1. An insurance company models claim sizes as having the following survival function. 25(x + 1) (x 2 + 2x + 5) 2 x 0. S(x) =

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1 ACSC/STAT 373, Actuarial Moels I Further Probability with Applications to Actuarial Science WINTER 5 Toby Kenney Sample Final Eamination Moel Solutions This Sample eamination has more questions than the actual final, in orer to cover a wier range of questions. Estimate times are provie after each question to help your preparation.. An insurance company moels claim sizes as having the following survival function S Calculate the TVaR of a loss at the 95% level. The VaR is given by solving S The solution is Now we calculate the T V ar as We have ] [ so the TVaR is

2 . An insurance company moels claim sizes as following a miture of two istributions. With probability.3, claims follow a Weibull istribution with τ an θ. With probability.7, claims follow a Pareto istribution with α 3 an θ 5. a Which of the following is the VaR of the istribution at the 9% level? i. ii iii 97.4 iv 3.6 The survival function is. The VaR is the solution to We try the values given: S.3e e e i..887 ii iii iv so iv 3.6 is the VaR. b Calculate the TVaR at the 9% level. The TVaR is given as We have e 3.6 an π Φ 3.6.3e [ so the TVaR is π Φ ]

3 3. An insurance company observes the following sample of claims:.,.6,.,3.5, 3.7 They use a Kernel ensity moel with a uniform kernel with banwith. What is the probability that a claim is between.3 an 3.3? The probability that a claim is between.3 an 3.3 is An insurance company observes the following sample of claims:.3,.6,.8, 3., 3.5 They use a Kernel ensity moel with kernel a gamma istribution with α an θ so the mean matches the observe ata point. What is the variance of a ranom claim uner this moel? The kernel ensity moel is a miture moel, so the variance is given by the law of total variance: VarX EVarX I + VarEX I where I inicates which of the ata points we are using. For a ata point, the epecte value of the corresponing kernel is, an the variance is. We therefore have VarX An insurance company observes the following sample of claims:.6,.,.4, 3., They use a Kernel ensity moel with a uniform kernel with banwith b. They calculate the VaR at the 9% level is 3.7. What banwith are they using? If the banwith is b, then the probability that a ecees 3.7 is ma, ma, ma, ma, + ma 5 b b b b Since 3.7 is larger than all the above values, there is no nee to consier the possibility that the above values coul be less than. In general, we woul take the minimum of each value above with zero. 3

4 We nee to solve this equal to.. For b small enough that only 3.5 contributes to the ensity at 3.7, this becomes b Multiplying by an rearranging gives.4 b which has no solutions in the interval. If we make.6 < b <.3, so that only 3.5 an 3. contribute, the equation becomes b Multiplying by an rearranging gives.8 b which gives b.8. This is in the interval, so this is the require banwith. 6. An insurance company observes the following sample of claims:.4,.9,.8,.9, 3.6 They use a Kernel ensity moel with a Gaussian normal kernel with stanar eviation. What is the moe claim size highest probability ensity? i.538 ii.69 iii.843 iv The ensity function is 5 e.4 + e.9 + e.8 + e.9 + e 3.6 π We want to fin where this is maimise. The erivative is proportional to.4e.4 +.9e.9 +.8e.8 +.9e e 3.6 We try the values given: 4

5 .4e.4 +.9e.9 +.8e.8 +.9e.9 + i ii iii iv The moe is i An insurance company assigns a risk factor Θ to each iniviual. These Θ follow a gamma istribution with α an θ 4. For an iniviual with risk factor Θ θ, the size of a claim follows an inverse gamma istribution with α 3 an this value of θ. What is the probability that a ranom iniviual makes a claim in ecess of $3,? The probability that a claim from an iniviual with risk factor θ ecees $3, is e θ 3 + θ 3 + θ 3 Therefore the probability that a claim from a ranomly chosen iniviual ecees $3, is the epecte value of this over the istribution of Θ. That is, e θ e 8.5θ 3 + θ 3 + θ 3 θ + θ 3 + θ3 3 4 Γ θ 4 Γ e θ 4 θ θ 8. An insurance company moels the sizes of claims as a miture istribution. Conitional on Θ θ, the claim size has survival function S θ4 4 θ The istribution of Θ is a gamma istribution with α 4 an θ 6. For two claims which share a value of Θ, what is the probability that both are larger than $7,? [Hint: You will nee to break into two cases: θ 7 an θ > 7. You may use 7 θ e 6 θ 6 Γ θ ] For a fie value of θ, the probability that two values are both larger than θ $7, is 4 7 θ 8 4 7, provie Θ < 7 an if Θ >

6 The probability of this is therefore the epectation of this over the istribution of Θ, which is 7 The probability that Θ > 7 is e 7 6 The first integral is θ 8 θ Γ4 e θ 6 θ + P Θ > Γ 7 θ e θ Γ4 6 Γ θ where the quantity being integrate is the ensity of a gamma istribution with α an θ 6. This probability is e ! The probability that X > 7 is therefore Γ 7 8 Γ An insurance company moels life epectancy as having hazar rate λt Θe.3t, where Θ varies between iniviuals following a normal istribution with µ.3 an σ.. Calculate the probability that an iniviual age 5 survives to age 8. For a fie value Θ θ, the probability of surviving to age is S e θ e.3t t e e.3.3 θ The probability of a ranomly chosen iniviual surviving to age is the epecte value of this over the istribution of Θ. That is M Θ e.3.3. Recall, for a normal istribution, we have Mt e µt+ σ t, so M Θ e.3 e.3e e So the probability that an iniviual lives to age 5 is S5 e.3e e an the probability that an iniviual lives to age 9 is S9 e.3e e

7 so the probability that an iniviual currently age 5 survives to age 9 is An insurance company moels life epectancy as having hazar rate λt Θ t, where Θ is uniformly istribute on the interval [.5,.35]. Calculate the epecte future lifetime of an iniviual age 6. The probability that an iniviual with Θ θ survives to age is e θ t t e θlog log θ. The epecte future lifetime for an iniviual age 6 with Θ θ is therefore θ 6 6 θ [ 6 6 θ 6 u θ 6 θ u u θ+ 6 θ θ + 6 θ + ] 6 The epecte future lifetime is therefore θ + θ [ ] u u.3 logu log An insurance company { moels an iniviual s lifetime as having ensity.5566 if < < function f Calculate the.36e.6 e.6 if > probability that an iniviual ies between ages 6 an 66. The probability that an iniviual ies between ages 6 an 66 is e.6 e e.6 [ 3 ] 6 + [ 6e e.6] e e.6 e e An insurance company moels the loss on a given policy as following a gamma istribution with α 3 an θ 5 for values less than 5, 7

8 an following a Pareto istribution with α 3 an θ for values greater than 5. 8% of claims are more than 5. Calculate the variance of a ranom claim. Conitional on being less than 5, the epecte value of a claim is 3 e e e e + + an the epecte value of the square is 5 4 e e e + + so the variance is Conitional on being more than 5, the epecte value is [ ] 5 an the left-shifte secon raw moment is [ ] So the variance is The epecte conitional variance is an the variance of the conitional epectation is , so the variance is

9 3. The number of incients in a year follows a Poisson istribution with λ 5. The number of claims resulting from an incient follows a negative binomial istribution with r. an β 3.4. Calculate the probability that there are eactly 3 claims in a given year. The istribution is a compoun Poisson-Negative binomial. The pgf is P z e z., so the probability of zero is P e We now use the recurrence relation We calculate f S k k i 5i k f Xif S k i 3.4 f S f S f S The number of policies sol in a year follows a binomial istribution with n an p.. The number of claims resulting from each policy sol follows a Poisson istribution with λ.. Calculate the variance of the total number of claims in a year. We can use the law of total variance VarS EVarS X+VarES X where X is the number of policies sol. We have VarS X.X, an ES X.X, so VarS.EX+.4 VarX The number of fires follows a Poisson istribution with λ 5, an the number of earthquakes follows a Poisson istribution with λ.4. The number of claims resulting from a fire follows a negative binomial with r an β.. The number of claims resulting from an earthquake follows a Poisson istribution with λ 8. Calculate the probability that there are eactly claims in a given year. The sum of compoun Poisson istributions is a compoun Poisson istribution. The primary istribution is a Poisson istribution with λ 5.4, an seconary istribution a miture with probability it is a negative binomial with r an β., an with probability it is a Poisson with parameter 8. This miture istribution has the following probabilities: 9

10 i f X i We use the recurrence We calculate f S k k i 5.4i k f Xif S k i f S e f S f S An insurance company moels the number of claims resulting from policies as following a compoun Poisson-Poisson istribution with parameters 3 an 8. The following year, the company sells 4 policies. What is the probability that there are eactly claims the following year? The following year, the number of claims shoul be moelle as a compoun Poisson-Poisson with parameters 3.5 an 8. This has p.g.f. P z e 3.5e8z We calculate P e 3.5e The recurrence relation is which gives f S k k i 3.5i 8i e 8 k i! f Sk f S 3.5 8e f S.75 8e e

11 7. An insurance company moels the number of claims resulting from 5 policies as following a compoun Poisson-Poisson istribution with parameters 5 an. How many policies shoul the company sell the following year in orer to make the probability of receiving at least claims at least.99? i 43 ii 64 iii 95 iv 74 If the company sells N policies the following year, the istribution of the number of claims is a compoun Poisson-Poisson istribution with parameters N N 3 an. The probability of zero claims is e 3 e, an the probability of one claim is N 3 e e N 3 e. We want to ensure that the probability of zero or one claim is at most.. That is: e N 3 e + e N. 3 e.886n +.935N. Evaluating the options given, we get: N e.886n +.935N i ii iii iv so iii N 95 is the number neee. 8. An insurance company moels loss size as following a Pareto istribution with α 4 an θ 6. The company introuces a euctible of $,. Calculate the epecte payment per claim after the euctible is introuce. 6 The probability that a loss ecees the euctible is S The survival function conitional on eceeing the euctible is S S so the loss per claim after the euctible is applie follows a Pareto istribution with α 4 an θ 7. The epecte payment per claim is therefore 7 4 $,

12 9. An insurance company moels loss size as following a Weibull istribution with τ an θ. The company wants to introuce a euctible so that the epecte payment per loss is $4. What euctible shoul it introuce? With a euctible of, the epecte payment per loss is given by S e π π Φ π e Setting this equal to $4 gives π Φ 4 Φ 4 π An insurance company moels loss size as following a log-logistic istribution istribution with γ an θ. The company wants to introuce a euctible with loss elimination ratio 3%. a What euctible shoul it introuce? The mean of the log-logistic istribution is Γ.5Γ.5 π, so to obtain a loss elimination ratio of 3%, the epecte payment per loss after the euctible must be 7π. The epecte payment per loss after a euctible of is introuce is S We therefore want to solve + [ tan ] + π tan

13 π tan 7π tan π 7π 3π tan 3π.5π tan.5π tan.5π 9.5 b In the following years, there is uniform inflation of 4% every year. How many years oes it take until the euctible calculate in a gives a loss elimination ratio of less than 5%? If we iscount for inflation, after n years, the euctible is equivalent to n. We want to etermine when this gives a loss elimination ration of 5%. As in part a, a loss elimination ratio of 5% means the epecte payment per loss is $75π, so to get the euctible, we solve: π tan 75π tan π 75π 5π tan 5π.5π tan.5π tan.5π We therefore want to solve n.4 n n log log.4 So the loss elimination ratio is below 5% after 6 years.. Losses follow a generalise Pareto istribution with α, τ 3, an θ 3. An insurance company introuces a euctible of $6. Calculate the loss elimination ratio of this euctible after inflation of %. 3

14 After inflation of %, losses follow a generalise Pareto istribution with α, τ 3 an θ 336. The epecte loss is therefore After the euctible, the epecte payment per loss is 6 Γ5 6 ΓΓ u u 336 u 5 u u 68u u [ u + 534u u u The loss elimination ratio is therefore Losses follow an inverse Pareto istribution with τ 4 an θ 6. a Calculate the epecte payment per claim with a policy limit of $,,. The epecte payment per claim is given by u 6 u u u u u 4 u ] 6 [4 logu u 6 3 u + 64 u log $97, 4.75 b Calculate the epecte payment per claim if there is 5% inflation the policy limit remains at $,,. 4

15 If there is 5% inflation, the loss istribution is an inverse Pareto istribution with τ 4 an θ 69. The epecte payment per claim is therefore u 69 u u u u u 4 u ] 69 [76 logu u 69 3 u u log $7, Losses follow an eponential istribution with θ 7. There is a euctible of $7, a policy limit of $5, an coinsurance such that the insurance pays 8% of the claim after the policy limit an euctible have been applie. Calculate the epecte payment per claim an the variance of the payment per claim. The epecte payment per claim is.8 5 e 7 7 e e u 7 u e 43 7 $ The epecte value of the square of the payment per claim is given by e 7.64 [ 4e 7 ] e e 43 7 e so the variance is Losses follow a Pareto istribution with α 3 an θ 5. There is a euctible of $. The insurance company wants to reuce the TVaR per claim for this policy at the 99.9% level to $6,. What policy limit shoul they set? 5

16 The VaR at the 99.9% level is the solution to The TVaR with a policy limit of u is therefore 54 + u u s 3 s [ 6 3 s ] u s u u + 5 The policy limit is therefore foun by solving u u u u 5 $9, Losses follow a Weibull istribution with τ 3 an θ 4. Loss frequency follows a Negative binomial istribution with r 6 an β 4. 6

17 The insurance company wants to reuce the epecte number of claims to. What euctible shoul it introuce in orer to achieve this? The epecte number of losses is rβ 4, so the probability that a loss leas to a claim shoul be 4. The euctible is therefore foun by solving S 4 e log log log 4 4 log 3 $, Losses follow a gamma istribution with α an θ. Loss frequency follows a Poisson istribution with λ 4. If the company introuces a euctible of $6, what is the probabilitiy that it receives more than claims after the euctible? The probability that a loss results in a claim after the euctible is e 6 e + 6.3e The claim frequency is therefore a Poisson istribution with parameter 4.3e.3, so the probability of receiving more than claims is e 5.e e e Losses follow a log-logistic istribution with γ an θ 6. Claim frequency with a euctible of $ is moelle as a negative binomial with r 5 an β.6. what woul be the istribution of the claim frequency if the euctible were remove? The probability that a loss leas to a claim is S If P z is the p.g.f. of the number of losses, then we have P. +.9z 3.6.6z 5 7

18 substituting w. +.9z, which gives z w 9, we get P w w w This is the p.g.f. of a negative binomial with β 6 9, an r 5. 8