View Full Version : Exam5: Krakowski Exhibit 5

Graduate

03-18-2004, 01:09 PM

"Revised" Exhibit5, Page310:

i) How the "Total Variance=1.012" is found?

ii) 2nd last line: Chosen Regional Damage ratio=0.69 Is this a given quantity or are we required to find it?

I would be grateful if someone can help me out, Thanks.

(1)1.012 is the total variance of the 66 figures given above.

(2)0.69 I think is chosen by the actuary according to the foregoing calculations and other considerations.(I am not sure)

Mel-o-rama

04-24-2004, 02:34 PM

Could some please fill in some more gaps for me? I don't seem to understand how the "Estimated Process variances" and the "Estimated VHM" were chosen.

It appears the the individual "Estimated Process variances" are a straight average between the individual "Process variances" and the overall "Average Process Variance". What is the reason for doing this? If I were doing this, I would have just used the "Average Process Variance" as the "expected" process variance in my "k" for all states. I suppose the purpose is to develop different credibilities for each of the states, so it would be inappropiate to apply 1 expected process variance to all states. But why average the individual actual process variances with the average process variance? That seems somewhat arbitrary and circular to me. Is this something I should just accept and move on?

Also, the "Estimated VHM" appears to be the mean between the actual variance of the means (0.098) and the difference between the total variance and the average process variance (1.012 - 1.007). I don't understand why we just don't use the variance of the means like we've been taught elsewhere. From what I can understand on the top of page 298, there appear to be 3 ways to estimate the "variance of the hypothetical means". The first method is what I've been taught before. I don't think I've heard the other 2 methods. Then Krawkowski seems to be saying that due to the data itself, it turns out that an average between the first 2 methods is the best way to go. Does this imply that if the data were different, that we might have to choose differently?

Trying to come back out of panic mode - if we see a question just like this on the exam, do you think it's safe to assume that

E(Var(X)) = the mean between [actual Var(X) by state] and [overall avg Var(X)]

and

Var(E(X)) = the mean between [Var(mean(X) by state)] and [Var(everything) minus overall avg Var(X)]?

horsePower

04-24-2004, 03:49 PM

(1)1.012 is the total variance of the 66 figures given above.

(2)0.69 I think is chosen by the actuary according to the foregoing calculations and other considerations.(I am not sure)

Yeh i am pretty sure it is a given. (I spent a LOT of time trying to figure this out).

vBulletin® v3.7.6, Copyright ©2000-2014, Jelsoft Enterprises Ltd.