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#1
11-03-2005, 10:05 PM
 ed999 Member CAS AAA Join Date: May 2002 Location: NYC Posts: 1,034
B and W constants

one question where they asked you to comment on the fact that B and W were constants. I didn't exactly reproduce the list from all 10 but simply said that if B and W were constants, the weight given to large insureds would be too high while the weight for smaller insureds to low...

do you think this makes sense???
#2
11-03-2005, 10:10 PM
 berad Join Date: May 2005 Posts: 4

that's what I put down. I tried to compare the B & W's in the prior plan to the newer one.
#3
11-03-2005, 10:17 PM
 Jack Divine Member Join Date: Nov 2003 Posts: 97

I agree - the loss variation in large insureds is greater than the law of large numbers indicates. Therefore, constant B&W values give too much weight to large insured experience, and not enough to small insureds.
#4
11-03-2005, 10:58 PM
 Avi Wiki ContributorSite Supporter Site Supporter CAS AAA Join Date: Aug 2002 Location: NY Studying for the rest of my life. College: Alumnus - Queens College - CUNY Favorite beer: Stone Ruination IPA Posts: 12,501 Blog Entries: 3

Being that Z = (E+B)/(E+W), I said that holding B&W fixed and letting E get large brings Z close to 1, which gives too much experience to large insureds.
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#5
11-03-2005, 11:04 PM
 lee Member Join Date: May 2004 Posts: 139

Quote:
 Originally Posted by Avi Being that Z = (E+B)/(E+W), I said that holding B&W fixed and letting E get large brings Z close to 1, which gives too much experience to large insureds.
Zp = E/E+B, Ze =E/E+K approaches 1, when E approaches infinity...
Large insured will be seft-insured.

Please check whether it is correct.

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