RMSC 2001 Introduction to Risk Management Tutorial 4 (2011/12) 1 February 20, 2012 Outline: 1. Failure Time 2. Loss Frequency 3. Loss Severity 4. Aggregate Claim ==================================================== 1 Failure Time 1. Hazard rate (force of mortality) - instantaneous failure rate - µ(x) = f(x) 1 F (x) 2. Models: (a) Exponential Distribution (constant hazard rate) f(t) = 1 λ e t λ F (x) = 1 e t λ t > 0 E[T ] = λ, V ar[t ] = λ 2 µ(t) = 1 λ memoryless property: P r(x > x + t X > x) = P r(x > t) constant hazard rate (b) Uniform Distribution (increasing hazard rate) f(t) = 1 θ F (t) = t θ t [0, θ] E[T ] = θ θ2, V ar[t ] = 2 12 µ(t) = 1 θ t If failure time follows uniform distribution with parameter θ and one is age x, remaining lifetime follows uniform distribution with parameter θ x. 1 All rights reserved @ 2012 by Wang Weiyin 1
3. Variance of insurance benefit and annuity E[X] = i x i p i V ar[x] = E[X 2 ] E[X] 2 (a) Insurance benefit E[X] = t 0 a 1 (t + 1)c(t + 1)P r(t < T t + 1) E[X 2 ] = t 0 [a 1 (t + 1)c(t + 1)] 2 P r(t < T t + 1) (b) Annuity g(t ) = T 1 t=0 a 1 (t)c(t) E[X] = t 0 g(t + 1)P r(t < T t + 1) E[X 2 ] = t 0 (g(t + 1))2 P r(t < T t + 1) Example 1 A special annuity for a 70 years old person will pay 10 at the end of every tenth year, starting from today, if he/she is alive at that point. Your are given: 10 p 70 = 0.8, 20 p 70 = 0.5, 30 p 70 = 0, i = 0.04 Calculate the variance of the present value of the payments on this annuity. Since 30 p 70 = 0, there are three possibilities: no payment: 10 q 70 = 0.2 one payment: 10 p 70 20 p 70 = 0.3 two payments: 20 p 70 30 p 70 = 0.5 E[Y ] = 0.3(10v 10 ) + 0.5(10(v 10 + v 20 )) = 3 1 + 5( + 1 ) = 7.68645 1.04 10 1.04 10 1.04 20 E[Y 2 ] = 0.3(10v 10 ) 2 + 0.5(10(v 10 + v 20 )) 2 = 30 1 + 50( + 1 ) 2 = 77.75727 1.04 20 1.04 10 1.04 20 V ar(y ) = 18.67579 2 Loss Frequency 1. Models: - used to model the number of payments - discrete random variable - probability function: p k = P r(n = k), k = 0, 1, 2... (a) Binomial Distribution p k = ( ) m p k (1 p) m k k E[N] = mp, V ar[n] = mp(1 p) 2
(b) Poisson Distribution p k = eλ λ k, k = 0, 1, 2... k! E[N] = λ, V ar[n] = λ Suppose there are n independent Poisson variables N i P oi(λ i ), then N = n i=1 N i P oi( n i=1 λ i) Example 2 Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins/minute. The denominations are randomly distributed - 60% of the coins are worth 1-20% of the coins are worth 5-20% of the coins are worth 10 Calculate the conditional expected value of the coins Tom found during his one-hour walk today, given that the among the coins his found exactly ten were worth 5 each. N P oi(0.5 60 = 30) N 1 P oi(18), N 2 P oi(6), N 3 P oi(6) E[X] = E[1 N 1 + 5 N 2 + 10 N 3 N 2 = 10] = 18 + 50 + 60 = 128 Remark: If we observe from the dataset that mean and variance are almost the same Poisson Distribution variance is smaller than mean Binomial Distribution variance is larger than mean Negative Binomial Distribution 3 Loss Severity 1. Models: - used to model the claim size - non-negative random variable (a) Normal Distribution (b) Gamma Distribution f(x) = 1 e (x µ)2 2 2 2π E[X] = µ, V ar[x] = 2 f(x) = βα Γ(α) xα 1 e βx, x 0 Γ(k) = (k 1)!, k = 1, 2, 3,... E[X] = α β, V ar[x] = α β 2 3
when α is an integer, gamma(α, β) is the sum of α independent exponential random variables with parameter β when α = 1, reduced to exp(β) (c) Pareto Distribution E[X] = f(x) = Tail Heaviness: Pareto > Gamma > Normal Example 3 αθ α (x + θ) ( α + 1), x 0 θ F (x) = 1 ( x + θ )α θ α 1, α > 1, V ar[x] = αθ 2 (α 1) 2 (α 2), α > 2 In 1993, the claim amounts for a certain line of the business were normally distributed with mean µ = 1000 and variance 2 = 10, 000. Inflation of 5% impacted all claims uniformly from 1993 to 1994. What is the distribution for claim amounts in 1994? (A) No longer a normal distribution (B) Normal with µ = 1000 and = 102.5 (C) Normal with µ = 1000 and = 105.0 (D) Normal with µ = 1050 and = 105.0 (E) Normal with µ = 1050 and = 102.5 If X is normal, then ax+b is normal as well. In particular, 1.05X is normal. E[1.05X] = 1.05E[X] = 1050, V ar[1.05x] = 1.05 2 V ar[x] = 10250, (1.05X) = 105.0 For any distribution, multiplying the distribution by a constant multiplies the mean and standard deviation by the same constant. 2. Coverage Modifications: (a) Deductible { 0, if X d < 0 Y = (X d) + = X d, if X d 0 E[(X d) + ] = (b) Percentage of Loss (Coinsurance) d (x d)f(x)dx = Y = αx, α (0, 1) E[αX] = αe[x] d [1 F (x)]dx 4
(c) Maximum Payment (Limit Loss) Y = X u = E[X u] = u E[X] = E[X u] + E[(X d) + ] { X, if X < u u, if X u xf(x)dx + u[1 F (u)] Example 4 Losses follow a Pareto distribution with α = 2, θ = 2000. Calculate the expected payment per loss on a coverage with ordinary deductible 500. For a Pareto with α > 1, E[X] E[X d] = θ α 1 ( θ θ + d )α 1 In our case, E[X] E[X 500] = 2000( 2000 2500 ) = 1600 4 Aggregate Claim 1. S = - N: modeled by loss frequency - X: modeled by loss severity X is are iid random variables - X and N are independent N i=1 X i (a) N fixed (b) N random E(S) = NE(X) V ar(s) = NV ar(x) E(S) = E(N)E(X) Compound Variance formula V ar(s) = (E(X)) 2 V ar(n) + E(N)V ar(x) Compound Variance formula for poisson: V ar(s) = λe(x 2 ) Example 5 The number of losses on an automobile comprehensive coverage has the following distribution: Loss sized follow a Pareto distribution with parameters α = 5 and θ = 1200 and are independent of loss counts and of each other. 5
Number of Losses Probability 0 0.4 1 0.3 2 0.2 3 0.1 Calculate the variance of aggregate losses. E(N) = 0.3 + 0.2(2) + 0.1(3) = 1 V ar(n) = 0.4(0 1) 2 + 0.2(2 1) 2 + 0.1(3 1) 2 = 1 E(X) = 1200/4 = 300 V ar(x) = 2(12002 ) 4 3 300 2 = 150, 000 V ar(s) = (1)(150, 000) + (1)(300 2 ) = 240, 000 Example 6 A claim severity distribution is exponential with mean 1000. An insurance company will pay the amount of each claim in excess of a deductible of 100. Calculate the variance of the amount paid by the insurance company for one claim, including the possibility that the amount paid is 0. You can treat it as a compound distribution with a Bernoulli frequency. The frequency of claims over 100 is e 100/1000 = 0.904837, and the severity of claims above 100 is exponential with mean 1000 since it is memoryless. By the compound variance formula: V ar(s) = 0.904837(1000 2 ) + (0.904837)(1 0.904837)(1000 2 ) = 990, 944 2. Normal Approximation and Premium Q: Suppose X N(µ, 2 ), we want Pr(Total Claim < Total Premium)= 1- α. How to determine the premium? Ans: P r(x x) = 1 α P r( X µ x µ ) = 1 α Φ( X µ ) = 1 α X µ = Φ 1 (1 α) x = µ + Φ 1 (1 α) Total Premium = E(S) + Φ 1 (1 α) (V ar(s)) Single Premium = Total Premium / no. of contracts Example 7 The number of claims for an insurance coverage has a Poisson distribution with mean λ. Claim size has the following distribution: 6
3000 F (x) = 1 ( 3000 + x )3, x 0 Claim counts and claim sizes are independent. Using the normal approximation, the probability that aggregate losses will be greater than 4000 is 0.2743. Determine λ. Claim size X is Perato with parameters α = 3, θ = 3000, so E[X] = 1500 and E[X 2 ] = 3000 2 E[S] = 1500λ, V ar[s] = (3000 2 )λ P r( X E(S) > 4000 E(S) ) = 1 Φ( 4000 E(S) ) = 0.02743 V ar(s) V ar(s) V ar(s) Hence: λ = 1.299 7