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#1
12-21-2011, 01:16 PM
 Ionic Order Member CAS SOA Join Date: Jan 2011 Location: Seattle Posts: 1,259
Question about compound variance (aggregate loss models)

ASM section 14.2 (11th edition) shows the derivation of the compound variance formula from the conditional variance formula. I understand all of the steps, except this one:

$E_N[Var_X(X)]=Var_X(X)$ only if X is independent of N, where N is the number of claims and X is the size of each claim

I think I'm missing a very fundamental rule in probability here, but I can't remember which one is it. Any ideas?
#2
12-21-2011, 02:56 PM
 mathmajor Member SOA Join Date: Dec 2010 Studying for DP Health College: B.S. Applied Math '09 Favorite beer: Crown 'n Coke Posts: 2,470

Well. Let me take a shot... forgive my lack of coding here.

I would turn En(Varx(x)) into En[Ex(x^2)-Ex(x)^2]... just expanding out the variance. Then you can separate (I think!) into En[Ex(x^2)]-En[Ex(x)^2]

I think if N and X are independent, you can say that with respect to N, those arguments inside En(*) can be treated as constants, and then you have expectations of constants which end up being the constants themselves. The variance probably could have been treated this way from the start, but it is clearer to me to break it into expectations.

Dr. W might be of more help here. Might be wrong.
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#3
12-21-2011, 04:38 PM
 Ionic Order Member CAS SOA Join Date: Jan 2011 Location: Seattle Posts: 1,259

Quote:
 Originally Posted by mathmajor Well. Let me take a shot... forgive my lack of coding here. I would turn En(Varx(x)) into En[Ex(x^2)-Ex(x)^2]... just expanding out the variance. Then you can separate (I think!) into En[Ex(x^2)]-En[Ex(x)^2] I think if N and X are independent, you can say that with respect to N, those arguments inside En(*) can be treated as constants, and then you have expectations of constants which end up being the constants themselves. The variance probably could have been treated this way from the start, but it is clearer to me to break it into expectations. Dr. W might be of more help here. Might be wrong.
Can you elaborate on the bold part a little? Are you referring to a particular principle?
#4
12-21-2011, 09:37 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,218

The statement does look puzzling, and I don't know why I said it - cross the sentence out. It's not in the latest edition.
#5
12-23-2011, 12:55 PM
 Ionic Order Member CAS SOA Join Date: Jan 2011 Location: Seattle Posts: 1,259

Do you maintain that $E_N[Var_X(X)]=Var_X(X)$ though? If not, how do you advance with the derivation?
#6
12-23-2011, 01:57 PM
 mathmajor Member SOA Join Date: Dec 2010 Studying for DP Health College: B.S. Applied Math '09 Favorite beer: Crown 'n Coke Posts: 2,470

What I meant was, given that N, X are independent, one can say En(Ex(X))=Ex(X) (or at least it comes out this way when you use the integral definition of En(*) and replace * with Ex(X))... and from there you can get to En(Varx(X))=Varx(X)
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#7
12-27-2011, 05:59 PM
 Ionic Order Member CAS SOA Join Date: Jan 2011 Location: Seattle Posts: 1,259

Quote:
 Originally Posted by mathmajor What I meant was, given that N, X are independent, one can say En(Ex(X))=Ex(X) (or at least it comes out this way when you use the integral definition of En(*) and replace * with Ex(X))... and from there you can get to En(Varx(X))=Varx(X)
I still don't get it. Could you write out the mathematical proof for En(Ex(X))=Ex(X)? Again, I feel like I'm missing some very important principle of probability here.

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