Example 1
The age-independent, five-year survival probability for a Scottish male diagnosed
with pancreatic cancer is 3.5% (Source: ISD Scotland press release, August 2004).
(i) What constant force of mortality is implied by this?
(ii) What is the one-year survival probability?
(iii) What distribution of future lifetime is implied by a constant force of mortality?
Solution
(i) Using the basic general result linking the survival probability to the force of
mortality:
Z t
µs ds
S(t) = exp −
0
Z t
= exp −
µds
0
= exp (−tµ)
So S(5) = e−5µ .
We are given that S(5) = 0.035.
Rearranging gives constant force of mortality, µ = − 15 log 0.035 = 0.670.
(ii)
S(1) = e−1.µ
= e−0.670
= 0.511
So, one-year survival probability is 51.1%.
(iii)
F (t) = 1 − S(t)
= 1 − e−tµ
which we recognise as the distribution function of the exponential distribution.
Example 2
The probability of a British male non-smoking doctor born in 1900–1909 surviving
from age 35 to age 70 was 76%. The same probability for an equivalent smoker
was 58% (Source: Doll et al, British Medical Journal, 2004).
(i) What is the constant additional force of mortality for smokers implied by this
over the age range 35–70?
(ii) The survival probabilities to age 90 were 18% for non-smokers and 4% for
smokers. What is the constant additional force of mortality for smokers implied
by this over the age range 35–90?
(iii) What is implied about our assumption of a constant additional force of
mortality?
Solution
(i) Recall the general result:
Z t
µx+s ds
t px = exp −
0
and assume that µsmoker
= µnonsmoker
+ λ.
x+s
x+s
Z t
smoker
µnonsmoker
+ λ ds
= exp −
t px
x+s
0
Z t
Z t
nonsmoker
= exp −
µx+s
ds . exp −
λds
=
⇒ e−tλ =
⇒λ=
0
nonsmoker −tλ
.e
t px
smoker
t px
nonsmoker
t px
smoker
1
t px
− log
nonsmoker
t
t px
We are given that 35 pnonsmoker
= 0.76 and
35
0.58
1
= 0.00772
So λ = − log
35
0.76
1
0.04
(ii) λ = − log
= 0.0273
55
0.18
smoker
35 p35
0
= 0.58.
(iii) The additional force of mortality due to smoking does not look to be constant.
Z t
µx+s ds
t px = exp −
0
Proof 1
t+dt px
Remember lim+
dt→0
= t px .dt px+t
= t px (1 − dt qx+t )
= t px (1 − dt.µx+t + o(dt))
(1)
o(dt)
= 0.
dt
Rearranging (1) gives:
t+dt px
− t px = −t px .dt.µx+t + o(dt)
and so:
− t px
o(dt)
= −t px µx+t +
dt
dt
t+dt px
Taking the limit as dt → 0:
∂
t px = −t px µx+t
∂t
This is a differential equation. Remember the general result:
∂
∂ G(t)
e
= eG(t) . G(t)
∂t
∂t
and recall the result from basic calculus that:
Z t
∂
g(s)ds = g(t)
∂t 0
Thus:
∂
∂t
Z
t
µx+s ds = µx+t
0
and therefore the solution to (2) is:
Z t
µx+s ds
t px = exp −
0
(2)
Z t
µx+s ds
t px = exp −
0
Proof 2
t px
=
t−1
Y
1 px+i
i=0
t
h −1
=
Y
h px+j
j=0
for some small h. Now h px+j = 1 − h qx+j and h qx+j ≈ h.µx+j for small h, so:
t
h −1
t px =
Y
h px+j
j=0
t
h −1
=
Y
(1 − h qx+j )
j=0
t
h −1
=
Y
(1 − h.µx+j ) + o(h)
j=0
Take natural logarithms of both sides, and note that log(1 − z) = −z + o(z) for
small z:
t
log t px =
h −1
X
log(1 − h.µx+j ) + o(h)
j=0
t
=
h −1
X
−h.µx+j + o(h)
j=0
Take the limit as h → 0+ :
Z
t
−µx+s ds
log t px =
0
which, after taking the exponent of each side, gives:
Z t
µx+s ds
t px = exp −
0