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A Reading of the Theory of Life Contingency Models A Preparation for the Actuarial Exam MLC/3L Marcel B. Finan Arkansas Tech University c



All Rights Reserved Answers Key



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The answer key manuscript is to help the reader to check his/her answers against mine. I am not in favor of providing complete and detailed solutions to every single problem in the book. The worked out examples in the book are enough to provide the user with the skills needed to tackle the practice problems. This manuscript should not be made public or shared with others. Best of wishes. Marcel B. Finan Russellville, Arkansas July 2011



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Section 18 18.1 No 18.2 III 18.3 A = 1, B = −1 18.4 0.033 18.5 0.9618 18.6 0.04757 18.7 s(0) = 1, s0 (x) < 0, s(∞) = 0 18.8 0.149 18.9 1 − e−0.34x , x ≥ 0 18.10



x2 , 100



x≥0



18.11 I 18.12 (a 0.3 (b) 0.3 18.13 1 −



x , 108



x≥0



18.14 (a) 18.15 (x + 1)e−x 18.16 0.34e−0.34x 18.17 λe−λx



5 18.18 f (x) =



 



7 , 16 3x , 8







0,



0 n so that Z 1 −n| Zx n K+1 Z 1 = ν n and n| Zx = ν K+1 . Thus, x:n = ν −ν = ν¨ a = n| Z¨x . x:n



d



d



K+1−n



37.18 This follows from the previous problem by taking expectation of both sides.



41



37.19 n| a ¨x = n Ex a ¨x+n =



ν n (px )n 1−e−(δ+µ)



=



e−n(δ+µ) . 1−e−(δ+µ)



37.20 0.4151 37.21 0.45 37.22 16.6087 37.23 15.2736 37.24 58.36 37.25 3.30467 37.26 We have ∞ X



ex =



k px =



∞ X



pxk−1 px+1 = px + px



k−1 px+1



= p(x)(1 + ex+1 ).



k=2



k=1



k=1



∞ X



(b) 0.0789. 37.27 ax =



e−(µ+δ) 1−e−(µ+δ)



37.28 We have ax =



∞ X



k



ν k Px = ∞ X



37.30 0.1782 37.31 ax:n = e 37.32 11.22



1−e−n(µ+δ) 1−e−(µ+δ)



ν k−1 k−1 px+1



k=1 ∞ X k=1



37.29 7.6







ν pxk−1 px+1 = νpx



∞ X



ν k−1 k−1 px+1 ) = νpx (1 +



k=2



−(µ+δ)



k



k=1



k=1



=νpx (1 +



∞ X







ν k k px+1 ) = νpx (1 + ax+1 )



42



Section 38 38.1 12.885 38.2 13.135 38.3 A80 = 0.8162 and a ¯80 = 2.5018 38.4 15.5 38.5 8.59 38.6 We have



(m) a ¨x:n



mn−1 1 X k = ν m k px m m k=0 mn



=



1 1 X k 1 + ν m k px − ν n n px m m m k=1 m (m)



=ax:n +



1 (1 − n Ex ). m



38.7 We have



(m)



a ¨x:n =¨ a(m) − n| a ¨(m) x x i − i(m) i d i − i(m) − a ¨ + x n Ex n| i(m) d i(m) d(m) i(m) d(m) i(m) d(m) i d i − i(m) = (m) (m) (¨ ax − n| a ¨x ) − (m) (m) (1 − n Ex ) i d i d (m) i d i−i = (m) (m) a ¨xn − (m) (m) (1 − n Ex ). i d i d



≈



i



d



a ¨ − (m) x



43 38.8 (a) We have a(m) =¨ a(m) − x x



1 m



i − i(m) 1 − (m) (m) (m) i d i d m i d i − i(m) 1 = (m) (m) (ax + 1) − (m) (m) − i d i d m 1 (m) i d i d i−i (1 − ν m )i(m) = (m) (m) ax + (m) (m) − (m) (m) − i d i d i d i(m) d(m) d(m) − d i d = (m) (m) ax + (m) (m) . i d i d



≈



i



d



a ¨ − (m) x



(b) We have (m)



1 (1 − n Ex ) m i d i − i(m) 1 ≈ (m) (m) a ¨xn − (m) (m) (1 − n Ex ) − (1 − n Ex ) i d i d m (m) i d i−i 1 = (m) (m) (1 − n Ex + ax:n ) − (m) (m) (1 − n Ex ) − (1 − n Ex ) i d i d m i d d(m) − d = (m) (m) ax:n (m) (m) (1 − n Ex ). i d i d (m)



ax:n =¨ ax:n −



38.9 (a) We have (m)



a ¨x:n =¨ a(m) − n| a ¨(m) x x m−1 m−1 − n| a ¨x + ≈¨ ax − n Ex 2m 2m m−1 (1 − n Ex ). =¨ ax:n − 2m (b) We have 1 m m−1 1 ≈¨ ax − − 2m m m−1 1 =ax + 1 − − 2m m m−1 =ax + . 2m



a(m) =¨ a(m) − x x



44 (c) We have (m) n| ax



(m)



=n Ex ax+n   m−1 ≈n Ex ax+n + 2m m−1 =n Ex ax+n + n Ex 2m m−1 =n| ax + n Ex . 2m



(d) We have (m)



ax:n =a(m) − n| ax x m−1 m−1 − n| ax − ≈ax + n Ex 2m 2m m−1 =ax:n + (1 − n Ex ). 2m 38.10 (a) We have (m)



(m)



a ¨x:n =¨ a(m) − n Ex a ¨x+n x m − 1 m2 − 1 ≈¨ ax − − (µ(x) + δ) 2m 12m m − 1 m2 − 1 − (µ(x + n) + δ) −n Ex (¨ ax+n − 2m 12m   m−1 m2 − 1 =¨ ax:n − (δ + µ(x) − n Ex (δ + µ(x + n))). (1 − n Ex ) − m 12m (b) The result follows by letting m → ∞ in the 3-term Woolhouse formula. (c) The result follows by letting m → ∞ in (a)



45



Section 39 39.1 218.79 39.2 8.56 39.3 5.1029 39.4 5.7341 39.5 5.3465 39.6 204.08 39.7 4.4561 39.8 (I¯ a)x =



R∞



39.9 (I¯ a)x =



Rn



0



0



dteν t i px dt



dteν t i px dt



39.10 (D¯ a)x:n =



Rn 0



dn − teν t i px dt



46



Section 40 40.1 0.2 ¯ x ) = 0.15940 40.2 P¯ (A¯75 ) = 0.02901 and Var(L 40.3 0.2 40.4 0.7125 40.5 0.1 40.6 0.05137 1 ¯ 1 ) = 0.1553 ) = 0.02 and Var(L 40.7 P¯ (A¯x:n 75:20 1 )= 40.8 P¯ (A¯x:n



1 δ



1 (1−e−nδ ) δ(ω−x) 1 1− δ(ω−x) (1−e−nδ )−e−nδ



(



n (1− ω−x ))



and



 ¯ 1 ) = 1 +  Var(L x:n



1 (1 δ(ω−x)



1−



1 (1 δ(ω−x)



−



2



− e−nδ )



e−nδ )



−



e−nδ



1−



n ω−x








 



1 1 (1 − e−nδ )2 × (1 − e−nδ ) − 2 2δ(ω − x) δ (ω − x)2 1 ¯ 1 ) = 0.13694 40.9 P¯ (A¯75:20 ) = 0.02402 and Var(L 75:20



40.10 0.25285 40.11 0.47355 40.12 0.04291 40.13 0.09998 40.14 We have



h P¯ (A¯x:n ) =



1 (1 δ(ω−x)



1−



−nδ



−e



1 (1 δ(ω−x)



)+e



−nδ



1−



n ω−x



− e−nδ ) − e−nδ 1 −




i



n ω−x



δ








47



¯ x:n ) = Var(L



1 (1 2δ(ω−x)




 1 n − e−2nδ ) + e−2nδ 1 − ω−x − δ(ω−x) (1 − e−nδ ) + e−nδ 1 − 
2 n 1 (1 − e−nδ ) − e−nδ 1 − ω−x 1 − δ(ω−x)



40.15 0.04498 40.16 0.10775 40.17 0.06626 40.18 0.4661 40.19 0.0229 40.20 0.42341 40.21 We have A¯1 A¯ 1 1 ) + P¯ (A¯x:n1 )A¯x+n = x:n + x:n A¯x+n P¯ (A¯x:n a ¯x:n a ¯x:n 1 ¯ A + n| A¯x ¯ = x:n Ax+n a ¯x:n A¯x = a ¯x:n =n P¯ (A¯x ) 40.22 We have A¯x:n A¯1 1 P¯ (A¯x:n ) + P¯ (A¯x:n )= + x:n a ¯x:n a ¯x:n 1 ¯ ¯ Ax:n + Ax:n = a ¯x:n 1 ¯ A = x:n = P¯ (A¯x:n1 ) a ¯x:n 40.23 0.01657 40.24 0.03363



n ω−x




2



48 40.25 0.0498 40.26 1.778 40.27 0.7696 40.28 −5.43 40.29 −14.09 40.30 0.005



49



Section 41 41.1 12381.06 41.2 124.33 41.3 P (Ax ) =



Ax a ¨x



=



qx qx +i 1+i qx +i



= νqx



41.4 16076.12 41.5 33.15 41.6



4 105



41.7 From the definition of P (Ax ) and the relation Ax + dax = 1 we can write 1 − dax Ax = ax ax P (Ax )ax =1 − dax ax (P (Ax ) + d) =1 1 ax = P (Ax ) + d P (Ax ) =



41.8 33.22 41.9 We have 1 1 Lx:n =Z x:n − P Y¨x:n   1 − Zx:n 1 =Z x:n − P d   1 − Z x:n1 1 − Z x:n 1 =Z x:n − P d   P P P 1 Z x:n + Z x:n1 − = 1+ d d d



41.10 0.317 41.11 2410.53



50



41.12 0.0368 41.13 281.88 41.14 −10877.55 41.15 261.14 41.16 0.2005 41.17 0.087 41.18 This follows easily by dividing 1 ) + Ax:n1 Ax:n = Ax:n



by a ¨x:n 41.19 We have 1 Ax:n A 1 + x:n Ax+n a ¨x:n a ¨x:n 1 A + Ax:n1 Ax+n = x:n a ¨x:n Ax = = n P (Ax ) a ¨x:n



1 P (Ax:n ) + P (Ax:n1 )Ax+n =



41.20 0.00435 41.21 0.03196 41.22 0.03524 41.23 0.51711



51 41.24 We have 



n| Lx



1 − Zx d








P P 1 =n| Zx − P = n| Zx − + Z x:n + n| Zx d d   P P 1 P = 1+ n|Zx + Z x:n − d d d



41.25 Note first that 1 K+1 Z x:n I(K ≥ n)ν K+1 I(K ≤ n − 1) = 0. n| Zx = ν



Thus, " E



n| Lx



+



P d



2 #



"  #  2 2 P P 1 =E (Z x:n )2 + 1 + (n| Zx )2 d d  2  2 P P 2 1 + 1+ (2 Ax:n = n| Ax d d



41.26 The loss random variable is ν K+1 I(K ≥ n) − P a ¨min (K+1,t) = n| Zx − P Y¨x:t . The actuarial present value is n| Ax



− Pa ¨x:t



41.27 The benefit premium which satisfies the equivalence principle is t P (n| Ax )



41.28 0.01567 41.29 13092.43 41.30 0.024969



=



n| Ax



a ¨x:t



52



Section 42 42.1 0.0193 42.2 0.0256 42.3 0.0347 42.4 This is the benefit premium for a 20-payment, semi-continuous whole life insurance issued to (40) with face value of 1000 42.5 0.04575 42.6 0.0193 42.7 0.0289 42.8 0.829 42.9 0.0069 42.10 11.183 42.11 −12972.51 42.12 0.0414 42.13 0.0620 42.14 0.0860



53 42.15 We have P (A¯x:n − n P (A¯x ) A¯x:n − A¯x = P (Ax:n1 ) Ax:n1 A¯1 + Ax:n1 − A¯x = x:n Ax:n1 A 1 − n A¯x = x:n 1 Ax:n A 1 − Ax:n1 A¯x+n = x:n = 1 − A¯x+n . Ax:n1 1 + Ax:n1 42.16 This follows from the formula A¯x:n = A¯x:n



42.17 0.0096 42.18 0.0092 42.19 We have Ax:n1 A¯x+n a ¨x Ax:n1 A¯x+n = . a ¨x:n + n Ex a ¨x+n



P (n| A¯x ) =



42.20 77079



54



Section 43 43.1 231.64 43.2 122.14 43.3 331.83 43.4 493.58 43.5 94.83 43.6 224.45 43.7 117.52 43.8 325.19 43.9 484.32



55



Section 44 44.1 7.747π 44.2 102 44.3 0.078π 44.4 0.88π 44.5 15.02 44.6 5.1 44.7 19.07 44.8 73.66 44.9 397.41 44.10 1.276 44.11 478.98 44.12 3362.51 44.13 900.20 44.14 17.346 44.15 3.007986 44.16 15513.82



56



Section 45 45.1 0.0363 45.2 0.0259 45.3 0.049 45.4 0.07707 45.5 0.02174 45.6 (a) E(Lx ) = bAx − π¨ ax (b) Var(Lx ) = b + 45.7 33023.89 45.8 27 45.9 0.208765 45.10 36.77




π 2 2 [ Ax d



− (Ax )2 ]



57



Section 46 46.1 We have ¯ (A¯x ) =A¯x+t − P¯ (A¯x )¯ ax+t   1 − δ¯ ax =(1 − δ¯ ax+t ) − a ¯x+t a ¯x a ¯x+t + δ¯ ax+t =1 − δ¯ ax+t − a ¯x a ¯x+t =1 − . a ¯x



tV



46.2 8.333 46.3 0.04 46.4 0.0654 46.5 1.6667 46.6 0.1667 46.7 0.47213 46.8 0.20 46.9 0.14375 46.10 0.3 46.11 0.1184 46.12 0.1667 46.13 0.1183 46.14 0.1183



58 46.15 we have ¯ (A¯x ) =A¯x+t − P¯ (A¯x )¯ ax+t  ¯ Ax+t − P¯ (A¯x ) =¯ ax+t a ¯x+t
=¯ ax+t P¯ (A¯x+t ) − P¯ (A¯x )



tV



46.16 0.1183 46.17 0.0654 46.18 The prospective formula is ¯ (A¯50 ) = A¯60 − P¯ (A¯50 )¯ a60 .



10 V



The retrospective formula is ¯ ¯ 10 V (A50 ) =



1 P¯ (A¯50 )¯ a50:10 − A¯50:10 10 E50



46.19 We have 1 P¯ (A¯x )¯ ax:n − A¯x:n ¯ ¯ V ( A ) = t x n Ex



=



P¯ (A¯x ) −



¯1 A x:n a ¯x:n



n Ex



a ¯x:n 1 P¯ (A¯x ) − P¯ (A¯x:n ) = 1 ¯ P (Ax:n )



46.20 True 46.21 0.0851



59 46.22 We have 1 ¯ (A¯x:n ) =A¯



tV



1 − P¯ (A¯x:n )¯ ax+t:n−t   µ 1 − A¯x:n (n−t)(µ+δ) = (1 − e )−µ µ+δ δ " # µ (n−t)(µ+δ) (n−t)(µ+δ) (1 − e ) − e 1 − µ µ+δ = (1 − e(n−t)(µ+δ) ) − µ µ+δ δ   µ µ2 µ − + (1 − e(n−t)(µ+δ) ) = µ+δ δ δ(µ + δ) 1 x+t:n−t



=0 × (1 − e(n−t)(µ+δ) ) = 0 46.23 Follows from the previous problem. 46.24 0.0294 1 1 1 46.25 t V¯ (A¯x:n )=a ¯x+t,n−t [P¯ (A¯x+t:n−t ) − P¯ (A¯x:n )]   ¯1 ) P¯ (A 1 1 x:n ¯ ¯ ¯ 46.26 t V (Ax:n ) = Ax+t:n−t 1 − P¯ (A¯ 1 ) x+t:n−t



46.27 0.4207 46.28 0.3317 46.29 Recall that a ¯x:n =



1 − A¯x:n . δ



Thus, ¯ (A¯x:n ) =A¯x+t:n−t − P¯ (A¯x:n )¯ ax+t:n−t A¯x:n =A¯x+t:n−t − a ¯ a ¯x:n x+t:n−t A¯x:n 1 − A¯x+t:n−t =A¯x+t:n−t − 1−A¯ · x:n δ



tV



δ



A¯ − A¯x:n = x+t:n−t ¯ 1 − Ax:n



60 46.30 0.3431 46.31 1 ¯ t V (Ax:n )



 =



1 Ax+t:n−t − P¯ (Ax:n1 )¯ ax+t:n−t , t < n 1, t = n.



46.32 0.7939 46.33 1 46.34 0.3088 46.35 0.2307 46.36 This is the 10th year benefit reserve for a fully continuous 20-year pure endowment of unit benefit issued to (75). 46.37 We have ¯ (Ax:n1 ) + 2A¯



tV



1 x+t:n−t



1 1 =Ax+t:n−t − P¯ (Ax:n1 )¯ ax+t:n−t + 2A¯x+t:n−t 1 1 1 =A¯x+t:n−t − A¯x+t:n−t − (P¯ (A¯x:n ) + P¯ (A¯x:n )¯ ax+t:n−t + 2A¯x+t:n−t =[A¯x+t:n−t − P¯ (A¯x:n )¯ ax+t:n−t ] + [A¯ 1 − P¯ (A¯1 )]¯ ax+t:n−t x+t:n−t



=t V¯ (A¯x:n ) +



¯ ¯1 t V (Ax:n ).



46.38 This follows from the previous problem and Problem 46.20. 46.39 24 46.40 4.6362 46.41 5.9055 46.42 14.2857 46.43 14.2857



x:n



61



Section 47 47.1 (a) 0.0533 (b) 0.1251 47.2 We have kV



(Ax ) =Ax+k − P (Ax )¨ ax+k (1 − d¨ ax ) a ¨x+k =1 − d¨ ax+k − a ¨x a ¨x+k =1 − d¨ ax+k − + d¨ ax+k a ¨x a ¨x+k =1 − a ¨x



kV



(Ax ) =Ax+k − P (Ax )¨ ax+k =P (Ax+k )¨ ax+k − P (Ax )¨ ax+k =(P (Ax+k ) − P (Ax ))¨ ax+k



47.3 0.053 47.4 We have



47.5 0.0534 47.6 We have kV



47.7 0.0534



(Ax ) =Ax+k − P (Ax )¨ ax+k   a ¨x+k =Ax+k 1 − P (Ax ) Ax+k   P (Ax ) =Ax+k 1 − P (Ax+k )



62 47.8 We have kV



(Ax ) =1 − =1 −



a ¨x+k a ¨x 1−Ax+k d 1−Ax d



1 − Ax+k 1 − Ax Ax+k − Ax = 1 − Ax =1 −



47.9 0.053 47.10 We have ¯ (Ax ) =Ax+k − P (Ax )¨ ax+k



kV



P (Ax )¨ ax − Ax =Ax+k − P (Ax )¨ ax+k + k Ex     a ¨x − k Ex a ¨x+k Ax − k Ex Ax+k =P (Ax ) − k Ex k Ex !   1 Ax:k a ¨x:k =P (Ax ) − k Ex k Ex =P (Ax )¨ sx:k −



1 Ax:k k Ex



47.11 0.053 47.12 We have P (Ax+k ) =



Ax+k Ax+k 1 =⇒ = . − Ax+k ) P (Ax+k ) P (Ax+k ) + d



d−1 (1



Thus,   P (Ax ) k V (Ax ) =Ax+k 1 − P (Ax+k ) [P (Ax+k ) − P (Ax )]Ax+k = P (Ax+k ) P (Ax+k ) − P (Ax ) = P (Ax+k ) + d



63 47.13 305.651 47.14 114.2984 47.15 0.0851 47.16 171.985 47.17 4420.403 47.18 0.0042 47.19 −0.0826 47.20 0.1587 47.21 0.2757 47.22 0.0138 47.23 629.89 47.24 528.48 47.25 (a) For a fully discrete n−year pure endowment, the insurer’s prospective loss at time k (or at age x + k) is: 1 k L(Ax:n )



= ν n−k I(K ≥ n) − P (Ax:n1 )¨ amin{(K−k+1,n−k)} , k < n



and n L(Ax:n1 ) = 1. (b) The prospective benefit reserve is  1 Ax+k:n−k − P (Ax:n1 )¨ ax+k:n−k 1 ) = V (A k x:n 1 47.26 0.23426 47.27 8119.54



k