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.