Claims on a particular type of insurance policy follow a compound Poisson process with an annual claim rate per policy of 0.4. Individual claim amounts are Exponentially distributed with mean 120. In addition, for a given claim and independent of its size, there is a probability of 20% that an extra claim handling expense of 30 is incurred. The insurer charges an annual premium of 60 per policy. Estimate, using a Normal approximation, the minimum number of policies to be sold so that the insurer has at least a 99% probability of making a profit. Let the individual total claim costs be denoted by X. Then X = Y + Z where Y is the cost of the claim and Z is the claim handling expense. Var(Y) = 120^2 = 14,400 Var(Z) = 0.2 * 30^2 - 6^2 = 144 Where did 6^2 comes from?
Hi, I would like to additionally ask, from what is E(Z^2) = 0.2*30^2 derived? Do we assume that Z~Bin(30,0.2), in which case E(Z^2) = 30*0.2*0.8 + (30*0.2)^2?
S=sum(X)~compound Poisson N~Poisson(lambda) X~?(parameter) Var(X)=E(X^2)-E^2(X)=>E(X^2)=Var(X)+E^2(X) Becasue E(N) & VAR(N) of Poisson are lambda, we could combine original equation of Var(X)=E(N)*Var(X)+Var(N)*E^2(X) become lambda*Var(X)+lambda*E^2(X)=lambda*E(X^2).
Hello Z is a discrete RV that can take the values 0 or 30 with probabilities 0.8 and 0.2, respectively. For a discrete RV, we have: E[h(Z)] = sum(z):[h(z) * P(Z = z)] So: E[Z^2] = 0^2 * 0.8 + 30^2 * 0.2 = 0.2 * 30^2. Hope that helps! Andy
