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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 sensitivity 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 General 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. 1
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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: 22108505. 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: 17936322. 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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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Co-Requisites: 1) STA 2395 Decision Theory and Bayesian Inference I 2) STA 2301 Tests of Hypotheses
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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
CHAPTER ONE RISK THEORY INTRODUCTION Definition Risk theory is essentially a branch of probability theory, devoted to decision-making under probabilistic uncertainty. Basic concepts of the theory are: risk, risk measure, risk price, and individual 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 policy’s 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 mathematical 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 insurance companies take over risks from customers. Insurers consider every available quantifiable 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
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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 assessment 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 revealedpreference 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 abbreviation 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 distribution 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 estimation. 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 magnitudes 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 es5
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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 lossdistribution 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 as viewed by individuals external to the measuring organization. =⇒ The highier the false accept risk, the greater the chances for returned goods, loss of reputation, litigation, and other undesirable outcomes. =⇒ In a commercial context, individuals external to a measuring organization are labeled “consumers”. For this reason, false accept risk has traditionally been called consumer’s 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 by individuals within to the measuring organization. =⇒ The highier the false reject risk, the greater the chances for unnecessary re-work and re-test. =⇒ In a commercial context, individuals within to a measuring organization are labeled “producer’s”. For this reason, false reject risk has traditionally been called producer’s risk. 6
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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 specific 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) P remium = p = E[X] = pure premium (ii) P remium = p = (1 + θ)E[X], is the pure premium with safety (security) loading , where θ ≥ 0. NB: θ = relative safety loading and θE[X] =total safety loading (iii) P remium = E[X] + aV ar(X) is the premium with variance loading principle. p (iv) P remium = E[X] + a V 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.
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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Utility theory and Insurance Under economics of insurance we examine the policyholder’s 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 µ(ω) to his wealth ω instead of just ω, where µ(.) is called his utility function. If a decision maker has to choose between random losses X and Y , then he compares E[µ(ω − X)] with E[µ(ω − Y )] and chooses the loss with the highest expected utility. With this model, the insured with wealth ω 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[µ(ω − X)] = µ(ω − P + ). At the equilibrium he doesn’t care, in terms of utility, whether he is insured or not. The insurer, with his own utility function and perhaps supplementary expenses will determine a minimum premium P − . If the insured’s maximum premium P + is larger than the insurer’s 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: µ(ω − p+ ) = pµ(ω) + (1 − p)µ(ω − 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 9
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
µ(ω − G) = pµ(ω) + (1 − p)µ(ω − 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 ω at probability p. We let µ(ω) = 0 and µ(0) = −1 be the starting values yielding the points (40000, 0) and (0, −1) when X = 40, 000 and say G = 28, 000 =⇒ µ(40, 000 − 28, 000) = 0.4µ(40, 000) + 0.6µ(0) = −0.6 i.e. µ(12, 000) = −0.6 =⇒ (12000, −0.6) is a third point on the utility function when X = 28, 000 and say G = 20, 000 =⇒ µ(40, 000 − 20, 000) = 0.4µ(40, 000) + 0.6µ(40000 − 28000) =⇒ µ(20, 000) = 0.4µ(40, 000) + 0.6µ(12000) = 0.4 × 0 + (−0.6) × 0.6 = −0.36 i.e. µ(20, 000) = −0.36 =⇒ (20000, −0.36) is a fourth point on the utility function when X = 20, 000 and say G = 12, 000 =⇒ µ(40, 000 − 12, 000) = 0.4µ(40, 000) + 0.6µ(40000 − 20000) =⇒ µ(28, 000) = 0.4µ(40, 000) + 0.6µ(20000) = 0.4 × 0 + (−0.36) × 0.6 = −0.216 i.e. µ(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
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Properties of µ(ω) 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 consistently between potential random losses X, then their exists a utility function µ(.) to appraise the decisions he makes are exactly the same as those resulting from comparing the losses X based on the expectation E[µ(ω − X)]. For comparison of X with Y , the utility function µ(x) and its linear transform aµ(x) + b for some a > 0 are equivalent, since they result in the same decision: E[µ(ω − X)] ≤ E[µ(ω − Y )] if and only if E[aµ(ω − X) + b] ≤ E[aµ(ω − Y ) + b] so from each class of equivalent utility function, 0
we can select one, for instance by requiring that µ(0) = 0 and µ(1) = 1. Assuming that µ (0) > 0, we could also use the utility function V (.) with V (0) = 0 and V 0 (0) = 0 such that V (x) =
µ(x) − µ(0) µ0 (x)
Theorem (Jensen’s inequality) If V (x) is a convex function and Y is a random variable then E[V (Y ) ≥ V (E[Y ])] with equality if and 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[µ(ω − X)] ≤ µ[E[(ω − X)] = µ(ω − E(x)) So this particular decision maker is rightly called risk a verse: he prefers to pay a risky amount X. Now, suppose that a risk averse insured with capital µ used the utility function µ(.). Assum11
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
ing he is insured against a loss X for a premium P , his expected utility will increase if E[µ(ω − X)] ≤ µE[(ω − X)] = µ(ω − E[X]) = µ(ω − P )
(1)
Since µ(.) is a non-decreasing continuous function, this is equivalent to P ≤ P + . It is the solution to the following utility equilibrium equation. E[µ(ω − X)] = µ(ω − P + )
(2)
Note: Jensen’s inequalities states that for a random variable X and function µ(ω). 00
(i) if µ (ω) < 0, then E[µ(X)] ≤ µ(E[X]) 00
(ii) if µ (ω) > 0, then E[µ(X)] ≥ µ(E[X])
Proof for case (i) µ(ω) in case (i) is a concave function with a maximum stationary point as shown below
The equation of a tangent function to the function µ(ω) at the point (µ, µ(µ)), 0
where µ = E[X] = ω is given as µ(ω) = µ(µ) + µ (µ)(ω − µ). 0
For all values of ω > 0, µ(ω) ≤ µ(µ)+µ (µ)(ω −µ) and replacing ω with X and taking expectation 12
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both sides we get; E[µ(X)] ≤ µ(E[X]). This first Jensen’s inequality describes the attitute of a risk averse individual, who prefers fixed loss over random losses.
Proof for case (ii) µ(ω) in case (ii) is a convex function with a minimum stationary point as shown below
The equation of a tangent function to the function µ(ω) at the point (µ, µ(µ)), 0
where µ = E[X] = ω is given as µ(ω) = µ(µ) + µ (µ)(ω − µ). 0
For all values of ω > 0, µ(ω) ≥ µ(µ)+µ (µ)(ω −µ) and replacing ω with X and taking expectation both sides we get; E[µ(X)] ≥ µ(E[X]). This second Jensen’s 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 µ(x), how can we approximate the maximum premium P + for a risk X?
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Solution Let u and σ 2 denote the mean and variance of X. Using the first terms in the series expansion of µ(.) in ω − µ we obtain
0
µ(ω − P + ) ≈ µ(ω − u) + (u − P + )µ (ω − u); 1 0 00 µ(ω − X) ≈ +µ(ω − u) + (u − X)µ (ω − u) + (u − X)2 µ (ω − u) 2
(3)
Taking expectations on both sides of the latter approximation yields
1 00 E[µ(ω − X)] ≈ µ(ω − u) + σ 2 µ (ω − u) 2
(4)
substituting 2 into 4, it follows from 3 that 1 00 0 µ (ω − u) ≈ (u − P + )µ (ω − u) 2 Therefore, the maximum premium P + for a risk x is approximately 00
1 µ (ω − u) P + ≈ u − σ2 0 2 µ (ω − u) This suggest the following definition: (absolute) risk aversion coefficient r(ω) of the utility function µ(.) of a wealth ω is given by
00
r(ω) =
−µ (ω) µ0 (ω)
Then the maximum premium P + to be paid for a risk X is approximately 1 P + ≈ u + r(ω − µ)σ 2 2
Exponential Premium The insurer with utility function U (.) and capital W, will insure the loss X for a premium P if E[µ(W + P − X) ≥ U (W )], hence P ≥ P − where P − is the minimum premium to be asked. This premium follows from solving the utility equilibrium equation reflecting the insurer’s position: U (W ) = E[U (W + P − − X)]
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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Question Suppose that an insurer has an exponential utility function with parameter α. 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) = −αe−αx yields P− =
1 log(mx (α)) α
(5)
Where mx (α) = E[eαx ] is the mgf of X at argument α. We observe that this exponential premium is independent of the insurer’s 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 α represents the risk aversion of the insured. Assume that the loss X is exponentially distributed with parameter β. Taking β = 0.01 yields E[X] =
1 β
= 100. If the insured’s utility function is
exponential with parameter α = 0.005, then 1 log(mx (α)) α β = 200log β−α = 200log(2)
P+ =
≈ 138.6 so the insured is willing to accept a sizable loading on the net premium E[X].
Quadratic utility Suppose that for ω < 5, the insured’s utility function is µ(ω) = 10ω − ω 2 . What is the maximum premium P + as a function of ω, ω ∈ [0, 5] for an insurance policy against a loss 1 with probability 1 2?
What happens to this premium is ω increases. Let E[X] =
1 2
and var(x) =
1 4
Solution We solve the equilibrium equation E[µ(ω − X)] = µ(ω − P + ) 15
(6)
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The expected utility after a loss X equals E[µ(ω − X)] = 11ω −
11 − ω2 2
(7)
and the utility after paying a premium P equals µ(ω − P ) = 10(ω − P ) − (ω − P )2
(8)
By the equilibrium equation 6, the right hand side of 7 and 8 should be equal, and after calculations r P = P (ω) =
(
11 1 − ω)2 + − (5 − ω), ω ∈ [0, 5) 2 4
Exercise A decision maker with an exponential utility function with risk aversion α > 0 wants to insure a gamma(n,1) distributed risk. Determine P + and prove that P + > n. When is P + = ∞ and what does that mean?
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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 can’t 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
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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 = P r[I = 1], 0 ≤ q ≤ 1. If I = 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 P r[X ≤ x|I = 0] = P r[X ≤ x|I = 1] just as well. Hence, the cdf of Z can be written as. F (s) = P r[S ≤ s] = P r[S ≤ s&I = 1] + P r[S ≤ s&I = 0] = P r[X ≤ s&I = 1] + P r[Y ≤ s&I = 0] = qP r[X ≤ s] + (1 − q)P r[Y ≤ s]
(10)
Now, let X be a discrete random variable and Y a continuous random variable. From (10), we 0
d get F (s) − f (s − 0) = qP r[X = s] and F (s) = (1 − q) ds P r[Y ≤ s] This construction yields a cdf F (s) 0
With steps where P r[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) = P r[S ≤ s] arises when a mixture of random variables S = IX + (1 − I)Y is used, where X is a discrete random variable, Y is a continuous random variable and I is a Bernoulli(q) random variable independent of X and Y . 18
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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 P r(I = 0) = 1 − q and P r(I = 1) = q, the mean and variance of I are q and q(1 − q), respectively, and the mean and variance of X are bq and b2 q(1 − q) . i.e. The p.f is 1−q fX (x) = P r(X = x) = q 0
f or x = 0 f or x = b elsewhere
and the d.f is
FX (x) = P r(X ≤ x) =
0
1−q 1
f or x < 0 f or 0 ≤ x < b
From the p.f and definitions of moments E[X] = bq
E[X 2 ] = b2 q
V ar(X) = b2 q(1 − q)
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f or x ≥ b
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CASE II The aggregate loss S = X, could also be several claims by an individual from a single policy.
Example
X
Accident cover due to
constant Amount paid
(
F ire
−→
b1
(
V ehicle
−→
b2
T errorism
−→
b3
(
N atural
−→
b4
(
Riot
−→
b5
−→ (
& B = Random variable %
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
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FX (x) = P r(X ≤ x) = P r(IB ≤ x) = P r(IB ≤ x|I = 0)P r(I = 0) + P r(IB ≤ x|I = 1)P r(I = 1)
Assignment Illustrate plots for pdf’s and cdf’s 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 1−q fX (x) = P r(X = x) = q 0
f or x = 0 f or x = b elsewhere
and the d.f is
FX (x) = P r(X ≤ x) =
0
1−q 1
f or x < 0 f or 0 ≤ x < b
From the p.f and definitions of moments E[X] = bq
E[X 2 ] = b2 q
V ar(X) = b2 q(1 − q) 21
f or x ≥ b
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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 P r(I = 0) = 1 − q and P r(I = 1) = q, the mean and variance of I are q and q(1 − q), respectively, and the mean and variance of X are bq and b2 q(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 P r(I = 0) = 0.85
P r(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 B 0 s distribution has a probability mass at the maximum claim size of 2000. Assume this probability mass is 0.1. Furthermore assume that
P r(B ≤ x|I = 1) =
x≤0
0 h
0.9 1 − 1 −
1
Note: The distribution Function for B, given I = 1 is 22
x 2000
2 i
0 < x < 2000 x ≥ 2000
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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) = P r(X ≤ x) = P r(IB ≤ x) = P r(IB ≤ x|I = 0)P r(I = 0) + P r(IB ≤ x|I = 1)P r(I = 1)
For x < 0, FX (x) = 0(0.85) + 0(0.15) = 0 h For 0 ≤ x < 2000, FX (x) = 1(0.85) + 0.9 1 − 1 −
2 x 2000
i
(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 P r(X = 0) = 0.85
P r(X = 2000) = 0.015 with p.d.f 23
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
0
fX (x) = FX (x) =
0.000135 1 −
x 2000
0
0 < x < 2000 elsewhere
Moments of X can be calculated by ˆ K
K
2000
xk fX (x)dx
E[X ] = 0 × P r(X = 0) + (2000) × P r(X = 2000) + 0
specifically, E[X] = 120
E[X 2 ] = 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
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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, µ = E[X] and σ 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] = nµ 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] = nσ 2
NOTE: Variance of aggregate claims=Number of insuring units times variance of claim severity.
St
M gf of S = MS (t) = E e
h
t[X1 +X2 +...+Xn ]
=E e
i
n Y tX n n = E e = {MX (t)} = MXi (t) i=1
Determination of probability distribution for S = X1 + X2 + ... + Xn using Central Limit Theorem Given that Xi0 s are independent random variables and n ≥ 30, then using normal approximation
S = X1 + X2 + ... + Xn =
n X
Xi ∼ N (E[S], V ar (S)) = N nµ, nσ 2
i=1
and
S − nµ S − E[S] p √ = ∼ N (0, 1) σ n V ar(S) also 25
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
n
1 X1 + X2 + ... + Xn 1X S= = Xi = S ∼ N n n n i=1
V ar(S) nµ n2 σ 2 σ2 E[S], =N = N µ, , n n n n
and
S − E[S] S−µ q = σ ∼ N (0, 1) √ n V ar(S)
Question Give an illustration that requires application of this theorem.
Determination of probability distribution for S = X1 + X2 + ... + Xn using Moment generating functions Since,
n i n h Y = {MX (t)}n = MXi (t) M gf of S = MS (t) = E eSt = E et[X1 +X2 +...+Xn ] = E etX i=1
Due to uniqueness theorem of moment generating functions. Which says that, mgf’s 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 product
Qn
i=1 MXi (t)
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; t −1
(a) If X ∼ poison(λ) then, MX (t) = eλ(e
) , E[X] = λ, and V ar(X) = λ
n (b) If X ∼ binomial(n, p) then, MX (t) = pet + (1 − p) , E[X] = np, and V ar(X) = npq. Where q = 1 − p. With a special case: 26
. During
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
When n = 1, then X ∼ Bernoulli(p). α −α θ (c) If X ∼ gamma(α, θ) then, MX (t) = θ−t = 1 − θt , E[X] = αθ , and V ar(X) =
α . θ2
With special cases: (i) When α = 1, then X ∼ exponential(θ). V 2
and θ = 12 , then X ∼ χ2V . r p then, E[X] = (d) If X ∼ N B(r, p) then, MX (t) = 1−(1−p)e t (ii) When α =
rq p,
and V ar(X) =
rq . p2
Where q = 1 − p. With a special case: When r = 1, then X ∼ geometric(p) 1 2 2 (e) If X ∼ N µ, σ 2 then, MX (t) = E[etX ] = eµt+ 2 σ t , E[X] = µ, V ar(X) = σ 2 . (f) If X ∼ U nif orm(a, b) then, MX (t) =
etb −eta tb−ta ,
E[X] =
a+b 2 ,
and
(b−a)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
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Hence the d.f of S is FS (s) = P r(S ≤ s) = P r(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) = P r[X + Y ≤ s] ˆ ∞ = P r[X + Y ≤ s|X ≤ x]dFX (x) −∞ ˆ ∞ = P r[Y ≤ s − x|X = x]dFX (x) −∞ ˆ ∞ = P r[Y ≤ s − x]dFX (x) −∞ ˆ ∞ FY (s − x)dFX (x) = −∞
= FX ∗ FY (s)
(11)
The cdf FX ∗ FY (.) is called the convolution of the cdf’s FX (.) and FY (.). For the density function we use the same notation. If X and Y are discrete random variables, we find X
FX ∗ FY (s) =
FY (s − x)fX (x)
x
and X
fX ∗ fY (s) =
fY (s − x)fX (x)
x
Where the sum is taken over all x with fX (x) > 0. If X and Y are continuous random variables, then
ˆ
∞
FY (s − x)fX (x)dx
FX ∗ FY (s) = −∞
and taking the derivatives under the integral sign, yields ˆ
∞
fX ∗ fY (s) =
fY (s − x)fX (x)dx −∞
Note: 1. For three cdf’s 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 ∗n 28
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Example: (convolution of discrete distributions) Let f1 (x) = 14 , 21 , 14 f or x = 0, 1, 2; f2 (x) = 12 , 21 f or x = 0, 2 and f3 (x) = 14 , 12 , 14 f or 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)
0
1 2
0
2
1 4 1 2 1 4
3
0
0
4
0
5
=
f1+2 (x)*
f3 (x) 1 4
0
1 8 2 8 2 8 2 8 1 8
0
0
0
0
6
0
0
0
0
7
0
0
0
0
8
0
0
0
0
1
1 2
=
f1+2+3 (x)
0 1 2
0 1 4
1 32 2 32 4 32 6 32 6 32 6 32 4 32 2 32 1 32
F1+2+3 (x) 1 32 3 32 7 32 13 32 19 32 25 32 29 32 31 32 32 32
Example: (Convolution of two uniform distributions) Suppose that X ∼ U nif orm(0, 1) and Y ∼ U nif orm(0, 2) are independent. What is the cdf of X +Y?
Solution We introduce the concept “indicator function". The indicator function of a set A is defined as follows
IA (x) =
1
if x ∈ A
0
if x ∈ /A
Indicator functions provide us with a concise notation for functions that are defined differently on some intervals. For all x, the cdf of X can be written as FX (x) = xI[0,1) (x) + I[1,∞) (x) 0
While for FY (y) = 12 I[0,2) (y) for all y, which leads to the differential 1 dFY (y) = I[0,2) (y)dy 2 29
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
The convolution formula (11), applied to Y + X rather than X + Y , then yields ˆ
ˆ
∞
2
FX (s − y)dFY (y) =
FY +X (s) = −∞
0
1 FX (s − y) dy, s ≥ 0 2
The interval of interest is 0 ≤ s < 3. Substituting it into [0,1), [1,2) and [2,3) yields ˆ FX+Y (s) = 0
s
ˆ s−1 ˆ s 1 1 1 (s − y) dy I[0,1) (s) + dy + (s − y) dy I[1,2) (s) 2 2 2 0 s−1 ˆ
s−1
+ 0
1 dy + 2
ˆ
2
1 (s − y) dy I[2,3) (s) 2 s−1
1 1 1 = S 2 I[0,1) (s) + (2s − 1)I[1,2) (s) + [1 − (3 − s)2 ]I[2,3) (s) 4 4 4 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 √1 e− 2 ( σ 2π
x−µ 2 σ
)
−∞ < x < ∞, −∞ < µ < ∞, σ 2 > 0
0
1
⇒ MX (t) = E[etX ] = eµt+ 2 σ
elsewhere
2 t2
0
E[X] = mX (t)|t=0 = µ V ar(X) = σ 2
Gamma distribution A random variable X possesses a gamma distribution with α and β =
30
1 θ
if its distribution is:
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
f (x) =
−x
xα−1 e β Γαβ α
0
⇒ MX (t) = (1 − βt)−α or
1 1−βt
0 < x < ∞, α > 0, β > 0 otherwise
α
E[X] = αβ V ar(X) = αβ 2 OR
f (x) =
β α xα−1 e−xβ Γαβ α
0
0 < x < ∞, α > 0, β > 0 otherwise
With E[X] =
α β
V ar(X) =
α β2
Beta distribution
f (x) =
xα−1 (1−x)β−1 B(α,β)
0 < x < 1, α > 0, β > 0
0
otherwise
With E[X] =
α α+β
V ar(X) =
αβ (α+β)2 (α+β+1)
Exponential distribution It is a special case for poison distribution with α = 1 and β =
=⇒ f (x) =
λe−λx
0
With MX (t) =
λ λ−t
31
1 λ
0 0
0
otherwise
Where MX (t) = does not exist
ˆ E[X] = = = = = = = =
∞
xαθα dx (x + θ)α+1 0 ˆ ∞ x αθα dx (x + θ)α+1 0 ˆ ∞ x + (θ − θ) αθα dx (x + θ)α+1 0 ˆ ∞ x+θ θ α αθ − dx (x + θ)α+1 (x + θ)α+1 0 ∞ 1 1 1 θ αθα . + dx α − 1 (x + θ)α+1 α (x + θ)α+1 0 1 1 θ α αθ . + dx α − 1 (θ)α+1 (αθ)α 1 αθα 1 − θα−1 α − 1 α θ α−1
and V ar(X) =
αθ2 (α−1)2 (α−2)
33
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Example In a Bayesian investigation in the insurance industry, a particular claim rate θ is to be estimated. To obtain a suitable prior distribution, an expert is consulted and he suggest that θ 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 expert’s information. Find the parameters for the appropriate gamma distribution.
Solution For a gamma distribution with parameters α and β, E[X] = αβ and V ar(X) = αβ 2 ⇒ αβ = 0.15 ⇒ αβ 2 = (0.05)2 ⇒α=
0.15 β , β(0.15)
= (0.05)2 ⇒ β = 0.0166666 and α = 9
Question Let X ∼ P oisson(λ) and Y ∼ P oisson(µ) be independent random variables. If S = 0, 1, 2, ..., find the 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
P r(X1 = x)
P r(X2 = x)
P r(X3 = x)
P r(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
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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, µk = E[X k ], P r(x) = P r[X ≤ x] and F (s) = P r[S ≤ s]. We can then calculate the expected value of S by using the conditional distribution of S, given N. i.e.
35
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
E[S] = E[E[S|N ]] =
∞ X
E[X1 + ... + XN |N = n]P r(N = n)
n=0
= = =
∞ X n=0 ∞ X n=0 ∞ X
E[X1 + ... + Xn |N = n]P r(N = n) E[X1 + ... + Xn ]P r(N = n) nµ1 P r(N = n) = µ1 E[N ]
n=0
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[N V ar(X)] + V ar(N µ1 ) = E[N ]V ar(X) + µ21 V ar(N ) mgf of S is MS (t) = E E[etS |N ] ∞ h i X = E et(X1 +...+XN ) |N = n P r(N = n) = =
n=0 ∞ X n=0 ∞ X
i h E et(X1 +...+Xn ) P r(N = n) {MX (t)}n P r(N = n) = E
h i N elogMX (t)
n=0
(13)
= MN (logMX (t))
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, which means t < −logq, we have MX (t) =
∞ X
ent pq n =
n=0
36
p 1 − qet
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Since X ∼ exponential(1), i.e. MX (t) = (1 − t)−1 equation (13) yields MS (t) = MN (logMX (t)) =
p p =1+q 1 − qMX (t) p−t
So the mgf of S is a mixture of the mgf’s of the constant 0 and of the exponential(p) distribution. Because of one-to-one correspondence of cdf’s and mgf’s we may conclude that the cdf of S is the mixture F (x) = p + q(1 − e−px ) = 1 − qepx f or 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 event’s 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, which means t < −logq, we have MX (t) =
∞ X
ent pq n =
n=0
p 1 − qet
Since X ∼ exponential(1), i.e. MX (t) = (1 − t)−1 equation (13) yields MS (t) = MN (logMX (t)) =
p p =1+q 1 − qMX (t) p−t
So the mgf of S is a mixture of the mgf’s of the constant 0 and of the exponential(p) distribution. Because of one-to-one correspondence of cdf’s and mgf’s we may conclude that the cdf of S is the mixture F (x) = p + q(1 − e−px ) = 1 − qepx f or x ≥ 0 This is a distribution function which has a jump of size p in 0 and is exponential otherwise.
37
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Distribution for the Number of claims To describe ‘rare event’s 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(λ) distributed number of accidents in one year. The parameter λ is unknown and different for every driver. We assume that λ is the outcome of a random variable Λ. The conditional distribution of N, the number of accidents in one year, given Λ = λ, is poison(λ).
(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 Λ ∼ gamma(α, β)
Solution Let U (λ) = P r(Λ ≤ λ) denote the distribution function of Λ. Then we can write the marginal distribution of N as
ˆ
∞
P r(N = n) =
P r(N = n|Λ = λ)dU (λ) ˆ
0
=
∞
e−λ
0
λn dU (λ) n!
while for mean and variance of N we have
E[N ] = E[E[N |Λ]] = E[Λ]
V ar(N ) = E[V ar(N |Λ)] + V ar[E[N |Λ]] = E[Λ] + V ar[Λ] ≥ E[N ] Now assume additionally that Λ ∼ gamma(α, β), then
38
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
MN (t) = E[E[e
where p =
β β+1 ,
tN
h
Λ(et −1)
|Λ]] = E e
i
t
= MΛ (e − 1) =
β β − (et − 1)
α
=
p 1 − (1 − p) et
α
β so N has a negative binomial(α, (β+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 n P r(X = x) = px (1 − p)n−x , x
x = 0, 1, 2, ...
The mgf MX (t) = E[ext ] n = ext px (1 − p)n−x x all x X n (pet )x (1 − p)n−x = all x x X
Recall
n ax bn−x = (a + b)n x x
X all
⇒ MX (t) = (pet + q)n and
0
E[X] = MX |t=0
= npet (pet + q)n−1 |t=0 = npe0 (pe0 + q)n−1 = np(p + q)n−1 but p + q = 1 = np 39
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
V ar = E[X 2 ] − (E[X])2 0
0
= MX (t)|t=0 − (MX (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 P r(X = x) = v+x−1 pv (1 − p)x−1 , v > 0, 0 < p < 1 x−1 (where v= the number of occurrences for the event) v p ⇒ MX (t) = 1−qe t
d ⇒ dt
p 1 − qet
v
= pv
d (1 − qet )−v dt
= pv (−v)(−qet )(1 − qet )−v−1 pv vqet pv vq = | = (1 − qet )v+1 t=0 (1 − q)v+1 pv vq = pv+1 vq = p
0
∴ E[X] = MX (t)|t=0 =
vq p
and V ar(X) =
vq p2
Suppose v = 1 then P r(X = x) = p(1 − p)x−1 , x = 1, 2, 3, ... which is geometric distribution with mgf, MX (t) =
p 1−qet ,
E[X] =
q p
and V ar(X) =
q p2
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)x−1 ,
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
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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[N E(X)] = E[N ]E[X] and V ar(X) = E[V ar(Z|N )] + V arE[Z|N ] = E[N V ar(X)] + V ar(N E[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 ] q = np. p = (120 × 0.02) × = 3.6
and
41
0.6 0.4
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
V ar(M ) = E[V ar(M |K)] + V ar(E[M |K]) = E[K]V ar(N ) + V ar(K)(E[N ])2 2 q q = np. 2 + npq p p 0.6 (0.6)2 = 120(0.02) + 120(0.02)(0.98) (0.4)2 (0.4)2 = 14.292
Theorem: Panjer’s 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 =
b a+ n
qn−1 ,
(14)
n = 1, 2, ...
for some real a and b. Then the following relations for probability of a total claim equal to s hold:
f (0) =
P r(N = 0)
if p(0) = 0
MN (logP (0))
f (s) =
p(0) > 0
s X bh 1 a+ p(h)f (s − h), 1 − ap(0) s
s = 1, 2, 3, ...
(15)
h=1
Proof: P r(S = 0) =
∞ X
P r(N = n)P n (0)
n=0
gives us starting value f (0). Write Tk = X1 + ... + Xk . First, note that because of symmetry. h i 1 E a + bX |T = a + kb . This expectation can be determined in the following way. k S
bX1 E a+ |Tk S
= =
s X h=0 s X h=0
bh a+ s
bh a+ s
P r(X1 = h|Tk = s)
42
P r(X1 = h)P r(Tk − X1 h|s − h) P r(Tk = s)
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Because (14) and the previous two equalities, we have for s = 1, 2, ... f (s) = = = = =
∞ X
P r(Tk = s) =
k=1 ∞ X
∞ X
qk−1
k=1
b a+ k
P r(Tk = s)
s X b qk−1 a+ P r(X1 = h)P r(Tk − X1 = s − h) k 0 k=1 s ∞ X X b a+ P r(X1 = h) qk−1 P r(Tk − X1 = s − h) k h=0 k=1 s X bh p(h)f (s − h) a+ s h=0 s X bh aP (0)f (s) + a+ p(h)f (s − h) s h=1
From which the second relation of (15) follows immediately
Distributions suitable for panjer’s recursion Only the following distributions satisfy b qn = a + q , n n−1
n = 1, 2, ...
1. Poison(λ) with a = 0 and b = λ ≥ 0, in this case (14) simplifies to
f (0) = e−λ(1−P (0)) s 1X f (s) = λhP (h)f (s − h) s
(16)
h=1
2. Negative binomial(r,p) with a = 1 − p and b = (1 − p)(r − 1), so 0 < a < 1 and a + b > 0. In this case (14) simplifies to
f (0) = pr f (s) = (1 − p)
s X h=1
h (r − 1) + 1 P (h)f (s − h) s
(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
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Y ∼ N B(r, p) and f (y) =
r+y−1 y
pr q y , 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−1 pr q x−r ., where x=0,1,2,... X ∼ N B(r, p) and f (x) = r−1 p and b = 3. Binomial(m,p) with a = − 1−p
(m+1)p (1−p)
. so a < 0, b = −a(m + 1) . In this case (14)
simplifies to
f (0) = (1 − p)m s X h p (m + 1) − 1 P (h)f (s − h) f (s) = (1 − p) s
(18)
h=1
Example: A compound poison distribution with λ = 4 and P r(X = 1, 2, 3) = 14 , 12 , 41 . What is the distribution of S = X1 + X2 + X3
Solution Equation (16) yields, with λ = 4, P (2) =
1 2
and P (1) = P (3) =
1 4
1 f (s) = [f (s − 1) + 4f (s − 2) + 3f (s − 3)], s
s = 1, 2, 3, ...
and the starting value is f (0) = e−4 ≈ 0.0183. We have f (1) = f (0) = e−4 f (2) = 12 [f (1)4f (0)] = 52 e−4 f (3) = 31 [f (2) + 4f (1) + 3f (0)] =
19 −4 6 e
and so on.
Question A compound Negative binomial distribution with r = 14, p = 0.7 and P r(X = 1, 2, 3) = 14 , 12 , 14 . What is the distribution of S = X1 + X2 + X3
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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Question A compound Binomial distribution with m = 15, p = 0.3 and P r(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 Panjer’s recursion theorem b qn = a + q , n n−1
n = 1, 2, ...
for Poisson, Negative binomial and Binomial distribution are as given in the discussion above.
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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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 insurer0 s 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 j th 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
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T
= min{t|t ≥ 0 &U (t) < 0} = ∞ if U (t) ≥ 0 f or 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 Ψ(u) = P r[T < ∞] NOTE: • Ruin is said to occur when U (t) is negative • Ruin is not equivalent to insolvency of an insurer • Infinite time probability of ruin is ψ(µ) = P r (U (t) < 0 f or some t ≥ 0). In this case ψ(µ) is the probability of ruin over an infinite horizon. ψ(µ) is an upper bound for ψ(µ, t). ψ(µ) is more trackable mathematically. • ψ(µ, t) = P r(T < t) is the probability of ruin before time t. • Let Tj denote the time when claim j occurs, such that T1 < T2 < T3 < ....... Then the random variable of the interarrival (or wait) time between claim j − 1 and j is defined as W1 = T1 and Wj = Tj − Tj−1 , j ≥ 2. With the following assumptions: (i) Premiums are collected at a constant rate c hence c(t) = ct (ii) The sequence {Tj }j≥1 forms an ordinary renewal process, provided Wj0 s are iid 47
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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 λ > 0, the increments of the process have the following property N (t + h) − N (t) ∼ P oisson(λh) 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(λh) distributed for every value of t.
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 smaller than any real number larger than 0. For the Poisson process we have P r[N (t + dt) − N (t) = 1|N (s); 0 ≤ s ≤ t] = e−λdt λdt = λdt P r[N (t + dt) − N (t) = 0|N (s); 0 ≤ s ≤ t] = e−λdt λdt = 1 − λdt P r[N (t + dt) − N (t) ≥ 2|N (s); 0 ≤ s ≤ t] = e−λdt λdt = 0 these equalities are only valid if we ignore terms of order (dt)2
Lumdberg’s 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 Ψ(u) ≤ e−Ru µ ≥ 0.
Assignment Proof the above theorem
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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 etX−tcW = 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. Find adjustment 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 insurer’s surplus at time n, where n = 0, 1, 2, 3, .... µ denotes initial surplus c denotes the constant premium received per unit time/period Sn denotes the aggregate claims of the n periods N ote : 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
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
independent, identically distributed random variables with µ = E[Xi ] < c. Un can also be written as; Un = µ + (c − X1 ) + (c − X2 ) + ... + (C − Xn ) and we define adjustment coefficient as the positive solution of the equation
h i MX−c (r) = E er(X−c) = e−rc MX (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 Xi0 s have a common distribution N (µ, σ 2 ).
Solution log [MX (r)] = µr + =⇒ R =
2(c−µ) σ2
σ2 r2 2
but logMX (r) = rc
where µ < c.
Generally Since for a random variable X d dt logMX (t)|t=0
= E[X] and
d2 logMX (t)|t=0 dt2
= V ar[X]
and using Maclaurin series expression logMX (r) = µr + 12 σ 2 r2 + ... where, σ 2 = V ar(X) =⇒Then R '
2(c−µ) σ2
Note: If X has a compound distribution and the relative security loading θ is given by c = (1 + θ)µ then R=
2{[µ+µθ]−µ} σ2
=
2θµ σ2
Let, pk = E[X]k = E[Loss]k then 50
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
R∼ =
2θp1 E[N ]
(p2 −p21 )E[N ]+p21 V 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 λ. b) N has a negative binomial distributed with parameters r and p.
Solution a) E[N ] = V ar(N ) = λ Therefore, R ∼ = b) E[N ] =
rq p,
2θp1 p2
V ar(N ) =
Therefore, R ∼ =
rq p2
2θp1 h
p2 +p21
Note; as p → 1, R ∼ =
1 p
i −1
2θp1 p2
CASE 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 λt Then R is the smallest positive root of
h i MS(t)−ct (r) = E er(S(t)−ct) = e−rct MS(t) (r) = e−rct eλt[MX (r)−1] = 1 Thus
λ [MX (r) − 1] = rc 51
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Substituting, c = (1 + θ)λp1 We get, 1 + (1 + θ)p1 r = MX (r). Which is a linear of r. Note: 1 + (1 + θ)p1 r = 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 β > 0.
Solution The adjustment coefficient is obtained from 1 + (1 + θ)p1 r = MX (r) as 1 +
(1+θ)r β
=
β β−r
=⇒ (1 + θ)r2 − θβr = 0 as r = 0 is a solution and R is smallest positive root =⇒ R =
θβ 1+θ
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 Ψ(u) =
e−Ru f or µ ≥ 0 E[e−RU (T ) |T < ∞]
where R is the smallest positive of 1 + (1 + θ)p1 r = MX (r).
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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Proof: Let R > 0 and t > 0. Then
E[e−RU (t) ] = E[e−RU (t) |T ≤ t]P r[T ≤ t] + E[e−RU (t) |T > t]P r[T > t]
(19)
The adjustment coefficient R has the property that E[e−RU (t) ] is constant in t. i.e. e−RU (t) is a martingale: and since U (t) = u + ct − S(t) and S(t) ∼Compound Poisson with parameter λt, we have, E[e−RU (t) ] = E[e−R{u+ct−S(t)} ] = e−Ru [e−Rc exp{λ(Mx (R) − 1)}]t = e−Ru (20) From equation 17 the left-hand side equals e−Ru . 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[e−RU (t) |T = v] = E[e−R{U (v)+c(t−v)−[S(t)−S(v)]} |T = v] = e−RU (v)|T =v e−Rc E[eR{S(t)−S(v)} |T = v] t−v = e−RU (v)|T =v e−Rc exp[λ(Mx (R) − 1)] h i = E e−RU (T ) |T = v (21) The total claims S(t) − S(v) between v and t has again a compound Poisson distribution. What happens 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 e−RU (t) |T ≤ v = E e−RU (T ) |T ≤ v also holds. Since P r[T ≤ t] ↑ P r[T < ∞] for t → ∞, it suffices to show that the last term in eqn 17 vanishes for t → ∞. For that purpose, we split the event 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 implies that we are not in ruin at time t, i.e. U (t) ≥ 0 so e−RU (t) ≤ 1. We have E[e−RU (t) |T > t]P r[T > t] = E[e−RU (t) |T > t & 0 ≤ U (t) ≤ uo (t)]P r[T > t & 0 ≤ U (t) ≤ uo (t)] + E[e−RU (t) |T > t & U (t) > uo (t)]P r[T > t & 0 ≤ U (t) > uo (t)] ≤ P r[U (t) ≤ uo (t)] + E[exp(−Ruo (t))] = 0 53
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
The second term vanishes if uo → ∞. For the first term in eqn 17, note that U(t) has an expected value U (t) = u + ct − λtu1 and a variance σ 2 (t) = λtu2 .
Question Calculate the probability of ruin in the case that the claim amount distribution is exponential with parameter β > 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 P r[U (T ) ∈ (−y − dy, −y) & T < ∞] =
λ [1 − p(y)]dy c
(22)
Proof: In a compound Poisson process, the probability of having a claim in the interval (t, t + dt) equals λdt. Which is independent of t and of the history of the process up to that time. So, between 0 and dt there is either no claim (with probability 1 − λdt), and the capital increases from u to u + cdt, or one claim with size X. In the latter case, there are two possibilities. If the claim size is less than u, then the process continues with capital u + cdt − X. Otherwise ruin occurs, but the capital at ruin is only larger than y if X > u + y. Defining G(u, y) = P r[U (T ) ∈ (−∞, −y) T < ∞|U (0) = u]
we can write ˆ G(u, y) = (1 − λdt)G(u + cdt, y) + λdt
ˆ
u
∞
G(u − x, y)dP (x) + 0
dP (x) .....................................................(∗)
u+y
If $G^{’}$ denotes the partial derivative of G with respect to u, then 0
G(u + cdt, y) = G(u, y) + cdtG (u, y)......................................................................................................................(∗∗) Substitute (**) into (*), subtract G(u,y) from both sides and divide by cdt. Then we get λ G (u, y) = c 0
ˆ
G(u, y) −
ˆ
u
∞
G(u − x, y)dP (x) − 0
u+y
54
dP (x)
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
integrating this over u ∈ [0, z] yields λ G(z, y) − G(0, y) = c
ˆ
ˆ
z
ˆ
z
ˆ
u
z
ˆ
∞
G(u − x, y)dP (x)du −
G(u, y)du − 0
0
0
0
dP (x)du .................................(∗ ∗ ∗)
u+y
The double integrals in (***) can be reduced to single integrals as follows. For the first double integral, exchange the order of integration, substitute v = u − x and again exchange the integration order. This leads to ˆ
z
ˆ
ˆ
u
z
ˆ
ˆ
z−v
G(u − x, y)dP (x)du = 0
0
0
z
G(v, y)P (z − v)dv
G(v, y)dP (x)dv = 0
0
In the second double integral in (***) we substitute v = u + y. Then ˆ
z
ˆ
ˆ
∞
z+y
[1 − P (v)]
dP (x)dv = 0
v+y
y
Hence λ G(z, y) − G(0, y) = c
ˆ
ˆ
z+y
[1 − P (v)]dv − 0
z+y
[1 − P (u)]du .............................................................(∗ ∗ ∗∗)
y
for z → ∞, the first term on both sides of (****) vanishes, leaving λ G(0, y) = c
ˆ
∞
[1 − P (u)]du y
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 P r[T < ∞], so regardless of P(.) we have λ Ψ(0) = c
ˆ
∞
[1 − p(y)]dy = 0
λ 1 u1 = c 1+θ
where, θis the relative security loading factor.
Reinsurance, individual claims Assume that the reinsurer pays an amount h(x) if the claims amount is x. In other words, the retained loss equals x − h(x). we consider two special cases h(x) = αx 0 ≤ α ≤ 1 for proportional reinsurance h(x) = (x − β)+ 0 ≤ β for excess of loss reinsurance 55
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Obviously, proportional reinsurance can be considered as a reinsurance on the usual adjustment coefficient Rh , which is the root of
ˆ
∞
er[x−h(x)] dP (x)
λ + (c − ch )r = λ
(23)
0
where ch denotes the re-insurance premium. The reinsurer uses a loading factor ξ on the net premium. Assume that λ = 1, and p(x) = 0.5 for x = 1 and x = 2. Furthermore, let c = 2, so that θ = 31 , and consider two values ξ =
1 3
and ξ = 25 . \par In case of proportional reinsurance h(x) = αx, the
premium equals 3 ch = (1 + ξ)λE[h(x)] = (1 + ξ) α 2 so, because of x − h(x) = (1 − α)x eqn 21 leads to the equation
3 1 1 1 + 2 − (1 + ξ) α r = er(1−α) + e2r(1−α) 2 2 2 For ξ = 13 , we have ch = 2α and Rh =
0.325 1−α ,
for ξ = 52 , we have ch = 2.1α.
Next, we consider the excess of loss reinsurance h(x) = (x − β)+ with 0 ≤ β ≤ 2. The reinsurance premium equals 1 1 1 ch = (1 + ξ)λ h(1) + h(2) = (1 + ξ) [(1 − β)+ + (2 − β)+ ] 2 2 2
while x − h(x) = min{x, β}, and therefore Rh is the root of i 1 1 h min{β,1}r 1 + 2 − (1 + ξ) [(1 − β)+ + (2 − β)+ ] r = e + emin{β,2}r 2 2
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 size 56
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
2. Increase of skewness of the individual claim size
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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 businessman 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 don’t know how much the claims arising from the policy over the following year will be. In fact they don’t 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 investor’s 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.
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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
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. Baye’s estimation
Maximum likelihood method The likelihood of a sample is the product of the probabilities of drawing the individual observations 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 λ. Calculate the maximum likelihood estimate of λ based on the above information.
Solution Let X be the random variable for the gross claim amounts. The likelihood function is L=
n Y
f (xi , λ)
i=1
but f (x, λ) =
1 −x eλ λ
,x > 0
n Y 1 −xi ⇒ e λ =L λ i=1 59
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
The chance that the 6 claims exceed the retention amount is ˆ
∞
P (X > 50000) = m
−m 1 −x e λ dx = e λ λ
where m is the retention amount of 50000 L=
4 Y 1 −x −m 6 eλ e λ λ i=1
=
1 − e λ4
P
xi
λ
e
−m λ
6
A general rule is to take logarithms and then maximize log L P ⇒ logL = −4logλ −
xi 6m − λ λ
P dL −4 xi 6m ⇒ = + 2 + 2 =0 dλ λ λ λ
P 4 xi 6m = + 2 λ λ2 λ X ⇒ 4λ = xi + 6m ⇒
ˆ= λ
P
xi + 6m 58681 + 300000 = = 89670.25 4 4
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 λ
years.
i) State the distribution of the random variable X whose value is x, the number of men who ˆ the M.L.E of λ. die within to years has been observed. Hence determine λ, ˆ and show that the value ii) Consider the case n = 1000, to = 20 years and x = 320. Evaluate λ 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
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Bayesian Estimation Loss function Let d be an estimate of a parameter, say θ. The loss function, defined by l(θ, 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 θ. The risk function of d is given by R(θ, d) = E[l(θ, 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(θ, d) = (d − θ)2 Note that this loss function is minimized when d is the mean of f (θ|x) the posterior distribution (ii) All or nothing loss function (0/1 loss function):Here the loss function takes the values 1 when d < < θ < 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 (θ|x) the posterior distribution (iii) Absolute error loss function:It is given by l(θ, d) = |d − θ| The loss function is minimized when d is the median of f (θ|x) the posterior distribution
Example A risk consists of 5 policies. On each policy in 1 month, there’s exactly 1 claim with probability θ and there’s a negligible probability of more than 1 claim in 1 month. The prior distribution of θ 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 θ 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 θ, i=1,2,3,...,12. The posterior density is given by f (θ|x) ∝ f (x|θ)h(θ) 61
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
∝θ
P12
i=1
xi
(1 − θ)60−
P12
i=1
xi
= θ10 (1 − θ)50
⇒ The posterior distribution is a beta distribution with parameter α = 11 and β = 51. Therefore under quadratic loss function,
E[X] =
α 11 = = 0.1774 α+β 11 + 51
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, θ in this case, equate the results to 0 and solve for for θ ie 10θ9 (1 − θ)50 − 50θ10 (1 − θ)49 = 0
⇒ 10θ9 (1 − θ)50 = 50θ10 (1 − θ)49 1 10θ9 (1 − θ) = 50θ10 ⇒ θˆ = = 0.167 6
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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 management of companies eg the general insurance firms etc.
NOTE: An efficient allocation of capital is a crucial finance function in modern times. It involves decisions to commit firms funds to the long-term’s assets. These decisions are important since they determine the firm’s value size by influencing its growth, profitability and risks. The investment decisions of a firm can also be known as capital budgeting or capital expenditure decisions. A capital budgeting decision of a firm can also be defined as the firm’s 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 firm’s investment decisions may include expansion, acquisition, modernization and replacement of long-term assets and also sale of a division or business (divestment). Other activities 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
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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 firm’s value. Firm’s value increases if investments are profitable and add to the shareholder’s wealth. This means that investments should be evaluated on the basis of a criterion with the objective of shareholder’s wealth maximization. An investment will add to the shareholder’s wealth if it yields benefits in excess of the minimum benefits as per the opportunity cost of the capital. In this topic we make the following assumptions.
1. The investment’s 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 firm’s 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 operating 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 irreversible. Once such capital items have been acquired, it’s 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 assessment 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
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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 shareholder’s 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 shareholder’s 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
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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 shareholder’s 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 N P V > 0 Consider a project that has the following cash inflows over its useful economic life. The project’s initial investment is I i.e.
Period
1
2
3
...
n
Cash flows
R1
R2
R3
...
Rn
Then NPV =
n X t=1
Rt −I (1 + i)t
Where i is the appropriate discount rate; n is the expected life of the investment; (1 + i)−t is the present value factor (PVF)
Example: Assume that project X costs $2500 now and expected to generate the year-end cash inflows of $900, $800, $700, $600 and $500 for 5 consecutive years. The opportunity cost of the capital (discount rate) may be assumed to be 10%. Compute the NPV for project X.
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Solution
NPV
=
n X t=1
Rt −I (1 + i)t
900 800 700 600 500 = + + + + 1 2 3 4 (1.1) (1.1) (1.1) (1.1) (1.1)5 = $2725.53 − $2500
− 2500
= $225.53 > 0 This shows that the net present value of project X (2725.53) is greater than the cash outflow (2500). Thus project X adds to the wealth of the owners and therefore should be accepted.
NB: The difficult part in calculation of present value of an investment project is the precise measurement of the discount rate.
Acceptance Rule: The acceptance rule using NPV method is to:-accept the investment project if N P V > 0 -reject the investment project if N P V < 0 -may accept the project if N P V = 0 NPV method can be used to select between mutually exclusive projects. The one with a higher NPV is selected.
Advantages of NPV method 1. It recognizes the time value of money ie a dollar received today is worth more than a dollar received tomorrow. 2. It uses all the cash flows occurring over the entire life of the project in calculating its worth. 3. It satisfies the value-additivity principle (eg NPV(A+B)=NPV A+NPV B) ie NPV’s of two or more projects can be added. This is an important criterion because it means that each project can be evaluated independently of the others on its own merit. 4. It is always consistent with the objective of maximizing the shareholder’s wealth. 67
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Disadvantages of NPV method 1. It requires estimates of cash flows which is a tedious task due to uncertainty prevailing 2. It is sensible to discount rates. Ranking of investment projects as per NPV rule is not independent of discount rates.
Illustration: Consider two projects, A and B, each consisting of $50. Project A returns $100 after 1 year and $25 after 2 years. Project B returns $30 after 1 year and $100 after 2 years. The NPV of the projects and their ranking at 5% and 10% discount rates are as follows NPV at 5%
Rank
NPV at 10%
Rank
Project A
67.92
II
61.57
I
Project B
69.27
I
59.91
II
The reason of change in ranking lies in the cash flows patterns. The impact of discounting becomes more severe for cash flows occurring later in the life the project. 3. It requires computation of opportunity cost of the capital especially when alternative projects with unequal lives or fund constraints are evaluated, which poses practical difficulties.
Internal Rate of Return (IRR) method It takes account of the magnitude of timing and cash flows. Assume that you deposit $10, 000 with a bank and you get back $10, 800 after 1 year. The true rate of return on this investment is Rate of return =
10, 800 − 10, 000 = 0.08 or 8% 10, 000
We observe that the rate of return of your investment (8%) makes the discounted value of your cashinflow (10,800) equal to 1 year investment (10,000). The rate of return (r) on an investment (co ) that generates a single cash flow after one period (c1 ) is given by
r
= =
c1 − co co c1 −1 c0
⇒ r+1=
c1 c0
(24)
From equation 1, we can define internal rate of return as the rate which equates investment 68
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outlay or outflow with the present value of inflow received after 1 period. In short, we can define IRR as the discount at which the project’s N P V = 0. Note that there’s no satisfactory way of defining the true rate of return of a long-term asset. The rate of return can be determined by solving the following equation for r. c0 =
c3 cn c1 c2 + + ... + + 2 3 1 + r (1 + r) (1 + r) (1 + r)n n X ct ⇒ c0 = (1 + r)t
(25)
t=1
Calculating IRR Case A: uneven cash flows Under uneven cash flows, we use trial and error method. The value of r in equation 2 can be found by trial and error method. The approach starts by selecting any discount rate to compute the present value of cash flows. If the calculated present value of the expected cash inflows is lower than the present value of cash outflows, a lower rate is tried. A higher rate is tried if the present value of cash inflows is higher than the present value of cash outflows. The process is repeated until N P V = 0.
Example A project costs $16000 and is expected to generate cash flows of $8000, $7000, and $6000 at the end of each year for the next 3 years. Compute the IRR.
Solution The IRR is the rate at which the NPV of the project is zero (0). We start by trying 20%.
NPV
=
3 X t=1
ct −I (1 + r)t
8000 7000 6000 = + + 2 1.2 (1.2) (1.2)3 = −$1000
69
− 16000
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⇒ The negative present value at 20% indicates that the project’s true IRR is lower than 20%. Next try 16%
NPV
=
3 X t=1
ct −I (1 + r)t
6000 8000 7000 + = + 2 1.16 (1.16) (1.16)3 = −$57
− 16000
Next, try 15%
NPV
=
3 X t=1
ct −I (1 + r)t
8000 7000 6000 = + + 2 1.15 (1.15) (1.15)3 = $200
− 16000
The true rate of return lies between 15% and 16%. We approximate IRR by method of linear interpolation, as follows
PV required $16000
Difference
PV at 15%$16200
$200
PV at 16%$15943
$257
⇒ r=15%+(16%-15%) 200 257 =15.8%
Case B: Level/Equal cash flows It is easy to compute IRR for a project with level or equal cash flows each period.
Illustration Assume an investment which costs $20000 and provides annual cash inflow of $5430 for 6 years. We want to compute the IRR, assuming opportunity cost of capital to be 10%. In this case, we start by computing the NPV. Note that $5430 is an annuity. NPV =
n X t=1
ct −I (1 + r)t
⇒ (1 + r)−t is the present value of annuity factor (PVAF) 70
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⇒ NPV
5430(1 − rn ) − 20000 i 1 6 5430 1 − 1.1 − 20000 = 0.1 = $3649 =
We compute IRR as follows, i.e. NPV P V AF
= −$20000 + $5430 × P V AF = 0 20000 = 5430 = 3.683
IRR is the rate which gives PVAF of 3.683 for 6years. From tables, IRR=16%. This implies that 16% is the IRR which equates the present value of the initial cash outlay with the constant annual cash inflows for 6 years.
Acceptance Rule -We accept the project if IRR is higher than the opportunity cost of capital ie r > k. -We reject the project if r < k -We may accept the project if r = k
NB: In case of independent projects, the IRR and the NPV rules will give the same results so long as the firm has no shortage of funds.
Advantages of IRR 1. It recognizes the time value of money \item It considers all the cash flows occurring over the entire life of the project 2. It is consistent with the shareholder’s wealth maximization objective 3. It is a true measure of profitability
Disadvantages of IRR 1. Does not hold the value-additivity principle 71
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2. It fails to indicate a correct choice between mutually exclusive projects under certain situations. 3. Requires estimates of cash flows which is a tedious task 4. Gives misleading and inconsistent results when the NPV of a project does not decline with discount rate.
Profitability Index (PI) It is a time-adjusted method of evaluating investment proposals. It is also called benefit cost ratio. It can be defined as the ratio of the present value of cash inflows at the required rate of return at the initial cash outflows of the investment. Ie PI =
P V of cash inf lows Initial cash outlay (cash outf low)
P V (ct ) co Pn ct
=
=
t=1 (1+r)t
co
Illustration The initial cash outlay of a project is $100000 and it can generate cash inflow of $40000, $30000, $50000, and $20000. Assume a 10% rate of discount and compute the profitability index.
Solution Pn PI =
ct t=1 (1.1)t
co 40000 × P V F1,0.1 + 30000 × P V F2,0.1 + 50000 × P V F3,0.1 + 20000 × P V F4,0.1 = 100000 112350 = 100000 = 1.1235 > 1
Acceptance Rule -We accept the project if P I > 1 72
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-We reject the project if P I < 1 We note that when P I > 1, the project will have a positive NPV
Advantages of Profitability Index 1. Recognizes the time value of money. 2. It considers all the cash flows occurring over the entire life of the project in calculation of its its worth. 3. It is consistent with the wealth maximization principle. 4. It is a relative measure of project’s profitability since the present value of the cash inflow is divided by the initial cash outflow.
Disadvantages of Profitability Index 1. It requires time estimates of cash flows which is a tedious task. 2. At times, it fails to indicate the correct choice between mutually exclusive projects.
Payback Method (PB) It is one of the most popular and widely recognized traditional method of evaluating investment projects. We can define payback as the number of years required to recover the original cash outlay invested in a project. If the project generates a constant annual cash inflow, then the payback should be equal to the initial investment divided by annual cash inflow. PB =
co Initial Investment = Annual cash inf low c
Illustration A project requires an outlay of $50000 and yields an annual cash inflow of $12500 for 7 years. Then PB is 50000 = 4 years 12500 In a case of unequal cash flows, the payback period is computed by adding the cash inflows until the total is the initial cash outlay
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Illustration A project requires a cash outlay of $20000 and generates cash inflows of $8000, $7000, $4000, and $3000 in 4 years. Adding the cash inflows for the first 3 years. we have $19000 of the original cash outlay. In the fourth year, we need only $1000. Time required to recover $1000 is 1000 3000
× 12 = 4 months.
The payback period is 3 years 4 months.
Acceptance Rule We accept the project if PB period is less than the standard PB period set by the management otherwise, we reject the project. Payback period can also be used as a method of ranking projects. The project with the shortest payback period is given the highest ranking while the one with the longest payback period is given the lowest rank. Thus, to choose among mutually exclusive projects, we select the one with the shortest payback period.
Advantages of PB 1. It is simple to understand and easy to calculate \item It costs less than the most sophisticated techniques which require a lot of the analysts’ time and use of computer. 2. Payback emphasizes on the early recovery of the investment, thus giving an insight into the liquidity of the project. 3. It is an easy and crude way to cope with risk. The riskiness of the project can be tackled by setting a shorter payback period as it may ensure a guarantee against loss.
Disadvantages of PB 1. It fails to take account of the cash inflows after the payback period 2. Not an appropriate method of measuring profitability of an investment project since it does not consider all cash flows yield by the project. 3. It ignores the time value of money 4. No objective way to determine the standard payback ie it is a subjective decision which may create administrative difficulties in setting it. 5. It is not consistent with the objective of maximizing the shareholders wealth. 74
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Accounting Rate of Return (ARR) method Sometimes we call it investment on return or return on investment (ROI). ARR is computed by dividing the average after tax profit by the average investment. The average investment will be equal to
1 2
of the original investment if it is depreciated consistently or it can be found by diving
the total of investment’s book value after depreciation by life of the project. Accounting Rate of Return is the Average Rate of Return ⇒ ARR = =
Average Income Average investment Pn t=1 [EBITt (1 − T )]/n (Io + In )/2
where EBITt =Earning before interest and taxes T=tax rate Io =Book value of investment in the beginning In =Book value of investment at the end of n years
Illustration A project will cost $40000 . Its stream of earnings before depreciation, interest and taxes (EBDIT) during the first year through 5 years is expected to be $10000, $12000, $14000,$16000 and $20000. Assuming a 50% tax rate and a depreciation of $8000 on straight line basis, compute the ARR.
Solution Period
1
2
3
4
5
Average
EBDIT(A)
10000
12000
14000
16000
20000
14400
Depreciation (B)
8000
8000
8000
8000
8000
8000
EBIT (C=A-B)
2000
4000
6000
8000
12000
6400
Taxes at 50% (D = 50%of C)
1000
2000
3000
4000
6000
3200
EBIT(1-T) (E=C-D)
1000
2000
3000
4000
6000
3200
Book value of investment at beginning
40000
32000
24000
16000
8000
24000
Book value of investment at end
32000
24000
16000
8000
-
16000
Average
36000
28000
20000
12000
4000
20000
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RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Where, EBIT (1 − T ) = Earning bef ore interest and af ter taxes. And
Pn ARR =
− T )]/n (Io + In )/2
t=1 [EBITt (1
3200 × 100 20000 = 16% =
Acceptance Rule The method accepts all those projects whose ARR is higher than the minimum rate established by the management and rejects the projects with ARR less than the minimum rate. The method gives the project the highest rank if it has the highest ARR and vice versa for projects with lowest ARR.
Advantages of ARR 1. It is easy to understand and calculate 2. It uses the accounting data with which executives are familiar with 3. It incorporates the entire stream of income in calculating the project’s profitability
Disadvantages of ARR 1. It uses accounting profits, not cash flows in appraising the projects. Accounting profits are based on arbitrary assumptions and choices and also include non-cash items. Thus the criterion is in appropriate to rely on in measuring acceptability of the investment projects. 2. It ignores the time value of money. 3. No objective way of determining the minimum acceptance rate of return.
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CHAPTER EIGHT RISK ANALYSIS IN CAPITAL BUDGETING Introduction Risk exists because of the inability of the decision maker to make a perfect forecast. Forecasts can’t be made with certainty because of uncertain future events they depend on. An investment is not risky if we can specify a unique sequence of cash flows for it. The problem is that cash flows can’t be forecased accurately and alternative sequences of cash flows can occur depending on the future events. Thus risk arises in investment evaluation because we cannot expect the occurrence of possible future events with certainty, making it difficult to make any correct prediction about the cash flow sequence. eg A firm is considering a proposal to commit its funds in a machine which will help to produce a new product. The demand for this product may be very sensitive to the general economic conditions. It may be very high under favorable economic conditions and very low under unfavorable economic conditions. This means that the investment is profitable in the former and unprofitable in the latter case. Uncertainty of economic conditions result to uncertainty about the cash flows associated with the investment.
Definition The risk associated with an investment may be defined as the variability that is likely to occur in the future returns of an investment. eg an investor may purchase the shares of a company. In this case it is not possible to estimate future returns accurately. In fact the returns may be negative, zero or large figure. This investment may be considered risky because of high degree of variability associated with the future returns. The greater the variability of the expected returns, the riskier the project. The most common measures of risk are standard deviation and coefficient of variation.
Techniques of handling/measuring risk in capital budgeting They include:
• Payback • Risk Adjusted Discount rate (RAD) • Certainty equivalent 77
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
• Sensitivity analysis • Statistical techniques • Decision tree
Payback It is a common method for explicitly recognizing risk associated with an investment project. Business firms using this method prefer short payback to longer payback. Longer term projects are viewed to be riskier. The payback period as a method of risk analysis is useful only in allowing for a special type of risk ie the risk that a project will go exactly as planned for a certain period and will then suddenly cease altogether and become worth nothing. It is suitable in assessment of risks of time future. This method ignores the time value of cash flows eg 2 projects with, say a 4 year payback period are at different risks if in one case the capital is recovered evenly over the 4 years while in the other it is recovered in the last year. The second project is more risky since anything can happen before the four years. If both cease after 3 years, the first project will have recovered
3 4
of its capital and a total loss in the second project.
Risk Adjusted Rate The more uncertain the returns in the future, the greater the risk and the greater the premium required. Based on this argument, it is proposed that the risk premium be incorporated into the capital budgeting analysis through the discount rate ie if the time prevalence for money is to be recognized by discounting estimated future cash flows at some risk free rate to their present value, then to allow for riskiness of those future cash flows, a risk premium rate may be added to risk free discount rate. Such a composite discount rate is called the Risk Adjusted Discount rate which allows for both time prevalence. We can define RAD rate as the sum of risk free rate and the premium rate reflecting the investor’s attitude towards risk. ⇒RAD is computed from the expression:Pn NPV =
t=0 N CF (1 + k)t
where k=Risk Adjusted Rate The RAD rate accounts for risk by varying the discount rate depending on the degree of risk of the investment projects. Higher rate will be used for riskier projects and a lower rate for less risky projects. The Net Present Value decreases with increase in k indicating that the riskier the project, the 78
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
less it’s likely to be accepted. If the Risk free rate is assumed to be 10%, some rate would be added to it, say 5% as a compensation for the risk of the investment and the composite 15% rate will be used to discount the cash flows.
Example An investment project will cost $50000 initially and is expected to generate cash flows in 4 years ie $25000, $20000, $10000 and $10000 respectively. What is the project’s NPV if it’s expected to generate certain cash flows? Assume 10% risk free rate.
Solution
NPV
=
n X t=1
Rt −I (1 + k)t
25000 20000 10000 10000 + + + − 50000 1.1 (1.1)2 (1.1)3 (1.1)4 = $3599
=
If the project is risky, higher rate should be used to allow for the perceived risk. Assuming $15\%$ rate, it implies that NPV
=
n X t=1
Rt −I (1 + k)t
25000 20000 10000 10000 + + + − 50000 2 3 1.15 (1.15) (1.15) (1.15)4 = −$845 =
We observe that the project would be accepted when no allowance for risk is granted but it will be rejected if risk premium is added to the discount rate.
Advantages of RAD 1. It is simple and can be easily understood. 2. It has a great deal of intuitive appeal for a risk averse businessman. 3. It incorporates an attitude towards uncertainty.
Disadvantages of RAD 1. There’s no easy way of deriving RAD rate. 79
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2. It is based on the assumption that investors are risk averse. 3. It does not make nay risk adjustments in the numerator for the cash flows that are forecased for the future years.
Certainty equivalent A common procedure for dealing with risk in capital budgeting is to reduce the forecasts of cash flows to some conservative level. For instance, an investor according to his best estimate expects a cash flow of $60000 next year. He will apply an intuitive correction factor and may work with $40000 to be on the safe side. Certainty equivalent approach may be expressed as
NPV =
n X αt N CFt (1 + kf )t t=0
where N CFt =forecasts of net cash flow without risk adjustment. αt =risk adjusted factor or certainty equivalent co-efficient kf =risk free rate assumed to be constant for all the periods The value of certainty equivalent coefficient αt , ranges between 0 and 1 and varies inversely with risk. If greater risk is perceived, lower αt is used and higher αt is used if lower risk is expected. The coefficients are subjective in nature and vary from one decision maker to another. We multiply the estimated cash flows with αt to get the certain cash flows. ⇒ αt = =
N CFt∗ N CFt certain net cashf low risky net cashf low
Illustration A project costs $6000 and has cash flows of $4000, $3000, $2000 and $1000 for 4 years consecutively. Assume that the associated αt factors are estimated to be αo = 1, α1 = 0.9, α2 = 0.7, α3 = 0.5, α4 = 0.3 and the risk free discount is 10%. We want to compute the NPV
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Solution
NPV
n X αt N CFt = (1 + kf )t t=0
= (1 × −6000) +
0.9 × 4000 0.7 × 3000 0.5 × 2000 0.3 × 1000 + + + 1.1 (1.1)2 (1.1)3 (1.1)4
= −$37 If IRR is used, we calculate the rate of discount which equates the present value of certainty equivalent cash inflows with present value of certainty equivalent of cash outflows. The project is accepted if the computed rate is greater than the minimum rate otherwise it is rejected.
Advantage 1. The approach recognizes risk explicitly
Disadvantages 1. The procedure for reducing forecased cash flows is implicit and may be inconsistent from one investment to another. 2. It may give wrong estimates since the forecaster may inflate them. 3. Exaggeration of original forecasts if the forecasts pass through several stages of management. 4. Focusing explicit attention only on gloomy outcomes may increase the chances of passing by some good investments.
RAD verses certainty equivalent Certainty equivalent approach recognizes risk in capital budgeting analysis by adjusting estimated cash flows and employs risk free rate to discount cash flows. On the other hand RAD rate adjusts for risk by adjusting the discount rate. The two approaches yield the same results if the risk free rate is constant and RAD rate is the same for all future periods. Thus n X αt N CFt N CFt = t (1 + kf ) (1 + k)t t=0
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Solving for αt , we have αt N CFt (1 + k)t = (1 + kf )t N CFt ⇒ αt = =
(1 + kf )t N CFt N CFt (1 + k)t (1 + kf )t (1 + k)t
For a period of say, 1+t, the above equation becomes αt+1 =
(1 + kf )t+1 (1 + k)t+1
If kf and k are constants for all future periods, then k > kf since 0 < αt < 1.
Example Suppose k = 10% and kf = 5% ⇒ α1 =
α2 =
(1.05)1 = 0.955 (1.1)1
(1.05)2 = 0.911 (1.1)2
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CHAPTER NINE SENSITIVITY ANALYSIS It is another technique used to handle risk in capital budgeting. In evaluation of an investment project, we work with the forecasts of cash flows which depend on expected revenue and costs. Further, the expected revenue is a function of sales volume and unit selling price. Again, the sales volume depend on the market size and the firm’s market shares. Cost is a function of sales volume and unit variable costs and fixed costs. The N.P.V and IRR of a project is determined by analyzing after-tax cash flows arrived at by combining forecasts of various variables. Uncertainty prevails in these variables. Reliability of NPV or IRR of the project will depend on their reliability of the forecasts of variables involved in estimating the net cash flows. To determine the reliability of NPV or IRR of the project, we can work out how much difference it makes if any of those forecasts goes wrong. Each forecast can be changed one at a time to at least 3 values ie pessimistic, expected and optimistic values. The NPV of the project is re-calculated under these different assumptions. These method of analyzing change in the projects NPV or IRR for a given change in one of the variables is called sensitivity analysis. The sensitivity analysis indicates how sensitive the project’s NPV or IRR is to changes in particular variables. The more sensitive the NPV, the critical is the variable.
Steps involved in sensitivity analysis 1. Identification of all variables which influence the project’s NPV or IRR. 2. Define the underlying relationship between the variables. 3. Analyze the impact of the change in each of the variables on project’s NPV
As the decision maker performs sensitivity analysis he computes the NPV for each forecast under 3 assumptions ie pessimistic, expected and optimistic assumptions. The approach examines the sensitivity of the variables underlying the computation of NPV/IRR rather than quantifying the risk. The shortcoming of sensitivity analysis is that it makes no statement about the probability that a low, baseline, or high value of a parameter might be obtained.
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Example A financial manager of a food processing company is considering the installation of a plant costing $10000 to increase it’s processing capacity. The expected values of the underlying variables are as follows: Investment ($10000), Sales volume(1000 units), Unit selling price($15), Unit variable cost($6.75), Annual fixed costs($4000), Depreciation(25%), Corporate tax rate(35%) and Discount rate(12%). Find the project’s after-tax cash flows over its expected life of 7 years.
The following table provides the project’s after-tax cash flows over its expected life of 7 years.
S/No.
Year
0
1
2
3
4
5
6
7
1.
Investment($)
-10,000
0
0
0
0
0
0
0
2.
Revenue
0
1500
1500
1500
1500
1500
1500
1500
3.
Variable cost
0
6750
6750
6750
6750
6750
6750
6750
4.
Fixed cost
0
4000
4000
4000
4000
4000
4000
4000
5.
Depreciation
0
2500
1875
1406
1055
791
593
445
6.
EBIT
0
1750
2375
2844
3195
3459
3659
3805
7.
Profit after tax
0
1137.5
1544
1849
2077
2248
2377
2473
8.
Net Cash flow
-10,000
3638
3419
3255
3132
3039
2970
2918
The NPV at 12% and IRR are 4829 and 26.8% respectively. Since N P V > 0, the project can be undertaken. Before the financial manager takes a decision, he may like to know whether the NPV changes if one forecasts goes wrong. Sensitivity analysis can be conducted with regard to volume, price, costs, etc. This is done by obtaining pessimistic and optimistic estimates of the underlying variables. Let us assume the following pessimistic and optimistic values for volume, price and costs
S/No.
Variable
Pessimistic
Expected
Optimistic
1.
Volume(Units)
750
1000
1250
2.
Unit selling price($)
12.75
15.00
16.50
3.
Unit variable cost($)
7.425
6.75
6.075
4.
Annual fixed costs($)
4800
4000
3200
The following table shows their re-calculated NPV where other variables don’t change. Sensitivity analysis under different assumptions 84
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S/No.
Variable
Pessimistic
Expected
Optimistic
1.
Volume(Units)
-1289
4829
10948
2.
Unit selling price($)
-184.5
4829
9729
3.
Unit variable cost($)
2827
4829
6832
4.
Annual fixed costs($)
2456
4829
7203
The above table shows the project’s NPV when each variable is set to it’s pessimistic and optimistic values. The most critical variable is sales volume followed by unit selling price. If the volume decreases by 25% ie 750, the NPV becomes -1289 and NPV is -184.5 if the selling price decreases by 15%
Discounted cash-flow break even analysis Sensitivity analysis is a variation of break even analysis. The break even point is a point where the project ceases to be profitable. ie when N P V < 0. Consider the expected values of the variables in the previous example. We can measure after-tax cash flows as follows:N CF = (Revenue − Expenses)(1 − T ) + (T ax × Depreciation)
(26)
The first part of the right hand expression is annuity ie [1000(15 − 6.75) − 4000](1 − 0.35) = 2763 and it’s present value at 12% discount rate is 2763 × 4.5638 = 12610 Next, we workout the present value of the series of depreciation tax shield as the discounted rate of the product’s tax rate and depreciation charges. ie T ax.Depreciation =
2500 × 0.35 1875 × 0.35 1406 × 0.35 1055 × 0.35 791 × 0.35 593 × 0.35 445 × 0.35 + + + + + + (1.12) (1.12)2 (1.12)3 (1.12)4 (1.12)5 (1.12)6 (1.12)7
⇒ PV =
7 X t=0
D = 2222 (1.12)t
The following expression is used to determine the break even point ie NPV
= P V of N CF − Investment = 0 = [((V (15 − 6.75) − 4000)0.65 × 4.5638 + 2222] − 10000 = 0 85
(27)
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Where V is the sales volume 4.5638 is the PV factor of a 7 year annuity (at 12%) 2222 is the PV of the series of depreciation tax shield. From equation 5, we solve for V, ie 24.473V − 11866 + 2222 − 10000 = 0 ⇒ V = 803. The project will start loosing money if the sales volume goes below 803 units ie if the sales decline by more than 19.7%. Similarly, we can work out the lowest selling price. Given other assumptions. how low the units selling price can go before the project’s NPV becomes negative? We solve the following equation ie N P V = [(1000(p − 6.75) − 4000)]0.65 × 4.5638 + 2222] − 10000 = 0; ⇒ P = 13.37 The selling price declined by more than 10.867% ⇒The sales volume variable is more critical than the selling price variable.
Advantages of sensitivity analysis 1. It enables the decision maker to identify the variables affecting the cash flow forecasts 2. It indicates the critical variables for which additional information may be obtained. 3. It helps to expose inappropriate forecasts hence guiding the decision maker to concentrate on relevant variables.
Disadvantages of sensitivity analysis 1. No clear cut results are provided since pessimistic and optimistic values are subjective thus the range of values suggested may be inconsistent. 2. It fails to focus on the inter-relationship between variables.
Statistical techniques for handling risk Statistical techniques can also be used to incorporate risk in capital budgeting These techniques are analytical in nature and enable the decision maker to make decisions under uncertainty. The 86
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
probability concept is fundamental to the use of the risk analysis techniques. The most critical information for capital budgeting decision is a forecast of future cash flows. So we make use of probability distributions to get the best estimate of such forecasts
Expected Net present value After getting the probability distribution of the future cash flows, we get the expected NPV ie EN P V =
n X EN CFt t=0
(1 + k)t
where EN CFt = N CFjt .Pjt
Example The following are possible net cash flows for projects X and Y and their associated probabilities over 1 year period. Both projects have a discount rate of 10%. Calculate the ENPV for each project. Which project is preferred? Assume initial cost of $5000 for each project.
Solution Project X
Project Y
Possible event
Cash flows($)
Probability
Cash flows($)
Probability
A
4000
0.1
12000
0.1
B
5000
0.2
10000
0.15
C
6000
0.4
8000
0.5
D
7000
0.2
6000
0.15
E
8000
0.1
4000
O.1
Project X: EN P V =
n X EN CFt t=0
(1 + k)t
but EN CFt = N CFjt .Pjt = 4000(0.1) + 5000(0.2) + 6000(0.4) + 7000(0.2) + 8000(0.1) = $6000 ⇒ EN P V = −$5000 +
$6000 1.1
= $454.5
Project Y: EN P V =
n X EN CFt t=0
(1 + k)t
but EN CFt = N CFjt .Pjt = 12000(0.1) + 10000(0.15) + 8000(0.5) + 6000(0.15) + 4000(0.1) = $8000 87
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
⇒ EN P V = −$5000 +
$8000 1.1
= $2272.7
⇒Project Y is preferred than project X. This can be extended to more than 1 year.
Standard deviation: An absolute measure of risk We have established that computation of ENPV incorporates the risk explicitly but we still have another better technique which will be obtained by measuring the variational dispersion of cash flows. The dispersion of cash flows indicates the degree of risk. The standard deviation of Net cash flow for each period can be expressed as; v uX u n σt = t (N CFjt − EN CFt )2 .Pjt j=1
where σt =standard deviation of net cash flow in period t Pjt =Probability associated with each cash flow Under the assumption of independence of cash flows over time, we use standard deviations of various periods to develop a measure of risk for the project. ie v uX u n σ=t j=1
σt2 (1 + kf )2t
where σ =standard deviation of probability distribution of possible net cash flows, σt2 =Variance of each period. When the cash flows are dependent over time, the standard deviation will be larger than under the assumption of independent cash flows. The greater the degree of correlation between the cash flows the larger will be the standard deviation. However, the expected Net Present Value (ENPV) remains unchanged irrespective of dependence or independence of cash flows. In case of perfect correlation, the formula for standard deviation is given by v uX u n σ=t j=1
σt (1 + kf )t
In case of moderate correlation (ie the cash are neither independent nor perfectly correlated over time) the above methods can’t be applied to measure the risk. Such a problem can be handled by making use of conditional probabilities and decision trees.
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Example Consider a project which costs $8, 000 at t = 0 and 3 years. The following table presents the cash flows and probability information for the project. Assuming a risk free discount rate of 10%. Calculate the expected value and the standard deviation of the probability distribution of possible NPV’s under the assumption of perfect correlation. Assuming a normal distribution, what is the probability of 0 or less, of 3500 or more. Compare the standard deviation with the one under assumption of independent of cash flows over time.
Year 1
Year 2
Year 3
Cash flows($)
Probability
Cash flows($)
Probability
Cash flows($)
Probability
6000
0.1
3000
0.15
6000
0.25
5000
0.4
4000
0.5
5000
0.2
4000
0.3
5000
0.25
4000
0.35
3000
0.2
6000
0.1
3000
0.2
Solution Work it out as an assignment
Coefficient of variation: A relative measure of risk The coefficient of variation is a relative measure of risk. We can define coefficient of variation as the standard deviation of probability distribution divided by its expected value ie CV =
standard deviation of distribution Expected value
It is a useful measure of risk when comparing the projects under the following situations 1. Same standard deviations but different expected values 2. Different standard deviations but same expected values 3. Different standard deviations and different expected values
Illustration Suppose project X has an expected value of $6000 and standard deviation of $1095.45 and Project Y has an expected value of $8000 and a standard deviation of $2097.62 89
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Project Y may be preferred because of larger expected NPV, but it is more risky as compared to project X. ie CV of project X =
1095.45 = 0.1826 6000
CV of project Y =
2097.62 = 0.2622 8000
The acceptance of project X and Y will depend on the investor’s attitude towards risk. He would prefer project Y if he is ready to assume more risk in order to obtain the higher expected monetary value. On the other hand, if he has a great aversion to risk, he would accept project X for it is less risky.
Decision trees Decision trees are useful tools for helping you to choose between several courses of action. They provide a highly effective structure within which you can explore options, and investigate the possible outcomes of choosing those options. They also help you to form a balanced picture of the risks and rewards associated with each possible course of action. This makes them particularly useful for choosing between different strategies, projects or investment opportunities, particulary when your resources are limited.
How to use Decision trees You start a decision tree with a decision that you need to make. Draw a small square to represent this on the left side of a large piece of paper, half way down the page. From this box draw out lines towards the right for each possible solution, and write a short description of the solution along the line. Keep the lines apart as far as possible so that you can expand your thoughts. At the end of each line, consider the results. If the result of taking that decision is uncertain, draw a small circle. If the rasult is another decision that you need to make, draw another square. Squares represent decisions, and circles represent uncertain outcomes. Write the decision or factor above the square or circle. If you have completed the solution at the end of the line, just leave it blank. Starting from the decision squares on your diagram, drwa out lines representing the options that you could select. From the circles draw lines representing possible outcomes. Again make a brief note on the line saying what it means. Keep on doing this until you have drawn out as many of the possible outcomes and decisions as you can see leading from the original decisions.
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Illustration A company has the options now of building a full-size plant or a small plant that can be expanded later. The decision depends primarily on future demands for the product the plant will manufacture. The construction of a full-size plant can be justfied economically if the level of demand is high. Otherwise, it may be advisable to construct a small plant now and then decide in two years whether it should be expanded. The multistage decision problem arises here because if the company decides to build a small plant now, a future decision must be made in two years regarding expansion. In other words, the decision process involves two stages: a decision now regarding the size of the plant, and a decision two years from now regarding expansion (assuming that it is decided to construct a small plant now).
The diagram above summarises the problem as a decision tree. It is assumed that the demand can be either high or low. The decision tree has two types of nodes: a square represents a decision point and a circle stands for a chance event. Thus, starting with node 1 (a decision point), we must make a decision regarding the size of the plant. Node 2 is a chance event from which two branches representing low and high demand emanate depending on the conditions of the market. These conditions will be represented by associating probabilities with each branch. Node 3 is also chance event from which two branches representing high and low demands emanate.
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Logically, the company will consider possible future expansion of the small plant only if the demand over the first two years turns out to be high. This is the reason node 4 represents a decision point with its two emanating branches representing the “expansion" and “no expansion" decisions. Again, nodes 5 and 6 are chance events, and the branches emanating fron each represent high and low demands. The data for the decision tree must include
1. The probabilities associated with the branches emanating from the chance events and 2. The revenues associated with different alternatives of the problem.
Suppose that the company is interested in studying the problem over a 10-year period. A market survey indicates that the probabilities of having high and low demands over the next 10 years are 0.75 and 0.25, respectively. The immediate construction of a large plant will cost $5 million and a small plant will cost only $1 million. The expansion of the small plant 2 years from now is estimated to cost $4.2 million. Estimates of annual income for each of the alternatives are given as follows.
1. Full-size plant and high (low) demand will yield $1, 000, 000 ($300, 000) annually. 2. Small plant and low demand will yield $200, 000 annually. 3. Small plant and high demand will yield $250, 000 for each of the 10 years. 4. Expanded small plant with high (low) demand will yield $900, 000 $(200, 000) annually. 5. Small plant with no expansion and high demand in the first two years followed by low demand will yield $200000 in each of the remaining 8 years. As a Financial Engineer assist the company in making the right decision
These data are summarised in the decision tree above. We are now ready to evaluate the alternatives. The final decision must tell us what to do at both of the decision nodes 1 and 4
E{net prof it|expansion} = (900000 × 0.75 + 200000 × 0.25) × 8 − 4200000 = $1, 600, 000 E{net prof it|no expansion} = (250000 × 0.75 + 200000 × 0.25) × 8 = $1,900,000 Thus, at node 4, the decision calls for no expansion, and the associated expected net profit is $1, 900, 000. We can now replace all the branches emanating from node 4 by a single branch with an expected net profit of $1, 900, 000, representing the net profit for the last 8 years. We now make 92
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
stage 1 computations corresponding to node 1 as follows: E{net prof it|large plant} = (1000000 × 0.75 + 300000 × 0.25) × 10 − 5000000 = $3,250,000 E{net prof it|small plant} = (1900000 + 500000) × 0.75 + 2000000 × 0.25 × −1000000 = $1, 300, 000 Thus the optimal decision at node 1 is to build a full-size plant now. Making this decision now obviously eliminates the need for considering the alternatives at node 4.
Exercise Suppose that demand during the last 8 years can be high, medium, or low, with probabilities 0.7, 0.2 and 0.1, respectively. The annual incomes are as follows. 1. Expanded small plant with high, medium, and low demands will yield annual income of $900, 000, $600, 000 and $200, 000. 2. Non expanded small plant with high, medium, and low demand will yield annual income of $400, 000, $280, 000 and $150, 000. i) Determine the optional decision at node 4. ii) Assist the company in making the right decision.
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CHAPTER TEN REVISION PAPERS Paper One DATE: PAPER 1
TIME: 2 HOURS
INSTRUCTIONS: Attempt question ONE and any other TWO questions . QUESTION ONE (30 MARKS) (a) Briefly explain what you understand about the following terms as used in Risk theory. (i) Risk matrix
[1 mark]
(ii) Sensitivity analysis
[1 mark]
(iii) Probability of ruin
[1 mark]
(iv) Risk process
[1 mark]
(v) Hypothesis testing
[1 mark]
(vi) Risk adjusted interest rate
[1 mark]
(b) An insurer with net worth 100 has accepted (and collected the premiums for) a risk X with the following probability distribution
P r(X = x) =
3 , f or x = 0 4 1 , f or x = 51 4
(i) What is the maximum a mount it should pay another insurer to accept 100% of this loss? Assume the first insurer’s utility function of wealth is µ(ω) = logω. (ii) An insurer with wealth 650 and the same utility function, µ(ω) = logω, is considering accepting the above risk. What is the maximum a mount this insurer would accept as a premium to cover 100% of the loss?
[5 marks]
(c) Given that Tj denotes the time when claim j occurs, such that T1 < T2 < T3 < .... Then the interarrival time between claim j −1 and j defined as Wj = Tj −Tj−1 for j ≥ 2 is exponentially distributed random variable with parameter β = 3 and also claim amount is exponentially distributed with parameter θ = 1. Using premium rate of 3.8, calculate the adjustment coefficient.
[4 marks]
(d) Derive the probability density function for the number of claims, given that the occurrence of a claim is a rare event that happens at the rate of T annually. Where T ∼ Gamma(λ, θ). 94
[6 marks]
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
(e) Define and proof Jensen’s inequalities. Hence, explain why the inequalities are appropriate for describing risk averse and risk lover preferences.
[8 marks]
QUESTION TWO (20 MARKS)
(a)
(i) Enumerate features of individual risk models for short term.
[2 marks]
(ii) Given that there are three fixed insured units in a portfolio and that each unit generates exponentially distributed claims with parameter i for i = 1, 2, 3. Using convolution and mgf technique determine the pdf and cdf of the aggregate claim amount for this portfolio. Assume that the claims are independent.
[11 marks]
(b) Discuss the relevancy of Panjer’s recursion theorem in risk theory. What does this theorem result to when the number of claims follows a negative binomial distribution with parameters (r, p).
[6 marks]
QUESTION THREE (20 MARKS)
(a) The loss random variable X has a pdf given by f (x) =
1 100 ,
0,
0 < x < 100 elsewhere
(i) Calculate E[X] and var(X).
[3 marks]
(ii) Consider a proportional policy where l(x) = kx,
0 < k < 1 and a stop-loss policy
where
ld (x) =
0 , f or x < d x − d, f or x ≥ d
Determine k and d such that the pure premium in each case is p = 12.5.
[4 marks]
(b) Determine first non-central, second central moments and moment generating function of the total claims if claims frequency is a random event. Hence workout the average and variation of the aggregate claims given that the random losses have Geometric(0.8) and number of claims have P oisson(Λ) distribution respectively. Where Λ ∼ Gamma(5, 0.5).
[13 marks]
QUESTION FOUR (20 MARKS)
(a) Given a Poisson claim number process with exponential(1) distributed interarrival times and exponential(2) distributed claim amounts. What is the probability of ruin? When the initial capital is 100 with fixed premium if 5.
[3 marks] 95
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(b) Consider the insurance portfolio that will produce 0, 1, 2 or 3 claims in a fixed time period with probability of 0.2, 0.2, 0.3 and 0.3 respectively. An individual claim will be of amount 1, 2 or 3 with probabilities 0.3, 0.6, 0.1 respectively. Find the variance of the aggregate claims for this portfolio. Sketch the probability density function of the total amount claimed on a Cartesian plane.
[17 marks]
Paper Two DATE: PAPER 2
TIME: 2 HOURS
INSTRUCTIONS: Attempt question ONE and any other TWO questions . QUESTION ONE (30 MARKS) (a) 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? [6 marks] (b) Define and state the uses of risk matrix in risk management? [3 marks] (c) Using moment generating function technique determine the distribution of Poisson claim frequency, with a random parameter that follows Gamma(2,0.05) distribution. [5 marks] (d) The initial outlay of a product is K Sh. 100,000 and it can generate cash inflow of K Sh. 40,000 after first year, K Sh. 30,000 after second year, K Sh. 50,000 after third year and K Sh. 20,000 after fourth year, before adjustments. Assuming that the rate of discount is 10% and risk free rate is 8%. Determine whether this project is profitable using; (i) Profitability index, (ii) IRR, and (iii) Certainty equivalent. [8 marks] (e) A loss random variable has an exponential distribution with mean θ. A random sample of 100 values is observed with the following results. 58 of the values lie between 0 and 100 32 of the the values lie between 100 and 200 10 of the values are in excess of 200 Find the maximum likelihood estimate for θ. 96
[4 marks]
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
(f) Losses from a portfolio of policies are believed to follow an exponential distribution with parameter θ, which is unknown. A reinsurance arrangement is in place, under which the reinsurer pays the amount of each loss in excess of K Sh. 800. The last four payments made by the reinsurer in respect of these policies were: 760, 954, 1201, 1158. Assuming that a suitable prior distribution for θ is an inverse gamma prior with parameters 2 and 2800 determine the Bayesian estimate of θ under 0/1 loss. [4 marks]
QUESTION TWO (20 MARKS)
(a) XYZ Ltd produces handmade children’s cots. Forecast sales for the next year are 10,000 cots at a selling price of K Sh. 60 per cot. Material costs are K Sh. 18 per cot and direct labour costs of K Sh. 22 per cot. Forecast fixed costs for the year are K Sh. 80,000. You are required to: (i) Calculate the percent change in each of the following variables which result in breakeven. - Unit selling price - Unit material costs - Unit labour costs - Volume i.e the margin of safety - Fixed costs [12 marks] (ii) Calculate and discuss the effect on profit of a 10% chance in each variable in the profit calculation. [3 marks] (b) Suppose that the initial capital of an insurer is 10 units, the claim interarrival time and the claim losses are exponentially distributed with parameters 0.5 and 0.2 respectively. Assuming that the relative safety loading factor is 0.5. (i) What is the upper bound probability of ruin for this insurer? (ii) What will be the probability of ruin in case the initial capital is zero? [5 marks]
QUESTION THREE (20 MARKS)
(a) Suppose a risk X has a compound Poisson(10) distributed with claim severity having a Gamma(1, 0.01) distribution and a decision maker with an exponential utility function with 97
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
insured risk aversion of 0.005, has wealth of K Sh. 30,000. Approximate the maximum premium this individual will be willing to pay for a complete insurance cover. [6 marks] (b) The net cash flows of two products X and Y, each with initial cost of K Sh. 110,000 and discount rate of 10% are distributed as follows. X = IB and Y = W1 + W2 + ... + WN , where I ∼ Binomial(20, 0.5), B ∼ Gamma(10, 0.001), W ∼ Exponential(0.002) and N ∼ N egative binomial(200, 0.5) (i) Using EN P V criteria, which project is preferred? (ii) With reasons, using coefficient of variation as a relative measure of risk which project is preferred? [14 marks]
QUESTION FOUR (20 MARKS)
(a) Discuss how the law of total probability can be used in determining the probability distributions of the open risk model and the closed risk model. Given that all claim severity are independent continuous random variables. [5 marks] (b) Suppose the loss random variable S is such that S = X1 + X2 + ... + XN , where Xi0 s are independent non negative integer valued random variables. With i = 1, 2, ..., N and P r(N =n) P r(N =n−1)
= a + nb .
Determine the real values of the constants a and b given that N ∼ N egative binomial(20, 0.5). Hence given that P r(X = 1, 2, 3) = 0.25, 0.5, 0.25, work out P r(S ≤ 2). [15 marks]
Paper Three DATE: PAPER 3
TIME: 2 HOURS
INSTRUCTIONS: Attempt question ONE and any other TWO questions . QUESTION ONE (30 MARKS)
(a) Define and explain what you understand by the terms risk process and probability of ruin. Hence state factors that may reduce or increase ruin probability
[6 marks]
(b) Given that for a property of 1 million shillings the insured and insurer utility function is µ(ω) = −e−3ω , show that both insured maximum premium and insurer minimum premium 98
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
will not depend on wealth. Is this utility function favourable to a risk averse decision maker?
[6 marks]
(c) Determine the pdf, mean and variance of the aggregate claims from three insurance units with exponential(1000) distribution
[6 marks]
(d) Find mean, variance and mgf of the aggregate gamma(3,2000) distributed claims from N poisson(10) distributed number of claims.
[7 marks]
(e) A compound poisson distribution has a parameter λ = 5 with P r(x1 ) = 15 , P r(x2 ) = 53 , P r(x3 ) = 1 5
. Find the copmound distribution of the sum of the claims X1 , X2 , X3
[5 marks]
QUESTION TWO (20 MARKS)
(a) Suppose that the random claims X1 , X2 , X3 from three portfolios are distributed as X1 ∼ poisson(3), X2 ∼ geometric(0.6), and X3 ∼ binomial(10, 0.5). Using convolution technique, find pdf of S = X1 + X2 + X3 for S = 0, 1, 2, 3, 4, 5, 6. Hence sketch the cdf of this aggregate claims. [14 marks] (b) X and Y are independent, normally distributed loss random variables. Given a random sample of 100 observations of each random variable with X = 957, Y = 975, standard deviation of X as 102.6 and standard deviation of Y as 67.5. Perform a suitable hypothesis test to decide whether the loss random variables have equal means. Use a 5% level of significance.
[6 marks]
QUESTION THREE (20 MARKS)
(a) Consider the insurance portfolio that will produce geometric(0.7) distributed claims, with individual claim amount of 1, 2, 3, or 4 with probabilities 0.3,0.4,0.1 and 0.2 respectively. Find pdf and cdf of the aggregate claims. Determine also the variance of these aggregate claims.
[14 marks]
(b) What will the Panjer’s recursion theorem reduce to in the case of the following three distributions. poison(λ), negative binomial(r,p) and binomial with parameters m and p. [6 marks]
QUESTION FOUR (20 MARKS)
(a) Suppose that data 5000, 16000, 24000, 13000, 26000, 40000, 50000, 43000, 75000, 82000, 60000, 32000, 18000, 20000 and 81000 in Kenyan shillings constitutes random claims expected on a certain risk for claims expected on a certain risk for any financial year. If the worth of the unit insured is K Sh. 95000, what maximum premium is the insured 99
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
likely to pay using an exponential utility function. Furthermore, if the initial capital of the insurer is K Sh. 300000. Using Lundbergs exponential bound theorem, find the upper bound ruin probability of the insurer.
[12 marks]
(b) Assume that a decision maker’s current wealth is ω = 10000 with utility function µ(ω) such that µ(0) = −1 and µ(w) = 0 . The decision maker would be willing to pay premium of up to P + for complete insurance, if he/she faces the risk of losing amount X at probability of 0.4 and retain the current wealth with probability 0.6. Determine the four values on the decision maker’s utility of wealth function µ(ω), given values of X are 10000, 6000, 3300 and 1700. Hence state whether the decision maker is risk averse, risk lover or risk neutral individual.
[8 marks]
Paper Four DATE: PAPER 4
TIME: 2 HOURS
INSTRUCTIONS: Attempt question ONE and any other TWO questions . QUESTION ONE (30 MARKS) (a) A decision maker has utility function u(x) =
√
x for x ≥ 0. He is given the choice between
two random amounts X and Y, in exchange for his entire present capital w. The probability distributions of X and Y are given by P r(X = 400) = P r(X = 900) = 0.5 and P r(Y = 100) = 1 − P r(Y = 1600) = 0.6. Explain which random amount between X and Y he would prefer. Can you think of utility functions with which he would prefer the rejected amount. [4 marks] (b) Suppose that the initial capital of an insurer is 10 units, the claim interarrival time and the claim losses are exponentially distributed with parameters 0.5 and 0.2 respectively. Assuming that the relative safety loading factor is 0.5. (i) What is the upper bound probability of ruin for this insurer? (ii) What will be the probability of ruin in case the initial capital is zero? [5 marks] (c) Briefly discuss what you understand about risk matrix in risk management. [4 marks] (d) A project will cost $50, 000. It’s stream of earnings before depreciation, interest and taxes during the first year through 6 years is expected to be $13, 000, $12, 000, $14, 000, $18, 000, $20, 000 and $22, 000. Assuming a 30% tax rate annually and a depreciation of 10%, compute the Accounting rate of return. 100
[9 marks]
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
(e) Determine probability for observing compound binomial distributed claims with parameters 10 and 0.5 greater or equal to 2. Given that claim severity are distributed as P r(X = 1, 2, 3) = 0.2, 0.5, 0.3 [8 marks] QUESTION TWO (20 MARKS) (a) Ken India Insurance company has categorized its policies into five homogeneous groups. During the year 2013, the groups produced claims distributed as follows. Homogeneous Group 1 2 3 4 5
Number of Policies 10 5 6 9 4
Claims distribution Normally distributed with parameters mean=100 and variance=20 Chi-Square distributed with 90 degrees of feedom Negative Binomial distributed with parameters 50 and 0.2 Exponential distributed with parameter 0.01 Uniform distributed over the interval (0,200)
Determine the measures of central tentancy and spread for the aggregate claims realized by Ken India insurance. Also find P r(Aggregate claims are atleast 3500) [12 marks] (b) A financial Engineer invested in a project that required initial capital worth KSh. 600. The project’s life is four years with expected cash in flows distributed as follows: 1 Exponential( 100 ) for the 1st year; Chi-Square with 100 degrees of freedom for the second
year; Uniform with parameters 0 and 300 for the 3rd year and Gamma(2,0.005) for the 4th year. Using ENPV and payback method discuss whether the project was worth undertaking, given that risk free discount rate is 10%. [8 marks] QUESTION THREE (20 MARKS) (a) Management accounting information for decision making is mainly composed of estimates simply because it is based upon predictions concerning the future. One solution to the consequent lack of precision is to use sensitivity analysis. (i) Describe Sensitivity analysis and illustrate how it is applied to a specific management accounting problem. (ii) How can Sensitivity analysis aid managers and what drawback and limitations does it have? 101
[7 marks]
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
(b) Determine coefficient of variation for aggregate claims with claim frequency probability equal to 0.1, 0.2,0.4,0.2,0.1 for frequencies 0,1,2,3,4 respectively and claim severity probabilities given as 0.2,0.5,0.3 for claims observed equal to 0,1,2 respectively. [13 marks]
QUESTION FOUR (20 MARKS)
(a) The data observed from three losses namely X,Y,Z by a financial Engineer for the last five months are as shown below. Monthly Observations Loss X Y Z
1 10 57 24
2 22 10 50
3 40 5 20
4 25 15 32
5 30 20 36
6 35 25 30
Test whether the losses have the same mean at 5% level of significance. [10 marks] (b) For a certain risk process, it is given that θ = 0.4 and p(x) = 21 (3e−3x + 7e−7x ). Which of the numbers 0,1 and 6 are roots of the adjustment coefficient equation 1 + (1 + θ)µ1 R = MX (R)? Which one is the real adjustment coefficient? One of the four expressions below is the ruin probability for this process; determine which expression is the correct one. (i) Ψ(µ) =
24 −µ 35 e
(ii) Ψ(µ) =
+
24 −µ 35 e
1 −6µ 35 e
+
11 −6µ 35 e
(iii) Ψ(µ) =
24 −0.5µ 35 e
+
1 −6.5µ 35 e
(iv) Ψ(µ) =
24 −0.5µ 35 e
+
1 −6.5µ 35 e
[10 marks]
Paper Five DATE: PAPER 5
TIME: 2 HOURS
INSTRUCTIONS: Attempt question ONE and any other TWO questions . QUESTION ONE (30 MARKS) 102
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
(a) What is risk theory?
[1 mark]
(b) The decision maker has a utility function µ(ω) = −e−αω for ω > 0, 0 < α < 21 , and is faced with a random loss that has a chi-square distribution with n degrees of freedom. Determine the maximum insurance premium the decision maker will pay, and prove that this premium is greater than n.
[8 marks]
(c) Let Xi for i = 1, 2 be independent and identically distributed with the pdf
P r(X = x) =
1 √ e−0.25(x−3) , 2 π
0,
f or − ∞ < x < ∞
otherwise
(i) Using moment generating function, find probability density function of S = X1 + X2 . [5 marks] (ii) What is the expectation and variance of S? (d)
[2 marks]
(i) Let S denote the number of people crossing a certain intersection by car in a given hour. How would you model S as a random sum?
[2 marks]
(ii) Suppose that the number of passengers in each car has a binomial distribution with parameters n and p. Find E[S], variance of S and mgf of S.
[5 marks]
(e) Suppose that Lundberg’s exponential bound theorem for the ruin probability is given as Ψ(µ) ≤ e−Rµ . Where R is the adjustment coefficient and µ = 10 is the initial surplus. Determine the upper bound probability of ruin when (i) The claims have a compound poisson with parameter λ = 0.2. (assume the safety loading factor is 4%).
[4 marks]
(ii) The claims have a compound negative binomial distribution with parameters r = 5 and p = 0.5. (assume the safety loading factor is 4%).
[3 marks]
QUESTION TWO (20 MARKS)
(a) Given that S has a compound Poisson distribution with λ = 2 and p(x) = 0.1x for x=1,2,3,4. Calculate probabilities that aggregate claims equal 0, 1, 2, ..., 16.
Hence determine E[S],
var(S). [15 marks] (b) The probability of a fire in a certain structure in a given period is 0.02. If a fire occurs, the damage to structure is uniformly distributed over the interval (0, a) where a is its total value. Calculate the mean and variance of fire damage to the structure within the time period.
[5 marks] 103
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QUESTION THREE (20 MARKS) (a)
(i) Determine the adjustment coefficient if the claim amount distribution is exponential with parameter β > 0. [4 marks] (ii) Calculate the probability of ruin in the case that the claim amount distribution is exponential with parameter β > 0.
(b)
[10 marks]
(i) An insurer with net worth 500 has accepted a risk X with the following probability distribution. P r(X = 0) = 0.4 and P r(X = 60) = 0.6 what is the maximum amount it should pay another insurer to accept 100% of this loss? Assume the first insurer’s utility function of wealth is µ(ω) = logω. [3 marks] (ii) An insurer, with wealth 1250 and the same utility function, µ(ω) = logω is considering accepting the above risk. What is the minimum amount this insurer would accept as a premium to cover 100% of the loss? [3 marks] QUESTION FOUR (20 MARKS)
(a) Compute for x=0,1,2,3,4,5 and P r(S ≥ 2) for the following three compound distributions, each with claim amount distribution given by
P r(X = x) =
0.2, f or x = 1 0.3, f or x = 2 0.5, f or x = 3 0, otherwise
(a) Poisson with parameter λ = 5 (b) Negative binomial with parameters r = 6 and p = 0.25 (c) Binomial with parameters m = 10 and p = 0.75 [15 marks] (b) The distribution of X, the claim severity from a portfolio of non-life insurance policies has a Pareto distribution with mean KSh. 350 and standard deviation of KSh. 452. If the insurer arranges excess of loss reinsurance with retention of KSh. 1200. Calculate the probability that a claim will involve the insurer. [5 marks] 104
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Paper Six DATE: PAPER 6
TIME: 2 HOURS
INSTRUCTIONS: Attempt question ONE and any other TWO questions . QUESTION ONE (30 MARKS)
(a) Briefly, what do you understand about the following terms as used in risk theory. (i) Risk process (ii) Utility function (iii) Jensen’s inequalities
[3 marks]
(b) Suppose X ∼ unif orm(0, 3) and Y ∼ unif orm(0, 4). Assuming X and Y are independent random variables. Find pdf of S = X + Y , using convolution technique. [6 marks] (c) The value a decision maker attaches to his/her wealth (ω) is given by a quadratic utility function with aversion coefficient (α) equal to 0.02 and ω < 50. Find the maximum insurance premium that the decision maker will pay for complete insurance. If he/she is willing to retain his/her wealth of amount ω = 20 units at a probability of 0.6 and suffer a loss of amount c = 10 units at aprobability of 0.4. Assuming that c ≤ ω < 50. [5 marks] (d) X and Y are independent, normally distributed loss random variables. You collect a random sample of 200 observations of each random variable and find that: X = 863, Y = 882, SX = 96.4 and SY = 58.5. Carry out a suitable hypothesis test to decide whether the loss random variables have equal means at 1% level of significance.
[4 marks]
(e) Given that the number of claims, N are geometric distributed with parameter p, for 0 < p < 1 and claims, X are exponentially distributed with parameter α = 2. What is the moment generating function of total amount of claims.
[5 marks]
(f) A portfolio consists of a total of 400 independent risks. On each risk, no more than one event can occur each year, and the probability of an event occurring is 0.035. When such an event occurs, the number of claims N has the following distribution. P (N = x) = 0.55(0.45)x−1 , x = 1, 2, ... Determine the mean and variance of the distribution of the number of claims which arise from this portfolio in one year.
[4 marks]
(g) An insurance company has an excess of loss reinsurance contract with retension of KSh. 40, 000. Over the last year, the insurer paid the following claims in shillings, 12, 000, 2, 700, 398 and 105
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
43, 567. In addition, the insurer paid the amount of KSh. 40, 000 on 7 claims with the excess being paid by the re-insurer. The insurer believes that distribution of gross claim amounts is exponential with mean λ. Calculate the maximum likelihood estimate of λ based on the above information.
[8 marks]
QUESTION TWO (20 MARKS)
(a) A decision maker’s utility function is given by µ(ω) =
√
ω. The decision maker has wealth
of ω = 12 and faces a random loss X with a uniform distribution on (0, 12). Show that this utility function satisfies requirements of a risk averse individual. Hence, What is the maximum amount this decision maker will pay for complete insurance against the loss?. [8 marks] (b) Consider an insurance portfolio that will produce zero, one, two, or three claims in a fixed time period with probabilities 0.1, 0.3, 0.4, and 0.2, respectively. An individual claim will be of amount 1,2, or, 3 with probabilities 0.5, 0.4, and 0.1, respectively. Calculate the variance of aggregate claims and P r(aggregate claims ≤ 6).
[12 marks]
QUESTION THREE (20 MARKS)
Suppose in a 1-year term life insurance paying an extra benefit in case of accidental death. If death occurs and is accidental, the benefit is 50, 000. For other causes of death, the benefit amount is 25, 000. In addition given that for the age, health, and occupation of aspecific individual, the probability of an accidental death within the year is 0.0009, while the probability of a non-accidental is 0.0025.
(a) Determine the conditional distribution of the total claim amount B, incurred during the period. Given that atleast a claim occurs. (b) Given that acertain automobile insurance provides collision coverage above a 250 deductable up to a maximum claim of 2, 000. If for a particular individual the probability of one claim in a period is 0.25 and the chance of more than one claim is 0. Assume that total claim’s (i.e. B) distribution from (a) above, has a probability mass at the maximum claim size of 2, 000, assume this probability mass is 0.2. Furthermore, assume that claim amounts between 0 and 2, 000 can be modeled by a continuous distribution with probability distribution funch 2 i x tion of total claims equal to 0 for x ≤ 0 and 0.8 1 − 1 − 2000 for 0 < x < 2, 000 and 1 for x ≥ 2, 000. Where X is the aggregate random claims in one period. (i) Sketch distribution function of the random amount claimed, X given a claim occurs. 106
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
(ii) Find the distribution of X. (iii) Derive an expression for determining the mean and variance of X, hence find their actual values. Without using the method of moments.
[20 marks]
QUESTION FOUR (20 MARKS)
(a) State and prove Panjer’s recursion theorem. Hence find the distribution of S = X1 + X2 + X3 + X4 given that S is a compound poison with parameter 10 and P r(X = 1, 2, 3, 4) = 0.2, 0.2, 0.5, 0.1.
[10 marks]
(b) Using Lundberg’s exponential bound theorem for the ruin probability with initial surplus as 10 units. Determine the upper bound probabilities of ruin when; (i) The claims are compound Poisson distributed with parameter λ = 0.2 and safety loading factor of 4%.
[5 marks]
(ii) The claims are compound negative binomial distributed with parameters r = 5, p = 0.5 and safety loading factor of 4%. [5 marks]
Paper Seven DATE: PAPER 7
TIME: 2 HOURS
INSTRUCTIONS: Attempt question ONE and any other TWO questions . QUESTION ONE (30 MARKS)
(a) Explain what you understand by the following terms (i) RAD (ii) Risk Theory [4 marks] (b) Briefly discuss the application of certainty equivalent and risk adjusted discount rate as techniques of handling risk in capital budgeting. (c) Suppose project M has an expected value of $6000 and a standard deviation of $1205 and project N has an expected value of $9800 and a standard deviation of $2297. Depending on the investor’s attitude towards risk, with reasons which project is preferable?
107
[5 marks]
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
(d) An investment project will cost $62, 000 initially and is expected to generate cash flows in 5 years i.e. $22, 000, $19, 000, $13, 500, $11, 600 and $10, 400 respectively. What is the project’s internal rate of return for the five years. [6 marks] (e) In a bayesian investigation in the financial industry, a particular claim rate θ is to be estimated. To obtain a suitable prior distribution, an expert is consulted and he suggest that θ will have a mean of 0.24 and a standard deviation of 0.06. It is required to construct a prior gamma distribution to fit the expert’s information. Find the parameters for the appropriate gamma distribution.
[3 marks]
(f) What is the return on portfolio and risk of portfolio for a two-asset portfolio comprising the following two assets if the correlation of their returns is 0.7? Asset A
Asset B
Expected return
15%
30%
Standard deviation of expected returns
5%
30%
$40000
$60000
Amount Invested
[5 marks]
QUESTION TWO (20 MARKS) (a) A project will cost $40, 000. It’s stream of earnings before depreciation, interest and taxes (EBDIT) during the first year through 5 years is expected to be $13, 000, $14, 500, $15, 000, $19, 000 and $24, 000. Assuming a 25% tax rate and a depreciation of $6, 000 on straight line basis, compute the Accounting rate of return.
[8 marks]
(b) Consider a project which costs $15, 000 at t = 0 and 3 years. The following table presents the cash flows and probability information for the project. Assuming a risk free discount rate of 18%. Calculate the expected value and the standard deviation of the probability distribution of possible NPV’s under the assumption of dependence. Assuming a normal distribution, what is the probability of 0 or less, of 3000 or more. Compare the standard deviation with the one under assumption of independent cash flows over time.
108
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Year 1
Year 2
Year 3
Cash flows($)
Probability
Cash flows($)
Probability
Cash flows($)
Probability
8500
0.2
3050
0.25
5600
0.2
7500
0.3
6050
0.4
6600
0.25
5500
0.25
4050
0.15
3600
0.45
3500
0.25
7050
0.2
4600
0.1 [12 marks]
QUESTION THREE (20 MARKS) A financial manager of cloth processing company is considering the installation of a plant costing $20, 000 to increase it’s processing capacity. The expected values of the underlying variables are as follows.
1.
Investment($)
→
20,000
2.
Sales Volume(units)
→
1,500
3.
Unit selling price($)
→
15
4.
Unit variable cost($)
→
8
5.
Annual fixed costs($)
→
5,000
6.
Depreciation(%)
→
20
7.
Corporate tax rate(%)
→
30
8.
Discount rate(%)
→
15
(a) Compute the project’s after-tax cash flows over it’s expected life of 7 years. Hence, based on N.P.V criteria, should the financial manager undertake the project or not. [6 marks] (b) Before the financial manager takes a decision, he/she may like to know whether the N.P.V changes if one of the forecast goes wrong. Perform sensitivity analysis with regard to volume, price and costs. Hence determine the most critical variable. Use the following pessimistic and optimistic values. [7 marks]
109
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
S/No.
Variable
Pessimistic
Expected
Optimistic
1.
Volume(Units)
90% of 1,500
1,500
110% of 1,500
2.
Unit selling price($)
90% of 15.00
15.00
110% of 15.00
3.
Unit variable cost($)
110% of 8.00
8.00
90% of 8.00
4.
Annual fixed costs($)
110% of 5,000
5,000
90% of 5,000
(c) Using the most critical variable obtained above and the forecasts under the expected assumptions, determine the variable’s break even point if sales volume is assumed to be a random variable.
[7 marks]
QUESTION FOUR (20 MARKS) (a) A company has the options now of building a full-size plant or a small plant that can be expanded later. The decision depends primarily on future demands for the product the plant will manufacture. The construction of a full-size plant can be justified economically if the level of demand is high. Otherwise, it may be advisable to construct a small plant now and then decide in two years whether it should be expanded. Suppose that the company is interested in studying the problem over a 12-year period. A market survey indicates that the probabilities of having high and low demands over the next 12 years are 0.55 and 0.45, respectively. The immediate construction of a large a large plant will cost $4.5 million and a small plant will cost only $0.95 millions. The expansion of the small plant 2 years from now is estimated to cost $4 million. Estimates of annual income for each of the alternatives are given as follows. (i) Full-size plant and high (low) demand will yield $1, 000, 500 ($300, 900) annually. (ii) Small plant and low demand will yield $200, 000 annually. (iii) Small plant and high demand will yield $255, 000 for each of the 12 years. (iv) Expanded small plant with high (low) demand will yield $907, 000 $(199, 500) annually. (v) Small plant with no expansion and high demand in the first two years followed by low demand will yield $200000 in each of the remaining 10 years. As a Financial Engineer assist the company in making the right decision.
[12 marks]
(b) Suppose that demand during the last 10 years can be high, medium, or low, with probabilities 0.5, 0.3 and 0.2, respectively. The annual incomes are as follows. (i) Expanded small plant with high, medium, and low demands will yield annual income of $900, 000, $600, 000 and $250, 000. (ii) Non expanded small plant with high, medium, and low demand will yield annual income of $400, 000, $290, 000 and $160, 000. 110
RISK THEORY NOTES PREPARED BY V.S.ANDIKA
Determine the optional decision if the company decides to build small plant due to high demand.
[8 marks]
111
