Bhattacharya R., Majumdar M. Random Dynamical Systems: Theory and Applications

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3.5 Splitting 253

Let x  z

, and denote by 

the set of all γ ∈ 

appearing within the

curly brackets in (5.1). Then, for all γ ∈ 

, ˜γ

−1

(−∞, x] = S, so that

for all µ and ν in P(S), one has µ( ˜γ

−1

(−∞, x]) − ν( ˜γ

−1

(−∞, x]) =

1 − 1 = 0. Therefore,

)

∗N

µ)(−∞, x] − (T

∗N

ν)(−∞, x]

)



[µ( ˜γ

−1

(−∞, x]) − ν( ˜γ

−1

(−∞, x])]Q

(dγ )

)



\

)



\

[(µ ◦ ˜γ

−1

)(−∞, x] − (ν ◦ ˜γ

−1

)(−∞, x]]Q

(dγ )

)

 (1 −χ

(µ, ν)(µ, ν ∈ P(S)). (5.7)

Note that, for the last inequality, we have used the facts

(i) Q

(

\

)

 1 − ˜χ

and

(ii) |µ( ˜γ

−1

(−∞, x]) − ν( ˜γ

−1

)(−∞, x])|  d

(µ, ν) (use (5.5) with

˜γ in place of γ ).

Similarly, if x < z

then, letting 

denote the set in curly brack-

ets in (5.2), we arrive at the same relations as in (5.7), expecting

that the extreme right-hand side is now (1 − ˜χ

(µ, ν). For this,

note that (µ ◦ ˜γ

−1

)(−∞, x] = µ(φ) = 0 = (ν ◦ ˜γ

−1

)(−∞, x] ∀γ ∈ 

Combining these two inequalities, we get

∗N

µ, T

∗N

ν)  (1 − ˜χ )d

(µ, ν)(µ, ν ∈ P(S)). (5.8)

That is, T

∗N

is a uniformly strict contraction and T

∗

is a contraction.

As a consequence, ∀n > N , one has

∗n

µ, T

∗n

ν) = d

∗N

∗(n−N)

µ), T

∗N

∗(n−N)

ν))

 (1 − ˜χ)d

∗(n−N)

µ, T

∗(n−N)

ν) ≤···

 (1 − ˜χ)

[n/N]

∗(n−[n/N ]N )

µ, T

∗(n−[n/N ]N )

ν)

 (1 − ˜χ)

[n/N]

(µ, ν). (5.9)

Now suppose that we are able to show that P(S)isacomplete metric

space under d

. Then by using (5.8) and Theorem 2.2 of Chapter 1, T

∗

254 Random Dynamical Systems

has a unique ﬁxed point π in P(S). Now take ν = π in (5.9) to get the

desired relation (5.4).

It remains to prove the completeness of (P(S), d

). Let µ

a Cauchy sequence in P(S), and let F

be the d.f. of µ

. Then

sup

x∈R

(x) − F

(x)|→0asn, m →∞. By the completeness of

R, there exists a function H (x) such that sup

x∈R

(x) − H(x)|→0.

It is simple to check that H is the d.f. of a probability measure ν,say,on

R (Exercise). We need to check that ν(S) = 1. For this, given ε>0 ﬁnd

n(ε) such that sup

(x) − H(x)| <

∀n  n(ε). Now ﬁnd y

∈ S

such that F

n(ε)

) > 1 −

. It follows that H (y

) > 1 − ε. In the same

manner, ﬁnd x

∈ S such that F

n(ε)

−) <ε/2, so that H(x

−) <ε.

Then ν(S)

 H(y

) − H (x

−) > 1 − 2ε ∀ε>0, so that ν(S) = 1.

Remark 5.4 Suppose that α

are strictly monotone a.s. Then if the initial

distribution µ is nonatomic (i.e., µ({x}) = 0 ∀x or, equivalently, the d.f.

of µ is continuous), µ ◦ γ

−1

is nonatomic ∀γ ∈  (outside a set of zero

P-probability). It follows that if X

has a continuous d.f., then so has X

and in turn X

has a continuous d.f., and so on. Since, by Theorem 5.1,

this sequence of continuous d.f.s (of X

(n  1)) converges uniformly

to the d.f. of π, the latter is continuous. Thus π is nonatomic if α

are

strictly monotone a.s.

Example 5.1 Let S = [0, 1] and  be a family of monotone nondecreas-

ing functions from S into S. As before, for any z ∈ S, let

(z) = α

···α

One can verify the following two results, following an argument in

Remark 5.1:

[R.1] P[X

(0) ≤ x] is nonincreasing in n and converges for each

x ∈ S.

[R.2] P[X

(1) ≤ x] is nondecreasing in n and converges for each

x ∈ S.

Write

(x) ≡ lim

n→∞

P(X

(0) ≤ x)

and

(x) = lim

n→∞

P(X

(1) ≤ x).

3.5 Splitting 255

Note that F

(x) ≤ F

(x) for all x. Consider the case when  ≡{f },

where

f (x) =











if 0 ≤ x <

≤ x ≤

< x ≤ 1

Exercise 5.1 Verify that f is a monotone increasing map from S into S,

but f is not continuous. Calculate that

(x) =



0if0≤ x <

1if

≤ x ≤ 1

(x) =



0if0≤ x <

1if

≤ x ≤ 1.

Neither F

nor F

is a stationary distribution function.

Example 5.2 Let S = [0, 1] and  ={f

, f

}. In each period f

is cho-

sen with probability

. f

is the function f deﬁned in Example 5.1, and

(x) =

, for x ∈ S.

Then

(x) = F

(x) =



0if0≤ x <

1if

≤ x ≤ 1

and F

(x) is the unique stationary distribution. Note that f









, i.e., f

and f

have a common ﬁxed point. Examples 5.1

and 5.2 are taken from Yahav (1975).

Exercise 5.2 Show that splitting (H) does not occur in Example 5.2

3.5.2 Splitting: A Generalization

Let S be a Borel subset of a complete separable metric space (or a Polish

space) and S its Borel sigmaﬁeld. Let A ⊂ S, and deﬁne

(µ, ν):= sup

AA

µ(A) −ν(A)

(µ, ν ∈ P(S)). (5.10)

Let  be a set of measurable maps on S, with a sigmaﬁeld



such that

the measurability assumption stated at the beginning of Section 3.2 is

satisﬁed. Consider a sequence of i.i.d. maps {α

: n ≥ 1}with distribution

Q on (,



256 Random Dynamical Systems

Consider the following hypothesis (H

(1) (P(S), d

) is a complete metric space; (5.11)

(2) there exists a positive integer N such that for all γ ∈ 

, one

has

(µ ˜γ

−1

,ν˜γ

−1

) ≤ d

(µ, ν)(µ, ν ∈ P(S)), (5.12)

where ˜γ is deﬁned in (3.6) with N in place of n;

(3) there exists ˆχ>0 such that ∀A ∈ A, and with N as in (2), one

has

P((α

···α

)

−1

A = S or ϕ) ≥ ˆχ. (5.13)

We shall sometimes refer to A as the splitting class.

Theorem 5.2 Assume the hypothesis (H

). Then there exists a unique

invariant probability π for the Markov process X

:= α

···α

, where

is independent of {α

:= n ≥ 1}. Also, one has

∗n

µ, π ) ≤ (1 − ˆχ )

[n/N]

(µ ∈ P(S)), (5.14)

where T

∗n

µ is the distribution of X

when X

has distribution µ, and

[n/N ] is the integer part of n/N.

Proof. Let A ∈ A. Then (5.13) holds, which one may express as

({γ ∈ 

: ˜γ

−1

A = S or ϕ}) ≥ ˆχ. (5.15)

Then, ∀µ, ν ∈ P(S),

)

∗N

µ)(A) −(T

∗N

ν)(A)

)



(µ( ˜γ

−1

A) − ν( ˜γ

−1

A))Q

(dγ )

)

. (5.16)

Denoting the set in curly brackets in (5.15) by 

, one then has

|(T

∗N

µ)(A) −(T

∗N

ν)(A)|

)



(µ( ˜γ

−1

A) − ν( ˜γ

−1

A))Q

(dγ )



\

(µ( ˜γ

−1

A) − ν( ˜γ

−1

A))Q

(dγ )

)



\

(µ( ˜γ

−1

A) − ν( ˜γ

−1

A))Q

(dγ )

)

(5.17)

3.5 Splitting 257

since on 

the set ˜γ

−1

A is S or ϕ, so that µ( ˜γ

−1

A) = 1 = ν( ˜γ

−1

A), or

µ( ˜γ

−1

A) = 0 = ν( ˜γ

−1

A). Hence, using (5.12) and (5.13),

|(T

∗N

µ)(A) −(T

∗N

ν)(A)|≤(1 − ˆχ)d

(µ, ν). (5.18)

Thus

∗N

µ, T

∗N

ν) ≤ (1 − ˆχ )d

(µ, ν). (5.19)

Since (P(S), d

) is a complete metric space by assumption (H

)(1),

and T

∗N

is a uniformly strict contraction on P(S) by (5.19), there exists

by Theorem 2.2 of Chapter 1, a unique ﬁxed point π of T

∗

, i.e., T

∗

π = π,

and

∗kN

µ, π ) = d

∗N

∗(k−1)N

µ), T

∗N

π)

≤ (1 − ˆχ )d

∗(k−1)N

µ, π ) ≤···

≤ (1 − ˆχ )

(µ, π ). (5.20)

Finally, since d

(µ, ν) ≤ 1 one has, with n = [n/N ]N + r,

∗n

µ, π ) = d

∗[n/N]N

∗r

µ, π ) ≤ (1 − ˆχ )

[n/N]

. (5.21)

This completes the proof.

We now state two corollaries of Theorem 5.2 applied to i.i.d. monotone

maps.

Corollary 5.1 Let S be a closed subset of

R. Suppose α

(n ≥ 1) is a

sequence of i.i.d. monotone maps on S satisfying the splitting condition



) There exist z

∈ S, a positive integer N, and a constant χ



> 0

such that

P (α

N −1

···α

x ≤ z

∀x ∈ S) ≥ χ



P (α

N −1

···α

x ≥ z

∀x ∈ S) ≥ χ



Then

(a) the distribution T

∗n

µ of X

:= α

···α

converges to a prob-

ability measure π on S exponentially fast in the Kolmogorov distance d

258 Random Dynamical Systems

irrespective of X

. Indeed,

∗n

µ, π ) ≤ (1 − χ



)

[n/N]

∀µ ∈ P(S), (5.22)

where [y] denotes the integer part of y.

(b) π in (a) is the unique invariant probability of the Markov process

Proof. To apply Theorem 5.2, let A be the class of all sets A =

(−∞, y] ∩ S, y ∈

R. Completeness of (P(S), d

) is established directly

(see Remark 5.5).

To check condition (2) of (H

), note that if γ is monotone nondecreas-

ing and A = (−∞, y] ∩ S, then γ

−1

((−∞, y] ∩ S) = (−∞, x] ∩ S or

(−∞, x) ∩ S, where x = sup{z: γ (z) ≤ y}. Thus,

|µ(γ

−1

A) − ν(γ

−1

A)|=|µ((−∞, x] ∩ S) − ν((−∞, x] ∩ S)

µ((−∞, x) ∩ S) −ν((−∞, x) ∩ S)

In either case, |µ(γ

−1

A) − ν(γ

−1

A)|≤d

(µ, ν), since µ((−∞, x −

1/n] ∩ S) ↑ µ((−∞, x) ∩ S) (and the same holds for ν). If γ is mono-

tone nonincreasing, then γ

−1

A is of the form [x, ∞) ∩ S or (x, ∞) ∩ S,

where x := inf {z: γ (z) ≤ y}. Again it is easily shown, |µ(γ

−1

A) −

ν(γ

−1

A)|≤d

(µ, ν). Finally, (5.13) holds for all A = (−∞, y] ∩ S,by



This completes the proof.

Remark 5.5 Let {µ

: n ≥ 1} be a Cauchy sequence in P(S) with respect

to the metric d

. This means, for µ

(n ≥ 1) considered as probability

measures on

R, the sequence of d.f.s {F

: n ≥ 1}is Cauchy with respect

to the supremum distance (or the Kolmogorov distance) on

R. Their uni-

form limit H is a d.f. of a probability measure µ on

R. This implies µ

converges weakly to µ on R. By Alexandrov’s theorem (see Theorem

C11.1, Chapter 2), µ(S) ≥ lim sup

n→∞

(S) = 1. That is, µ(S) = 1.

In particular, µ

(S ∩ (−∞, x]) = F

(x) → H(x) = µ(S ∩ (−∞, x])

uniformly in x ∈

R. Hence d

(µ

,µ) → 0. A more general proof

of completeness of (P(S), d

) may be found in Complements and

Details.

3.5 Splitting 259

To state the next corollary, let S be a closed subset of R



. Deﬁne A

to be the class of all sets of the form

A ={y ∈ S: φ(y) ≤ x},

φ continuous and monotone on S into



, x ∈



. (5.23)

Again by “ ≤ ” we mean the partial order: x = (x

,...,x



) ≤ y =

,...,y

)iffx

≤ y

∀i = 1, 2,...,. In the following corollary we

will interpret (H) to hold with this partial order ≤, and with “x ≥ y”

meaning y ≤ x.

Corollary 5.2 Let S be a closed subset of



and A be the class of sets

(5.23). If α

(n ≥ 1) is a sequence of i.i.d. monotone maps which are

continuous Q a.s., and the splitting condition (H) holds, then there exists

a unique invariant probability π and

∗n

µ, π ) ≤ (1 − ˜χ )

[n/N]

∀µ ∈ P(S), n ≥ 1, (5.24)

where d

(µ, ν):= sup{

µ(A) −ν(A)

: A ∈ A}, ˜χ = min{˜χ

, ˜χ

}, and

[n/N ] is the integer part of n/N.

Proof. The completeness of (P(S), d

), i.e., (5.11) is proved in Com-

plements and Details (see Lemma C5.1).

Condition (2) in (H

), or (5.12), is immediate. For if A ={y: φ(y) ≤

x}, γ is continuous and monotone, then ˜γ

−1

A ={y:(φ ◦ ˜γ)y ≤ x}∈A

since φ ◦ ˜γ is monotone and continuous.

It remains to verify (H

) (3), i.e., (5.13). We write x

in place of

in (H). Let A in (5.23) be such that φ is monotone nondecreasing. If

φ(x

) ≤ x, then, by the splitting condition (H),

˜χ ≤ P(α

···α

z ≤ x

∀z ∈ S) ≤ P(φα

···α

z ≤ φ(x

)∀z ∈ S)

≤ P(φα

···α

z ≤ x∀z ∈ S) = P(a

···α

z ∈ A∀z ∈ S)

= P((α

···α

)

−1

A = S). (5.25)

If x in the deﬁnition of A in (5.23) is such that φ(x

)  x (i.e., at least

one coordinate of φ(x

) is larger than the corresponding coordinate of

x), then

˜χ ≤ P(α

···α

z ≥ x

∀z ∈ S) ≤ P(φα

···α

z ≥ φ(x

)∀z ∈ S)

≤ P(φα

···α

z  x∀z ∈ S) ≤ P(α

···α

z ∈ A

∀z ∈ S)

= P((α

···α

)

−1

A = ϕ). (5.26)

260 Random Dynamical Systems

Now let φ in the deﬁnition of A in (5.23) be monotone decreasing. If

φ(x

) ≤ x, then

˜χ ≤ P(α

···α

z ≥ x

∀z ∈ S)

≤ P(φα

···α

z ≤ φ(x

)∀z ∈ S) ≤ P(φα

···α

z ≤ x∀z ∈ S)

= P(α

···α

z ∈ A∀z ∈ S). (5.27)

If φ(x

)  x, then

˜χ ≤ P(α

···α

z ≤ x

∀z ∈ S) ≤ P(φα

···α

z ≥ φ(x

)∀z ∈ S)

≤ P(φα

···α

z  x∀z ∈ S) = P(α

···α

z ∈ A

∀z ∈ S).

(5.28)

Thus (H

) (3) is veriﬁed for all A ∈ A. This completes the proof.

Remark 5.6 Throughout Subsections 3.5.1, 3.5.2, we have tacitly as-

sumed that the sets within parenthesis in (5.15) (and (5.1) and (5.2)) are

measurable, i.e., they belong to



⊗N

. If this does not hold, we assume

that these sets have measurable subsets for which the inequality (5.15)

holds.

Remark 5.7 In Remark 5.1, it was shown that for monotone nondecreas-

ing maps α

on a compact interval S = [a, b], splitting is necessary for

the existence of a unique nontrivial (i.e., nondegenerate) invariant prob-

ability. A similar result holds in multidimension (see Bhattacharya and

Lee 1988).

3.5.3 The Doeblin Minorization Theorem Once Again

As an application of Theorem 5.2, we provide a proof of the important

Doeblin minorization theorem (Chapter 2, Theorem 9.1) when the state

space S is a Borel subset of the complete separable metric space.

Corollary 5.3 Let p(x, A) be a transition probability on a Borel subset

S of a complete, separable metric space with S as the Borel sigmaﬁeld of

S. Suppose there exists a nonzero measure λ on S and a positive integer

m such that

(m)

(x, A) ≥ λ(A) ∀x ∈ S, A ∈ S. (5.29)

3.5 Splitting 261

Then there exists a unique invariant probability π for p(·, ·) such that

sup

x, A

(n)

(x, A) − π (A)|≤(1 − ¯χ



)

[n/m]

, ¯χ



:= λ(S). (5.30)

Proof. We will show that (5.30) is an almost immediate consequence

of Theorem 5.2. For this, express p

(m)

(x, A)as

(m)

(x, A) = λ(A) + (p

(m)

(x, A) − λ(A))

= ¯χ



(A) +(1 − ¯χ



(x, A), (5.31)

where, assuming 0 <λ(S) ≡



< 1 (see Exercise 5.3),

(A):=

λ(A)

¯χ



, q

(x, A) =

(m)

(x, A) − λ(A)

1 − ¯χ



. (5.32)

Let β

(n ≥ 1) be an i.i.d. sequence of maps on S constructed as fol-

lows. For each n, with probability ¯χ



let β

≡ Z

, where Z

is a ran-

dom variable with values in S and distribution λ; and with probabil-

ity 1 − ¯χ



let β

= α

, where α

is a random map on S such that

P(α

x ∈ A) = q

(x, A) (see Proposition C1.1, Complements and De-

tails). Then Theorem 5.2 applies to the transition probability p

(m)

(x, A)

(for p(x, A)) with A = S, N = 1. Note that P(β

−1

A = S or φ) ≥

P(β

(·) ≡ Z

) = ¯χ



. Hence (5.13) holds. Since A = S in this exam-

ple, completeness of (P(S), d

) and the condition (5.12) obviously

hold.

Exercise 5.3 Suppose λ(S) = 1 in (5.29). Show that λ is the unique

invariant probability for p, and that (5.30) holds, with the conven-

tion 0

◦

= 1. [Hint: p

(m)

(x, A) = λ(A) ∀A ∈ S, ∀x ∈ S. For every k ≥

0, p

m+k

(x, A) = λ(A)∀A ∈ S, ∀x ∈ S, since the conditional distribu-

tion of X

m+k

,givenX

, is λ (no matter what X

is).]

As the corollaries indicate, the signiﬁcance of Theorem 5.2 stems

from the fact that it provides geometric rates of convergence in ap-

propriate metrics for different classes of irreducible as well as nonirre-

ducible Markov processes. The metric d

depends on the structure of the

process.

262 Random Dynamical Systems

3.6 Applications

3.6.1 First-Order Nonlinear Autoregressive Processes (NLAR(1))

There is a substantial and growing literature on nonlinear time series

models (see Tong 1990; Granger and Terasvirta 1993; and Diks 1999).

A stability property of the general model

n+1

= f (X

,ε

n+1

)

(where ε

is an i.i.d. sequence) can be derived from contraction assump-

tions (see Lasota and Mackey 1989). As an application of Corollary

5.3, we consider a simple ﬁrst-order nonlinear autoregressive model

(NLAR(1)).

Let S =

R, and f :

R → [a, b] a (bounded) measurable function.

Consider the Markov process

n+1

= f (X

) + ε

n+1

, (6.1)

where

[A.1] {ε

} is an i.i.d. sequence of random variables whose common

distribution has a density

φ (with respect to the Lebesgue measure),

which is bounded away from zero on interval [c, d] with d − c >

b − a.

Let X

be a random variable (independent of {ε

}). The transition

probability of the Markov process X

has the density

p(x, y) =

φ(y − f (x)).

We deﬁne ψ(y)as

ψ(y) ≡



≡ inf{

φ(z): c ≤ z ≤ d} for y ∈ [c + b, d + a],

0 for y ∈ [c + d, d + a]

(6.2)

It follows that for all x ∈

p(x, y) ≡

φ(y − f (x)) ≥ ψ(y). (6.3)

Proposition 6.1 If [A.1] holds then there exists a unique invariant prob-

ability π and (5.30) holds with m = 1 and ¯χ



ψ( y)dy > 0.

Proof. In view of (6.3) and the fact that ψ(y) ≥ χ



> 0on[c + b , d +

a], one has ¯χ



ψ(y)dy ≥ (d − c − (b − a))χ



> 0. Since (5.29)