

FlashChat  Actuarial Discussion  Preliminary Exams  CAS/SOA Exams  Cyberchat  Around the World  Suggestions 
DW Simpson International Actuarial Jobs 
ShortTerm Actuarial Math Old Exam C Forum 

Thread Tools  Search this Thread  Display Modes 
#31




Question #274
The answered provided by SOA is as follows:
S(x) = e^H(x) H(x) = 0.5534 (I got this part.) NelsonAalen Estimate = 1/50 + 3/49 + 5/k + 7/12 = 0.5534 > k = 36. Why the last term is NOT 7/21? Did I miss something? Please let me know at your earliest convenience thanks. L
__________________
BBA, MA, SOA Exams: P/1 (Passed in Nov 2015) FM/2 (Passed in Feb 2016) MLC /3M (Wrote in May, waiting for result) C/4 (Taking the challenge next week) MFE/3F (Taking the challenge next month) 
#32




Quote:
__________________
ASA 
#35




SOA Question 2:
The number of claims has a Poisson Dist. Claims have a Pareto dist with parameters theta = .5 and alpha =6. The number of claims and claim sizes are independent. The observed pure premium should be within 2% of the expected pure premium 90% of the time. Calculate the expected number of claims needed for full credibility. What key words in this question indicate that we are looking for number of claims needed for full credibility of aggregate losses?
__________________

#36




pure premium

#37




SOA Question 4:
How is a * 3^(a1) * a * 6^(a1) * a * 14^(a1) * (25^a)^2 proportional to: a^3 * (3*6*14*625)^a I understand how we get a^3, but I am not sure for the second piece how the exponent for all numbers can be a when for three of those values, the exponent was (a1)?
__________________

#38




I think you have a type in your last term.

#39




Quote:
So now we have to deal with 3^{a1} * 6^{a1} * 14^{a1} * 25^{2a} = 3^{1} * 3^{a} * 6^{a} * 6^{1} * 14^{a} * 14^{1} * 625^{a} Everything that is not raised to the a'th (or a'th) power is a constant so hence we can say it's proportional to and you are left with (3*6*14*625)^{a} 
#40




Quote:
__________________

Thread Tools  Search this Thread 
Display Modes  

