Course 4, ASM, Problem 70.10

[J-Man]

I have a question regarding 70.10. For this question, there is an insurance portfolio of 2 classes of insureds, classes A and B, m_j is the number of exposures in a class for a specific year, and the probablility that a risk from class A is selected is 2/3, whereas for B it is 1/3.

For class A, the mean loss per exposure is 2100 with variance 15000+10000/m_j.
For class B, the mean loss per exposure is 3000 with variance 30000+40000/m_j.

My question amounts to how to determine the overall variance for a risk drawn from these classes. The manual's author computes it to be
v=(2/3)(15000+10000/m_j) + (1/3)(30000+40000/m_j), so v=20000+20000/m_j. But is the overall variance for these risks equal to the expected value of the process variance?

What I did first was to back out E(A^2) (from Var(A) and E(A)^2) and
E(B^2), weighed these to get E(Tot^2), then subtracted E(Tot)^2 to get Var(Tot). This gives an answer different from v in the previous paragraph. (Using this method, I got 200000+20000/m_j.) Why is this wrong?

Thanks for helping!

[Daisy]

isn't total variance = VHM + EVPV? I don't have the manual in front of me...but try computing the VHM and the EVPV and go from there...or if the problem is really asking for the EVPV...then their solution is correct. The defn of the EVPV is the weighted average of the process variance for each group....in this case using 1/3 and 2/3 as weights for each of the process variances. I apologize if this is incorrect...as I said before, I don't have the book in front of me.

[M.]

v is just EPV. He calculates "a" later in the problem, which is VHM. The sum of the two would be total variance.