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ATW316 Semester Test 3 [57 marks] [2 hours] 26 April 2012 Question 1 [16] Consider aggregate claims over a period of 1 year, π, on a portfolio of general insurance policies: π + π1 + π2 + β― + ππ The number of claims each year, π, has a Poisson distribution with mean 12. π1 , π2 , β¦ are assumed to be random variables, independent of each other and independent of π, with the following distribution: π(π₯) = 0.01π β0.01π₯ 0 < π₯ < π 200 π(π = π 200) = π β2 The insurance company calculates premiums using a premium loading of 40% and is considering entering into one of the following re-insurance arrangements: (A) No reinsurance. (B) Individual excess of loss insurance with retention 100 with a reinsurance company that calculates premiums using a premium loading of 55%. (C) Proportional reinsurance with retention 75% with a reinsurance company that calculates premiums using a premium loading of 45%. i) Find the insurance companyβs expected profit under (A). [3] πΈ(ππππππ‘) = πΈ(ππππππ’π ππππππ β πΆπππππ ππ’π‘ππ) = πΈ[1.4πΈ(π) β πΈ(π)] = 1.4πΈ(π) β πΈ(π) πΈ(π) = πΈ(π)πΈ(π) = 12πΈ(π) 200 = 12 [β« π₯(0.01)π β0.01π₯ ππ₯ + 200π β2 ] 0 200 = 12 [βπ₯π β0.01π₯ |200 + β« π β0.01π₯ ππ₯ + 200π β2 ] 0 0 = 12 [β200π β0.01(200) β 1 β0.01(200) 1 π + + 200π β2 ] 0.01 0.01 = 1037.59766 πΈ(ππππππ‘) = 0.4(1037.59766) = 415.039064 ii) Find the premium charged by reinsurance company under (B), assuming that the reinsurer knows nothing of claims below the retention level. [4] ππππππ’π ππ πππππ π’πππππ πππππππ¦ = (1 + π)πΈ(ππ ) 1.55πΈ(π)πΈ(π) = π(π > 100) 200 1.55(12) = [ β« (π₯ β 100)π(π₯)ππ₯ + 100π(π = 200)] π(π > 100) 100 200 = 1.55(12) [ β« (π₯ β 100)0.01π β0.01π₯ ππ₯ + 100π β2 ] π(π > 100) 100 200 1.55(12) 200 200 = [βπ₯π β0.01π₯ |100 + β« π β0.01π₯ ππ₯ β 100(βπ β0.01π₯ )100 + 100π β2 ] π(π > 100) 100 = 1.55(12) 1 β0.01(200) 1 β0.01(100) [β200π β0.01(200) + 100π β0.01(100) + (β π + π ) π(π > 100) 0.01 0.01 β 100(βπ β0.01(200) + π β0.01(100) ) + 100π β2 ] = 1.55(12) 1 β2 1 β1 [β200π β2 + 100π β1 β π + π + 100(π β2 ) β 100π β1 β2 0.01 0.01 +π ] 200 [(βπ β0.01π₯ )100 + 100π β2 ] = 1.55(12) (23.25441579) [0.367879441] = 1175.744239 iii) Find the probability that the insurer makes a profit of less than 500 under (C) using a Normal approximation. [9] π[ππππππ‘ < 500] = π[1.4πΈ(π) β 1.45πΈ(ππ ) β ππΌ < 500] = π[ππΌ > 1.4πΈ(π) β 1.45(0.25)πΈ(π) β 500] πΈ(π) = 12πΈ(π) = 1037.59766 π[ππΌ > 576.5075723] = π [π > 576.5075723 β πΈ(ππΌ ) βπππ(ππΌ ) ] πΈ(ππΌ ) = 0.75(1037.59766) = 778.198245 πππ(ππΌ ) = πππ(0.75π) = 0.752 πππ(π) = 0.752 (12)πΈ(π 2 ) 200 = 0.752 (12) [β« π₯ 2 (0.01)π β0.01π₯ ππ₯ + 2002 π β2 ] 0 200 (0.01)3 3β1 β0.01π₯ π€(3) = 0.752 (12) [ β« π₯ π ππ₯ + 2002 π β2 ] 0.012 π€(3) 0 π€(3) {π[π < 200]} + 2002 π β2 ] π€βπππ π~πΊππππ(3,0.01) 0.012 π€(3) 2 = 0.752 (12) [ {π[π2(3) < 4]} + 2002 π β2 ] 0.012 π€(3) {0.3233} + 2002 π β2 ] = 0.752 (12) [ 0.012 = 80,186.02647 = 0.752 (12) [ π [π > 576.5075723 β 778.198245 β80,186.02647 ] = π[π < 0.712256576] = 0.76115 Question 2 [10] An insurance firm models hailstorm claims according to the following assumptions: The number of hailstorms in a year follows a Poisson (4) distribution. The number of claims arising from the π π‘β hailstorm is modelled as a Poisson(ππ ) distribution. The parameters ππ are independently and identically distributed random variables with mean 4 and standard deviation 2. The individual claim amounts from each storm follow a Pareto(5, ππ ) distribution. The mean claim amounts ππ = 0.25ππ are assumed to be independently and identically distributed with mean π 3000 and standard deviation π 600. ππ and ππ are independently distributed. Use this information to answer the following questions: i) Calculate the mean and the variance of the annual aggregate claims outgo from all storms. [8] πΈ(π) = πΈ[πΈ(π|π©π , π¬π )] = πΈ[πΈ(πΎ)πΈ(ππ |π©π , π¬π )] = πΈ[πΈ(πΎ)πΈ(ππ |π©π )πΈ(πππ |π¬π )] = πΈ[4π©π π¬π ] = 4(4)(3,000) = 48,000 πππ(π) = πΈ(πππ(π|π©π , π¬π )) + πππ(πΈ(π|π©π , π¬π )) = πΈ[πΈ(πΎ)πΈ(ππ2 |π©π , π¬π )] + πππ[πΈ(πΎ)πΈ(ππ |π©π , π¬π )] = πΈ[4{πππ(ππ |π©π , π¬π ) + πΈ(ππ |π©π , π¬π )2 }] + πππ[4π©π π¬π ] 2 = πΈ[4{πΈ(ππ |π©π )πΈ(πππ |π¬π ) + (π©π π¬π )2 }] + 42 πππ[π©π π¬π ] 2 = 4πΈ [{π©π [πππ(πππ |π¬π ) + πΈ(πππ |π¬π ) ] + π©π2 π¬2π }] + 42 {πΈ(π©π2 π¬2π ) β πΈ(π©π π¬π )2 } 5 2 = 4πΈ [{π©π [ π¬2π + π¬2π ] + π©π2 π¬2π }] + 42 {πΈ(π©π2 )πΈ(π¬2π ) β (4(3,000)) } 3 8 2 = 4πΈ [{π©π [ π¬2π ] + π©π2 π¬2π }] + 42 {{22 + 42 }{6002 + 30002 } β (4(3,000)) } 3 4(8) = [πΈ(π©π )πΈ[π¬2π ]] + 4[πΈ(π©π2 )πΈ(π¬2π )] + 691,200,000 3 4(8) [(4){6002 + 30002 }] + 4[{22 + 42 }{6002 + 30002 }] + 691,200,000 3 = 399,360,000 + 748,800,000 + 691,200,000 = 1,839,360,000 = ii) By using the normal approximation, approximate the probability that the annual aggregate claims outgo from all storms will exceed π 120 000. [2] π[π > 120,000] = π [π > 120,000 β 48,000 β1,839,360,000 = 1 β 0.95352 = 0.04648 ] Question 3 [5] An insurance company has insured a fleet of cars for the last four years. For year π (π = 1, β¦ ,4), let ππ and ππ be the total amount claimed and the number of cars in the fleet, respectively. Let ππ = ππ /ππ be the average amount claimed per car in year π. Assume that the distribution of ππ depends on a risk parameter π and that the conditions of Empirical Bayes Credibility Theory Model 2 are satisfied. The company has insured 10 similar fleets over the last four years. Using the data from these years, π, πΈ(π 2 (π)) and π(π(π)) are estimated to be 62.8, 106.32 and 5.8 respectively. Calculate next yearβs credibility premium for a fleet of cars with claims over the last four years given below, if the fleet will have 16 cars next year. Year 1 2 3 4 Total amount claimed 1,000 1,200 1,500 1,400 Number of cars 15 16 18 15 Also explain how and why the credibility factor would be affected if the estimate of π(π(π)) increases and comment on the effect on the credibility premium. π = 62.8 πΈ(π 2 (π)) = 106.32 π(π(π)) = 5.8 πΜ = 64 πΜ = 65.625 64 = 0.777349639 106.32 64 + 5.8 π(65.625) + (1 β π)(62.8) = 64.99601273 π= If π(π(π)) increases then π will increase. If π increases the credibility premium will also increase, since the weight placed on data from the risk itself is higher and the mean of the data from the risk itself is higher than the overall mean. Question 4 [26] Using the formula for the adjustment coefficient π πππ (π ) = π + ππ 4.1 Derive a simple upper and lower bound for π . In both cases state clearly what the conditions are for these bounds to exist. Claims on a portfolio of insurance policies arrive as a Poisson process with parameter 100. Individual claim amounts follow a normal distribution with mean 30 and variance 52 . The insurer calculates premiums using a premium loading of 20% and has initial surplus of 100. 4.2 Define carefully the ruin probabilities π(100), π(100,1) and π1 (100,1). [3] π(100) = π[π(π‘) < 0 π€βπππ 0 < π‘ < β] π(100,1) = π[π(π‘) < 0 π€βπππ 0 < π‘ β€ 1] π1 (100,1) = π[π(1) < 0] 4.3 Show that for this portfolio the value of the adjustment coefficient π is 0.011 correct to 3 decimal places. [5] 1 + (1 + π)π1 π = ππ (π ) 1 2 )π 2 1 + 1.2(30)π = π 30π +2(5 πΏ. π». π = 1 + 1.2(30)(0.011) = 1.396 1 2 )(0.011)2 π . π». π = π 30(0.011)+2(5 = 1.39307356 πΏ. π». π β π . π». π 4.5 Calculate an upper bound for π(100) and an estimate of π1 (100,1). [5] π(100) β€ π βπ (100) = 0.332871083 π1 (100,1) = π[π(1) < 0] = π[100 + 1.2(30)(100) β π(1) < 0] = π[π(1) > 3700] 3700 β πΈ[π(1)] = π [π > ] βπππ[π(1)] πΈ[π(1)] = πΈ[π(1)]πΈ(π) = 100(30) = 3000 πππ[π(1)] = πΈ[π(1)]πΈ(π 2 ) = 100(52 + 302 ) = 92,500 π[π > 2.301585822] = 1 β 0.98928 = 0.01072 Suppose that the insurer now takes out a proportional reinsurance contract with retention level πΌ. The premium charged by the reinsurer has a loading of 30%. 4.6 State the conditions that must hold in respect of the retention level πΌ to ensure that the net premium income of the insurer is positive and exceeds the expected aggregate claims per unit time [2], and hence: {Counts 4} ππππ‘ > 0 (1 + π)πΈ(π) β (1 + π)πΈ(ππ ) > 0 1.2(30)π > 1.3(1 β πΌ)(30)π πΌ > 0.076923076 And ππππ‘ > πΈ(ππΌ ) 1.2(100)(30) β 1.3(100)(30)(1 β πΌ) > (30)(100)πΌ πΌ> 1 3 4.7 Maximise the adjustment coefficient R under this proportional reinsurance contract and state the upper bound for the insurerβs ultimate probability of ruin. [5] {Only counts 3} 1 From 2.4 we can see that πΌ must lie in the interval, (3 , 1]. At some value for πΌ in this interval, it is reasonable to suppose that π will be maximised. However substituting the appropriate values into the equation π + ππππ‘ π = πππ (π ) gives: 100 + (1.2(30)100 β 1.3(30)(100)(1 β πΌ))π = 100ππ (π ) Now 1 2 )(πΌπ )2 ππ (π ) = πΈ(π π π ) = πΈ(π π (πΌπ) ) = ππ (πΌπ ) = π 30πΌπ +2(5 So 1 2 2 100 + (β200 + 3900πΌ)π = π 30πΌπ +2(5 )(πΌπ ) This must now be written in the form π = π(πΌ) and then the derivative must be taken with respect to πΌ to find the value of πΌ for which π is maximised, i.e. ππ ππΌ = π β² (πΌ) = 0. If this value 1 for πΌ lies within the required interval, (3 , 1], then we can say that there exists a retention level πΌ for which the adjustment coefficient is maximised, in other words, there exists a reinsurance agreement under the premium loading structure specified which would improve the βsafety coefficientβ of the company and hence reduce the probability of ruin. However, there is no explicit expression of the form π = π(πΌ) in this case and numerical methods need to be employed.