April 2010, Q4 (i)

Why is the probability that an individual claim involves reinsurer not the:

Integral (from M to infinity) of (X-M) times f(x)dx?

 

Did you mean to ask why reinsurer's part is : integral (from M to infinity) of (X-M).f(x).dx ?

 

Actually the solution says that the probability involving reinsurer is:

Integral (from M to infinity) of fx.dx

But I think it should be:

Integral (from M to infinity) of (X-M) times f(x)dx


Can someone correct me if Im wrong?

 

What you are thinking is theoretically correct if it is asked to find out reinsurer's expected amount of claim. It takes part when claim amount (X) exceeds M (retention). But it also depends on what exactly the question asked to find. Can you please post screenshots of both question and solution?

 

Question:

The number of claims N on a portfolio of insurance policies follows a binomial
distribution with parameters n and p. Individual claim amounts follow an exponential
distribution with mean 1/λ. The insurer has in place an individual excess of loss
reinsurance arrangement with retention M.
(i) Derive an expression, involving M and λ, for the probability that an individual
claim involves the reinsurer.

Solution:

Let X represent the distribution of individual claims. Let denote the
probability that an individual claim involves the reinsurer. Then

P(X>M) =

 

contd.:

P(X>M) = Integral (from M to infinity) of pdf of exponential distribution.

 

Yes it is correct. They are finding out the probability of a claim that involves reinsurer. If they would ask to find expected amount of claim handled by reinsurer then you had to use integral (from M to infinity) f(x).(X-M).dx

 

Thanks!