developed in Chapter2to a select survival model. This leads to the following definitions:
S[x]+s(t)=Pr[a life currently agedx+swho was select at agexsurvives to agex+s+t],
tq[x]+s =Pr[a life currently agedx+swho was select at agexdies before age x+s+t],
tp[x]+s =1− tq[x]+s≡S[x]+s(t),
µ[x]+sis the force of mortality at agex+sfor an individual who was select at agex,
µ[x]+s=limh→0+
1−S[x]+s(h) h
.
From these definitions we can derive the following formula
tp[x]+s=exp
− t
0
µ[x]+s+udu
.
This formula is derived precisely as in Chapter2. It is only the notation which has changed.
For a select survival model with a select periodd and fort≥d, that is, for durations at or beyond the select period, the values ofµ[x−t]+t, sp[x−t]+tand
u|sq[x−t]+t do not depend ont, they depend only on the current agex. So, for t ≥d we drop the more detailed notation,µ[x−t]+t, sp[x−t]+t andu|sq[x−t]+t, and writeµx, spx andu|sqx. For values oft < d, we refer to, for example, µ[x−t]+tas being in theselectpart of the survival model and fort≥dwe refer toµ[x−t]+t(≡µx)as being in theultimatepart of the survival model.
3.9 Select life tables
For an ultimate survival model, as discussed in Chapter2, the life table{lx} is useful since it can be used to calculate probabilities such ast|uqx for non- negative values oft,uandx. We can construct aselect life tablein a similar way but we need the table to reflect duration as well as age, during the select period. Suppose we wish to construct this table for a select survival model for ages at selection from, say,x0(≥0). Letd denote the select period, assumed to be an integer number of years.
The construction in this section is for a select life table specified at all ages and not just at integer ages. However, select life tables are usually presented at integer ages only, as is the case for ultimate life tables.
First we consider the survival probabilities of those individuals who were selected at leastd years ago and hence are now subject to the ultimate part of the model. The minimum age of these people isx0+d. For these people,
future survival probabilities depend only on their current age and so, as in Chapter2, we can construct an ultimate life table,{ly}, for them from which we can calculate probabilities of surviving to any future age.
Letlx0+d be an arbitrary positive number. Fory≥x0+d we define ly= (y−x0−d)px0+dlx0+d. (3.10) Note that(y−x0−d)px0+d =(y−x0−d)p[x0]+dsince, given that the life was select at least d years ago, the probability of future survival depends only on the current age,x0+d. From this definition we can show that fory>x≥x0+d
ly= y−xpxlx. (3.11)
This follows because
ly = (y−x0−d)px0+d
lx0+d
= y−xp[x0]+x−x0
(x−x0−d)p[x0]+d
lx0+d
= y−xpx
(x−x0−d)px0+d
lx0+d
= y−xpxlx.
This shows that within the ultimate part of the model we can interpretlyas the expected number of survivors to ageyout oflx lives currently agedx(<y), who were select at leastd years ago.
Formula (3.10) defines the life table within the ultimate part of the model.
Next, we need to define the life table within the select period. We do this for a life select at agexby ‘working backwards’ from the value oflx+d. Forx≥x0
and for 0≤t≤d, we define
l[x]+t=lx+d/d−tp[x]+t (3.12) which means that if we hadl[x]+tlives agedx+t, selectedtyears ago, then the expected number of survivors to agex+d islx+d. This defines the select part of the life table.
Example 3.9Fory≥x+d >x+s>x+t≥x≥x0, show that
y−x−tp[x]+t = ly
l[x]+t
(3.13) and
s−tp[x]+t= l[x]+s
l[x]+t
. (3.14)
3.9 Select life tables 61 Solution 3.9First,
y−x−tp[x]+t= y−x−dp[x]+d d−tp[x]+t
= y−x−dpx+d d−tp[x]+t
= ly
lx+d
lx+d
l[x]+t
= ly
l[x]+t
, which proves (3.13). Second,
s−tp[x]+t = d−tp[x]+t/d−sp[x]+s
= lx+d
l[x]+t
l[x]+s
lx+d
= l[x]+s
l[x]+t
,
which proves (3.14). 2
This example, together with formula (3.11), shows that our construction pre- serves the interpretation of thels as expected numbers of survivors within both the ultimate and the select parts of the model. For example, suppose we have l[x]+t individuals currently agedx+t who were select at agex. Then, since
y−x−tp[x]+t is the probability that any one of them survives to agey, we can see from formula (3.13) thatlyis the expected number of survivors to agey.
For 0≤t ≤s≤ d, formula (3.14) shows thatl[x]+s can be interpreted as the expected number of survivors to agex+sout ofl[x]+tlives currently agedx+t who were select at agex.
Example 3.10Write an expression for 2|6q[30]+2 in terms ofl[x]+tandlyfor appropriatex,t, andy, assuming a select period of five years.
Solution 3.10Note that 2|6q[30]+2 is the probability that a life currently aged 32, who was select at age 30, will die between ages 34 and 40. We can write this probability as the product of the probabilities of the following events:
• a life aged 32, who was select at age 30, will survive to age 34, and,
• a life aged 34, who was select at age 30, will die before age 40.
Table 3.5. An extract from the US Life Tables,
2002, Females.
x lx
70 80 556
71 79 026
72 77 410
73 75 666
74 73 802
75 71 800
Hence,
2|6q[30]+2= 2p[30]+2 6q[30]+4
=l[30]+4
l[30]+2
1−l[30]+10
l[30]+4
=l[30]+4−l40
l[30]+2
.
Note thatl[30]+10 ≡l40since 10 years is longer than the select period for this
survival model. 2
Example 3.11A select survival model has a select period of three years. Its ultimate mortality is equivalent to the US Life Tables, 2002, Females. Somelx
values for this table are shown in Table3.5.
You are given that for all agesx≥65,
p[x]=0.999, p[x−1]+1=0.998, p[x−2]+2=0.997.
Calculate the probability that a woman currently aged 70 will survive to age 75 given that
(a) she was select at age 67, (b) she was select at age 68, (c) she was select at age 69, and (d) she is select at age 70.
Solution 3.11 (a) Since the woman was select three years ago and the select period for this model is three years, she is now subject to the ultimate part of the survival model. Hence the probability she survives to age 75 isl75/l70,
3.9 Select life tables 63 where thels are taken from US Life Tables, 2002, Females. The required probability is
l75/l70=71 800/80 556=0.8913.
(b) In this case, the required probability is
5p[68]+2=l[68]+2+5/l[68]+2=l75/l[68]+2=71 800/l[68]+2. We can calculatel[68]+2by noting that
l[68]+2p[68]+2=l[68]+3=l71=79 026.
We are given thatp[68]+2=0.997. Hence,l[68]+2=79 264 and so
5p[68]+2=71 800/79 264=0.9058.
(c) In this case, the required probability is
5p[69]+1=l[69]+1+5/l[69]+1=l75/l[69]+1=71 800/l[69]+1. We can calculatel[69]+1by noting that
l[69]+1p[69]+1p[69]+2=l[69]+3=l72=77 410.
We are given that p[69]+1 = 0.998 and p[69]+2 = 0.997. Hence, l[69]+1=77 799 and so
5p[69]+1=71 800/77 799=0.9229.
(d) In this case, the required probability is
5p[70]=l[70]+5/l[70]=l75/l[70]=71 800/l[70]. Proceeding as in parts (b) and (c), we have
l[70]p[70]p[70]+1p[70]+2=l[70]+3=l73=75 666, giving
l[70]=75 666/(0.997×0.998×0.999)=76 122.
Hence
5p[70]=71 800/76 122=0.9432.
2
Table 3.6. CMI (Table A5) extract: mortality rates for male non-smokers who have whole life
or endowment policies.
Age,x
Duration 0 q[x]
Duration 1 q[x−1]+1
Duration 2+
qx 60 0.003469 0.004539 0.004760 61 0.003856 0.005059 0.005351 62 0.004291 0.005644 0.006021 63 0.004779 0.006304 0.006781
· · · ·
70 0.010519 0.014068 0.015786 71 0.011858 0.015868 0.017832 72 0.013401 0.017931 0.020145 73 0.015184 0.020302 0.022759 74 0.017253 0.023034 0.025712 75 0.019664 0.026196 0.029048
Example 3.12CMI (Table A5) is based on UK data from 1999 to 2002 for male non-smokers who are whole life or endowment insurance policyholders.
It has a select period of two years. An extract from this table, showing values of q[x−t]+t, is given in Table3.6. Use this survival model to calculate the following probabilities:
(a) 4p[70], (b) 3q[60]+1, and (c) 2|q73.
Solution 3.12Note that CMI (Table A5) gives values ofq[x−t]+tfort=0 and t =1 and also fort≥2. Since the select period is two yearsq[x−t]+t ≡qxfor t ≥2. Note also that each row of the table relates to a mancurrentlyagedx, wherexis given in the first column. Select life tables, tabulated at integer ages, can be set out in different ways – for example, each row could relate to a fixed age at selection – so care needs to be taken when using such tables.
(a) We calculate4p[70]as
4p[70]=p[70]p[70]+1p[70]+2p[70]+3
=p[70]p[70]+1p72p73
=(1−q[70]) (1−q[70]+1) (1−q72) (1−q73)
=0.989481×0.984132×0.979855×0.977241
=0.932447.
3.9 Select life tables 65 (b) We calculate3q[60]+1as
3q[60]+1=q[60]+1+p[60]+1q62+p[60]+1p62q63
=q[60]+1+(1−q[60]+1)q62+(1−q[60]+1) (1−q62)q63
=0.005059+0.994941×0.006021
+0.994941×0.993979×0.006781
=0.017756.
(c) We calculate2|q73as
2|q73= 2p73q75
=(1−q73) (1−q74)q75
=0.977241×0.974288×0.029048
=0.027657.
2 Example 3.13A select survival model has a two-year select period and is specified as follows. The ultimate part of the model follows Makeham’s law, so that
µx=A+Bcx
whereA = 0.00022,B = 2.7×10−6andc = 1.124. The select part of the model is such that for 0≤s≤2,
µ[x]+s =0.92−sµx+s. Starting withl20=100 000, calculate values of (a) lxforx=21, 22,. . ., 82,
(b) l[x]+1forx=20, 21,. . ., 80, and, (c) l[x]forx=20, 21,. . ., 80.
Solution 3.13First, note that
tpx=exp
−At− B
logccx(ct−1)
and for 0≤t≤2,
tp[x]=exp
− t
0
µ[x]+sds
=exp
0.92−t
1−0.9t
log(0.9)A+ ct−0.9t log(0.9/c)Bcx
. (3.15)
Table 3.7. Select life table with a two-year select period.
x l[x] l[x]+1 lx+2 x+2 x l[x] l[x]+1 lx+2 x+2
100 000.00 20 50 98 552.51 98 450.67 98 326.19 52 99 975.04 21 51 98 430.98 98 318.95 98 181.77 53 20 99 995.08 99 973.75 99 949.71 22 52 98 297.24 98 173.79 98 022.38 54 21 99 970.04 99 948.40 99 923.98 23 53 98 149.81 98 013.56 97 846.20 55 22 99 944.63 99 922.65 99 897.79 24 54 97 987.03 97 836.44 97 651.21 56 23 99 918.81 99 896.43 99 871.08 25 55 97 807.07 97 640.40 97 435.17 57 24 99 892.52 99 869.70 99 843.80 26 56 97 607.84 97 423.18 97 195.56 58 25 99 865.69 99 842.38 99 815.86 27 57 97 387.05 97 182.25 96 929.59 59 26 99 838.28 99 814.41 99 787.20 28 58 97 142.13 96 914.80 96 634.14 60 27 99 810.20 99 785.70 99 757.71 29 59 96 870.22 96 617.70 96 305.75 61 28 99 781.36 99 756.17 99 727.29 30 60 96 568.13 96 287.48 95 940.60 62 29 99 751.69 99 725.70 99 695.83 31 61 96 232.34 95 920.27 95 534.43 63 30 99 721.06 99 694.18 99 663.20 32 62 95 858.91 95 511.80 95 082.53 64 31 99 689.36 99 661.48 99 629.26 33 63 95 443.51 95 057.36 94 579.73 65 32 99 656.47 99 627.47 99 593.83 34 64 94 981.34 94 551.72 94 020.33 66 33 99 622.23 99 591.96 99 556.75 35 65 94 467.11 93 989.16 93 398.05 67 34 99 586.47 99 554.78 99 517.80 36 66 93 895.00 93 363.38 92 706.06 68 35 99 549.01 99 515.73 99 476.75 37 67 93 258.63 92 667.50 91 936.88 69 36 99 509.64 99 474.56 99 433.34 38 68 92 551.02 91 894.03 91 082.43 70 37 99 468.12 99 431.02 99 387.29 39 69 91 764.58 91 034.84 90 133.96 71 38 99 424.18 99 384.82 99 338.26 40 70 90 891.07 90 081.15 89 082.09 72 39 99 377.52 99 335.62 99 285.88 41 71 89 921.62 89 023.56 87 916.84 73 40 99 327.82 99 283.06 99 229.76 42 72 88 846.72 87 852.03 86 627.64 74 41 99 274.69 99 226.72 99 169.41 43 73 87 656.25 86 555.99 85 203.46 75 42 99 217.72 99 166.14 99 104.33 44 74 86 339.55 85 124.37 83 632.89 76 43 99 156.42 99 100.80 99 033.94 45 75 84 885.49 83 545.75 81 904.34 77 44 99 090.27 99 030.10 98 957.57 46 76 83 282.61 81 808.54 80 006.23 78 45 99 018.67 98 953.40 98 874.50 47 77 81 519.30 79 901.17 77 927.35 79 46 98 940.96 98 869.96 98 783.91 48 78 79 584.04 77 812.44 75 657.16 80 47 98 856.38 98 778.94 98 684.88 49 79 77 465.70 75 531.88 73 186.31 81 48 98 764.09 98 679.44 98 576.37 50 80 75 153.97 73 050.22 70 507.19 82 49 98 663.15 98 570.40 98 457.24 51
(a) Values oflxcan be calculated recursively from
lx=px−1lx−1 for x=21, 22,. . ., 82.
(b) Values ofl[x]+1can be calculated from
l[x]+1=lx+2/p[x]+1 for x=20, 21,. . ., 80.
(c) Values ofl[x]can be calculated from
l[x]=lx+2/2p[x] for x=20, 21,. . ., 80.
The values are shown in Table3.7. 2