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The strong deviation theorem for discrete-time and continuous-state nonhomogeneous Markov chains

Abstract

In this paper, the notion of asymptotic average log-likelihood ratio, as a measure of the difference between the sequence of random variables and Markov chains, is introduced, and by constructing a nonnegative martingale, the strong deviation theorem for discrete-time and continuous-state nonhomogeneous Markov chains is established.

1 Introduction

Let (Ω,F,P) be the probability space, and let { X n ,n0} be a sequence of continuous random variables taking values in R and with the joint density function f n ( X 0 , X 1 X n ), n=1,2, . Let Q be another probability measure on (Ω,F), and { X n ,n0} be an independent random sequence on the measure Q, with the joint density function g n ( X 0 , X 1 X n ), n=1,2, .

Let

r n ( ω ) = g n ( X 0 , X 1 X n ) f n ( X 0 , X 1 X n ) , r ( ω ) = lim inf n 1 n ln r n ( ω ) ( ln 0 = ) ,
(1)

where ω is a sample point. In statistical terms, r n (ω) and r(ω) are called the likelihood ratio and the asymptotic average log-likelihood ratio, respectively [1]. Obviously, if f n ( x 0 , x 1 x n )= g n ( x 0 , x 1 x n ), n1, then r n (ω)0, a.s. So r(ω) can be used as a measure of deviation between f n ( x 0 , x 1 x n ) and g n ( x 0 , x 1 x n ) when n tends to infinity. The smaller r(ω) is, the smaller the deviation is.

Definition 1 [2]

Let { X n ,n0} be a nonhomogeneous Markov chain with the initial distribution u(x), xR, and the transition probability density p n = p n (x,y), x,yR, n1. If

P( X 0 B)= B u(x)dx,P( X n + 1 B| X n =x)= B p n (x,y)dy,

this Markov chain is called a discrete-time and continuous-state nonhomogeneous Markov chain.

Let { X n ,n0} be a discrete-time and continuous-state nonhomogeneous Markov chain on the measure Q with the initial distribution density u(x), xR and the transition probability density p n = p n (x,y), x,yR, n1. Then for any Borel, set B

Q ( X 0 B ) = B u ( x ) d x , Q ( X n + 1 B | X n = x ) = B p n ( x , y ) d y ,

then

g n ( x 0 , x 1 ,, x n )=u( x 0 ) p 0 ( x 0 , x 1 ) p n 1 ( x n 1 , x n )=u( x 0 ) k = 1 n p k 1 ( x k 1 , x k ),

so

r n (ω)= u ( X 0 ) k = 1 n p k 1 ( X k 1 , X k ) f n ( X 0 , X 1 X n ) .
(2)

There have been some works on deviation theorem, a kind of strong limit theorem represented by inequalities. Liu and Yang [3] have studied the limit properties of a class of averages of functions of two variables of arbitrary information sources. Liu and Yang [4] investigated the strong deviation theorems for arbitrary information source relative to Markov information source. Liu [5] discussed a class of strong deviation theorems for an arbitrary stochastic sequence with respect to the marginal distribution by using generating function method, and also studied the problem above by means of Laplace transform [6]. Liu and Wang [7] have studied a strong limit theorem expressed by inequalities for the sequences of absolutely continuous random variables. Recently, Fan [8] has studied some strong deviation theorems for dependent continuous random sequence.

In this paper, by using the notion of asymptotic log-likehood and the martingale convergence theorem, and extending the analytic technique proposed by Liu [9], Liu and Yang [7] to the case of discrete-time and continuous-state nonhomogeneous Markov chains, we obtain the strong deviation theorem for discrete-time and continuous-state nonhomogeneous Markov chains.

2 Main result

Theorem 1 Let { X n ,n0} be a discrete-time and continuous-state nonhomogeneous Markov chain on the measure Q, r(ω) and r n (ω) be defined by (1) and (2), respectively. Let { B n ,n1} be a sequence of Borel set of the real line, and I B n be the indicative function of B n . Let

a(ω)= lim sup n 1 n k = 1 n B k p k 1 ( X k 1 , x k )d x k b,ωΩ,
(3)

and

D 1 = { ω : r ( ω ) b } , D 2 = { ω : r ( ω ) b } .

Then

(a) lim sup n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] 2 b r ( ω ) +r(ω)a.s.;
(4)
(b) lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] 2 b r ( ω ) a.s. on  D 1 ,
(5)

and

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] br(ω)a.s. on  D 2 .
(6)

Proof Let λ be a nonnegative constant, and let

h k ( x k 1 , x k )= { λ p k 1 ( x k 1 , x k ) 1 + ( λ 1 ) B k p k 1 ( x k 1 , x k ) d x k , x k B k ; p k 1 ( x k 1 , x k ) 1 + ( λ 1 ) B k p k 1 ( x k 1 , x k ) d x k , x k B k .
(7)

It is easy to see that u( x 0 ) k = 1 n h( x k 1 , x k ) is a density function of n+1 variables. Let

t n (λ,ω)= u ( X 0 ) k = 1 n h k ( X k 1 , X k ) f n ( X 0 , X 1 X n ) ,
(8)

then t n (λ,ω) is a nonnegative supermartingale that converges a.s. Hence there exists A(λ)F, P(A(λ))=1 such that

lim sup n 1 n ln t n (λ,ω)0,ωA(λ).
(9)

Letting λ=1 in (9), we obtain

lim sup n 1 n ln r n (ω)0,ωA(1).
(10)

This implies that

r(ω)0,ωA(1).
(11)

We have by (7)

k = 1 n h k ( X k 1 , X k ) = k = 1 n λ I B k ( X k ) p k 1 ( X k 1 , X k ) 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k = λ k = 1 n I B k ( X k ) k = 1 n p k 1 ( X k 1 , X k ) 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k .
(12)

It follows from (2), (8), and (12) that

ln t n (λ,ω)= k = 1 n I B k ( X k )lnλ k = 1 n ln [ 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ] +ln r n (ω).
(13)

By (9) and (13), we have

lim sup n 1 n ( k = 1 n I B k ( X k ) ln λ + ln r n ( ω ) k = 1 n ln [ 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ] ) 0 , ω A ( λ ) .
(14)

(a) Let λ>1. Dividing the two sides of (14) by lnλ, we obtain

lim sup n 1 n ( k = 1 n I B k ( X k ) + ln r n ( ω ) ln λ k = 1 n ln [ 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ] ln λ ) 0 , ω A ( λ ) .
(15)

By (1) and (15), we have

lim sup n 1 n ( k = 1 n I B k ( X k ) k = 1 n ln [ 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ] ln λ ) r ( ω ) ln λ , ω A ( λ ) .
(16)

By (3), (16), the property of the superior limit

lim sup n ( a n b n )d lim sup n ( a n c n ) lim sup n ( b n c n )+d,

and the inequality 0ln(1+x)x (x0), we have

lim sup n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] lim sup n 1 n k = 1 n [ ln [ 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ] ln λ B k p k 1 ( X k 1 , x k ) d x k ] + r ( ω ) ln λ lim sup n 1 n k = 1 n [ ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ln λ B k p k 1 ( X k 1 , x k ) d x k ] + r ( ω ) ln λ b ( λ 1 ln λ 1 ) + r ( ω ) ln λ , ω A ( λ ) .
(17)

By using the inequality 1 1 λ <lnλ (λ>1), we have by (17)

lim sup n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] b(λ1)+ λ r ( ω ) λ 1 ,ωA(λ).
(18)

Let Q be the set of rational numbers in the interval (1,+), and let

A = λ Q A(λ),g(λ,r)=b(λ1)+ λ r λ 1 .
(19)

Then we have by (18),

lim sup n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] g ( λ , r ( ω ) ) ,ω A ,λ Q .
(20)

Let b>0. It is easy to see if r>0, g(λ,r) as a function of λ attains its smallest value g(1+ r b ,r)=2 b r +r on the interval (1,+), and g(λ,0) is increasing on the interval (1,+) and lim λ 1 + 0 g(λ,0)=0. For each ω A A(1) if r(ω), take λ n (ω) Q , n=1,2, such that λ n (ω)1+ r ( ω ) b , we have

lim n g ( λ n ( ω ) , r ( ω ) ) =2 b r ( ω ) +r(ω).
(21)

By (20), we have

lim sup n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] g ( λ n ( ω ) , r ( ω ) ) ,n=1,2,.
(22)

By (21) and (22), we have

lim sup n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] 2 b r ( ω ) + r ( ω ) , ω A A ( 1 ) .
(23)

If r(ω)=, (23) holds obviously. Since P( A A(1))=1, (4) holds by (23) when b>0.

When b=0, letting λ=e in (20), we have

lim sup n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] r(ω),ωA(e).
(24)

Since P(A(e))=1, (4) also holds by (24) when b=0.

(b) Let 0<λ<1. Dividing the two sides of (14) by lnλ, we have

lim inf n 1 n ( k = 1 n I B k ( X k ) k = 1 n ln [ 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ] ln λ + ln r n ( ω ) ln λ ) 0 , ω A ( λ ) .
(25)

By (1) and (25), we have

lim inf n 1 n ( k = 1 n I B k ( X k ) k = 1 n ln [ 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ] ln λ ) r ( ω ) ln λ , ω A ( λ ) .
(26)

By (26), (3), the property of the inferior limit

lim inf n ( a n b n )d lim inf n ( a n c n ) lim inf n ( b n c n )+d,

and the inequality ln(1+x)x (1<x0), we have

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] lim inf n 1 n k = 1 n [ ln [ 1 + ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ] ln λ B k p k 1 ( X k 1 , x k ) d x k ] + r ( ω ) ln λ lim inf n 1 n k = 1 n [ ( λ 1 ) B k p k 1 ( X k 1 , x k ) d x k ln λ B k p k 1 ( X k 1 , x k ) d x k ] + r ( ω ) ln λ b ( λ 1 ln λ 1 ) + r ( ω ) ln λ , ω A ( λ ) .
(27)

By using the inequality 1 1 λ <lnλ<0 and lnλ<λ1<0 (0<λ<1), we have by (27)

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] b ( λ 1 ) + r ( ω ) λ 1 , ω A ( λ ) A ( 1 ) .
(28)

Let Q be the set of rational numbers in the interval (0,1), and let

A = λ Q A(λ),h(λ,r)=b(λ1)+ r λ 1 .
(29)

Then we have by (28)

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] h ( λ , r ( ω ) ) , ω A A ( 1 ) , λ Q .
(30)

Let b>0. It is easy to see that if 0<r<b, then h(λ,r) as a function of λ attains its largest value h(1 r b ,r)=2 b r on the interval (0,1), and h(λ,0) is increasing on the interval (0,1) and lim λ 1 0 h(λ,0)=0, and h(λ,b)=b(λ1+ 1 λ 1 ) is decreasing on the interval (0,1) and lim λ 0 + h(λ,b)=2b. For each ω A A(1) D 1 , take τ n (ω) Q , n=1,2, such that τ n (ω)1 r ( ω ) b . Then we have

lim n h ( τ n ( ω ) , r ( ω ) ) =2 b r ( ω ) .
(31)

By (30), we have

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] h ( τ n ( ω ) , r ( ω ) ) ,n=1,2,.
(32)

By (31) and (32),

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] 2 b r ( ω ) , ω A A ( 1 ) D 1 .
(33)

Since P( A A(1))=1, (5) holds by (33) when b>0.

When b=0, r(ω)=0 for ω D 1 A(1), hence we have by (30)

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] 0,ωA(λ)A(1) D 1 ,
(34)

since P(A(λ)A(1))=1, (5) also holds by (34) when b=0.

It is easy to see that when 0b<r, h(λ,r) as a function of λ is decreasing on the interval (0,1) and lim λ 0 + h(λ,r)=(r+b). For each ω A A(1) D 2 , when r(ω), take λ n (ω) Q , n=1,2, , such that λ n (ω)0. We have

lim n h ( λ n ( ω ) , r ( ω ) ) =r(ω)b.
(35)

By (30), we have

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] h ( λ n ( ω ) , r ( ω ) ) ,n=1,2,.
(36)

It follows from (35) and (36) that

lim inf n 1 n k = 1 n [ I B k ( X k ) B k p k 1 ( X k 1 , x k ) d x k ] r ( ω ) b , ω A A ( 1 ) D 2
(37)

when r(ω)=, (37) also holds obviously. Since P( A A(1))=1, (6) follows from (37) directly. □

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Acknowledgements

This work is supported by the National Natural Science Foundations of China (11071104, 11226210), and the Research Foundation for Advanced Talents of Jiangsu University (11JDG116).

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Correspondence to Bei Wang.

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BW carried out the study of strong deviation theorem and drafted the manuscript. ZS participated in the proof of theorem. All authors read and approved the final manuscript.

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Wang, B., Shi, Z. The strong deviation theorem for discrete-time and continuous-state nonhomogeneous Markov chains. J Inequal Appl 2013, 462 (2013). https://doi.org/10.1186/1029-242X-2013-462

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