A life table{lx}x≥xoprovides exactly the same information as the corresponding survival distribution,Sxo. However, we commonly use the term ‘life table’ to
3.3 Fractional age assumptions 61 mean thelx function tabulated at integer ages only. This does not contain all the information in the corresponding survival model, as it is not sufficient for calculating probabilities involving non-integer ages or durations. For example, the life table gives us 1p30 = l31/l30, but not 1p30.5 (non-integer age) or
0.75p30(non-integer duration), or0.75p30.5(non-integer age and duration). So, if we only have values of lx at integer ages from the life table, we need an additional assumption, or some further information, to calculate probabilities involving non-integer ages and durations. Specifically, we need to make some assumption about the distribution of the future lifetime random variable between integer ages.
We use the termfractional age assumptionto describe such an assumption.
It may be specified in terms of the force of mortality function or the survival or mortality probabilities.
In this section we assume that a life table is specified at integer ages only and we describe the two most useful fractional age assumptions.
3.3.1 Uniform distribution of deaths
The uniform distribution of deaths (UDD) assumption is the most common fractional age assumption. It can be formulated in two different, but equivalent, ways as follows.
UDD1 For integerx, and for 0≤s<1, assume that
sqx=sqx. (3.6)
That is, for any integerx, the mortality probability overs<1 years is stimes the one-year mortality probability.
UDD2 For a life (x), where x is an integer, with future lifetime random variable,Tx, and curtate future lifetime random variable, Kx, define a new random variableRx to represent the fractional part of the future lifetime of (x) lived in the year of death, so that Tx = Kx +Rx. We assume
Rx∼U(0, 1), independent ofKx.
So this assumption states thatRx has a uniform distribution on (0,1), regardless of the distribution of Kx. Recall that if X∼U(0, 1), then Pr(X≤u)=ufor 0≤u≤1 (seeAppendix A).
The equivalence of these two assumptions is demonstrated as follows. First, assume that UDD1 is true. Then for integerx, and for 0≤s<1,
Pr[Rx≤s]=∞
k=0
Pr[Rx≤sandKx=k]
= ∞ k=0
Pr[k≤Tx≤k+s]
= ∞ k=0
kpx sqx+k
=∞
k=0
kpxs(qx+k) using UDD1
=s ∞ k=0
kpxqx+k
=s ∞ k=0
Pr[Kx=k]
=s.
This proves that Rx ∼ U(0, 1). To prove the independence of Rx andKx, note that
Pr[Rx≤sandKx=k]=Pr[k≤Tx≤k+s]
= kpx sqx+k
=skpxqx+k
=Pr[Rx≤s] Pr[Kx=k]
sinceRx∼U(0, 1). This proves that UDD1 implies UDD2.
To prove the reverse implication, assume that UDD2 is true. Then for integerx, and for 0≤s<1,
sqx=Pr[Tx≤s]
=Pr[Kx=0 andRx≤s]
=Pr[Rx≤s] Pr[Kx=0]
asKxandRxare assumed independent. Thus,
sqx=s qx. (3.7)
The UDD2 derivation of the result explains why this assumption is called the Uniform Distribution of Deaths, but in practical applications of this assumption, formulation UDD1 is the more useful of the two.
3.3 Fractional age assumptions 63 An immediate consequence is that
lx+s =lx−s dx 0≤s<1. (3.8) This follows because for 0≤s<1
sqx= lx−lx+s
lx
and substitutings qx=s dx/lxforsqx, we have sdx
lx = lx−lx+s
lx
. Hence
lx+s=lx−s dx for 0≤s≤1.
Thus, UDD implies that lx+s is a linearly decreasing function of sbetween integer ages. In practice this can be very useful, as it allows us to use linear interpolation to calculate survival probabilities for non-integer terms, provided the starting age is an integer. Thus, iftandxare both integers, and 0<s<1, then under UDD
t+spx= lx+t+s
lx
= lx+t−s dx+t
lx
= lx+t−s(lx+t−lx+t+1) lx
= (1−s)lx+t+s lx+t+1
lx
(3.9)
⇒t+spx=(1−s)×tpx+s×t+1px. (3.10) Note that both (3.9) and (3.10) are linear interpolations.
Differentiatingequation (3.6)with respect tos, we obtain d
ds sqx= d
dss qx=qx, 0≤s<1.
We also know that
d
ds sqx=spxμx+s, s>0,
because the left-hand side is the derivative of the distribution function forTx, which is equal to the density function on the right-hand side. So, between integer ages, the density function is constant, and more specifically,
spxμx+s =qx for 0≤s<1. (3.11)
Since qx is constant with respect to s, and spx is a decreasing function ofs, we can see thatμx+sis an increasing function ofs, which is appropriate for ages of interest to insurers. However, if we apply the approximation over successive ages, we obtain a discontinuous function for the force of mortality, with discontinuities occurring at integer ages, as we illustrate inExample 3.4.
Although this is undesirable, it is not a serious drawback in practice.
Example 3.2 Given thatp40=0.999473, calculate0.4q40.2under the assump- tion of a uniform distribution of deaths.
Solution 3.2 It is important to remember that the UDD result, thatsqx=s qx, requiresxto be in integer and requires 0<s<1. So the first step in applying UDD to this problem is to restate the required probability in terms that involve survival probabilities from age 40 as
0.4q40.2=1− 0.4p40.2=1−0.6p40
0.2p40 =1−1−0.6q40
1−0.2q40
. Now we can apply UDD to the numerator and denominator giving
0.4q40.2=1−1−0.6q40
1−0.2q40 =0.000211.
Example 3.3 Use the life table extract inTable 3.1,with the UDD assumption, calculate (a)1.7q33and (b)1.7q33.5.
Solution 3.3 (a) Because the starting age is an integer, we can use the linear interpolation approach from equation (3.9), with x = 33,t = 1, and s=0.7, giving
1.7q33=1−1.7p33 =1−0.3l34+0.7l35
l33 =1−0.991808
=0.008192.
(b) To calculate 1.7q33.5, we first need to express it in terms of survival probabilities from age 33, as
1.7q33.5=1−1.7p33.5=1−2.2p33
0.5p33 =1−0.8l35+0.2l36
0.5l33+0.5l34
=0.008537.
Alternatively, using (3.8),
1.7q33.5=1−1.7p33.5=1−l35.2
l33.5
=1−l35−0.2d35
l33−0.5d33
=0.008537.
3.3 Fractional age assumptions 65 Example 3.4 Under the assumption of a uniform distribution of deaths, calculate lim
t→1−μ40+t using p40 =0.999473, and calculate lim
t→0+μ41+t using p41 =0.999429.
Solution 3.4 From formula (3.11), we haveμx+t=qx/tpxfor 0≤t<1.
Settingx=40 yields
t→lim1−μ40+t =q40/p40 =5.27×10−4, while settingx=41 yields
t→lim0+μ41+t=q41=5.71×10−4.
3.3.2 Constant force of mortality
A second fractional age assumption is that the force of mortality is constant between integer ages. Thus, for integerxand 0≤s<1, we assume thatμx+s
does not depend ons, and we denote itμ∗x. We can obtain the value ofμ∗xfrom the life table by using the fact that
px=exp
⎧⎨
⎩− 1
0
μx+sds
⎫⎬
⎭.
So, ifμx+s =μ∗x for 0≤ s<1 thenpx =e−μ∗x andμ∗x = −logpx. Further, givenμ∗x, andr<1, we have
rpx=exp
⎧⎨
⎩− r
0
μ∗xds
⎫⎬
⎭=e−rμ∗x =(px)r. Similarly, forr,t>0 andr+t<1,
rpx+t=exp
⎧⎨
⎩− r
0
μ∗xds
⎫⎬
⎭=(px)r. (3.12) Thus, under the constant force assumption, the probability of surviving for a period of r < 1 years from age x + t is independent of t, provided that r +t<1.
The assumption of a constant force of mortality between integer ages leads to a step function for the force of mortality over successive years of age, whereas we would expect the force of mortality to increase smoothly. However, if the true force of mortality increases slowly over the year of age, the constant force of mortality assumption is reasonable.
Example 3.5 Given thatp40=0.999473, calculate0.4q40.2under the assump- tion of a constant force of mortality.
Solution 3.5 We have0.4q40.2=1−0.4p40.2=1−(p40)0.4=2.108×10−4. Example 3.6 Given that q70 = 0.010413 and q71 = 0.011670, calculate
0.7q70.6under the assumption of a constant force of mortality.
Solution 3.6 We will work from the survival probability. We need to separate the survival probabilities applying in different age-years to use (3.12). That is
0.7p70.6= 0.4p70.6×0.3p71
=p0.470 ×p0.371 =0.9895870.4×0.9883300.3
=0.992321
⇒0.7q70.6=0.007679.
Example 3.7 Using the life table extract inTable 3.1,with the constant force of mortality assumption, calculate (a)1.7q33and (b)1.7q33.5.
Solution 3.7 (a) We write1.7q33in terms of survival probabilities over whole and fractional years as
1.7q33=1−1.7p33=1−p33 0.7p34=1−p33p0.734 =1−0.991805
=0.008195.
(b) Similarly,
1.7q33.5=1−1.7p33.5=1−2.2p33
0.5p33 =1−2p33 0.2p35 0.5p33
=1−2p33p0.235 p0.533
=0.008537.
Note that inExamples 3.2and3.5,and inExamples 3.3and3.7, we have used two different methods to solve the same problems, and the solutions agree to at least five decimal places. It is generally true that the assumptions of a uniform distribution of deaths and a constant force of mortality produce very similar solutions to problems. The reason for this is that under the constant force of mortality assumption
qx=1−e−μ∗x ≈μ∗x