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Examensarbete

Large deviations on longest runs

Yurong Zhu

LITH - MAT - INT - G - - 2016/01 - - SE

Large deviations on longest runs

Mathematical Statistics, Linkopings Universitet

Yurong Zhu

LITH - MAT - INT - G - - 2016/01 - - SE

Examensarbete: 15 ECTS

Level: G

Supervisor: Xiangfeng Yang,Mathematical Statistics, Linkopings Universitet

Examiner: Xiangfeng Yang,Mathematical Statistics, Linkopings Universitet

Linkoping: June 2016

Abstract

The study on the longest stretch of consecutive successes in “random” trialsdates back to 1916 when the German philosopher Karl Marbe wrote a paperconcerning the longest stretch of consecutive births of children of the same sexas appearing in the birth register of a Bavarian town. The result was actuallyused by parents to “predict” the sex of their children. The longest stretch ofsame-sex births during that time in 200 thousand birth registrations was ac-tually 17 ≈ log2(200 × 103). During the past century, the research of longeststretch of consecutive successes (longest runs) has found applications in variousareas, especially in the theory of reliability.

The aim of this thesis is to study large deviations on longest runs in the settingof Markov chains. More precisely, we establish a general large deviation princi-ple for the longest success run in a two-state (success or failure) Markov chain.Our tool is based on a recent result regarding a general large deviation for thelongest success run in Bernoulli trails. It turns out that the main ingredient inthe proof is to implement several global and local estimates of the cumulativedistribution function of the longest success run.

Keywords: Large deviation principle, Markov chain, Reliability theory, K-out-of-n system.

URL for electronic version:

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Yurong Zhu, 2016. v

Acknowledgements

Foremost, I would like to express my deepest gratitude and appreciation to mysupervisor Professor Xiangfeng Yang. His great patience, persistent support,ingenious guidance and valuable advices benefited me a lot in the completionand success of my bachelor’s degree thesis study. How lucky I am having sucha good supervisor.

I would like to thank my beloved parents and brother for giving me couragethroughout writing this thesis. Besides, I’m grateful to my friend Dandan whohas accompanied me all the time and cheered me up. My thanks are alsoaddressed to another important friend Daniel for his continuous encouragementand suggestions on how to use LaTex.

Yurong Zhu, 2016. vii

Contents

1 Introduction 11.1 Background and purpose . . . . . . . . . . . . . . . . . . . . . . . 11.2 Structure of the thesis . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Markov chains 32.1 Discrete-time Markov chains . . . . . . . . . . . . . . . . . . . . . 32.2 Continuous-time Markov chains . . . . . . . . . . . . . . . . . . . 52.3 Applications of Markov chains . . . . . . . . . . . . . . . . . . . . 6

3 Large deviations on longest runs 93.1 The main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Auxiliary formulas . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Proof of Theorem 3.1.1 134.1 An important limit . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Bryc’s inverse Varadhan lemma . . . . . . . . . . . . . . . . . . . 164.3 End of proof of Theorem 3.1.1 . . . . . . . . . . . . . . . . . . . . 17

5 Conclusions 25

Yurong Zhu, 2016. ix

x Contents

Chapter 1

Introduction

1.1 Background and purpose

A consecutive-k-out-of-n: F system is a sequence of n ordered components wherethe system fails if and only if at least k consecutive components are failed. Thissystem has a wide range of applications, such as the design of integrated circuity[2], telecommunications and pipeline pumping stations [3], etc.

The longest success run L(n), namely the longest stretch of consecutive suc-cesses, has been applied in many fields, for instance: hypothesis testing, systemreliability, quality control and DNA sequences (cf. [1]). Therefore considerableattention has been attracted to study the distribution and limiting behaviorsof longest success runs in cases of Bernoulli trials, Markov dependent trials,and exchangeable binary trials (see for instance [5], [6], [7], [8], [9], [10], [11]and [18]), meanwhile we also refer to [12], [13], [14], [15] and [16] for the latestprogress on this aspect.

Since the longest runs in Bernoulli trails have been extensively and fully stud-ied, in this thesis we will focus on longest runs in Markov dependent trails -Markov chains. Let {Xk}1≤k≤n be the first n steps in a time-homogeneous two-state (success and failure) Markov chain with ‘1’ and ‘0’ representing ‘success’and ‘failure’ respectively. Moreover, we suppose that the initial distribution isP(X1 = 0) = p0,P(X1 = 1) = p1 = 1− p0, and the transition matrix is:

A =

( F S

F p00 p01S p10 p11

)We assume 0 < p0 < 1 and 0 < pij < 1(i, j = 0, 1), in order to avoid triviality.Let us define the longest success run L(n) as the longest stretch of consecutivesuccesses in the first n step of the chain. Although there is an explicit formulato express the exact distribution of L(n) (see [11]), we can hardly derive thelimiting distribution of L(n) as n → ∞ due to the complexity. Even for theIID (independent and identical distribution) case, there is no information wecan get as n → ∞. Therefore, offering appropriately probability estimating ofL(n) has been of great importance. In [14], a general large deviation principle

Yurong Zhu, 2016. 1

2 Chapter 1. Introduction

(LDP) was fully investigated for the longest success run L(n) in a sequence ofindependent Bernoulli trails. Likewise the aim of this thesis is to generalize theresults in [14] and establish a general LDP of L(n) in a Markov chain definedabove. To this end, let us recall several facts on L(n).

We firstly recall a law of large numbers (cf. [19] for example):

L(n)

log1/p11 n→ 1, as n→∞. (1.1.1)

The above convergence is in the sense of convergence in probability, that is,

P

(∣∣∣∣∣ L(n)

log1/p11 n− 1

∣∣∣∣∣ < ε

)→ 1, as n→∞ (1.1.2)

for any ε > 0. It is natural that such a law of large numbers suggests to study

the large deviation probabilities P(

L(n)log1/p11

n ∈ A), where A does not contain

the most probable point ‘1’. Such large deviation probabilities have been com-prehensively investigated in [15, Theorem 1.1].

Furthermore, the above law of large numbers also naturally leads to study LDP

for the family of normalized longest success runs { L(n)log1/p11

n}. Actually, such a

LDP, in particular, when p10 ≤ p00 + p11, has been investigated in [15] basedon the moment generating function of L(n). In [15], a LDP for another family

of normalized longest success runs {L(n)n } has been obtained as well. This LDP

corresponds to the law of large numbers L(n)n → 0 which is directly from (1.1.1).

It is obvious that log1/p11 n is the most accepted speed in LDP in view of (1.1.1),however, other speeds provide useful information as well, which can be seen from[15]. Based on this perspective, it makes sense to ask if a LDP can be establishfor other families of normalized longest success runs in a Markov chain? Acomplete answer to this question in Bernoulli trails is recently given in [14]. Inthis thesis, our aim to give a complete answer to the same question in the settingof Markov chains. More specifically, we will present a LDP for the family of

normalized longest success runs {L(n)α(n)} with a general speed α(n), which not

only includes those two LDPs derived in [15], but also gives new LDPs. Literally,all possible LDPs for normalized longest success runs in Markov chains can berecovered and unified in this thesis.

1.2 Structure of the thesis

Markov chains are the basic objects of the thesis, therefore in Chapter 2 we willmake a summary review on Markov chains in the setting of both discrete-timeand continuous-time, together with their possible applications. In Chapter 3the main result of the thesis will be presented, whose complete proof will begiven in Chapter 4. At the end of the thesis some conclusion remarks will begiven.

Chapter 2

Markov chains

Markov chains are the basic objects used in this thesis, and in this chapter webriefly recall the definitions and several properties of them. Loosely speaking,a Markov chain is a stochastic process that starts from one state and movessuccessively to another on a state space, and the current state can affect andonly affect the outcome of next state. Furthermore, it must possess a propertywhich is called Markov property (memorylessness): the probability distributionof next state depends only on current state and not on the sequence of eventsthat preceded it. In the literature, Markov chains consist of two different kinds:discrete-time Markov chain and continuous-time Markov chain. While usuallythe time parameter is discrete, i.e. a discrete-time Markov chain, the statespace may be arbitrary. However, in most situations, Markov chains employfinite or countably infinite (that is, discrete) state space, which have a morestraightforward statistical analysis. What we focus on in this thesis is exactlythe discrete-time, discrete state-space Markov chain. Since the system changesrandomly, it is generally impossible to predict with certainty the state of aMarkov Chain at a given point in the future. However, the statistical propertiescan be predicted, and these statistical properties are still of a lot of importance.

2.1 Discrete-time Markov chains

The definition of a Markov chain (discrete-time, discrete-state space Markovchain) can be simply given as follows: A Markov chain is a sequence of randomvariables X1, X2, X3, . . . Xi . . . with the Markov property. All possible valuesof Xi form a countable set S called the state space of the chain. The changesof states of the system are called transitions. If the current state is i in thechain, then it moves to the next state j with a probability denoted by pij , theprobabilities associated with state changes are called transition probabilities. Itshould be remarked that the transition probability depends only on the presentstate of the system and not on any previous states. Mathematically speaking, aMarkov chain is characterized by a state space, a transition matrix describing theprobabilities of particular transitions, and an initial state (or initial distribution)across the state space. Below is the formal definition of a Markov chain.

Definition 1. For a stochastic process {Xn, n ∈ N} to be a Markov chain, it isrequired that for any n ∈ N and any states i0, i1, . . . , in+1 ∈ S, the conditional

Yurong Zhu, 2016. 3

4 Chapter 2. Markov chains

probabilities satisfy:

P[Xn+1 = in+1|Xn = in, Xn−1 = in−1, · · · , X0 = i0] = P[Xn+1 = in+1|Xn = i]

where the conditional probabilities are well defined, i.e.

P(X0 = i0, X1 = i1, ..., Xn = in) > 0.

Based on the above definition, we can now define the following general transitionprobabilities as

Definition 2 (transition probabilities). Normally, the conditional probabilitiesare designated as transition probabilities, that is:

pij(n) = P{Xn+1 = j|Xn = i} where i, j ∈ S,

which denote the transition probability at time n.

In general, the transition probability is not only related to states i, or j, butalso associated with time n. If the transition probabilities in a Markov chain areindependent of n, then the chain is said to be time-homogeneous, i.e, pij(n) =pij . throughout the thesis, we will be restricted to time-homogeneous Markovchains. If we put all the transition probabilities together as a matrix, then wereach the so-called transition matrix:

A =

p11 p12 · · · p1j · · ·p21 p22 · · · p2j · · ·...

.... . .

......

pi1 pi2 · · · pij · · ·...

......

......

From the above definition, we can also derive the joint distribution as follows.

Property 1. Using the initial distribution of the chain and the transition prob-abilities, we can easily compute the probability distribution of the random vector(X0, · · · , Xn):

P(X0 = i0, · · · , Xn = in)

= P(Xn = in|X0 = i0, · · · , Xn−1 = in−1)

= P(Xn = in|Xn−1 = in−1)P(X0 = i0, · · · , Xn−1 = in−1)

= · · ·= P(Xn = in|Xn−1 = in−1)P(Xn−1 = in−1|Xn−2 = in−2) · · ·

P(X1 = i1|X0 = i0)P(X0 = i0)

We remark here that the transition probabilities should satisfy the followingrequirements.

Property 2. For transition probabilities, it holds that

pij > 0, i, j ∈ S,∑j∈S

pij = 1, i ∈ S

2.2. Continuous-time Markov chains 5

In summary, it is clear that we have three different ways to describe a Markovchain which can be seen from the following simple example.

• States transfer diagram

Figure 2.1: a Markov chain process

• Transition probabilities matrix

P =

1 0 0 0

1/3 1/3 1/3 00 1/3 1/3 1/30 0 1 0

• Function expression

pij = f(i, j)

where f(i, j) are given explicitly, for instance f(1, 1) = 1, f(2, 3) = 1/3,and so on.

2.2 Continuous-time Markov chains

The other type of Markov chains is continuous-time Markov chains. As before,loosely speaking a continuous-time Markov chain is a mathematical model whichtakes values in some finite state space and for which the time spent in eachstate takes non-negative real values and has an exponential distribution. It is acontinuous-time random process and also holds Markov property, which meansthe future outcomes of the model depends only on the current state of the modeland not on any historical outcomes. Below is a formal definition.

Definition 3. Suppose that {X(t), t ≥ 0} is a continuous-time stochastic pro-cess taking values in a state space S = {in, n ≥ 0}. For {X(t), t ≥ 0} to be atime-continuous Markov chain, it is required that for any i1, . . . , in+1 ∈ S andany 0 ≤ t1 < t2 < . . . < tn+1, it holds

P{X(tn+1) = in+1|X(t1) = i1, X(t2) = i2, . . . , X(tn) = in}=P{X(tn+1) = in+1|X(tn) = in}.


The transition probabilities can be similarly given as

Definition 4 (transition probabilities). For s, t ≥ 0, the transition probabilitiespij(t+ s, t) are defined as

pij(t+ s, t) = P(X(t+ s) = j|X(t) = i).

As the discrete-time case, if the transition probabilities pij(t + s, t) dependonly on the difference (t + s) − t = s, then the Markov chain is called time-homogeneous. Similarly, the joint distribution of a continuous-time Markovchains can be obtained as follows.

Property 3. Using the transition probabilities, we can easily compute the prob-ability distribution of the random vector (X(t0), · · · , X(tn)):

P(X(t0) = i0, · · · , X(tn) = in)

= P(X(tn) = in|X(t0) = i0, · · · , X(tn−1) = in−1)

= P(X(tn) = in|X(tn−1) = in−1)P(X(t0) = i0, · · · , X(tn−1) = in−1)

= · · ·= P(X(tn) = in|X(tn−1) = in−1)P(X(tn−1) = in−1|X(tn−2) = in−2) · · ·

P(X(t1) = i1|X(t0) = i0)P(X(t0) = i0)

= P(X(t0) = i0) · pi0i1(t1 − t0) . . . pin−1in(tn − tn−1).

2.3 Applications of Markov chains

Because of the Markov property: the future depends only on current (not onthe past), nowadays Markov chains find a wide range of applications includ-ing physics, chemistry, information science, economics and finance, statistics,medicine, music, game theory and sports. Here we would like to introduce oneof the applications which will in the theory of reliability. Such an applicationalso has close connections with the longest runs which are the main topic of thisthesis.

In a signal station, according to the outcome of current station, the outcomeof next station can be adjusted to ensuring successful signal transmissions inthe most effective way. Suppose that one station takes into account informationonly from the closest previous station, then the signal transmission process is aMarkov chain; see the following graph for such a scenario.

Figure 2.2: A Markov chain model

2.3. Applications of Markov chains 7

The Markov chain is determined by the current state it, (t = 0, 1, · · · , n) and thetransition matrix A. For this time-homogeneous two-state (success and failure)Markov chain model, the state for each signal station Xt, (t = 0, 1, · · · , n) onlytakes two possibilities, i.e.

it =

{0, failure,

1, success.

Then the transition matrix of {it} can be written as

A =

( F S

F p00 p01S p10 p11

).

Chapter 3

Large deviations on longestruns

Let {Xk}1≤k≤n be the first n steps in a time-homogeneous two-state (successand failure) Markov chain with ‘1’ and ‘0’ representing ‘success’ and ‘failure’respectively. Suppose that the initial distribution is P(X1 = 0) = p0,P(X1 =1) = p1 = 1− p0, and the transition matrix is:

A =

( F S

F p00 p01S p10 p11

)We assume 0 < p0 < 1 and 0 < pij < 1(i, j = 0, 1), in order to avoid triviality.In this chapter, we focus on the longest success run L(n) which is the longeststretch of consecutive successes in the first n step of the chain. More precisely,we will derive a general LDP which recovers the two special LDPs recentlyobtained in [15]. The main result of the thesis is formulated as follows.

3.1 The main result

Theorem 3.1.1. We assume p10 ≤ p00 + p11, and suppose that the speed α(n)satisfies the following two conditions:

(1). log1/p11 n ≤ α(n) ≤ n;(2). the limit limn→∞ ln(n)/α(n) =: β exists.

Then the family of normalized longest success runs {L(n)/α(n)} satisfies theLDP with the speed α(n) and a good rate function S(x) defined as

S(x) =

{x · ln(1/p11)− β, x ∈ D,+∞, x /∈ D,

(3.1.1)

where the interval D is given by

D = {x ∈ R : β/ ln(1/p11) ≤ x ≤ lim supn→∞

n/α(n)}.

That is,

Yurong Zhu, 2016. 9

10 Chapter 3. Large deviations on longest runs

(i) for any open set O ⊆ R,

lim infn→∞

1

α(n)ln P (L(n)/α(n) ∈ O) ≥ − inf

x∈OS(x); (3.1.2)

(ii) for any closed set F ⊆ R,

lim supn→∞

1

α(n)ln P (L(n)/α(n) ∈ F ) ≤ − inf

x∈FS(x). (3.1.3)

A few special cases of Theorem 3.1.1 are listed as follows, together with theirconnections with the LDPs obtained in [15].

• α(n) = log1/p11 n. In this case the interval D = {x : x ≥ 1}, β = ln(1/p11)and S(x) = (x− 1) ln(1/p11) for x ≥ 1. This LDP has been proved in [15].

• α(n) = n. The interval D = {x : 0 ≤ x ≤ 1}, β = 0 and S(x) = x ln(1/p11)for 0 ≤ x ≤ 1. This LDP has also been obtained in [15].

• α(n) = nα with 0 < α < 1. This is a new type of LDP and the speedis between log1/p11 n and n. The interval D = {x : x ≥ 0}, β = 0 andS(x) = x ln(1/p11) for x ≥ 0.

3.2 Auxiliary formulas

The proof of Theorem 3.1.1 will be based on the following simple auxiliaryformulas whose proofs are given explicitly for the purpose of clarity.

Lemma 3.2.1.

lim supα(n)→∞

1

α(n)ln (A+B) = max

[lim supα(n)→∞

1

α(n){ln(A), ln(B)}

].

Proof of Lemma 3.2.1. It holds that

ln (A+B) = ln {exp [ln(A)] + exp [ln(B)]}

= ln

{exp [ln(A)] ·

[1 +

exp [ln(B)]

exp [ln(A)]

]}= ln {exp [ln(A)] + ln [1 + exp [ln(B)− ln(A)]]}= ln(A) + ln {1 + exp [− (ln(A)− ln(B))]}= ln(B) + ln [1 + exp [− (ln(B)− ln(A))]]

= max{ln(A), ln(B)}+ ln {1 + exp [−| ln(A)− ln(B)|]},

therefore,

lim supα(n)→∞

1

α(n)ln (A+B)

= lim supα(n)→∞

1

α(n){max{ln(A), ln(B)}+ ln [1 + exp (−| ln(A)− ln(B)|)]}

= max

[lim supα(n)→∞

1

α(n){ln(A), ln(B)}

]

3.2. Auxiliary formulas 11

+ lim supα(n)→∞

1

α(n)ln {1 + exp [−| ln(A)− ln(B)|]}.

Trivially −| ln(A) − ln(B)| ≤ 0, which implies 0 ≤ exp [−| ln(A)− ln(B)|] ≤ 1.Thus,

lim supα(n)→∞

1

α(n)ln {1 + exp [−| ln(A)− ln(B)|]} → 0, as α(n)→ 0,

which yields that

lim supα(n)→∞

1

α(n)ln (A+B) = max

[lim supα(n)→∞

1

α(n){ln(A), ln(B)}

].

Lemma 3.2.2. For every a > b > 0, it holds that

ln(a− b) ≥ ln(a)− b

a− b.

Proof of Lemma 3.2.2. From Taylor series,

f(x) = f(c) +f′(c)

1!(x− c) +

f′′(c)

2!(x− c)2 + ...

We have, when f(x) = lnx,

lnx = ln c+1

c(x− c)− 1

2c2(x− c)2 + ...

≤ ln c+1

c(x− c)

It follows from replacing x, c with a, (a− b) respectively that

ln a ≤ ln(a− b) +b

a− b

=⇒ ln(a− b) ≥ ln(a)− b

a− b.

12 Chapter 3. Large deviations on longest runs

Chapter 4

Proof of Theorem 3.1.1

The proof is based on Bryc’s Inverse Varadhan Lemma (cf. [4]) and an appro-priate estimate on the limit of normalized longest success runs {L(n)/α(n)}.

4.1 An important limit

First of all, we would like to introduce an important limit which will be con-stantly used in the following sections.

Lemma 4.1.1. Under the assumption p10 ≤ p00 + p11, it holds that

limn→∞

1

α(n)ln P (L(n)/α(n) ≥ x) = −S(x),

where x > β/ ln(1/p11).

Proof of Lemma 4.1.1. If x > lim supn→∞ n/α(n), then P (L(n)/α(n) ≥ x) ≤P (L(n) > n) = 0, as n→∞, then Lemma 4.1.1 trivially holds. Therefore, nowwe assume x ∈ D and x 6= β/ ln(1/p11). Let k = [xα(n)], the following estimateshold (see [15, section 2.2])

1− (1− C4pk11)n−k+1 ≤ P (L(n) ≥ k) ≤ 1− (1− C5p

k11)n−k+1 + e(n), (4.1.1)

where

e(n) =C2(p01 + p10)

p01(p00 − p10)n−1

+C2(p01 + p10)

p11· (p00 − p10)n − pk−111 (p00 − p10)n−k

p00 − p10 − p11.

According to the properties of a Markov chain, we have

p00 + p01 = 1, p10 + p11 = 1

=⇒ p00 − p10 − p11 = −p01

=⇒ e(n) =C2(p01 + p10)

p01×

Yurong Zhu, 2016. 13

14 Chapter 4. Proof of Theorem 3.1.1

×

[(p00 − p10)n−1 − (p00 − p10)n − pk−111 (p00 − p10)n−k

p11

]

=⇒ e(n) =C2(p01 + p10)

p01×

×[(

1

p00 − p10− 1

p11)(p00 − p10)n + p−211 p

k11(p00 − p10)n−k

],

for C2 = p01(p0p01−p1p10)(p01+p10)2

. We use absolute value of e(n) and obtain

|e(n)| = |p0p01 − p1p10|p01 + p10

×

×(| 1

p00 − p10− 1

p11||p00 − p10|n + p−211 p

k11|p00 − p10|n−k

).

Obviously |e(n)| ≤ 1, and there must exist const1, const2 > 0 such that

|e(n)| ≤ const1 · |p00 − p10|n + const2·k11 |p00 − p10|n−k < 1.

It now follows that

lim supn→∞

1

α(n)ln |e(n)|

≤ lim supn→∞

1

α(n)ln(const1 · |p00 − p10|n + const2 · pk11|p00 − p10|n−k

)≤ max{− ln

1

|p00 − p10|, −x ln(1/p11)}

Since p10 ≤ p00 + p11,

=⇒ |p00 − p10| ≤ p11 ≤ 0

=⇒ 1

|p00 − p10|≥ 1

p11≥ 1

=⇒ ln1

|p00 − p10|≥ ln

1

p11

=⇒ ln1

|p00 − p10|≥ x ln

1

p11, where 0 < x < 1

=⇒ − ln1

|p00 − p10|≤ −x ln

1

p11

Finally, we obtain

lim supn→∞

1

α(n)ln |e(n)| ≤ −x ln(1/p11). (4.1.2)

On the other hand, by using 1− ax = −aθ · ln(a) · x with θ ∈ [0, x], we obtain

1− (1− C5pk11)n−k+1

=1−(

(1− C5pk11)1/(C5p

k11))C5p

k11(n−k+1)

4.1. An important limit 15

=−(

(1− C5pk11)1/(C5p

k11))θn· ln(

(1− C5pk11)1/(C5p

k11))· C5p

k11(n− k + 1)

where θn ∈ [0, C5pk11(n− k + 1)]. Notice that

pk11(n− k + 1) ≤ pxα(n)−111 · n =1

p11· pxα(n)−log1/p11

n

11

=1

p11· pα(n)(x−

log1/p11n

α(n))

11 .

With the assumpation x > β/ ln(1/p11), we can have

x > β/ ln(1/p11) =⇒ x > limn→∞

ln(n)/α(n)

ln(1/p11)=⇒ x >

log1/p11 n

α(n).

Namely, x− log1/p11n

α(n) > 0, for large enough n. This implies that

limn→∞

pk11(n− k + 1) = 0.

Then we can derive

1− (1− C5pk11)n−k+1

=−(

(1− C5pk11)1/(C5p

k11))θn· ln(

(1− C5pk11)1/(C5p

k11))· C5p

k11(n− k + 1)

=const4·k11 (n− k + 1)

Similarly, we can prove that

P (L(n) ≥ k) ≥ 1− (1− C4pk11)n−k+1 = const3 · pk11(n− k + 1).

It now follows that

limn→∞

1

α(n)ln {const3·k11 (n− k + 1)}

≤ limn→∞

1

α(n)P(L(n)

α(n)≥ x

)≤ limn→∞

1

α(n)ln {const4·k11 (n− k + 1) + e(n)};

x ln p11 + β

≤ limn→∞

1

α(n)P(L(n)

α(n)≥ x

)≤ max{(x ln p11 + β), x ln p11};

x ln p11 + β

≤ limn→∞

1

α(n)P(L(n)

α(n)≥ x

)≤ x ln p11 + β.

Now, we can easily take the limit and obtain

limn→∞

1

α(n)P(L(n)

α(n)≥ x

)= x ln p11 + β = −S(x),

which prove the lemma. Likewise, we can also prove

limn→∞

1

α(n)ln P (L(n)/α(n) > x) = −S(x)


for x > β/ ln(1/p11). To see this, we rewrite the equation (4.1.1) using k + 1instead of k when L(n)/α(n) > x. It follows now that

1− (1− C4pk+111 )n−k+2 ≤ P (L(n)/α(n) > x) = P (L(n) > k + 1)

≤ 1− (1− C5pk11 + 1)n−k+2 + e(n),

where

e(n) =C2(p01 + p10)

p01(p00 − p10)n−1

+C2(p01 + p10)

p11· (p00 − p10)n − pk11(p00 − p10)n−k−1

p00 − p10 − p11.

Then the conclusion trivially holds after taking n→∞.

4.2 Bryc’s inverse Varadhan lemma

Lemma 4.2.1 (Bryc’s Inverse Varadhan Lemma). If the family of probabilitymeasures of the normalized longest success runs {L(n)/α(n)} is exponentiallytight (whose definition can be found in Lemma 4.3.1 below), and assume thefollowing limit exists

Λ(φ) := limn→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}]for all continuous and bounded functions φ(x). Then the family {L(n)/α(n)}satisfies a LDP with a good rate function S(x) defined as

S(x) = supx∈R

[φ(x)− Λ(φ)] .

Furthermore we can derive

Λ(φ) = supx∈R

[φ(x)− S(x)] .

To pass from a continuous and bounded function φ to a probability, a roughand intuitive idea can be found in [17, Theorem 3.2]. Namely, the limit

limn→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}]= sup

x∈R[φ(x)− S(x)]

gives

E[exp

{φ

(L(n)

α(n)

)· α(n)

}]≈∑x∈R

exp[α(n)

(φ(x)− S(x)

)]for large enough n. So it follows∑

x∈Rexp[α(n)φ(x)] · P

(L(n)

α(n)= x

)≈∑x∈R

exp[α(n)

(φ(x)− S(x)

)]=⇒

∑x∈R

P(L(n)

α(n)= x

)≈∑x∈R

exp(− α(n)S(x)

),

It now trivially follows that

limn→∞

1

α(n)ln P

(L(n)

α(n)= x

)≈ − inf

x∈RS(x).

4.3. End of proof of Theorem 3.1.1 17

4.3 End of proof of Theorem 3.1.1

According to Bryc’s Inverse Varadhan Lemma, we now need to prove the expo-nentially tightness property for the family of probability measures of the nor-malized longest success runs {L(n)/α(n)}.

Lemma 4.3.1 (exponential tightness). Under the assumption p10 ≤ p00 + p11,the family of probability measures {µL(n)/α(n)}n≥1 is exponentially tight, that is,for any constant 0 < a <∞, there is a compact set Ka such that

lim supn→∞

1

α(n)ln P (L(n)/α(n) ∈ Kc

a) ≤ −a

where Kca denoting the complement of Ka.

Proof of Lemma 4.3.1. For any given constant 0 < a <∞, we define a compactset as

Ka = [0, max {(1 + β)/ ln(1/p11), (β + |a|)/ ln(1/p11)}] .

It follows from Lemma 4.1.1 that

limn→∞

1

α(n)ln P (L(n)/α(n) ∈ Kc

a)

= limn→∞

1

α(n)ln P

(L(n)

α(n)≥ max

{1 + β

ln(1/p11),

β + |a|ln(1/p11)

})+ 0

= −S(max

{1 + β

ln(1/p11),

β + |a|ln(1/p11)

})

≤ −[max

{1 + β

ln(1/p11),

β + |a|ln(1/p11)

}· ln(1/p11)− β

]≤ −|a| ≤ −a.

On the other hand, the proof of Theorem 3.1.1 is the limit in Lemma 4.2.1 asformulated below.

Lemma 4.3.2 (Laplace transform). Under the assumption p10 ≤ p00 + p11, itholds that for every continuous and bounded function φ(x),

limn→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}]= sup

x∈R[φ(x)− S(x)] .

Proof of Lemma 4.3.2. To achieve the limit, we prove the lower bound and theupper bound respectively.

Proof of the lower bound

It suffices to to prove that for every x ∈ R,

lim infn→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}]≥ [φ(x)− S(x)] . (4.3.1)


If x /∈ D, then S(x) = +∞, and the inequality (4.3.1) holds trivially. Whenx ∈ D, it is clear that for a small δ > 0, there is an ε > 0 such that

infy∈Ox

φ(y) ≥ φ(x)− δ, where Ox := (x− ε, x+ ε).

We consider the following estimates

lim infn→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}]≥ lim inf

n→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}· 1{L(n)/α(n)∈Ox}

]≥φ(x)− δ + lim inf

n→∞

1

α(n)ln P

(L(n)

α(n)∈ Ox

)=φ(x)− δ + lim inf

n→∞

1

α(n)ln

[P(L(n)

α(n)> x− ε

)− P

(L(n)

α(n)≥ x+ ε

)].

(i) For x = β/ ln(1/p11) = ln(n)/ (α(n) · ln(1/p11)) = log(1/p11) n/α(n), it isthen obviously that S(x) = 0, and based on equation (1.1.2),

P(L(n)

α(n)> x− ε

)= P

(L(n)

α(n)>

log(1/p11) n

α(n)− ε)

= P

(L(n)

log(1/p11) n> 1− ε · α(n)

log(1/p11) n

)

≥ P

(L(n)

log(1/p11) n= 1

)≈ 1.

Likewise,

P(L(n)

α(n)≥ x+ ε

)= P

(L(n)

α(n)≥

log(1/p11) n

α(n)+ ε

)= P

(L(n)

log(1/p11) n≥ 1 +

ε · α(n)

log(1/p11) n

)

= 1− P

(L(n)

log(1/p11) n< 1 +

ε · α(n)

log(1/p11) n

)

≤ 1− P

(L(n)

log(1/p11) n= 1

)≈ 0.

Therefore,

lim infn→∞

1

α(n)ln

[P(L(n)

α(n)> x− ε

)− P

(L(n)

α(n)≥ x+ ε

)]= lim inf

n→∞

1

α(n)ln (1− 0)


=0.

Furthermore, we obtain

lim infn→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}]≥φ(x)− δ → [φ(x)− S(x)] (with δ → 0).

(ii) For x > β/ ln(1/p), we can then apply Lemma 4.1.1 and the inequality inLemma 3.2.2, and derive

lim infn→∞

1

α(n)ln

[P(L(n)

α(n)> x− ε

)− P

(L(n)

α(n)≥ x+ ε

)]≥ lim inf

n→∞

1

α(n)

(ln

[P(L(n)

α(n)> x− ε

)]

−P(L(n)α(n) ≥ x+ ε

)P(L(n)α(n) > x− ε

)− P

(L(n)α(n) ≥ x+ ε

)

≥ lim infn→∞

1

α(n)ln

[P(L(n)

α(n)> x− ε

)]− lim sup

n→∞

1

α(n)

1

e(2ε ln(1/p11)−ε1−ε2)α(n)−1,

where the fact

P(L(n)α(n) ≥ x+ ε

)P(L(n)α(n) > x− ε

)− P

(L(n)α(n) ≥ x+ ε

) ≤ 1

e(2ε ln(1/p11)−ε1−ε2)α(n)−1(4.3.2)

comes from

P(L(n)α(n) ≥ x+ ε

)P(L(n)α(n) > x− ε

)− P

(L(n)α(n) ≥ x+ ε

) =1

P(L(n)α(n)

>x−ε)P(L(n)

α(n)≥x+ε)

− 1

.

More precisely, according to Lemma 4.1.1, it trivially holds that

limn→∞

ln P(L(n)

α(n)> x− ε

)= α(n) · [−S(x− ε)] = α(n) · [−(x− ε) · ln(1/p11) + β] ;

limn→∞

ln P(L(n)

α(n)≥ x+ ε

)= α(n) · [−S(x+ ε)] = α(n) · [−(x+ ε) · ln(1/p11) + β] ;

∃ ε1 makes limn→∞

ln P(L(n)

α(n)> x− ε

);

≥ α(n) · [−(x− ε) · ln(1/p11)− ε1 + β]

∃ ε2 makes limn→∞

ln P(L(n)

α(n)≥ x+ ε

)


≤ α(n) · [−(x+ ε) · ln(1/p11) + ε2 + β] .

Henceforth,

limn→∞

P(L(n)

α(n)> x− ε

)≥ eα(n)·[−(x−ε)·ln(1/p11)−ε1+β],

limn→∞

P(L(n)

α(n)≥ x+ ε

)≤ eα(n)·[−(x+ε)·ln(1/p11)+ε2+β].

Finally we arrive

limn→∞

P(L(n)α(n) > x− ε

)P(L(n)α(n) ≥ x+ ε

) ≥ eα(n)·[−(x−ε)·ln(1/p11)−ε1+β]

eα(n)·[−(x+ε)·ln(1/p11)+ε2+β]

= e(2ε ln(1/p11)−ε1−ε2)α(n).

Then the estimate (4.3.2) follows.

Now for small enough ε1 > 0 and ε2 > 0 such that 2ε ln(1/p11)− ε1− ε2 > 0, itfollows

lim supn→∞

1

α(n)

1

e(2ε ln(1/p11)−ε1−ε2)α(n)−1→ 0.

Therefore we obtain

lim infn→∞

1

α(n)ln

[P(L(n)

α(n)> x− ε

)− P

(L(n)

α(n)≥ x+ ε

)]≥ lim inf

n→∞

1

α(n)ln

[P(L(n)

α(n)> x− ε

)]=− S(x− ε)→ −S(x) (with ε→ 0).

Furthermore, the lower bound follows that

lim infn→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}]≥φ(x)− δ − S(x)→ [φ(x)− S(x)] (with δ → 0).

Proof of the upper bound

Now, we focus on proving the upper bound in Lemma 4.3.2, that is,

lim supn→∞

1

α(n)ln E

[exp

{φ

(L(n)

α(n)

)· α(n)

}]≤ sup

x∈R[φ(x)− S(x)] . (4.3.3)

For a constant a > 0 and a fixed small δ > 0, let us consider an interval

Iε :=

[β

ln(1/p11)− ε

2, max

{a+ β

ln(1/p11),

β + 1

ln(1/p11)

}],

and its partition

x0 =β

ln(1/p11)− ε

2,


x1 =β

ln(1/p11)+ε

2,

...

xN(ε) = max

{a+ β

ln(1/p11),

β + 1

ln(1/p11)

},

for some N(ε) depending on ε, where the small constant ε is chosen such that

φ(xi)− δ ≤ maxy∈[xi,xi+1]

φ(y) ≤ φ(xi) + δ, i = 0, 1, . . . , N − 1.

It now follows that,

E[exp

{φ

(L(n)

α(n)

)· α(n)

}]≤e‖φ‖·α(n) · P

(L(n)

α(n)∈ Icε

)+

N−1∑i=0

E[exp

{φ

(L(n)

α(n)

)· α(n)

}1{L(n)/α(n)∈[xi,xi+1]}

]

≤e‖φ‖·α(n) · P(L(n)

α(n)∈ Icε

)+

N−1∑i=0

e(φ(xi)+δ)α(n) · P(L(n)

α(n)≥ xi

).

Now Lemma 4.3.1 implies

lim supn→∞

1

α(n)ln P

(L(n)

α(n)∈ Icε

)= lim sup

n→∞

1

α(n)ln

[P(L(n)

α(n)<

β

ln(1/p11)− ε

2

)+P(L(n)

α(n)> max

{a+ β

ln(1/p11),

β + 1

ln(1/p11)

})]= max

{lim supn→∞

1

α(n)ln P

(L(n)

α(n)<

β

ln(1/p11)− ε

2

), −a

}= max{−∞, −a}=− a.

We remark here that negative infinity limit above is based on the fact that

P(L(n)

α(n)<

β

ln(1/p11)− ε

2

)= 0

for β = 0, and

P(L(n)

α(n)<

β

ln(1/p11)− ε

2

)≤ P

(L(n)

log1/p11 n< 1− ε

4

(β

ln(1/p11)+ 1

)−1)for β > 0. The results in [15] show that for each 0 < x < 1, it holds that

limn→∞

1

log1/p11 nln

[− ln P

(L(n)

log1/p11 n≤ 1− x

)]= x · ln(1/p11),


from which we can derive

=⇒ limn→∞

ln

[− ln P

(L(n)

log1/p11 n≤ 1− x

)]= ln(1/p11)x·log1/p11

n

=⇒ − limn→∞

ln P

(L(n)

log1/p11 n≤ 1− x

)= p

x·logp11 n11

=⇒ limn→∞

P

(L(n)

log1/p11 n≤ 1− x

)= e−p

logp11nx

11 = e−nx

.

Thus, with trivially 0 < ε4

(β

ln(1/p11)+ 1)−1

< 1,

lim supn→∞

1

α(n)ln P

(L(n)

α(n)<

β

ln(1/p11)− ε

2

)≤ lim sup

n→∞

1

α(n)ln e−n

x

= limn→∞

−nx

α(n)→ −∞.

This also implies P(L(n)α(n) <

βln(1/p11)

− ε2

)→ 0, which yields

P(L(n)

α(n)≥ x0

)= 1− P

(L(n)

α(n)< x0

)→ 1

=⇒ lim supn→∞

1

α(n)ln P

(L(n)

α(n)≥ x0

)= 0 = S(

β

ln(1/p11)).

Now the the finite sum can be estimated as, based on Lemma 4.1.1,

lim supn→∞

1

α(n)ln

[N−1∑i=0

e(φ(xi)+δ)α(n) · P(L(n)

α(n)≥ xi

)]

≤max{φ(x0) + δ + lim sup

n→∞

1

α(n)ln P

(L(n)

α(n)≥ x0

),

φ(xi) + δ − S(xi), i = 1, . . . , N − 1}

= max

{φ(x0) + δ − S(

β

ln(1/p11)), φ(xi) + δ − S(xi), i = 1, . . . , N − 1

}≤max

{φ(

β

ln(1/p11)) + 2δ − S(

β

ln(1/p11)),

φ(xi) + δ − S(xi), i = 1, . . . , N − 1}

≤ supx∈R

[φ(x)− S(x)] + 2δ → supx∈R

[φ(x)− S(x)], as δ → 0.

Now the proof follows

lim supn→∞

1

α(n)ln E

[exp{φ

(L(n)

α(n)

)· α(n)}

]≤max{ −a , sup

x∈R[φ(x)− S(x)] }

= supx∈R

[φ(x)− S(x)], as a→∞.


The upper bound is thus proved. Finally, Theorem 3.1.1 gets proved directlybased on the above Bryc’s Inverse Varadhan Lemma.

Chapter 5

Conclusions

In this thesis we first made a summary review on discrete-time and continuous-time Markov chains and some of their applications (see Chapter 2). Then weconsidered a time-homogenous two-state Markov chain, and focused on the mainobject of the thesis: the longest success run L(n), and formulated a LDP with ageneral speed which can recover two special LDPs recently derived in [15] (seeChapter 3). The proof of our main result (the above mentioned LDP with ageneral speed) was given in Chapter 4 based on Bryc’s Inverse Varadhan Lemma.

We want to stress that an assumption was imposed throughout the thesisp10 ≤ p00 + p11 in order to overcome several technical difficulties. In termsof the structure of the Markov chain, this assumption simply means that theprobability that the chain moves from ‘1’ to ‘0’ should be less than or equal tothe probability that the chain stays still. In this sense, this assumption seemsto be completely unnecessary. This would be one of my further research topics.


26 Chapter 5. Conclusions

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