So I'm having a lot of trouble understanding a particular aspect of this problem.

Here it is:

The annual number of losses incurred by a policyholder of an auto insurance policy, N, is geometrically distributed with parameter

Solution: let X=amount of a single loss (set

X=1

5

Because

E(S)= E(N)*E(X) (intuitive)

Var(S)= Var(N)*E^2(X) + E(N)*Var(X) (not intuitive)

(THIS is where I get hopelessly lost)

You should memorize the above formulas.

Note in the second formula above, the first item on the right hand side is Var(N)*E^2(X),

For E(X) and Var(X), plug values fr X into the BA II Plus. So we get 3.4K and 3.84K^2.

Since N starts from 0, not 1 (can have 0 losses) we treat N as the number if failures and get E(N)= 1/p - 1 = .25, Var(N)=1/p^2 - 1/p = .3125

Sooo E(S)=.85K=850, Var(S)=4.5725K^2=4,572,500($^2).

All right. Anyways, I understand that when X and N are independent of each other, E(X*N)=E(X)*E(N). BUT we were told S does NOT =X*N, but instead

Then on to the variance. I have no idea. Again, relying on that equation and the independence, it would seem to me we would get Var(X1+X2+...+Xn)=Var(X1)+Var(X2)+...+Var(Xn).

Am I totally missing something here? I would try to memorize it if I understood it..

Here it is:

The annual number of losses incurred by a policyholder of an auto insurance policy, N, is geometrically distributed with parameter

*p*= 0.8. If losses do occur, the amount of losses is either $1,000 with probability of 0.4 or $5,000 with probability of 0.6. Assume the number of losses and amounts of losses are independent. Let*S*=annual aggregate loss incurred by the policyholder. Find*E*(*S*) and*Var*(*S*).Solution: let X=amount of a single loss (set

*K*=$1,000 to make our calculation easier):X=1

*K*with probability 0.45

*K*with probability 0.6*S*=*X1*+*X2*+...+*Xn*. Please note that*S*does NOT =*NX*. Though*X1*,*X2*,...,*Xn*are independent identically distributed, it's not necessarily true that*X1*=*X2*=*X3*=...=*Xn*. This is similar to the following idea. If you throw 2 coins, you may get a head in the first coin and get a tail in the second coin, even though the number of the head or tail you get from the first coin and the number of the head or tail you get from the second coin are independent identically distributed Bernoulli random variable.Because

*N*and*X*are independent, we haveE(S)= E(N)*E(X) (intuitive)

Var(S)= Var(N)*E^2(X) + E(N)*Var(X) (not intuitive)

(THIS is where I get hopelessly lost)

You should memorize the above formulas.

Note in the second formula above, the first item on the right hand side is Var(N)*E^2(X),

*not*Var(N)*E(X). To see why, notice that Var(S=X1 + X2 + ... + Xn) is $^2 (variance of a dollar amount*S*is dollar squared); E(N)Var(X) is dollar squared. If you use Var(N)E(X) as opposed to the correct term Var(N)E^2(X), then Var(N)E(X) is a dollar amount, which will not make the equation hold.For E(X) and Var(X), plug values fr X into the BA II Plus. So we get 3.4K and 3.84K^2.

Since N starts from 0, not 1 (can have 0 losses) we treat N as the number if failures and get E(N)= 1/p - 1 = .25, Var(N)=1/p^2 - 1/p = .3125

Sooo E(S)=.85K=850, Var(S)=4.5725K^2=4,572,500($^2).

All right. Anyways, I understand that when X and N are independent of each other, E(X*N)=E(X)*E(N). BUT we were told S does NOT =X*N, but instead

*S*=*X1*+*X2*+...+*Xn*. So I understand that E(X1+X2+...+Xn)=E(X1)+E(X2)+...+E(Xn) whether there is independence or not. But how do we get from there to E(X)*E(N)? Assuming that we are relying on S as being equal to that sum. Which the solution tells us we are.Then on to the variance. I have no idea. Again, relying on that equation and the independence, it would seem to me we would get Var(X1+X2+...+Xn)=Var(X1)+Var(X2)+...+Var(Xn).

Am I totally missing something here? I would try to memorize it if I understood it..