Problem 1. Losses are individually modelled by Exponential Random Variables Xi with mean 400. The frequency of losses is modeled by a zero-modified Poisson Random Variable N with parameter λ = 20 and
(1) Assume no deductible. What is the mean number of claims?
(2) What would we need to set the deductible to be such that the expected number of payouts is 10? If you could not compute the first part, assume the mean is 20.
Hint: Use the probability generating function of the Poisson/truncated Poisson to compute moments of the frequency distribution.Add
Problem 2. Policyholders are divided into two distinct (mutually exclusive) groups. Claims for each group are modelled as a poisson distributions with parameters λ1 = 5 and λ2 = 20. Group 1, with claims modelled as Poisson(λ1 = 5) comprises 40% of all policyholders. Assuming you do not know which group a randomly selected policyholder is in, what is the mean and variance for the number of claims the holder will submit?
Problem 3. For a nursing home insurance policy, you are given that the average length of stay is 500 days and 30% of the stays are terminated in the first 30 days. These terminations are distributed uniformly during that period.( f(x)=0.01 for x less than 30). The policy pays 50 per day for the first 30 days and 100 per day thereafter. Determine the expected benefits payable for a single stay.
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Problem 4. For a compound Poisson distribution, λ = 5 and individual losses have . Some of the pf values for the aggregate distribution S are given in Table below. Determine fS(6).
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z 0 +0.01 +0.02 +0.03 +0.04 +0.05 +0.06
0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454
0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764
0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315
1 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770
1.2 1.3 1.4 1.5 1.6 1.7 1.8
1.9
2
2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
2.9
3
0.8849As
0.9032
0.9192
0.9332
0.9452
0.9554
0.9641
0.9713
0.9772
0.9821
0.9861
0.9893
0.9918
0.9938
0.9953
0.9965
0.9974
0.9981
0.9987 sign 0.8869me 0.8888nt oje0.8925ct xam0.89
0.9049 0.9066 0.9099 0.9131
0.9207 0.9222 Pr0.8907 0.9251 E0.8944 0.9279
0.9082 0.9115
0.9345htt ps:/0.9357 0.9236 wco0.9382 0.9265 co0.9406
0.9463 0.9564 0.9474 0.9573 0.9495 0.9515 0.960
0.9649 dd /po0.9370 0.9591 der.0.93940.9686
A 9656 0.9671 0.9505 wcod
0.9719 9726 0.9484 0.9599 0.9678 0.97
0.9778 0.9826 0.9783 0.9830 0.9582 hat
0.9664 0.9738
0.9864 0.9896 0.9868 0.9898 po0.9744 0.9803 0.984
0.9920 0.9940 0.9922 0.9941WeC 0.97320.9793 0.9798 0.9881 0.990
0.9955 0.9966 0.9956 0.9967 0.9838 0.9931 0.994
0.9975 0.9982 0.9976 0.9982 0.9788 0.9875 0.9842 0.9961 0.997
0.9987 0.9987 0.9834 0.9904 0.9878 0.9979 0.998
0.9871 0.9927 0.9906 0.9989
0.9901 0.9945 0.9929 0.9925 0.9959 0.9946 0.9943 0.9969 0.9960 0.9957 0.9977 0.9970
0.9968 0.9984 0.9978
0.9977 0.9988 0.9984
0.9983 0.9989
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Standard Normal Distribution Table Φ(−z) = 1 −
Φ(z)
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Solutions
Problem 1.
(1) No deductible means that the claim distribution does not matter. The mean for a zero-modified Poisson can be found as the derivative of the Probability Generating function PM(z) at z = 1. Note that we have:
Theterm is constant with respect to z, and therefore has derivative 0. We therefore have that our mean is given by: E[claims] = Add
(2) We want to find x such that the surival function for our loss distribution if the first part was not completed). This is a straightforward computation. First, following the correct answer from the first part:
and if 20 was used:
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Problem 2. First, we compute the mean. Let N be the number of claims for the policyholder. We have:
E[N] = E[E[N|λ]] = E[λ] = 0.4(5) + 0.6(20) = 14 For Variance, we must consider the conditioning as well: Var(N) = E[N2] − E[N]2
= E[E[N2|λ]] − 142
= E[λ + λ2] − 142
= E[λ] + E[λ2] − 142
= 14 + 0.4(52) + 0.6(202) − 142 = 68 Alternatively, you can apply the Variance decomposition: Var(N) = E[Var(N|λ)] + Var(E[N|λ])
= E[λ] + Var(λ)
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= 14 + (0.4(52) + 0.6(202) − 142) = 68
Problem 3.
E(x) =500,F(30) = 0.3, and f(x) = 0.01 for 0 < x ≤ 30
Problem 4. Formula from Page 173(9.23)
), for x = 1,2······m
We have λ = 5 m = 7
f
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