Ultimate Ruin Probability in the Dependent Claim Sizes and Claim Occurrence Times Models

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2013-07-10 Ultimate Ruin Probability in the Dependent Claim Sizes and Claim Occurrence Times Models

Thompson, Emmanuel

Thompson, E. (2013). Ultimate Ruin Probability in the Dependent Claim Sizes and Claim Occurrence Times Models (Unpublished doctoral thesis). University of Calgary, Calgary, AB. doi:10.11575/PRISM/28541 http://hdl.handle.net/11023/801 doctoral thesis

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Ultimate Ruin Probability in the Dependent Claim Sizes and Claim Occurrence Times

Models

Emmanuel Thompson

A THESIS

SUBMITTED TO THE FACULTY OF GRADUATE STUDIES

IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE

DEGREE OF DOCTOR OF PHILOSOPHY

DEPARTMENT OF MATHEMATICS AND STATISTICS

CALGARY, ALBERTA

JULY, 2013

c Emmanuel Thompson 2013 Abstract

In the literature of risk theory, two risk models have emerged to have been studied extensively.

These are the Classical Compound Poisson risk model and the Continuous Time Renewal

risk model also referred to as Sparre-Andersen model. The focus of this thesis is the latter

and in the Sparre-Andersen risk model, the distribution of claim inter-arrival times (T ) and

the claim sizes (amounts) (X) are assumed to be independent. This assumption is not only

too restrictive but also inappropriate for some real world situations like collision coverages

under automobile insurance in cities with hash weather conditions.

In this thesis, the assumption of independence between T and X for diﬀerent classes of

bivariate distributions is relaxed prior to computing the ultimate ruin probability. Also, the

eﬀect of correlation on ruin probability is investigated. Correlation is introduced through

the use of the Moran and Downton’s bivariate Exponential distribution. The underlying

method in the entire modeling process is the Wiener-Hopf factorization technique, details of

which are covered in chapter 2.

The main results are covered in chapters 3, 4, 5, 6 and 7. In all of these chapters, we

assume T to follow an Exponential distribution while X is Gamma distributed with shape

parameter 2 in chapter 3. In chapter 4, X follows Gamma distribution with shape parameter

(m) where m is greater than 2. In chapter 5, a mixture of Exponential (Hyper Exponential) distributions is used to model X. The Pareto distribution is used to model X in chapter

6, in the modeling process however, the cummulative distribution function (CDF) of W =

X − cT where (T,X) is Exponential-Pareto distribution is approximated by the CDF of

Y = X − cT where (T,X) is Exponential-Hyper Exponential distribution and Exponential-

Gamma (m) distribution with the main estimation approach being method of moments estimation. Chapter 7 follows somewhat a diﬀerent approach by employing asymmetric

ﬁnite mixture of Double Exponential distributions and the EM algorithm prior to computing

i the ruin probability. In some of these situations, an explicit expression for the ultimate ruin probability is obtained and in chapters 3 and 5, we show that increasing correlation between T and X diminishes the impact of ruin probability. Also, in situations where an explicit expression for the ultimate ruin probability is not possible, excellent approximation is obtained. Chapter 8 summarizes the entire thesis, provides concluding remarks and future research. Acknowledgments

This thesis would not have been possible without the guidance and the help of several indi- viduals who in one way or another contributed and extended their valuable assistance in the preparation and completion of this study. First and foremost, my utmost gratitude to my su- pervisor, advisor and teacher, Professor Rohana S. Ambagaspitiya whose guidance, sincerity and encouragement I will never forget. I am grateful to my supervisory committee members, particularly Professor David Scollnik and Professor Alexandru Badescu for their concern and deep seated advice. I am highly indebted to Professor Abraham Olatunji Fapojuwo from

Faculty of Engineering, University of Calgary who accepted to serve as internal/external examiner and my profound appreciation also goes to Professor Cary Tsai from Simon Fraser university actuarial science program who accepted to serve as an external examiner to this thesis.

Professor Renate Scheidler, director of graduate studies in the department of Mathematics and Statistics had kind concern and consideration regarding my academic requirements to whom I say a big thank you. I am also thankful to the many wonderful people I came into contact with during both my Masters and PhD degree. Of particular reference is Niroshan

Withanage, Beilei Wu my colleagues, and all my oﬃce mates Shuang Lu, Xia Bing and Shiva

Gopal Wagle.

Last but not the least, my lovely wife Sandra Senam Addison, my parents and the one above all of us, the omnipresent God, for answering my prayers for giving me the strength to plod on despite my constitution wanting to give up and throw in the towel, thank you so much

Dear Lord.

iii Table of Contents

Abstract ...... i Acknowledgments ...... iii Table of Contents ...... iv List of Tables ...... vii List of Figures ...... viii List of Symbols ...... ix 1 Introduction ...... 1 1.1 Collective Risk Theory ...... 1 1.1.1 Historical Perspective ...... 1 1.2 Risk Models ...... 3 1.2.1 The Compound Binomial Model ...... 4 1.2.2 The Compound Poisson Model ...... 5 1.2.3 The Compound Mixed Poisson Model ...... 6 1.2.4 The Compound Negative Binomial Model ...... 6 1.3 Claim Size Distributions ...... 7 1.3.1 Light-Tailed Claim Distribution ...... 7 1.3.2 Heavy-Tailed Claim Distribution ...... 8 1.4 Arrival Process ...... 8 1.5 Risk Surplus Process and Ruin Probability ...... 10 1.5.1 Classical Compound Poisson Risk Model ...... 11 1.5.2 Continuous Time Renewal Process ...... 11 1.6 Mathematical Preliminaries ...... 14 1.6.1 Moran and Downton’s Bivariate Exponential Distribution ...... 14 1.6.2 Laplace Transform ...... 15 1.6.3 Inverse Laplace Transform ...... 15 1.6.4 Rational Function ...... 16 1.6.5 Locating Zeros of a Certain Polynomial ...... 16 2 Wiener-Hopf Factorization Method ...... 18 2.1 Introduction ...... 18 2.2 Ladder Processes ...... 18 2.3 Renewal Functions ...... 20 2.4 Maximum and Minimum ...... 24 2.5 Survival Probability ...... 27 3 Ruin Probability for Correlated Poisson Claim Count and Gamma (2) Risk Process ...... 30 3.1 Introduction ...... 30 3.1.1 Poisson Claim ...... 30 3.1.2 Relationship Between Poisson and Exponential Distribution . . . . . 31 3.2 Joint Moment Generating Function of (T,X)...... 31 3.3 Characteristic Function of Y and its Related Transform ...... 32 3.4 Zeros of the Polynomials g(s) and g(s) − z ...... 33 3.5 Factors and Properties of the Transform of Y ...... 36

iv 3.6 Survival Probability ...... 38 3.7 Eﬀect of Correlation on Ultimate Ruin Probability ...... 40 3.8 Numerical Examples ...... 40 4 Ruin Probability for Correlated Poisson Claim Count and Gamma (m) Risk Process ...... 44 4.1 Introduction ...... 44 4.2 Joint Moment Generating Function of (T,X)...... 44 4.3 Characteristic Function of Y and its Related Transform ...... 45 4.4 Zeros of the Polynomial g(s) − z ...... 46 4.4.1 Application of Rouche’s Theorem ...... 50 4.5 Factors and Properties of the Transform of Y ...... 53 4.6 Survival Probability ...... 54 4.7 Numerical Examples ...... 56 5 Ruin Probability for Correlated Poisson Claim Count and Hyper Exponential Risk Process ...... 59 5.1 Introduction ...... 59 5.1.1 Hyper Exponential Distribution ...... 59 5.2 Joint Moment Generating Function of (T,Xj) ...... 62 5.3 Characteristic Function of Y and Related Transform ...... 62 5.3.1 Zeros of a Certain m + 1 Degree Polynomial ...... 65 5.4 Survival Probability ...... 71 5.5 Numerical Examples ...... 73 6 Ruin Probability for Correlated Poisson Claim Count and Pareto Risk Process 77 6.1 Introduction ...... 77 6.1.1 Pareto Distribution ...... 77 6.1.2 Method of Moments Matching ...... 78 6.2 The Bivariate Exponential-Pareto Distribution ...... 79 6.3 Approximating Distributions ...... 83 6.3.1 Mixture of Exponential Distributions ...... 83 6.3.2 Gamma (m) Distribution ...... 84 6.4 Numerical Results ...... 86 7 Asymmetric Double Exponential Distributions and its Finite Mixtures . . . . 92 7.1 Introduction ...... 92 7.1.1 Double Exponential Distribution ...... 92 7.1.2 Finite Mixture Models ...... 93 7.2 Asymmetric Double Exponential Distribution ...... 94 7.2.1 MLE of Parameters ...... 94 7.3 Finite Mixture ...... 96 7.4 Numerical Results ...... 97 8 Summary, Concluding Remarks and Future Research ...... 115 8.1 Summary ...... 115 8.1.1 Chapter 2: Wiener-Hopf Factorization Method ...... 115 8.1.2 Chapter 3: Ruin Probability for Correlated Poisson Claim Count and Gamma (2) Risk Process ...... 116

v 8.1.3 Chapter 4: Ruin Probability for Correlated Poisson Claim Count and Gamma (m) Risk Process ...... 118 8.1.4 Chapter 5: Ruin Probability for Correlated Poisson Claim Count and Hyper Exponential Risk Process ...... 119 8.1.5 Chapter 6: Ruin Probability for Correlated Poisson Claim Count and Pareto Risk Process ...... 120 8.1.6 Chapter 7: Asymmetric Double Exponential Distributions and its Fi- nite Mixtures ...... 121 8.2 Concluding Remarks ...... 121 8.3 Future Research ...... 122 Bibliography ...... 123 A Maple and Matlab codes ...... 128 A.1 Maple Codes ...... 128 A.2 Matlab Codes ...... 148

vi List of Tables

1.1 Examples of light-tailed claim distributions ...... 8 1.2 Examples of heavy-tailed claim distributions ...... 8

5.1 βj against aj ...... 66 5.2 βj against bj ...... 66 7.1 Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 2 and simulated sample size = 10000. . . 99 7.2 Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 2 and simulated sample size = 15000. . . 102 7.3 Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 2 and simulated sample size = 20000. . . 105 7.4 Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 3 and simulated sample size = 15000. . . 108 7.5 Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 3 and simulated sample size = 20000. . . 111

vii List of Figures and Illustrations

3.1 Graph of g(s) and 0 < z < 1...... 36 3.2 Eﬀect of correlation on ruin probability ...... 43

4.1 Graph of g(s) for the case m = 4, 6, ...... 48 4.2 Graph of g(s) for the case m = 3, 5, ...... 49

5.1 Graph of two and three component Hyper Exponential distributions . . . . . 61 5.2 Eﬀect of correlation on ruin probability in a two-component Hyper Exponen- tial distribution ...... 75 5.3 Eﬀect of correlation on ruin probability in a three-component Hyper Expo- nential distribution ...... 76

6.1 Graph of Pareto density function ...... 78 6.2 Comparison of Simulated ruin probability and Approximated ruin probability in scenario 1...... 87 6.3 Comparison of Simulated ruin probability and Approximated ruin probability in scenario 2...... 88 6.4 Comparison of Simulated ruin probability and Approximated ruin probability in scenario 3...... 89 6.5 Comparison of Simulated ruin probability and Approximated ruin probability in scenario 4...... 90

7.1 Comparison of Simulated ruin probability and Approximated ruin probability for a two component when simulation sample size is 10000...... 100 7.2 Comparison of Simulated ruin probability and Approximated ruin probability for a two component when simulation sample size is 15000...... 103 7.3 Comparison of Simulated ruin probability and Approximated ruin probability for a two component when simulation sample size is 20000...... 106 7.4 Comparison of Simulated ruin probability and Approximated ruin probability for a three component when simulation sample size is 15000...... 109 7.5 Comparison of Simulated ruin probability and Approximated ruin probability for a three component when simulation sample size is 20000...... 112

viii List of Symbols, Abbreviations and Nomenclature

Symbol Deﬁnition

ATM Automated teller machine

CDF Cummulative distribution function cf Characteristic function i.i.d. Independent and identically distributed

Im(ω) Imaginary part of ω

Re(ω) Real part of ω max Maximum mgf Moment generating function min Minimum PDF Probability density function

U of C University of Calgary

Γ(γ, β) Gamma distribution with shape γ and rate β

MLE Maximum likelihood estimate

ix Chapter 1

Introduction

1.1 Collective Risk Theory

Broadly speaking, the business of insurance is exposed to two major risks (uncertainties).

Those risks attributed to general economic fluctuations and poor investments are classified as commercial risks. However, risk fluctuations due to the difference between actual claim sizes

(amounts) and expected claim sizes (amounts) are referred to as insurance risk. To analyze the random movements and to investigate the related mathematical risks, European actuaries have developed a considerable body of mathematical knowledge known as the theory of risk, which consequently seeks to prescribe how an insurance business may be protected from the adverse eﬀects of these stochastic movements (Kahn, 1962).

There are two angles of view from which risk theory may be studied and understood, the collective and the individual. To examine the gain or loss on a whole portfolio, individual risk theory proceeds initially by obtaining the gain or loss on each individual policy; then by summing these individual gain or loss it furnishes information about the aggregate gains or losses on all the policies in the whole portfolio. In collective risk theory however, the emphasis is to examine directly the risk of the insurance undertaking as a whole. Essentially, interest is concentrated on the size (amount) of the aggregate claims or the aggregate gains or losses emanating from all the policies in the portfolio under review.

1.1.1 Historical Perspective

The collective risk theory was ﬁrst introduced into the actuarial literature by Cramer and

Lundberg in early 1900s (Lundberg (1909) and Cramer (1930)). The classical theory emerged as a Compound Poisson Risk model, which embodies an initial capital plus collected pre-

1 miums less paid claims. The theory assumes that the time of occurrence of claims are independent of the claim sizes (amounts). The picture being painted is that the time until the next claim occurs is unrelated to the size of that claim. Another basic but equally necessary assumption is the net-proﬁt condition. This necessary condition states that, on the average the time value of collected premiums must be bigger than the claims paid. The classical model was simple enough to calculate probabilities of interest, however with the passage of time, it has witnessed tremendous improvement. In spite of this positive stride, the classical model is still important because it serves not only as a building block for ruin problems and related quantities but also, provides an explanation to the two major causes of big losses: frequent claims and large claims. Realistically, most of the techniques developed for the Cramer-Lundberg (classical) model are useful for the more realistic renewal risk model.

In an insurance risk model, the ruin probability is the subject of analysis. However, explicit expressions for the ruin probability are very difficult to obtain hence, bounds and approx- imations to the probability of ruin quite often than not are rather investigated. At the beginning, the primary emphasis of investigation was on specific conditions on the claim size distribution (Cramer (1930) and Gerber (1973)). In subsequent years, some studies extended the focus by examining the effect of adding a perturbation model by a diffusion on the classical model. Ma and Sun (2003), Yuen et al. (2004) and Pergamenshchikov and

Zeitouny (2006) are among few references.

In 1957, Sparre-Andersen proposed a more realistic process, by using a more general distribution to model the inter-arrival times. The interest then swinged towards the choice of the distribution of the inter-arrival times and this marked the genesis of a new theory.

Sparre-Andersen’s model permits a more general distribution for the inter-arrival times but retains the Cramer-Lundberg assumptions on the claim sizes. Bounds and asymptotics on the time value of ruin were derived under this new assumption regarding the model. For

2 example, it was shown that for any inter-arrival time distribution, the probability of ruin still has an exponential bound (Andersen (1957)). Further Lundberg type bounds were obtained for the joint distribution of the surplus immediately before and at ruin, by building an exponential martingale (Ng and Yang (2005)). For particular distributions for the inter-arrival times, the papers of Dickson and Hipp (1998), Dickson (2002), Li and Garrido (2004) and

Gerber and Shiu (2005) analyzed either the asymptotic behavior of the probability of ruin or the moments of the time of ruin. The subject of the most recent analyses is the eﬀect of a perturbation. Assume that a model is perturbed by an independent diﬀusion process and the inter-arrival time is Erlang(n), a generalization of the defective renewal equation is presented in Li and Garrido (2003).

Also, Gerber and Shiu (1998) paper which saw a uniﬁed approach of computing ruin probabilities through the discounted penalty function also extended the study of some classes of the Sparre-Andersen model. Dickson and Hipp (2001), Gerber and Shiu (2005) and Li and

Garrido (2004, 2005) are among few references.

1.2 Risk Models

In this section, we review the Compound Binomial, Compound Poisson, Compound Mixed

Poisson and the Compound Negative Binomial risk models.

We proceed by ﬁrst considering in a collective case an insurance contract in some ﬁxed time period (0,T ]. We let N denote the number of claims in (0,T ]. Then,

N X S = Xi (1.1) i=1 is the aggregate claims. The following are however assumed.

• N and (X1,X2, ...) are independent.

• X1,X2, ... are independent.

• X1,X2, ... are identically distributed with distribution function, say F .

3 • The random sum (S) has a distribution function F ∗n(s).

rXi n Let MX (r) = E[e ] and µn = E[X1 ]. The distribution of S is given as:

P [S ≤ s] = E[P (S ≤ s|N)] ∞ X = P [S ≤ s|N = n]P [N = n] n=0 ∞ X = P [N = n]F ∗n(s). (1.2) n=0 The distribution of the random sum (S) is called compound distribution. Applying condi- tioning and the law of iterative expectation, it is easy to see that the ﬁrst moment E(S) and

2 variance V ar(S) are E(S) = E(N)E(X1) and V ar(S) = V ar(N)[E(X1)] + E(N)V ar(X1) respectively. The moment generating function MS(r) is given as:

rS MS(r) = E e

h r PN X i = E e i=1 i

" N # Y = E erXi i=1 " " N ## Y = E E erXi |N i=1 " N # Y = E MX (r) i=1 N = E (MX (r))

= E eNlog(MX (r))

= MN [log(MX (r))] . (1.3)

Note that MN (r) is the moment generating function of N.

1.2.1 The Compound Binomial Model

The Compound Binomial model is of the form shown in (1.3) when the following assumptions are made:

4 • the entire risk consists of several single independent risks.

• the time interval in question can be partitioned into inﬁnitely many indepen-

dent smaller intervals.

• there is at most one claim per single risk and interval.

Assume that the probability of a single claim in an interval is p. Based on the above

assumptions, it then follows that N ∼ B(n, p) for some n ∈ N. In this setup too, it is obvious that the ﬁrst moment, the variance and the moment generating function of S can be respectively shown as:

E(S) = npE(X1)

2 2 V ar(S) = np[E(X1 ) − pE(X1) ]

n MS(r) = [pMX (r) + 1 − p] .

1.2.2 The Compound Poisson Model

The Compound Poisson model results when we consider the Compound Binomial model

alongside the following assumption:

• for n → ∞ as p → 0 and λ = np remains inﬁnitely large.

If we let λ = np, then

λ lim pet + (1 − p)n = lim [1 − (1 − et)]n n→∞ n→∞ n − λ 1−et = e n ( ). (1.4)

As a consequence of (1.4)

N ∼ P oi(λ).

5 The ﬁrst moment, variance and moment generating function of S can be derived as:

E(S) = λE(X1)

2 V ar(S) = λE(X1 )

λ(MX (r)−1) MS(r) = e .

1.2.3 The Compound Mixed Poisson Model

One main weakness of the Compound Poisson model is that its distribution is always skewed to the right. The Compound Mixed Poisson model overcomes this disadvantage by allowing the parameter λ to be stochastic. Assume G represents the distribution function of λ. Then,

P [N = n] = E[P (N = n|λ)] λn = E e−λ n! Z ∞ vn = e−vdG(v). (1.5) 0 n!

The ﬁrst moment, variance and the moment generating function are given respectively as:

E(S) = E(λ)E(X1)

2 2 V ar(S) = V ar(λ)[E(X1)] + E(λ)E(X1 )

MS(r) = Mλ(MX (r) − 1).

1.2.4 The Compound Negative Binomial Model

Consider the case of a Compound Mixed Poisson model where λ is Gamma distributed with shape γ and scale β i.e. λ ∼ Γ(γ, β). It follows that,

r MN (r) = Mλ(e − 1) β γ = β − (er − 1) β !γ = β+1 . (1.6) β r 1 − (1 − β+1 )e

6 In this setup, N is a Negative Binomial random variable. The recognition that a Poisson distribution does not always ﬁt a real data soley due to over-dispersion (the situation where variance is greater than the mean). Therefore estimating parameters using the Negative

Binomial distribution yields robust estimates and better results (Jong and Heller (2008)).

The ﬁrst moment, variance and moment generating functions are well presented in most standard textbooks on mathematical statistics. We have however provided them to ease reference. Assume N ∼ NB(a, p), then

a(1 − p) E(S) = E(X ) p 1 a(1 − p)E(X )2 a(1 − p) V ar(S) = 1 + E(X2) − E(X )2 p2 p 1 1 p a MS(r) = . 1 − (1 − p)MX (r)

1.3 Claim Size Distributions

In this section, we consider classes of distributions that have been used to model claim sizes (amounts). Now, we distinguish between two types of distributions:- light-tailed and heavy-tailed claim distributions.

1.3.1 Light-Tailed Claim Distribution

A distribution function FX (x) is described as light-tailed, if there exist constants u > 0, v > 0 such that

c −vx FX (x) = 1 − FX (x) ≤ ue .

Or equivalently, if there exist z > 0, such that MX (z) < ∞, where MX (z) denote the moment

c generating function of X. Note that FX (x) denote the survival function. Table 1.1 provides some examples of light-tailed claim distributions.

7 Table 1.1: Examples of light-tailed claim distributions

Name Parameters fX (x); x ≥ 0 Exponential β > 0 βexp(−βx) βα α−1 Gamma α > 0, β > 0 Γ(α) x exp(−βx) Weibull β > 0, τ ≥ 1 βτxτ−1exp(−βxτ ) Pn Pn Mixed Exponential βi > 0, i=1 ai = 1 i=1 aiβiexp(−βix)

Table 1.2: Examples of heavy-tailed claim distributions

Name Parameters fX (x); x ≥ 0 Weibull β > 0, 0 < τ < 1 βτxτ−1exp(−βxτ )) ln(x−µ)2 Log-normal µ ∈ R, σ > 0 √ 1 exp − 2πσx 2σ2 α λ α Pareto α > 0, λ > 0 λ+x λ+x 0 ατλαxτ−1 Burr α > 0, λ > 0 τ > 0 (λ+xτ )(α+1)

1.3.2 Heavy-Tailed Claim Distribution

c Distribution FX (x) is described as heavy-tailed, if for all u > 0, v > 0 such that FX (x) >

−vx ue , or equivalently, if for all z > 0, MX (z) = ∞. Some examples of well known heavy- tailed claim distributions are presented in Table 1.2. Both light-tailed and heavy-tailed claim distributions relevant in this thesis will be discussed in some greater details in subsequent chapters where applicable.

1.4 Arrival Process

Aside from the claim size distribution, another equally important process in risk modeling is the claim arrival process or claim count process. A claim count process is a counting process

{N(t); t ≥ 0} and is formally deﬁned as a stochastic process with the following properties

(Lefebvre (2007)):

• N(t) is a random variable with possible values as 0,1,...

8 • The function N(t) is nondecreasing: N(t2) − N(t1) ≥ 0 if t2 > 0, t1 ≥ 0.

• N(t2) − N(t1) is the number of events that occurred in the interval (t1, t2].

The arrival of a car at a washing bay, the arrival of a bank customer at an ATM queue,

the arrival of a new claim at an insurance company are examples of events which counting

processes can easily model. The Compound Poisson process is the most extensively studied

risk model in the actuarial literature where N(t) the claim count process is a Poisson process.

A Poisson process with rate λ > 0 (λ is constant) is a counting process {N(t) ≥ 0} having

independent increments for which

• N(0) = 0 and

• N(τ + t) − N(τ) ∼ P oi(λt) ∀ τ, t ≥ 0.

• Also, stationary increments as the distribution of N(τ + t) − N(τ) is indepen-

dent of τ.

Therefore, if {N(t) ≥ 0} is a process with independent and stationary increments, then the following situations hold.

P [N(s) = 1] = λs + o(s) and

P [N(s) = 0] = 1 − λs + o(s) where o(s) is such that o(s) lim = 0. s↓0 s This implies that the probability that there will be exactly one event in an interval of length s

must be proportional to the length of the interval in addition to a term that is inﬁnitesimally

negligible if s is suﬃciently small. Moreover, we have

P [N(s) ≥ 2] = 1 − {P [N(s) = 0] + P [N(s) = 1]}

= o(s).

9 Now, we have enough background to state the risk surplus process {U(t); t ≥ 0}.

1.5 Risk Surplus Process and Ruin Probability

The risk surplus process {U(t); t ≥ 0} is deﬁned as a model for the time evolution of the surplus of an insurance company. It can best be described as a model of accumulation of an insurer’s capital. In practice, an insurance company commences business with a start-up capital of u, usually called the initial capital (surplus). It collects premiums of c per unit period at the end of each period. Claim size (amount) Xi is paid out for i = 1, ..., n. Xi’s are assumed to be independent and identically distributed (i.i.d.) random variables. In this set-up we have a discrete-time model for which the surplus at time t with initial capital of u is given as

n(t) X U(t) = u + ct − Xi, t = 1, 2, ... (1.7) i=1 The insurance company is said to be in ruin if the U(t) ≤ 0 at a period t ≥ 1. In the discrete case, we deﬁne the time of ruin random variable T (u) as

T (u) = min{t ≥ 1 : U(t) ≤ 0|U0 = u}. (1.8)

Therefore, given an initial surplus u, the probability of ruin denoted ψ(u) also referred to as inﬁnite horizon probability of ruin is given as

ψ(u) = P (T (u) < ∞). (1.9)

Given an initial capital u, the ﬁnite horizon probability of ruin by time t is

ψt(u) = P (T (u) ≤ t). (1.10)

In the continuous-time model, the insurance company concerned continuously collects premi-

ums and losses in respect of claims may happen at any time. Here, we represent the number

10 of claims by a counting process N(t) in the interval (0, t] with claim amounts X1, ..., XN(t) are i.i.d. random variables. Therefore the risk surplus process becomes N(t) X U(t) = u + ct − Xi (1.11) i=1 The definition of both infinite and finite horizon probability of ruin are analogous to the ones under the discrete-time model with just some slight modification. We only change t from discrete to continuous. That is to say we have t > 0 instead of t ≥ 1. In this thesis our concern is the continuous-time model. Broadly speaking, two particular type of risk surplus processes have received extensive attention in the field of actuarial science. These are the classical Compound Poisson and the Continuous Time Renewal process also called the Sparre-Andersen model.

1.5.1 Classical Compound Poisson Risk Model

We now consider a unique case of the risk surplus process U(t) in continuous time as given by (1.11). In the classical theory, it is assumed that the number of claims N(t) is a Poisson process as indicated earlier and it is independent of the claim size (amount) Xi. Both N(t) and Xi are assumed to be i.i.d. random variables. Complying with these assumptions leads to the Classical Compound Poisson risk model.

1.5.2 Continuous Time Renewal Process

Pi Let us define T1 as the time epoch of first claim and k=1 Tk as the time epoch of the ith claim; i.e. we are defining T1,T2,...,Ti as the inter-arrival (interclaim) times. We therefore can view the classical risk process as the processes with exponentially distributed interclaim times. Andersen (1957) relaxed the exponentially distributed interclaim times assumption to accommodate more general classes of distributions. In this context, the risk surplus process is of the form: n n X X U(t) = u + (cTi − Xi) + c(t − Ti) (1.12) i=1 i=1

11 where T1 + T2 + ... + Tn ≤ t < T1 + T2 + ... + Tn+1, with U(t) = u + ct if t < T1. Andersen (1957) considered the corresponding ruin probability for inﬁnite time period and derived an integral equation in respect of that. Ever since, this model is known as the Sparre-Andersen model or renewal risk process and it has been studied extensively by many researchers. For instance, Dufresne and Gerber (1988), Dickson (1992, 1993) , Gerber and Shiu (1997), and

Dickson and Hipp (1998) are among few references where ruin probabilities and many related quantities such as the marginal and joint distributions of the time to ruin, the deﬁcit at ruin, surplus prior to ruin and claim size causing ruin have been studied within the general framework of the Sparre-Andersen model.

In the Sparre-Andersen model, the distribution of Ti and Xi are assumed to be independent. This assumption is viewed to be too restrictive and quite apart from that, there are certain real world situations where it is questionable if not inappropriate. A credible example where the independence assumption is contestable is the case of collision coverages under automobile insurance in districts, cities, communities, regions, provinces or countries with severe weather conditions. Recent developments clearly indicate that prominent researchers have sought to extend the classical Compound Poisson risk model by relaxing the independence assumption. Albrecher and Boxma (2004) have proposed that the distribution of the time between two successive claims depends on the previous claim size. Albrecher and Teugels

(2006) sought to introduce an arbitrary dependence structure among the interclaim time and the claim size prior to deriving the asymptotic ﬁnite time and inﬁnite time ruin probabilities.

Boudreault et al. (2006) proposed that the distribution of the next claim size depends on the time elapsed since the last claim. Ambagaspitiya (2009) considered two classes of bivariate distribution to model the joint distribution of the interclaim time and the claim size. This thesis also sought to relax the independence assumption by following completely a diﬀerent approach. To be more speciﬁc, our objectives are set at extending dependency between the

T and X to diﬀerent types of distributions, using diﬀerent marginals. We then derive the

12 exact expressions for the ultimate ruin probability and then establish the eﬀect of correlation on ruin probability. Ultimate ruin probability is the inﬁnite horizon probability of ruin given in (1.9).

In our derivations as it may be seen, we introduce dependence in a form of a bivariate exponential distribution for simplicity. A number of bivariate exponential distributions exist in the literature. For instance, Chapter 47 of Kotz, Balakrishnan and Johnson (2000) provides

12 diﬀerent distributions. In this thesis, we chose Moran and Downton’s bivariate exponential distribution because it has the explicit moment generating function.

Here is how the rest of this thesis is organized. In chapter 2 we discuss the Wiener-Hopf factorization technique. In chapter 3 we highlight the results for the case where T is Expo- nentially distributed (Poisson claim) and X is Gamma distributed with shape parameter 2.

In chapter 4 we provide the results for the case where T is Exponentially distributed (Pois- son claim) and X is Gamma distributed with shape parameter m where m > 2. We further show the results for where T is Exponential but X follows a Hyper Exponential distribution in chapter 5. Chapter 6 discusses the case where T is Exponentially distributed but X is a

Pareto distribution, in the modeling process however, the cummulative distribution function

(CDF) of W = X − cT where (T,X) is Exponential-Pareto distribution is approximated by the CDF of Y = X − cT where (T,X) is Exponential-Hyper Exponential distribution and

Exponential-Gamma (m) distribution with the main estimation approach being method of moments estimation. Chapter 7 follows somewhat a diﬀerent approach by employing asymmetric ﬁnite mixture of Double Exponential distributions and the EM algorithm prior to computing the ruin probability. Summary, concluding remarks and future research are given in chapter 8.

13 1.6 Mathematical Preliminaries

Prior to delving deeper, we introduce some mathematical deﬁnitions and results central to

this thesis.

1.6.1 Moran and Downton’s Bivariate Exponential Distribution

When two marginal distributions both being exponential are modeled jointly then we have

a bivariate exponential distribution.

The joint density function of (T,X) is given as: √ θ θ 2 ρθ θ tx θ t + θ x f (t, x) = 1 2 I 1 2 exp − 1 2 (1.13) T,X 1 − ρ 0 1 − ρ 1 − ρ

( z )2r where t, x > 0, θ , θ > 0, 0 ≤ ρ ≤ 1 and I (z) = P∞ 2 is the well known modiﬁed 1 2 0 r=0 (r!)2 Bessel function of the ﬁrst kind of order 0. The parameter ρ measures the positive correlation

between Ti and Xi with ρ = 0 implies independence between Ti and Xi. In this distribution,

estimating the parameters θ1 and θ2 are quite straightforward, as the marginal distributions

of T and X are exponentially distributed with parameters θ1 and θ2 respectively. However, Kotz, Balakrishnan and Johnson (2000) speciﬁes a number of estimators for ρ.

The bivariate random variable (T,X) are generated by ﬁrst setting

1 2 2 T = (U1 + U2 ) 2θ1

and

1 2 2 X = (U3 + U4 ) 2θ2

where Ui (i = 1, 2, 3, 4) is a standard normal random variable, and (U1,U3) and (U2,U4) each has a bivariate normal distribution with correlation coeﬃcient ρ. Then the distribution of

(T,X) is Moran and Downton’s bivariate exponential and the moment generating function

14 takes the following form:

t1T +t2X MT,X (t1, t2) = E[e ] θ θ = 1 2 . (1.14) (θ1 − t1)(θ2 − t2) − ρt1t2

1.6.2 Laplace Transform

In defining the Laplace transform, we first recall the definition of an improper integral from calculus. If δ is an integral over the interval [a, T ] for every T > a, then the improper integral of δ over [a, ∞) is defined as Z ∞ Z T δ(u)du = lim δ(u)du. (1.15) a T →∞ a We say that the improper integral converges if the limit of (1.15) exists; otherwise it diverges or does not exist (Trench (2000)).

Suppose that δ(u) is deﬁned for all u ≥ 0. The Laplace transform of δ is the function Z ∞ δ∗(s) = e−suδ(u)du. (1.16) 0 Theorem 8.1.2 (page 375) in Trench (2000) presents an important property of the Laplace transform. A less complex version is also shown in Asmar (2000) under Theorem 2 (page

380). Both theorems emphasize the linearity property of the Laplace transform.

∗ If δi is deﬁned for s > 0 and let c1, c2, ..., cn be constants. Then

∗ ∗ ∗ (c1δ1 + ... + cnδn) = c1δ1 + ... + cnδn. (1.17)

1.6.3 Inverse Laplace Transform

It is important to recognize that δ(u) can be recovered and uniquely determined from its

Laplace transform δ∗(s). Since we have already deﬁned the Laplace transform of δ(u) as shown in (1.16), we can comfortably and conﬁdently say that δ is an inverse Laplace transform of δ∗ and write

δ = `−1[δ∗] (1.18)

15 where `−1 is the inverse Laplace transform and is also a linear operator.

1.6.4 Rational Function

If P (s) and Q(s) are two polynomial functions such that

P (s) δ∗(s) = (1.19) Q(s) is a proper rational function of s where the degree of P is less than the degree of Q. Consider a rational function of the form

∗ P (s) δ (s) = m m (1.20) (s − s1) 1 ...(s − sk) k then the associated partial fraction expression is of the form

∗ A1 A2 B1 B2 δ (s) = + 2 + ... + + 2 + ... (1.21) (s − s1) (s − s1) (s − s2) (s − s2) where A1,A2..., B1,B2, ... are constants and s1, s2, ... are the zeros of Q(s). It follows that applying the linear operator to (1.21), we recover δ(u) as shown.

δ(u) = `−1[δ∗(s)] (1.22)

s1u s1u = A1e + A2ue + ... (1.23)

Thus every proper rational function of s has an inverse Laplace transform.

1.6.5 Locating Zeros of a Certain Polynomial

In determining the zeros of polynomials to be derived in this thesis, we employ the concept of

Rouche’s theorem. We make particular reference to Theorem 1.6 (page 181) in Lang (1991) which provides an elegant approach to the theorem and also details of the proof.

Let ℘ be a closed path homologous to 0 in H and assume that ℘ has an interior. Let f, g be analytic on H and

|g(s) − f(s)| < |g(s)| (1.24)

16 for s on ℘. Then f and g have the same number of zeros in the interior of ℘. Here, we provide a simpliﬁed approach to facilitate the use of the theorem in our case. In particular, we consider rather |g(s)| > 1 as it is obvious that we can re-write (1.24) as

f(s) − 1 < 1. (1.25) g(s)

Also, for the inequality in (1.25) to hold, we should have

f(s) |f(s)| = < 1. (1.26) g(s) |g(s)|

Finally for the theorem to work, |g(s)| > 1 from (1.26). We provide a simpliﬁed example of

the theorem.

Example

Let f(z) = z13 − 7z5 + z − 1. We want to ﬁnd the number of zeros of this polynomial inside

the unit circle. Let

g(z) = −7z5

for |z| = 1, it is clear that

|g(z) − f(z)| = | − 7z5 − (z13 − 7z5 + z − 1)|

= | − z13 − z + 1| < | − 7z5| = 7.

Hence g and f have the same number of zeros inside the unit circle, and this number is

clearly equal to 5 since 7z5 = 0.

17 Chapter 2

Wiener-Hopf Factorization Method

2.1 Introduction

Andersen (1957) uses the Wiener-Hopf Factorization method to obtain ﬁnite time ruin probabilities in the renewal risk model. Thorin (1971, 1974, 1982) uses the same method to obtain the ﬁnite time ruin probabilities for the Sparre-Andersen model. Wiener-Hopf factorization in general can be used to obtain transforms of quantities related to a random walk.

Diﬀerent texts uses rather diﬀerent notation to present the same result. For instance Section

6.4 of Rolski et al. (1998) has a somewhat diﬀerent presentation from that of Prabhu (1980).

Asmussen (2000) also presents the fundamentals of Wiener-Hopf factorizations. As we have been directed by Kalashnikov (1998) to Prahbu (1980), we are presenting a uniﬁed discussion of Wiener-Hopf Factorization in this chapter. The material is mainly from Prahbu (1980), except Theorem (2.1), which is from Feller (1971). As Prahbu (1980) is an out of print text with somewhat brief explanations in certain sections, it is our utmost conviction that readers will beneﬁt tremendously from having a one stop description of the Wiener-Hopf

Factorization method.

2.2 Ladder Processes

Let

S0 ≡ 0, and

Sn = X1 + X2 + ... + Xn (n ≥ 1).

18 We assume that the Xi (i ≥ 1) are mutually independent random variables with the charac-

teristic function φXi (ω). ¯ Let us deﬁne the random variables {Nk, k ≥ 0} and {Nk, k ≥ 0} as follows:

¯ N0 ≡ 0, ¯ N1 = min{n > 0 : Sn ≤ 0},

¯ ¯ ¯ Nk = min{n > Nk−1 : Sn ≤ SNk−1 } (k ≥ 2),

N0 ≡ 0,

N1 = min{n : Sn > 0},

Nk = min{n : Sn > SNk−1 }, (k ≥ 2).

¯ ¯ The random variable Nk is called the kth descending ladder epoch and SNk the corresponding ladder height. Similarly, Nk and SNk are called the kth ascending ladder epoch and height ¯ ¯ respectively. For convenience, let us denote N1 = N and N1 = N. Once a ladder point is reached, the random walk starts from the scratch in the sense that

given (N1,SN1 ) the random variables Sn − SN1 (n > N1) depend only on Xk (N1 < k ≤ n)

but not on Xk (k ≤ N1). It follows that the variables (Nk − Nk−1,SNk − SNk−1 )(k ≥ 1) form a renewal process in two dimensions. In other words, these pairs of random variables

are independent and have the same distribution as (N,SN ). Therefore the pair (Nk,SNk ) is

¯ ¯ ¯ ¯ the k−fold convolution of (N,SN ). Similarly, (Nk,SNk ) is the k−fold convolution of (N,SN ).

Theorem 2.1 For n ≥ 1, x > 0 we have

n X 1 1 P [N = n, S ≤ x] = P [0 < S ≤ x]. (2.1) k k n n n k=1 Proof:

Feller (1971) pages 413-414 (equations 7.1 to 7.7) provides an elegant proof for the following:

∞ X 1 1 P [N = n, S > 0] = P [S > 0]. (2.2) k k n n n k=1

19 As Feller (1971) notes similar arguments can be used to prove

∞ X 1 1 P [N = n, 0 < S ≤ x] = P [0 < S ≤ x]. (2.3) k k n n n k=1

It must be noted that P [Nk = n, Sn ≤ x] = 0 for k > n, as it is not possible to have more ladder epochs than the number of partial sums. However both Feller (1971) and Prabhu

(1980) write summation from k = 1 to ∞ for notational convenience in subsequent sections.

2.3 Renewal Functions

Now, let us deﬁne two renewal functions.

un(x) = P [Sn > Sm(0 ≤ m ≤ n − 1),Sn ≤ x)(n ≥ 1, x > 0),   0 for x < 0, u0(x) =  1 for x ≥ 0.

vn(x) = P [Sn ≤ Sm(0 ≤ m ≤ n − 1),Sn > x)(n ≥ 1, x ≤ 0),   0 for x ≤ 0, v0(x) =  1 for x > 0.

Since the event {Sn > Sm(0 ≤ m ≤ n − 1)} occurs if and only if n is an ascending ladder epoch,

un(x) = P [Nk = n, SNk ≤ x for some k ≥ 1], (2.4) n X = P [Nk = n, SNk ≤ x]. (2.5) k=1 Similarly

n X ¯ ¯ vn(x) = P [Nk = n, SNk > x]. (2.6) k=1 Note that since it is not possible to have more ladder epochs than the number of partial

¯ ¯ sums as mentioned previously, the probabilities P [Nk = n, SNk ≤ x] and P [Nk = n, SNk > x]

20 will be zero for k > n. Also these probabilities will be zero for k = 0. Therefore we may write

∞ X un(x) = P [Nk = n, SNk ≤ x], (2.7) k=0

∞ X ¯ ¯ vn(x) = P [Nk = n, SNk > x], (2.8) k=0 and we will see in subsequent development that this form has some advantages. We deﬁne the transforms of these renewal functions as

∞ Z ∞ ∗ X n iωx u (z, ω) = z e un(dx), (2.9) n=0 0−

∞ Z 0+ ∗ X n iωx v (z, ω) = z (−1) e vn(dx). (2.10) n=0 −∞ Let us assume the random walk under consideration is induced by the distribution function

K(x).

Let

Kn(x) = P [Sn ≤ x], (n ≥ 1, −∞ < x < ∞)

with K1(x) = K(x). Let

¯ χ(z, ω) = E(zN eiωSN ), χ¯(z, ω) = E(zN eiωSN¯ ). (2.11)

Theorem 2.2 For the transforms deﬁned by (2.9)-(2.11) we have the following results:

( ∞ ) X zn Z ∞ χ(z, ω) = 1 − exp − eiωxK (dx) , (2.12) n n n=1 0+

( ∞ ) X zn Z 0+ χ¯(z, ω) = 1 − exp − eiωxK (dx) , (2.13) n n n=1 −∞

( ∞ ) X zn Z ∞ u∗(z, ω) = exp eiωxK (dx) , (2.14) n n n=1 0+

21 ( ∞ ) X zn Z 0+ v∗(z, ω) = exp eiωxK (dx) . (2.15) n n n=1 −∞ Proof: This is Theorem 1 in page 49 of Prabhu (1980).

Using Theorem 2.1 we have

∞ ∞ ∞ X zn Z ∞ X Z ∞ X 1 eiωxK (dx) = eiωxzn P [N = n, S ∈ dx], n n k k n n=1 0+ n=1 0+ k=1 exchanging the order of summations, we have

∞ ∞ ∞ X zn Z ∞ X 1 X Z ∞ eiωxK (dx) = zn eiωxP [N = n, S ∈ dx]. n n k k n n=1 0+ k=1 n=1 0+

The second summation expression is the joint transform of the random variables (Nk,SNk ), we then have

∞ n Z ∞ ∞ X z iωx X 1 N iωS e K (dx) = E[z k e Nk ], n n k n=1 0+ k=1

since the distribution of (Nk,SNk ) is the k−fold convolution of the distribution of (N,SN ) with itself,

∞ ∞ X zn Z ∞ X 1 eiωxK (dx) = [χ(x, ω)]k n n k n=1 0+ k=1 = −log[1 − χ(z, ω)].

This leads to (2.12). Also, since

∞ X un(x) = P [Nk = n, SNk ≤ x] k=0 we obtain

∞ ∞ ∗ X N iωS X k 1 u (z, ω) = E[z k e Nk ] = [χ(z, ω)] = , 1 − χ(z, ω) k=0 k=0 hence (2.14).

We can prove (2.13) and (2.15) using similar arguments.

22 Theorem 2.3 Let φ(ω) = E(eiωX1 ). Then

1. 1 − zφ(ω) = [1 − χ(z, ω)][1 − χ¯(z, ω)]

2. Let D(z, ω) and D¯(z, ω) be functions such that for ﬁxed z in (0,1), they are bounded analytic functions for Im(ω) ≥ 0 and Im(ω) ≤ 0 respectively, bounded away from zero,

D(z, ω) → 1 as Im(ω) → ∞, and 1 − zφ(ω)= D(z, ω)D¯(z, ω) (0 < z < 1, ω real).

Proof: This is the theorem 2 in page 50 of Prahbu (1980).

1. Using the results of (2.12) and (2.13) we have that

( ∞ ) X zn Z ∞ [1 − χ(z, ω)][1 − χ¯(z, ω)] = exp − eiωxK (dx) n n n=1 −∞ ( ∞ ) X zn = exp − φ(ω)n n n=1 = exp{ln(1 − zφ(ω))}

= 1 − zφ(ω).

2. We have

[1 − χ(z, ω)][1 − χ¯(z, ω)] = D(z, ω)D¯(z, ω)

for real ω, since generating functions are deﬁned on the real ω. Let us deﬁne the function

Φ(ω) for complex value ω as follows:   1−χ(z,ω)  D(z,ω) for Im(ω) ≥ 0, Φ(ω) =  D¯(z,ω)  1−χ¯(z,ω) for Im(ω) ≤ 0.

We notice that the function Φ(ω) is a bounded analytic function on the whole complex plane;

from Liouville’s theorem we can conclude that this function is constant. Let ω = ω1 + iω2

(ω1, ω2 real), we ﬁnd that

|χ(z, ω)| ≤ E[zN |eiωSN |] < E[e−ω2SN ] → 0

as ω2 → ∞. Therefore Φ(ω) → 1 as Im(ω) → ∞. Since Φ(ω) is constant this proves that Φ(ω) ≡ 1 and concludes the proof.

23 2.4 Maximum and Minimum

Let us consider the random variable

Mn = max(0,S1,S2, ..., Sn). (2.16)

Theorem 2.4 The joint distribution of Mn and Mn − Sn takes the form

n X P [Mn ≤ x, Mn − Sn ≤ y] = um(x)vn−m(−y)(n ≥ 0, x ≥ 0, y ≥ 0). (2.17) m=0 Proof:

This is the Theorem 5 in page 25 of Prabhu (1980).

Let N(n) = max{i : Ni ≤ n}, so that N(n) is the number of ascending ladder epochs in the time interval (0, tn]. Then

n X P [Mn ≤ x, Mn − Sn ≤ y] = P [N(n) = k, Mn ≤ x, Mn − Sn ≤ y]. (2.18) k=0

Now consider the probability [N(n) = k, Mn ≤ x, Mn − Sn ≤ y]; this implies kth ascending ladder epoch can take any value between k and n; and Mn will be the kth ladder height; the next ascending ladder epoch will be greater than n. Using this we can write

P [N(n) = k, Mn ≤ x, Mn − Sn ≤ y] = n X P [Nk = m, Nk+1 > n, SNk ≤ x, SNk − Sn ≤ y]. (2.19) m=k By using the total probability rule, we can write

P [Nk = m, Nk+1 > n, SNk ≤ x, SNk − Sn ≤ y] = Z x

P [Nk = m, SNk ∈ dz]P [Nk+1 > n, SNk − Sn ≤ y|Nk = m, SNk = z]. 0− But the conditional probability can be simpliﬁed as follows:

P [Nk+1 > n, SNk − Sn ≤ y|Nk = m, SNk = z] =

P [Nk+1 − Nk > n − m, Sn − SNk ≥ −y|Nk = m, SNk = z]

= P [N1 > n − m, Sn−m ≥ −y],

24 since (Nk+1 − Nk,Sn − SNk ) is independent of the ladder point (Nk,SNk ) and has the same

distribution as (N1,SN1 ).

Now the event N1 > n − m implies all the partial sums up to and including n − m, i.e.

S1,S2, ..., Sn−m, are less than or equal to zero. Since X1,X2, ..., Xn are i.i.d. random variables

Sn−m − Sr contains sum of n − m − r + 1 Xi’s, i.e. less than n − m i.i.d. random variables. Therefore

P [N1 > n − m, Sn−m ≥ −y] = P [Sr ≤ 0 (0 ≤ r ≤ n − m),Sn−m ≥ −y]

= P [Sn−m − Sr ≤ 0 (0 ≤ r ≤ n − m),Sn−m ≥ −y]

= P [Sn−m ≤ Sr ≤ 0 (0 ≤ r ≤ n − m),Sn−m ≥ −y]

= vn−m(−y).

Collecting all the terms we have

Z x

P [Nk = m, Nk+1 > n, SNk ≤ x, SNk − Sn ≤ y] = P [Nk = m, SNk ∈ dz]vn−m(−y) 0−

= P [Nk = m, SNk ≤ x]vn−m(−y), substituting this in (2.19) and then substituting the result in (2.18) we have

n n X X P [Mn ≤ x, Mn − Sn ≤ y] = P [Nk = m, SNk ≤ x]vn−m(−y), (2.20) k=0 m=k by changing the order of summation

n n X X P [Mn ≤ x, Mn − Sn ≤ y] = P [Nk = m, SNk ≤ x]vn−m(−y), (2.21) m=0 k=0 and we conclude the proof by using the deﬁnition of um(x) as given in (2.5).

Theorem 2.5 The joint characteristic function of Mn,Mn − Sn will take the form

∗ ∗ u (z, ω1)v (z, −ω2) Proof:

This is a slight variation of Theorem 4 in page 52 of Prabhu (1980).

25 Using the joint distribution function for Mn,Mn − Sn in (2.17), we can write the joint characteristic function as

n Z ∞ Z 0 iω1Mn+iω2(Mn−Sn) X iω1x −iω2y E[e ] = e um(dx) e vn−m(dy). m=0 0 −∞ Therefore

∞ ∞ n Z ∞ Z 0 X n iω1Mn+iω2(Mn−Sn) X n X iω1x −iω2y z E[e ] = z e um(dx) e vn−m(dy) n=0 n=0 m=0 0 −∞ ∞ n Z ∞ Z 0 X X n iω1x −iω2y = z e um(dx) e vn−m(dy) n=0 m=0 0 −∞ ∞ ∞ Z ∞ Z 0 X X n iω1x −iω2y = z e um(dx) e vn−m(dy) m=0 n=m 0 −∞ ∞ Z ∞ ∞ Z 0 X m iω1x X n−m −iω2y = z e um(dx) z e vn−m(dy) m=0 0 n=m −∞ ∗ ∗ = u (z, ω1)v (z, −ω2), (2.22)

∗ ∗ where the transforms u (z, ω1) and v (z, −ω2) are deﬁned in (2.9) and (2.10). Since

∞ X lim E[eiωMn ] = lim(1 − z) znE[eiωMn ], n→∞ z→1 n=0 let,

lim E[eiωMn ] = φ(ω). n→∞

Then,

iωMn E[e ] = φ(ω) + φn(ω)(φn(ω) → 0 as n → ∞), ∞ ∞ ∞ X n iωMn X n X n z E[e ] = φ(ω) z + z φn(ω) n=0 n=0 n=0 ∞ φ(ω) X = + znφ (ω). 1 − z n n=0

26 Hence,

∞ ∞ X n iωMn X n (1 − z) z E[e ] = φ(ω) + (1 − z) z φn(ω) n=0 n=0 ∞ ∞ X n iωMn X n lim(1 − z) z E[e ] = φ(ω) + lim(1 − z) z φn(ω) z→1 z→1 n=0 n=0 ∞ X lim(1 − z) znE[eiωMn ] = φ(ω). z→1 n=0 Therefore,

∞ X lim E[eiωMn ] = lim(1 − z) znE[eiωMn ], n→∞ z→1 n=0 we have

lim E[eiωMn ] = lim(1 − z)u∗(z, ω)v∗(z, 0). n→∞ z→1

Therefore if we can evaluate the transforms u∗(z, ω) and v∗(z, ω) we will have the desired expected value.

We can summarize the classical technique as follows:

1. Factorize 1 − zφ(ω) into two factors D(z, ω) and D¯(z, ω) with the following properties a. D(z, ω) should be analytical and bounded away from zero for Im(ω) ≥ 0, b. D¯(z, ω) should be analytical and bounded away from zero for Im(ω) ≤ 0, c. and D(z, ω) → 1 as Im(ω) → ∞.

2. Set

1 − z lim E[eiωMn ] = lim . n→∞ z→1 D(z, ω)D¯(z, 0)

2.5 Survival Probability

Survival probability is the likelihood that the insurer’s surplus at any positive time will be above zero.

The Sparre-Andersen model shown in the introductory part of this thesis can be reduced to

27 the study of a random walk as follows:

Let

n X Sn = (Xi − cTi), i = 1, 2, ... i=1 n X = Yi, i = 1, 2, ... (2.23) i=1

We are to note that EY < 0, S0 = 0 and Sn = {S1,S2, ...} for n ≥ 1.

If we deﬁne another random variable Mn as:

Mn = max{S0,S1,S2, ..., Sn}.

Then, the maximal aggregate loss for the surplus process becomes L as follows:

L = lim Mn. (2.24) n→∞

We ﬁnd the characteristic function of L from the characteristic function of Y by following the summarized classical technique shown under section (2.4) of this chapter.

The survival probability is denoted by δ(u) and it relates to the maximal aggregate loss in the following way (Ambagaspitiya, 2009).

1 δ∗(s) = M (−s) (2.25) s L

∗ where δ (s) represents the Laplace transform of δ(u) and ML(.) is the moment generating function of L. Once δ∗(s) is known, the inverse Laplace transform can be taken to obtain the exact function of δ(u). Also, (2.25) can be derived as follows:

Z ∞ δ∗(s) = e−suδ(u)du 0 P [L = 0] Z ∞ = + e−suP [0 < L ≤ u]du s 0 1 Z ∞ d = P [L = 0] + e−su P [0 < L ≤ u]du s 0 du 1 = E[e−sL]. s

28 The ultimate ruin probability is therefore obtained by taking the complement of δ(u) as shown:

ψ(u) = 1 − δ(u). (2.26) where ψ(u) is the ultimate ruin probability.

29 Chapter 3

Ruin Probability for Correlated Poisson Claim Count

and Gamma (2) Risk Process

3.1 Introduction

This chapter focuses on the derivation and the result of the ultimate ruin probability when the claim number (counts) process is assumed to be Poisson distributed where as the claim sizes (amounts) are assumed to be Gamma distributed with shape parameter 2. The case where the claim counts and claim sizes are assumed to be independent has been considered by Dickson and Hipp (1998). In this thesis however, we relax the independence assumption by introducing dependency between the claim counts and the claim sizes. Also, the eﬀect of correlation on ultimate ruin probability is established in this chapter.

3.1.1 Poisson Claim

Since the claim count process in this chapter is Poisson distributed, it is only fair to shed light on the Poisson distribution and how it relates to the Exponential distribution. By deﬁnition, a random variable N is said to follow a Poisson distribution with mean λ if its probability mass function (pmf) is given as:

e−λλn P [N = n] = ; λ > 0, n = 0, 1, ... (3.1) n!

In general, the Poisson distribution arises as the number of arrivals of claims in a ﬁxed time period in which the expected number of arrivals is known and arrivals are independent of each other.

1 Also, a random variable T is said to have an Exponential distribution with mean λ and

30 density given as:

−λt fT (t) = λe ; t ≥ 0, λ > 0. (3.2)

One important interpretation of the random variable T is that, it is the waiting time until the ﬁrst arrival of a claim when arrivals are such that the number of arrivals in the time interval (0, t] is Poisson λt for each t.

3.1.2 Relationship Between Poisson and Exponential Distribution

The relationship between Poisson and Exponential distribution can be established by ﬁrst considering a sequence of claim arrivals over time. We denote T as the time until the arrival of a claim and for each t, we let N(t) represents the number of arrivals (claims) in the time interval [0, t]. Since N(t) is Poisson λt for each t, then for t ≥ 0,

P [T > t] = P [ﬁrst claim arrives after time t]

= P [No claim arrives in [0, t]]

= e−λt. (3.3)

The survival and the density functions of T result from (3.2).

The moment generating function (mgf) of T is given as:

Z ∞ st MT (s) = e fT (t)dt 0 λ = , s < λ. (3.4) λ − s

3.2 Joint Moment Generating Function of (T,X)

Assume X1 and X2 are Moran and Downton bivariate Exponential as there are many bivariate Exponential with correlation coeﬃcient ρ ≥ 0 and the following joint moment generating

31 function (mgf):

X1t1+X2t2 MX1,X2 (t1, t2) = E[e ] 1 = . (3.5) (1 − t1)(1 − t2) − ρt1t2 i.e. X1 and X2 each are Exponential with mean 1. Let us introduce the following two random variables T,X. X T = 1 β1 X + Z X = 2 , β2 where Z is independent of X1 and X2. Also, Z is Exponential with mean 1. With this notation, T and X are correlated random variables. The mgf of it can be derived as follows:

h X1t1 X2t2+Zt2 i β + β MT,X (t1, t2) = E e 1 2

t1 t2 t2 = MX1,X2 ( , )MZ ( ) β1 β2 β2 1 = h i . (3.6) 1 − t1 1 − t2 − ρ t1t2 1 − t2 β1 β2 β1β2 β2 Note that T and X are bivariate where T is Exponential and X is Gamma with shape parameter 2.

3.3 Characteristic Function of Y and its Related Transform

Assume T represents the inter-arrival time between claims and X is the claim size. For the purpose of motivating and applying the Wiener-Hopf factorization method, we need the mgf of Y = X − cT .

Y s MY (s) = E[e ]

= E[eXs−csT ]

= MT,X (−cs, s) 1 = h i . (3.7) 1 + cs 1 − s + ρ cs2 1 − s β1 β2 β1β2 β2

32 The characteristic function of Y is obtained by replacing s with iω in the preceding equation.

Now, we require the factor 1 − zφY (ω) which is given as: z 1 − zφY (ω) = 1 − h i 1 + iωc 1 − iω − ρ cω2 1 − iω β1 β2 β1β2 β2

h 2 i 1 + iωc 1 − iω − ρ cω 1 − iω − z β1 β2 β1β2 β2 = h i . (3.8) 1 + iωc 1 − iω − ρ cω2 1 − iω β1 β2 β1β2 β2 For notational convenience, we let the denominator of (3.7) be cs s cs2 s g(s) = 1 + 1 − + ρ 1 − . β1 β2 β1β2 β2 Note the reversion back to s.

3.4 Zeros of the Polynomials g(s) and g(s) − z

To obtain the zeros of g(s) and g(s) − z, we proceed by ﬁrst expanding g(s) as follows: c c 1 s g(s) = − (1 − ρ)s2 + − s + 1 1 − β1β2 β1 β2 β2 c 3 c 1 c 1 2 c 2 = 2 (1 − ρ)s − (1 − ρ) + − s + − s + 1, (3.9) β1β2 β1β2 β2 β1 β2 β1 β2 we have 1 M (s) = . Y g(s) Note that g(s) is a polynomial of degree 3 in s. dM (s) g0(s) Y = − ds g(s)2 where g0(s) is the derivative of g(s) with respect to s. dM (s) E[Y ] < 0 ⇒ Y | < 0 ⇒ g0(0) > 0. ds s=0 But c 2 g0(0) = − β1 β2 c 2 ⇒ > . β1 β2

33 Now let us look at the zeros of g(s) − z when 0 < z < 1. Note g(s) = 0 has one positive

zero s = β2 and two other zeros. As the leading coeﬃcient of the polynomial g(s) (or the

3 c(1−ρ) coeﬃcient of s ) is 2 > 0 we know that as β1β2

lim g(s) → ∞, s→∞

and

lim g(s) → −∞. s→−∞

Also g(0) = 1 and g0(0) > 0 as s → 0 , therefore g(s) is an increasing function of s at s = 0.

Also note that 0 c 2 c 1 c 1 c 2 g (s) = 3 2 (1 − ρ)s − 2 (1 − ρ) + − s + − β1β2 β1β2 β2 β1 β2 β1 β2 0 c 2 c 1 c 1 c 2 g (β2) = 3 2 (1 − ρ)β2 − 2 (1 − ρ) + − β2 + − β1β2 β1β2 β2 β1 β2 β1 β2 ρc = − < 0 β1

as s → β2, g(s) is a decreasing function of s. Therefore g(s) must have another zero in the

interval (β2, ∞) and the remaining zero must be negative. Let us denote −s1 as the negative

zero with (s1 > 0) and s2 = β2 and s3 is largest zero. Then we can write g(s) as

c(1 − ρ) g(s) = 2 (s + s1)(s − s2)(s − s3), (3.10) β1β2

where from (3.9) s1 and s3 are respectively

r 2 − c − 1 + c − 1 + 4(1 − ρ) c β1 β2 β1 β2 β1β2 s = (3.11) 1 2(1 − ρ) c β1β2 and

r 2 c − 1 + c − 1 + 4(1 − ρ) c β1 β2 β1 β2 β1β2 s = . (3.12) 3 2(1 − ρ) c β1β2 Now consider the zeros of g(s) − z for 0 < z < 1. These will be the values of s at the point of intersection for y = g(s) and y = z. Therefore g(s) − z will have one negative zero and

34 two positive zeros which are all functions of z. Let us identify these zeros as follows:

lim s1(z) → s1, z→0

lim s2(z) → s2 = β2, z→0 and

lim s3(z) → s3. z→0

By the shape of the graph of g(s) relationships for the zeros of g(s) − z.

s1(z) < s1,

s2(z) < β2, and

s3(z) > s3.

Also we can obtain the following limiting relationships.

lim s1(z) → 0, z→1

lim s2(z) → s2(1) < β2, z→1 and

lim s3(z) → s3(1) > s3. z→1

Therefore we can write g(s) − z as

c(1 − ρ) g(s) − z = 2 (s + s1(z))(s − s2(z))(s − s3(z)). (3.13) β1β2

Figure 3.1 gives a quick visual impression of the relationship between g(s) and g(s) − z. In particular, it further enhances our understanding of the behaviour of the zeros of g(s) − z vis-a-vis g(s).

35 Figure 3.1: Graph of g(s) and 0 < z < 1.

3.5 Factors and Properties of the Transform of Y

An interesting fact is that, if we let s = 0 in (3.13), the following results.

c(1 − ρ) 1 − z = 2 s1(z)s2(z)s3(z). β1β2

36 Similarly, by putting s = 0 in (3.10) we establish this relationship

c(1 − ρ) 2 s1s2s3 = 1. β1β2

Now,

s (z)s (z)s (z) 1 − z = 1 2 3 , s1s2s3 and

1 − z s (1)s (1) lim = 2 3 . z→1 s1(z) s1s2s3

Therefore,

g(s) − z (s + s (z))(s − s (z))(s − s (z)) = 1 2 3 . (3.14) g(s) (s + s1)(s − s2)(s − s3)

Now let us write

(iω − s (z))(iω − s (z)) D(z, ω) = 2 3 (iω − s2)(iω − s3) and

iω + s (z) D¯(z, ω) = 1 . iω + s1

We can also rewrite D(z, ω) and D¯(z, ω) in the following form:

s − s (z) s − s (z) D(z, ω) = 1 − 2 2 1 − 3 3 s2 − iω s3 − iω and

s − s (z) D¯(z, ω) = 1 − 1 1 . s1 + iω and by letting ω = ω1 + iω2 where ω1 and ω2 are real we have

s − s (z) s − s (z) D(z, ω) = 1 − 2 2 1 − 3 3 s2 − iω1 + ω2 s3 − iω1 + ω2

37 and

s − s (z) D¯(z, ω) = 1 − 1 1 . s1 + iω1 − ω2

We discover that ¯ 1. D(z, ω) is a bounded function, bounded away from zero for ω2 < 0.

2. D(z, ω) is a bounded function, bounded away from zero for ω2 > 0.

3. limω2→∞ D(z, ω) = 1. We conclude that D(z, ω) and D¯(z, ω) are the required factors and therefore in deriving the ultimate ruin probability they are conditions precedent.

3.6 Survival Probability

To obtain an explicit expression for the ultimate ruin probability, we ﬁrst need to derive the survival probability. The characteristic function of the maximal aggregate loss L is obtained from the characteristic function of Y by (2.24) and the ﬁfth equation on page 27 as follows:

1 1 − z φL(ω) = lim D(1, ω) z→1 D¯(z, 0)

(iω − s2)(iω − s3) 1 − z = lim s1 (iω − s2(1))(iω − s3(1)) z→1 s1(z) (iω − s2)(iω − s3) s2(1)s3(1) = s1 (iω − s2(1))(iω − s3(1)) s1s2s3 (iω − s )(iω − s ) s (1)s (1) = 2 3 2 3 . (3.15) (iω − s2(1))(iω − s3(1)) s2s3

In chapter 2, we have provided detail information and valuable references as to how to arrive

at an expression leading to (3.15).

Therefore the Laplace transform of the survival probability δ∗(s) from (2.25) is given by:

1 (s + s )(s + s ) s (1)s (1) δ∗(s) = 2 3 2 3 s (s + s2(1))(s + s3(1)) s2s3 A B C = + + . (3.16) s s + s2(1) s + s3(1)

38 A, B and C are obtained by successively substituting s = 0, s = −s2(1) and s = −s3(1) on both side of following equation

s2(1)s3(1) A(s + s2(1))(s + s3(1)) + Bs(s + s3(1)) + Cs(s + s2(1)) = (s + s2)(s + s3). s2s3

We have A = 1,

s2(1)s3(1) −Bs2(1)(s3(1) − s2(1)) = (s2 − s2(1))(s3 − s2(1)) s2s3

and

s (1) (s − s (1))(s − s (1)) B = − 3 2 2 3 2 s2s3 s3(1) − s2(1)

s2(1) s2(1) 1 − s 1 − s = − 2 3 . 1 − s2(1) s3(1) Similarly by symmetry

s3(1) s3(1) 1 − s 1 − s C = − 2 3 . 1 − s3(1) s2(1)

Therefore the survival probability is obtained by inverting δ∗ (s) which results in   1 − s2(1) 1 − s2(1) 1 − s3(1) 1 − s3(1) s2 s3 −s (1)u s2 s3 −s (1)u δ (u) = 1 −  e 2 + e 3  .(3.17) 1 − s2(1) 1 − s3(1) s3(1) s2(1)

The ultimate ruin probability becomes   1 − s2(1) 1 − s2(1) 1 − s3(1) 1 − s3(1) s2 s3 −s (1)u s2 s3 −s (1)u ψ (u) =  e 2 + e 3  . (3.18) 1 − s2(1) 1 − s3(1) s3(1) s2(1)

39 3.7 Eﬀect of Correlation on Ultimate Ruin Probability

To establish the eﬀect of correlation on ruin probability, we let

∗ s3 → f3 (ρ) and s3(1) → f3 (ρ) where

r 2 c − 1 + c − 1 + 4(1−ρ)c β1 β2 β1 β2 β1β2 f (ρ) = . (3.19) 3 2 (1 − ρ) c β1β2 However, we observe the following relationships.

lim f3(ρ) → β2 ρ→0

lim f3(ρ) → +∞ ρ→1

∗ lim f3 (ρ) > lim f3(ρ) → β2 ρ→0 ρ→0

∗ lim f3 (ρ) > lim f3(ρ) → +∞. ρ→1 ρ→1

As ρ → [0, 1], s3 → [β2, +∞) and s3 (1) → (β2, +∞). Note that,

1. s3 is an increasing function of ρ;

2. s3 (1) is an increasing function of ρ;

3. s2 (1) < β2 is neither an increasing nor a decreasing function of ρ; 1− s2(1) 1− s2(1) s2 s3 4. s2(1) is a decreasing function of ρ; 1− s (1) 3 1− s3(1) 1− s3(1) 5. s2 s3 is a decreasing function of ρ. 1− s3(1) s2(1) Therefore as ρ increases ψ (u) decreases.

3.8 Numerical Examples

In this section, we give some numerical examples. Our choice of parameters depends on

satisfying the condition EY < 0 and this applies to all the examples that will be given in the proceeding chapters in respect of the exact expressions for the ruin probability.

Example 1

1 Consider the case where c = 3, β1 = 2, β2 = 4 and ρ = 5 . By employing (3.11) and (3.12),

40 we solve for

25 1 √ s = − 1105, 1 12 12 = −0.6868

s2 = 4 and 25 1 √ s = + 1105 3 12 12 = 4.8535.

Also using (3.13) by letting s = 0 and z = 1, we have

s1(1) = 0, 49 1 √ s (1) = − 481 2 12 12 = 2.2557 and 49 1 √ s (1) = + 481 3 12 12 = 5.9110.

Substituting all the above in (3.18), the ultimate ruin probability results as follows:

ψ(u) = 0.37745e−2.2556u − 0.064235e−5.9110u.

Example 2

2 Consider the case where c = 5, β1 = 2, β2 = 3 and ρ = 5 . By employing (3.11) and (3.12), we solve for

13 1√ s = − 241, 1 6 6 = −0.4207

s2 = 3 and 13 1√ s = + 241 3 6 6 = 4.7540.

41 Also using (3.13) by letting s = 0 and z = 1, we have

s1(1) = 0, 11 1√ s (1) = − 22 2 3 3 = 2.1032 and 11 1√ s (1) = + 22 3 3 3 = 5.2301.

Substituting all the above in (3.18), the ultimate ruin probability results as follows:

ψ(u) = 0.27879e−2.1032u − 0.050048e−5.2302u.

Example 3

Again, we consider the situation where c = 5, β1 = 2 and β2 = 3. In Figure 3.2, we display ultimate ruin probability for different values of the correlation coefficient (ρ). It is clear and understandable from the graph that, positive correlation relates inversely with ultimate ruin probability i.e. positive correlation lessens the effect of ultimate ruin probability. This graphical evidence goes to support the proof for the effect of correlation on ruin probability as established previously under section 3.7.

42 Figure 3.2: Eﬀect of correlation on ruin probability

43 Chapter 4

Ruin Probability for Correlated Poisson Claim Count

and Gamma (m) Risk Process

4.1 Introduction

This chapter extends the derivation of the ultimate ruin probability considered in the pre-

ceding chapter. The emphasis of this chapter therefore, is to work out the exact expression

for the ultimate ruin probability when the claim number (counts) is assumed to be Poisson

distributed whiles the claim sizes (amounts) are assumed to be Gamma distributed with

shape parameter m where m > 2. Here too, we relax the independence assumption by intro-

ducing dependency between the claim counts and the claim sizes. The claim count process

is the same as the process considered in chapter 3.

4.2 Joint Moment Generating Function of (T,X)

Here too, assume X1 and X2 are Moran and Downton bivariate Exponential with correlation coeﬃcient ρ ≥ 0 and the joint mgf as shown in (3.5) with the same set of assumptions

as before. Now, we introduce two random variables T,X with X having another form as

compared with the X studied in chapter 3 as follows:

X T = 1 β1 X + Z X = 2 , β2

where Z is a Gamma random variable with shape (m − 1) and is independent of X1 and

X2. With this notation, it is obvious that T and X are correlated random variables. The

44 associated mgf can be obtained as follows:

h X1t1 X2t2+Zt2 i β + β MT,X (t1, t2) = E e 1 2 (4.1)

t1 t2 t2 = MX1,X2 ( , )MZ ( ) (4.2) β1 β2 β2 1 = . (4.3) h i m−1 1 − t1 1 − t2 − ρ t1t2 1 − t2 β1 β2 β1β2 β2

Note here that, T and X are bivariate where T is Exponential and X is Gamma with shape

parameter m.

4.3 Characteristic Function of Y and its Related Transform

Assume T represents the inter-arrival time and X is the claim size. To enable us utilize the

Wiener-Hopf factorization method, we require the mgf of Y = X − cT .

Y s MY (s) = E[e ] (4.4)

= E[eXs−csT ] (4.5)

= MT,X (−cs, s) (4.6) 1 = . (4.7) h i m−1 1 + cs 1 − s + ρ cs2 1 − s β1 β2 β1β2 β2

The characteristic function of Y is obtained by substituting s with iω in (4.7). Also, we need the related factor 1 − zφY (ω) which is obtained as follows:

z 1 − zφ (ω) = 1 − (4.8) Y h i m−1 1 + iωc 1 − iω − ρ cω2 1 − iω β1 β2 β1β2 β2 m−1 h iωc iω cω2 i iω 1 + β 1 − β − ρ β β 1 − β − z = 1 2 1 2 2 . (4.9) h i m−1 1 + iωc 1 − iω − ρ cω2 1 − iω β1 β2 β1β2 β2

For ease of notation and derivation, we let the denominator of (4.7) be

cs s cs2 s m−1 g(s) = 1 + 1 − + ρ 1 − . β1 β2 β1β2 β2

45 As we saw in Chapter 3, we can factorize the term inside the square bracket in the above expression resulting in

s s m−1 s g(s) = 1 + 1 − 1 − (4.10) s1 s2 s3 where s3 > β2. Therefore g(s) has m + 1 zeros comprising of one negative and m positive zeros. We denote −s1 as the negative zero with (s1 > 0), s2 = β2 as the m − 1 repeated positive zeros and s3 as the largest positive zero. Now, the task is to ﬁnd the location of m + 1 zeros of g(s) − z for 0 < z < 1. In the proceeding section, we apply Rouche’s Theorem to locate the neighborhood of those zeros.

4.4 Zeros of the Polynomial g(s) − z

In order to arrive at the zeros of g(s) and g(s) − z, we ﬁrst need to expand g(s) as follows:

1 M (s) = . Y g(s)

Note that g(s) is a polynomial of degree m + 1 in s.

dM (s) g0(s) Y = − . ds g(s)2 where g0(s) is the derivative of g(s) with respect to s.

dM (s) E[Y ] < 0 ⇒ Y | < 0 ⇒ g0(0) > 0, ds s=0 but

c m g0(0) = − , β1 β2 c m ⇒ > . β1 β2

When m is an even positive integer, we have,

m c(1−ρ) 1. the leading (−1) m > 0; β1β2 2. the highest power of g (s) is odd;

46 m−1 3. 1 − s has odd power; β2 4. g (0) = 1 and g0 (0) > 0 which implies that g0 (s) is an increasing function of s at s = 0.

0 5. Since g (β2) = 0, we have a saddle point. Therefore we need to determine the behaviour of g(s) in the neighborhood of β2. If we bound s such that s < s3 and for some > 0,

m−1 m−1 (s − β2) < 0 but g(s) > 0 when s is s = β2 − . However, (s − β2) > 0 but g(s) < 0 when s is s = β2 + . In the neighborhood of β2, g(s) changes sign. Figure 4.1 shows the graph for the case where m takes only even positive integers.

47 Figure 4.1: Graph of g(s) for the case m = 4, 6, ....

When m is an odd positive integer, we have,

m c(1−ρ) 1. the leading (−1) m < 0; β1β2 2. the highest power of g (s) is even; m−1 3. 1 − s has even power; β2 4. g (0) = 1 and g0 (0) > 0 which implies that g0 (s) is an increasing function of s at s = 0.

48 m−1 5. if we bound s such that s < s3 and for some > 0, (s − β2) > 0 and g(s) > 0 when

m−1 s is s = β2 − . Similarly, (s − β2) > 0 and g(s) > 0 when s is s = β2 + . In the neighborhood of β2, g(s) is positive. Figure 4.2 reveals the graph for the case where m takes only odd positive integers.

Figure 4.2: Graph of g(s) for the case m = 3, 5, ....

49 4.4.1 Application of Rouche’s Theorem

Let C1 and C2 denote two circles in the complex plane; C1 centered at −s1 with radius s1

and C2 centered at s3 with radius s3. The points on circle C1 and C2 can be denoted by

iθ iθ −s1(1 − e ) and s3(1 + e ). With the idea of complex number analysis, we can show the following:

Since s1 < s2 < s3,

s1 iθ 1 − 1 − e = 1 s1

s1 iθ 1 + 1 − e ≥ 1 s2

s1 iθ 1 + 1 − e ≥ 1 s3 and

s3 iθ 1 + 1 + e ≥ 1 s1

s3 iθ 1 − 1 + e ≥ 1 s2

s3 iθ 1 − 1 + e = 1. s3

However,

s3 iθ 1 − 1 + e ≥ 1 s2

is not explicit as the others. To prove this, we ﬁrst let

eiθ = cosθ + isinθ,

50 we have

2 2 s3 iθ s3 1 − 1 + e = 1 − (1 + cosθ + isinθ) s2 s2 2 s3 s3 = 1 − (1 + cosθ) − i sinθ s2 s2 s 2 s 2 = 1 − 3 (1 + cosθ) + − 3 sinθ s2 s2 s s s 2 = 1 − 2 3 − 2 3 cosθ + 3 s2 s2 s2 s 2 s 2 s 2 +2 3 cosθ + 3 cos2θ + 3 sin2θ s2 s2 s2 " # s s 2 s s 2 = 1 − 2 3 + 2 3 − 2cosθ 3 − 3 s2 s2 s2 s2 s s 2 s s = 1 − 2 3 + 2 3 + 2 3 cosθ 3 − 1 s2 s2 s2 s2 minimum of this occurs at cos θ = −1 s s 2 s s ≥ 1 − 2 3 + 2 3 − 2 3 3 − 1 s2 s2 s2 s2 2 s3 iθ 1 − 1 + e ≥ 1. s2

It is obvious from these inequalities that on C1 and C2, |g(s)| ≥ 1. Therefore, by Rouche’s theorem g(s) and g(s) − z have the same number of zeros inside each circle for |z| < 1.

Consequently, g(s) − z has one zero inside the circle C1 and m zeros inside the circle C2. Let us now identify these roots as follows:

lim s1(z) → s1, z→0

lim s2,j(z) → s2 = β2, j = 1, ..., m − 1, z→0 and

lim s3(z) → s3. z→0

51 The nature of the graph of g(s) relates with the zeros of g(s) − z in two ways. We ﬁrst consider the case where m is even.

|s1(z)| < s1,

|s2,j(z)| < β2,

and

|s3(z)| > s3.

Also we can obtain the following limiting relationships.

lim s1(z) → 0, z→1

lim s2,j(z) → s2,j(1) < β2, z→1

and

lim s3(z) → s3(1) > s3. z→1

We consider the case where m is odd.

s1(z) < s1,

s2,j(z) < β2 ∪ s2,j(z) > β2,

and

s3(z) < s3.

In the limiting situation, we have

lim s1(z) → 0, z→1

lim s2,j(z) → s2,j(1) < β2 ∪ z→1

lim s2,j(z) → s2,j(1) > β2, z→1

52 and

lim s3(z) → s3(1) < s3. z→1

Therefore, we can write g(s) − z as

m c(1 − ρ) m−1 g(s) − z = (−1) m (s + s1(z))(s − s2(z)) (s − s3(z)). (4.11) β1β2

Then,

m c(1−ρ) Qm−1 g(s) − z (−1) β βm (s + s1(z)) j=1 (s − s2(z))(s − s3(z)) = 1 2 , (4.12) m c(1−ρ) m−1 g(s) (−1) m (s + s1)(s − s2,j) (s − s3) β1β2 " Qm−1 # ( j=1 (s2,j(z) − s))(s3(z) − s) s + s1(z) = m−1 . (4.13) (s2 − s) (s3 − s) s + s1

4.5 Factors and Properties of the Transform of Y

We therefore at this point can write the factors of 1 − zφY (ω) as follows: " Qm−1 # ( j=1 (s2,j(z) − iω))(s3(z) − iω) iω + s1(z) 1 − zφY (ω) = m−1 . (4.14) (s2 − iω) (s3 − iω) iω + s1

We let

"Qm−1 # j=1 (s2,j(z) − iω)(s3(z) − iω) D(z, ω) = m−1 (s2 − iω) (s3 − iω) and

iω + s (z) D¯(z, ω) = 1 . iω + s1

We further express D(z, ω) and D¯(z, ω) in the following form:

m−1 Y s2 − s2,j(z) s3 − s3(z) D(z, ω) = 1 − 1 − s − iω s − iω j=1 2 3 and

s − s (z) D¯(z, ω) = 1 − 1 1 . s1 + iω

53 By replacing ω with ω = ω1 + iω2 where ω1 and ω2 are real, we obtain the following:

m−1 Y s2 − s2,j(z) s3 − s3(z) D(z, ω) = 1 − 1 − s − iω + ω s − iω + ω j=1 2 1 2 3 1 2

and

s − s (z) D¯(z, ω) = 1 − 1 1 . s1 + iω1 − ω2

It is clear at this stage that ¯ 1. D(z, ω) is a bounded function, bounded away from zero for ω2 < 0.

2. D(z, ω) is a bounded function, bounded away from zero for ω2 > 0.

3. limω2→∞ D(z, ω) = 1. Therefore the required factors are D¯(z, ω) and D(z, ω).

4.6 Survival Probability

Here, the intention is to derive the exact expression for the ultimate ruin probability. Prior to doing that, we may ﬁrst need to establish the link between the characteristic function of the maximal aggregate loss (L) and the characteristic function of Y . We therefore proceed by obtaining the characteristic function of L as shown:

1 1 − z φL(ω) = lim D(1, ω) z→1 D¯(z, 0) m−1 (iω − s2) (iω − s3) 1 − z = h i lim s1 . (4.15) Qm−1 z→1 s1(z) j=1 (iω − s2,j(1)) (iω − s3(1))

However, m−1 c(1 − ρ) Y 1 − z = s (z) (s (z)) s (z). (4.16) β βm 1 2,j 3 1 2 j=1 Also

c(1 − ρ) m−1 m s1s2 s3 = 1. (4.17) β1β2

54 We therefore obtain

Qm−1 1 − z j=1 (s2,j(z)) s3(z) = m−1 (4.18) s1(z) s1s2 s3 hence

m−1 Qm−1 (s2 − iω) (s3 − iω) j=1 (s2,j(z)) s3(z) φL(ω) = lim (4.19) h m−1 i z→1 m−1 Q s2 s3 j=1 (s2,j(1) − iω) (s3(1) − iω)

m−1 Qm−1 (s − iω) (s − iω) (s2,j(1)) s3(1) = 2 3 j=1 . (4.20) h m−1 i m−1 Q s2 s3 j=1 (s2,j(1) − iω) (s3(1) − iω)

Writing s for iω, the Laplace transform of the survival probability and the partial fractions expansion are given as follows:

1 δ∗(s) = M (−s), (4.21) s L m−1 Qm−1 (s + s ) (s + s ) (s2,j(1)) s3(1) = 2 3 j=1 , (4.22) Qm−1 m−1 s j=1 (s + s2,j(1))(s + s3(1)) s2 s3 m−1 A X Bj C = + + . (4.23) s s + s (1) s + s (1) j=1 2,j 3

A, Bj and C are obtained by successively substituting s = 0, s = −s2,j(1) for j = 1, ..., m−1 and s = −s3(1) in both sides of the following equation

m−1 Qm−1 s3(1)(s + s2) (s + s3) k=1 s2,k(1) m−1 = s2 s3

m−1 m−1 m−1 Y Y Y A(s + s3(1)) (s + s2,k(1)) + Bjs(s + s3(1)) (s + s2,j) + Cs (s + s2,k(1)) k=1 k=1,k6=j k=1

m−1 m−1 m−1 m−1 Y X Y Y = A(s + s3(1)) (s + s2,k(1)) + Bjs(s + s3(1)) (s + s2,j) + Cs (s + s2,k(1)). k=1 j=1 k=1,k6=j k=1 We therefore have A = 1,

s (1)(s − s (1))m−1(s − s (1)) Qm−1 s (1) −B = 3 2 2,j 3 2,j k=1 2,k , j Qm−1 s2,j(1)(s3(1) − s2,j(1)) k=1,k6=j [s2,k(1) − s2,j(1)]

55 and

(s − s (1))m−1(s − s (1)) Qm−1 s (1) −C = 2 3 3 3 k=1 2,k . Qm−1 k=1 (s2,k(1) − s3(1))

Inverting δ∗(s) or taking inverse Laplace transform, we have

m−1 X −s2,j (1)u −s3(1)u δ(u) = 1 − Bje − Ce , (4.24) j=1 and the ultimate ruin probability becomes,

m−1 X −s2,j (1)u −s3(1)u ψ(u) = Bje + Ce . (4.25) j=1

4.7 Numerical Examples

We provide some numerical examples in this section.

Example 1

2 Consider the case where m is odd say m = 3 with c = 3, β1 = 2, β2 = 4 and ρ = 5 . Equating (4.10) to zero and solving, we obtain

s1 = −0.7094,

s2 = 4 (2 repeated roots) and

s3 = 6.2650.

Also using (4.11) by letting s = 0 and z = 1, we have

s1(1) = 0,

s2,1(1) = 1.3491,

s2,2(1) = 6.1032 + 1.5109i and

s3(1) = 6.1032 − 1.5109i

56 Putting all the above in (4.25) we arrive at the ultimate ruin probability as follows:

ψ(u) = 0.54748 e−1.3491 u − (0.079536 cos (1.5109 u) + 0.082248 sin (1.5109 u)) e−6.1032 u.

Example 2

2 Consider the case where m is even say m = 4 with c = 3, β1 = 2, β2 = 4 and ρ = 5 . Equating (4.10) to zero and solving, we obtain

s1 = −0.7094,

s2 = 4 (2 repeated roots) and

s3 = 6.2650.

Also using (4.11) by letting s = 0 and z = 1, we have

s1(1) = 0,

s2,1(1) = 0.6753,

s2,2(1) = 4.8016 − 2.4258i,

s2,3(1) = 4.8016 + 2.4258i and

s3(1) = 7.2770.

Putting all the above in (4.25) we arrive at the ultimate ruin probability as follows:

ψ(u) = 0.71333 e−0.6753 u − (0.021889 e−7.2770 u + (0.046138 cos (2.4258 u) +

0.062149 sin (2.4258 u))e−4.8016 u.

Example 3

2 Consider the case where m is odd say m = 5 with c = 3, β1 = 2, β2 = 4 and ρ = 5 .

57 Equating (4.10) to zero and solving, we obtain

s1 = −0.7094,

s2 = 4 (2 repeated roots) and

s3 = 6.2650.

Also using (4.11) by letting s = 0 and z = 1, we have

s1(1) = 0,

s2,1(1) = 0.2650,

s2,2(1) = 3.7985 − 2.7551i,

s2,3(1) = 3.7985 + 2.7551i,

s2,4(1) = 6.8468 + 1.3693i and

s3(1) = 6.8468 − 1.3693i.

Putting all the above in (4.25) we arrive at the ultimate ruin probability as follows:

ψ(u) = 0.86139 e−0.26496 u − (0.021097 cos (1.3693 u) + 0.011023 sin (1.3693 u)) e−6.8468 u −

(0.017638 cos (2.7551 u) + 0.034102 sin (2.7551 u)) e−3.7985 u.

58 Chapter 5

Ruin Probability for Correlated Poisson Claim Count

and Hyper Exponential Risk Process

5.1 Introduction

The aim of this chapter is to obtain an explicit expression for the ultimate ruin probability when the claim number (counts) is still Poisson distributed but the claim sizes (amounts) are assumed to be Hyper Exponentially distributed and the two random variables are correlated.

5.1.1 Hyper Exponential Distribution

The Hyper Exponential distribution is a mixture of k exponentials for some k. The probability density function in the univariate sense is of the form

k X −λj t h(t) = pjλje , λj > 0, t ≥ 0, (5.1) j=1 where pj ≥ 0 for all j and p1 + ... + pk = 1. Our choice of a Hyper Exponential in this circumstance stems from the fact that it is analytically easy in comparative terms and also belongs to a special class of distributions referred to as phase-type distributions. Given our modeling set-up, it has been found in Neuts (1981) that phase-type distributions are better in respect of tractability. Phase-type distribution is a probability distribution that ensues due to one or more inter-dependent Poisson processes occurring in sequence. It is important to emphasize that Hyper Exponential distributions have explicit Laplace transforms. The

59 Laplace transform of (5.1) is given as

Z ∞ h∗(s) = e−sth(t)dt (5.2) 0 Z ∞ = e−stdH(t) (5.3) 0 k X piλj = . (5.4) λ + s j=1 j

The example shown in Figure 5.1 gives both the two component (two mixture) and three component (three mixture) Hyper Exponential distribution.

60 Figure 5.1: Graph of two and three component Hyper Exponential distributions

61 5.2 Joint Moment Generating Function of (T,Xj)

This time, we assume (T,Xj) for j = 1, 2, . . . , n are Moran and Downton’s bivariate Expo- nential with mgf given as

1 MT,Xj (t1, t2) = h i. (5.5) 1 − t1 1 − t2 − ρ t1t2 β βj ββj

Consider the bivariate distribution of (T,X) constructed by the following

n X fT,X (t, x) = pjfT,Xj (t, x). (5.6) j=1

It is a mixture of Moran and Downton’s bivariate Exponential distribution where pj ≥ 0 and p1 + ... + pn = 1. The mgf of (T,X) is given by

n X pj MT,X (t1, t2) = h i. (5.7) t1 t2 t1t2 j=1 1 − 1 − − ρ β βj ββj

5.3 Characteristic Function of Y and Related Transform

Here too, T is assumed to be the inter-arrival time and Xj for j = 1, 2, ..., n is assumed to be the claim size. Again, to allow the usage of the Wiener-Hopf factorization technique, we

need the mgf of Y = X − cT . Prior to that, we require the mgf of Yj = Xj − cT which is given as follows:

Yj s MYj (s) = E e (5.8)

= E eXj s−csT (5.9)

= MT,Xj (−cs, s) (5.10) 1 = . (5.11) 2 1 + cs 1 − s + cρs β βj ββj

62 Therefore the mgf of Y becomes

n X MY (s) = pjMT,Xj (s) (5.12) j=1 n X pj = . (5.13) cs s cρs2 j=1 1 + 1 − + β βj ββj

We represent the denominator of (5.13) by gj(s) and further expand it to obtain c(1 − ρ) 2 c 1 gj(s) = − s + − s + 1. (5.14) ββj β βj

Since gj(s) is quadratic in s, we can factorize it as follows:

c(1 − ρ) gj(s) = − (s + aj)(s − bj) ββj c(1 − ρ) = (s + aj)(bj − s) ββj s s = 1 + 1 − (5.15) aj bj where aj and bj are respectively

r 2 − c − 1 + c − 1 + 4(1 − ρ) c β βj β βj ββj a = (5.16) j 2(1 − ρ) c ββj

and r 2 c − 1 + c − 1 + 4(1 − ρ) c β βj β βj ββj b = . (5.17) j 2(1 − ρ) c ββj

63 The characteristic function of Y is obtained by replacing s with iω in (5.13). Beyond that,

we require the related factor 1 − zφY (ω) which is obtained as follows:

n X pj 1 − zφ (s) = 1 − z (5.18) Y g (s) j=1 j p p p = 1 − z 1 + 2 + ... + n (5.19) g1(s) g2(s) gn(s) " # p1g2(s)...gn(s) + ... + png1(s)...gn−1(s) = 1 − z Qn (5.20) j=1 gj(s) "Pn Qn # k=1 pk j=1,j6=k gj(s) = 1 − z Qn (5.21) j=1 gj(s) Qn hPn Qn i j=1 gj(s) − z k=1 pk j=1,j6=k gj(s) = Qn . (5.22) j=1 gj(s)

Note the reversion back to s, this is done to deal expeditiously with algebraic derivations.

Since

d E[Y ] = M (s) | ds Y s=0 " n # d X pj = | ds g (s) s=0 j=1 j n X −2 0 = − pj [gj(s)] gj(s) |s=0 j=1 n X −2 0 = − pj [gj(0)] gj(0) j=1

and gi(0) = 1; n X 0 E(Y ) = − pjgj(0) j=1 but 0 −2c c 1 gj(s) = (1 − ρ)s + − ββj β βj and

0 c 1 gj(0) = − . β βj

64 Therefore E[Y ] < 0 implies that

n X c 1 p − > 0 j β β j=1 j n n c X X pj p − > 0 β j β j=1 j=1 j n c X pj − > 0 β β j=1 j n c X pj > . β β j=1 j At this point we can rewrite the denominator of (5.22) as:

n n Y Y c(1 − ρ) h(s) = g (s) = (s + a )(b − s) (5.23) j ββ j j j=1 j=1 j and the numerator as:

n " n n # n Y X Y Y c(1 − ρ) h(s, z) = gj(s) − z pk gj(s) = (s + aj(z))(bj(z) − s) (5.24) ββj j=1 k=1 j=1,j6=k j=1

where aj(z) and bj(z) are the complex roots for j = 1, 2, ..., n and 0 < z < 1. Since the behaviour of the zeros of h(s) as z → 0 which is the same as the zeros of h(s, 0)

has 2n zeros where n of them are positive zeros and the other n are negative zeros. Now,

we need the zeros of h(s, z) for 0 < z < 1. To do this, we employ Rouche’s Theorem in the

proceeding sub-section.

5.3.1 Zeros of a Certain m + 1 Degree Polynomial

For ﬁxed values of c, β, ρ and βj+1 > βj for j = 1, ..., n, we have aj+1 < aj for j = 1, ..., n.

For instance, given c = 3, β = 2, ρ = 0.5, we have a table of βj for j = 1, ..., 10 and the

corresponding aj as shown in Table 5.1.

Similarly if βj+1 > βj for j = 1, ..., n, we have bj+1 > bj for j = 1, ..., n. By way of example,

given c = 3, β = 2, ρ = 0.5, we have a table of βj and the corresponding bj as indicated in

Table 5.2. It is obvious at this stage that a1 is the largest of a1, ..., an and bn is the largest

of b1, ..., bn.

65 Table 5.1: βj against aj

βj aj 2 0.7748517731 2.5 0.7540291165 3.0 0.7398481518 3.5 0.7295930552 4.0 0.7218416878 4.5 0.7157813545 5.0 0.7109153165 5.5 0.7069233785 6.0 0.7035900918 6.5 0.7007652598 7.0 0.6983410372

Table 5.2: βj against bj

βj bj 2 3.441518439 2.5 4.420695782 3.0 5.406514819 3.5 6.396259719 4.0 7.388508351 4.5 8.382448018 5.0 9.377581984 5.5 10.37359004 6.0 11.37025675 6.5 12.36743192 7.0 13.36500770

66 To prove this mathematically we need to rewrite aj and bj as follows:

2 aj = r 2 c − 1 + c − 1 + 4(1 − ρ) c β βj β βj ββj (cβ − β) + p(cβ − β)2 + 4(1 − ρ)cββ b = j j j . j 2(1 − ρ)c

c 1 In the expression for aj we notice that − is the overriding factor and β βj

1 c 1 ↑ βj ⇒↓ ⇒↑ − ⇒↓ aj. βj β βj

From the expression for bj , it is obvious that it increases as βj increases. Now, we need to show by Rouche’s theorem that

n " n n #

Y X Y gj(s) > −z pj gj(s) . (5.25) j=1 k=1 j=1,j6=k

The preceding inequality can be deemed proven if the following condition is met without the slightest reservation.

| gj(s)| ≥ 1, (j = 1, ..., n) (5.26) s s 1 + 1 − ≥ 1, (j = 1, ..., n). (5.27) aj bj

Let C1 and C2 denote two circles deﬁned in the complex plane; C1 has center at −a1 with

radius a1. C2 has center at bn with radius bn. The points on C1 and C2 are respectively

iθ iθ −a1(1 − e ) and bn(1 + e ).

iθ Let s= −a1(1 − e ), then s s |gj(s)| = 1 + 1 − aj bj ! ! a 1 − eiθ a 1 − eiθ 1 1 = 1 − 1 + . aj bj Consider

a 1 − eiθ 1 1 + ≥ 1. bj

67 However,

a 1 − eiθ 1 1 − ≥ 1 aj is not straight forward as the other. To prove this, we let

eiθ = cosθ + isinθ.

iθ 2 2 a1(1 − e ) a1(1 − cosθ − isinθ) 1 − = 1 − aj aj a 2 a 2 = 1 − 1 (1 − cosθ) + 1 sinθ aj aj a a a 2 = 1 − 2 1 + 2 1 cosθ + 1 aj aj aj a 2 a 2 a 2 −2 1 cosθ + 1 cos2θ + 1 sin2θ aj aj aj a a 2 a a = 1 − 2 1 + 2 1 − 2 1 cosθ 1 − 1 . aj aj aj aj Since the minimum occurs at cosθ =1, then iθ 2 a1(1 − e ) 1 − ≥ 1. aj

iθ Let s= bn(1 + e ), then s s |gj(s)| = 1 + 1 − aj bj ! ! b 1 + eiθ b 1 + eiθ n n = 1 + 1 − . aj bj Consider

b 1 + eiθ n 1 + ≥ 1. aj However,

b 1 + eiθ n 1 − ≥ 1 bj is not straight forward as the other. To prove this, we let

eiθ = cosθ + isinθ.

68 iθ 2 2 bn(1 + e ) bn(1 + cosθ + isinθ) 1 − = 1 − bj bj b 2 b 2 = 1 − n (1 + cosθ) + n sinθ bj bj b b b 2 = 1 − 2 n − 2 n cosθ + n bj bj bj b 2 b 2 b 2 +2 n cosθ + n cos2θ + n sin2θ bj bj bj b b 2 b b = 1 − 2 n + 2 n + 2 n cosθ n − 1 . bj bj bj bj Since the minimum occurs at cosθ = −1, then iθ 2 bn(1 + e ) 1 − ≥ 1. bj

By the theorem | gj(s)| ≥ 1. Therefore h(s, z) has the same number of zeros inside each circle |z| < 1. As a consequence, h(s, z) has n zeros inside C1 and n zeros inside C2. In conclusion, h(s, z) has the same number of zeros as h(s).

Therefore the behaviour of the zeros of h(s, 0) is equivalent to the zeros of h(s) as z → 0 is

given by

lim aj(z) → aj(0) = aj, for j = 1, 2, ..., n, z→0

and

lim bj(z) → bj(0) = bj, for j = 1, 2, ..., n. z→0

Also, the behaviour of the zeros of of h(s, 1) equivalent to the zeros of h(s, z) as z → 1 is

also given by

lim a1(z) → a1(1) = 0, z→1

lim aj(z) → aj(1), for j = 2, ..., n, z→1

and

lim bj(z) → bj(1), for j = 1, 2, ..., n. z→1

69 By replacing s with iω, we have

hQn c(1−ρ) i (iω + aj(z))(bj(z) − iω) j=1 ββj 1 − zφY (ω) = (5.28) hQn c(1−ρ) i (iω + aj)(bj − iω) j=1 ββj " n #" n # Y iω + aj(z) Y bj(z) − iω = . (5.29) iω + a b − iω j=1 j j=1 j Now let us write n Y bj(z) − iω D(z, ω) = b − iω j=1 j and n Y iω + aj(z) D¯(z, ω) = . iω + a j=1 j

We rewrite D(z, ω) and D¯(z, ω) as n Y bj − bj(z) D(z, ω) = 1 − b − iω j=1 j and n Y aj − aj(z) D¯(z, ω) = 1 − a + iω j=1 j

and by substituting ω = ω1 + iω2 where ω1 and ω2 are real, we have n Y bj − bj(z) D(z, ω) = 1 − b − iω + ω j=1 j 1 2 and n Y aj − aj(z) D¯(z, ω) = 1 − . a + iω − ω j=1 j 1 2 It is clear at this stage that ¯ 1. D(z, ω) is a bounded function, bounded away from zero for ω2 < 0.

2. D(z, ω) is a bounded function, bounded away from zero for ω2 > 0.

3. limω2→∞ D(z, ω) = 1. Therefore D(z, ω) and D¯(z, ω) are the required factors.

70 5.4 Survival Probability

1 1 − z φL(ω) = lim . D(1, ω) z→1 D¯(z, 0)

Now

n 1 Y bj − iω = D(1, ω) b (1) − iω j=1 j and

1 − z 1 − z = D¯(z, 0) Qn aj (z) j=1 aj n Y aj = (1 − z) . (5.30) a (z) j=1 j

Note that from (5.24)

n " n n # n Y X Y Y aj(z)bj(z)c s s gj(s) − z pk gj(s) = (1 − ρ) 1 + 1 − ; ββj aj(z) bj(z) j=1 k=1 j=1,j6=k j=1 (5.31) when s = 0, (5.31) reduces to

n Y aj(z)bj(z)c 1 − z = (1 − ρ). (5.32) ββ j=1 j

Also from (5.32) as z = 0, we have

n Y ajbjc(1 − ρ) = 1. (5.33) ββ j=1 j

Putting (5.33) into (5.32) and after re-arrangement, we obtain the following:

" n # Y aj(z)bj(z) 1 − z = . (5.34) a b j=1 j j

By (5.30),(5.34) and taking the limit, we obtain

" n # n Y aj Y bj(1) lim(1 − z) = . (5.35) z→1 a (z) b j=1 j j=1 j

71 Now, " n # Y bj(1) bj − iω φ (ω) = (5.36) L b b (1) − iω j=1 j j " n 1 − iω !# Y bj = . (5.37) 1 − iω j=1 bj (1) The Laplace transform of the survival probability is given as

1 δ∗(s) = M (−s) s L h i Qn 1 + s j=1 bj = h i s Qn 1 + s j=1 bj (1) that can be expanded into partial fractions as follows:

A B B = + 1 + ... + n s 1 + s 1 + s b1(1) bn(1) n A X Bi = + . (5.38) s 1 + s i=1 bi(1)

A and Bj are obtained by successively substituting s = 0 and s = −bj(1) for j = 1, ..., n in both sides of the following equation

" n # " n # n " n # Y s Y s X Y s 1 + = A 1 + + s Bi 1 + . bj bj(1) bj(1) j=1 j=1 i=1 j=1,j6=i We have A = 1 and h i Qn 1 − bi(1) j=1 bj Bi = − h i. Qn bi(1) bi(1) 1 − j=1,j6=i bj (1) Therefore h i n Qn bi(1) j=1 1 − ∗ 1 X bj 1 δ (s) = − h i s Qn bi(1) s i=1 bi(1) 1 − 1 + j=1,j6=i bj (1) bi(1) h i n Qn 1 − bi(1) 1 X j=1 bj 1 = − h i . (5.39) s Qn bi(1) (b (1) + s) i=1 1 − i j=1,j6=i bj (1)

72 By taking the inverse Laplace transform of (5.39), we obtain h i n Qn bi(1) j=1 1 − X bj −b (1)u δ(u) = 1 − h ie i . (5.40) Qn bi(1) i=1 1 − j=1,j6=i bj (1)

Hence the ultimate ruin probability becomes h i n Qn bi(1) j=1 1 − X bj −b (1)u ψ(u) = h ie i . (5.41) Qn bi(1) i=1 1 − j=1,j6=i bj (1)

5.5 Numerical Examples

In this section, we give some numerical examples.

Example 1

1 Consider the case where m = 2, c = 3, β = 2, β1 = 2.2, β2 = 2.4, ρ = 5 , p1 = 0.5 and

p2 = 0.5. By employing (5.16) and (5.17), we obtain

a1 = −0.6981

a2 = −0.7005

b1 = 2.6172

b2 = 2.8648.

Also using (5.24) by letting z = 1 and solving, we have

a1(1) = 0

a2(1) = −0.6993

b1(1) = 2.0202

b2(1) = 2.7624.

Substituting all the above in (5.41), the resulting ultimate ruin probability is given as

ψ(u) = 0.2503e−2.0202u + 0.0054e−2.7624u.

73 Example 2

2 Consider the case where m = 3, c = 3, β = 2, β1 = 2.5, β2 = 3.0, β3 = 3.5, ρ = 5 , p1 = 0.3,

p2 = 0.33 and p3 = 0.37. By employing (5.16) and (5.17), we obtain

a1 = −0.7152

a2 = −0.7229

a3 = −0.7332

b1 = 3.7888

b2 = 4.6120

b3 = 5.4375.

Also using (5.24) by letting z = 1 and solving, we have

a1(1) = 0

a2(1) = −0.71890

a3(1) = −0.72940

b1(1) = 4.38194

b2(1) = 5.27234

b3(1) = 3.46104.

Substituting all the above in (5.41), the resulting ultimate ruin probability is given as

ψ(u) = 0.10866e−3.46104u + 0.033729e−4.38194u + 0.016013e−5.27234u

Example 3

Again, we consider the situation where m = 2, c = 3, β = 2, β1 = 2.2, β2 = 2.4, p1 = 0.5, p2 = 0.5. In Figure 5.2, we display ultimate ruin probability for diﬀerent values of the correlation coeﬃcient (ρ). It is clear and understandable from the graph that, positive correlation relates inversely with ultimate ruin probability i.e. positive correlation diminishes

74 the magnitude of ultimate ruin probability. This observation is attributed to the fact that, larger claims occur later than smaller claims.

Figure 5.2: Eﬀect of correlation on ruin probability in a two-component Hyper Exponential distribution

Example 4

Also, we look at the case where m = 3, c = 3, β = 2, β1 = 2.5, β2 = 3.0, β3 = 3.5,

75 p1 = 0.3, p2 = 0.33 and p3 = 0.37. In Figure 5.3, we display ultimate ruin probability for varying values of the correlation coeﬃcient (ρ). It is evident from the graph that, positive correlation relates inversely with ultimate ruin probability i.e. positive correlation reduces the eﬀect of ultimate ruin probability. This observation is attributed to the fact that, larger claims occur later than smaller claims.

Figure 5.3: Eﬀect of correlation on ruin probability in a three-component Hyper Exponential distribution

76 Chapter 6

Ruin Probability for Correlated Poisson Claim Count

and Pareto Risk Process

6.1 Introduction

The aim of this chapter is to obtain approximation for the ultimate ruin probability when the claim number (counts) is still Poisson distributed but the claim sizes (amounts) are assumed to be Pareto distributed and the two random variables are correlated. In the literature, the Pareto distribution belongs to a very important class of distributions called heavy-tailed claim (long-tailed) distributions which has already been mentioned in details in the introductory part of the thesis. To advance the discussion, it may not be superﬂuous to

ﬁrst deﬁne the Pareto distribution.

6.1.1 Pareto Distribution

A random variable X is said to be a Pareto distribution (Pareto type II distribution with

µ = 0) with parameters α and λ if its distribution function is given by

λ α F (x) = 1 − , x ≥ 0. (6.1) X λ + x

The equivalent density function is given as

αλα fX (x) = , x ≥ 0. (6.2) (λ + x)α+1

Here, both α and λ are positive real numbers. The Pareto distribution is much similar to the

Exponential distribution and this is not magical because the Pareto is a continuous mixture of Exponentials with Gamma mixing weights. A possible graph of fX is given in Figure 6.1. For any Exponential-Pareto pair of distributions with the same means, the tail of the

77 Pareto distribution will always be heavier that the tail of the corresponding Exponential distribution.

Figure 6.1: Graph of Pareto density function

6.1.2 Method of Moments Matching

Also, in the derivations and computations to come later, it is necessary to have higher moments for joint and correlated random variables of the Exponential and Pareto distributions

78 as well as Laplace transforms of these distributions. Evidently, no convenient expression for

the Laplace transform of the Pareto distribution exist. However, eﬀective approaches are in

place to evade this complication. A signiﬁcant contribution in this thesis is to indicate a pos-

sibility of approximating a Pareto distribution by a suitable light tailed claim distributions.

One approach we have sought to follow is the method of moments matching. The approach

is such that, we ﬁrst ﬁx the parameters of the bivariate Exponential-Pareto distribution,

then calculate the kth moments. We then match these computed moments to the moments of the approximating distribution to estimate its parameters.

Mathematically, a method of moments estimate of π is any solution of p equations

0 0 µk(π) =µ ˆk, k = 1, 2, ...p. (6.3)

6.2 The Bivariate Exponential-Pareto Distribution

We may have observed from all preceding chapters starting from chapter 3 that the consid-

eration has been how to model T and X jointly in the presence of correlation and of the form

W = X − cT . In this section, our interest is still in furtherance of this idea but rather make

the assumption that T which denotes the claim inter-arrival time follows an Exponential

distribution whereas X follows a Pareto distribution. Going forward, we require the joint

cummulative distribution function (CDF) of (T,X). Since one of our assumptions too is

that T and X are correlated, we begin our setup by deﬁning the following random variables,

T ∼ Exp(θ1), θ1 > 0;

Y ∼ Exp(θ2), θ2 > 0;

X ∼ P areto(α, λ), α > 0, λ > 0;

where (T,Y ) is Moran and Downton’s bivariate Exponential distribution as in the form

shown in (1.13).

79 Let

θ1 = θ,

θ2 = 1,

−y FY (y) = 1 − e , λ α F (x) = 1 − . X λ + x

We solve for y by equating FY (y) to FX (x) as follows.

λ α 1 − e−y = 1 − , λ + x λ α e−y = , λ + x λ + xα y = ln . (6.4) λ

Since X → (0, ∞) as Y → (0, ∞), we therefore have the joint probability density function

(PDF) of (T,X) as follows:

−1 dy fT,X (t, x) = f (t, y) T,Y dx " λ+x α # " λ+x α 1 # θα θt + ln 2(ρθtln ) 2 = exp − λ I λ . (6.5) (1 − ρ)(λ + x) 1 − ρ 0 1 − ρ

The CDF becomes

" λ+v α # " λ+v α 1 # Z t Z x 2 θα θu + ln λ 2(ρθuln λ ) FT,X (t, x) = exp − I0 dudv. 0 0 (1 − ρ)(λ + v) 1 − ρ 1 − ρ (6.6)

( z )2r where I (z) = P∞ 2 is the well-known modiﬁed Bessel function of the ﬁrst kind of 0 r=0 (r!)2 order 0. In (6.6), T and X follow respectively Exponential and Pareto distribution.

Now, the moments of T (E(T k)) for k = 1, 2, ..., n can easily be obtained as follows:

k! E(T k) = . (6.7) θk

80 Also, the moments of X (E(Xk)) for k = 1, 2, ..., n can be obtained as follows: k!λk(α − k − 1)! E(Xk) = . (6.8) (α − 1)! The joint moment of T and X from (6.6) appears very complicated, so we ﬁrst need to work out X from (6.4) as shown

Y X = λ e α − 1 (6.9) then we use X as represented by (6.9) to obtain expressions involving (Y,T ). As it were,

(Y,T ) forms the Moran and Downton’s bivariate Exponential distribution, so we can write the joint mgf as

T t1+Y t2 MY,T (t1, t2) = E e 1 = (6.10) t1 ρt1t2 1 − θ (1 − t2) − θ then take the kth derivative as shown ∂k ∂k 1 E eT t1+Y t2 = . (6.11) k k t1 ρt1t2 ∂t1 ∂t1 1 − θ (1 − t2) − θ

Depending on the joint moment of interest, we evaluate both sides of (6.11) at t1 = 0 and

a t2 = α to obtain k h k aY i ∂ 1 a E T e α = evaluate at t1 = 0, t2 = . (6.12) k t1 ρt1t2 α ∂t1 1 − θ (1 − t2) − θ These derivatives can easily be obtained from MAPLE. We derive the ﬁrst joint moment between T and X as follows:

h Y i E (XT ) = E λT e α − 1

Y = λE T e α − λE (T ) λα (α − 1 + ρ) λ = − . (6.13) θ (α − 1)2 θ Note that h Y i ∂ 1 1 E T e α = evaluate at t1 = 0, t2 = ∂t t1 ρt1t2 α 1 1 − θ (1 − t2) − θ α (α − 1 + ρ) = . (6.14) θ (α − 1)2

81 The correlation coeﬃcient between T and X is obtained as follows

cov(T,X) ρ = σT σX E (XT ) − E (X) E (T ) = σT σX α (α−1+ρ) λ 2 − = θ (α−1) θ(α−1) λ p α θ(α−1) α−2 α(α − 1 + ρ) − λ(α − 1) = p α . (6.15) λ(α − 1) α−2

Also, note that s λ2 λ2 σX = 2 − (α − 1) (α − 2) (α − 1)2 s λ2α = (α − 1)2 (α − 2) λ r α = . (6.16) (α − 1) α − 2

Now, we need to calculate the kthmoments of W = (X − cT ) where (T,X) is of the form shown by (6.6).

Let

W k = (X − cT )k, k = 1, 2, ...n. (6.17)

As there are 6 parameters in this situation, the moments for k = 1, 2, 3, 4, 5, 6 are obtained

as follows:

E(W k) = E(X − cT )k k X k k−i i = E (X) E (−cT ) . (6.18) i i=0 The ﬁrst and second moments of W are respectively

E(W ) = E(X) − cE(T ) λ c = − (6.19) α − 1 θ

82 and

E(W 2) = E(X2) − 2cE(XT ) + c2E(T 2) 2λ2 λα (α − 1 + ρ) λ 2c2 = − 2c − + . (6.20) (α − 1)(α − 2) θ (α − 1)2 θ θ2

Higher moments can be obtained by utilizing (6.7) to (6.12) and (6.18).

6.3 Approximating Distributions

The two approximating distributions we are giving consideration in this section as far as the

Pareto distribution is concerned in respect of modeling the claim sizes are:

• Mixture of Exponential (Hyper Exponential) distributions.

• Gamma (m) distribution.

6.3.1 Mixture of Exponential Distributions

We commence by letting Y = X − cT , then

Y s MY (s) = E e

= E es(X−cT ) n X pj = cs s ρcs2 j=1 1 + 1 − + β βj ββj n X pjajbj 1 1 = + . (6.21) a + b b − s a + s j=1 j j j j

Note that aj and bj are the same as in (5.16) and (5.17). Now, taking the inverse of (6.21),

we obtain the density function of Y i.e. fY (y) as

 n p a b  P j j j e−bj y y > 0,  j=1 aj +bj    n p a b  P j j j eaj y y < 0. j=1 aj +bj

83 The next stage is to derive the moments of Y .

n Z ∞ Z 0 X pjajbj E(Y k) = yke−bydy + ykeaydy (6.22) a + b j=1 j j 0 −∞ n Z ∞ Z 0 X pjajbj = yke−bydy − (−x)ke−axdx (x = −y) (6.23) a + b j=1 j j 0 ∞ n Z ∞ Z ∞ X pjajbj = yke−bydy + (−1)k xke−axdx (6.24) a + b j=1 j j 0 0 n " # X pjajbj Γ(k + 1) Γ(k + 1) = + (−1)k (6.25) a + b k+1 k+1 j=1 j j bj aj n " # X pjajbj 1 1 = + (−1)k k!. (6.26) a + b k+1 k+1 j=1 j j bj aj

We now match moments of Y with the moments of W according to the deﬁnition of matching moments to obtain parameter estimates for the approximating distribution.

6.3.2 Gamma (m) Distribution

X1 X2+Z Here too we proceed on the premise that, T = and X = and (X1,X2) form the β1 β2 Moran and Downton’s bivariate Exponential. Z follows a Gamma distribution with shape parameter m − 1 and is independent of X1 and X2. The joint mgf becomes

X1 X2+Z t1 β +t2 β MT,X (t1, t2) = E e 1 2

X1 X2 Z h t1 +t2 i h t2 i = E e β1 β2 E e β2 .

We write

Y = X − cT X + Z X = 2 − c 1 β2 β1 c 1 1 = −X1 + X2 + Z . β1 β2 β2

84 Therefore, the mgf of Y becomes

cs s s −X1 β +X2 β Z β MY (s) = E e 1 2 E e 2 1 = , (6.27) m−1 1 + s 1 − s 1 − s a b β2

where a and b are given as (3.11) and (3.12) respectively.

From the foregoing we can partition Y into Y1 ((X1,X2) form Moran and Downton’s bivariate

Exponential distribution) and Y2 (Gamma distribution with shape parameter m − 1) and

both sum up to Y . Also, it is increasingly evident from (6.21) that Y1 has PDF of the form

  ab e−by y > 0,  a+b 

  ab ay  a+b e y < 0.

Since Y2 ∼Gamma(m − 1, β2), we write its moments as follows:

k Γ(m − 1 + k) E Y2 = k . (6.28) Γ(m − 1)β2

It is easy to show this from the standard Gamma distribution.

Z ∞ k k E Y2 = y fY2 (y)dy 0 Z ∞ βm−1 = yk 2 y(m−1)−1e−β2ydy 0 Γ(m − 1) Z ∞ βm−1 = 2 y(m−1+k)−1e−β2ydy 0 Γ(m − 1) βm−1 Γ(m + k − 1) Z ∞ βm+k−1 2 2 (m−1+k)−1 −β2y = m+k−1 y e dy Γ(m − 1) β2 0 Γ(m + k − 1) Γ(m + k − 1) = k . β2 Γ(m − 1)

Therefore the moments of Y are obtained as follows:

k X k E Y k = E Y i E Y k−i , i 1 2 i=0

85 where

ab 1 1 E Y k = + (−1) k! 1 a + b bk+1 ak+1 and

k (m + k − 2)! E Y2 = k . (m − 2)!β2

We now match moments of Y with the moments of W according to the deﬁnition of matching

moments to obtain parameter estimates for the approximating distribution.

6.4 Numerical Results

In this section, we present the numerical results for both the simulated and the approxi-

mated ruin probabilities. In our comparison, we only show the results for the case of the

Exponential-Hyper Exponential distribution and the Exponential-Pareto distribution. The

results are presented in four scenarios.

For the simulated ruin probability, we ﬁxed the parameters of the Exponential-Pareto dis-

9 tribution as follows: θ = 3, c = 5, α = 13, λ = 13 and ρ = 10 . In the case of the approximated ruin probability, we employ the nonlinear least squares curve

ﬁtting method to obtain starting estimates. We then match moments of the two distribu-

tions to estimate the parameters of the approximating distribution.

By employing (5.24) as z = 1 and (5.41), the resulting ultimate ruin probability is obtained.

We repeated this procedure under four scenarios by varying the values of the parameters of the Exponential-Pareto distribution.

86 Scenario 1 0.4 Simulated Ruin Probability 0.35 Approximated Ruin Probability

0.3

0.25

0.2

Ruin Probability 0.15

0.1

0.05

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 u (Initial Capital)

Figure 6.2: Comparison of Simulated ruin probability and Approximated ruin probability in scenario 1.

87 Scenario 2 0.25 Simulated Ruin Probability Approximated Ruin Probability

0.2

0.15

0.1 Ruin Probability

0.05

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 u (Initial Capital)

Figure 6.3: Comparison of Simulated ruin probability and Approximated ruin probability in scenario 2.

88 Scenario 3 0.45 Simulated Ruin Probability 0.4 Approximated Ruin Probability

0.35

0.3

0.25

0.2 Ruin Probability 0.15

0.1

0.05

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 u (Initial Capital)

Figure 6.4: Comparison of Simulated ruin probability and Approximated ruin probability in scenario 3.

89 Scenario 4 0.4 Simulated Ruin Probability 0.35 Approximated Ruin Probability

0.3

0.25

0.2

Ruin Probability 0.15

0.1

0.05

0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 u (Initial Capital)

Figure 6.5: Comparison of Simulated ruin probability and Approximated ruin probability in scenario 4.

90 Figures 6.2, 6.3, 6.4 and 6.5 show the results of this comparison and it can be seen that, the Exponential-Hyper Exponential distribution is not a good approximation for the

Exponential-Pareto distribution when the method of moment matching is used as a method of estimating the parameters of the approximating distribution. We also looked at the behaviour of Exponential-Gamma (m) distribution also as an approximation of the Exponential-

Pareto distribution. It is imperative to mention that it did not provide a good ﬁt either, details of that results have been omitted as it has no eﬀect in arriving at new results in this thesis.

In the next chapter, we give consideration to another form of approximating distribution in addition to an alternative parameter estimation procedure.

91 Chapter 7

Asymmetric Double Exponential Distributions and its

Finite Mixtures

7.1 Introduction

In the preceding chapter, we have sought to use both the Exponential-Hyper Exponential and

Exponential-Gamma (m) distributions to approximate the Exponential-Pareto distribution by method of moment matching prior to computing the ruin probability. Indeed, in as much as the derivations and the algebra were accurately carried out, which presuppose that theoretically the two distributions should be equivalent, rather the numerical results clearly indicate a poor fit in both cases as we have seen earlier. In this chapter therefore, our interest is to propose another approach which is by far better than what we have already discussed under Chapter 6. To be more precise, we will be discussing another estimating method that exploits the capabilities of the EM algorithm by fitting a finite mixture of asymmetric Double

Exponential distributions. In this regard, it may not be out of place to ﬁrst and foremost introduce the Double Exponential distribution and further discuss ﬁnite mixture models.

7.1.1 Double Exponential Distribution

The Double Exponential distribution as shown in Casella and Berger (2002) is created by reﬂecting the Exponential distribution around its mean and its PDF is given as

1 − |x−µ| f(x|µ, σ) = e σ , −∞ < x < ∞, −∞ < µ < ∞, σ > 0. (7.1) 2σ

Note that µ and σ are the location and scale parameters respectively. The Double Expo- nential distribution is not bell-shaped and the absolute value sign can pose problems during

92 integration, it is therefore imperative to divide the region of integration into two parts around x = µ.

7.1.2 Finite Mixture Models

Finite Mixture of distributions have gained enormous significance with the emergence of high speed and low cost computers. Also in the last few years, the literature has grown to such a point that practical applications are being felt and continued to be used in many fields such as medicine, engineering and social science to mention but a few. The first documented major analysis regarding finite mixture models was by Karl Pearson in his paper, Pearson

(1894) which involved a moment-based approach to ﬁtting of a normal mixture model.

Many distributions of continuous type can easily be approximated randomly well by a finite mixture of normal densities with common variance. Mixture models therefore, provide an efficient, effective and adaptable semi-parametric approach in modeling unidentifiable distri- butional shapes no matter what the aim may be. Finite mixture models are very flexible and convenient in dealing with situations where a single parametric family lacks the capacity to provide satisfactory model for local variations in an observed data.

In the 1960’s, the fitting of finite mixture models by maximum likelihood (ML) became widespread. However, the seminal paper of Dempster, Laird, and Rubin (1977) on the EM algorithm sparked the interest to use finite mixture models to study heterogeneous data.

McLachlan and Peel (2000) is among few excellent reference on finite mixture models. The use of EM algorithm for the fitting of finite mixture models, especially mixtures from the exponential family of distributions has been demonstrated by Hasselblad (1969). The class of mixtures and the estimation procedure employed in this thesis are synonymous with

Hasselblad (1969). The ”EM” in EM algorithm is an acronym which stands for expectation- maximization and the EM algorithm is an iterative statistical procedure for searching for the maximum likelihood estimates of parameters of a model that relies on unobserved latent variables. The iterative process always involves two steps, we have the expectation (E) step

93 which is the expectation of the log-likelihood and the maximization (M) step which maxi-

mizes the expectation of the log-likelihood performed during the E step. In the next section,

we delve further into asymmetric Double Exponential distribution.

7.2 Asymmetric Double Exponential Distribution

Let Xi, i = 1, 2 are two independent Exponentially distributed random variables with pa-

rameters θi, i = 1, 2 respectively. Consider the distribution of Y = X2 − X1. The PDF of Y can be obtained by using the transformed method or by inverting the mgf.  θ θ  eθ1y y < 0, f (y) = 1 2 Y θ + θ 1 2  e−θ2y y > 0.

Another way to write this distribution is

θ1θ2 fY (y) = exp(θ1T1(y) − θ2T2(y)) θ1 + θ2

where T1(y) and T2(y) are given as follows:   y y < 0, T1(y) = (7.2)  0 y > 0,   0 y < 0, T2(y) = (7.3)  y y > 0.

7.2.1 MLE of Parameters

Let y1, y2, . . . , yn be a random sample from the above distribution. Let us denote

n 1 X y¯ = T (y) (7.4) − n 1 i=1 and

n 1 X y¯ = T (y). (7.5) + n 2 i=1

94 Let the joint PDF of Y1, ..., Yn be a function of (θ1, θ2). Let us also denote the likelihood function L(θ1, θ2; y1, ..., yn) and is given as n Y θ1θ2 L(θ , θ ; y , ..., y ) = exp(θ T (y ) − θ T (y )) (7.6) 1 2 1 n θ + θ 1 1 i 2 2 i i=1 1 2 n n n ! θ1θ2 X X = exp θ T (y ) − θ T (y ) . (7.7) θ + θ 1 1 i 2 2 i 1 2 i=1 i=1 The corresponding log-likelihood function becomes

n n X X logL(θ1, θ2; y1, ..., yn) = n log(θ1θ2) − n log(θ1 + θ2) + θ1 T1(yi) − θ2 T2(yi)(7.8) i=1 i=1 and (7.8) is maximized by setting its ﬁrst derivative with respect to θ1 and θ2 to 0. That is solving the following two equations

∂logL(θ , θ ; y , ..., y ) 1 2 1 n = 0 ∂θ1 and ∂logL(θ , θ ; y , ..., y ) 1 2 1 n = 0. ∂θ2 ˆ For θ1, we have n nθ2 n X − = − T (y ) θ θ θ + θ 1 i 1 2 1 2 i=1 1 1 − = −y¯− θ1 θ1 + θ2 1 1 θ2 − = −y¯−, r = θ1 θ1(1 + r) θ1 ˆ r 1 θ1 = − . (r + 1) y¯− ˆ For θ2, we have n nθ1 n X − = T (y ) θ θ θ + θ 2 i 1 2 1 2 i=1 1 1 − =y ¯+ θ2 θ1 + θ2 1 r θ2 − =y ¯+, r = θ2 θ2(1 + r) θ1 ˆ 1 1 θ2 = . (r + 1) y¯+

95 Note that, by letting θ2 = rθ1, we also have

ˆ 1 1 θ1 = . r(r + 1) y¯+

θ2 Also by letting θ1 = r , we have 2 ˆ r 1 θ2 = − . (r + 1) y¯− However,

θˆ r = 2 ˆ θ1 1 1 y¯ (r + 1) = − − (r + 1) y¯+ r y¯ r2 = − − y¯+ r y¯ r = − − . y¯+ Then by maximizing the log-likelihood functions we can obtain MLEs as follows:

ˆ r 1 1 1 θ1 = − = , (7.9) (r + 1) y¯− r(r + 1) y¯+ and

2 ˆ r 1 1 1 θ2 = − = (7.10) (r + 1) y¯− (r + 1) y¯+ where r is given by θˆ r y¯ r = 2 = − − . ˆ θ1 y¯+

7.3 Finite Mixture

We now approximate the CDF of Y = X − cT where (T,X) is Exponential-Pareto distri-

bution with the CDF of (7.11) where (T,X) is Exponential-Hyper Exponential distribution.

The complete data is represented by Y whereas the ﬁnite mixture of asymmetric Double

Exponential random variables are the latent variables.

K X fX (x) = pifX (x, ai, bi). (7.11) i=1

96 Here

aibi fX (x, ai, bi) = exp(aiT1(x) − biT2(x)) ai + bi PK i.e. each component is asymmetric Double Exponential PDF and i=1 pi = 1. Estimating parameters of this distribution is quite diﬃcult as it contains K − 1 + 2K parameters, So

we employ EM algorithm. We used the original method outlined in Hasselblad (1969). For

the initial estimates of ai, bi we calculate MLEs assuming all a = ai and b = bi and then by

setting ai = a ∗ random(0, 1), and bi = b ∗ random(0, 1) and for p we just used a random number scaled to 1.

The algorithm works as follows.

K X v v v g(xi) = pj fX (xi, aj , bj ), j=1 f (x ,av,bv)T (x ) PN X i j j s i i=1 g(xi) tsj = f (x ,av,bv) , s = 1, 2, j = 1, 2, ..K., PN X i j j i=1 g(xi) s t1j rj = − , j = 1, 2,...,K, t2j

v+1 1 1 aj = , j = 1, 2,...,K, rj(rj + 1) t2j v+1 v bj = rjaj j = 1, 2,...,K, and

pvf (x ,av,bv) PN j X i j j i=1 pv+1 = g(xi) j = 1, 2,...,K. j N

The algorithm was veriﬁed using Monte Carlo methods.

7.4 Numerical Results

In this section, we perform numerical analysis to compare ruin probability based on the ex-

act distribution to the ruin probability based on the approximated distribution. We refer to

the ruin probability emanating from the exact distribution as the simulated ruin probability

97 whereas the one from the approximated distribution as the approximated ruin probability.

Note that, the exact distribution is referred to as the Exponential-Pareto distribution and

that of the approximated distribution is called the asymmetric ﬁnite mixture of Double Ex-

ponential Distributions.

For the case of simulated ruin probability, we use (6.6) with parameter values ﬁxed at:

ρ = 0.5, θ = 5, λ = 3, α = 4, c = 6. Also, we set the number of simulations at 1000 and the

simulated number of claims per run also at 1000.

In the case of the approximated ruin probability, we ﬁrst draw a sample of size N from the distribution of Y = X −cT where (T,X) is Exponential-Pareto distribution. Then, we again assume that sample of the same size is from the distribution Y = X − cT however, T is

Exponential and X is Hyper Exponential distribution then we estimate the parameters.

Case 1

For K = 2 and simulated sample size = 10000, we have

i pi ai bi

1 0.2250 0.8563 0.5409

2 0.7750 0.9353 1.7045

Also using the idea of (5.24) as z = 1 and (5.41), the resulting ultimate ruin probability

is given as

ψ(u) = 0.73717 e−0.16177 u + 0.059234 e−1.1603 u.

98 Table 7.1: Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 2 and simulated sample size = 10000. Initial capital Simulated ruin probability Approximated ruin probability Diﬀerence 0 0.8190 0.7964 0.0226 2 0.5790 0.5392 0.0398 4 0.4260 0.3865 0.0395 6 0.3190 0.2793 0.0397 8 0.2440 0.2021 0.0419 10 0.1890 0.1462 0.0428 12 0.1500 0.1058 0.0442 14 0.1140 0.0766 0.0374 16 0.0970 0.0554 0.0416 18 0.0750 0.0401 0.0349 20 0.0610 0.0290 0.0320 22 0.0480 0.0210 0.0270 24 0.0390 0.0152 0.0238 26 0.0310 0.0110 0.0200 28 0.0190 0.0080 0.0110 30 0.0120 0.0058 0.0062 32 0.0070 0.0042 0.0028 34 0.0060 0.0030 0.0030 36 0.0060 0.0022 0.0038 38 0.0060 0.0016 0.0044 40 0.0050 0.0011 0.0039 42 0.0050 0.0008 0.0042 44 0.0040 0.0006 0.0034 46 0.0040 0.0004 0.0036 48 0.0030 0.0003 0.0027 50 0.0030 0.0002 0.0028

99 Number of Components = 2 and Simulation Sample Size = 10000 0.9 Simulated Ruin Probability 0.8 Approximated Ruin Probability

0.7

0.6

0.5

0.4 Ruin Probability 0.3

0.2

0.1

0 0 5 10 15 20 25 30 35 40 45 50 u (Initial Capital)

Figure 7.1: Comparison of Simulated ruin probability and Approximated ruin probability for a two component when simulation sample size is 10000.

100 Case 2

For K = 2 and simulated sample size = 15000, we have

i pi ai bi

1 0.8724 0.9278 1.5148

2 0.1276 0.9707 0.4414

Also using the idea of (5.24) as z = 1 and (5.41), the resulting ultimate ruin probability is given as

ψ(u) = 0.7245191391 e−0.1463183169 u + 0.08308906227 e−0.8791753099 u.

101 Table 7.2: Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 2 and simulated sample size = 15000. Initial capital Simulated ruin probability Approximated ruin probability Diﬀerence 0 0.8190 0.8076 0.0114 2 0.5790 0.5550 0.0240 4 0.4260 0.4060 0.0200 6 0.3190 0.3016 0.0174 8 0.2440 0.2248 0.0192 10 0.1890 0.1677 0.0213 12 0.1500 0.1252 0.0248 14 0.1140 0.0934 0.0206 16 0.0970 0.0697 0.0273 18 0.0750 0.0520 0.0230 20 0.0610 0.0388 0.0222 22 0.0480 0.0290 0.0190 24 0.0390 0.0216 0.0174 26 0.0310 0.0161 0.0149 28 0.0190 0.0120 0.0070 30 0.0120 0.0090 0.0030 32 0.0070 0.0067 0.0003 34 0.0060 0.0050 0.0010 36 0.0060 0.0037 0.0023 38 0.0060 0.0028 0.0032 40 0.0050 0.0021 0.0029 42 0.0050 0.0016 0.0034 44 0.0040 0.0012 0.0028 46 0.0040 0.0009 0.0031 48 0.0030 0.0006 0.0024 50 0.0030 0.0005 0.0025

102 Number of Components = 2 and Simulation Sample Size = 15000 0.9 Simulated Ruin Probability 0.8 Approximated Ruin Probability

0.7

0.6

0.5

0.4 Ruin Probability 0.3

0.2

0.1

0 0 5 10 15 20 25 30 35 40 45 50 u (Initial Capital)

Figure 7.2: Comparison of Simulated ruin probability and Approximated ruin probability for a two component when simulation sample size is 15000.

103 Case 3

For K = 2 and simulated sample size = 20000, we have

i pi ai bi

1 0.1556 0.8960 0.4565

2 0.8444 0.9515 1.6247

Also using the idea of (5.24) as z = 1 and (5.41), the resulting ultimate ruin probability is given as

ψ(u) = 0.07605012370 e−0.9930760987 u + 0.7342279565 e−0.1416931789 u.

104 Table 7.3: Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 2 and simulated sample size = 20000. Initial capital Simulated ruin probability Approximated ruin probability Diﬀerence 0 0.8190 0.8103 0.0087 2 0.5790 0.5635 0.0155 4 0.4260 0.4180 0.0080 6 0.3190 0.3140 0.0050 8 0.2440 0.2364 0.0076 10 0.1890 0.1780 0.0110 12 0.1500 0.1341 0.0159 14 0.1140 0.1010 0.0130 16 0.0970 0.0761 0.0209 18 0.0750 0.0573 0.0177 20 0.0610 0.0432 0.0178 22 0.0480 0.0325 0.0155 24 0.0390 0.0245 0.0145 26 0.0310 0.0184 0.0126 28 0.0190 0.0139 0.0051 30 0.0120 0.0105 0.0015 32 0.0070 0.0079 -0.0009 34 0.0060 0.0059 0.0001 36 0.0060 0.0045 0.0015 38 0.0060 0.0034 0.0026 40 0.0050 0.0025 0.0025 42 0.0050 0.0019 0.0031 44 0.0040 0.0014 0.0026 46 0.0040 0.0011 0.0029 48 0.0030 0.0008 0.0022 50 0.0030 0.0006 0.0024

105 Number of Components = 2 and Simulation Sample Size = 20000 0.9 Simulated Ruin Probability 0.8 Approximated Ruin Probability

0.7

0.6

0.5

0.4 Ruin Probability 0.3

0.2

0.1

0 0 5 10 15 20 25 30 35 40 45 50 u (Initial Capital)

Figure 7.3: Comparison of Simulated ruin probability and Approximated ruin probability for a two component when simulation sample size is 20000.

106 Case 4

For K = 3 and simulated sample size = 15000, we have

i pi ai bi

1 0.5091 0.9499 2.1458

2 0.0317 0.8257 0.2788

3 0.4592 0.9290 0.8986

Also using the idea of (5.24) as z = 1 and (5.41), the resulting ultimate ruin probability is given as

ψ(u) = 0.1335323931 e−0.4283067134 u+0.02097336689 e−1.820970367 u+0.6512863430 e−0.1338620741 u.

107 Table 7.4: Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 3 and simulated sample size = 15000. Initial capital Simulated ruin probability Approximated ruin probability Diﬀerence 0 0.8190 0.8058 0.0132 2 0.5790 0.5556 0.0234 4 0.4260 0.4054 0.0206 6 0.3190 0.3019 0.0171 8 0.2440 0.2275 0.0165 10 0.1890 0.1726 0.0164 12 0.1500 0.1314 0.0186 14 0.1140 0.1003 0.0137 16 0.0970 0.0766 0.0204 18 0.0750 0.0586 0.0164 20 0.0610 0.0448 0.0162 22 0.0480 0.0343 0.0137 24 0.0390 0.0262 0.0128 26 0.0310 0.0201 0.0109 28 0.0190 0.0153 0.0037 30 0.0120 0.0117 0.0003 32 0.0070 0.0090 -0.0020 34 0.0060 0.0069 -0.0009 36 0.0060 0.0053 0.0007 38 0.0060 0.0040 0.0020 40 0.0050 0.0031 0.0019 42 0.0050 0.0024 0.0026 44 0.0040 0.0018 0.0022 46 0.0040 0.0014 0.0026 48 0.0030 0.0011 0.0019 50 0.0030 0.0008 0.0022

108 Number of Components = 3 and Simulation Sample Size = 15000 0.9 Simulated Ruin Probability 0.8 Approximated Ruin Probability

0.7

0.6

0.5

0.4 Ruin Probability 0.3

0.2

0.1

0 0 5 10 15 20 25 30 35 40 45 50 u (Initial Capital)

Figure 7.4: Comparison of Simulated ruin probability and Approximated ruin probability for a three component when simulation sample size is 15000.

109 Case 5

For K = 3 and simulated sample size = 20000, we have

i pi ai bi

1 0.0147 0.8770 0.2081

2 0.3571 0.9165 0.7508

3 0.6282 0.9686 2.0387

Also using the idea of (5.24) as z = 1 and (5.41), the resulting ultimate ruin probability is given as

ψ(u) = 0.5848015623 e−0.119100579 u + 0.1919203873 e−0.3182409455 u

+0.03211654587 e−1.606497962 u.

110 Table 7.5: Comparison of Simulated ruin probability and Approximated ruin probability when the number of components = 3 and simulated sample size = 20000. Initial capital Simulated ruin probability Approximated ruin probability Diﬀerence 0 0.8190 0.8088 0.0102 2 0.5790 0.5637 0.0153 4 0.4260 0.4170 0.0090 6 0.3190 0.3146 0.0044 8 0.2440 0.2406 0.0034 10 0.1890 0.1857 0.0033 12 0.1500 0.1443 0.0057 14 0.1140 0.1126 0.0014 16 0.0970 0.0882 0.0088 18 0.0750 0.0692 0.0058 20 0.0610 0.0543 0.0067 22 0.0480 0.0427 0.0053 24 0.0390 0.0336 0.0054 26 0.0310 0.0265 0.0045 28 0.0190 0.0209 -0.0019 30 0.0120 0.0164 -0.0044 32 0.0070 0.0129 -0.0059 34 0.0060 0.0102 -0.0042 36 0.0060 0.0080 -0.0020 38 0.0060 0.0063 -0.0003 40 0.0050 0.0050 0.0000 42 0.0050 0.0039 0.0011 44 0.0040 0.0031 0.0009 46 0.0040 0.0024 0.0016 48 0.0030 0.0019 0.0011 50 0.0030 0.0015 0.0015

111 Number of Components = 3 and Simulation Sample Size = 20000 0.9 Simulated Ruin Probability 0.8 Approximated Ruin Probability

0.7

0.6

0.5

0.4 Ruin Probability 0.3

0.2

0.1

0 0 5 10 15 20 25 30 35 40 45 50 u (Initial Capital)

Figure 7.5: Comparison of Simulated ruin probability and Approximated ruin probability for a three component when simulation sample size is 20000.

112 Now, we make a comparison between the approximated ruin probability and the simulated ruin probability when the simulation sample sizes are pegged at 10000, 15000 and

20000 taking cognisance that the CDF of the approximating distribution is the two and the three component asymmetric Double Exponential distributions.

We ﬁrst consider the two component as we increase the simulation sample size. Both Table

7.1 and Figure 7.1 compare the approximated ruin probability to the simulated ruin probability when the approximated CDF is a two component asymmetric Double Exponential distribution and the simulated sample size is 10000. It is clear from the table and the ﬁgure that, the approximated ruin probability is very close to the simulated ruin probability. As the simulation sample size increases from 10000 to 15000, the closeness between the two ruin probabilities improves tremendously as exhibited in both Table 7.2 and Figure 7.2 and that the approximated CDF is the two component asymmetric Double Exponential distribution.

Again, further improvement is observed in both Table 7.3 and Figure 7.3 in the degree of closeness between the approximated ruin probability and the simulated ruin probability as the simulation sample size increases from 15000 to 20000.

In a similar fashion, we compare the approximated ruin probability to the simulated ruin probability when the approximated CDF this time around is the three component asymmetric Double Exponential distribution. The case where the simulation sample size is 10000, the approximation is not too good therefore omitted at this stage of the analysis. However, the approximated ruin probability gets closer to the simulated ruin probability as the simulation sample size increases from 10000 to 15000 by taking into consideration the fact that the approximated CDF is the three component asymmetric Double Exponential distribution as illustrated in both Table 7.4 and Figure 7.4. Similarly, the approximation improves further as the simulation sample size surges from 15000 to 20000 considering the situation that the approximated CDF is still the three component asymmetric Double Exponential distribution as shown in both Table 7.5 and Figure 7.5.

113 From the foregoing analysis, clearly the approximated ruin probability ﬁts neatly the simulated ruin probability and therefore constitutes an excellent approximation.

114 Chapter 8

Summary, Concluding Remarks and Future Research

This chapter reviews and presents the main summary, concluding remarks emanating from the thesis and future research.

8.1 Summary

Herein, we provide a summary and key highlights of issues discussed in the various chapters of the thesis.

8.1.1 Chapter 2: Wiener-Hopf Factorization Method

Section 2.1

This section focused on the uses of the Wiener-Hopf factorization method in obtaining the

ﬁnite time ruin probabilities within the Sparre-Andersen risk model framework. The section also cited reference sources such as Section 6.4 of Rolski et al. (1998), Asmussen (2000),

Prabhu (1980) and Feller (1971) for the Wiener-Hopf Factorization Method.

Section 2.2

This section commenced with the deﬁnition of a random walk and the two ladder processes:

¯ ¯ descending ladder point (Nk,SNk ) and ascending ladder point (Nk,SNk ). The section further provided the ﬁrst theorem (Theorem 2.1) which aided in the derivation of the Wiener-Hopf factorization method.

Section 2.3

∗ ∗ The two renewal functions un(x) and vn(x) and their transforms u (z, ω), v (z, ω), χ(z, ω) =

¯ E(zN eiωSN ) andχ ¯(z, ω) = E(zN eiωSN¯ ) were deﬁned in this section. These transforms led to some results which were put together as theorem 2.2. Also, Theorem 2.3 was highlighted in

115 this section.

Section 2.4

The maximum of the sequence {0,S1, ...Sn} denoted Mn was deﬁned in this section. Also the joint distribution and the joint characteristic function of Mn − Sn were also put together in a form of theorems 2.4 and 2.5 respectively.

In summary, chapter 2 provided a concise approach to the classical technique as follows:

2. Set

1 − z lim E[eiωMn ] = lim . n→∞ z→1 D(z, ω)D¯(z, 0)

Once, 1 and 2 are met, then the maximal aggregate loss for the surplus process was given as

L. Through L, the Laplace transform of the survival probability δ∗(s) was determined and the inverse Laplace transform resulted in the survival probability (δ(u)). The ultimate ruin probability (ψ(u)) was obtained in this section by taking the complement of δ(u).

8.1.2 Chapter 3: Ruin Probability for Correlated Poisson Claim Count and Gamma (2)

Risk Process

Section 3.1

This section introduced the main objective of the chapter. It focused on the derivation of the ultimate ruin probability when claim number process was assumed to be Poisson distributed while the claim sizes were assumed to be Gamma distributed with shape parameter 2 in the presence of dependency. The section also introduced the Poisson distribution and then established its linkage with the Exponential distribution.

Section 3.2

116 This section ﬁrst deﬁned the joint moment generating function of X1 and X2 where the joint distribution function is the Moran and Downton’s bivariate Exponential distribution. Also,

the section deﬁned additional random variables T and X, where T is the claim inter-arrival times and X is the claim sizes (amounts).

Section 3.3

The characteristic function of Y = X − cT was derived in this section. To motivate the application of the Wiener-Hopf factorization method, there was the need to obtain the factor

1 − zφY (ω) which adequate coverage was provided for in this section. Section 3.4

g(s)−z The factor 1 − zφY (ω) resulted in the rational function g(s) . This section highlighted the various procedures in obtaining the zeros of g(s) and g(s) − z. The zeros of g(s) were

−s1, s2 = β2 and s3, however the zeros of g(s) − z as z = 1 were obtained as s1(1) = 0,

s2(1) < s2 = β2 and s3(1) > s3. Section 3.5

This section covered the factors and properties of the transform of Y . These factors D(z, ω)

and D¯(z, ω) were shown to have satisﬁed the conditions precedent, hence the required factors

prior to computing the survival probability.

Section 3.6

This section clearly showed how the survival probability was arrived at through the applica-

tion of step 2 of the classical technique, partial fractions and the Laplace transform. It was

further shown that the compliment of the survival probability resulted in the ultimate ruin

probability.

Section 3.7

It was indicated in this section that as ρ increases between 0 and 1, ψ(u) decreases. Hence,

ultimate ruin probability was associated with decreasing ρ.

Section 3.8

117 This section provided some numerical examples.

8.1.3 Chapter 4: Ruin Probability for Correlated Poisson Claim Count and Gamma (m)

Risk Process

Section 4.1

This section introduced the generalization of the results for the ultimate ruin probability

shown in chapter 3. The extension was in the claim size (amount) distribution where claim

sizes were assumed to be Gamma distributed with shape parameter m, where m is greater

than 2.

Section 4.2

This section deﬁned the joint mgf of X1 and X2 as stated in section 3.2. The section introduced further two random variables T and X where T was Exponentially distributed with

1 mean = β and X was Gamma distributed with shape parameter m. Section 4.3

To facilitate the application of the Wiener-Hopf factorization method, The section high-

lighted the joint characteristic function of Y = X − cT and subsequently linked it to the factor 1 − zφY (ω). Section 4.4

The factor 1 − zφY (ω) was reduced to the ratio of two polynomials g(s) and g(s) − z in this section. Both polynomials were shown to have both odd and even powers alongside their graphical display in this section. Also, the section employed Rouche’s theorem in determining and locating the zeros of the polynomial g(s) − z. It was shown that g(s) and g(s) − z for |z| < 1 has the same number of zeros.

Section 4.5

This section focused on the factors and properties of the transformation of Y . These factors

D(z, ω) and D¯(z, ω) for 0 < z < 1, ω ∈ R were found to have satisﬁed the required conditions under the classical techniques.

118 Section 4.6

This section displayed the characteristic function of the maximal aggregate loss, the Laplace

transform of the survival probability and the corresponding survival probability.

Section 4.7

This section provided some numerical examples.

8.1.4 Chapter 5: Ruin Probability for Correlated Poisson Claim Count and Hyper Expo-

nential Risk Process

Section 5.1

This section introduced the main chapter as it highlighted an explicit expression for the

ultimate ruin probability when the claim number (counts) was still Poisson distributed but

the claim sizes (amounts) were assumed to be Hyper Exponentially distributed and the two

random variables were correlated.

This section deﬁned the Hyper Exponential distribution as a mixture of k exponentials for some k and the corresponding Laplace transform in the univariate sense. The section concluded with a graphical example of the two and three component Hyper Exponential distribution.

Section 5.2

The section proceeded with the deﬁnition of the joint mgf of (T,Xj) and that of (T,X) for

j = 1, 2, ..., n which were given as MT,Xj (t1, t2) and MT,X (t1, t2) respectively. In this section,

the joint mgf MT,Xj (t1, t2) was set-up such that it formed the Moran and Downton’s bivari-

ate Exponential distribution whiles MT,X (t1, t2) formed a mixture of Moran and Downton’s bivariate Exponential distribution.

Section 5.3

This section presented the mgf, characteristic function of Y = X − cT and its related trans-

h(s,z) form which was given as a rational function h(s) . By the application of Rouche’s theorem, the zeros of the two polynomials were obtained and further decomposed into two factors

119 D¯(z, ω) and D(z, ω) with the required factors. Evidence from Rouche’s theorem pointed to

the fact that, h(s) and h(s, z) for |z| < 1 had the same number of zeros.

Section 5.4

In this section, the survival probability was derived from the characteristic function of the

maximal aggregate loss via partial fractions and the Laplace transform of the survival prob-

ability. The compliment of the survival probability resulted in the ultimate ruin probability.

Section 5.5

This section provided some numerical examples.

8.1.5 Chapter 6: Ruin Probability for Correlated Poisson Claim Count and Pareto Risk

Process

Section 6.1

This section highlighted the objective of the chapter. It also introduced the Pareto distri-

bution (Pareto type II distribution with µ = 0) and the main estimating procedure in the

chapter which was the method of moments estimation.

Section 6.2

The joint PDF and CDF of the bivariate Exponential-Pareto distribution were derived in

this section. The section further showed the moments of T and X and the joint moment

XT . The section drew the curtain by deriving the moments of W = X − cT .

Section 6.3

This section introduced the two approximating distributions. These were distributions resulting from Exponential-Hyper Exponential and Exponential-Gamma (m) distributions.

The section expanded the discussion by deriving the moments of Y = X − cT , where (T,X)

were Exponential-Hyper Exponential and Exponential-Gamma (m) distributions.

Section 6.4

This section highlighted the main numerical results.

120 8.1.6 Chapter 7: Asymmetric Double Exponential Distributions and its Finite Mixtures

Section 7.1

This section commented on the lack of goodness of ﬁt between the two approximating distributions and the Exponential-Pareto distribution discussed under chapter 6. The section also, provided an overview of the Double Exponential distribution and further shed light on

ﬁnite mixture models.

Section 7.2

The section commenced by introducing the asymmetric Double Exponential distribution and then estimated the parameters of the distribution via the MLE approach. The section further expanded the discussion to cover ﬁnite mixture of asymmetric Double Exponential distributions and then linked the estimation of the mixture’s parameter estimation with the

EM algorithm. The section concluded by stating the algorithm in a clear and concise man- ner.

Section 7.3

This section highlighted the main numerical results.

8.2 Concluding Remarks

In this thesis, we have seen diﬀerent classes of bivariate distribution in modeling the joint distribution of the claim sizes (amounts) and the claim inter-arrival times. Speciﬁcally, chapter 3 concentrated on a bivariate Exponential-Gamma (2) distribution. chapter 4 studied the situation of a bivariate Exponential-Gamma (m) distribution. Another class of bivariate

Exponential-Mixture of Exponential distribution was given attention in chapter 5. Chap- ter 6 considered the situation where two classes of distributions were used to approximate

Exponential-Pareto distribution by following the method of moments estimation. Chapter 7 followed somewhat a diﬀerent approach by considering asymmetric ﬁnite mixtures of Double

Exponential distributions and then capitalizing on the capabilities of the EM algorithm in

121 the estimation process prior to computing the ruin probability. In all of these considera-

tions, we have adduced ample evidence to substantiate the fact that, analytically the exact

expressions in respect of the ultimate ruin probability were possible. However, in the case

of heavy-tailed claims, the asymmetric ﬁnite mixtures of Double Exponential distributions

coupled with the strength of EM algorithm provided a better approximation in computing

the ruin probability as compared with distributions emanating from Exponential-Hyper Ex-

ponential and Exponential-Gamma (m) based on method of moments estimation as we have witnessed in the preceding chapters.

In the literature, Dickson and Hipp (1998) had already dealt with the case where claim counts and claim sizes were assumed to be independent. In our case, we have sought to extend the study by somewhat introducing dependency. Relaxing the independence assumption by way of introducing correlation between the claim sizes (amounts) and claim inter-arrival times.

As a consequence, we have observed that, increasing correlation between the two random variables led to a reduction in the ultimate ruin probability. This exposition maybe informed by the positive correlation between claim sizes and claim inter-arrival times. Also, we have tested the capabilities of the EM algorithm in arriving at an approximated ultimate ruin probability and the results obtained were great and promising.

8.3 Future Research

So far, the purview of this thesis had been bivariate distributions, positive correlation, Moran and Downton’s bivariate Exponential distribution and approximating a heavy-tailed claim distribution with a mixture of light-tailed claim distributions. We are yet to observe ex- tensions from bivariate to multivariate distributions, the eﬀect of negative correlation and other variations of bivariate distributions such as the forms presented in Chapter 47 of Kotz,

Balakrishnan and Johnson (2000).

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127 Appendix A

Maple and Matlab codes

We used Maple and Matab for the computations. In particular, we used Maple for verifying algebra in certain instances especially getting moments of Y = X −cT in the case of bivariate

Exponential-Pareto distribution.

A.1 Maple Codes

Figure 3.1 Chapter 3: c := 3 ; beta [ 1 ] := 2 ; beta [ 2 ] := 4 ; rho := 2/5; g(s):=((1+(c∗ s )/( beta [1] ))∗(1 −( s )/( beta [ 2 ] ) ) + ( rho ∗c∗ s ˆ ( 2 ) )

/( beta [ 1 ] ∗ beta [2]))∗(1 − s /( beta [ 2 ] ) ) ; plot ([g(s), 1], tickmarks = [[ −2 = ””, 0= ”0”, 2= ””,

4 = ””, 6 = ””, 8 = ””], default]);

Example 1 Chapter 3: c := 3 ; beta [ 1 ] := 2 ; beta [ 2 ] := 4 ; rho := 1/5; g(s):=((1+(c∗ s )/( beta [1] ))∗(1 −( s )/( beta [ 2 ] ) ) + ( rho ∗c∗ s ˆ ( 2 ) )

/( beta [ 1 ] ∗ beta [2]))∗(1 − s /( beta [ 2 ] ) ) ;

128 g(s, 0) := solve(g(s)); s[1] := g(s, 0)[3]; s[2] := g(s, 0)[1]; s[3] := g(s, 0)[2]; g(s, 1) := solve(g(s) −1); s[11] := g(s, 1)[1]; s[21] := g(s, 1)[3]; s[31] := g(s, 1)[2];

D i g i t s := 5 ; ruin := evalf((1− s[21]/s[2])∗(1 − s[21]/s[3]) ∗ exp(−s [ 2 1 ] ∗ u)

/(1− s[21]/s[31])+(1 − s[31]/s[2])∗(1 − s[31]/s[3]) ∗ exp(−s [ 3 1 ] ∗ u)

/(1− s[31]/s[21]));

Example 2 Chapter 3: c := 5 ; beta [ 1 ] := 2 ; beta [ 2 ] := 3 ; rho := 2/5; g(s):=((1+(c∗ s )/( beta [1] ))∗(1 −( s )/( beta [ 2 ] ) ) + ( rho ∗c∗ s ˆ ( 2 ) )

/( beta [ 1 ] ∗ beta [2]))∗(1 − s /( beta [ 2 ] ) ) ; g(s, 0) := solve(g(s)); s[1] := g(s, 0)[3]; s[2] := g(s, 0)[1]; s[3] := g(s, 0)[2]; g(s, 1) := solve(g(s) −1); s[11] := g(s, 1)[1]; s[21] := g(s, 1)[3]; s[31] := g(s, 1)[2];

D i g i t s := 5 ; ruin := evalf((1− s[21]/s[2])∗(1 − s[21]/s[3]) ∗ exp(−s [ 2 1 ] ∗ u)

/(1− s[21]/s[31])+(1 − s[31]/s[2])∗(1 − s[31]/s[3]) ∗ exp(−s [ 3 1 ] ∗ u)

/(1− s[31]/s[21]));

129 Example 3 chapter 3 (Figure 3.2):

We increae rho from 1/5, ..., 4/5.

beta [ 1 ] := 2 ;

beta [ 2 ] := 4 ;

rho := 1/5;

g(s):=((1+(c∗ s )/( beta [1] ))∗(1 −( s )/( beta [ 2 ] ) ) + ( rho ∗c∗ s ˆ ( 2 ) )

/( beta [ 1 ] ∗ beta [2]))∗(1 − s /( beta [ 2 ] ) ) ;

g(s, 0) := solve(g(s));

s[1] := g(s, 0)[3]; s[2] := g(s, 0)[1]; s[3] := g(s, 0)[2];

g(s, 1) := solve(g(s) −1);

s[11] := g(s, 1)[1]; s[21] := g(s, 1)[3];

s[31] := g(s, 1)[2];

D i g i t s := 5 ;

r[1] := evalf((1− s[21]/s[2])∗(1 − s[21]/s[3]) ∗ exp(−s [ 2 1 ] ∗ u)

/(1− s[21]/s[31])+(1 − s[31]/s[2])∗(1 − s[31]/s[3]) ∗ exp(−s [ 3 1 ] ∗ u)

/(1− s[31]/s[21]));

plot ([r[1], r[2], r[3], r[4]]);

Figure 4.1 Chapter 4:

beta [ 1 ] := 2 ;

beta [ 2 ] := 4 ;

rho := 2/5; m := 4 ;

g(s):=((1+(c∗ s )/( beta [1] ))∗(1 −( s )/( beta [ 2 ] ) ) + ( rho ∗c∗ s ˆ ( 2 ) )

/( beta [ 1 ] ∗ beta [2]))∗(1 − s /( beta [ 2 ] ) ) ˆ (m−1) ;

plot ([g(s), 1], tickmarks = [[ −2 = ””, 0= ”0”, 2= ””,

4 = ””, 6 = ””, 8 = ””], default]);

130 Figure 4.2 Chapter 4:

beta [ 1 ] := 2 ;

beta [ 2 ] := 4 ;

rho := 2/5; m := 5 ;

g(s):=((1+(c∗ s )/( beta [1] ))∗(1 −( s )/( beta [ 2 ] ) ) + ( rho ∗c∗ s ˆ ( 2 ) )

/( beta [ 1 ] ∗ beta [2]))∗(1 − s /( beta [ 2 ] ) ) ˆ (m−1) ;

plot ([g(s), 1], tickmarks = [[ −2 = ””, 0= ”0”, 2= ””,

4 = ””, 6 = ””, 8 = ””], default]);

Example 1 Chapter 4: m:=3;

c :=3;

beta1 :=2;

beta2 :=4;

rho :=2/5;

ETX :=C/ beta1 − m/ beta2 ;

ETX1 := subs (C=c ,ETX) ;

i f ETX1 < 0 then

c := round( solve (ETX=0,C))+1;

print (”Given premium rate is too low”);

print (”c=”,c,”is selected”);

f i ;

s3 := ((c/beta1 −1/beta2 ) +

( ( c/ beta1 −1/beta2)ˆ2+4∗(1− rho )∗ c /( beta1 ∗ beta2))ˆ(1/2))/(2∗(1 − rho )

∗c /( beta1 ∗ beta2 ) ) ;

s2 := beta2 ;

131 s1 := (−(c/ beta1 −1/beta2 ) +

( ( c/ beta1 −1/beta2)ˆ2+4∗(1− rho )∗ c /( beta1 ∗ beta2))ˆ(1/2))/(2∗(1 − rho )

∗c /( beta1 ∗ beta2 ) ) ;

gs :=simplify((1+s/s1)∗(1− s / s2 )ˆ(m−1)∗(1− s / s3 ) −1.0);

s3r := evalf(s3);

Digits := 100;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

od :

PRD := (1+s/s2)ˆ(m−1)∗(1+s/s3r)/(s ∗PRD) ;

DEL :=collect(simplify(evalf(invlaplace(PRD,s,u))) ,[ exp , sin , cos ]);

D i g i t s := 5 ;

DEL := fnormal(evalf(DEL));

print (DEL) ;

Example 2 Chapter 4: m:=4;

c :=3;

beta1 :=2;

beta2 :=4;

rho :=2/5;

132 ETX :=C/ beta1 − m/ beta2 ;

ETX1 := subs (C=c ,ETX) ;

i f ETX1 < 0 then

c := round( solve (ETX=0,C))+1;

print (”Given premium rate is too low”);

print (”c=”,c,”is selected”);

f i ;

s3 := ((c/beta1 −1/beta2 ) +

( ( c/ beta1 −1/beta2)ˆ2+4∗(1− rho )∗ c /( beta1 ∗ beta2))ˆ(1/2))/(2∗(1 − rho )

∗c /( beta1 ∗ beta2 ) ) ;

s2 := beta2 ;

s1 := (−(c/ beta1 −1/beta2 ) +

( ( c/ beta1 −1/beta2)ˆ2+4∗(1− rho )∗ c /( beta1 ∗ beta2))ˆ(1/2))/(2∗(1 − rho )

∗c /( beta1 ∗ beta2 ) ) ;

gs :=simplify((1+s/s1)∗(1− s / s2 )ˆ(m−1)∗(1− s / s3 ) −1.0);

s3r := evalf(s3);

Digits := 100;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

od :

133 PRD := (1+s/s2)ˆ(m−1)∗(1+s/s3r)/(s ∗PRD) ;

DEL :=collect(simplify(evalf(invlaplace(PRD,s,u))) ,[ exp , sin , cos ]);

D i g i t s := 5 ;

DEL := fnormal(evalf(DEL));

print (DEL) ;

Example 3 Chapter 4: m:=5;

c :=3;

beta1 :=2;

beta2 :=4;

rho :=2/5;

ETX :=C/ beta1 − m/ beta2 ;

ETX1 := subs (C=c ,ETX) ;

i f ETX1 < 0 then

c := round( solve (ETX=0,C))+1;

print (”Given premium rate is too low”);

print (”c=”,c,”is selected”);

f i ;

s3 := ((c/beta1 −1/beta2 ) +

( ( c/ beta1 −1/beta2)ˆ2+4∗(1− rho )∗ c /( beta1 ∗ beta2))ˆ(1/2))/(2∗(1 − rho )

∗c /( beta1 ∗ beta2 ) ) ;

s2 := beta2 ;

s1 := (−(c/ beta1 −1/beta2 ) +

( ( c/ beta1 −1/beta2)ˆ2+4∗(1− rho )∗ c /( beta1 ∗ beta2))ˆ(1/2))/(2∗(1 − rho )

∗c /( beta1 ∗ beta2 ) ) ;

gs :=simplify((1+s/s1)∗(1− s / s2 )ˆ(m−1)∗(1− s / s3 ) −1.0);

134 s3r := evalf(s3);

Digits := 100;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

od :

PRD := (1+s/s2)ˆ(m−1)∗(1+s/s3r)/(s ∗PRD) ;

DEL :=collect(simplify(evalf(invlaplace(PRD,s,u))) ,[ exp , sin , cos ]);

D i g i t s := 5 ;

DEL := fnormal(evalf(DEL));

print (DEL) ;

Figure 5.1 Chapter 5:

p [ 2 ] := . 3 3 ;

p [ 3 ] := 1−p[1] −p [ 2 ] ;

beta [ 1 ] := 2 ;

beta [ 2 ] := 2 . 5 ;

beta [ 3 ] := 3 ;

g := p [ 1 ] ∗ beta [ 1 ] ∗ exp(−beta [ 1 ] ∗ x)+p [ 3 ] ∗ beta [ 3 ] ∗ exp(−beta [ 3 ] ∗ x ) ;

h := p [ 1 ] ∗ beta [ 1 ] ∗ exp(−beta [ 1 ] ∗ x)+p [ 2 ] ∗ beta [ 2 ] ∗ exp(−beta [ 2 ] ∗ x )

+p [ 3 ] ∗ beta [ 3 ] ∗ exp(−beta [ 3 ] ∗ x ) ;

plot ( [ g , h ] , x ) ;

135 Table 5.1 Chapter 5:

beta := 2 ;

rho := . 5 ;

a [ j ] := (−c/beta+1/k+sqrt ( ( c/beta−1/k)ˆ2+(4∗(1− rho ) )

∗c /( beta∗k ) ) ) ∗ beta∗k/((2∗(1 − rho ))∗ c ) ;

for k from 2 by .5 to 7 do eval ( a [ j ] ) end do ;

Table 5.2 Chapter 5:

beta := 2 ;

rho := . 5 ;

b [ j ] := ( c/beta−1/k+sqrt ( ( c/beta−1/k)ˆ2+(4∗(1− rho ) )

∗c /( beta∗k ) ) ) ∗ beta∗k/((2∗(1 − rho ))∗ c ) ;

for k from 2 by .5 to 7 do eval (b [ j ] ) end do ;

Example 1 Chapter 5:

c :=3;

beta :=2;

n :=2;

betav := ([2.2,2.4]);

prop := ([0.5,1 − 0 . 5 ] ) ;

rho : = 0 . 2 ;

ETX := −C/beta ;

for i from 1 to n do

ETX :=ETX+ prop[ i]/betav[ i ];

od ;

ETX1 := subs (C=c ,ETX) ;

136 i f ETX1 > 0 then

c := round( solve (ETX=0,C))+1;

print (”Given premium rate is too low”);

print (”c=”,c,”is selected”);

f i ;

for i from 1 to n do

s2[i] := ((c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

s1 [ i ] := (−(c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

od ;

prds :=1;

for i from 1 to n do

prds := prds ∗(1+s/s1[ i])∗(1 − s / s2 [ i ] ) ;

od :

prsum :=0;

for i from 1 to n do

prsum := prsum + prds/(1+s/s1[i])/(1 − s / s2 [ i ] ) ∗ prop [ i ]

od :

gs := prds−prsum ;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

137 PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

od :

PRDS := product((1+s/s2[ ii ]) , ii=1..n)/(s ∗PRD) ;

DEL :=collect(simplify(evalf(invlaplace(PRDS,s,u))) ,[ exp , sin , cos ]);

D i g i t s := 5 ;

DEL := fnormal(evalf(DEL));

Ruin := 1− DEL;

Example 2 Chapter 5:

c :=3;

beta :=2;

n :=3;

betav := ([2.5, 3.0, 3.5]);

prop := ([.3, .33, 1 −.3 −.33]);

rho : = 0 . 4 ;

ETX := −C/beta ;

for i from 1 to n do

ETX :=ETX+ prop[ i]/betav[ i ];

od ;

ETX1 := subs (C=c ,ETX) ;

i f ETX1 > 0 then

c := round( solve (ETX=0,C))+1;

print (”Given premium rate is too low”);

print (”c=”,c,”is selected”);

138 f i ;

for i from 1 to n do

s2[i] := ((c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

s1 [ i ] := (−(c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

od ;

prds :=1;

for i from 1 to n do

prds := prds ∗(1+s/s1[ i])∗(1 − s / s2 [ i ] ) ;

od :

prsum :=0;

for i from 1 to n do

prsum := prsum + prds/(1+s/s1[i])/(1 − s / s2 [ i ] ) ∗ prop [ i ]

od :

gs := prds−prsum ;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

139 od :

PRDS := product((1+s/s2[ ii ]) , ii=1..n)/(s ∗PRD) ;

DEL :=collect(simplify(evalf(invlaplace(PRDS,s,u))) ,[ exp , sin , cos ]);

D i g i t s := 5 ;

DEL := fnormal(evalf(DEL));

Ruin := 1− DEL;

Example 3 chapter 5 (Figure 5.2):

We increae rho from 1/5, ..., 4/5.

c :=3;

beta :=2;

n :=2;

betav := ([2.2,2.4]);

prop := ([0.5,1 − 0 . 5 ] ) ;

rho : = 0 . 2 ;

ETX := −C/beta ;

for i from 1 to n do

ETX :=ETX+ prop[ i]/betav[ i ];

od ;

ETX1 := subs (C=c ,ETX) ;

i f ETX1 > 0 then

c := round( solve (ETX=0,C))+1;

print (”Given premium rate is too low”);

print (”c=”,c,”is selected”);

f i ;

for i from 1 to n do

s2[i] := ((c/beta−1/betav[i]) +

140 ( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

s1 [ i ] := (−(c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

od ;

prds :=1;

for i from 1 to n do

prds := prds ∗(1+s/s1[ i])∗(1 − s / s2 [ i ] ) ;

od :

prsum :=0;

for i from 1 to n do

prsum := prsum + prds/(1+s/s1[i])/(1 − s / s2 [ i ] ) ∗ prop [ i ]

od :

gs := prds−prsum ;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

od :

PRDS := product((1+s/s2[ ii ]) , ii=1..n)/(s ∗PRD) ;

DEL :=collect(simplify(evalf(invlaplace(PRDS,s,u))) ,[ exp , sin , cos ]);

141 D i g i t s := 5 ;

DEL := fnormal(evalf(DEL));

Ruin := 1− DEL;

plot ( Ruin ) ;

Example 4 chapter 5 (Figure 5.3):

We increae rho from 1/5, ..., 4/5.

c :=3;

beta :=2;

n :=3;

betav := ([2.5, 3.0, 3.5]);

prop := ([.3, .33, 1 −.3 −.33]);

rho : = 0 . 4 ;

ETX := −C/beta ;

for i from 1 to n do

ETX :=ETX+ prop[ i]/betav[ i ];

od ;

ETX1 := subs (C=c ,ETX) ;

i f ETX1 > 0 then

c := round( solve (ETX=0,C))+1;

print (”Given premium rate is too low”);

print (”c=”,c,”is selected”);

f i ;

for i from 1 to n do

s2[i] := ((c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

142 s1 [ i ] := (−(c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

od ;

prds :=1;

for i from 1 to n do

prds := prds ∗(1+s/s1[ i])∗(1 − s / s2 [ i ] ) ;

od :

prsum :=0;

for i from 1 to n do

prsum := prsum + prds/(1+s/s1[i])/(1 − s / s2 [ i ] ) ∗ prop [ i ]

od :

gs := prds−prsum ;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

od :

PRDS := product((1+s/s2[ ii ]) , ii=1..n)/(s ∗PRD) ;

DEL :=collect(simplify(evalf(invlaplace(PRDS,s,u))) ,[ exp , sin , cos ]);

D i g i t s := 5 ;

DEL := fnormal(evalf(DEL));

143 Ruin := 1− DEL;

plot ( Ruin ) ;

Figure 6.1 Chapter 6:

lambda := 10;

alpha[1] := 3;

alpha[2] := 5;

alpha[3] := 7;

f := alpha[1] ∗ lambdaˆ{ alpha [ 1 ] } /( lambda+x)ˆ{ alpha [1] −1};

g := alpha[2] ∗ lambdaˆ{ alpha [ 2 ] } /( lambda+x)ˆ{ alpha [2] −1};

h := alpha[3] ∗ lambdaˆ{ alpha [ 3 ] } /( lambda+x)ˆ{ alpha [ 3 ] − 1 } ;;

plot ([f,g, h], x=0..100);

Figure 6.2 Chapter 6:

Step 4: We use the estimates from step 3 to compute the approximated ruin probability

c :=7;

beta :=5;

n :=2;

betav := ([0.9416,4.231886019]);

prop := ([0.9248872359,1 −0.9248872359]);

rho :=0.707642686;

ETX := −C/beta ;

for i from 1 to n do

ETX :=ETX+ prop[ i]/betav[ i ];

od ;

ETX1 := subs (C=c ,ETX) ;

144 i f ETX1 > 0 then

c := round( solve (ETX=0,C))+1;

print (”Given premium rate is too low”);

print (”c=”,c,”is selected”);

f i ;

for i from 1 to n do

s2[i] := ((c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

s1 [ i ] := (−(c/beta−1/betav[i]) +

( ( c/beta−1/betav[ i])ˆ2+4∗(1 − rho )∗ c /( beta∗ betav[i]))ˆ(1/2))

/(2∗(1 − rho )∗ c /( beta∗ betav[i ]));

od ;

prds :=1;

for i from 1 to n do

prds := prds ∗(1+s/s1[ i])∗(1 − s / s2 [ i ] ) ;

od :

prsum :=0;

for i from 1 to n do

prsum := prsum + prds/(1+s/s1[i])/(1 − s / s2 [ i ] ) ∗ prop [ i ]

od :

gs := prds−prsum ;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

145 PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

od :

PRDS := product((1+s/s2[ ii ]) , ii=1..n)/(s ∗PRD) ;

DEL :=collect(simplify(evalf(invlaplace(PRDS,s,u))) ,[ exp , sin , cos ]);

D i g i t s := 5 ;

DEL := fnormal(evalf(DEL));

Ruin := 1− DEL;

Figure 7.1 Chapter 7:

Step 2: We use estimates from step 1 to obtain the approximated ruin probability

n :=2; # for 3 component we let n:=3

s1[1] := 0.8563;

s1[2] := 0.9353;

s2[1] := 0.5409;

s2[2] := 1.7045;

prop[1] := 0.225;

prop[2] := 0.775;

prds :=1;

for i from 1 to n do

prds := prds ∗(1+s/s1[ i])∗(1 − s / s2 [ i ] ) ;

146 od :

prsum :=0;

for i from 1 to n do

prsum := prsum + prds/(1+s/s1[i])/(1 − s / s2 [ i ] ) ∗ prop [ i ]

od :

gs := prds−prsum ;

AIS := evalf(allvalues(RootOf(gs,s)));

PRD :=1;

for x in AIS do

i f Im( x ) >0 and Re( x ) >0 then

PRD := PRD∗(1 +(s ˆ2+2∗Re( x )∗ s )/ abs ( x ) ˆ 2 ) ;

e l i f x >0 then

PRD := PRD∗(1+ s /x ) ;

f i

od :

PRDS := product((1+s/s2[ ii ]) , ii=1..n)/(s ∗PRD) ;

PRDS := convert(PRDS,parfarc ,s);

DEL := invlaplace(PRDS,s,u);

#DEL :=collect(simplify(evalf(invlaplace(PRDS,s,u)))

,[ exp , sin , cos ]);

#For final results round everything to 5 digits.

#Digits := 5;

# Clean up the expression by deleting the terms with

# small coefficients as a result of floating point computations

#DEL := fnormal(evalf(DEL));

147 Ruin := 1− DEL;

A.2 Matlab Codes

Figure 6.2 Chapter 6:

Step 1: Generate CDF of (T,X) i.e. bivariate Exponential-Pareto distribution

% rho correlation coefficient

rho = 0 . 5 ;

% 1/theta 1 is mean of exponential claim epochs time

Theta1 = 5 ;

% Parameters of Pareto part

Lambda = 3 ;

Alpha =4;

% c is the security loading

% u is the initial surplua

c =6;

u=1;

% Nsim: number of simulations runs

% Nclaims: simulated number of claims per run

Nsims =1;

Ntimes = 200;

% Nrans is the total number of bivariate random numbers needed

148 Nrans = Nsims∗Ntimes ;

% First simulate two sets of bivariate standard normal random

%variables with given correlation

SIGMA = [1,rho; rho,1];

r 1 = mvnrnd([0 ,0] ,SIGMA,Nrans);

r 2 = mvnrnd([0 ,0] ,SIGMA,Nrans);

% Time and Y are the bivariate exponential random vectors

Time = ( r 1(:,1).ˆ2+r 2(:,1).ˆ2)/(2 ∗ Theta1 ) ;

Y = ( r 1(:,2).ˆ2+r 2(:,2).ˆ2)/2;

% Convert Y into X, Pareto using the transformation

X = Lambda∗(exp(Y/Alpha ) −1);

% Convert columns into matrices

% each column of TimeMat is one simulation run of the claim

%occurrence times each column of ClaimMat is one simulation

%run of the claim sizes

TimeMat = reshape (Time,Nsims ,Ntimes);

ClaimMat = reshape (X,Nsims,Ntimes);

% calculate the cumulative sum of each column

149 Y= ClaimMat−c∗TimeMat ;

[f,uu] = ecdf(Y);

Step 2: Use Nonlinear Least Squares (Curve Fitting) to obtain starting values for the

next step

function F = how1(x,xdata)

%I 1 holds negative xdata values

% i 2 holds positive xdata values

I 1 = xdata(xdata <0);

I 2 = xdata(xdata >0);

%CDF of bivarate Exponential−Hyper Exponential

F 1 = ( x (1)∗ x (2)∗ x ( 3 ) / ( x ( 2 ) ∗ (x(2)+x(3)))) ∗ exp( x (2)∗ I 1 ) + . . .

((1−x ( 1 ) ) ∗ x (4)∗ x ( 5 ) / ( x ( 4 ) ∗ (x(4)+x(5))) ∗ exp( x (4)∗ I 1 ) ) ;

F 2 = ( x (1)∗ x (2)∗ x(3)/(x(2)+x(3)) ∗ (1/x(2)+1/x(3)...

∗(1−exp(−x (3)∗ I 2 ))))+((1 −x ( 1 ) ) ∗ x (4)∗ x(5)/(x(4)+x(5))...

∗(1/x(4)+1/x(5)∗(1 −exp(−x (5)∗ I 2 ) ) ) ) ;

F = [ F 1 F 2 ] ; end

% Run this after the function to obtain starting

%values for the moment matching

uu= []; f= [];

x0= [0.7 0.7 3 4 0.9]; % starting values

options=optimset( ’TolFun’ ,1e −12,’TolX’ ,1e −12, ’MaxFunEvals’ ...

,1e12, ’MaxIter’ ,100000);

150 [x,resnorm,residual ,exitflag] = lsqcurvefit(@how1,x0,uu,f ,...

[0.1 0.5 0.5 0.5 2],[0.3 10 10 10 10 ]);

Step 3: Use fsolve to match the moments of the bivariate Exponential-Pareto and

Exponential-Hyper Exponential distributions

%Run this function

function f = DexpMoments( k,Param )

p = Param(1);

a1 = Param(2);

b1 = Param(3);

a2 = Param(4);

b2 = Param(5);

f = factorial(k)∗( p∗a1∗b1 /( a1+b1 )∗((1/b1)ˆ(k+1)−(−1/a1)ˆ(k+1))...

+(1−p)∗ a2∗b2 /( a2+b2 )∗((1/b2)ˆ(k+1)−(−1/a2)ˆ(k+1))); end

%Run this function

function F = momentmatch(x)

F = [DexpMoments(1,x)+2/5;

DexpMoments(2 ,x)− 37/30 ;

DexpMoments(3 ,x)+407311/250 ;

DexpMoments(4 ,x)+16846993298/78125 ;

DexpMoments(5 ,x)+958863813877/46875];

end

%Run this function to obtain the parameter estimates of

151 %the Exponential−Hyper Exponential distribution

x0 = [.9990;.9392;.8545;.5;2];

options=optimset( ’Display ’ , ’iter ’ , ’MaxFunEvals’ ,1e12 ,...

’MaxIter’ ,2e12);

[x,fval ,exitflag ,output] = fsolve(@momentmatch,x0,options);

Step 5: We compare the approximated ruin probability with the simulated ruin prob-

ability. Note that the approximated ruin probability is based on the Exponential-Hyper

Exponential and the simulated ruin probability is based on Exponential-Pareto.

% simulated ruin probability from the following

% rho correlation coefficient

rho = 0 . 5 ;

% 1/theta 1 is mean of exponential claim epochs time

Theta1 = 5 ;

% Parameters of Pareto part

Lambda = 3 ;

Alpha =4;

% c is the security loading

% u is the initial surplua

c =6;

u=1;

% Nsim: number of simulations runs

% Nclaims: simulated number of claims per run

152 Nsims =1000;

Ntimes = 1000;

Iseed = 1234567;

rng(Iseed);

% Nrans is the total number of bivariate random numbers needed

Nrans = Nsims∗Ntimes ;

% first simulate two sets of bivariate standard normal random

% variables with given correlation

SIGMA = [1,rho; rho,1];

r 1 = mvnrnd([0 ,0] ,SIGMA,Nrans);

r 2 = mvnrnd([0 ,0] ,SIGMA,Nrans);

% Time and Y are the bivariate exponential random vectors

Time = ( r 1(:,1).ˆ2+r 2(:,1).ˆ2)/(2 ∗ Theta1 ) ;

Y = ( r 1(:,2).ˆ2+r 2(:,2).ˆ2)/2;

% Convert Y into X, Pareto using the transformation

X = Lambda∗(exp(Y/Alpha ) −1);

% Convert columns into matrices

% each column of TimeMat is one simulation run of the claim

% occurrence times each column of ClaimMat is one simulation

153 % run of the claim sizes

TimeMat = reshape (Time,Ntimes ,Nsims);

ClaimMat = reshape (X,Ntimes ,Nsims);

% calculate the cumulative sum of each column

i =0;

for u=0:0.01:10

i= i +1;

surplus = u + c∗cumsum(TimeMat,1) −cumsum(ClaimMat ,1);

% any(surplus <0) will look at one column at a time of surplus

% and give 0 if that column has no negative values and 1 if

% i t does

RuinProbability(i) = sum(any( surplus <0))/Nsims ;

end

u =0:0.01:10;

% Ruin probability from step 4

Ruin1 = 0.3395746371∗ exo ( −2.600720391∗u ) −0.01041981885...

∗exp( −16.96582096∗u)

Ruin2 = .1818462645∗exp( −4.570803962∗u) −0.1548178042e − 1 . . .

∗exp( −18.27246128∗u)

Ruin3 = .2602384481∗exp( −3.975202257∗u) −0.3435028441e − 1 . . .

∗exp( −11.38638847∗u)

154 Ruin4 = .1388783737∗exp( −10.52869050∗u)+0.5319816345e − 3 . . .

∗exp( −161.4517198∗u)

figure ; subplot ( 2 , 2 , 1 ) plot (u,RuinProbability , ’−r’,u,Ruin1, ’ −.b ’ ) t i t l e (’C = 2 and Ntimes = 10000 ’ ) xlabel ( ’u ( I n i t i a l Capital)’); ylabel ( ’ Ruin Probability ’); l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’); set (leg , ’Location’,’NorthEast’); set (leg , ’Interpreter ’,’none’);

subplot ( 2 , 2 , 2 ) plot (u,RuinProbability , ’−r’,u,Ruin2, ’ −.b ’ ) t i t l e (’C = 2 and Ntimes = 15000 ’ ) xlabel ( ’u ( I n i t i a l Capital)’); ylabel ( ’ Ruin Probability ’); l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’); set (leg , ’Location’,’NorthEast’); set (leg , ’Interpreter ’,’none’);

subplot ( 2 , 2 , 3 ) plot (u,RuinProbability , ’−r’,u,Ruin3, ’ −.b ’ )

155 t i t l e (’C = 2 and Ntimes = 20000 ’ )

xlabel ( ’u ( I n i t i a l Capital)’);

ylabel ( ’ Ruin Probability ’);

l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’);

set (leg , ’Location’,’NorthEast’);

set (leg , ’Interpreter ’,’none’);

subplot ( 2 , 2 , 4 )

plot (u,RuinProbability , ’−r’,u,Ruin4, ’ −.b ’ )

t i t l e (’C = 3 and Ntimes = 10000 ’ )

xlabel ( ’u ( I n i t i a l Capital)’);

ylabel ( ’ Ruin Probability ’);

l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’);

set (leg , ’Location’,’NorthEast’);

set (leg , ’Interpreter ’,’none’);

Figure 7.1 Chapter 7:

Step 1: Estimate the parameters of the asymmetric ﬁnite mixture Double Exponential

%This is the root file

rho = 0 . 5 ;

% 1/theta 1 is mean of exponential claim epochs time

Theta1 = 5 ;

% parameters of Pareto part

Lambda = 3 ;

156 Alpha =4;

% c is the security loading

% u is the initial surplua

c =6;

% Nsim: number of simulations runs

% Nclaims: simulated number of claims per run

Nsims =1;

% do not change line 18

% you may increase Ntimes

Ntimes = 10000;

Iseed = 1234567;

rng(Iseed);

% Nrans is the total number of bivariate random numbers needed

Nrans = Nsims∗Ntimes ;

% first simulate two sets of bivariate standard normal

% random variables with given correlation

SIGMA = [1,rho; rho,1];

r 1 = mvnrnd([0 ,0] ,SIGMA,Nrans);

r 2 = mvnrnd([0 ,0] ,SIGMA,Nrans);

% Time and Y are the bivariate exponential random vectors

157 Time = ( r 1(:,1).ˆ2+r 2(:,1).ˆ2)/(2 ∗ Theta1 ) ;

Y = ( r 1(:,2).ˆ2+r 2(:,2).ˆ2)/2;

% Convert Y into X, Pareto using the transformation

X = Lambda∗(exp(Y/Alpha ) −1);

% Convert columns into matrices

% each column of TimeMat is one simulation run

% of the claim occurrence times

% each column of ClaimMat is one simulation

% run of the claim sizes

TimeMat = reshape (Time,Nsims ,Ntimes);

ClaimMat = reshape (X,Nsims,Ntimes);

% calculate the cumulative sum of each column

Y= ClaimMat−c∗TimeMat ;

[simCDF,xvals] = ecdf(Y);

% you may try C=2 and 3

C=2;% for three component we change C=3

[ a est , b est , w est, counter, difference] = ...

doublexp mixture model(Y, C, 1.0e −3);

a est ’

b est ’

158 w est

EY0 = sum( w est ’ . / ( a e s t+b e s t ) . ∗ ( a e s t . / b est −b e s t . / a e s t ) )

% Run this function

function pdfg = pdfofg(x,C,p,a,b)

% pdf of mixture of doublexp

% INPUTS:

% x =a vector of points at which to calculate

% thenormalCDF

% p = mixing proportions (row vector)

% a = parameter vector for negative side

% b= parameter vector for positive side

temp = zeros (1 , length ( x ) ) ;

for j =1:C

temp= temp + p(j)∗ doublexp density(x,a(j),b(j)); end ;

pdfg = temp ;

function [ a est , b est] = mledoublexp(t1,t2)

%t1 is the mean of the negative values

%(the total of negative /(total # of observations)

159 %t2 is the mean of the positive values

r a t i o = (−t1/t2)ˆ0.5;

b est = 1/(ratio+1)/t2;

a e s t = b e s t / r a t i o ;

% Run this function

function [ a est , b est , p est, counter, difference] = ...

doublexp mixture model(values , C, epsilon)

% Inputs :

% values = row vector of the observed values

% notethisonlyworksfor1Ddata

% C =number of classes (mixtures),

% whichshouldbefairlysmall

% epsilon = precision for convergence

% Outputs :

% a est = vectors a of each class

% b est = vector of b for each class

% p est = class membership probability estimates

% counter = number of iterations required

% difference= total absolute difference in parameters

% ateachiterationtoget

% anideaofconvergencerate

160 % A Double exponential mixture model means that each

% data point is drawn (randomly) from one of C

% classes of data, with probability p i of being

% drawn from class i, and each class is

% distributed as double exponential with exponential

% parameter on negative side is a i and the exponential

% parameter on the positive side b i

% get and check inputs

N = length ( values ) ;

i f ( nargin < 3)

epsilon = 1.0e −6; end

% initialize

counter = 0 ;

yminus = sum(values(values <0))/N;

yplus = sum(values(values >0))/N;

[ a est1 , b est1] = mledoublexp(yminus,yplus);

temp = rand(C, 1 ) ;

p e s t = temp/sum( temp ) ;

161 for j =1:C

a e s t ( j ) =a e s t 1 ∗ p e s t ( j ) ;

b e s t ( j ) = b e s t 1 ∗ p e s t ( j ) ; end

difference = epsilon;

% now iterate

while (difference >= epsilon & counter < 25000)

a e s t o l d = a e s t ;

b e s t o l d = b e s t ;

p e s t o l d = p e s t ;

%M step: ML estimate the parameters of each

% c l a s s ( a e s t and b e s t )

for j =1:C

Denom = sum( doublexp density(values , a e s t o l d ( j ) , b e s t o l d ( j ) ) . /

...

pdfofg(values ,C,p e s t o l d , a e s t o l d , b e s t o l d ) ) ;

xminus = sum( doublexp density(values(values <0) , a e s t o l d ( j ) . . .

, b e s t o l d ( j ) ) . ∗ values(values <0)./pdfofg(values(values < 0 ) . . .

,C, p e s t o l d , a e s t o l d , b e s t o l d ) ) /Denom ;

162 xplus = sum( doublexp density(values(values >0) , a e s t o l d ( j ) . . .

, b e s t o l d ( j ) ) . ∗ values(values >0)./pdfofg(values(values > 0 ) . . .

,C, p e s t o l d , a e s t o l d , b e s t o l d ) ) /Denom ;

[ a e s t ( j ) , b est(j)] = mledoublexp(xminus,xplus);

p e s t ( j ) = sum( p e s t o l d ( j )∗ doublexp density(values ,...

a e s t o l d ( j ) , b e s t old(j))./pdfofg(values ,C,p e s t o l d , . . .

a e s t o l d , b e s t o l d ))/N;

end

difference(counter+1) = sum( abs ( a e s t o l d − a e s t ) ) + . . .

sum( abs ( b e s t o l d − b e s t ) ) + . . .

sum( abs ( p e s t o l d − p e s t ) ) ;

i f (mod(counter , 10)==0)

fprintf (’iteration %d , d i f f e r e n c e = %.12 f \n ’ ,

counter , difference(counter+1));

end

counter = counter + 1; end

function pdfdb = doublexp density(x, a, b)

% Compute density of double exponential distribution function

163 %

%file: doublexp d e n s i t y .m

% Calculate double exponential density in 1D.

% INPUTS:

% x =a vector of points at which to calculate

% the double exponentialPDF

% a = exponential parameter on the negative side

% b = exponential parameter on the positive side

pdfdb (x<0) = a∗b/( a+b)∗exp( a∗x (x <0));

pdfdb (x>0) = a∗b/( a+b)∗exp(−b∗x (x >0));

Step 3: We compare approximated ruin probability from step 2 to simulated ruin prob-

ability

% 1/theta 1 is mean of exponential claim epochs time

Theta1 = 5 ;

% parameters of Pareto part

Lambda = 3 ;

Alpha =4;

% c is the security loading

% u is the initial surplua

164 c =6;

u=1;

% Nsim: number of simulations runs

% Nclaims: simulated number of claims per run

Nsims =1000;

Ntimes = 1000;

Iseed = 1234567;

rng(Iseed);

% Nrans is the total number of bivariate random numbers needed

Nrans = Nsims∗Ntimes ;

% first simulate two sets of bivariate standard normal random

% variables with given correlation

SIGMA = [1,rho; rho,1];

r 1 = mvnrnd([0 ,0] ,SIGMA,Nrans);

r 2 = mvnrnd([0 ,0] ,SIGMA,Nrans);

% Time and Y are the bivariate exponential random vectors

Time = ( r 1(:,1).ˆ2+r 2(:,1).ˆ2)/(2 ∗ Theta1 ) ;

Y = ( r 1(:,2).ˆ2+r 2(:,2).ˆ2)/2;

% Convert Y into X, Pareto using the transformation

165 X = Lambda∗(exp(Y/Alpha ) −1);

% Convert columns into matrices

% each column of TimeMat is one simulation run of the

% claim occurrence times

% each column of ClaimMat is one simulation run of

% the claim sizes

TimeMat = reshape (Time,Ntimes ,Nsims);

ClaimMat = reshape (X,Ntimes ,Nsims);

% calculate the cumulative sum of each column

i =0;

for u=0:0.01:10

i= i +1;

surplus = u + c∗cumsum(TimeMat,1) −cumsum(ClaimMat ,1);

% any(surplus <0) will look at one column at a time of surplus

% and g i v e 0

% if that column has no negative values and 1 if it does

RuinProbability(i) = sum(any( surplus <0))/Nsims ;

end

u =0:0.01:10;

166 % In here add the line for ruin probability from Maple program

% being cautios to place ∗

Ruin1 = 0.73717 ∗exp( −0.16177∗ u) + 0.059234∗ exp( −1.1603∗u ) ;

Ruin2 = 1.∗10ˆ( −10)+.7245191391∗exp( −.1463183169∗u ) + . . .

0.8308906227e−1∗exp( −.8791753099∗u ) ;

Ruin3 = 0.7605012370e−1∗exp( −.9930760987∗u ) + . . .

0.7342279565∗exp( −.1416931789∗u ) ;

Ruin4 = 3.∗10ˆ( −10) −0.9420810197e−5∗exp( −1.577453067∗u ) . . .

+1.002163633∗exp( −0.4541159560e−3∗u) −0.1103439679e − 3 . . .

∗exp( −.3989384190∗u) −0.2043868562e−2∗exp( −.1485553040∗u ) ;

Ruin5 = .1335323931∗exp( −.4283067134∗u ) . . .

+0.2097336689e−1∗exp( −1.820970367∗u ) + . . .

.6512863430∗exp( −.1338620741∗u ) ;

Ruin6 = ( −1∗10ˆ( −9))+0.5848015623∗exp( −0.119100579∗u ) . . .

+0.1919203873∗exp( −0.3182409455∗u)+0.03211654587∗...

exp( −1.606497962∗u ) ;

D1 = RuinProbability−Ruin1 ;

D2 = RuinProbability−Ruin2 ;

D3 = RuinProbability−Ruin3 ;

D4 = RuinProbability−Ruin4 ;

D5 = RuinProbability−Ruin5 ;

D6 = RuinProbability−Ruin6 ;

figure ;

subplot ( 3 , 2 , 1 )

167 plot (u,RuinProbability , ’−r’,u,Ruin1, ’ −.b ’ ) t i t l e (’C = 2 and Ntimes = 10000 ’ ) xlabel ( ’u ( I n i t i a l Capital)’); ylabel ( ’ Ruin Probability ’); l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’); set (leg , ’Location’,’NorthEast’); set (leg , ’Interpreter ’,’none’);

subplot ( 3 , 2 , 2 ) plot (u,RuinProbability , ’−r’,u,Ruin2, ’ −.b ’ ) t i t l e (’C = 2 and Ntimes = 15000 ’ ) xlabel ( ’u ( I n i t i a l Capital)’); ylabel ( ’ Ruin Probability ’); l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’); set (leg , ’Location’,’NorthEast’); set (leg , ’Interpreter ’,’none’);

subplot ( 3 , 2 , 3 ) plot (u,RuinProbability , ’−r’,u,Ruin3, ’ −.b ’ ) t i t l e (’C = 2 and Ntimes = 20000 ’ ) xlabel ( ’u ( I n i t i a l Capital)’); ylabel ( ’ Ruin Probability ’); l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’);

168 set (leg , ’Location’,’NorthEast’); set (leg , ’Interpreter ’,’none’);

subplot ( 3 , 2 , 4 ) plot (u,RuinProbability , ’−r’,u,Ruin4, ’ −.b ’ ) t i t l e (’C = 3 and Ntimes = 10000 ’ ) xlabel ( ’u ( I n i t i a l Capital)’); ylabel ( ’ Ruin Probability ’); l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’); set (leg , ’Location’,’NorthEast’); set (leg , ’Interpreter ’,’none’);

subplot ( 3 , 2 , 5 ) plot (u,RuinProbability , ’−r’,u,Ruin5, ’ −.b ’ ) t i t l e (’C = 3 and Ntimes = 15000 ’ ) xlabel ( ’u ( I n i t i a l Capital)’); ylabel ( ’ Ruin Probability ’); l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’); set (leg , ’Location’,’NorthEast’); set (leg , ’Interpreter ’,’none’);

subplot ( 3 , 2 , 6 ) plot (u,RuinProbability , ’−r’,u,Ruin6, ’ −.b ’ ) t i t l e (’C = 3 and Ntimes = 20000 ’ )

169 xlabel ( ’u ( I n i t i a l Capital)’); ylabel ( ’ Ruin Probability ’); l e g = legend ( ’Simulated Ruin Probability ’,

’Approximated Ruin Probability ’); set (leg , ’Location’,’NorthEast’); set (leg , ’Interpreter ’,’none’);

170