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DeaconMet
09-07-2005, 08:53 AM
I actually have two questions regarding expected losses.
The first one is #10 from Broverman, Problem Set 4.
An insurance company issued the following fully discrete n-year endowment insurance policies, all at issue age x:
Face Amount Annual Premium Number of Policies
2 .16 100
4 .32 100

The solution for the loss when the face amount is 2 is given as follows:
2(Ax:n) -.16(a(double dot)x:n).
I understand this until I compare it to another problem, #12 in Broverman's manual, that goes as follows:
L is the loss at issue RV for a fully discrete whole life insurance of 1 on (x). The annual premium charged for this insurance is .044. You are given:
Ax =.4 a(double) x =10. Var L =.12
An insurer has a portfolio of 100 such insurances, 80 having a death benefit of 4 and the other 20 having a death benefit of 1.
On this one, the solution states as follows:
4Ax- 4(.044)(a(double dot)x).
Why is it that in this problem you multiply the annual premium by the death benefit, while in the previous one you don't?
Thanks!

3tac
09-07-2005, 09:43 AM
B/c in #10, you are given the premium for a policy with with face value of 2 is 0.16 while in #12, you are given information about a policy with face value of 1.

Does that help?

DeaconMet
09-07-2005, 04:05 PM
so when the face value of premium is 1, you multiply the death benefit by (a(double dot)?

wat?
09-07-2005, 04:14 PM
so when the face value of premium is 1, you multiply the death benefit by (a(double dot)?

I think you're missing the point. What 3tac's trying to say is that for the first question, you're given that the premium for that given insurance is 0.16. The "0.16" already factors in that the insurance has a face amount of 2. For the second question, the premium 0.044 is given for an insurance with a face amount of 1. So, to get the premium for the same insurance with a face amount of 4, you multiply the premium by 4 to get 0.176.

DeaconMet
09-08-2005, 02:50 PM
It just clciked for me! Thanks!