Actuarial Outpost SOA Exam STAM #328
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 Short-Term Actuarial Math Old Exam C Forum

#1
06-08-2019, 01:53 PM
 joelcheung SOA Join Date: Mar 2018 Location: Singapore Studying for Exam STAM College: Singapore Management University, sophomore Posts: 9
SOA Exam STAM #328

The solution states that 'For losses in AY4 the projection is 2.333 years', and 'For losses in AY3 the projection is 3.333 years'.

May I check what is the start and end points of the abovementioned durations?

Taking the case of the losses in AY4, is the 2.333 years measured from...

1. July 1 AY4 to Nov 1 AY6 (average accident date for policies sold from Nov 1 AY5)?

2. July 1 AY5 (the average accident year of policies sold from July 1 AY4 that will be in effect through July 1 AY6 [This is based on ACTEX STAM-841 Section 41*]) to Nov 1 AY7?

*:This section states that policies can become effective from Jan 1 Year Z to Dec 31 Year Z and complete data for policy year Z is not available until Dec 31 of Year Z+1, and thus average accident date is Dec 31 Year Z.

Thank you!
#2
08-24-2019, 09:58 PM
 gauchodelpaso Member SOA Join Date: Feb 2012 College: Eastern Michigan U Posts: 180

The way that I understand it , but I may be wrong, is that accident year four, goes from January 1 CY4 to December 31 CY4, with midpoint for all accident dates of July 1 CY4. I normally write this time as 4.5. It does not matter when the policy was written, as long as the accident date is within CY4.

Now the midpoint for the accidents of the new rate is November 1 CY6 or t=6 5/6=6.8333...

So the difference is 6.833-4.5=2.333
Same for AY3, and 3.333 as the difference between their midpoints.
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 Tags #328, average accident date, exam stam, ratemaking