HI, In chapter 7,Exam type question-The number of claims,N which aries in a year from a group of policies, has a negativebinomial distribution, where P(N=n)=(n+2)c(n).9^3x.1^n .The claime amount( in1000's) are independent and identically distributed as a Gamma(62) and are also independent of N. Let Y be the total claim amount arising from these policies. Calculate the standard deviation of the total claim amount. My problem is this.We know that Mean of negative binomial(k,P) distribution is K/p.but in the solution Mean is calculated by K(1-p)/p.why?
Dukerio is correct. There are two negative binomial distributions in the Tables. The PF given matches the type 2 negative binomial on page 9 of the Tables and so we need to use the mean and variance of that distribution. Most of the time in CT3 we use the type 1 negative binomial which counts the number of trials up to and including the kth success. However, since for this distribution n = k, k+1, k+2, ... it is not very useful for modelling the number of claims! Hence the use of the type 2 where n = 0, 1, 2, ... Bottom line - if they give you a PF check which distribution it matches before quoting the mean and variance.
