• Research
• Open Access

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

Journal of Inequalities and Applications20132013:462

https://doi.org/10.1186/1029-242X-2013-462

• Accepted: 9 September 2013
• Published:

## 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.

## Keywords

• discrete-time and continuous-state nonhomogeneous Markov chain
• asymptotic average log-likelihood ratio
• transition probability density
• strong deviation theorem

## 1 Introduction

Let $\left(\mathrm{\Omega },\mathcal{F},P\right)$ be the probability space, and let $\left\{{X}_{n},n\ge 0\right\}$ be a sequence of continuous random variables taking values in R and with the joint density function ${f}_{n}\left({X}_{0},{X}_{1}\cdots {X}_{n}\right)$, $n=1,2,\dots$ . Let Q be another probability measure on $\left(\mathrm{\Omega },\mathcal{F}\right)$, and $\left\{{X}_{n},n\ge 0\right\}$ be an independent random sequence on the measure Q, with the joint density function ${g}_{n}\left({X}_{0},{X}_{1}\cdots {X}_{n}\right)$, $n=1,2,\dots$ .

Let
$\begin{array}{c}{r}_{n}\left(\omega \right)=\frac{{g}_{n}\left({X}_{0},{X}_{1}\cdots {X}_{n}\right)}{{f}_{n}\left({X}_{0},{X}_{1}\cdots {X}_{n}\right)},\hfill \\ r\left(\omega \right)=-\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}ln{r}_{n}\left(\omega \right)\phantom{\rule{1em}{0ex}}\left(ln0=-\mathrm{\infty }\right),\hfill \end{array}$
(1)

where ω is a sample point. In statistical terms, ${r}_{n}\left(\omega \right)$ and $r\left(\omega \right)$ are called the likelihood ratio and the asymptotic average log-likelihood ratio, respectively . Obviously, if ${f}_{n}\left({x}_{0},{x}_{1}\cdots {x}_{n}\right)={g}_{n}\left({x}_{0},{x}_{1}\cdots {x}_{n}\right)$, $n\ge 1$, then ${r}_{n}\left(\omega \right)\equiv 0$, a.s. So $r\left(\omega \right)$ can be used as a measure of deviation between ${f}_{n}\left({x}_{0},{x}_{1}\cdots {x}_{n}\right)$ and ${g}_{n}\left({x}_{0},{x}_{1}\cdots {x}_{n}\right)$ when n tends to infinity. The smaller $r\left(\omega \right)$ is, the smaller the deviation is.

Definition 1 

Let $\left\{{X}_{n},n\ge 0\right\}$ be a nonhomogeneous Markov chain with the initial distribution $u\left(x\right)$, $x\in R$, and the transition probability density ${p}_{n}={p}_{n}\left(x,y\right)$, $x,y\in R$, $n\ge 1$. If
$P\left({X}_{0}\in B\right)={\int }_{B}u\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,\phantom{\rule{2em}{0ex}}P\left({X}_{n+1}\in B|{X}_{n}=x\right)={\int }_{B}{p}_{n}\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy,$

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

Let $\left\{{X}_{n},n\ge 0\right\}$ be a discrete-time and continuous-state nonhomogeneous Markov chain on the measure Q with the initial distribution density $u\left(x\right)$, $x\in R$ and the transition probability density ${p}_{n}={p}_{n}\left(x,y\right)$, $x,y\in R$, $n\ge 1$. Then for any $Borel$, set B
$\begin{array}{c}Q\left({X}_{0}\in B\right)={\int }_{B}u\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,\hfill \\ Q\left({X}_{n+1}\in B|{X}_{n}=x\right)={\int }_{B}{p}_{n}\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy,\hfill \end{array}$
then
${g}_{n}\left({x}_{0},{x}_{1},\dots ,{x}_{n}\right)=u\left({x}_{0}\right){p}_{0}\left({x}_{0},{x}_{1}\right)\cdots {p}_{n-1}\left({x}_{n-1},{x}_{n}\right)=u\left({x}_{0}\right)\prod _{k=1}^{n}{p}_{k-1}\left({x}_{k-1},{x}_{k}\right),$
so
${r}_{n}\left(\omega \right)=\frac{u\left({X}_{0}\right){\prod }_{k=1}^{n}{p}_{k-1}\left({X}_{k-1},{X}_{k}\right)}{{f}_{n}\left({X}_{0},{X}_{1}\cdots {X}_{n}\right)}.$
(2)

There have been some works on deviation theorem, a kind of strong limit theorem represented by inequalities. Liu and Yang  have studied the limit properties of a class of averages of functions of two variables of arbitrary information sources. Liu and Yang  investigated the strong deviation theorems for arbitrary information source relative to Markov information source. Liu  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 . Liu and Wang  have studied a strong limit theorem expressed by inequalities for the sequences of absolutely continuous random variables. Recently, Fan  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 , Liu and Yang  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 $\left\{{X}_{n},n\ge 0\right\}$ be a discrete-time and continuous-state nonhomogeneous Markov chain on the measure Q, $r\left(\omega \right)$ and ${r}_{n}\left(\omega \right)$ be defined by (1) and (2), respectively. Let $\left\{{B}_{n},n\ge 1\right\}$ be a sequence of $Borel$ set of the real line, and ${I}_{{B}_{n}}$ be the indicative function of ${B}_{n}$. Let
$a\left(\omega \right)=\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\le b,\phantom{\rule{1em}{0ex}}\mathrm{\forall }\omega \in \mathrm{\Omega },$
(3)
and
${D}_{1}=\left\{\omega :r\left(\omega \right)\le b\right\},\phantom{\rule{2em}{0ex}}{D}_{2}=\left\{\omega :r\left(\omega \right)\ge b\right\}.$
Then
$\text{(a)}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\le 2\sqrt{br\left(\omega \right)}+r\left(\omega \right)\phantom{\rule{1em}{0ex}}\mathit{\text{a.s.}};$
(4)
(5)
and
(6)
Proof Let λ be a nonnegative constant, and let
${h}_{k}\left({x}_{k-1},{x}_{k}\right)=\left\{\begin{array}{ll}\frac{\lambda {p}_{k-1}\left({x}_{k-1},{x}_{k}\right)}{1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({x}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}},& {x}_{k}\in {B}_{k};\\ \frac{{p}_{k-1}\left({x}_{k-1},{x}_{k}\right)}{1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({x}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}},& {x}_{k}\notin {B}_{k}.\end{array}$
(7)
It is easy to see that $u\left({x}_{0}\right){\prod }_{k=1}^{n}h\left({x}_{k-1},{x}_{k}\right)$ is a density function of $n+1$ variables. Let
${t}_{n}\left(\lambda ,\omega \right)=\frac{u\left({X}_{0}\right){\prod }_{k=1}^{n}{h}_{k}\left({X}_{k-1},{X}_{k}\right)}{{f}_{n}\left({X}_{0},{X}_{1}\cdots {X}_{n}\right)},$
(8)
then ${t}_{n}\left(\lambda ,\omega \right)$ is a nonnegative supermartingale that converges a.s. Hence there exists $A\left(\lambda \right)\in \mathcal{F}$, $P\left(A\left(\lambda \right)\right)=1$ such that
$\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}ln{t}_{n}\left(\lambda ,\omega \right)\le 0,\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).$
(9)
Letting $\lambda =1$ in (9), we obtain
$\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}ln{r}_{n}\left(\omega \right)\le 0,\phantom{\rule{1em}{0ex}}\omega \in A\left(1\right).$
(10)
This implies that
$r\left(\omega \right)\ge 0,\phantom{\rule{1em}{0ex}}\omega \in A\left(1\right).$
(11)
We have by (7)
$\begin{array}{rcl}\prod _{k=1}^{n}{h}_{k}\left({X}_{k-1},{X}_{k}\right)& =& \prod _{k=1}^{n}\frac{{\lambda }^{{I}_{{B}_{k}}\left({X}_{k}\right)}{p}_{k-1}\left({X}_{k-1},{X}_{k}\right)}{1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}}\\ =& {\lambda }^{{\sum }_{k=1}^{n}{I}_{{B}_{k}}\left({X}_{k}\right)}\prod _{k=1}^{n}\frac{{p}_{k-1}\left({X}_{k-1},{X}_{k}\right)}{1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}}.\end{array}$
(12)
It follows from (2), (8), and (12) that
$ln{t}_{n}\left(\lambda ,\omega \right)=\sum _{k=1}^{n}{I}_{{B}_{k}}\left({X}_{k}\right)ln\lambda -\sum _{k=1}^{n}ln\left[1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]+ln{r}_{n}\left(\omega \right).$
(13)
By (9) and (13), we have
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\left(\sum _{k=1}^{n}{I}_{{B}_{k}}\left({X}_{k}\right)ln\lambda +ln{r}_{n}\left(\omega \right)-\sum _{k=1}^{n}ln\left[1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le 0,\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).\hfill \end{array}$
(14)
(a) Let $\lambda >1$. Dividing the two sides of (14) by lnλ, we obtain
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\left(\sum _{k=1}^{n}{I}_{{B}_{k}}\left({X}_{k}\right)+\frac{ln{r}_{n}\left(\omega \right)}{ln\lambda }-\sum _{k=1}^{n}\frac{ln\left[1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]}{ln\lambda }\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le 0,\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).\hfill \end{array}$
(15)
By (1) and (15), we have
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\left(\sum _{k=1}^{n}{I}_{{B}_{k}}\left({X}_{k}\right)-\sum _{k=1}^{n}\frac{ln\left[1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]}{ln\lambda }\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{r\left(\omega \right)}{ln\lambda },\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).\hfill \end{array}$
(16)
By (3), (16), the property of the superior limit
$\underset{n\to \mathrm{\infty }}{limsup}\left({a}_{n}-{b}_{n}\right)\le d\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{limsup}\left({a}_{n}-{c}_{n}\right)\le \underset{n\to \mathrm{\infty }}{limsup}\left({b}_{n}-{c}_{n}\right)+d,$
and the inequality $0\le ln\left(1+x\right)\le x$ ($x\ge 0$), we have
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[\frac{ln\left[1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]}{ln\lambda }-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\\ \phantom{\rule{2em}{0ex}}+\frac{r\left(\omega \right)}{ln\lambda }\\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[\frac{\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}}{ln\lambda }-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]+\frac{r\left(\omega \right)}{ln\lambda }\\ \phantom{\rule{1em}{0ex}}\le b\left(\frac{\lambda -1}{ln\lambda }-1\right)+\frac{r\left(\omega \right)}{ln\lambda },\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).\end{array}$
(17)
By using the inequality $1-\frac{1}{\lambda } ($\lambda >1$), we have by (17)
$\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\le b\left(\lambda -1\right)+\frac{\lambda r\left(\omega \right)}{\lambda -1},\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).$
(18)
Let ${Q}^{\ast }$ be the set of rational numbers in the interval $\left(1,+\mathrm{\infty }\right)$, and let
${A}^{\ast }=\bigcap _{\lambda \in {Q}^{\ast }}A\left(\lambda \right),\phantom{\rule{2em}{0ex}}g\left(\lambda ,r\right)=b\left(\lambda -1\right)+\frac{\lambda r}{\lambda -1}.$
(19)
Then we have by (18),
$\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\le g\left(\lambda ,r\left(\omega \right)\right),\phantom{\rule{1em}{0ex}}\omega \in {A}^{\ast },\lambda \in {Q}^{\ast }.$
(20)
Let $b>0$. It is easy to see if $r>0$, $g\left(\lambda ,r\right)$ as a function of λ attains its smallest value $g\left(1+\sqrt{\frac{r}{b}},r\right)=2\sqrt{br}+r$ on the interval $\left(1,+\mathrm{\infty }\right)$, and $g\left(\lambda ,0\right)$ is increasing on the interval $\left(1,+\mathrm{\infty }\right)$ and ${lim}_{\lambda \to 1+0}g\left(\lambda ,0\right)=0$. For each $\omega \in {A}^{\ast }\cap A\left(1\right)$ if $r\left(\omega \right)\ne \mathrm{\infty }$, take ${\lambda }_{n}\left(\omega \right)\in {Q}^{\ast }$, $n=1,2,\dots$ such that ${\lambda }_{n}\left(\omega \right)\to 1+\sqrt{\frac{r\left(\omega \right)}{b}}$, we have
$\underset{n\to \mathrm{\infty }}{lim}g\left({\lambda }_{n}\left(\omega \right),r\left(\omega \right)\right)=2\sqrt{br\left(\omega \right)}+r\left(\omega \right).$
(21)
By (20), we have
$\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\le g\left({\lambda }_{n}\left(\omega \right),r\left(\omega \right)\right),\phantom{\rule{1em}{0ex}}n=1,2,\dots .$
(22)
By (21) and (22), we have
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\sqrt{br\left(\omega \right)}+r\left(\omega \right),\phantom{\rule{1em}{0ex}}\omega \in {A}^{\ast }\cap A\left(1\right).\hfill \end{array}$
(23)

If $r\left(\omega \right)=\mathrm{\infty }$, (23) holds obviously. Since $P\left({A}^{\ast }\cap A\left(1\right)\right)=1$, (4) holds by (23) when $b>0$.

When $b=0$, letting $\lambda =e$ in (20), we have
$\underset{n\to \mathrm{\infty }}{limsup}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\le r\left(\omega \right),\phantom{\rule{1em}{0ex}}\omega \in A\left(e\right).$
(24)

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

(b) Let $0<\lambda <1$. Dividing the two sides of (14) by lnλ, we have
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\left(\sum _{k=1}^{n}{I}_{{B}_{k}}\left({X}_{k}\right)-\sum _{k=1}^{n}\frac{ln\left[1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]}{ln\lambda }+\frac{ln{r}_{n}\left(\omega \right)}{ln\lambda }\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge 0,\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).\hfill \end{array}$
(25)
By (1) and (25), we have
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\left(\sum _{k=1}^{n}{I}_{{B}_{k}}\left({X}_{k}\right)-\sum _{k=1}^{n}\frac{ln\left[1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]}{ln\lambda }\right)\hfill \\ \phantom{\rule{1em}{0ex}}\ge \frac{r\left(\omega \right)}{ln\lambda },\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).\hfill \end{array}$
(26)
By (26), (3), the property of the inferior limit
$\underset{n\to \mathrm{\infty }}{liminf}\left({a}_{n}-{b}_{n}\right)\ge d\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{liminf}\left({a}_{n}-{c}_{n}\right)\ge \underset{n\to \mathrm{\infty }}{liminf}\left({b}_{n}-{c}_{n}\right)+d,$
and the inequality $ln\left(1+x\right)\le x$ ($-1), we have
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\\ \phantom{\rule{1em}{0ex}}\ge \underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[\frac{ln\left[1+\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]}{ln\lambda }-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\\ \phantom{\rule{2em}{0ex}}+\frac{r\left(\omega \right)}{ln\lambda }\\ \phantom{\rule{1em}{0ex}}\ge \underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[\frac{\left(\lambda -1\right){\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}}{ln\lambda }-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]+\frac{r\left(\omega \right)}{ln\lambda }\\ \phantom{\rule{1em}{0ex}}\ge b\left(\frac{\lambda -1}{ln\lambda }-1\right)+\frac{r\left(\omega \right)}{ln\lambda },\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right).\end{array}$
(27)
By using the inequality $1-\frac{1}{\lambda } and $ln\lambda <\lambda -1<0$ ($0<\lambda <1$), we have by (27)
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\ge b\left(\lambda -1\right)+\frac{r\left(\omega \right)}{\lambda -1},\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right)\cap A\left(1\right).\hfill \end{array}$
(28)
Let ${Q}_{\ast }$ be the set of rational numbers in the interval $\left(0,1\right)$, and let
${A}_{\ast }=\bigcap _{\lambda \in {Q}_{\ast }}A\left(\lambda \right),\phantom{\rule{2em}{0ex}}h\left(\lambda ,r\right)=b\left(\lambda -1\right)+\frac{r}{\lambda -1}.$
(29)
Then we have by (28)
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\ge h\left(\lambda ,r\left(\omega \right)\right),\phantom{\rule{1em}{0ex}}\omega \in {A}_{\ast }\cap A\left(1\right),\lambda \in {Q}_{\ast }.\hfill \end{array}$
(30)
Let $b>0$. It is easy to see that if $0, then $h\left(\lambda ,r\right)$ as a function of λ attains its largest value $h\left(1-\sqrt{\frac{r}{b}},r\right)=-2\sqrt{br}$ on the interval $\left(0,1\right)$, and $h\left(\lambda ,0\right)$ is increasing on the interval $\left(0,1\right)$ and ${lim}_{\lambda \to 1-0}h\left(\lambda ,0\right)=0$, and $h\left(\lambda ,b\right)=b\left(\lambda -1+\frac{1}{\lambda -1}\right)$ is decreasing on the interval $\left(0,1\right)$ and ${lim}_{\lambda \to {0}^{+}}h\left(\lambda ,b\right)=-2b$. For each $\omega \in {A}_{\ast }\cap A\left(1\right)\cap {D}_{1}$, take ${\tau }_{n}\left(\omega \right)\in {Q}_{\ast }$, $n=1,2,\dots$ such that ${\tau }_{n}\left(\omega \right)\to 1-\sqrt{\frac{r\left(\omega \right)}{b}}$. Then we have
$\underset{n\to \mathrm{\infty }}{lim}h\left({\tau }_{n}\left(\omega \right),r\left(\omega \right)\right)=-2\sqrt{br\left(\omega \right)}.$
(31)
By (30), we have
$\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\ge h\left({\tau }_{n}\left(\omega \right),r\left(\omega \right)\right),\phantom{\rule{1em}{0ex}}n=1,2,\dots .$
(32)
By (31) and (32),
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\ge -2\sqrt{br\left(\omega \right)},\phantom{\rule{1em}{0ex}}\omega \in {A}_{\ast }\cap A\left(1\right)\cap {D}_{1}.\hfill \end{array}$
(33)

Since $P\left({A}_{\ast }\cap A\left(1\right)\right)=1$, (5) holds by (33) when $b>0$.

When $b=0$, $r\left(\omega \right)=0$ for $\omega \in {D}_{1}\cap A\left(1\right)$, hence we have by (30)
$\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\ge 0,\phantom{\rule{1em}{0ex}}\omega \in A\left(\lambda \right)\cap A\left(1\right)\cap {D}_{1},$
(34)

since $P\left(A\left(\lambda \right)\cap A\left(1\right)\right)=1$, (5) also holds by (34) when $b=0$.

It is easy to see that when $0\le b, $h\left(\lambda ,r\right)$ as a function of λ is decreasing on the interval $\left(0,1\right)$ and ${lim}_{\lambda \to {0}^{+}}h\left(\lambda ,r\right)=-\left(r+b\right)$. For each $\omega \in {A}_{\ast }\cap A\left(1\right)\cap {D}_{2}$, when $r\left(\omega \right)\ne \mathrm{\infty }$, take ${\lambda }_{n}\left(\omega \right)\in {Q}_{\ast }$, $n=1,2,\dots$ , such that ${\lambda }_{n}\left(\omega \right)\to 0$. We have
$\underset{n\to \mathrm{\infty }}{lim}h\left({\lambda }_{n}\left(\omega \right),r\left(\omega \right)\right)=-r\left(\omega \right)-b.$
(35)
By (30), we have
$\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\ge h\left({\lambda }_{n}\left(\omega \right),r\left(\omega \right)\right),\phantom{\rule{1em}{0ex}}n=1,2,\dots .$
(36)
It follows from (35) and (36) that
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{liminf}\frac{1}{n}\sum _{k=1}^{n}\left[{I}_{{B}_{k}}\left({X}_{k}\right)-{\int }_{{B}_{k}}{p}_{k-1}\left({X}_{k-1},{x}_{k}\right)\phantom{\rule{0.2em}{0ex}}d{x}_{k}\right]\hfill \\ \phantom{\rule{1em}{0ex}}\ge -r\left(\omega \right)-b,\phantom{\rule{1em}{0ex}}\omega \in {A}_{\ast }\cap A\left(1\right)\cap {D}_{2}\hfill \end{array}$
(37)

when $r\left(\omega \right)=\mathrm{\infty }$, (37) also holds obviously. Since $P\left({A}_{\ast }\cap A\left(1\right)\right)=1$, (6) follows from (37) directly. □

## Declarations

### 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).

## Authors’ Affiliations

(1)
Faculty of Science, Jiangsu University, Zhenjiang, 212013, China

## References 