(1) An insurer offers group term insurance on 250 mutually independent lives for
ID: 3051445 • Letter: #
Question
(1) An insurer offers group term insurance on 250 mutually independent lives for a premium of 285. The probability of a claim is 0.02 for each life. The distribution of the number of lives by benefit amount is:
Benefit Amount Number Covered
20 100
50 100
100 50
(a) Reinsurance is purchased which costs 120% of expected claims above a retention limit of 40 per life. Using the normal approximation, determine the probability that the total of retained claims and reinsurance premiums will exceed the premium. (Ans 0.348)
(b) Calculate the expected value of the claims paid by a stop-loss reinsurance coverage if there is a retention limit of 40 per life and the stop-loss deductible amount on the business retained is 300. (Ans 0.8408, using Excel functions PI, SQRT, EXP, NORMSDIST)
(c) Calculate the retention limit that minimizes the probability that the total of retained claims and reinsurance premiums will exceed the premium. Assume that the limit is between 20 and 50. (Ans 32)
Explanation / Answer
Benefit 20:
2% chance of paying 20, 98% chance of paying 0
E(X) = (.02)(20) = .4
E(X2 ) = (.02)(20)2 = 8
Var(X) = 8 - .42 = 7.84
Premium = 0
Benefit 50:
2% chance of paying 40, 98% chance of paying 0
E(X) = (.02)(40) = .8
E(X2 ) = (.02)(40)2 = 32
Var(X) = 32 - .82 = 31.36
Premium = (1.2)(.02)(10) = .24
Benefit 100:
2% chance of paying 40, 98% chance of paying 0
E(X) = (.02)(40) = .8
E(X2 ) = (.02)(40)2 = 32
Var(X) = 32 - .82 = 31.36
Premium = (1.2)(.02)(60) = 1.44
Let T = The sum of all premiums received minus claims and premiums paid
E(T) = 285 - (100(.4+0) + 100(.8 + .24) + 50(.8 + 1.44)) = 29
Var(T) = 100(7.84) + 100(31.36) + 50(31.36) = 5488
P(T<0) = P(Z < (0 - 29)/Sqrt[5488]) = P(Z<-.3915) = .347714
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