A CONTINUOUS SEVERITY EXAMPLE
In the car insurance example, we assumed that repair or replacement costs could take only a fixed number of values. In this section we repeat some of the concepts and calculations introduced in prior sections but in the context of a continuous severity distribution.
Consider an insurance policy that reimburses annual hospital charges for an insured individual. The probability of any individual being hospitalized in a year is 15%. That is, P H( = ) = .1 015. Once an individual is hospitalized, the charges X have a probability density function (p.d.f.) f x H e X x== − 1 01 01c h . . for x > 0.
Determine the expected value, the standard deviation, and the ratio of the standard deviation to the mean (coefficient of variation) of hospital charges for an insured individual.