# Aggregate Loss Models

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1 Aggregate Loss Models Chapter 9 Stat Loss Models Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 1 / 22

2 Objectives Objectives Individual risk model Collective risk model Computing the aggregate loss models Approximate methods Effect of policy modifications Chapter 9 (sections ; exclude and examples 9.9 and 9.11) Assignment: read section 9.7 Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 2 / 22

3 Individual risk model Individual risk model Consider a portfolio of n insurance policies. Denote the loss, for a fixed period, for each policy i by X i, for i = 1,..., n. Assume these lossses are independent and (possibly) identically distributed. The aggregate loss, S, is defined by the sum of these losses: S = X 1 + X X n. It is possible that here a policy does not incur a loss so that each X i has a mixed distribution with a probability mass at zero. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 3 / 22

4 Individual risk model Alternative representation Alternative representation Assume that the losses are also identically distributed say as X. Then we can write X as the product of a Bernoulli I and a positive (continuous) random variable Y : X = IY I = 1 indicates there is a claim, otherwise I = 0 means no claim. Let Pr(I = 1) = q and hence Pr(X = 0) = 1 q. In addition, assume E(Y ) = µ Y and Var(Y ) = σ 2 Y. Typically it is assumed that I and Y are independent so that E(X) = E(I)E(Y ) = qµ Y and Var(X) = q(1 q)µ 2 Y + qσ 2 Y. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 4 / 22

5 Individual risk model Mean and variance Mean and variance in the individual risk model The mean of the aggregate loss can thus be written as and its variance is E(S) = nqµ Y Var(S) = nq(1 q)µ 2 Y + nqσ 2 Y. Example: Consider a portfolio on 1,000 insurance policies where each policy has a probability of a claim of When a claim occurs, the amount of claim has a Pareto distribution with parameters α = 3 and θ = 100. Calculate the mean and variance of the aggregate loss. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 5 / 22

6 Individual risk model Illustrative examples Illustrative examples Example 1: An insurable event has a 10% probability of occurring and when it occurs, the amount of the loss is exactly 1,000. Market research has indicated that consumers will pay at most 115 for insuring this event. How many policies must a company sell in order to have a 95% chance of making money (ignoring expenses)? Assume Normal approximation. Example 2: Exercise 9.20 of textbook. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 6 / 22

7 Individual risk model Exact distribution using convolution Exact distribution of S using convolution Consider the case of S = X 1 + X 2. The density of S can be computed using convolution as: f 2 (s) = f S (s) = s 0 f 1 (s y)f 2 (y)dy = s 0 f 2 (s y)f 1 (y)dy To extend to n dimension, do it recursively. Let f (n 1) (s) be the (n 1)-th convolution so that f n (s) = f S (s) = = s 0 s 0 f (n 1) (s y)f n (y)dy f n (s y)f (n 1) (y)dy Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 7 / 22

8 Individual risk model Mixed random variable The case of the mixed random variable Often, X i has a probability mass at zero so that its density function is given by { qi, for x = 0, f i (x) = (1 q i )f Y (x), for x > 0, where f Y ( ) is a legitimate density function of a positive continuous random variable. Here, q i is interpreted as the probability that the loss is zero. In this case, use the cumulative distribution function: F 2 (s) = F S (s) = s 0 F 1 (s y)df 2 (y) = To extend to n dimension, do it recursively. s 0 F 2 (s y)df 1 (y) Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 8 / 22

9 Individual risk model The case of the Exponential The case of the Exponential distribution Consider the case where X i has the density function expressed as { 0.10, for x = 0, f i (x) = 0.90f Y (x), for x > 0, where Y is an Exponential with mean 1, for i = 1, 2. Derive expressions for the density and distribution functions of the sum S = X 1 + X 2. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 9 / 22

10 Individual risk model Approximation by CLT Approximating the individual risk model For large n, according to the Central Limit Theorem, S can be approximated with a Normal distribution. Use the results on slides p. 5 to compute the mean and the variance of the aggregate loss S in the individual risk model. Then, probabilities can be computed using Normal as follows: [ ] S E(S) Pr(S s) = Pr S E(S) Var(S) Var(S) Pr = Φ [ ] Z S E(S) Var(S) ( ) S E(S) Var(S) Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 10 / 22

11 Individual risk model Illustrative example Illustrative example 1 An insurer has a portfolio consisting of 25 one-year life insurance policies grouped as follows: Insured amount b k Number of policies n k The probability of dying within one year is q k = 0.01 for each insured, and the policies are independent. The insurer sets up an initial capital of \$1 to cover its future obligations. Using Normal approximation, calculate the probability that the insurer will be able to meet its financial obligation. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 11 / 22

12 Collective risk model Collective risk model Let X i be the claim payment made for the ith policyholder and let N be the random number of claims. The insurer s aggregate loss is S = X X N = N X i. i=1 This is called the Collective Risk Model. If N, X 1, X 2,... are independent and the individual claims X i are i.i.d., then S has a compound distribution. N: frequency of claims; X: the severity of claims. Central question is finding the probability distribution of S. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 12 / 22

13 Collective risk model Properties Properties of the collective risk model X is called the individual claim and assume has moments denoted by µ k = E(X k ). Mean of S: E(S) = E(X)E(N) = µ 1 E(N) Variance of S: Var(S) = E(N)Var(X) + Var(N)µ 2 1 MGF of S: M S (t) = M N [log M X (t)] PGF of S: P S (t) = P N [P X (t)] CDF of S: Pr(S s) = n=0 Pr(S s N = n)pr(n = n) Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 13 / 22

14 Collective risk model Exponential/Geometric Example - Exponential/Geometric Suppose X i Exp(1) and N Geometric(p). MGF of N: M N (t) = Mean of S: E(S) = q/p p 1 qe t ; MGF of X: M X(t) = 1 1 t Variance of S: Var(S) = q/p + q/p 2 = (q/p)(1 + 1/p) p MGF of S: M S (t) = 1 qm X (t) = p 1 q/(1 t) = p + q p p t Observe that the mgf of S can be written as a weighted average: p mgf of r.v. 0 + q mgf of Exp(p) r.v. Thus, F S (s) = p + q(1 e ps ) = 1 qe ps for s > 0. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 14 / 22

15 Collective risk model Poisson number of claims Poisson number of claims If N Poisson(λ), so that λ is the average number of claims, then the resulting distribution of S is called a Compound Poisson. MGF of N: M N (t) = exp [ λ ( e t 1 )] It can then be shown that: Mean of S: E(S) = λe(x) = λµ 1 Variance of S: Var(S) = λe(x 2 ) = λµ 2 MGF of S: M S (t) = exp [λ (M X (t) 1)] Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 15 / 22

16 Collective risk model Illustrative example Illustrative example 2 Suppose S has a compound Poisson distribution with λ = 0.8 and individual claim amount distribution x Pr(X = x) Compute f S (s) = Pr(S = s) for s = 0, 1,..., 6. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 16 / 22

17 Collective risk model Panjer s recursion formula Panjer s recursion formula Let aggregate claims S have a compound distribution with integer-valued non-negative claims with pdf p(x) for x = 0, 1, 2,.... Let the probability of n claims satisfies the recursion relation p k = (a + b n )p k 1 for k = 1, 2,... for some real a and b. Recall this is the (a, b, 1) class of distribution. Then, for the probability of total claims s, we have f S (s) = 1 1 ap(0) For s = 0, we have f S (0) = s h=1 ( a + bh ) p(h)f S (s h). s { Pr(N = 0), if p(0) = 0 M N [log p(0)], if p(0) > 0. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 17 / 22

18 Collective risk model The case of Compound Poisson Panjer s recursion for Compound Poisson This is the case where the number of claims is Poisson. Starting value: With Poisson(λ): f S (0) = { Pr(N = 0), if p(0) = 0 M N [log p(0)], if p(0) > 0. f S (s) = 1 s s λhp(h)f S (s h). h=1 Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 18 / 22

19 Collective risk model Illustrative example Back to illustrative example 2 Same as previous example, but now we solve using Panjer s recursion. Effectively, the recursion formula boils down to f S (s) = 1 s [0.2f S(s 1) + 0.6f S (s 2) + 0.9f S (s 3)]. [Why??] Initial value is: f S (0) = Pr(N = 0) = e 0.8 = Successive values satisfy: f S (1) = 0.2f S (0) = 0.2e 0.8 = f S (2) = 1 2 [0.2f S(1) + 0.6f S (0)] = 0.32e 0.8 = f S (3) = 1 3 [0.2f S(2) + 0.6f S (1) + 0.9f S (0)] = e 0.8 = and f S (4) = , f S (5) = , and f S (6) = [calculated by hand - verify!] Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 19 / 22

20 Collective risk model Approximating the compound distribution Use CLT to approximate the compound distribution If the average number of claims is large enough, we may use the Normal approximation to estimate the distribution of S. All you need are the mean and variance of the aggregate loss [slides p. 13] Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 20 / 22

21 Collective risk model Closed under convolution Compound Poisson is closed under convolution Theorem 9.7 of book. If S 1, S 2,..., S m are independent compound Poisson random variables with Poisson parameters λ i and claim distributions (CDF) F i, for i = 1, 2,..., m, then S = S 1 + S S m also has a compound Poisson distribution with parameter λ = λ 1 + λ λ m and individual claim (severity) distribution F (x) = m i=1 λ i λ F i(x). You add the number of claim parameters, and the individual claim distribution is a weighted sum. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 21 / 22

22 Collective risk model Illustrative example Illustrative example 3 You are given S = S 1 + S 2, where S 1 and S 2 are independent and have compound Poisson distributions with λ 1 = 3 and λ 2 = 2 and individual claim amount distributions: x p 1 (x) p 2 (x) Determine the mean and variance of the individual claim amount for S. Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 22 / 22

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