Let X1,X2,...XN be independent identically distributed random variables, where N is a
non-negative integer valued random variable. Let Z = X1 + X2 + : : + XN, (assuming that Z = 0 if N = 0). Find E(Z) and show that
var(Z) = var(N)E(X1)^2 + E(N) var (X1)
Not a clue on this one! I know Expectation of Z is NE(X1)
but how do I show this result?
Many Many thanks
This is a problem of a random sum of random variables, which is very useful in actuarial science.
First, I saw u already know E(sum Xi）= N*E（Xi) (A). keep this result, it's gotta be useful later.
Second, remember var(sum Xi)=E(var(sum Xi|N=n))+var(E(sum Xi|N=n)) (B).
We will work out the second part in the right side of equation (B);
var(E(sum Xi|N=n))=var(E(sum Xi)) since Xi's and N are independent
OK, substitute the E(sum Xi) result (A）in,
and var(E(sum Xi|N=n))=var(N*E(Xi))=var(N)*(E(Xi))^2
hew...
Here come the first part in the right side of equation (B),
E(var(sum Xi|N=n))=E(N*var(X)) since Xi's are independent and X and N are indep.
then, E(var(sum Xi|N=n))=E(N)*var(X)
Combine two parts, u get the result u expected. u need to show a bit more detail when u prove the first part in equation (B). I skipped some steps. If you want to be an actuary, we should talk more!
N is a random variable, known as a stopping time.
So does not make sense, the left side is constant while the right is random.
I don't know if we need independence or not between the N and the
's in the mean case, but we may need it, in obtaining the variance.
Most likely .
I just looked in up in my advisor's advisor's book.
This is correct, it's known as Wald's Equation.
proof is in...
http://en.wikipedia.org/wiki/Wald's_equation
I also found the proof to the variance on page 139 of that book.