

FlashChat  Actuarial Discussion  Preliminary Exams  CAS/SOA Exams  Cyberchat  Around the World  Suggestions 
D.W. Simpson & Company International Actuary Jobs 

Thread Tools  Display Modes 
#1




Multiple lives question
peewee,
This comes from ASM 36.20. For an insurance on (55:55) that pays a death benefit of 1000 at the moment of the first death, you are given: (i) Future lifetimes are independent. (ii) Mortality on both lives follow de Moivre's law with w=100. (iii) force of interest is .04 Calculate the net single premium for the insurance. I know this is not a very difficult problem, especially in the single life. But with multiple lives the integral gets very ugly. Is there another way to do this problem other than integrals? Thanks. Have a great Easter weekend. 
#2




Hey DRat,
Sorry for not checking this for a few days, but I spent this weekend at the beach. Here's how I would do this problem (no evaluation of integrals): The integral that would give this NSP is 1000 * Int(from 0 to 45) [v^t * t_p_xy * mu_xy(t)] dt. Notice that because of independence, t_p_xy = t_p_x * t_p_y, and mu_xy(t) = mu_x(t) + mu_y(t). So when you multiply t_p_xy * mu_xy(t), you'll get the sum of two terms: t_p_x * t_p_y * mu_x(t) + t_p_x * t_p_y * mu_y(t). Now recognize that under DML, t_p_x * mu_x(t) = 1/(wx) = 1/45 in this case. Likewise t_p_y * mu_y(t) = 1/45. So the sum of the two terms in the last paragraph is 1/45 * t_p_y + 1/45 * t_p_x = 2/45 * t_p_55 since both x and y are 55. Factoring out the constant 2/45, we get the NSP = 1000 * 2/45 * Int(from 0 to 45) [v^t * t_p_55] dt. Now recognize the integral as being a(bar)_55 = (1  A(bar)_55) / delta. Using DML, A(bar)_55 = 1/45 * a_angle45 = 1/45 * (1e^(45*.04))/.04 = .463723. So I get NSP = 1000 * 2/45 * (1  .463723)/.04 = 596. P.S. Is this the manual's answer? Please let me know what their answers to the questions are from now on, so that I'll know if I've miscalculated something. 
Thread Tools  
Display Modes  

