Actuarial Outpost Var (S )=Var (NX )=Var (N )E2 ( X ) + E (N)Var ( X )
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#1
10-22-2008, 03:59 AM
 javen Member Join Date: Nov 2007 Posts: 64
Var (S )=Var (NX )=Var (N )E2 ( X ) + E (N)Var ( X )

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 ).

I can not understand the formular in the title...who will help me ,thanks.
#2
10-22-2008, 07:05 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,459

The formula you're asking about is the conditional variance formula. You could find more about it by Googling "conditional variance".

Here is one AO post explaining where it comes from. You should also read post 2 in that thread discussing what it means.

In particular, when you write Var(S) = Var(NX), that particular expression is misleading (I don't think you misunderstand; but you wrote it poorly).

S is the sum of N independent random variables. S = X1+X2+X3+...+Xn, where the number of terms in the sum is a random variable. So while X1 and X2 have the same distribution, X1 and X2 may have different values.

The random variable NX would be different. That would say determine N; determine X (one time). Multiply them together. That process is a random variable, but it is not the random variable you want in this problem. It would have the same mean as in this problem, but not the right variance for this problem (it would have a higher variance than in this problem).
#3
10-26-2008, 03:23 PM
 javen Member Join Date: Nov 2007 Posts: 64

Thanks a lot

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