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Short-Term Actuarial Math Old Exam C Forum

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Old 06-01-2018, 11:38 PM
kklin kklin is offline
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Default ASM Example 51E

I do not understand the reasoning behind the 2 different ways in obtaining the Buhlmann Z (i.e. when ‘an individual is selected’ for part 1 vs when ‘a group is selected’ for part 2).

Precisely, why does part 2 treat the process as a mixed distribution? Furthermore, how do I know in the future when to treat the process as a mixed distribution?

Any help would be much appreciated!
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Last edited by kklin; 06-02-2018 at 01:59 PM..
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Old 06-06-2018, 07:26 PM
Academic Actuary Academic Actuary is online now
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As no one else seems to respond, I will give my input. This solution is needlessly complicated. If we are looking at the entire population the are 4 Poissons with lambdas
(.2,.4,.6,1.2) with probabilities (.375,.375,.125,.125). By creating two lists, the calculator with give you the mean (EVPV) and the variance (VHM) which will give us the k for case 1.

In case 2, we have group one with lambdas (.2,.4) and conditional probabilities (.5,.5) while for group two we have (.6,1.2) with probabilities (.5,.5). The the mean and variance for group one are (.3,.31) where the second term is the mean plus the variance of the lambdas while for group 2 it is (.9,.99). Using the probabilities (.75,.25) you can get the mean of the variances and variances of the mean.

In case one it is for the same risk while for case two it is for a risk from the same group.
I wouldn't get hung up on the term mixed distribution as I would consider case 1 also a mixed distribution.
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