Statistics > QUESTIONS & ANSWERS > University of California, Santa Barbara - PSTAT 172PSTAT172B Multiple Decrement HW solution (All)

University of California, Santa Barbara - PSTAT 172PSTAT172B Multiple Decrement HW solution

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1. Given the following: (i) l63 ( )   500 (ii) q63 (1)  0.05 (iii) q63 (2)  0.50 (iv) 1| 63 q(1)  0.07 (v) 2| 63 q(1)  0.042 (vi) 2 63 q(2)  0.60 (vii) l66 ( )   0 Calculate (2) d65 . SOLUTION (1) (2) (1) d d d 63 63 64     500x0.05 25; 500x0.50=250; 500x0.07=35 (2) d65  500x0.042=21 We can complete the double decrement table: x tx l 1 x d 2 x d 63 500 25 250 64 225 35 (a) 65 (b) 21 (c) 66 0 By (i) and (vi), d d 63 64 (2) (2)   500x0.6=300and (a) = 50; (b)= 225-35-50=140 and (c) = 140-21=119. 2. A life (x) is subject to two decrements: a. The associated single decrement distribution for decrement 1 is a DeMoivre distribution with ω = 5. b. The associated single decrement distribution for decrement 2 is a DeMoivre distribution with ω = 10. Find t px ()3 SOLUTION The "associated single decrement table" refers to the survival function when the decrement is acting alone. This is represented as t px !(1) , t px !(2) . t px !(1) = (5 - t)/5 0 ≤ t ≤ 5 and t px !(2) = (10 - t)/10 0 ≤ t ≤ 10. t px () = t px ! (1) . t px ! (2) = (5 - t)/5 . (10 - t)/10 (0 ≤ t ≤ 5) and = 0 (5 < t ≤ 10) 3. Using the data from Question 2, Find q x(1) and qx(2) How do q x(1) and qx(2) compare with q x!(1) and qx!(2) ? SOLUTION To derive q x(1) and qx(2): The force of mortality, µx(1) (t) = 1/(5-t) and µx(2) (t) = 1/(10-t) f (t, 1) = t p x () . µx(1) (t) = (5 - t) . (10 - t) . (1 ) 5 10 (5 - t) f (t, 1) = 10 - t ; f (t, 2) = (5 - t) 50 50 qx (1) = ʃ01 (10 - t) .dt = 0.19 and 504 qx (2) = ʃ01 (5 - t) .dt = 0.09 50 Associated single decrements: qx !(1) = 1/5 = 0.20 and qx !(2) = 1/10 = 0.10 So: qx !(1) > qx (1) etc. Note that this is intuitively reasonable: If the independent rates of decrement are uniformly distributed, then 10% of the population is (on average) not exposed to decrement (2) because it has been removed by decrement (1). So the effect of decrement (2) when working in combination with decrement (1) is 0.1 * 0.9 or 0.09, which is what we got by integration. Similarly for decrement 1. 4. An employee aged 50 is subject to 2 decrements, mortality and retirement. a. Associated single decrement mortality table follows De Moivre with ω = 100. b. Retirement, with probability µ(2) 50 (t) = 0.05. Calculate the probability that this employee remains employed at least 5 years but less than 10 years. SOLUTION Find Pr [ 5 ≤ T < 10] = 5 p50() - 10 p50() Decrement 1 is De Moivre with ω = 505 Decrement 2 is constant force of mortality of 0.05 t p50!(1) = 50 - t 50 t p50!(2) = e-0.05t So 5 p50() - 10 p50() = 45 e-0.25t - 40 e-0.50t 50 50 = 0.216 5. In a 3 decrement model, qx(1) = 0.02, qx(2) = 0.05, qx(3) = 0.03. Find px!(1), px!(2), px!(3)

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