Actuarial Outpost Severity modification with discrete severity
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#1
06-23-2017, 08:01 PM
 mattcarp Member CAS Join Date: Sep 2016 Studying for Exam 6 College: UC Berkeley Posts: 381
Severity modification with discrete severity

Suppose frequency is neg.bin and severity is geometric. If I want to eliminate the probability of X=0 do I multiply beta in the frequency rv by 1-P (X=0)? Then if I have N* then P(S=1)=P(N*=1)P(X=1) where X is the unmodified geometric. Is this how it works?
#2
06-23-2017, 08:38 PM
 aNYthing Member CAS Join Date: Aug 2014 Location: NYC Studying for Exam 5 Posts: 524

No, you need the payment per payment variable for the Geometric RV.
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#3
06-23-2017, 09:01 PM
 mattcarp Member CAS Join Date: Sep 2016 Studying for Exam 6 College: UC Berkeley Posts: 381

Quote:
 Originally Posted by aNYthing No, you need the payment per payment variable for the Geometric RV.
Do you know how to find the payment per payment rv when I am given an aggregate deductible? I'm given that the aggregate deductible is 2, and I'm trying to calculate P(S=0) and P(S=1). If I have N* then P(S=0) = P(N*=0). Are you saying that P(S=1) = P(N*=1)P(Y^p = 1). The concept of payment per payment is confusing in the context of aggregate deductible.

Last edited by mattcarp; 06-23-2017 at 09:07 PM..
#4
06-23-2017, 09:08 PM
 aNYthing Member CAS Join Date: Aug 2014 Location: NYC Studying for Exam 5 Posts: 524

Quote:
 Originally Posted by mattcarp Do you know how to find the payment per payment rv when I am given an aggregate deductible?
Do it like you'd do any other problem with an aggregate deductible and discrete severity distribution, but when finding P(S=1), P(S=2), etc, use the modified frequency distribution and the payment per payment distribution (divide each severity probability by P[X>0]).
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#5
06-23-2017, 09:11 PM
 aNYthing Member CAS Join Date: Aug 2014 Location: NYC Studying for Exam 5 Posts: 524

Quote:
 Originally Posted by mattcarp Are you saying that P(S=1) = P(N*=1)P(Y^p = 1). The concept of payment per payment is confusing in the context of aggregate deductible.
Yeah, where P[Y^p=1] = P[X=1]/P[X>0]. It confused me too.
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#6
06-24-2017, 08:39 AM
 mattcarp Member CAS Join Date: Sep 2016 Studying for Exam 6 College: UC Berkeley Posts: 381

Wow this really helps a lot. Thank you!