRRRR R B ˜MYPk_δ x µY B ˜M CqkYδx µY.

(18)

5.5. The Central Limit Theorem

5.5 The Central Limit Theorem

Now we are in a position to formulate and prove the main theorem of this paper.

Theorem 5.5.1. (Central Limit Theorem) Let P be a Markov operator that satisfies (A1) - (A3) and letXn be a Markov chain corresponding to P . Assume that X0 x for x> X.

Then for any bounded Lipschitz function g X _{R such that} _R_Xgdµ 0, where µ is the unique invariant distribution for P , the sequence of random vectors Xn satisfies

lim n ªP gX0 gX1 . . . gXn º n @ a 1 º 2πσ2 S a ªe y2 2σ2dy for all a> R, if σ2 lim n ªEµ gX0 gX1 . . . gXn º n 2 A 0. Otherwise, if σ 0, the sequence

gX0 gX1 gXn

º

n for nC 1

converges in distribution to 0.

Proof. To prove the theorem we show that for any x> X

lim n ªExexpit gX0 gX_º1 . . . gXn n e σ2_t2_~2 for t> R.

Without loss of generality we may assume that g is bounded by 1 and its Lipschitz constant is also bounded by 1. We will make use of the following formula:

eia 1 ia a2~2 Raa2, where SRaS B 1 and lima 0Ra R0 0.

Fix x> X, t > R 0 and ε A 0. Let k0 > N and η A 0 be such that

S1 σ2

0t2~2kk eσ

2_t2_~2

S @ ε for Sσ0 σS @ η and k C k0.

Set D sup_n_C1_R_X_R_Xρu, zPn_δ

(19)

to Proposition 5.2.1 DB sup nC1 SX ρu, x0Pnδxdu S X ρz, x0µdz @ ª.

For given x, t, ε, by Lemma 5.4.3, we choose K A 0 and N0 > N such that

EPn_δ x <@ @@ @> gX0 . . . gXm º m 2 χK,ªW gX0 . . . gXm º m W =A AA A?B ε~2t2 (5.9) for nC N0 and m> N.

Fix lC N0 such that

qlCDStS1 q1 MStS @ ε and qlC@ 1, (5.10) where the constants q, C are given by condition (5.1) and the constant M is given by Lemma 5.4.2. Since sup m,nC1E Pn_δ x <@ @@ @> gX0 . . . gXm º m 2=A AA A?@ ª one may choose kC k0 such that

EPn_δ x gX0 . . . gXm º m 2 WR ºθt k gX0 . . . gXm º m W χ0,KW gX0 . . . gX_º m m W B ε~2t

2 _{for all m, n}_{C 1 and θ > 0, 1,}

(5.11)

by the fact that SRaS 0 as a 0 and Sσm σS @ η for m C k, where

σ2_m _Eµ gX0 . . . gX_º ml m 2 Eµ gXl . . . gX_º m m 2 .

This can be done due to the fact that σm tends to σ as m ª. For positive integers u, v

and w, vC l, where l is such that condition (5.10) holds, we set

Iu,v,w Exexpit

gXl . . . gXw . . . gXu1wl . . . gXuw

º

(20)

5.5. The Central Limit Theorem

and

Iw Exexpit

gX0 . . . gX_º w

w .

Note that limn ªSIn Im,m, n~mS 0 for every m C 1. So let us consider in further detail

the terms Ij,k,m for given k, j B k and m C k. In the following we will use the notation

ESXj1m E SFj1m. We have Ij,k,m Exexpit gXl . . . gXm . . . gXj1ml . . . gXjm º km Ex exp it gXl . . . gXm . . . gXj2ml . . . gXj1m º km ESXj1mexp it gX_j1ml . . . gXjm º km Ex exp it gXl . . . gXm . . . gXj2ml . . . gXj1m º km 1 it º kmESXj1mgXj1ml . . . gXjm t2 2kESXj1m gX_j1ml . . . gXjm º m 2 t2 kESXj1m <@ @@ @> gX_j1ml . . . gXjm º m 2 Rºt kmgXj1ml . . . gXjm =A AA A? . (5.12) Since EµgXi 0, by (A2) we obtain

SEugXl . . . gXmS SEPl_δugX₀ . . . gX_ml E_µgX₀ . . . gX_mlS BQm i l SEPi_δugX₀ E_µgX₀S B m Q i l YPi_δ x µY B ql_C_{1 q}1_Yδ u µY B qlC1 q1S Xρu, zµdz

and hence we have

ExW it º kmESXj1mgXj1ml . . . gXjmW B q l_C_{1 q}1_ºStS kmExSX ρX_j1m, zµdz B ql_C_{1 q}1_ºStS kmSXSX ρu, zµdzPj1mδxdu B qlCD1 q1 StS º km.

(21)

On the other hand, by Lemma 5.4.2 and (A2) we have RRRRR RRRRR REu gXl . . . gXm º m 2 Eµ gXl . . . gXm º m 2RRRRR RRRRR R RRRRR RRRRR REP l_δ_u gX0 . . . gXml º m 2 Eµ gX0 . . . gXml º m 2 RRRRR RRRRR R B ql_CMm l m Yδu µY B q l_CM S_Xρu, zµdz, where M is given in Lemma 5.4.2 and consequently we obtain

ExRRRRRRRRRR R t2 kESXj1m gX_j1ml . . . gXjm º m 2 t2 kEµ gXl . . . gX_º m m 2 RRRRR RRRRR R B ql_Mt2 kExSX ρX_j1m, zµdz B qlCMt 2 k SXSX ρu, zµdzPj1mδxdu B ql_{CM D}t2 k.

Finally conditions (5.9) and (5.11) will allow us to evaluate the term

ExRRRRRRRRRR R t2 kESXj1m <@ @@ @> gX_j1ml . . . gXjm º m 2 Rºt kmgXj1ml . . . gXjm =A AA A?RRRRRRRRRRR . Indeed, we have ExRRRRRRRRRR R t2 kESXj1m <@ @@ @> gX_j1ml . . . gXjm º m 2 Rºt kmgXj1ml . . . gXjm =A AA A?RRRRRRRRRRR B t2 kEx <@ @@ @>ESXj1mRRRRRRRRRR R gX_j1ml . . . gXjm º m 2 Rºt kmgXj1ml . . . gXjm RRRRR RRRRR R =A AA A? t2 kEx <@ @@ @>RRRRRRRRRRR gX_j1ml . . . gXjm º m 2 Rºt kmgXj1ml . . . gXjm RRRRR RRRRR R =A AA A?

(22)

5.6. Example t2 kEPj1mlδx <@ @@ @> gX0 . . . gXml º m 2 W R ºt kmgX0 . . . gXmlW =A AA A? B t2 kEPj1mlδx <@ @@ @> gX0 . . . gXml º m l 2 χK,ªW gX0 . . . gXml º m l W =A AA A? t2 kEPj1mlδx gX0 . . . gX_º ml m l 2RRRRR RRRRR RRR ¾ m l m t º k gX0 . . . gX_º ml m l RRRRRRRRRR_RR χ0,KW gX0 . . . gXml º m l W by (5.9) and (5.11) B t2 kε~2t 2t2 kε~2t 2_ε~k.

Now from (5.12) it follows that

SIj,k,m Ij1,k,m1 σ2mt2~2kS B qlCD1 q1 StS º km q l_{CM D}t2 k ε k B 2ε~k for j 1, . . . , k, by condition (5.10) and the fact that mC k. Iterating this formula k-times we obtain

SIk,k,m 1 σm2t

2_~2kk_{S B 2ε}

and consequently

SIk,k,m eσ

2_t2_~2

S B 3ε for all m sufficiently large.

Since ε A 0 was arbitrary, we obtain that limn ªSIk,k, n~k InS 0, which completes the

proof.

5.6 Example

Let X, ρ be a Polish space and let T, A be a measurable space. Let ν A 0,ª be some measure on T . Let p T X 0,ª be a measurable function such that

S_Tpt, xνdt 1 for x> X.

We shall assume that pt, X 0,ª for t > T is a Lipschitzean function. Denote its Lipschitz constant by kt.

Let πt X BX 0, 1, t > T , be a transition probability and let Pt and Ut denote the

corresponding Markov operator and its dual, respectively. We shall consider the Markov chain Xn on some space Ω, F, P corresponding to the action of randomly chosen

(23)

tran-sition probabilities with probability depending on potran-sition, i.e. the Markov chain Xn

given by the formula:

P Xn1> SXn x πx, S

T pt, xπtx, νdt for x > X and n C 1. (5.13)

This model generalizes some random dynamical systems studied in [HSl16].

If the distribution of X0 is equal to µ, then the distribution of Xn is given by Pnµ, where

P M M is of the form P µA S

TSXpt, xπtx, Aµdxνdt for A> BX. (5.14)

Theorem 5.6.1. Assume that πtfor t> T are transition probabilities and there exists q @ 1

such that for the Markov operators Pt corresponding to πt we have

YPtµ1 Ptµ2Y B qYµ1 µ2Y for µ1, µ2 > M11.

Moreover, for some x0 > X and the operator Ut dual to Pt, t> T , we have

Ut ρ , x02x B a ρx, x02 b

for some a@ 1 and b A 0. Finally set γt sup

x>XSX

ρz, x0πtx, dz

for some x0> X and assume that

S_Tγtktνdt @ 1 q and

S_Tpt, xγtνdt @ ª for x> X.

Let Xn be a Markov chain given by (5.13). Assume that X0 x for x > X. Then for

any bounded Lipschitz function g X _{R such that} _R_Xgdµ 0, where µ is the unique invariant distribution for P , the sequence of random vectors Xn satisfies

lim n ªP gX0 gX_º1 . . . gXn n @ v 1 º 2πσ2S v ªe y2 2σ2dy for v> R,

(24)

5.6. Example if σ2 _lim n ªEµ gX0 gX1 . . . gXn º n 2 A 0. Otherwise, if σ 0, the sequence

gX0 gX1 gXn

º

n for nC 1

converges in distribution to 0.

Proof. From Theorem 5.5.1 it follows that to finish the proof it is enough to show that the Markov operator P given by (5.14) satisfies conditions (A1) - (A3).

Observe that P is a Feller operator and its dual is of the form U fx S

TSX

pt, xfzπtx, dzνdt for f > CbX. (5.15)

The operator U may be extended to an arbitrary positive unbounded function f and then it is also given by formula (5.15). The operator U can be, in fact, extended to the space of all Lipschitz functions. To show this, observe that for x0> X we have

U ρ , x0x S T pt, x S X ρz, x0πtx, dzνdt B S T pt, xγtνdt B S_T Spt, x pt, x0Sγtνdt S T pt, x0γtνdt B S_T ktγtνdt S T pt, x0γtνdt @ ª.

Therefore for any Lipschitz function f X _{R with Lipschitz constant L and x > X we} have

SUfxS B USfSx B Sfx0S LUρ , x0x @ ª,

by the fact that the Lipschitz constant of the function SfS is bounded by L.

Set ˆq R_T γtktνdt q @ 1 and let f X _{R be an arbitrary Lipschitz function with} Lipschitz constant bounded by 1 and such that fx0 0. We show that then Uf is a

(25)

Lipschitz function with Lipschitz constant bounded by ˆq. Indeed, for any x, y> X we have SUfx UfyS VS_T S_Xpt, xfzπtx, dzνdt S TSX pt, yfzπty, dzνdtV B S_T S_XSpt, x pt, ySSfzSπtx, dzνdt S_T pt, y VS X fzπtx, dz S X fzπty, dzV νdt B S_T kt S X ρz, x0πtx, dz νdt ρx, y q S T pt, yνdt ρx, y B S_T ktγtνdt q ρx, y ˆqρx, y.

To verify (A1) we compute S_Xρx, x0P µdx S

X

U ρ , x0xµdx

S_XUρ , x0x Uρ , x0x0µdx Uρ , x0x0

B ˆqS_Xρx, x0µdx Uρ , x0x0 B S X

ρx, x0µdx Uρ , x0x0 @ ª

for µ> M1 1.

Observe that for any µ1, µ2 > M11 we have

Yµ1 µ2Y sup VS X

fxµ1dx S X

fxµ2dxV ,

where the supremum is taken over all Lipschitz functions f X _{R with Lipschitz constant} bounded by 1 and such that fx0 0. Indeed for any f X R with Lipschitz constant

bounded by 1, we have VS_Xfxµ1dx S

X

fxµ2dxV VS

Xfx fx0µ1dx SXfx fx0µ2dxV .

To show that (A2) holds fix µ1, µ2> M11. Then we have

YP µ1 P µ2Y sup VS X fxP µ1dx S X fxP µ2dxV ˆ q supVS X ˆ q1U fxµ1dx S X ˆ q1U fxµ2dxV B ˆqsup VS X fxµ1dx S X fxµ2dxV ˆ qYµ1 µ2Y

(26)

5.6. Example

Here the supremum is taken over all Lipschitz functions f X _{R with Lipschitz constant} bounded by 1 and such that fx0 0, by the fact that ˆq1U f is a Lipschitz function with

Lipschitz constant bounded by 1. Hence (A2) holds.

Finally, we show that (A3) holds with µ δx0. To do this it is enough to prove that U ρ , x0x2B a ρx, x02 b (5.16)

for some a@ 1 and b A 0. Indeed, iterating the above formula we have

Un ρ , x02x B an ρx, x02 b1 a1 for nC 1 and x > X,

consequently, sup nC1 SX ρx, x02Pnδx0dx B sup nC1a n S_X ρx, x02δx0dx b1 a1 b1 a1@ ª. and (A3) will hold. To verify condition (5.16) we compute

U ρ , x0x2 S T pt, x0 S X ρz, x02πtx0, dzνdt S T pt, x0Ut ρ , x02xνdt B S_Tpt, x0a ρx, x02 bνdt a ρx, x02 b.

(27)