RMSC 2001 Introduction to Risk Management

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1 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 2012 by Wang Weiyin 1

2 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. 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) (b) Poisson Distribution p k = eλ λ k, k = 0, 1, 2... k! E[N] = λ, V ar[n] = λ 2

3 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. 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 f(x) = 1 e (x µ) π E[X] = µ, V ar[x] = 2 (b) Gamma Distribution 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 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) 3

4 (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 What is the distribution for claim amounts in 1994? (A) No longer a normal distribution (B) Normal with µ = 1000 and = (C) Normal with µ = 1000 and = (D) Normal with µ = 1050 and = (E) Normal with µ = 1050 and = 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) (c) Maximum Payment (Limit Loss) d (x d)f(x)dx = Y = αx, α (0, 1) Y = X u = E[X u] = E[αX] = αe[x] u d { X, if X < u u, if X u xf(x)dx + u[1 F (u)] [1 F (x)]dx 4

5 E[X] = E[X u] + E[(X d) + ] Example 4 Losses follow a Pareto distribution with α = 2, θ = Calculate the expected payment per loss on a coverage with ordinary deductible 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: Number of Losses Probability Loss sized follow a Pareto distribution with parameters α = 5 and θ = 1200 and are independent of loss counts and of each other. Calculate the variance of aggregate losses. 5

6 Example 6 A claim severity distribution is exponential with mean 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 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: 3000 F (x) = 1 ( 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 Determine λ. 6

RMSC 2001 Introduction to Risk Management

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