For an ultimate survival model the life table{lx}is useful since it can be used to calculate probabilities such ast|uqxfor values oft,uandx. We can construct aselect life tablein a similar way but we need the table to reflect duration since selection, 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, xo (≥ 0). Let d 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 for the life tables introduced earlier in this chapter.
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 lives is xo+d. For these people, future survival probabilities depend only on their current age and so, as in Chapter 2,we can construct an ultimate life table,{ly}, for them from which we can calculate probabilities of surviving to any future age.
Letlxo+dbe an arbitrary positive number. Fory>x≥xo+d, we define
ly=y−xpx lx. (3.13)
Formula (3.13) defines the life table within the ultimate part of the model, which applies after the select period has expired. Within the ultimate part of the model we can interpretlyas the expected number of survivors to ageyout oflxlives currently agedx(<y), who were select at leastdyears previously.
Next, we define the life table within the select period. We do this for a life who was select at agexby working backwards from the value oflx+d. For x≥xoand for 0≤t≤d, we definel[x]+tas
d−tp[x]+t= lx+d
l[x]+t ⇒l[x]+t= lx+d d−tp[x]+t
, (3.14)
which means that if we hadl[x]+tlives agedx+t, who were selecttyears ago, then the expected number of survivors to agex+dislx+d.
3.9 Select life tables 77 Example 3.9 A life currently aged 52 was select at age 50. The survival model has a select period of 10 years. Write down the following probabilities in terms ofl[x]+tandlyfor appropriatex,t, andy:
(a) the probability that the life survives to age 58, and (b) the probability that the life survives to age 62.
Solution 3.9 (a) As the select period is 10 years, we are interested in the life surviving within the select period from [50]+2 to [50]+8, so the required probability is
6p[50]+2=l[50]+8
l[50]+2
.
(b) We are now interested in the life surviving from [50]+2 for a period of 10 years, which takes the life beyond the end of the select period, so the required probability is
10p[50]+2=l[50]+12
l[50]+2 = l62
l[50]+2
.
The construction of the select life table preserves the interpretation of thelx
andl[x]+tfunctions as expected numbers of survivors within the ultimate and the select parts of the model respectively. For example, suppose we havel[x]+t
individuals currently agedx+twho were select at agex.
Forusuch thatu+t≥d, we have
lx+t+u=up[x]+t l[x]+t
demonstrating thatlx+t+uis the expected number of survivors to agex+t+u from thel[x]+tlives agedx+twho were select at agex.
Forusuch thatu+t<d, we have
l[x]+t+u=up[x]+tl[x]+t
demonstrating thatl[x]+t+uis the expected number of survivors to agex+t+u out ofl[x]+tlives currently agedx+t, who were select at agex.
Example 3.10 Write an expression for 2|6q[30]+2 in terms ofl[x]+tandlyfor appropriatex,t, andy, assuming a select period of five years.
Solution 3.10 Note that2|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 Extract from 2002 US Life Tables, 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.
Example 3.11 A 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 inTable 3.5.
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 is l75/l70, where thelxvalues are taken from US Life Tables, 2002, Females.
The required probability is
5p70 =71 800/80 556=0.8913.
3.9 Select life tables 79 (b) We have
5p[68]+2= l[68]+2+5
l[68]+2 = l75
l[68]+2 = 71 800 l[68]+2
. We calculatel[68]+2by noting that
l[68]+2×p[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=0.9058.
(c) We have
5p[69]+1= l[69]+1+5
l[69]+1 = l75
l[69]+1 = 71 800 l[69]+1
. We calculatel[69]+1by noting that
l[69]+1×p[69]+1×p[69]+2=l[69]+3=l72=77 410.
We are given thatp[69]+1 =0.998 andp[69]+2=0.997. Hence,l[69]+1 = 77 799 and so5p[69]+1=0.9229.
(d) We have
5p[70]= l[70]+5
l[70] = l75
l[70] =71 800 l[70]
. Proceeding as in parts (b) and (c),
l[70]×p[70]×p[70]+1×p[70]+2=l[70]+3=l73=75 666
⇒l[70]=75 666/(0.997×0.998×0.999)=76 122
⇒5p[70]=0.9432.
Example 3.12 An extract from CMI Table A5 is given inTable 3.6.This table 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.
(a) Use the table to calculate the following probabilities:
(i)4p[70], (ii)3q[60]+1, (iii)2|q73. (b) You are given thate72=13.3122.
(i) Calculatee[70].
(ii) Calculate◦e[70]assuming UDD between integer ages.
Solution 3.12 CMI Table A5 gives values ofq[x−t]+tfort=0 andt=1 and also fort ≥ 2. Since the select period is two yearsq[x−t]+t ≡qx fort ≥ 2.
Table 3.6 CMI Table A5: male non–smokers who have whole life or endowment policies.
Duration 0 Duration 1 Duration 2+
Age,x q[x] q[x−1]+1 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
Note also that each row of the table relates to a lifecurrentlyagedx, wherex is 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) (i) We calculate4p[70]as
4p[70]=p[70]×p[70]+1×p[70]+2×p[70]+3
=p[70]×p[70]+1×p72×p73
=(1−q[70])×(1−q[70]+1)×(1−q72)×(1−q73)
=0.989481×0.984132×0.979855×0.977241
=0.932447.
(ii) 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
=0.017756.
(iii) We calculate2|q73as
2|q73= 2p73q75
=(1−q73) (1−q74)q75
=0.027657.
3.9 Select life tables 81 (b) (i) Recall fromExercise 2.2thatex=px(1+ex+1).
Similarly,e[70]=p[70](1+e[70]+1)and e[70]+1 =p[70]+1(1+e72), giving
e[70]=p[70]+2p[70](1+e72)
=0.989481+0.973780(1+13.3122)
=14.9264.
(ii) We want to calculate
e◦[70]= ∞ 0
tp[70]dt,
and as UDD allows us to calculate survival probabilities from integer ages for non-integer terms, we restate this integral in terms of a sum of one-year integrals as
◦e[70]= 1 0
tp[70]dt+ 2 1
tp[70]dt+ 3 2
tp[70]dt+ · · ·. Now consider each term in this sum:
1 0
tp[70]dt= 1 0
(1−tq[70])dt= 1 0
(1−t q[70])dt=1−12q[70]
=1−12(1−p[70])=12(1+p[70]), 2
1
tp[70]dt=p[70]
1 0
tp[70]+1dt=p[70]1
2(1+p[70]+1)
=12(p[70]+2p[70]), 3
2
tp[70]dt=2p[70]
1 0
tp72dt=2p[70]1
2(1+p72)=12(2p[70]+3p[70]), and so on. Collecting the terms together gives
◦e[70]= 12
1+p[70]+p[70]+2p[70]+2p[70]
+3p[70]+3p[70]+4p[70]+ · · ·
= 12+p[70]+2p[70]+3p[70]+ · · ·
=e[70]+12
=15.4264.
Example 3.13 A 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 3.13 First, 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) (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.
Sample values are shown inTable 3.7.The full table up to age 100 is given in
Table D.1inAppendix D.
This model is used extensively throughout this book for examples and exercises. We call it theStandard Select Survival Modelin future chapters.
The ultimate part of the model, which is a Makeham model with A = 0.00022,B = 2.7×10−6 and c = 1.124, is also used in many examples and exercises where a select model is not required. We call this theStandard Ultimate Survival Model.