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Modern Actuarial Risk Theory

Rob Kaas, August 4, 2003

Note: this pdf-file contains the slides such as they are used in my classes at the

University of Amsterdam. They are made available to teachers using Modern

Actuarial Risk Theory as is, without any warranty or pretense. Users are urged to

share any criticism with the author of this text, at R.Kaas@UvA.nl, to the benefit of

all users.

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Summary Ch. 1

The expected utility hypothesis explains why insureds pay so much

more than the net premium that insurance companies can exist; also it

gives a way to compute sensible premiums.

Various classes of different utility functions exist that each have

different desirable properties.

E.g., quadratic utility requires only mean and variance of the risk,

exponential utility leads to a premium independent of the (possibly

random or unknown) current capital.

Stop-loss reinsurance (more general, insurance with a deductible), is

optimal because the retained loss has the lowest variance when the net

premiums are equal. In fact, it also leads to the best resulting expected

utility. In other situations, other types of coverage may be optimal.

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Left as exercises:

1: As above for d [2,3]; give numerical answers for B = 405 andd = 2 /d = 3.

2: Optimize over d [2,3]; also for B =4043: When d = 2, compute the survival probability using NP

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E If the claims are exponential() then v,t:

Pr[U(T) >y U(T0)= v, T= t] =

Pr[X> v+y X> v] = e(v+y) / ev = ey

U(T) T

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The process of a driver moving through this scale can be described by a

Markov system. After the effects of the initial scaling have worn off,

we get the steady state distribution; it is an eigenvector of the transition

matrix.

One may assess the efficiency of the bonus-malus system by using the

elasticity of the average premium paid in the steady state to the mean

number of claims (Loimaranta).

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Choice of model: multiplicative model tariff

Example:

Usage(1) Private basis: 100% 1 = 1Usage(2) Business +15%

2= 1.15

Weight(1) 900 kg basis: 100% 1 = 1Weight(2) 1000 kg +10% 2 =1.1Weight(3) 1100 kg +20% 3 =1.2

Sijkl = i j kl + "purely random error" log Sijkl log + log i +...+log l (log-linear model)

Estimation: determine values , i, j, ... such thatestimated values i j kl observations sijkl

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Completing the triangle by predictions :Xij

ij

1 2 3 4 5

1 A A A B E2 A A A B3 C C C X

344 D D D* X445 F

The element just outside the observed data is computed as followsX34

(here (C) denotes the total of all elements C in the table):

= = =X34 34

3(

1+

2+

3)

4(

1+

2)

(1

+ 2) (

1+

2+

3)

(C) (B)

(A)

since (C) and (B) are marginal totals, and (A) = R1 + R2 K4 K5.

Also, = = .X44(D) (B)

(A)

(D) {(B)+ X34

}

{(A)+(C)}

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10.5 Incomplete information

For a risk Y we assume that it is known that

E[Y] = Pr[0Y b] = 1Which risk Y leads to the extremal stop-loss premium in a given

priority d?

Minimal risk: Maximal risk:

X Z: Pr[Z=b] = b = 1Pr[Z=0](concentration) (dispersion)

See book Fig. 10.3, page 244.

These same extremal risks lead to extreme ruin probabilities and

extreme compound Poisson stop-loss premiums.

Both lower and upper bound have a degenerate claim size distribution

(adjusting the Poisson parameter for the upper bound from to /b).See also exer 10.5.7.

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