reviewed risk theory notes

RISK THEORY FOR ACTUARIAL SCIENCES AND

FINANCIAL ENGINEERING STUDENTS

V. S. ANDIKA

January 25, 2015

RISK THEORY NOTES PREPARED BY V.S.ANDIKA

Course Objective

At the end of the course students will able apply statistics to risk measurement, perform sensi-

tivity analysis and test risk models

Course Outline

- Introduction to risk theory

- Economics of insurance

- Individual risk models for a short term

- Collective risk models for single and extended periods

- Appraisal techniques

- Analysis of risk/uncertainty and situation measurement of risk in precise terms

- Sensitivity analysis of loss distribution

- Statistical inference for loss distributions

- Ruin theory and applications

- Risk matrix

Reference Books

1) Daykin, C & Pentikainen, Practical Risk Theory for Actuaries, Chapman and Hall, ISBN:

9780412428500, 1993.

2) N.L. Bowers, et al, Actuarial Mathematics, 2nd edition, Society of Actuaries, ISBN: 9780938959465,

1997.

3) IB Hossack, JH Pollard, & B Zehnwirth, Introductory Statistics with Applications in Gen-

eral Insurance, Cambridge University Press, ISBN: 9780521655347, 1999.

4) Kluggman S A, and Panjer, H H, et al, Loss Models: From Data to Decisions, 3rd edition,

John-Wiley and Sons, ISBN: 9780470391334, 2012.

5) Lam, J., Enterprise Risk Management from Incentives to Controls, Wiley Finance, ISBN:

9780471430001, 2003.

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6) Bessis, J., Risk Management in Banking, 3rd edition, John-Wiley, ISBN: 9780470689851,

2011 .

7) Dowd, K., Beyond Value at Risks: The New Science of Risk Management,. Wiley, ISBN:

9780471976226, 2008.

8) Rob Kaas, Marc Goovaerts, Jan Dhaene, Michel Denuit ; Modern Actuarial Risk Theory,

Kluwer Academic publishers, 2001

Reference Journals

1) Journal of Risk and Uncertainty, ISSN: 0895-5646, ESSN: 1573-0476.

2) The Journal of Risk Analysis and Crisis Response (JRACR), ISSN: 2210-8491, ESSN: 2210-

8505.

3) Journal of Risk and Insurance, ISSN: 0022-4367, ESSN: 1539-6975.

4) Risk and Decision Analysis, ISSN: 1569-7371, ESSN: 1875-9173.

5) The Geneva Risk and Insurance, ISSN: 1554-964X, ESSN: 1554-9658.

6) International Journal of Theoretical and Applied Finance, ISSN: 0219-0249, ESSN: 1793-

6322.

7) The Mathematical Scientist, ISSN: 0312-3685.

8) Stochastic Analysis and Applications, ISSN: 0736-2994.

9) Journal of Modern Applied Statistical Methods, ISSN: 1538-9472.

10) The Annals of Statistics (Ann. Stat.), ISSN: 0090-5364.

Pre-Requisites:

1) STA 2300 Theory of Estimation

2) STA 2200 Probability and Statistics II

3) STA 2105 Calculus for Statistics II

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Co-Requisites:

1) STA 2395 Decision Theory and Bayesian Inference I

2) STA 2301 Tests of Hypotheses

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CHAPTER ONE

RISK THEORY INTRODUCTION

Definition

Risk theory is essentially a branch of probability theory, devoted to decision-making under prob-

abilistic uncertainty. Basic concepts of the theory are: risk, risk measure, risk price, and indi-

vidual attitude to risk.

Risk theory also connotes the study usually by actuaries and insurers of the financial impact

on a carrier of a portfolio of insurance policies. For example, if the carrier has 100 policies that

insures against a total loss of $1000, and if each policys chance of loss is independent and has

a probability of loss of p then the loss can be described by a binomial variable. With a large

enough portfolio however, we can use the Poisson function for the frequency of loss variable

where l is used as the mean equal to the number of policies multiplied by p.

Risk theory is a theory of decision-making under probabilistic uncertainty. From mathemat-

ical point of view it is a branch of probability theory, while its applications cover all aspects of

life. Financial applications are most advanced, including banking, insurance, managing market

and credit risks, investments and business risks. To name just a few, there are also applications

to managing risks of health hazard, environment pollution, engineering and ecological risks.

Terms used in risk theory

1. Risk management:- can be defined as the culture, processes, and structures that are

directed towards the effective management of potential opportunities and adverse effects

This is a broad definition that can quite rightly apply in nearly all fields of management

from financial and human resources management through to environmental management.

However in the context of contaminated sites, risk management can be taken to mean the

process of gathering information to make informed decisions to minimize the risk of adverse

effects to people and the environment.

Risk management is very important for insurance industry. Insurance means that insur-

ance companies take over risks from customers. Insurers consider every available quan-

tifiable factors to develop profiles of high and low insurance risk. Level of risk determines

insurance premiums. Generally, insurance policies involving factors with greater risk of

claims are charged at a higher rate. With much information at hand, insurers can evaluate

risk of insurance policies at much higher accuracy. To this end, insurers collect a vast

amount of information about policyholders and insured objects. Statistical methods and

4


tools based on data mining techniques can be used to analyze or to determine insurance

policy risk levels.

2. Risk assessment:- involves estimating the level of risk estimating the probability of an

event occurring and the magnitude of effects if the event does occur. Essentially risk assess-

ment lies at the heart of risk management, because it assists in providing the information

required to respond to a potential risk.

3. Acceptable risk:- The term "acceptable risk" describes the likelihood of an event whose

probability of occurrence is small, whose consequences are so slight, or whose benefits

(perceived or real) are so great, that individuals or groups in society are willing to take

or be subjected to the risk that the event might occur. The concept of acceptable risk

evolved partly from the realization that absolute safety is generally an unachievable goal,

and that even very low exposures to certain toxic substances may confer some level of

risk. The notion of virtual safety corresponding to an acceptable level of risk emerged as

a risk management objective in cases where such exposures could not be completely or

cost-effectively eliminated.

Two proxy measures have been used to determine acceptable risk levels. The revealed-

preference approach assumes that society, through trial and error, has achieved a nearly

optimal, and thus acceptable, balance of risks and benefits. The expressed-preference

approach uses opinion surveys and public consultations to obtain information about risk

levels warranting mitigation action.

Although regulatory authorities are reluctant to define a precise level of acceptable risk,

lifetime risks in the order of one in a million have been discussed in regulatory applications

of the acceptable risk concept. This level of risk is considered to be de minimis, an abbre-

viation of the legal concept de minimus non curatlex (the law does not concern itself with

trifles). Attempts have also been made to establish benchmarks, such as the risk of being

hit by lightning, to help interpret such small risks. Higher levels of risk might be tolerated

in the presence of offsetting health or economic benefits, when the risk is voluntary rather

than involuntary, or when the population at risk is small.

Although conceptually attractive, application of the concept of acceptable risk is fraught

with difficulty, ultimately involving consideration of social values. Inequities in the dis-

tribution of risks and benefits across society further complicate the determination of an

acceptable level of risk.

4. Risk determination:- involves the related processes of risk identification and risk estima-

tion. Risk identification is the process of observation and recognition of new risk parameters

or new relationships among existing risk parameters, or perception of a change in the mag-

nitudes of existing risk parameters. Risk, at the general level, involves two major elements:

the occurrence probability of an adverse event and the consequences of the event. Risk es-

5


timation, consequently, is an estimation process, starting from the occurrence probability

and ending at the consequence values.

5. Risk evaluation:- is a complex process of developing acceptable levels of risk to individuals,

groups, or the society as a whole. It involves the related processes of risk acceptance and

risk aversion.

6. Risk acceptance:- implies that a risk taker is willing to accept some risks to obtain a gain

or benefit, if the risk cannot possibly be avoided or controlled. The acceptance level is a

reference level against which a risk is determined and then compared. If the determined

risk level is below the acceptance level, the risk is deemed acceptable. If it is deemed

unacceptable and avoidable, steps may be taken to control the risk or the activity should

be ceased. The perception and the acceptance of risks vary with the nature of the risks and

depend upon many underlying factors. The risk may involve a "dread" hazard or a common

hazard, be encountered occupationally or non-occupationally, have immediate or delayed

effects and may effect average or especially sensitive people or systems.

7. Risk aversion:- is the control action, taken to avoid or eliminate the risk, regulate or

modify the activities to reduce the magnitude and/or frequency of adverse affects, reduce

the vulnerability of exposed persons, property or in this case urban systems, develop and

implement mitigation and recovery procedures, and institute loss-reimbursement and loss-

distribution schemes.

8. False accept risk:- is the probability that out-of-tolerance product or other parameters are

perceived to be in-tolerance.

= False accept risk constitutes a measure of the quality of a measurement process asviewed by individuals external to the measuring organization.

= The highier the false accept risk, the greater the chances for returned goods, loss ofreputation, litigation, and other undesirable outcomes.

= In a commercial context, individuals external to a measuring organization are labeledconsumers. For this reason, false accept risk has traditionally been called consumers

risk.

9. False reject risk:- is the probability that in-tolerance product or other parameters are

perceived to be out-of-tolerance.

= False reject risk is a measure of the quality of a measurement process as viewed byindividuals within to the measuring organization.

= The highier the false reject risk, the greater the chances for unnecessary re-work andre-test.

= In a commercial context, individuals within to a measuring organization are labeledproducers. For this reason, false reject risk has traditionally been called producers risk.

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Types of risks includes:

1. Systematic Risks (Systematic errors):- are classified as those whose sign and magnitude

remain fixed over a specified period of time or whose values change in a predictable way

under specified conditions. Example of this risk is market risk.

2. Unsystematic Risks (Random errors): - are those whose sign and/or magnitude may

change randomly over a specified period of time or whose values are unpredictable, given

randomly changing conditions. Examples of this risks includes; Stock and company spe-

cific risk

3. Credit or Default Risk :- can they pay the interest on the load or the dividend on that

stock

4. Country Risk :- Measure of political and economic stability

5. Foreign Exchange Risk: - will your money depreciate when you buy an asset in another

country?

6. Interest Rate Risk :- will the central bank raise interest rates

7. Political Risk :- Political stability of a country

8. Execution risk: - the time between when you see your price and when the trade actually

goes to the market.

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CHAPTER TWO

ECONOMICS OF INSURANCE

Introduction

The insurance industry exists because people are willing to pay for being insured which is highier

than their expected claims. As a result, an insurer collects a premium that is larger than the

expected claim size.

Let X be a loss random variable then empirical or pragmatic rules of premium are

(i) Premium = p = E[X] = pure premium

(ii) Premium = p = (1 + )E[X], is the pure premium with safety (security) loading , where 0.NB: = relative safety loading and E[X] =total safety loading

(iii) Premium = E[X] + aV ar(X) is the premium with variance loading principle.

(iv) Premium = E[X] + aV ar(X) is the premium with standard deviation loading principle.

Risk theory refers to a body of techniques to model and measure the risk associated with a

portfolio in insurance contracts.

A first approach consists modeling the distribution of total claims over a fixed period of time

using the classical collective model of risk theory. A second input of interest to the actuary is

the evolution of the surplus of the insurance company over many periods of time. In ruin theory,

the main quantity of interest is the probability that the surplus becomes negative, in which case

technical ruin of the insurance company occurs.

Modern life is characterized by risks of different kind: some threatening all persons and

some restricted to the owners of property, motor cars, etc, while still others are typical for some

individuals or for special occupations.

QUESTION

Define risk theory and discuss any two areas of interest in risk theory to an actuary.

8


Utility theory and Insurance

Under economics of insurance we examine the policyholders utility with and without insurance

and relate their preference to take out insurance or bear the risk themselves to the nature of

their utility function. We show that if an agent prefers more to less wealth and is risk a verse

then the agent will take out insurance.

The insurance industry exists because people are willing to pay for being insured which is

higher than their expected claims. As a result, an insurer collects a premium that is larger than

the expected claim size.

Generally without being a ware a decision maker, attaches a value () to his wealth instead

of just , where (.) is called his utility function. If a decision maker has to choose between

random losses X and Y , then he compares E[( X)] with E[( Y )] and chooses the losswith the highest expected utility. With this model, the insured with wealth is able to determine

the maximum premium P+ he is prepared to pay for a random loss X. This is done by solving

the equilibrium equation E[( X)] = ( P+). At the equilibrium he doesnt care, in termsof utility, whether he is insured or not.

The insurer, with his own utility function and perhaps supplementary expenses will deter-

mine a minimum premium P. If the insureds maximum premium P+ is larger than the in-

surers minimum P, both parties involved increase their utility if the premium is between P

and P+.

Determination of utility function

The utility function of a decision maker can be obtained using the utility model: ( p+) =p() + (1 p)( X), as illustrated in the example below.

Example

Describe how you will determine five points on the utility function of an individual whose wealth

is K Sh. 40,000. Assuming that probability of a risk occurring is 0.4 and the probability of

retaining the entire wealth is 0.6. Hence using the sketched utility function how would you

classify this type of risk individual and why?

Solution

Using the utility model

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( G) = p() + (1 p)( X)

and by asking the question(s);

How much are you willing to pay as maximum premium G for complete cover of loss X likely

to occur at probability 1 p and retain your wealth at probability p.

We let () = 0 and (0) = 1 be the starting values yielding the points (40000, 0) and (0,1)

when X = 40, 000 and say G = 28, 000

= (40, 000 28, 000) = 0.4(40, 000) + 0.6(0) = 0.6

i.e. (12, 000) = 0.6 = (12000,0.6) is a third point on the utility function

when X = 28, 000 and say G = 20, 000

= (40, 000 20, 000) = 0.4(40, 000) + 0.6(40000 28000)

= (20, 000) = 0.4(40, 000) + 0.6(12000) = 0.4 0 + (0.6) 0.6 = 0.36

i.e. (20, 000) = 0.36 = (20000,0.36) is a fourth point on the utility function

when X = 20, 000 and say G = 12, 000

= (40, 000 12, 000) = 0.4(40, 000) + 0.6(40000 20000)

= (28, 000) = 0.4(40, 000) + 0.6(20000) = 0.4 0 + (0.36) 0.6 = 0.216

i.e. (28, 000) = 0.216 = (28000,0.216) is a fifth point on the utility function

Since this function is a concave function then the individual is a risk averse individual.10


Properties of ()

1. Should be a non-decreasing function, since more wealth generally implies a large utility

level.

2. Should accommodate the risk averse decision makers. Since they prefer a fixed loss over a

random loss that has the same expected value.

The expected utility model

Imagine that an individual runs the risk of losing an amount B with probability 0.01. He can

insure himself against this loss, and is willing to pay a premium p for this insurance policy. How

are B and P related? If B is very small, then P will be hardly larger than 0.01B. However, if B is

somewhat larger, say 500, then P will be a little larger than 5. If B is very large, P will be a lot

larger than 0.01B, since this loss could result in bankruptcy. So clearly, the premium for a risk

is not homogeneous, i.e. not proportional to the risk.

In economics, the model developed by Von Neumann & Morgenstern (1947) describes how

decision makers choose between uncertain prospects. If a decision maker is able to choose con-

sistently between potential random losses X, then their exists a utility function (.) to appraise

the decisions he makes are exactly the same as those resulting from comparing the losses X

based on the expectation E[( X)].

For comparison of X with Y , the utility function (x) and its linear transform a(x) + b for

some a > 0 are equivalent, since they result in the same decision: E[( X)] E[( Y )] ifand only if E[a( X) + b] E[a( Y ) + b] so from each class of equivalent utility function,we can select one, for instance by requiring that (0) = 0 and (1) = 1. Assuming that

(0) > 0,

we could also use the utility function V (.) with V (0) = 0 and V (0) = 0 such that

V (x) =(x) (0)

(x)

Theorem (Jensens inequality)

If V (x) is a convex function and Y is a random variable then E[V (Y ) V (E[Y ])] with equality ifand only if V (.) is linear on the support of Y or V ar(Y ) = 0.

From this inequality it follows that for a concave utility function E[( X)] [E[( X)] =( E(x)) So this particular decision maker is rightly called risk a verse: he prefers to pay arisky amount X.

Now, suppose that a risk averse insured with capital used the utility function (.). Assum-

11


ing he is insured against a loss X for a premium P , his expected utility will increase if

E[( X)] E[( X)] = ( E[X]) = ( P ) (1)

Since (.) is a non-decreasing continuous function, this is equivalent to P P+. It is thesolution to the following utility equilibrium equation.

E[( X)] = ( P+) (2)

Note:

Jensens inequalities states that for a random variable X and function ().

(i) if () < 0, then E[(X)] (E[X])

(ii) if () > 0, then E[(X)] (E[X])

Proof for case (i)

() in case (i) is a concave function with a maximum stationary point as shown below

The equation of a tangent function to the function () at the point (, ()),

where = E[X] = is given as () = () + ()( ).

For all values of > 0, () ()+()() and replacing with X and taking expectation12


both sides we get; E[(X)] (E[X]).

This first Jensens inequality describes the attitute of a risk averse individual, who prefers

fixed loss over random losses.

Proof for case (ii)

() in case (ii) is a convex function with a minimum stationary point as shown below

The equation of a tangent function to the function () at the point (, ()),

where = E[X] = is given as () = () + ()( ).

For all values of > 0, () ()+()() and replacing with X and taking expectationboth sides we get; E[(X)] (E[X]).

This second Jensens inequality describes the attitute of a risk lover individual, who prefers

random loss over fixed losses. If the decision maker is neither risk averse or risk lover then he

is referred to as risk neutral individual. i.e. he is indifferent towards random and fixed losses.

Risk averse coefficient

Given the utility function (x), how can we approximate the maximum premium P+ for a risk X?

13


Solution

Let u and 2 denote the mean and variance of X. Using the first terms in the series expansion of

(.) in we obtain

( P+) ( u) + (u P+)( u);( X) +( u) + (uX)( u) + 1

2(uX)2( u) (3)

Taking expectations on both sides of the latter approximation yields

E[( X)] ( u) + 122

( u) (4)

substituting 2 into 4, it follows from 3 that

1

2( u) (u P+)( u)

Therefore, the maximum premium P+ for a risk x is approximately

P+ u 122( u)

( u)

This suggest the following definition: (absolute) risk aversion coefficient r() of the utility func-

tion (.) of a wealth is given by

r() =()()

Then the maximum premium P+ to be paid for a risk X is approximately

P+ u+ 12r( )2

Exponential Premium

The insurer with utility function U(.) and capital W, will insure the loss X for a premium P if

E[(W + P X) U(W )], hence P P where P is the minimum premium to be asked. Thispremium follows from solving the utility equilibrium equation reflecting the insurers position:

U(W ) = E[U(W + P X)]

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Question

Suppose that an insurer has an exponential utility function with parameter . What is the

minimum premium P to be asked for a risk X?

Solution

Solving the equilibrium equation U(W ) = E[U(W + P X)] with U(x) = ex yields

P =1

log(mx()) (5)

Where mx() = E[ex] is the mgf of X at argument . We observe that this exponential premium

is independent of the insurers current wealth W, in line with risk aversion coefficient being a

constant.

The expression for the maximum premium P+ is the same as 5, but now of course rep-

resents the risk aversion of the insured. Assume that the loss X is exponentially distributed

with parameter . Taking = 0.01 yields E[X] = 1 = 100. If the insureds utility function is

exponential with parameter = 0.005, then

P+ =1

log(mx())

= 200log

(

)

= 200log(2)

138.6

so the insured is willing to accept a sizable loading on the net premium E[X].

Quadratic utility

Suppose that for < 5, the insureds utility function is () = 10 2. What is the maximumpremium P+ as a function of , [0, 5] for an insurance policy against a loss 1 with probability12? What happens to this premium is increases. Let E[X] =

12 and var(x) =

14

Solution

We solve the equilibrium equation

E[( X)] = ( P+) (6)

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The expected utility after a loss X equals

E[( X)] = 11 112 2 (7)

and the utility after paying a premium P equals

( P ) = 10( P ) ( P )2 (8)

By the equilibrium equation 6, the right hand side of 7 and 8 should be equal, and after calcu-

lations

P = P () =

(11

2 )2 + 1

4 (5 ), [0, 5)

Exercise

A decision maker with an exponential utility function with risk aversion > 0 wants to insure a

gamma(n,1) distributed risk. Determine P+ and prove that P+ > n. When is P+ = and whatdoes that mean?

16


CHAPTER THREE

THE INDIVIDUAL RISK MODEL FOR SHORT TERMS

Introduction

We focus on the total claim amount S for the portfolio of an insurer.

- The individual risk model is defined as S = X1 +X2 + ...+Xn where Xi is the loss on insured

unit i.

- n is the number of risk units insured and S is the random variable which denotes the sum

of random loss on a segment of insuring organization risks.

- Under individual risks model one does not recognize the time value of money. Thus the

title individual risk model for short terms.

- Also in this case we discuss only closed models; that is the number of insured units n in

S = X1 +X2 + ...+Xn is known and fixed at the beginning of the period.

Note:

If we postulate about migration in and out of the insurance system, we have an open model.

In the Insurance practice, risks usually cant be modeled by purely discrete random variables,

nor by purely continuous random variables. Thus assuming that the risks in a portfolio are

independent random variables, the distribution of their sum can be calculated by making use

of convolution, moment generating functions, characteristic functions, probability generating

functions (pgf) and cumulant generating functions (cgf). Sometimes it is possible to recognize

the mgf of a convolution and consequently identify the distribution.

Mixed distributions and risks

Many distributions functions that are employed to model insurance payments have continuously

increasing parts, but also some positive steps.

Let Z represent the payment on three contract. Then, as a rule, there are three possibilities.

1. The contract is claim-free, hence S = 0.

17


2. The contract generates a claim which is larger than the maximum sum insured, say M .

Then, S = M .

3. The contract generates a normal claim, hence 0 < S < M .

Apparently, the cdf of S has steps in 0 and M . For the part in between we could use a discrete

distribution, since the payment will be some entire multiple of the monetary unit. This would

be a very large set of possible values, each of them with a very small probability, so using a

continuous cdf seems more convenient.

Note:

The following two-staged model allows us to construct a random variable with a distribution that

is a mixture of a discrete and a continuous distribution.

Let I be an indicator random variable, with values I = 1 or I = 0. Where I = 1 indicates that

some event has occurred. Suppose that the probability of the event is q = Pr[I = 1], 0 q 1. IfI = 1, the claim Z is drawn from the distribution of X, if I = 0, then from Y . This means that

S = IX + (1 I)Y (9)

If I = 1, then S can be replaced by X, if I = 0 can be replaced with Y . Note that we can consider

X and Y to be stochastically independent of I, since given I = 0 the value of X is irrelevant, so

we can take Pr[X x|I = 0] = Pr[X x|I = 1] just as well. Hence, the cdf of Z can be written as.

F (s) = Pr[S s]= Pr[S s&I = 1] + Pr[S s&I = 0]= Pr[X s&I = 1] + Pr[Y s&I = 0]= qPr[X s] + (1 q)Pr[Y s] (10)

Now, let X be a discrete random variable and Y a continuous random variable. From (10), we

get F (s) f(s 0) = qPr[X = s] and F (s) = (1 q) ddsPr[Y s] This construction yields a cdf F (s)With steps where Pr[X = s] > 0, but it is not a step function, since F

> 0 on the range of Y .

Remark:

A mixed continuous/discrete cdfS(s) = Pr[S s] arises when a mixture of random variablesS = IX + (1 I)Y is used, where X is a discrete random variable, Y is a continuous randomvariable and I is a Bernoulli(q) random variable independent of X and Y .

18


Types of individual risk models

CASE I

The aggregate loss S = X could be a claim by an individual against a single policy where the

insurer agrees to pay amount X = b if the event insured against occurs at probability q and

X = 0 if the event does not occur at probability 1 q.

In this case; S = X = Ib.

where b is the constant amount payable in the event of death and I is the random variable

that is 1 for the event of death and 0 otherwise. Thus Pr(I = 0) = 1 q and Pr(I = 1) = q, themean and variance of I are q and q(1 q), respectively, and the mean and variance of X are bqand b2q(1 q) .

i.e.

The p.f is

fX(x) = Pr(X = x) =

1 q for x = 0q for x = b

0 elsewhere

and the d.f is

FX(x) = Pr(X x) =

0 for x < 0

1 q for 0 x < b1 for x b

From the p.f and definitions of moments

E[X] = bq

E[X2] = b2q

V ar(X) = b2q(1 q)

19


CASE II

The aggregate loss S = X, could also be several claims by an individual from a single policy.

Example Accident cover due to constant Amount paid

( Fire b1( V ehicle b2

X ( Terrorism b3 B = Random variable( Natural b4 ( Riot b5

In this case: S = X = IB

where, B gives the total claim amount incurred during the period, and I is the indicator for

the event that at least one claim has occurred.

and we have

E[S] = E[E[S|I]] = E[E[IB|I]]= E[IE[B|I]]= E[I]E[B|I]= E[I]E[B]

NOTE: The expected claim total equals expected value of indicator variable times expected

claim size.

V ar(S) = E[V ar(S|I)] + V ar(E[S|I]) = E[V ar(IB|I)] + V ar(E[IB|I])= E[IV ar(B|I)] + V ar(IE[B|I])= E[I]V ar(B|I) + (E[B|I])2 V ar(I)

cdf of S is

Letting FS(s) be the d.f of S = X we have

20


FX(x) = Pr(X x) = Pr(IB x)= Pr(IB x|I = 0)Pr(I = 0)+ Pr(IB x|I = 1)Pr(I = 1)

Assignment

Illustrate plots for pdfs and cdfs of S.

Application of Mixed distributions in modeling individual random variables

In a one-year term life insurance the insurer agrees to pay an amount b if the insured dies within

a year of policy issue and to pay nothing if the insured survives the year. The probability of a

claim during the year is denoted by q. The claim random variable, X, has a distribution that can

be described by either its probability function(p.f) or distribution function(d.f)

The p.f is

fX(x) = Pr(X = x) =

1 q for x = 0q for x = b

0 elsewhere

and the d.f is

FX(x) = Pr(X x) =

0 for x < 0

1 q for 0 x < b1 for x b

From the p.f and definitions of moments

E[X] = bq

E[X2] = b2q

V ar(X) = b2q(1 q)21


These formulas can also be obtained by writing

X = Ib

where b is the constant amount payable in the event of death and I is the random variable that

is 1 for the event of death and 0 otherwise. Thus Pr(I = 0) = 1 q and Pr(I = 1) = q, the meanand variance of I are q and q(1 q), respectively, and the mean and variance of X are bq andb2q(1 q) as above.

Extending X = Ib, we postulate that where X is the claim random variable for the period, B

gives the total claim amount incurred during the period, and I is the indicator for the event that

at least one claim has occurred.

Example

Assume that for a particular individual the probability of one claim in a period is 0.15 and the

chance of more than one claim is 0. Then

Pr(I = 0) = 0.85

Pr(I = 1) = 0.15

Now consider an automobile insurance providing collision coverage above a 250 deductible up

to a maximum claim of 2000.

Since B is the claim incurred by the insurer, rather than the amount of damage to the car,

we can infer two characteristics of I and B. First, the event I = 0 includes those collisions in

which the damage is less than 250 deductible. The other inference is that Bs distribution has

a probability mass at the maximum claim size of 2000. Assume this probability mass is 0.1.

Furthermore assume that

Pr(B x|I = 1) =

0 x 0

0.9[1 (1 x2000)2] 0 < x < 2000

1 x 2000

Note:

The distribution Function for B, given I = 1 is

22


Now, First we derive the distribution of X and use it to calculate E[X] and V ar(X). Letting

FX(x) be the d.f of X we have

FX(x) = Pr(X x) = Pr(IB x)= Pr(IB x|I = 0)Pr(I = 0)+ Pr(IB x|I = 1)Pr(I = 1)

For x < 0, FX(x) = 0(0.85) + 0(0.15) = 0

For 0 x < 2000, FX(x) = 1(0.85) + 0.9[1 (1 x2000)2] (0.15)

For x 2000, FX(x) = 1(0.85) + 1(0.15) = 1

This is a mixed distribution. It has both probability masses and a continuous part as can be

seen in the graph below

Corresponding to this d.f is a combination p.f and p.d.f given by

Pr(X = 0) = 0.85

Pr(X = 2000) = 0.015

with p.d.f

23


fX(x) = FX(x) =

0.000135(1 x2000

)0 < x < 2000

0 elsewhere

Moments of X can be calculated by

E[XK ] = 0 Pr(X = 0) + (2000)K Pr(X = 2000) + 2000

0xkfX(x)dx

specifically,

E[X] = 120

E[X2] = 150, 000

Thus V ar(X) = 135, 600

CASE III

The aggregate loss S = X could be also claim amount from n insuring units of an insuring

organization, or a single portfolio.

In this case:

S = X1 +X2 + ...+Xn

where;

24


Xi; i = 1, 2, ....n, denotes the payment on policy i.

The risk Xi are assumed to be independent random variables and denotes the total claim

amount on ith insuring unit .

We have;

Let, = E[X] and 2 = V ar(X). Where, X is the claim severity.

Then

E[S] = E[X1 +X2 + ...+Xn] = E[X1] + E[X2] + ...+ E[Xn] = nE[X] = n

NOTE: Expected value of aggregate claims=Number of insuring units times expected value of

claim severity.

V ar (X) = V ar[X1 +X2 + ...+Xn] = V ar[X1] + V ar[X2] + ...+ V ar[Xn] = nV ar[X] = n2

NOTE: Variance of aggregate claims=Number of insuring units times variance of claim sever-

ity.

Mgf of S = MS(t) = E[eSt]

= E[et[X1+X2+...+Xn]

]={E[etX]}n

= {MX(t)}n =ni=1

MXi(t)

Determination of probability distribution for S = X1 + X2 + ... + Xn using Central

Limit Theorem

Given that X is are independent random variables and n 30, then using normal approximation

S = X1 +X2 + ...+Xn =

ni=1

Xi N (E[S], V ar (S)) = N(n, n2

)and

S E[S]V ar(S)

=S nn N(0, 1)

also

25


1

nS =

X1 +X2 + ...+Xnn

=1

n

ni=1

Xi = S N(E[S],

V ar(S)

n

)= N

(n

n,n22

n

)= N

(,2

n

)

and

S E[S]V ar(S)

=S n

N(0, 1)

Question

Give an illustration that requires application of this theorem.

Determination of probability distribution for S = X1 + X2 + ... + Xn using Moment

generating functions

Since,

Mgf of S = MS(t) = E[eSt]

= E[et[X1+X2+...+Xn]

]={E[etX]}n

= {MX(t)}n =ni=1

MXi(t)

Due to uniqueness theorem of moment generating functions. Which says that, mgfs uniquely

determines distributions. i.e. variables with similar distributions have the same mgf function.

This means that one can deduce the pdf of S by examining the productni=1MXi(t) . During

examination one tries to find out if the product is mgf of a known distribution.

Recall:

Suppose that X is a random variable with pdf f(x) and mgf MX(t).

Then;

(a) If X poison() then, MX(t) = e(et1), E[X] = , and V ar(X) =

(b) If X binomial(n, p) then, MX(t) =[pet + (1 p)]n , E[X] = np, and V ar(X) = npq.

Where q = 1 p.With a special case:

26


When n = 1, then X Bernoulli(p).

(c) If X gamma(, ) then, MX(t) =(

t)

=(1 t

), E[X] = , and V ar(X) =

2.

With special cases:

(i) When = 1, then X exponential().(ii) When = V2 and =

12 , then X 2V .

(d) If X NB(r, p) then, MX(t) =(

p1(1p)et

)rthen, E[X] = rqp , and V ar(X) =

rqp2

.

Where q = 1 p.With a special case:

When r = 1, then X geometric(p)

(e) If X N (, 2) then, MX(t) = E[etX ] = et+ 122t2 , E[X] = , V ar(X) = 2.(f) If X Uniform(a, b) then, MX(t) = etbetatbta , E[X] = a+b2 , and (ba)

2

12

Determination of probability distribution for S = X1+X2+ ...+Xn using convolution

In the risk model we are interested in the distribution of the total S of the claims on a number

of policies, with S = X1 + X2 + ... + Xn, where Xi; i = 1, 2, ....n, denotes the payment on policy i.

The risk Xi are assumed to be independent random variables.

Let us consider the sum of two random variables S = X + Y , with the sample space shown

below (Representing the event {X + Y s})

The line X + Y = s and the region below the line represent the event [X + Y s]

27


Hence the d.f of S is FS(s) = Pr(S s) = Pr(X + Y s)

The operation Convolution calculates the distribution function of X + Y from those of two

independent random variables X and Y , as follows.

FX+Y (s) = Pr[X + Y s]=

Pr[X + Y s|X x]dFX(x)

=

Pr[Y s x|X = x]dFX(x)

=

Pr[Y s x]dFX(x)

=

FY (s x)dFX(x)= FX FY (s) (11)

The cdf FX FY (.) is called the convolution of the cdfs FX(.) and FY (.). For the density functionwe use the same notation. If X and Y are discrete random variables, we find

FX FY (s) =x

FY (s x)fX(x)

and

fX fY (s) =x

fY (s x)fX(x)

Where the sum is taken over all x with fX(x) > 0. If X and Y are continuous random variables,

then

FX FY (s) =

FY (s x)fX(x)dx

and taking the derivatives under the integral sign, yields

fX fY (s) =

fY (s x)fX(x)dx

Note:

1. For three cdfs i.e. for cdf of X + Y + Z

(FX FY ) FZ FX (FZ FZ) FX FZ FZ

i.e it does not matter in which order we do the convolutions. \item For the sum of n

independent and identically distributed random variables with marginal cdf, the cdf is the

n-fold convolution power of F , written as

F F ... F = F n28


Example: (convolution of discrete distributions)

Let f1(x) = 14 ,12 ,

14 for x = 0, 1, 2; f2(x) =

12 ,

12 for x = 0, 2 and f3(x) =

14 ,

12 ,

14 for x = 0, 2, 4.

Let f1+2 denote the convolution of f1 and f2 and f1+2+3 denote the convolution of f1, f2 and

f3. To calculate F1+2+3, we need to compute the values as shown in the table below.

X f1(x)* f2(x) = f1+2(x)* f3(x) = f1+2+3(x) F1+2+3(x)

0 1412

18

14

132

132

1 12 028 0

232

332

2 1412

28

12

432

732

3 0 0 28 0632

1332

4 0 0 1814

632

1932

5 0 0 0 0 6322532

6 0 0 0 0 4322932

7 0 0 0 0 2323132

8 0 0 0 0 1323232

Example: (Convolution of two uniform distributions)

Suppose that X Uniform(0, 1) and Y Uniform(0, 2) are independent. What is the cdf ofX + Y ?

Solution

We introduce the concept indicator function". The indicator function of a set A is defined as

follows

IA(x) =

1 if x A0 if x / AIndicator functions provide us with a concise notation for functions that are defined differ-

ently on some intervals. For all x, the cdf of X can be written as

FX(x) = xI[0,1)(x) + I[1,)(x)

While for FY (y) =

12I[0,2)(y) for all y, which leads to the differential

dFY (y) =1

2I[0,2)(y)dy

29


The convolution formula (11), applied to Y +X rather than X + Y , then yields

FY+X(s) =

FX(s y)dFY (y) = 2

0FX(s y)1

2dy, s 0

The interval of interest is 0 s < 3. Substituting it into [0,1), [1,2) and [2,3) yields

FX+Y (s) =

{ s0

(s y)12dy

}I[0,1)(s) +

{ s10

1

2dy +

ss1

(s y)12dy

}I[1,2)(s)

+

{ s10

1

2dy +

2s1

(s y)12dy

}I[2,3)(s)

=1

4S2I[0,1)(s) +

1

4(2s 1)I[1,2)(s) + [1

1

4(3 s)2]I[2,3)(s)

Notice that X + Y is symmetrical around s = 1.5

Models for individual claim amount (claim severity)

In this section we consider continuous distributions to model individual claim amount. This is

justified on the grounds that the monetary amounts are continuous rather than discrete. The

following are the most commonly used continuous distributions.

Normal distribution

A continuous random variable X is said to follow a normal distribution if its pdf is given by

f(x) =

1

2pie

12(

x )

2

< x


f(x) =

x1e

x

0 < x 0, > 00 otherwise

MX(t) = (1 t) or(

11t

)E[X] =

V ar(X) = 2

OR

f(x) =

x1ex

0 < x 0, > 00 otherwise

With

E[X] =

V ar(X) = 2

Beta distribution

f(x) =

x1(1x)1

B(,) 0 < x < 1, > 0, > 0

0 otherwise

With

E[X] = +

V ar(X) = (+)2(++1)

Exponential distribution

It is a special case for poison distribution with = 1 and = 1

= f(x) =

ex 0 < x


E[X] = 1

V ar(X) = 12

Chi-Square distribution

It is a special case for gamma distribution where = r2 and = 2, r > 0

f(x) =

xr21e

x2

( r2

)2r2

0 < x


but Y = lnx, dy = dxx dx = xdy = eydy

E[X] = 1

2pi

e1

22[(y)222y]dy

= e+1

22

e1

22[(y(+2)]2

2pidy

= e+1

22

With V ar(X) = e(2+ 2)[e(2 1)]

Pareto distribution

f(x) =

x

(x+)+1x > 0, > 0, > 0

0 otherwise

Where

MX(t) = does not exist

E[X] =

0

x

(x+ )+1dx

=

0

x

(x+ )+1dx

=

0

x+ ( )(x+ )+1

dx

=

0

(x+

(x+ )+1

(x+ )+1

)dx

= [

1

1 .1

(x+ )+1+

1

(x+ )+1

]0

dx

= [

1

1 .1

()+1+

()

]dx

=

1

[1

1 1

]=

1

and

V ar(X) = 2

(1)2(2)

33


Example

In a Bayesian investigation in the insurance industry, a particular claim rate is to be estimated.

To obtain a suitable prior distribution, an expert is consulted and he suggest that will have

a mean of 0.15 and a standard deviation of 0.05. It is required to consult a prior gamma

distribution to fit the experts information. Find the parameters for the appropriate gamma

distribution.

Solution

For a gamma distribution with parameters and , E[X] = and V ar(X) = 2

= 0.15 2 = (0.05)2

= 0.15 , (0.15) = (0.05)2 = 0.0166666 and = 9

Question

Let X Poisson() and Y Poisson() be independent random variables. If S = 0, 1, 2, ..., findthe distribution of X + Y = S using the convolution method.

Question

Independent random variables Xk for four lives have the discrete probability functions given

below.

X Pr(X1 = x) Pr(X2 = x) Pr(X3 = x) Pr(X4 = x)

0 0.6 0.7 0.6 0.9

1 0.0 0.2 0.0 0.0

2 0.3 0.1 0.0 0.0

3 0.0 0.0 0.4 0.0

4 0.1 0.0 0.0 0.1

Use a convolution process on the non-negative values of x to obtain FS(x) for X = 0, 1, 2, ...

where S = X1 + X2 + X3 + X4, hence calculate the mean, variance, skewness and kurtosis of

the distribution of S.

34


CHAPTER FOUR

COLLECTIVE RISK MODELS (or Compound risk models)

Introduction

Under collective risk models we calculate the distribution of the total claim amount in a certain

time period, but now we regard the portfolio as a collective that produces a claim at random

points in time. We write

S = X1 +X2 + ...+XN (12)

Where N denotes the number of claims and Xi is the ith claim, and by convention, we take S = 0

if N = 0. So the terms of S in (12) correspond to actual claims. But for individual risk model

there are many terms equal to zero, corresponding to the policies which do not produce a claim.

The number of claims N is a random variable, and we assume that the individual claims Xi are

independent and identically distributed. In the special case that N is poison distributed. S has

a compound poison distribution. If N is (negative) binomial distributed, then S has a compound

(negative) binomial distribution.

Note:

- In collective risk models we require the claim number N and the claim amounts Xi to be

independent.

- Collective risk model is a computationally efficient model, which is also rather close to

reality.

- In collective models, some policy information is ignored.

Compound distribution

Assume that S = X1+X2+...+XN is a compound distribution, Xi are distributed as X, k = E[Xk],

Pr(x) = Pr[X x] and F (s) = Pr[S s]. We can then calculate the expected value of S by usingthe conditional distribution of S, given N.

i.e.

35


E[S] = E[E[S|N ]] =n=0

E[X1 + ...+XN |N = n]Pr(N = n)

=n=0

E[X1 + ...+Xn|N = n]Pr(N = n)

=

n=0

E[X1 + ...+Xn]Pr(N = n)

=n=0

n1Pr(N = n) = 1E[N ]

NOTE: The expected claim total equals expected claim frequency times expected claim size.

V ar(S) = E[V ar(S|N)] + V ar(E[S|N ])= E[NV ar(X)] + V ar(N1)

= E[N ]V ar(X) + 21V ar(N)

mgf of S is

MS(t) = E[E[etS |N ]]

=

n=0

E[et(X1+...+XN )|N = n

]Pr(N = n)

=n=0

E[et(X1+...+Xn)

]Pr(N = n)

=n=0

{MX(t)}n Pr(N = n) = E[(elogMX(t)

N)]

= MN (logMX(t)) (13)

Example (on compound distribution with closed form)

Let N geometric(p),

for 0 < p < 1 and X exponential(1). What is the mgf of S?

Solution

Write q = 1 p. First, we compute the mgf of S, and then we try to identify it. For qet < 1, whichmeans t < logq, we have

MX(t) =

n=0

entpqn =p

1 qet

36


Since X exponential(1), i.e. MX(t) = (1 t)1 equation (13) yields

MS(t) = MN (logMX(t)) =p

1 qMX(t) = 1 + qp

p t

So the mgf of S is a mixture of the mgfs of the constant 0 and of the exponential(p) distribution.

Because of one-to-one correspondence of cdfs and mgfs we may conclude that the cdf of S is

the mixture

F (x) = p+ q(1 epx) = 1 qepx for x 0

This is a distribution function which has a jump of size p in 0 and is exponential otherwise.

Distribution for the Number of claims

To describe rare events the poison distribution which has only one parameter is always the first

choice. If the model for the number of claims exhibits a larger spread around the mean value,

one may use the negative binomial distribution instead.

Example (on compound distribution with closed form)

Let N geometric(p), 0 < p < 1 and X exponential(1). What is the mgf of S?

Solution

Write q = 1 p. First, we compute the mgf of S, and then we try to identify it. For qet < 1, whichmeans t < logq, we have

MX(t) =

n=0

entpqn =p

1 qet

Since X exponential(1), i.e. MX(t) = (1 t)1 equation (13) yields

MS(t) = MN (logMX(t)) =p

1 qMX(t) = 1 + qp

p t

So the mgf of S is a mixture of the mgfs of the constant 0 and of the exponential(p) distribution.

Because of one-to-one correspondence of cdfs and mgfs we may conclude that the cdf of S is

the mixture

F (x) = p+ q(1 epx) = 1 qepx for x 0

This is a distribution function which has a jump of size p in 0 and is exponential otherwise.

37


Distribution for the Number of claims

To describe rare events the poison distribution which has only one parameter is always the first

choice. If the model for the number of claims exhibits a larger spread around the mean value,

one may use the negative binomial distribution instead.

Question

Assume that some car driver causes a poison() distributed number of accidents in one year. The

parameter is unknown and different for every driver. We assume that is the outcome of a

random variable . The conditional distribution of N, the number of accidents in one year, given

= , is poison().

(a) What is the marginal distribution of N?

(b) What is the mean and variance of N?

(c) What is the distribution of N if gamma(, )

Solution

Let U() = Pr( ) denote the distribution function of . Then we can write the marginaldistribution of N as

Pr(N = n) =

0

Pr(N = n| = )dU()

=

0

en

n!dU()

while for mean and variance of N we have

E[N ] = E[E[N |]] = E[]

V ar(N) = E[V ar(N |)] + V ar[E[N |]]= E[] + V ar[] E[N ]

Now assume additionally that gamma(, ), then

38


MN (t) = E[E[etN |]] = E

[e(e

t1)]

= M(et 1) =

(

(et 1))

=

(p

1 (1 p) et)

where p = +1 , so N has a negative binomial(,

(+1)) distribution.

Models for the timing and occurrence of events

These models use discrete variables since they count the number of events occurring within a

specified time interval. The following are the most commonly used distributions used to model

such events.

Binomial distribution

Pr(X = x) =

nx

px(1 p)nx, x = 0, 1, 2, ...The mgf

MX(t) = E[ext]

=all x

ext

nx

px(1 p)nx=

all x

nx

(pet)x(1 p)nxRecall

all x

nx

axbnx = (a+ b)n

MX(t) = (pet + q)n

and

E[X] = MX |t=0 = npet(pet + q)n1|t=0

= npe0(pe0 + q)n1

= np(p+ q)n1 but p+ q = 1

= np

39


V ar = E[X2] (E[X])2

= MX(t)|t=0 (M

X(t))

2|t=0= npq

Negative binomial distribution

A random variable X is said to have a negative binomial distribution if pdf is given by

Pr(X = x) =(v+x1x1

)pv(1 p)x1, v > 0, 0 < p < 1

(where v= the number of occurrences for the event)

MX(t) =(

p1qet

)v

ddt

(p

1 qet)v

= pvd

dt(1 qet)v

= pv(v)(qet)(1 qet)v1

=pvvqet

(1 qet)v+1 |t=0 =pvvq

(1 q)v+1

=pvvq

pv+1

=vq

p

E[X] = M X(t)|t=0 =vq

p

and

V ar(X) =vq

p2

Suppose v = 1 then Pr(X = x) = p(1 p)x1, x = 1, 2, 3, ... which is geometric distribution withmgf,

MX(t) =p

1qet , E[X] =qp and V ar(X) =

qp2

Example

A portfolio consists of a total of 120 independent risks. On each risk, no more than 1 event can

occur each year, and the probability of an event occurring is 0.02. When such an event occurs,

the number of claims N has the following distribution. P (N = x) = 0.4(0.6)x1, X = 1, 2, 3, ...

Determine the mean and variance of the distribution of the number of claims which arise

from this portfolio in one year.40


Solution

Recall that if X and Y are independent random variables then E[Y ] = E[E[Y |X]]. By convention,let the random variable Z = X1 +X2 + ...+Xn denote the aggregate claim amount where Xi, i =

1, 2, 3, ..., n are iid, random variables, independent of N, the number of claims on a risk in 1 year.

Then

E[Z] = E[[E(Z|N)]]= E[NE(X)]

= E[N ]E[X]

and

V ar(X) = E[V ar(Z|N)] + V arE[Z|N ]= E[NV ar(X)] + V ar(NE[X])

= E[N ]V ar(X) + V ar(N)(E[X])2

Now let M be the total number of claims, also let K be the number of risks for which the events

occurs.

M = N1 +N2 + ...+NK , where N geometric, k B(n, p) = B(120, 0.02)

Then mean is

E[M ] = E[E[M |K]]= E[K]E[N ]

= np.q

p

= (120 0.02) 0.60.4

= 3.6

and

41


V ar(M) = E[V ar(M |K)] + V ar(E[M |K])= E[K]V ar(N) + V ar(K)(E[N ])2

= np.q

p2+ npq

(q

p

)2= 120(0.02)

0.6

(0.4)2+ 120(0.02)(0.98)

(0.6)2

(0.4)2

= 14.292

Theorem: Panjers recursion

Consider a compound distribution with inter-valued non-negative claims with pdf P (x), x =

0, 1, 2, ... for which the probability qn of having n claims satisfies the following recursion relation

qn =

(a+

b

n

)qn1, n = 1, 2, ... (14)

for some real a and b. Then the following relations for probability of a total claim equal to s hold:

f(0) =

Pr(N = 0) if p(0) = 0MN (logP (0)) p(0) > 0

f(s) =1

1 ap(0)s

h=1

(a+

bh

s

)p(h)f(s h), s = 1, 2, 3, ... (15)

Proof:

Pr(S = 0) =

n=0

Pr(N = n)Pn(0)

gives us starting value f(0). Write Tk = X1 + ... + Xk. First, note that because of symmetry.

E[a+ bX1S |Tk

]= a+ bk . This expectation can be determined in the following way.

E

[a+

bX1S|Tk]

=s

h=0

(a+

bh

s

)Pr(X1 = h|Tk = s)

=s

h=0

(a+

bh

s

)Pr(X1 = h)Pr(Tk X1h|s h)

Pr(Tk = s)

42


Because (14) and the previous two equalities, we have for s = 1, 2, ...

f(s) =k=1

Pr(Tk = s) =k=1

qk1(a+

b

k

)Pr(Tk = s)

=k=1

qk1s0

(a+

b

k

)Pr(X1 = h)Pr(Tk X1 = s h)

=s

h=0

(a+

b

k

)Pr(X1 = h)

k=1

qk1Pr(Tk X1 = s h)

=s

h=0

(a+

bh

s

)p(h)f(s h)

= aP (0)f(s) +

sh=1

(a+

bh

s

)p(h)f(s h)

From which the second relation of (15) follows immediately

Distributions suitable for panjers recursion

Only the following distributions satisfy

qn =

(a+

b

n

)qn1 , n = 1, 2, ...

1. Poison() with a = 0 and b = 0, in this case (14) simplifies to

f(0) = e(1P (0))

f(s) =1

s

sh=1

hP (h)f(s h) (16)

2. Negative binomial(r,p) with a = 1 p and b = (1 p)(r 1), so 0 < a < 1 and a+ b > 0. In thiscase (14) simplifies to

f(0) = pr

f(s) = (1 p)s

h=1

[(r 1)h

s+ 1

]P (h)f(s h) (17)

NOTE: The constants a and b for Negative binomial distribution are obtained using the

distribution that models the number of failures before rth success. I.e. if Y is the number

of failures before rth success then;

43


Y NB(r, p) and f(y) = r + y 1

y

prqy, where y=0,1,2...Also note that;

If X denotes the bernoulli trial at which the rth success occurs, where r is fixed. Then;

X NB(r, p) and f(x) = x 1

r 1

prqxr., where x=0,1,2,...3. Binomial(m,p) with a = p1p and b = (m+1)p(1p) . so a < 0, b = a(m + 1) . In this case (14)

simplifies to

f(0) = (1 p)m

f(s) =p

(1 p)s

h=1

[(m+ 1)

h

s 1]P (h)f(s h) (18)

Example:

A compound poison distribution with = 4 and Pr(X = 1, 2, 3) = 14 ,12 ,

14 . What is the distribution

of S = X1 +X2 +X3

Solution

Equation (16) yields, with = 4, P (2) = 12 and P (1) = P (3) =14

f(s) =1

s[f(s 1) + 4f(s 2) + 3f(s 3)], s = 1, 2, 3, ...

and the starting value is f(0) = e4 0.0183. We have

f(1) = f(0) = e4

f(2) = 12 [f(1)4f(0)] =52e4

f(3) = 13 [f(2) + 4f(1) + 3f(0)] =196 e4 and so on.

Question

A compound Negative binomial distribution with r = 14, p = 0.7 and Pr(X = 1, 2, 3) = 14 ,12 ,

14 . What

is the distribution of S = X1 +X2 +X3

44


Question

A compound Binomial distribution with m = 15, p = 0.3 and Pr(X = 1, 2, 3) = 14 ,12 ,

14 . What is the

distribution of S = X1 +X2 +X3

Question

Prove that in deed the values of the constants a and b in Panjers recursion theorem

qn =

(a+

b

n

)qn1 , n = 1, 2, ...

for Poisson, Negative binomial and Binomial distribution are as given in the discussion above.

45


CHAPTER FIVE

RUIN THEORY

Under ruin theory we focus again on collective risk models, but now in the long term. We

consider the development in time of the capital U(t) of an insurer. This is a stochastic process

which increases continuously because of the earned premiums, and decreases stepwise because

of the payment of claims. When the capital becomes negative, we say that ruin has occurred.

The risk process

A stochastic process consists of related random variables, indexed by the time t. We define the

surplus process or risk process as follows:

U(t) = u+ ct S(t), t 0

where

U(t) = the insurers capital at time t

u = U(0) = the initial capital

c = the (constant)premium income per unit of time

S(t) = X1 +X2 + ...+XN(t), with

N(t) = the number of claims up to time t and

Xi = the size of the ith claim, assumed to be non negative.

The figure below shows a typical realization of risk process

The random variables T1, T2, ...denote the time points at which a claim occurs. The slope of

the process is c if there are no claims; if however, t = Tj for some j, then the capital drops by Xj,

which is the size of the jth claim. Since in the figure above, at time T4 the total of the incurred

claims X1 + X2 + X3 + X4 is larger than the initial capital u plus the earned premium CT4, the

remaining surplus U(T4) is less than 0. This state of the process is called ruin and the point in

time at which this occurs for the first time is denoted by T. So

46


T = min{t|t 0 &U(t) < 0}= if U(t) 0 for all t

The probability that ruin ever occurs, i.e, the probability that T is finite, is called the ruin

probability. It is written as follows

(u) = Pr[T


Poisson Process

Before we turn to the claim process S(t), i.e. the total claims up to time t, we first look at the

process N(t) of the number of claims up to t. We assume that N(t) is a Poisson process.

The process N(t) is a poison process if for some intensity > 0, the increments of the process

have the following property

N(t+ h)N(t) Poisson(h)

for all t > 0, h > 0 and each history N(s), s t

As a result, a Poisson process has the following properties:

The increments are independent: If the intervals (ti, ti + h), i=1,2,... are disjoint, then the

increments i.e. N(ti + hi)N(ti) are independent

The increments are stationary: N(t + h)N(t) is Poisson(h) distributed for every value oft.

Next to this global definition of the claim number process, we can also consider infinitesimal

increments N(t + dt) N(t), where the infinitesimal number" dt again is positive, but smallerthan any real number larger than 0. For the Poisson process we have

Pr[N(t+ dt)N(t) = 1|N(s); 0 s t] = edtdt = dt

Pr[N(t+ dt)N(t) = 0|N(s); 0 s t] = edtdt = 1 dt

Pr[N(t+ dt)N(t) 2|N(s); 0 s t] = edtdt = 0

these equalities are only valid if we ignore terms of order (dt)2

Lumdbergs exponential bound theorem for the ruin probability

For a compound Poisson risk process with an initial capital u, a premium per unit of time c,

claims with cdf P(.) and mgf Mx(t), and an adjustment coefficient R, we have the following

inequality for the ruin probability (u) eRu 0.

Assignment

Proof the above theorem

48


How to determine adjustment coefficient (R)

CASE I: Given Claim amount and Interarrival time random variables

The adjustment coefficient is defined as the smallest strictly positive solution (if it exists) of

the Lundberg equation

h(t) = E[etXtcW

]= 1

Where X is the claim random variable, W is the interarrival time random variable, c is the

constant premium rate that satisfies the condition E[x cW ] < 0. If X and W are independent,as in the most common models, then the equation can be re-written as;

h(t) = MX(t)MW (tc) = 1

Question

If W and X are independent and W exp(2), X exp(1) and premium rate is c = 2.4. Findadjustment coefficient.

CASE II: A Discrete time model

For a discrete time model the surplus process is given as

Un = u+ nc Sn

where;

Un denotes the insurers surplus at time n, where n = 0, 1, 2, 3, ....

denotes initial surplus

c denotes the constant premium received per unit time/period

Sn denotes the aggregate claims of the n periods

Note :

Un is viewed by examining amount of surplus on a periodic basis. Since insurance managers

submit financial report on a yearly, semi annual, quarterly or monthly basis.

Assume, Sn = X1 +X2 + ...+Xn. Where Xi is the sum of the claims in period i. X1, X2, ... are

49


independent, identically distributed random variables with = E[Xi] < c.

Un can also be written as; Un = + (cX1) + (cX2) + ...+ (C Xn) and we define adjustmentcoefficient as the positive solution of the equation

MXc(r) = E[er(Xc)

]= ercMX(r) = 1 or logMX(r) = rc

Where X denotes a random variable with the distribution of the annual claims.

Example

Derive an expression for R in the special case where the X is have a common distribution N(, 2).

Solution

log [MX(r)] = r +2r2

2 but logMX(r) = rc

= R = 2(c)2

where < c.

Generally

Since for a random variable X

ddt logMX(t)|t=0 = E[X] and d

2

dt2logMX(t)|t=0 = V ar[X]

and using Maclaurin series expression

logMX(r) = r +12

2r2 + ...

where, 2 = V ar(X)

=Then R ' 2(c)2

Note:

If X has a compound distribution and the relative security loading is given by c = (1 + ) then

R = 2{[+]}2

= 22

Let, pk = E[X]k = E[Loss]k

then

50


R = 2p1E[N ](p2p21)E[N ]+p21V ar(N)

Where N is a random variable distributed as the number of claims in a period.

Example

Approximate R if;

a) N has a poison distributed with parameter .

b) N has a negative binomial distributed with parameters r and p.

Solution

a) E[N ] = V ar(N) =

Therefore, R = 2p1p2b) E[N ] = rqp , V ar(N) =

rqp2

Therefore, R = 2p1p2+p

21

[(1p

)1]

Note; as p 1, R = 2p1p2CASE III: A Continuous time model

For a continuous time series, the surplus process is defined as;

U(t) = U + c(t) S(t)

and,

considering a period of length t > 0 where the a mount of premiums collected is ct and the

claim distribution is compound poison with expected number of claims t Then R is the smallest

positive root of

MS(t)ct(r) = E

[er(S(t)ct)

]= erctMS(t)(r) = erctet[MX(r)1] = 1

Thus

[MX(r) 1] = rc

51


Substituting, c = (1 + )p1

We get, 1 + (1 + )p1r = MX(r). Which is a linear of r.

Note: 1 + (1 + )p1r = MX(r) has two solutions, A side from the trivial solution r = 0, there is

a positive solution r = R which is defined as adjustment coefficient.

Example

Determine the adjustment coefficient is the claim distribution is exponential with parameter

> 0.

Solution

The adjustment coefficient is obtained from 1 + (1 + )p1r = MX(r)

as 1 + (1+)r =r

= (1 + )r2 r = 0

as r = 0 is a solution and R is smallest positive root

= R = 1+

Question

Calculate the adjustment coefficient if all claims are of size 1.

Approximation of Ruin probability in the case of compound Poisson aggregate

claims process

Here we give an expression for the ruin probability which involves the mgf of U(T) i.e the capital

at the moment of ruin, conditionally given the event that ruin occurs in a finite time period.

This expression enables us to give an exact expression for the ruin probability in case of an

exponential distribution. i.e the ruin probability for u 0 satisfies

(u) =eRu

E[eRU(T )|T


Proof:

Let R > 0 and t > 0. Then

E[eRU(t)] = E[eRU(t)|T t]Pr[T t] + E[eRU(t)|T > t]Pr[T > t] (19)

The adjustment coefficient R has the property that E[eRU(t)] is constant in t. i.e. eRU(t) is a

martingale: and since U(t) = u + ct S(t) and S(t) Compound Poisson with parameter t, wehave,

E[eRU(t)] = E[eR{u+ctS(t)}]

= eRu[eRcexp{(Mx(R) 1)}]t

= eRu

(20)

From equation 17 the left-hand side equals eRu. For the first conditional expectation in eqn 17

we take v [0, t] and write, using U(t) = U(v) + c(t v) [S(t) S(v)] see also equation 18

E[eRU(t)|T = v] = E[eR{U(v)+c(tv)[S(t)S(v)]}|T = v]= eRU(v)|T=veRcE[eR{S(t)S(v)}|T = v]= eRU(v)|T=v

{eRcexp[(Mx(R) 1)]

}tv= E

[eRU(T )|T = v

](21)

The total claims S(t) S(v) between v and t has again a compound Poisson distribution. Whathappens after v is independent of what happened before v, so U(v) and S(t)S(v) are independent.The term in curly brackets equals 1. Equality in eqn 19 holds for all v t, so E [eRU(t)|T v] =E[eRU(T )|T v] also holds.Since Pr[T t] Pr[T t according to the size of U(t). More precisely,we consider the cases U(t) uo(t) and U(t) > uo(t) for some function uo. Notice that T > t impliesthat we are not in ruin at time t, i.e. U(t) 0 so eRU(t) 1. We have

E[eRU(t)|T > t]Pr[T > t] = E[eRU(t)|T > t & 0 U(t) uo(t)]Pr[T > t & 0 U(t) uo(t)]+ E[eRU(t)|T > t & U(t) > uo(t)]Pr[T > t & 0 U(t) > uo(t)] Pr[U(t) uo(t)] + E[exp(Ruo(t))]= 0

53


The second term vanishes if uo . For the first term in eqn 17, note that U(t) has an expectedvalue U(t) = u+ ct tu1 and a variance 2(t) = tu2.

Question

Calculate the probability of ruin in the case that the claim amount distribution is exponential

with parameter > 0.

Theorem: Distribution of the capital at time of ruin

If the initial capital u equals 0, then for all capital y > 0 we have

Pr[U(T ) (y dy,y) & T u+ y. Defining

G(u, y) = Pr[U(T ) (,y) T


integrating this over u [0, z] yields

G(z, y)G(0, y) = c

{ z0G(u, y)du

z0

u0G(u x, y)dP (x)du

z0

u+y

dP (x)du

}.................................( )

The double integrals in (***) can be reduced to single integrals as follows. For the first double in-

tegral, exchange the order of integration, substitute v = ux and again exchange the integrationorder. This leads to

z0

u0G(u x, y)dP (x)du =

z0

zv0

G(v, y)dP (x)dv =

z0G(v, y)P (z v)dv

In the second double integral in (***) we substitute v = u+ y. Then

z0

v+y

dP (x)dv =

z+yy

[1 P (v)]

Hence

G(z, y)G(0, y) = c

{ z+y0

[1 P (v)]dv z+yy

[1 P (u)]du}.............................................................( )

for z , the first term on both sides of (****) vanishes, leaving

G(0, y) =

c

y

[1 P (u)]du

which completes the proof.

Approximation of ruin probability when initial capital is zero

The ruin probability at 0 depends on the safety loading only. Integrating eqn 20 for y (0,)yields Pr[T


Obviously, proportional reinsurance can be considered as a reinsurance on the usual adjust-

ment coefficient Rh, which is the root of

+ (c ch)r =

0er[xh(x)]dP (x) (23)

where ch denotes the re-insurance premium. The reinsurer uses a loading factor on the net

premium.

Assume that = 1, and p(x) = 0.5 for x = 1 and x = 2. Furthermore, let c = 2, so that = 13 ,

and consider two values = 13 and =25 . \par In case of proportional reinsurance h(x) = x, the

premium equals

ch = (1 + )E[h(x)] = (1 + )3

2

so, because of x h(x) = (1 )x eqn 21 leads to the equation

1 +

[2 (1 + )3

2

]r =

1

2er(1) +

1

2e2r(1)

For = 13 , we have ch = 2 and Rh =0.3251 , for =

25 , we have ch = 2.1.

Next, we consider the excess of loss reinsurance h(x) = (x )+ with 0 2. The reinsur-ance premium equals

ch = (1 + )

[1

2h(1) +

1

2h(2)

]=

1

2(1 + ) [(1 )+ + (2 )+]

while x h(x) = min{x, }, and therefore Rh is the root of

1 +

(2 1

2(1 + ) [(1 )+ + (2 )+]

)r =

1

2

[emin{,1}r + emin{,2}r

]

NOTE:

Assuming that other factors remain constant, the following reduces the probability of ruin.

1. Increase of C

2. Increase of initial surplus

3. Increase of re-insurance

The following increases the probability of ruin

1. Increase of variance of the individual claim size56


2. Increase of skewness of the individual claim size

57


CHAPTER SIX

RISK AND MODELLING

The uncertainty regarding future cash flows is of three types ie Amount, occurrence and timing.

The uncertainty regarding the amount of cash flow is witnessed when for instance, a business-

man buys shares and plans to sell them within the year. The amount realized from selling the

shares would be determined by the market price of the shares within the year which is not

known at the moment. Similarly, when a general insurance company provides comprehensive

motor insurance to a motorist, they dont know how much the claims arising from the policy over

the following year will be. In fact they dont know whether there will be any claims arising from

the policy which brings us to the notion of uncertainty arising from the occurrence of events.

The final dimension of uncertainty regards the timing of cash flows. In a whole life policy, policy

holders pay regular premiums until death, and on death, a nominated beneficiary is paid a fixed

sum assured. Though the amount is known in advance, the timing of crucial event (death) is

not known and this is a source of uncertainty.

Financial Engineers solve financial problems by modeling cash flows, projecting these cash

flows and then using the cash flows to obtain a solution. In finance, financial economists use the

standard deviation of a cash flow as an indicator of its riskiness. However, standard deviation

does not take into account the special circumstances of the investor. Investors have different

attitudes towards risk and different objectives.

Financial risk takes into account these different objectives being defined as the risk that the

investors objectives may be met. It is actually expressed as a probability.

Statistical models

From the fact that cash flows have three dimensions ie time, occurrence and the amount, we

describe some of the models used to represent these three dimensions with particular emphasis

on applications in general insurance. We will first consider models used to represent timing and

occurrence of events associated with cash flows, proceed to models representing the amount of

cash flows and then aggregate the amount arising from several events. Aggregation will be the

total claim amount paid out under a single policy or a portfolio of policies over a specified period

of time eg 1 year. Modeling the aggregate claims will guide us on the correct premium to charge.

It will also enable us to decide what reinsurance arrangements will be best for our needs.

58


ESTIMATION OF MODEL PARAMETERS

We have various methods of estimating model parameters. They include:

1. Maximum Likelihood

2. Least square method

3. Method of moments

4. Bayes estimation

Maximum likelihood method

The likelihood of a sample is the product of the probabilities of drawing the individual observa-

tions from the population. In this method, we look for the value of the population parameter that

maximizes the likelihood function by differentiating the function with respect to the population

parameter of interest and equating it to 0 and then solving the equation.

Example

An insurance company has an excess of loss reinsurance contract with retention of 50,000 US

dollars. Over the last year, the insurer paid the following claims in dollars 12372, 2621,389 and

43299. In addition, the insurer paid the amount of 50,000 dollars on 6 claims with the excess

being paid by the re-insurer. The insurer believes that distribution of gross claim amounts is

exponential with mean . Calculate the maximum likelihood estimate of based on the above

information.

Solution

Let X be the random variable for the gross claim amounts. The likelihood function is

L =

ni=1

f(xi, )

but

f(x, ) =1

ex , x > 0

ni=1

1

exi = L

59


The chance that the 6 claims exceed the retention amount is

P (X > 50000) =

m

1

ex dx = e

m

where m is the retention amount of 50000

L =4i=1

1

ex

(em

)6

=1

4e xi

(em

)6A general rule is to take logarithms and then maximize log L

logL = 4logxi 6m

dLd

=4

+

xi

2+

6m

2= 0

4

=

xi

2+

6m

2

4 =

xi + 6m

=

xi + 6m

4=

58681 + 300000

4= 89670.25

Assignment Question

Consider a group of n males, each aged 30 years living in a particular society. Their lives may

be assumed to be independent. Suppose that x of the n men die by the end of a subsequent

period of duration to years and that n-x survive the period. Suppose that the lifetime of these

men from age 30 is to be modeled as a random variable with an exponential distribution with

mean 1 years.

i) State the distribution of the random variable X whose value is x, the number of men who

die within to years has been observed. Hence determine , the M.L.E of .

ii) Consider the case n = 1000, to = 20 years and x = 320. Evaluate and show that the value

of its standard error is approximately 0.00108.

iii) Calculate an approximate 95% confidence interval for the mean life time for age 30 of such

men

60


Bayesian Estimation

Loss function

Let d be an estimate of a parameter, say . The loss function, defined by l(, d) is defined to be

a real valued function (non-negative) which measures the loss in taking d as the decision on the

value of . The risk function of d is given by R(, d) = E[l(, d)]. The best estimator is the one with

minimum risk. We have various types of loss functions.

(i) Squared error loss function (Quadratic loss function):-

It is given by l(, d) = (d )2

Note that this loss function is minimized when d is the mean of f(|x) the posterior distri-bution

(ii) All or nothing loss function (0/1 loss function):-

Here the loss function takes the values 1 when d < < < d + and 0 otherwise. is a

small number that may be allowed to tend to 0. The loss function is minimized when d is

the mode of f(|x) the posterior distribution

(iii) Absolute error loss function:-

It is given by l(, d) = |d |The loss function is minimized when d is the median of f(|x) the posterior distribution

Example

A risk consists of 5 policies. On each policy in 1 month, theres exactly 1 claim with probability

and theres a negligible probability of more than 1 claim in 1 month. The prior distribution of

is uniform on 0 and 1. There are a total of 10 claims on this risk over a 12 month period.

i) Derive the posterior distribution

ii) Determine the Bayesian estimate of under quadratic loss and all or nothing loss functions.

Solution

Let Xi be the number of claims in month i, so that xi has a binomial distribution with parameters

5 and , i=1,2,3,...,12. The posterior density is given by

f(|x) f(x|)h()61


12i=1 xi (1 )60

12i=1 xi = 10(1 )50

The posterior distribution is a beta distribution with parameter = 11 and = 51.

Therefore under quadratic loss function,

E[X] =

+ =

11

11 + 51= 0.1774

Under all or nothing loss function, the Bayesian estimate is the posterior mode. To get the mode,

we differentiate the posterior density with respect to the parameter of interest, in this case,

equate the results to 0 and solve for for ie

109(1 )50 5010(1 )49 = 0

109(1 )50 = 5010(1 )49

109(1 ) = 5010 = 16

= 0.167

62


CHAPTER SEVEN

NATURE OF INVESTMENT DECISION

Introduction

The fundamental problem in economics is to choose what to produce and how to produce it.

This translates for financial engineers into choosing between the set of cash flows. An important

aspect of this regards the criteria to be applied in making such choices. There are two basic

types of commercial and economic values. \par The first is investment in the economic sense of

the term and involves the actual creation or the significant modification of assets. The result is

a capital project and techniques have been developed for appraising such projects.

The second type involves purchasing already existing assets and thereafter being entitled to

the proceeds from the assets. The objective in the latter case will be defined as maximizing

returns subject to an acceptable degree of risk. The former could include the financial manage-

ment of companies eg the general insurance firms etc.

NOTE:

An efficient allocation of capital is a crucial finance function in modern times. It involves deci-

sions to commit firms funds to the long-terms assets. These decisions are important since they

determine the firms value size by influencing its growth, profitability and risks.

The investment decisions of a firm can also be known as capital budgeting or capital expen-

diture decisions. A capital budgeting decision of a firm can also be defined as the firms decision

to invest its current funds most efficiently in the long term assets in anticipation of an expected

flow of benefits over a series of years.

The firms investment decisions may include expansion, acquisition, modernization and re-

placement of long-term assets and also sale of a division or business (divestment). Other activi-

ties include change in the methods of sales distribution advertisement campaigns, research and

development programs.

Features of investment decisions:

1. Exchange of current funds for future benefits

2. The funds are invested in long-term assets

3. The future benefits will occur in the firm over a series of years.63


It is emphasized that the expenditures and benefits of an investment should be measured

in cash. In investment analysis, it is cash flow which is important and not the accounting

profits. Investment decisions also affect the firms value. Firms value increases if investments

are profitable and add to the shareholders wealth. This means that investments should be

evaluated on the basis of a criterion with the objective of shareholders wealth maximization.

An investment will add to the shareholders wealth if it yields benefits in excess of the min-

imum benefits as per the opportunity cost of the capital. In this topic we make the following

assumptions.

1. The investments project opportunity cost of capital is known

2. The expenditure and benefits of the investment are known with certainty

Importance of investment decisions:

Investment decisions require special attention because of the following reasons:-

1. They influence the firms growth in the long-run: Investment decision effects extend into

the future and have to be endured for longer periods than the consequences of current

operating expenditures. Unwanted or unprofitable expansion of assets results in heavy op-

erating costs. Inadequate investment of assets will make it difficult for the firm to compete

successfully and maintain its market shares.

2. They affect the risk of the firm: A long term commitment of funds may change the risk

complexity of the firm. If the adoption of an investment increases average gain but causes

frequent changes in its earnings, the firm will become more risky.

3. They involve commitment of large amounts of funds: Investment decisions involve large

amounts of funds which makes it essential for the firm to plan its investment programs very

carefully and make an advance arrangement for procuring/obtaining finances internally

and externally.

4. They are irreversible or reversible at substantial loss: Most investment decisions are ir-

reversible. Once such capital items have been acquired, its difficult to find a market for

them. The firm incurers heavy losses if such assets are scrapped.

5. They are among the most difficult decisions to make: Investment decisions are an as-

sessment of future events which are difficult to predict. It is very complex to correctly

estimate the future cash flows of an investment. The uncertainty in cash flows is caused

by economic, social, political and technological forces.

64


Investment evaluation criteria (Appraisal techniques)

The following are steps involved in evaluation/appraised of an investment:

1. Estimation of cash flows

2. Estimation of required rate of return

3. Application of decision rule for making the choice.

In this topic we assume the the first two and concentrate on the third step. In specific

we focus on merits and demerits of various decision rules. Investment decision rules may be

referred to as capital budgeting techniques or investment criteria. A good appraised technique

should be used to measure the economic worth of an investment project. A good investment

evaluation criteria should possess the following characteristics:-

1. It should maximize the shareholders wealth

2. It should consider all cash flows to determine the true profitability of the project.

3. It should be in a position to separate good projects from bad projects.

4. It should be in a position to rank projects according to their true profitability.

5. It should recognize the fact that bigger cash flows are preferred to smaller cash flows, and

early cash flows are preferred to later cash flows.

6. It should help to choose more mutually exclusive projects, the projects which maximize the

shareholders wealth.

7. It should be applicable to any conceivable investment project independent of others.

We have various investment criteria for appraising the worth of an investment project. We

group them into two categories:-

1. Discounted Cash Flow (DCF) criteria

- Net Present Value (NPV)

-Internal Rate of Return (IRR)

-Profitability Index (PI)

-Discounted Payback Period

2. Non-discounted cash flow criteria

-Payback Period

-Accounting Rate of Return (ARR)65


Discounted cash flow criterion (DCF)

Net Present Value (NPV) method

It is the most valid technique of evaluating an investment project. It is generally consistent with

the objective of maximizing the shareholders wealth. It also recognizes the time value of money.

Steps involved in calculating of NPV

1. Forecast the cash flows of the investment project on realistic assumptions.

2. Identify an appropriate discount rate to discount the forecasted cash flows.

3. Calculate the present value of cash flows using the opportunity cost of the capital as the

discount rate

4. Net present value is computed by subtracting present value of cash out flows from the

present value of cash in flows.

The Net Present value is accepted if the NPV > 0

Consider a project that has the following cash inflows over its useful economic life. The

project

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