2017 Q18
I'm at a loss here as to how to tie part a to the source text.
- What does the given "Exposure distribution" represent?:
- it's not G(x) since it's not an increasing function (and also G(x) is the objective of this question :) )
- it's not G'(x) otherwise 1/G'(0) would be equal to the mean and 1/80% is not the mean (normalized or not normalized)
- it's also not F(x) or f(x), since the examiner's report says we should have transformed the "exposure distribution" into a distribution of losses
- in the sample answer #2, the accepted definition of G(x) is E[x;1]/E[x]. since x is the normalized loss, this implies that E[x;1] = E[x], and therefore the accepted definition of G(x) yields G(x) = 1. How does this tie to the definition of G(d) on page 101, where G(d) = E[min(x;d)]/E[x].
In my opinion this is a key problem to understand the basics of this paper, but maybe it's notation, or a poorly written CAS question, or a code-18... but there is a lot that could be clarified here.
Thanks for any clarification you can provide
Comments
The given exposure distribution is the distribution of possible losses including the possibility of no loss occurring. The exposure distribution used in the Bernegger paper is the severity distribution, which is why we condition on a loss occurring to get the Bernegger exposure distribution.
In sample answer 2 the CAS has yet another typo. E[x;1] = E[x] because d = X/M so d=1 is the largest possible value and is equivalent to not limiting the losses to a value below the maximum possible loss.
The CAS should have written G(d) = E[x;d]/E[x].
You can see the CAS has a typo because they compute E[x;0.25], E[x;0.5], etc. and then calculate g(0.25), g(0.5), etc. by varying the numerator but not the denominator. If x was truly the variable in their equation for G(x) then both the numerator and denominator would be varying with x.