![]() |
|
|
|||||||
| FlashChat | Actuarial Discussion | Preliminary Exams | CAS/SOA Exams | Cyberchat | Around the World | Suggestions |
![]() |
|
|
Thread Tools | Display Modes |
|
#1
|
|||
|
|||
|
The question is as follows:
"The following table was calculated based on loss amounts for a group of motorcycle insurance policies: cj dj uj xj Pj 250 6 0 1 0 500 6 0 2 5 1000 7 1 4 9 2750 0 1 7 11 5500 0 1 1 3 6000 0 0 1 1 10,000 0 0 0 0 You are given α= 1 and β= 0. Using the procedure in the Loss Models text, estimate the probability that a policy with a deductible of 500 will have a claim payment in excess of 5500." The solution computed S(500) by taking the Kaplan-Meier estimate at the 250 deductible line but not including the 500 deductible line. I had 3 questions: 1. Why wasn't the information at the 500 line incorporated in calculating S(500)? 2. Is there a way to word the question differently so that the information at the 500 line would be included? 3. If I faced a similar situation with a Nelson-Aalen equivalent, would this still be the case? Thanks in advance. |
|
#2
|
|||
|
|||
|
1. given d=500 means data at 500 are truncated.
so, the $500 losses assumed to happen between 500 and 1000 and you include the $500 losses information in the next interval S(1000) (or S(2750)...,etc) . i'll save the latter two questions for the next person ![]() ![]() |
|
#4
|
|||
|
|||
|
Thanks to the both of you for the help
|
|
#5
|
|||
|
|||
|
I still can't see why S(500) isn't 5/6 * 9/11
|
|
#6
|
|||
|
|||
|
I would tend to agree that the definition of S(500) is Pr(t>500) rather than Pr(t>=500) so it seems strange that they would not include the deaths at time 500 to calculate S(500). Anyone have insights?
__________________
|
|
#8
|
||||
|
||||
|
http://www.aceyourexams.net/errata/E...009Changes.pdf
Quote:
|
|
#10
|
|||
|
|||
|
Digging into this, I found that, ironically, Stu liked this problem enough to put it exactly in the exercises of the new LM edition (14.35). I have the solutions manual, but it sheds no more light than Broverman.
Apparently large data sets with ranges (cj , cj+1), S(cj) stops at (cj-1 , cj). As the saying goes, with mathematics, you don't learn things, you just get used to them. |
![]() |
| Thread Tools | |
| Display Modes | |
|
|