RM2_Practice_Question_Set2_solution

STAT3058/6058 Practice Questions Solutions Topic 2 - Aggregate Claims Modelling 1. Evaluate the MGF of the aggregate claim amount at t = 0 . 01 , if the number of claims is binomial with parameters n = 100 and p = 0 . 01 , and individual claims sizes are Gamma distributed with parameters α = 10 and θ = 5 . Solution: Since N ∼ Bin(100 , 0 . 01) and X ∼ Gamma(10 , 5) , we have: M N ( t ) = ( 0 . 99 + 0 . 01 e t ) 100 and M X ( t ) = (1 − 5 t ) − 10 So: M S ( t ) = M N [ln M X ( t )] = h 0 . 99 + 0 . 01 e ln M X ( t ) i 100 = [0 . 99 + 0 . 01 M X ( t )] 100 = 0 . 99 + 0 . 01(1 − 5 t ) − 10 100 and: M S (0 . 01) = 0 . 99 + 0 . 01(1 − 5(0 . 01)) − 10 100 = 1 . 95 2. The random variable S has a compound Poisson distribution with Poisson parameter 4 . The individual claim amounts are either 1 , with probability 0.3, or 3 , with probability 0.7. Calculate the probability that S = 4 . Solution: We can get an aggregate claim amount of 4 in only two ways: a) 2 claims, with one claim of amount 1 and one claim of amount 3 , or b) 4 claims, with each of amount 1 . The probability of this happening is: P ( S = 4) = P ( N = 2) P ( X 1 = 1) P ( X 2 = 3) + P ( N = 2) P ( X 1 = 3) P ( X 2 = 1) + P ( N = 4) P ( X 1 = 1) P ( X 2 = 1) P ( X 3 = 1) P ( X 4 = 1) The X ’s are identical, so this simplifies to: P ( S = 4) = 2 P ( N = 2) P ( X = 1) P ( X = 3) + P ( N = 4) P ( X = 1) 4 = 2 × e − 4 4 2 2! × 0 . 3 × 0 . 7 + e − 4 4 4 4! × 0 . 3 4 = 0 . 06312 1

3. Let S 1 and S 2 be independent random variables having compound Poisson distributions with parameters λ 1 = 3 and F 1 ( x ) and λ 2 = 1 and F 2 ( x ) , respectively. Suppose that the support of both F 1 ( x ) and F 2 ( x ) is the set { 1 , 2 , 3 , 4 , 5 } and the CDFs are given in the following table: x 1 2 3 4 5 F 1 ( x ) 0 . 1 0 . 3 0 . 3 0 . 7 1 . 0 F 2 ( x ) 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0 Note the support of a random variable, and therefore its CDF, is the set of its possible outcomes. (a) What is the distribution of S = S 1 + S 2 ? Find E [ S ] , V [ S ] and the mgf of S . (b) Find P ( S ≤ 3) . Solution: (a) From course notes, we know that S has a compound Poisson distribution with parameters Λ = λ 1 + λ 2 = 4 and F ( x ) = 3 4 F 1 ( x ) + 1 4 F 2 ( x ) , so that F ( x ) is given by the table: x 1 2 3 4 5 F ( x ) 0 . 125 0 . 325 0 . 375 0 . 725 1 . 0 From this information, we can calculate the pmf of a random variable with a CDF of F ( x ) as p ( x ) = F ( x ) − F ( x − 1) : x 1 2 3 4 5 p ( x ) 0 . 125 0 . 2 0 . 05 0 . 35 0 . 275 So, the mean and raw second moment of a random variable X with pmf p ( x ) are: µ 1 = E [ X ] = 1(0 . 125) + 2(0 . 2) + 3(0 . 05) + 4(0 . 35) + 5(0 . 275) = 3 . 45 µ 2 = E [ X 2 ] = 1 2 (0 . 125) + 2 2 (0 . 2) + 3 2 (0 . 05) + 4 2 (0 . 35) + 5 2 (0 . 275) = 13 . 85 . Therefore, we have E [ S ] = Λ µ 1 = 4(3 . 45) = 13 . 8 and V [ S ] = Λ µ 2 = 4(13 . 85) = 55 . 4 . Finally, we know that the mgf of S is given by m S ( t ) = exp [Λ { m X ( t ) − 1 } ] where m X ( t ) = E ( e tX ) = 0 . 125 e t + 0 . 2 e 2 t + 0 . 05 e 3 t + 0 . 35 e 4 t + 0 . 275 e 5 t . So, we get m S ( t ) = exp ( 0 . 5 e t + 0 . 8 e 2 t + 0 . 2 e 3 t + 1 . 4 e 4 t + 1 . 1 e 5 t − 4 ) . (b) Since the minimum value of the support of both F 1 ( x ) and F 2 ( x ) is 1 , it must be the case that S ≥ N . Therefore, S ≤ 3 implies N ≤ 3 . Hence, we can see that: P ( S ≤ 3) = ∞ X n =0 P ( S ≤ 3 | N = n ) P ( N = n ) = 3 X n =0 P ( S ≤ 3 | N = n ) e − 4 4 n n ! = P ( S = 0 | N = 0) e − 4 + 3 X n =1 P n X i =1 X i ≤ 3 N = n ! e − 4 4 n n ! = e − 4 + 4 P ( X 1 ≤ 3) e − 4 + 8 P ( X 1 + X 2 ≤ 3) e − 4 + 32 3 P ( X 1 + X 2 + X 3 ≤ 3) e − 4 = 1 3 e − 4 n 3 + 12 F (3) + 24 F ∗ (2) (3) + 32 F ∗ (3) (3) o . 2

the X i ’s and follows a distribution with pmf : p N ( n ) = P( N = n ) = 2 n n !6 . 389 , n = 1 , 2 , 3 , . . . The aggregate claim amount is given by the random sum S = ∑ N i =1 X i . Find E ( S ) and V ( S ) . Solution: Here we have: E ( N ) = ∞ X n =1 n 2 n n !6 . 389 = 2 6 . 389 ∞ X n =1 2 n − 1 ( n − 1)! = 2 6 . 389 ∞ X m =0 2 m m ! = 2 6 . 389 e 2 = 2 . 313 , since ∑ ∞ m =0 2 m m ! = e 2 = 7 . 389 . Similarly, we have E { N ( N − 1) } = ∞ X n =1 n ( n − 1)2 n n !6 . 389 = 2 2 6 . 389 ∞ X n =2 2 n − 2 ( n − 2)! = 4 6 . 389 ∞ X m =0 2 m m ! = 4 6 . 389 e 2 = 4 . 626 , which implies that V( N ) = E ( N 2 ) − { E ( N ) } 2 = E { N ( N − 1) } + E ( N ) − { E ( N ) } 2 = 4 . 626 + 2 . 313 − 2 . 313 2 = 1 . 589 . Therefore, we get E ( S ) = E ( N ) E ( X 1 ) = 2 . 313 θ, and V( S ) = E ( N ) V ( X 1 ) + { E ( X 1 ) } 2 V( N ) = 2 . 313 θ 2 + 1 . 589 θ 2 = 3 . 902 θ 2 . 6. Suppose that we have a portfolio of insurance policies for which claim amounts are always of size $100 , 000 . Further, suppose that N , the number of claims made in a specified period, has an expectation of E ( N ) = 10 . We are trying to decide whether to purchase an excess-of-loss reinsurance scheme with retention level M , the premium charge of which is set equal to Premium = E ( S Z ) 1 + 10000 M , where Z i = ( X i − M ) I ( X i >M ) represents the amount of the i th claim for which the reinsurer is liable and E ( S Z ) = E ∑ N i =1 Z i . In other words, the premium charged is equal to the expected liability of the reinsurer plus a surcharge which depends on the size of the retention level. (a) What is the expected claim amount for which we are liable if we employ the reinsurance? What is our total expected cost, including the reinsurance premium? Does it seem worthwhile to purchase the reinsurance on the grounds of reducing our expected costs? (b) Suppose we have a capital reserve of $1 , 150 , 000 . What level of reinsurance should we choose in order to minimise the chance that our total costs exceed our reserve? 4

Solution: (a) As long as M < 100000 , the amount of any claim for which we are liable is Y i = M and we see that S Y = ∑ N i =1 Y i = NM . Thus, our expected liability is E ( Y i ) = M and our expected total liability is E ( S Y ) = E ( N ) × M = 10 M . Of course, if M ≥ 100000 then Y i = 100000 , E ( Y i ) = 100000 and S Y = 100000 N and E ( S Y ) = 1000000 . On the other hand, E ( S Z ) = E ∑ N i =1 Z i = E { N (100000 − M ) } = 1000000 − 10 M , provided M < 100000 [ and E ( S Z ) = 0 otherwise]. Thus, the total cost to us under a reinsurance scheme with a retention level of M < 100000 is 10 M + (1000000 − 10 M ) 1 + 10000 M = 1000000 + 1000000 M − 100000 > 1000000 since M < 100000 implies 1000000 M > 100000 . Thus, since 1000000 is the expected cost for no rein- surance (or, equivalently, reinsurance schemes with M ≥ 100000 ), there is no benefit to reinsurance from the perspective of expected costs. (b) We have: P ( Total Cost > Reserve ) = P NM + E ( S Z ) 1 + 10000 M > 1150000 = P NM > 1150000 − (1000000 − 10 M ) 1 + 10000 M = P N > 10 + 250000 M − 10 10 M 2 . So, we need to make 10+ 250000 M − 10 10 M 2 as large as possible (since such a choice will make the probability in question as small as possible). Taking derivatives and setting to zero yields the equation: − 250000 M 2 + 2 × 10 10 M 3 = 0 = ⇒ M = 80000 So, it appears we should choose a reinsurance retention level of M = $80 , 000 . [NOTE: The fact that this critical point is a maximum instead of a minimum can easily be seen by taking second derivatives and noting that they are negative at the point in question.] However, since the distribution of N is discrete and 10 + 250000 80000 − 10 10 80000 2 = 11 . 5625 , we see that any value of M which makes 10 + 250000 M − 10 10 M 2 > 11 will actually produce the same chance that the total cost will exceed our reserve in this case. Therefore, as 10 + 250000 50000 − 10 10 50000 2 = 11 and 10 + 250000 100000 − 10 10 100000 2 = 11 . 5 > 11 , we see that any 50000 < M ≤ 100000 will be equivalent in this case. Thus, it again appears that the choice of no reinsurance is the best option, as it has minimum risk and minimum cost in this case. 7. Consider a portfolio of insurance policies where the number of claims has a binomial distribution with pa- rameters n = 1 , 000 and p = 0 . 01 . Claim severities are independent of claim numbers, and the claim severity distribution is assumed to be exponential with mean $100 . The insurer takes out an individual excess of loss reinsurance contract with a retention level of $200 . Calculate the mean of S Y , where S Y is the aggregate claims paid by the insurer net of reinsurance (i.e., after the XoL scheme is applied). Solution: 5

Thus, we obtain P ( S ≤ 4 . 5) = P S − 3 √ 0 . 6 ≤ 4 . 5 − 3 √ 0 . 6 ≈ Φ 4 . 5 − 3 √ 0 . 6 = Φ(1 . 9365) = 0 . 9735968 . As given in the question, the "exact" value is 0 . 9654 , indicating that the normal approximation is rather accurate in this case. Moreover, we see that the approximate probability in this case is noticeably higher than in the case of negative binomial claim numbers, despite the fact that the expected number of claims is the same in each case. This is due to the greater variance in claim numbers under the negative binomial model, which indicates that larger total claim amounts are somewhat more likely in this model than under the Poisson claim number model. 9. The employees of a company are covered by group life insurance which pays a specified benefit amount if an employee dies. There are two types of employee, and the life insurance benefits differ by employee type. For junior employees, the benefit amount is $20 , 000 , and the probability of dying during a year is 0 . 012 . There are 250 junior employees. For senior employees, the benefit amount is $50 , 000 , and the probability of dying during a year is 0 . 008 . There are 1,250 senior employees. The aggregate claims payable during the year are assumed to conform to the individual risk model. Calculate the probability that the aggregate claims payable in a given year will exceed $900 , 000 , using a normal approximation. Solution: We need to first find E ( S ) and V ( S ) : E ( S ) = X i q i µ i = 1 , 250 × 50 , 000 × 0 . 008 + 250 × 20 , 000 × 0 . 012 = 560 , 000 V ( S ) = X i q i ( σ 2 i + (1 − q i ) µ 2 i ) = X i q i (1 − q i ) µ 2 i = 1 , 250 × 50 , 000 2 × 0 . 008 × 0 . 992 + 250 × 20 , 000 2 × 0 . 012 × 0 . 988 = 2 . 59856 × 10 10 We now assume: S ∼ N (560 , 000 , 2 . 59856 × 10 10 ) . So we have P ( S > 900 , 000) = P N (0 , 1) > 900 , 000 − 560 , 000 √ 2 . 59856 × 10 10 = 1 − P ( N (0 , 1) < 2 . 109175) = 0 . 0175 . 7

10. Suppose that X has a Gamma distribution with parameters α and θ . (a) Suppose that k = 2 α is a positive integer. Show that Y = 2 X/θ has a chi-squared distribution with k degrees of freedom. Note that the chi-squared distribution pdf is as follows: f ( x ) = 1 2 k/ 2 Γ( k/ 2) x ( k/ 2) − 1 e − x/ 2 . (b) Suppose that you wanted to approximate the probability that a compound distributed quantity S ex- ceeded some value using the translated Gamma method, but you only had chi-squared tables at your disposal. Discuss how your result from part (a) would be useful. (c) Suppose that E ( S ) = 10 , V ( S ) = 30 and Skew( S ) = 270 . Use the translated Gamma approximation and chi-squared tables to estimate P ( S ≤ 20 . 9) . Also, estimate the value of s which solves the equation P ( S ≤ s ) = 0 . 99 . Solution: (a) We have Y = 2 θ − 1 X , so that X = 1 2 θY . Thus, a simple change of variable shows that: f Y ( y ) = f X 1 2 θy dx dy = 1 θ α Γ( α ) 1 2 θy α − 1 e − θ − 1 ( 1 2 θy ) 1 2 θ = 1 θ α Γ( α ) 1 2 α θ α y α − 1 e − 1 2 y = 1 2 α Γ( α ) y α − 1 e − 1 2 y = 1 2 k/ 2 Γ( k/ 2) y ( k/ 2) − 1 e − 1 2 y where k = 2 α . Thus, Y has a chi-squared distribution with k = 2 α degrees of freedom. (b) As noted in lectures, if Y ∼ G ( α g , θ g ) and X = Y + k is the translated Gamma approximation to S , we have: P ( S ≤ s ) ≈ P ( X ≤ s ) = P ( Y ≤ s − k ) = P ( 2 θ − 1 g Y ≤ 2 θ − 1 g ( s − k ) ) = P ( χ 2 2 α ≤ 2 θ − 1 g ( s − k ) ) (c) From the given information, we can calculate the coefficient of skewness for S as ρ S = 270 30 3 / 2 = 1 . 643 . Thus, the appropriate translated Gamma distribution for use in approximating S has parameters k, α g and θ g which solve the equations: 1 . 643 = ρ S = 2 √ α g , 30 = α g θ 2 g , 10 = k + α g θ g Solving the first equation shows that α g = 4 / (1 . 643) 2 = 1 . 482 , which implies that θ g = p 30 / 1 . 482 = 4 . 5 and k = 10 − 4 . 5(1 . 482) = 3 . 333 . So, letting Y ∼ G (1 . 482 , 4 . 5) , we have P ( S ≤ 20 . 9) ≈ P ( Y + 3 . 333 ≤ 20 . 9) = P ( Y ≤ 17 . 567) = P 2 4 . 5 Y ≤ 7 . 808 ≈ P ( χ 2 3 ≤ 7 . 808 ) ≈ 0 . 95 [NB: The actual value of P ( Y ≤ 17 . 567) is 0 . 9512 .] Similarly, the 99th percentile of S is the value s 0 . 99 which satisfies: 0 . 99 = P ( S ≤ s 0 . 99 ) ≈ P ( Y ≤ s 0 . 99 − 3 . 333) ≈ P χ 2 3 ≤ 2 4 . 5 ( s 0 . 99 − 3 . 333) 8

V ( S ) = n X i =1 q i ( σ 2 i + (1 − q i ) µ 2 i ) ⇒ V ( S ) = 500 X A 0 . 03 ( σ 2 A + (0 . 97)270 . 426 2 ) + 300 X B 0 . 02 ( σ 2 B + (0 . 98)300 2 ) + 100 X C 0 . 05(0 . 95)300 2 σ 2 A = exp ( 2 µ + 2 σ 2 ) − 270 . 4264 2 = 169 , 671 . 2 σ 2 B = αθ 2 = 0 . 15(2000) 2 σ 2 C = 0 ⇒ V ( S ) = 500 X A 0 . 03(240607 . 5) + 300 X B 0 . 02 ( 0 . 15(2000) 2 + (0 . 98)(300) 2 ) + 427 , 500 ⇒ V ( S ) = 3 , 609 , 112 + 300 X B 0 . 02 ( 0 . 15(2000) 2 + (0 . 98)(300) 2 ) + 427 , 500 = 8 , 165 , 812 Hence, we get P ( S > 12 , 000) = 1 − P ( S ≤ 12 , 000) = 1 − P Z ≤ 12 , 000 − E ( S ) p V ( S ) ! = 1 − P ( Z ≤ 1 . 625) = 0 . 052 . 13. A portfolio of term-life assurances is composed of 10,000 independent policies, each with probability of death during the period of cover of q i = 0 . 001 . For 4000 policies, the death benefit is $100 , 000 and the premium charged is $150 . On the remaining 6000 policies, the death benefit is $500 , 000 and the premium charged is $750 . Let S be the total claim amount made on the entire portfolio. (a) Calculate E ( S ) and V ( S ) . Also, use the normal approximation to estimate the probability that S exceeds the total premiums charged, 4000($150) + 6000($750) = $5 , 100 , 000 . (b) Suppose you have available only χ 2 -tables. Use the translated Gamma method applied to e S , the col- lective risk model approximation to S , to estimate the chance that the aggregate claim amount exceeds $5 , 100 , 000 . Discuss whether this estimate is reliable. (c) Based on the estimate of part ( a ) , the risk of the portfolio is deemed too high. To combat this, premiums are to be raised by $ P for policies with benefit $500 , 000 and by $2 P for policies with benefit $100 , 000 . However, policyholders with benefit $500 , 000 will be given the option to switch their policy to the $100 , 000 benefit level. It is believed that for every $9 increase in the premium, 450 policyholders will choose to switch their policy type, while 170 policyholders with the $100 , 000 benefit and 110 policyholders with the $500 , 000 benefit will discontinue their policies. Find the value of P which will reduce the chance of S exceeding total premiums to 5% (as measured by normal approximation method). Solution: (a) We have: E ( S ) = ∑ 10000 i =1 q i µ i = 4000(0 . 001)(100000) + 6000(0 . 001)(500000) = 3400000 V ( S ) = 10000 X i =1 q i σ 2 i + (1 − q i ) µ 2 i = 4000(0 . 001)(0 . 999) ( 100000 2 ) + 6000(0 . 001)(0 . 999) ( 500000 2 ) = 1 . 53846 × 10 12 . 10

Therefore, P ( S > 5100000) = P S − 3400000 √ 1 . 53846 × 10 12 > 5100000 − 3400000 √ 1 . 53846 × 10 12 ≈ 1 − Φ 5100000 − 3400000 √ 1 . 53846 × 10 12 = 1 − Φ(1 . 37) ≈ 0 . 0853 . So, there is approximately an 8 . 5% chance that the total claim costs will exceed the total premiums collected. (b) The collective risk model approximation ˜ S is a compound Poisson quantity with rate Q = ∑ 10000 i =1 q i = 4000(0 . 001) + 6000(0 . 001) = 4 + 6 = 10 and an individual term distribution which has first three moments: E ( Z ) = 0 . 4(100000) + 0 . 6(500000) = 340000 E ( Z 2 ) = 0 . 4 ( 100000 2 ) + 0 . 6 ( 500000 2 ) = 1 . 54 × 10 11 E ( Z 3 ) = 0 . 4 ( 100000 3 ) + 0 . 6 ( 500000 3 ) = 7 . 54 × 10 16 . So, E ( ˜ S ) = Q × E ( Z ) = 3400000 , V ( ˜ S ) = Q × E ( Z 2 ) = 1 . 54 × 10 12 and ρ ˜ S = Skew( ˜ S ) { V ( ˜ S ) } 3 / 2 = Q × E ( Z 3 ) { 1 . 54 × 10 12 } 3 / 2 = 0 . 39454 . Solving the system of equations: ρ ˜ S = 2 / √ α g ; V ( ˜ S ) = α g θ 2 g ; E ( ˜ S ) = k + α g θ g leads to the parameters α g = 25 . 6968 , θ g = 244805 . 195 and k = − 2890716 . 18 for the approximating translated Gamma random variable X = Y + k . Thus, we have P ( S > 5100000) ≈ P ( ˜ S > 5100000) ≈ P ( Y > 5100000 + 2890716 . 18) = P (2 Y/ 244805 . 195 > 2(7990716 . 18) / 244805 . 195) ≈ P χ 2 (51 . 4) > 65 . 28 Now, there are various ways of getting at this probability. One is to just use the χ 2 (50) distribution, note that 65 . 28 is almost exactly half-way between 63 . 17 and 67 . 50 , and thus approximate the desired probability using a linear interpolation of 0.075. An alternative, more accurate, linear interpolation, would first find the 0 . 9 -quantiles and 0 . 95 -quantiles for 51 . 4 degrees of freedom as 63 . 17 + 51 . 4 − 50 60 − 50 (74 . 40 − 63 . 17) = 64 . 74 and 67 . 50 + 51 . 4 − 50 60 − 50 (79 . 08 − 67 . 50) = 69 . 12 . Thus, the linear interpolation for the actual probability is: 0 . 05 + 65 . 28 − 64 . 74 69 . 12 − 64 . 74 (0 . 1 − 0 . 05) = 0 . 938 . Alternatively, we could use a normal approximation to the χ 2 (51 . 4) distribution to see that: P χ 2 (51 . 4) > 65 . 28 ≈ P N (0 , 1) > 65 . 28 − 51 . 4 √ 102 . 8 = Φ(1 . 37) = 0 . 0853 , 11

(d) Suppose that we believe that the two claim probabilities, q 1 and q 2 are actually random, such that E ( q 1 ) = 0 . 007 and E ( q 2 ) = 0 . 021 while V ( q 1 ) = V ( q 2 ) = 0 . 000004 . Under this assumption, discuss how the values of E ( S ) and V ( S ) will change from those calculated in part (a) (NOTE: you need not actually calculate the new mean or variance). Solution: (a) We have: E ( S ) = 10000 X i =1 q i µ i = 6000(0 . 007)(2)(1000) + 4000(0 . 021)(4)(600) =285600 V ( S ) = 10000 X i =1 q i σ 2 i + (1 − q i ) µ 2 i =6000(0 . 007) (2) ( 1000 2 ) + (0 . 993) ( 2 2 ) ( 1000 2 ) + 4000(0 . 021) (4) ( 600 2 ) + (0 . 979) ( 4 2 ) ( 600 2 ) =845463360 . Therefore, P ( S > 400000) = P S − 285600 √ 84546330 > 400000 − 285600 √ 845463360 ≈ 1 − Φ 400000 − 285600 √ 845463360 = 1 − Φ(3 . 93) ≈ 0 . So, there is essentially no chance that S will exceed $400 , 000 . (b) The collective risk model approximation ˜ S is a compound Poisson quantity with rate Q = ∑ 10000 i =1 q i = 6000(0 . 007) + 4000(0 . 021) = 42 + 84 = 126 and an individual term distribution which has first three moments: E ( Z ) = 42 126 E ( X 1 ) + 84 126 E ( X 2 ) = 1 3 (2)(1000) + 2 3 (4)(600) = 2266 . 67 E ( Z 2 ) = 42 126 E ( X 2 1 ) + 84 126 E ( X 2 2 ) = 1 3 (2)(3) ( 1000 2 ) + 2 3 (4)(5) ( 600 2 ) = 6800000 E ( Z 3 ) = 42 126 E ( X 3 1 ) + 84 126 E ( X 3 2 ) = 1 3 (2)(3)(4) ( 1000 3 ) + 2 3 (4)(5)(6) ( 600 3 ) = 2 . 528 × 10 10 . So, E ( ˜ S ) = Q × E ( Z ) = 285600 , V ( ˜ S ) = Q × E ( Z 2 ) = 856800000 and ρ ˜ S = Skew( ˜ S ) { V ( ˜ S ) } 3 / 2 = Q × E ( Z 3 ) { 856800000 } 3 / 2 = 0 . 127 . (c) Solving the system of equations: ρ ˜ S = 2 / √ α g ; V ( ˜ S ) = α g θ 2 g ; E ( ˜ S ) = k + α g θ g leads to the parameters α g = 247 . 972 , θ g = 1858 . 82353 and k = − 175336 . 19 for the approximating translated Gamma random variable X = Y + k . Thus, we have P ( S > 300000) ≈ P ( ˜ S > 300000) ≈ P ( Y > 300000 + 175336 . 19) = P Y − 247 . 972(1858 . 82353) 1858 . 82353 × √ 247 . 972 > 475336 . 19 − 247 . 972(1858 . 82353) 1858 . 82353 × √ 247 . 972 ≈ P ( Z > 0 . 49) = 0 . 3121 , 13

where Z has a standard normal distribution. [NOTE: The actual probability that a Gamma dis- tribution with shape 247 . 972 and scale 1858 . 82353 exceeds 300000 + 175336 . 19 is 0.3057.] Finally, we note that ∑ 10000 i =1 q 2 i µ 2 i = 11336640 , which is just 1 . 34% of V ( S ) , so the collective risk model approximation is likely to be reasonably accurate. (d) N/A (The content in this question is no longer examinable.) 15. An insurance portfolio is composed of 7000 independent policies each of which can make at most one claim during the period of cover. For 3000 policies, the probability of a claim being made is q 1 = 0 . 009 and the claim amount is Weibull distributed with parameters γ 1 = 0 . 5 and θ 1 = 50 . For the remaining 4000 policies, the probability of a claim being made is q 2 = 0 . 03 and the claim amount is Weibull distributed with parameters γ 2 = 1 and θ 2 = 1500 . Let S be the total claim amount made on the entire portfolio. (a) Using the individual risk model, calculate E ( S ) and V ( S ) . (b) We are considering changing the composition of our portfolio in order to decrease its chance of a large total claim amount. To do so, we plan to decrease the number of the (riskier) first type of policies. However, we still want our portfolio to contain 7000 overall policies. Using the normal approximation to the individual risk model, determine the number of policies of each type we should include in the modified portfolio so that the approximate probability of the total claim amount exceeding $450 , 000 is 1% . Why do we consider the first type of policy riskier even though it has the lower probability of making a claim? (c) Let ˜ S be the collective risk model approximation to S (based on the original portfolio of 3000 policies of the first type and 4000 policies of the second type). Use the translated Gamma method applied to ˜ S to estimate the chance that the total aggregate claim amount made on the portfolio exceeds $250 , 000 . Do you think that this estimate is reliable in this instance? Why or why not? Solution: (a) The expected value and variance of a Weibull distribution with γ = 0 . 5 and θ = 50 can be calculated as E ( X i ) = 2(50) 2 = 5000 and V ( X i ) = 24(50) 4 − (5000) 2 = 1 . 25 × 10 8 . Also, a Weibull distribution with γ = 1 and θ = 1500 is just an exponential distribution with θ = 1500 , and thus has E ( X i ) = 1500 and V ( X i ) = 2 . 25 × 10 6 . So, E ( S ) =3000(0 . 009)(5000) + 4000(0 . 03)(1500) = 315000 V ( S ) =3000(0 . 009) 1 . 25 × 10 8 + (1 − 0 . 009) ( 5000 2 ) + 4000(0 . 03) 2 . 25 × 10 6 + (1 − 0 . 03) ( 1500 2 ) =4575825000 (b) The expected claim amount on each policy is (0 . 009)(5000) = (0 . 03)(1500) = 45 , thus changing the make-up of the portfolio will not effect the expected value of S . Now, we would like to choose our portfolio make-up so that 0 . 01 = P ( S > 450000) ≈ 1 − Φ 450000 − 315000 √ v , where v =(3000 − n )(0 . 009) 1 . 25 × 10 8 + (1 − 0 . 009) ( 5000 2 ) + (4000 + n )(0 . 03) 2 . 25 × 10 6 + (1 − 0 . 03) ( 1500 2 ) 14

(a) Use the normal approximation to estimate the probability that S exceeds 750,000. (b) Use the normal approximation to estimate the 98th percentile of the distribution of S . (c) Repeat parts (a) and (b) using the translated Gamma approximation method. Solution: (a) We first note that for a Weibull distribution we saw: E ( X k i ) = θ k/γ Γ k γ + 1 Thus, in this case, we have µ 1 = E ( X i ) = θ 2 Γ(3) = ( 100 2 ) (2!) = 20000 and µ 2 = E ( X 2 i ) = θ 4 Γ(5) = ( 100 4 ) (4!) = 2400000000 . Therefore, E ( S ) = λµ 1 = 400000 and V ( S ) = λµ 2 = 48000000000 . So, using the normal approximation, we have: P ( S > 750000) ≈ 1 − Φ 750000 − 400000 √ 48000000000 = 1 − 0 . 9449 = 0 . 0551 (b) We know that the 98th percentile of the standard normal distribution is z 0 . 98 = 2 . 05375 . Therefore, the approximate 98th percentile of S in this case is: s 0 . 98 ≈ E ( S ) + z 0 . 98 p V ( S ) = 400000 + 2 . 05375 √ 48000000000 = 849954 (c) In order to use the translated Gamma approximation, we will need µ 3 = E ( X 3 i ) = θ 6 Γ(7) = ( 100 6 ) (6!) = 7 . 2 × 10 14 . This implies that the coefficient of skewness of S is ρ S = Skew( S ) V ( S 3 / 2 ) = λµ 3 ( λµ 2 ) 3 / 2 = µ 3 µ 3 / 2 2 √ λ = 7 . 2 × 10 14 (24 × 10 8 ) 3 / 2 √ 20 = 7 . 2 × 10 2 24 3 / 2 √ 20 = 1 . 369 . Therefore, the appropriate parameters for the translated Gamma approximation are α g = 4 1 . 369 2 = 2 . 133 , θ g = p 48000000000 / 2 . 133 = 150000 and k = 400000 − 2 . 133(150000) = 80000 . So, P ( S > 750000) ≈ P ( Y + k > 750000) = P ( Y > 670000) = 0 . 0742 . [NB: P ( χ 2 4 > 2 ( 150000 − 1 ) (670000) ) = P ( χ 2 4 > 8 . 933 ) = 0 . 0628 and P ( χ 2 5 > 8 . 933 ) = 0 . 1118 , and since 2 α g = 4 . 266 , we might approximate P ( Y > 670000) by interpolation as 0 . 0628+0 . 266(0 . 1118 − 0 . 0628) = 0 . 0758 .] Similarly, the 98th percentile of S, s 0 . 98 , satisfies: 0 . 98 = P ( S ≤ s 0 . 98 ) ≈ P ( Y + k ≤ s 0 . 98 ) = P ( Y ≤ s 0 . 98 − 80000) Now, the 98th percentile of Y can be calculated as 910078.8. [NB: χ 2 4 (0 . 98) = 11 . 668 and χ 2 5 (0 . 98) = 13 . 388 , so that we could approximate the 98th percentile of Y by interpolation as: 1 2 × 150000 × (11 . 668 + 0 . 266 × (13 . 388 − 11 . 668)) = 909414 .] Therefore, s 0 . 98 ≈ 910078 . 8 + 80000 = 990078 . 8 . Note that both the probability and the quantile are somewhat different than the estimates based on the normal approximation, and this is not surprising since the coefficient of skewness for S in this case was quite large, indicating that the distribution of S is quite skewed and thus not amenable to very accurate normal approximation. 16

17. An insurance company offers two types of policy. Claims are binomially distributed, and all policies are assumed to be independent. For policy type A, there are 100 policies, claims amounts follow a Gamma distribution with parameters α = 24355 and θ = 0 . 25 , and the probability of a claim being made is q A = 0 . 02 . For policy type B , there are 200 policies, each claim is either $1000 with probability 60% , or $3000 with probability 40% , and the probability of a claim being made is q B = (0 . 015) . Let S be the total claim amount on the entire portfolio and assume that we have a cash reserve of $ K . You are given that the third raw moment for a Gamma distributed random variable is: αθ 3 ( α + 1)( α + 2) . (a) Using the individual risk model and the normal approximation, estimate K , such that S will exceed K with probability 1% . (b) Let ˜ S be the collective risk model approximation to S . Calculate the mean, variance, and standardised coefficient of skewness of ˜ S . (c) Use the translated Gamma method applied to ˜ S to estimate K , such that the chance that the total aggregate claim amount made on the portfolio exceeds K is 1% . Suppose you only have chi-square tables available. Solution: (a) In order to use the normal approximation, first we need to find the mean and variance of S . E ( S ) = n X i =1 q i µ i = 100 X A 0 . 02(24355)(0 . 25) + 200 X B 0 . 015(1000(0 . 6) + 3000(0 . 4)) = 100 X A 0 . 02(6088 . 75) + 200 X B 0 . 015(1800) = $17 , 577 . 5 σ 2 A = αθ 2 = 24355 ( 0 . 25 2 ) E X 2 B = ( 1000 2 ) 0 . 6 + ( 3000 2 ) 0 . 4 = 4 , 200 , 000 σ 2 B = E X 2 B − ( E [ X B ]) 2 = 4 , 200 , 000 − 1800 2 = 960 , 000 V ( S ) = n X i =1 q i σ 2 i + (1 − q i ) µ 2 i = 100 X A 0 . 02 ( (24355)0 . 25 2 + (1 − 0 . 02) ( 24355 2 ) 0 . 25 2 ) + 200 X B 0 . 015 ( 960 , 000 + (1 − 0 . 015)1800 2 ) = 100 X A 726 , 658 . 8244 + 200 X B 62 , 271 = $85 , 120 , 082 . 44 . 17

18. (a) A compound Poisson distribution has λ = 5 and claim amount distribution Pr(100) = 0 . 7 , Pr(500) = 0 . 21 , Pr(1000) = 1 − 0 . 7 − 0 . 21 . Determine the probability that the aggregate claims will be exactly 600 . (b) For an insurance portfolio, the number of claims is a compound binomial distribution with m = 100 , and q = 0 . 1 . Claim amounts have a Weibull distribution with θ = 2 , and γ = 1 . Find the variance of the aggregate claims. (c) Aggregate claims from a risk have a compound Poisson distribution with λ = 200 and an individual claim amount distribution with is Pareto with parameters α = 3 and δ = 305 . The insurer has taken out excess-of-loss reinsurance with a retention level of 400 . Calculate the mean and variance of the reinsurer’s aggregate claims. Solution: (a) We know P ( S = 600) = P ( X 1 = 600 | N = 1) P ( N = 1) + P ( X 1 + X 2 = 600 | N = 2) P ( N = 2) + . . . Now, given the information, there are only 2 ways to get 600 exactly: 2 claims : 100, 500 (or 500, 100) 6 claims : 100, 100, 100, 100, 100, 100 P ( S = 600) = P ( X 1 + X 2 = 600 | N = 2) P ( N = 2) + P 6 X i =1 X i = 600 | N = 6 ! P ( N = 6) where P ( N = n ) = 5 n e − 5 n ! . Hence, P ( X 1 + X 2 = 600 | N = 2) P ( N = 2) = [(0 . 7)(0 . 21) + (0 . 21)(0 . 7)] 5 2 e − 5 2! = 0 . 024762 P 6 X i =1 X i = 600 | N = 6 ! P ( N = 6) = (0 . 7) 6 e − 5 5 6 6! = 0 . 017203 P ( S = 600) = 0 . 024762 + 0 . 017203 ≈ 0 . 0420 (b) V [ S ] = mq ( µ 2 − qµ 2 1 ) V [ S ] = 100(0 . 1) ( µ 2 − (0 . 1) µ 2 1 ) For Weibull distribution, we have:: E θ,γ ( X k ) = θ k/γ Γ k γ + 1 = 2 k Γ( k + 1) E θ,γ ( X ) = 2Γ(2) = 2 E θ,γ ( X 2 ) = 2 2 Γ(3) = 8 V [ S ] = 100(0 . 1)(8 − (0 . 1)4) = 76 (c) Let S R be the reinsurer’s loss distribution. Then, E ( S R ) = E ( N ) E ( Z ) = 200 Z ∞ M ( x − M ) f X ( x ) dx = 200 Z ∞ M ( x − M ) (3)305 3 ( x + 305) 4 dx 19

Let y = x − M E ( S R ) = 200 Z ∞ 0 y (3)305 3 ( y + M + 305) 4 dy = 305 3 × 200 (305 + M ) 3 Z ∞ 0 y (3)(305 + M ) 3 ( y + M + 305) 4 dy = 305 3 × 200 (305 + M ) 3 E ( Y ) = 305 3 × 200 (305 + M ) 3 305 + M 2 = 5 , 708 . 49 where the third and second last equality follows from Y ∼ Pareto (3 , 305 + M ) . Hence, Z ∞ 0 y (3)(305 + M ) 3 ( y + M + 305) 4 dy = E ( Y ) = 305 + M 2 V ( S R ) = 305 3 (305 + M ) 3 200 Z ∞ 0 y 2 (3)(305 + M ) 3 ( y + M + 305) 4 dy = 305 3 200 (305 + M ) 3 E ( Y 2 ) = 305 3 200 (305 + M ) 3 2(305 + M ) 2 2(1) = 8 , 048 , 971 . 63 Again for Y ∼ Pareto (3 , 305 + M ) , we have used Z ∞ 0 y 2 (3)(305 + M ) 3 ( y + M + 305) 4 dy = E ( Y 2 ) = 2(305 + M ) 2 2(1) . 20