When a deductible of d is introduced, the frequency is reduced to p[1-Fx(d)].
This seems apparant --- the frequency would be reduced by the cumulative
distribution percent of claims below the deductible. This is true for a
straight deductible, a franchise deductible, and a disappearing deductible.
The problem givens p=0.1 (frequency before a deductible) and p[1-Fx(100)] =
..025 (frequency after $100 franchise deductible).
The ratio is then p[1-Fx(x)]/p = .025/0.1 = .25 = [1-Fx(100)] which is telling
us that 25% of claim occurrences are above the $100 deductible, and 75% below
it.
SDugan@kemperinsurance.com ("Dugan, Stephen") on 07/15/98 05:28:01 AM
To: studygroup9@lists.casact.org ("'studygroup9@lists.casact.org'") @ INTERNET
cc: (bcc: Gwendolyn L. Anderson)
Subject: Hogg & Klugman
I have a question about Hogg & Klugman from the 1996 exam, question
37.....
When there is no deductible, the expected frequency is 0.1 and the
expected severity is $200.
With a franchise deductible of $100, the expected frequency becomes
0.025 and the expected severity becomes $600.
If X is a random variable representing a claim (prior to imposing the
deductible), then what is the limited expected value E[X;100]?
Solution:
E[X;d] = E[X] - (E[W] - d)(1 - Fx(100))
Where (1 - Fx(100)) = (Frequency with Franchise Deductible)/p =
0.025/0.1 = 0.25
E[X;d] = 200 - (600 - 100)(0.25) = 75
My Question:
It is not obvious to me that (1 - Fx(100)) is simply the ratio of the
two frequencies. Since the solution hinges on knowing this "fact", can
anyone explain in words why this is? Thanks.
SD.