often practical to introduce a limiting age in calculations, at some point where the probability of surviving longer is negligible. We will see examples later in this chapter.
2.3 The force of mortality 39 From standard results in probability theory, we know that the probability density function for the random variable Tx, which we denote fx, is related to the distribution functionFxand the survival functionSxby
fx(t)= d
dtFx(t)= −d dtSx(t). So, it follows fromequation (2.9)that
μx= f0(x) S0(x).
We can also relate the force of mortality function at any age x+t,t > 0, to the lifetime distribution ofTx. Assume xis fixed and t is variable. Then d(x+t)=dtand so
μx+t = − 1 S0(x+t)
d
d(x+t)S0(x+t)
= − 1 S0(x+t)
d
dtS0(x+t)
= − 1 S0(x+t)
d
dt(S0(x)Sx(t))
= − S0(x) S0(x+t)
d dtSx(t)
= −1 Sx(t)
d dtSx(t).
Hence
μx+t= fx(t)
Sx(t). (2.10)
This relationship gives a way of findingμx+tgiven the survival functionSx. We can also use equation (2.9)to develop a formula forSx(t)in terms of the force of mortality function,μx+s, for 0≤s≤t. We use the fact that for a functionhwhose derivative exists, and whereh(x) >0 for allx,
d
dxlogh(x)= 1 h(x)
d dxh(x), so, fromequation (2.9), we have
μx= −d
dxlogS0(x),
and integrating this identity over(0,y)yields y
0
μxdx= −
logS0(y)−logS0(0) .
As logS0(0)=log Pr[T0>0]=log 1=0, we obtain
S0(y)=exp
⎧⎨
⎩− y
0
μxdx
⎫⎬
⎭, from which it follows that
Sx(t)= S0(x+t) S0(x) =exp
⎧⎨
⎩−
x+t
x
μrdr
⎫⎬
⎭=exp
⎧⎨
⎩− t
0
μx+sds
⎫⎬
⎭. (2.11)
This means that if we know μy for all y ≥ 0, then we can calculate all the survival probabilitiesSx(t), for anyxandt. In other words, the force of mortality function fully describes the lifetime distribution, just as the function S0does. In fact, it is often more convenient to describe the lifetime distribution using the force of mortality function than the survival function.
Example 2.2 As inExample 2.1, let
F0(x)=1−(1−x/120)1/6 for 0≤x≤120. Derive an expression forμx.
Solution 2.2 AsS0(x)=(1−x/120)1/6, it follows that d
dxS0(x)=16(1−x/120)−5/6
−1201 , and so
μx= −1 S0(x)
d
dxS0(x)= 7201 (1−x/120)−1= 1 720−6x. As an alternative, we could use the relationship
μx= − d
dxlogS0(x)= − d dx
1
6log(1−x/120)
= 1
720(1−x/120)
= 1 720−6x.
2.3 The force of mortality 41 2.3.1 Mortality laws
We saw inequation (2.11)that the full distribution ofTx can be determined if we know the force of mortality μy for all y ≥ x. This has led to several important distributions for future lifetime being derived by assuming a mathematical function for the force of mortality; historically, these are called mortality laws.
Two of the most useful mortality laws areGompertz’ law, given by μx=B cx whereB>0, c>1,
andMakeham’s law, which is a generalization of Gompertz’ law, given by μx=A+B cx whereA,B>0,c>1.
The force of mortality under Gompertz’ law increases exponentially with age sincec>1; we haveB>0 since the force of mortality must be positive.
The force of mortality under Makeham’s law adds a constant term which was designed to reflect the risk of accidental death. This term has more impact at younger ages, when the age-related force of mortality is very small. At older ages, the exponential term is the dominant one. These two formulae are very similar (and a simple way to remember which is which is that the letter ‘a’
appears in both Makeham’s name and his mortality law).
We will see in the next chapter that the force of mortality for most populations is not an increasing function of age over the entire age range.
Nevertheless, both models often provide a good fit to mortality data over certain age ranges, particularly from middle age to early old age.
Example 2.3 Derive expressions for Sx(t) for (a) Gompertz’ law, and (b) Makeham’s law.
Solution 2.3 (a) For Gompertz’ law, usingequation (2.11), we have Sx(t)=exp
⎧⎨
⎩−
x+t
x
Bcrdr
⎫⎬
⎭. Writingcras exp{rlogc},
x+t
x
Bcrdr=B
x+t
x
exp{rlogc}dr
= B
logcexp{rlogc}
x+t
x
= B logc
cx+t−cx ,
giving
Sx(t)=exp
− B
logccx(ct−1)
. (b) Similarly to (a), we have
x+t
x
(A+Bcr)dr=At+B
x+t
x
exp{rlogc}dr
=At+ B logc
cx+t−cx , giving
Sx(t)=exp
−At− B
logccx(ct−1)
. (2.12)
We remark that this is often written as Sx(t)=stgcx(ct−1),
wheres=e−Aandg=exp{−B/logc}.
One of the earliest mortality laws proposed wasDe Moivre’s law which states that μx = 1/(ω −x) for 0 ≤ x < ω. This is a very unrealistic model, and therefore impractical for human populations. Under De Moivre’s law,Txis uniformly distributed on the interval(0,ω−x), which means that the probability that a life currently agedxdies between agesx+tandx+t+dtis the same for allt>0, as long asx+t< ω.
Thegeneralized De Moivre’s law states thatμx = α/(ω−x)for some α >0, and for 0≤x< ω. We have already met an example of this mortality law in Example 2.2 where ω = 120 and α = 1/6. Again, it does not in any way represent human mortality, and so is not useful in practice. Under the generalized De Moivre’s law,Txhas a beta distribution on(0,ω−x)for 0≤x< ω.
Another simple mortality law that is very unrealistic for modelling human mortality is theconstant force of mortalityassumption which statesμx=μ for all x ≥ 0. Under this model, Tx has an exponential distribution. (See Exercise 2.7.)
Although the De Moivre and constant force models are not useful for overall mortality for humans, they may be used in other contexts, such as modelling the failure time of machine components. In addition, inChapter 3,we use these models in a limited sense, to model mortality between integer ages.
2.3 The force of mortality 43 Example 2.4 Calculate the survival function and probability density function forTxusing Gompertz’ law of mortality, withB =0.0003 andc= 1.07, for x= 20,x=50 andx=80. Plot the results and comment on the features of the graphs.
Solution 2.4 Forx = 20, the force of mortality isμ20+t = Bc20+t and the survival function is
S20(t)=exp
− B
logcc20(ct−1)
. The probability density function is found from (2.10):
μ20+t = f20(t) S20(t)
⇒f20(t)=μ20+tS20(t)=Bc20+texp
− B
logcc20(ct−1)
.
Figure 2.1shows the survival functions for ages 20, 50 and 80, andFigure 2.2 shows the corresponding probability density functions. These figures illustrate some general points about lifetime distributions.
First, we see an effective limiting age, even though, in principle there is no age at which the survival probability is exactly zero. Looking atFigure 2.1, we see that althoughSx(t) >0 for all combinations ofxandt, survival beyond age 120 is very unlikely.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0 10 20 30 40 50 60 70 80 90 100
Time, t
Survival probability
Figure 2.1 Sx(t)forx=20 (bold), 50 (solid) and 80 (dotted).
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07
0 10 20 30 40 50 60 70 80 90 100
Time, t fx(t)
Figure 2.2 fx(t)forx=20 (bold), 50 (solid) and 80 (dotted).
Second, we note that the survival functions are ordered according to age, with the probability of survival for any given value oftbeing highest for age 20 and lowest for age 80. For survival functions that give a more realistic representation of human mortality, this ordering can be violated, but it usually holds at ages of interest to insurers. An example of the violation of this ordering is that S0(1)may be smaller thanSx(1)for x ≥ 1, as a result of perinatal mortality.
Looking atFigure 2.2,we see that the densities for ages 20 and 50 have similar shapes, but the density for age 80 has a quite different shape. For ages 20 and 50, the densities have their respective maximums at (approximately) t = 60 and t = 30, indicating that death is most likely to occur around age 80. The decreasing form of the density for age 80 also indicates that death is more likely to occur at age 80 than at any other age for a life now aged 80. A further point to note about these density functions is that although each density function is defined on(0,∞), the spread of values of fx(t)is much greater forx=20 than forx=50, which, as we will see inTable 2.1,results in a greater variance of future lifetime forx=20 than forx=50.