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

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3.6 Applications 263

holds with λ having density ψ , the Doeblin stability result Corollary

5.3 applies directly, with m = 1.

For monotone f we can relax the assumption on the distribution of

the noise term ε

n+1

in (6.1). Let supp(Q) denote the support of the

distribution Q of ε

Proposition 6.2 Suppose f is monotone on

R into [a, b]. Let the inﬁmum

and supremum of supp(Q) be c and d, respectively. If d − c > b − a,

then X

(n ≥ 0) has a unique invariant probability π , and the distribution

of X

converges to π exponentially fast in the Kolmogorov distance as

n →∞, uniformly with respect to all initial distributions.

Proof. Choose z

∈ (c + b, d + a). Then X

(x) ≡ f (x) +ε

satisﬁes

P(X

(x) ≤ z

∀x ∈ R) = P(ε

≤ z

− f (x) ∀x ∈ R)

≥ P(ε

≤ z

− b) = ˜χ

,say,˜χ

> 0,

since z

− b > c. Similarly,

P(X

(x) ≥ z

∀x ∈ R) ≥ P(ε

≥ z

− a) = ˜χ

, say, ˜χ

> 0,

since z

− a < d. Hence Theorem 5.1 applies with N = 1.

Generalizations of Proposition 6.1 and 6.2 to higher order NLAR(k),

k > 1, are given in Chapter 4.

3.6.2 Stability of Invariant Distributions in Models

of Economic Growth

Models of descriptive as well as optimal growth under uncertainty have

led to random dynamical systems that are stable in distribution. We look

at a “canonical” example and show how Theorem 5.1 can be applied. A

complete list of references to the earlier literature, which owes much to

the pioneering efforts of Brock, Mirman, and Zilcha (Brock and Mirman

(1972, 1973); Mirman and Zilcha (1975)), is in Majumdar, Mitra, and

Nyarko (1989).

Recall that, for any function h on S into S, we write h

for the nth

iterate of h.

264 Random Dynamical Systems

Example 6.1 Consider the case where S = R

, and  ={F

, F

,...,

,...,F

}, where the distinct laws of motion F

satisfy

[F.1] F

is strictly increasing, continuous, and there is some r

> 0

such that F

(x) > xon(0, r

) and F

(x) < x for x > r

. Note that

) = r

for all i = 1,...,N . Next, assume

[F.2] r

= r

for i = j.

In other words, the unique positive ﬁxed points r

of distinct laws

of motion are all distinct. We choose the indices i = 1, 2,...,N

so that r

< r

< ····< r

. Let Prob(α

= F

) = p

> 0(1≤ i ≤ N ).

Consider the Markov process {X

(x)} with the state space (0, ∞).

If y ≥ r

, then F

(y) ≥ F

) > r

for i = 2,...N , and F

) = r

so that X

(x) ≥ r

for all n ≥ 0ifx ≥ r

. Similarly, if y ≤ r

, then

(y) ≤ F

) < r

for i = 1,...,N − 1 and F

) = r

, so that

(x) ≤ r

for all n ≥ 0ifx ≤ r

. Hence, if the initial state x is in

, r

], then the process {X

(x): n ≥ 0} remains in [r

, r

] forever. We

shall presently see that for a long-run analysis, we can consider [r

, r

]

as the effective state space.

We shall ﬁrst indicate that on the state space [r

, r

] the splitting

condition (H) is satisﬁed. If x ≥ r

, F

(x) ≤ x, F

(x) ≡ F

(x)) ≤

(x), etc. The limit of this decreasing sequence F

(x) must be a ﬁxed

point of F

, and therefore must be r

. Similarly, if x ≤ r

, then F

(x)

increases to r

. In particular,

lim

n→∞

) = r

, lim

n→∞

) = r

Thus, there must exist a positive integer n

such that

) < F

This means that if z

∈ [F

), F

)], then

Prob(X

(x) ≤ z

∀x ∈ [r

, r

])

≥ Prob (α

= F

for 1 ≤ n ≤ n

) = p

> 0,

Prob(X

(x) ≥ z

∀x ∈ [r

, r

])

≥ Prob(α

= F

for 1 ≤ n ≤ n

) = p

> 0.

3.6 Applications 265

Hence, considering [r

, r

] as the state space, and using Theorem

5.1, there is a unique invariant probability π, with the stability property

holding for all initial x ∈ [r

, r

Now ﬁx the initial state x ∈ (0, r

), and deﬁne m(x) = min

i=1,...,N

(x).

One can verify that (i) m is continuous, (ii) m is strictly increasing, and

(iii) m(r

) = r

and m(x) > x for x ∈ (0, r

), and m(x) < x for x > r

Let x ∈ (0, r

). Clearly m

(x) increases with n, and m

(x) ≤ r

. The

limit of the sequence m

(x) must be a ﬁxed point of m and is, therefore,

. Since F

) > r

for i = 2,...,N , there exists some ε>0 such that

(y) > r

(2 ≤ i ≤ N ) for all y ∈ [r

− ε, r

]. Clearly there is some n

such that m

(x) ≥ r

− ε.Ifτ

= inf{n ≥ 1: X

(x) > r

}then it follows

that for all k ≥ 1

Prob(τ

> n

+ k) ≤ p

Since p

goes to zero as k →∞, it follows that τ

is ﬁnite almost surely.

Also, X

(x) ≤ r

, since for y ≤ r

, (i) F

(y) < F

) for all i and

(ii) F

) < r

for i = 1, 2,...,N − 1 and F

) = r

. (In a single

period, it is not possible to go from a state less than or equal to r

to one

larger than r

.) By the strong Markov property, and our earlier result,

τ +m

(x) converges in distribution to π as m →∞for all x ∈ (0, r

Similarly, one can check that as n →∞, X

(x) converges in distribution

to π for all x > r

Exercise 6.1 Consider the case where S = R

and  ={F, G} where

(i) F is strictly increasing, continuous, F(0) > 0 and there is

x such

that F(x) > x on (0,

x) and F(x) < x for x >

(ii) G is strictly increasing, continuous, G(x) < x for all x > 0.

Let Prob(α

= F) = p, Prob(α

= G) = 1 − p. Analyze the long-

run behavior of the process (2.2). In this context, see the paper by Foley

and Hellwig (1975). In their model uncertainty arises out of the possibility

of unemployment.

The assumption that  is ﬁnite can be dispensed with if one has addi-

tional structures in the model.

266 Random Dynamical Systems

Example 6.2 (Multiplicative shocks) Let F : R

→ R

satisfy

[F.1] F is strictly increasing and continuous.

We shall keep F ﬁxed.

Consider  = [θ

,θ

], where 0 <θ

<θ

, and assume the fol-

lowing concavity and “end point” conditions:

[F.2] F(x)/x is strictly decreasing in x > 0,

F(x



)



< 1 for some x



F(x



)



> 1 for some x



> 0.

Since

θ F(x)

is also strictly decreasing in x, [F.1] and [F.2] imply that

for each θ ∈ , there is a unique x

> 0 such that

θ F(x

)

= 1, i.e.,

θ F(x

) = x

. Observe that

θ F(x)

> 1 for 0 < x < x

θ F(x)

< 1 for x >

.Now,θ



>θ



implies x



> x





F(x



)



F(x



)



= 1 =



F(x



)



, im-

plying “x



> x



.”

Write  ={f : f = θ F,θ∈ }, and f

≡ θ

F, f

≡ θ

Assume that θ

(n)

is chosen i.i.d. according to a density function g(θ)

on , which is positive and continuous on .

In our notation, f

) = x

; f

) = x

.Ifx ≥ x

, and

has a density g, then f (x) ≡

θ F(x) ≥ f (x

) ≥ f

) = x

. Hence

(x) ≥ x

for all n ≥ 0ifx ≥ x

.Ifx ≤ x

f (x) ≤ f (x

) ≤ f

) = x

Hence, if x ∈ [x

, x

] then the process X

(x) remains in [x

, x

]

forever. Now, lim

n→∞

) = x

and lim

n→∞

) = x

, and

there must exist a positive integer n

such that f

) < f

by the arguments given for F

, F

in the preceding example. Choose

some z

∈ ( f

), f

)). There exist intervals [θ

,θ

+ δ], [θ

−

δ, θ

] such that for all θ ∈ [θ

,θ

+ δ] and

θ ∈ [θ

− δ, θ

(θ F)

) < z

< (

θ F)

Then the splitting condition holds on the state space [x

, x

]. Now

ﬁx x such that 0 < x < x

, then

θ F(x) ≥ θ

F(x) > x.

Let m be any given positive integer. Since (θ

(x) → x

n →∞, there exists n



≡ n



(x) such that (θ

(x) > x

−

for all

n >



. This implies that X

(x) > x

−

for all n ≥ n



. Therefore,

3.6 Applications 267

lim inf

n→∞

(x) ≥ x

. We now argue that with probability 1,

lim inf

n→∞

(x) > x

. For this, note that if we choose η =

−θ

and

ε>0 such that x

− ε>0, then min{θ F(y) − θ

F(y): θ

≥ θ ≥ θ

η, y ≥ x

− ε}=ηF(x

− ε) > 0. Write η



≡ ηF(x

− ε) > 0. Since

with probability 1, the i.i.d. sequence {θ

(n)

: n = 1, 2,...} takes values in

[θ

+ η, θ

] inﬁnitely often, one has lim inf

n→∞

(x) > x

−

+ η



Choose m so that

<η



. Then with probability 1 the sequence X

(x)

exceeds x

. Since x

= f

(n)

) ≥ X

(x) for all n, it fol-

lows that with probability 1 X

(x) reaches the interval [x

, x

] and

remains in it thereafter. Similarly, one can prove that if x > x

then with

probability 1, the Markov process X

(x) will reach [x

, x

] in ﬁnite

time and stay in the interval thereafter.

Remark 6.1 Recall from Section 1.5 that in growth models, the condi-

tion [F.1] is often derived from appropriate “end point” or Uzawa–Inada

condition. It should perhaps be stressed that convexity assumptions have

not appeared in the discussion of this section so far.

3.6.3 Interaction of Growth and Cycles

We go back to the end of Section 1.9.5 where we indicated that some

variations of the Solow model may lead to complex behavior.

One can abstract from the speciﬁc “sources” of complexity and from

variations of the basic ingredients of the model, and try to study the

long-run behavior of an economy in which qualitatively distinct laws of

motion arise with positive probability. This is the motivation behind the

formal analysis of a rather special case.

Consider a random dynamical system with S =

and with two pos-

sible laws of motion, denoted by F and G (i.e.,  ={F, G}) occurring

with probabilities p and 1 − p (0 < p < 1). The law of motion F has

the strong monotonicity and stability property with an attracting posi-

tive ﬁxed point (see [G.1]); however, the other law G triggers cyclical

forces and has a pair of locally attracting periodic points of period 2

(and a repelling ﬁxed point: see [P.1]–[P.4]). One may interpret F as

the dominant long-run growth law (the probability p is “large”), while

G represents short-run cyclical interruptions. Our main result identiﬁes

conditions under which Theorems 5.1 and 5.2 are applicable.

268 Random Dynamical Systems

The law of motion that generates the growth process is represented by

a continuous increasing function F :[0, 1] → [0, 1]. We assume that

[G.1] F has a ﬁxed point r > 1/2 such that

F(x) > xfor0 < x < r,

F(x) < x for x > r.

Whether or not F(0) = 0 is not relevant for our subsequent analysis.

Note that the trajectory from any initial x

> 0 converges to r; indeed, if

0 < x

< r , the sequence F

) increases to r ; whereas if x

> r , the

sequence F

) decreases to r.

The law of motion that triggers cyclical forces is denoted by a contin-

uous map G :[0, 1] → [0, 1]. We assume

[P.1] G is increasing on [0, 1/2] and decreasing on [1/2, 1].

[P.2] G(x) > xon[0, 1/2].

Suppose that the law of motion F appears with probability p > 0

and G appears with probability 1 − p > 0. We can prove the following

boundedness property of the trajectories:

Lemma 6.1 Let x ∈ (0, 1) and x

= x. Write M = G(1/2) and

x = max[x, M, F (1)],

= min[x, 1/2, G(

x)].

Then for all n

 0

 X



x. (6.4)

Proof. Note that x

≤

x >

. The range of F on [x

x]is[F(x

x)] ⊂ [x

x], since F(x

) ≥ x

(recall that F(x)  x on [0,

]) and

x) ≤ F(1)



x. The range of G on [x

x] ≡ [x

] ∪ [

x]is

[G(x

), M] ∪ [G(

x), M]. Now G(x

) ≥ x

(since G(x) > x on [0,

] and

x) ≥ x

(by deﬁnition of x

). Therefore, the range of G on [x

x]is

contained in [x

x]. Note that M ≤

x. Thus if (6.4) holds for some n,

then x

≤ F(X

) ≤

x, x

≤ G(X

) ≤

x, i.e., x

≤ X

n+1

≤

x. Since (6.4)

holds for n = 0, it holds for all n.

The simplest way of capturing “cyclical forces” is to make the follow-

ing assumption:

3.6 Applications 269

[P.3]G has two periodic points of period 2 denoted by {β

,β

}, and a

repelling ﬁxed point x

∗

, and no other ﬁxed point or periodic point.

Moreover, {β

,β

} are locally stable ﬁxed points of G

≡ G ◦ G.

We shall also assume

[P.4] G has an invariant interval [a, b](1/2 ≤ a < b < 1); a <β

∗

<β

< b. Also r ∈ (a, b), r /∈{β

, x

∗

,β

Note that G

is increasing on [a, b]. By invariance of [a, b], G

(a) ≥

a, G

(b) ≤ b.

By using the fact that G

is increasing on [a, b], we get that the se-

quence G

(a) is nondecreasing and converges to some limit, say, z

By continuity of G

, z

is a ﬁxed point of G

. Since a  β

, it follows

that z

 β

and by assumption [P.3], z

must be β

. Hence, G

(a)

increases to the ﬁxed point β

. Similarly, G

(b) decreases to the ﬁxed

point β

. Since neither a nor b is a ﬁxed point of G

,itisinfactthe

case that G

(a) − a > 0, G

(b) − b < 0. Hence, G

(x) − x > 0 for all

x ∈ [a,β

) and, G

(x) − x < 0 for all x ∈ (β

, b]. (If, for some z in

[a,β

), G

(z) − z ≤ 0, by continuity of G

, there is a ﬁxed point of G

between a and z, a contradiction.)

It follows that for all x in [a,β

), G

(x) increases to β

and for all x

in (β

, b], G

(x) decreases to β

as n →∞.

Observe, next, that G

x − x cannot change sign on either (β

, x

∗

)or

(β

, x

∗

). For otherwise, G

has a ﬁxed point other than β

, β

, x

∗

Suppose, if possible, that G

(x) − x > 0on(β

, x

∗

); then from any

y ∈ (β

, x

∗

), G

(y) will increase to some ﬁxed point; this ﬁxed point

must be x

∗

;butx

∗

is repelling, a contradiction; hence, G

(x) − x < 0

on (β

, x

∗

). A similar argument ensures that G

(x) − x > 0on(x

∗

,β

It follows that G

(x) decreases to β

for all x in (β

, x

∗

) and G

(x)

increases to β

for all x in (x

∗

,β

) (as n →∞). To summarize

For all x in [a,β

),G

(x) − x > 0 and G

(x) increases to β

,as

n ↑∞. For all x in (β

, b],G

(x) − x < 0 and G

(x) decreases to

, as n ↑∞. Also, G

(x) − x < 0on (β

, x

∗

) and G

(x) − x > 0 on

∗

,β

). Hence, G

(x) decreases to β

on (β

, x

∗

) and increases to β

on (x

∗

,β

) as n ↑∞.

Further, from [G.1] and [P.4], [a, b] is invariant under F. Consider

n+1

= α

n+1

), where α

= F with probability p and α

= G with

270 Random Dynamical Systems

probability 1 − p. In what follows, the initial X

= x ∈ (0, 1). Using

Lemma 6.1 and the structure of F, it is clear that the process X

(x)

lands in [a, b] with probability 1 after a ﬁnite number of steps. For the

process lands in [a, b] from any initial x ∈ (0, 1),F occurs in a ﬁnite

(but sufﬁciently large) number of consecutive periods; i.e., there is a

long “run” of F. It is useful to distinguish four cases depending on the

location of the attractive ﬁxed point r of F.

Case (i). a < r <β

< x

∗

<β

< b. Here, choose δ small enough so

that r + δ<β

; there is some N

such that F

[a, b] ⊂ (r − δ, r + δ)

for all n

 N

. Choose ε>0 so that β

− ε>r + δ. Recall that

(a) increases to β

; let N

be such that G

(a) >β

− ε for

all n

 N

. Then G

(x) >β

− ε for all x ∈ [a, b] and for all

 N

Now choose some z

satisfying r + δ<z

<β

− ε. Let N be

some even integer greater than max(2N

, N

). Then

Prob(α

···α

x  z

∀x ∈ [a, b])

 (1 − p)

This follows from the possibility of G occurring in N consecutive

periods. Also,

Prob(α

···α

x  z

∀x ∈ [a, b])  p

which follows from the possibility of F occurring in N consecutive

periods. This veriﬁes the splitting condition (H), in Subsection 3.5.1.

Case (ii). a <β

< r < x

∗

<β

< b. Here choose ε, δ positive

so that a <β

+ ε<r − δ<r + δ<x

∗

<β

< b. Again, there is

some N

such that F

[a, b] ⊂ (r − δ, r + δ) ∀n  N

. There is N

such that G

(r + δ)  β

+ ε for all n  N

, so that G

[a, r + δ] ⊂

[a,β

+ ε] ∀n ≥ N

Consider the composition G

◦ F

. It is clear that for all

x ∈ [a, b], G

◦ F

(x)  β

+ ε. Let N be 2N

+ N

and choose

some z

satisfying β

+ ε<z

< r − δ. Then F

[a, b] ⊂ [z

, b],

◦ F

[a, b] ⊂ [a, z

], so that

Prob(α

···α

x  z

∀x ∈ [a, b])  p

Prob(α

···α

x  z

∀x ∈ [a, b])  p

(1 − p )

Thus the splitting condition is veriﬁed in this case also.

3.6 Applications 271

Case (iii). a <β

< x

∗

< r <β

< b. Choose δ>0, ε>0 such

that x

∗

< r − δ<r + δ<β

− ε. There exists N

such that

[a, b] ⊂ (r − δ, r + δ), ∀n ≥ N

and G

(r − δ) >β

− ε so

that G

[r − δ, b] ⊂ (β

− ε, b]. Hence, splitting occurs with z

∈

(r + δ, β

− ε) and N = 3N

: P(α

···α

x ≤ z

∀x ∈ [a, b]) ≥ p

P(α

···α

x ≥ z

∀x ∈ [a, b]) ≥ p

(1 − p )

Case (iv). a <β

< x

∗

<β

< r < b. Choose δ>0,ε > 0 such that

+ ε<r − δ. There exists N

such that F

[a, b] ⊂ (r − δ, r +

δ), and G

(b) <β

+ ε so that G

[a, b] ⊂ [a,β

+ ε). Hence

splitting occurs with z

∈ (β

+ ε, r − δ) and N = 2N

P(α

···α

x ≤ z

∀x ∈ [a, b]) ≥ (1 − p)

P(α

···α

x ≥ z

∀x ∈ [a, b]) ≥ p

We now prove the following result:

Theorem 6.1 Under the hypotheses [G.1], [P.1]–[P.4], the distribution

of X

(x) converges exponentially fast in the Kolmogorov distance to a

unique invariant probability, for every initial x ∈ (0, 1).

Proof. In view of the preceding arguments, Theorem 5.1 guaran-

tees exponential convergence of the distribution function H

(z): =

Prob(X

(x)  z) to the distribution function F

(z) of the invariant prob-

ability π with an error no more than (1 − δ)

[n/N]

, provided the initial state

x ∈ [a, b]. If x ∈ (0, 1) ∩ [a, b]

then, by Lemma 6.1, there exists 0 < x

x < 1 such that X

(x) ∈ [x

x]∀n  0. Since F

(y) ↑ r as n ↑∞for

0 < y < r , and F

(y) ↓ r as n ↑∞for r < y < 1, it follows that there

exists an integer n

such that F

(y) ∈ [a, b]∀n  n

and ∀y ∈ [x

x].

Therefore, Prob(X

(x) /∈ [a, b])  (1 − p

)

[m/n

]

∀m  n

. For the

probability of not having an n

-long run of F in any of the consecutive

[m/n

] periods of length, n

is (1 − p

)

[m/n

]

. Now let n  2n

, m

[n/2], m

= n − m

. Then ∀z ∈ R one has

(z) = Prob(X

(x)  z, X

(x) ∈ [a, b])

+Prob(X

(x)  z, X

(x) /∈ [a, b])

[a,b]

Prob(X

n−m

(y)  z) p

)

(x, dy)

+Prob(X

(x)  z, X

(x) /∈ [a, b]).

272 Random Dynamical Systems

Since splitting (H) holds, there exist ˜χ>0 and N such that

(1) the integrand above lies between F

(z) ± (1 − ˜χ )

/N ]

(2) 1



[a,b]

)

(x, dy)  1 − (1 − p

)

]

, and

(3) Prob(X

(x)  z, X

(x) /∈ [a, b])

 (1 − p

)

]

it follows that

(z) − F

(z)|  (1 − ˜χ )

/N ]

+ 2(1 − p

)

]

Example 6.3 We now consider an interesting example of {F, G} satis-

fying the hypotheses [G.1], [P.1]–[P.4], so that Theorem 6.1 holds.

Consider the family of maps F

:[0, 1] → [0, 1] deﬁned as

(x) = θ x(1 − x).

For 1 <θ ≤ 4, the ﬁxed points of F

are 0, x

∗

= 1 −

.Forθ>2,

∗

> 1/2. Let us try to ﬁnd an interval [a, b] on which F

is monotone and

which is left invariant under F

, i.e., F

([a, b]) ⊂ [a, b]. One must have

1/2 ≤ a < b ≤ 1, and one can check that [1/2,θ/4] is such an interval,

provided F

(θ/4) ≡ F

(1/2) ≥ 1/2. This holds if 2 ≤ θ ≤ 1 +

√

5. For

any such θ, F

is strictly decreasing and F

is strictly increasing on

[1/2,θ/4].

Consider 3 <θ≤ 1 +

√

5. In this case, |F



∗

)|=θ − 2 > 1 so that

∗

is a repelling ﬁxed point of F

. One can show that there are two

locally attracting periodic points of period 2, say, β

and β

, and

< x

∗

<β

(see Section 1.7). Since F

is a fourth-degree polyno-

mial, {0,β

, x

∗

,β

} are the only ﬁxed points of F

. Also, F

(x) − x

does not change sign on (β

, x

∗

)oron(x

∗

,β

). Take θ = 3.1. Then the

invariant interval is [1/2, 0.775] and x

∗

= 0.677. One can compute (see,

e.g., Sandefur 1990, pp. 172–181), that when θ ∈ (3, 1 +

√

6),β

and

are given by

√

θ + 1[(

√

θ + 1 ±

√

θ − 3/2θ]. For θ = 3.1, one has

= 0.558014125,β

= 0.76456652, F



(β

)

∼

−0.3597, F



(β

) =

−1.6403, |F



(β



(β

∼

0.5904 < 1.

As an example of functions of the kind considered in this sec-

tion, let F(x) = (3/4)x

1/2

and G(x) = F

(x) as above, with θ = 3.1.

Let Prob(α

= F) = p, Prob(α

= G) = 1 − p(0 < p < 1). Then with

a = 1/2, b = θ/4 = (3.10)/4, [a, b] is an invariant interval for F

and G. The ﬁxed point of F in [a, b]is(3/4)

= 9/16 = r. The re-

pelling ﬁxed point of G is x

∗

= 0.677 ... and the period-two orbit

is {β

= 0.558 ...,β

= 0.764 ...}. Thus the hypotheses imposed on