Math 630 Notes
Survival Models
• Furure Life Time Tx and Its Distribution
– (x): a life aged x
– Tx : the future life time of (x)
Example: Consider a life (55). If T55 = 30, then (55) dies at age 55 + T55 = 85.
– The CDF of Tx : Fx (t) = P (Tx ≤ t), the probability that (x) dies within t years
– Survival function of (x): Sx (t) = 1 − Fx (t) = P (Tx > t), the probability that (x)
survives for at least t years
– Notation: t px = Sx (t) = P (Tx > t), t qx = Fx (t) = P (Tx ≤ t). For t = 1, we
drop the front subscript and use px = 1 px and qx = 1 qx
– An important formula P (Tx ≤ t) = P (T0 ≤ x + t|T0 > x).
– Similarly, P (Tx > t) = P (T0 > x + t | T0 > x)
– Exercise Show that
1) Fx (t) =
F0 (x + t) − F0 (x)
S0 (x)
2) Sx (t) =
S0 (x + t)
or S0 (x + t) = S0 (x)Sx (t)
S0 (x)
3) Sx (t + u) = Sx (t)Sx+t (u)
– Conditions on Sx (t): Sx (0) = 1,
lim Sx (t) = 0, Sx (t) is non-increasing.
t→∞
lim t2 Sx (t) = 0 ( ⇒ lim tSx (t) = 0.)
t→∞
√
1
– Example (Exercise 2.3) Given the survival function S0 (x) = 10
100 − x for
0 ≤ x ≤ 100, find the probability that (0) will die between ages 19 and 36.
– Assumptions: Sx (t) is smooth,
t→∞
– Example. You are given the following survival data of a group of 100 people,
where lx is the number of people in the group who survive to age x.
x
lx
0
100
1
97
2 ···
94 · · ·
50
32
51
28
52
26
53
23
Find F0 (53), S0 (51), f0 (51), S2 (50), and the probability that someone age 2 will
die between 51 and 53.
P (51 < T0 ≤ 53 | T0 > 2) =
S0 (51) − S0 (53)
= P (49 < T2 ≤ 51)
S0 (2)
or just count how many among the 94 survived to age 2 die between 51 and 53.
• The Force of Mortality
1
1
1
P (Tx ≤ ∆x) = lim
P (T0 ≤ x + ∆x | T0 > x) = lim
Fx (∆x)
– µx = lim
∆x→0 ∆x
∆x→0 ∆x
∆x→0 ∆x
– µx ∆x ≈ P (Tx ≤ ∆x)
S0 (x + ∆x)
S0 (x) − S0 (x + ∆x)
=
implies that
S0 (x)
S0 (x)
1 S0 (x) − S0 (x + ∆x)
−1 dS0 (x)
−S00 (x)
µx = lim
=
=
∆x→0 ∆x
S0 (x)
S0 (x) dx
S0 (x)
– Fx (∆x) = 1 − Sx (∆x) = 1 −
−S00 (x)
d
= − ln S0 (x)
S0 (x)
dx
– Let f0 (x) be the density function of T0 . Then −S00 (x) = F00 (x) = f0 (x), and so
– µx =
– µx =
f0 (x)
S0 (x)
– Let x be fixed and t be variable. Then µx+t = · · · =
– µx+t =
−Sx0 (t)
−1 d
Sx (t) =
Sx (t) dt
Sx (t)
−Sx0 (t)
fx (t)
d
=
= − ln Sx (t)
Sx (t)
Sx (t)
dt
d
– Integrating µx+s = − ds
ln Sx (s) from s = 0 to s = t, we get
Z t
Z
Sx (t) = exp −
µx+s ds = exp −
0
x+t
µr dr .
x
– µx ⇐⇒ Sx
– Example. (Exercise 2.1 (a)–(d))
Let F0 (t) = 1 − (1 − t/105)1/5 for 0 ≤ t ≤ 105. Calculate
(a) the probability that a newborn dies before age 60.
(b) the probability that a life aged 30 survives to at least age 70.
(b) the probability that a life aged 20 dies between ages 90 and 100.
(a) the force of mortality at age 50.
– Example. (Exercise 2.5 (a) and (b))
Let F0 (t) = 1 − e−λt , where λ > 0.
(a) Show that Sx (t) = e−λt .
(b) Show that µx = λ.
– Special Mortality Laws
∗
∗
∗
∗
Constant mortality: µx = c
Gompertz’ Law: µx = Bcx (See Example 2.3)
Makeham’s Law: µx = A + Bcx
1
De Moivre’s Law: µx = ω−x
for 0 ≤ x < ω
– Example
Given that µx is constant µ and that the probability that a life aged 60 survives
to age 80 is 0.1, find µx .
Z 20
µx+t dt = 0.1 ⇒ e−20µ = 0.1 ⇒ µ = 0.11513.
20 p60 = 0.1 ⇒ exp −
0
– Example. DML with ω = 100.
1
µx =
for 0 ≤ x < 100 ⇒ for 0 ≤ t < 100 − x,
100 − x
Z t
1
100 − (x + t)
ds =
,
t px = exp −
100 − x
0 100 − (x + s)
t qx
=
t
.
100 − x
(Use a time line to express these expressions.)
• More on Notations
– t px = Sx (t) = P (Tx > t), t qx = Fx (t) = P (Tx ≤ t),
u|t qx = P (u < Tx ≤ u + t) = Sx (u) − Sx (u + t) = Fx (u + t) − Fx (u)
= u pxt · qx+u = u px − u+t px = u+t qx − u qx , u+t px = u px · t px+u
1 d
1 d
– µx = −
(x p0 ), µx+t = −
(t px )
x p0 dx
t px dt
Z t
d
– fx (t) = Fx (t) = t px µx+t , t qx =
s px µx+s ds
dt
0
–
u|t qx
• Mean and Variance of Tx
Z ∞
Z ∞
Z ∞
◦
tt px µx+t dt = · · · =
tfx (t) dt =
– ex = E(Tx ) =
t px dt
0
0
0
Z ∞
Z ∞
2
2
– E(Tx ) =
t fx (t) dt = · · · = 2
t · t px dt
0
V(Tx ) =
E(Tx2 )
0
2
− [E(Tx )] =
E(Tx2 )
◦ 2
− ex
– Example
For constant force of mortality µx = 0.3, t px = e−0.3t ⇒
Z ∞
1
1
◦
ex =
=
.
e−0.3t dt =
0.3
µx
0
Z ∞
2
2
E(Tx ) = 2
t · e−0.3t dt = · · · =
⇒
0.32
0
2
2
1
1
V(Tx ) =
−
=
.
2
0.3
0.3
0, 32
In general, for constant µx = µ,
◦
ex = E(Tx ) =
1
,
µ
V(Tx ) =
1
.
µ2
– Example. (DML with ω = 100)
100−x
Z
◦
ex =
0
t px
=
100−(x+t)
100−x
⇒
100 − (x + t)
100 − x
dt =
.
100 − x
2
It can be computed that
V(Tx ) =
(100 − x)2
.
12
In general
◦
ex =
ω−x
,
2
V(Tx ) =
(ω − x)2
.
12
• Curtate Life Time Kx
– Definition: Kx = bTx c, the integer part of Tx .
– P (Kx = k) = P (k ≤ Tx < k + 1) = k| qx = k px − k+1 px = k px · qx+k
– ex = E(Kx ) =
∞
X
kP (Kx = k) =
k=1
– Similarly,
E(Kx2 )
V(Kx ) = 2
∞
X
∞
X
k(k px − k+1 px ) = · · · =
k=1
= ··· = 2
∞
X
k · k px −
k=1
∞
X
k px
k=1
∞
X
k px
=2
∞
X
k · k px − ex and so
k=1
k=1
k · k px − ex − e2x
k=1
– Example. Find e50 for DML with ω = 100.
– Exercise. Find e50 for CFM with µx = 0.3.
◦
◦
– Relation between ex and ex : ex ≈ ex + 21 .
• Suggested Exrecises
– From the text:
Exercises 2.1–2.3, 2.6–2.7, 2.10–2.11 (Exercises 2.9, 2.14, and 2.15 require more
calculus; they are strongly recommended.)
100 2
.
– Find 5| q40 if S0 (t) = 100+t
– Given µx =
2
100−x
for 0 ≤ x < 100. Calculate
10| q65 .
– Under DML with ω = 100, calculate the probability that (30) will die in his 70’s.
◦
0.01 if t ≤ 5
– Given that µ70+t =
, calculate e70 .
0.02 if t > 5
(Hint: Find explicit expressions for t p70 for t ≤ 5 and for t > 5.)
R∞
◦
Appendix It is shown in the text that ex = 0 t px dt using integration by parts. In order
to use integration by parts, it is assumed that lim tSx (t) = 0. Here is an alternative proof
t→∞
without making the further assumption. First a general resulty in probability.
Lemma If X is a non-negative continuous random variable with CDF F (x) and if E(X)
exists, then
Z ∞
E(X) =
(1 − F (t)) dt.
(1)
0
Proof. Let f (x) = F 0 (x), the density function of X. Then, we have
Z ∞ Z s Z ∞
dt · f (s) ds
s · f (s) ds =
E(X) =
0
Z0 ∞ Z s
Z ∞0Z ∞
=
f (s) dt ds =
f (s) ds dt
0
0
0
t
Z ∞Z ∞
Z ∞
=
f (s) ds dt =
P (X > t) dt
0
t
0
Z ∞
=
(1 − F (t)) dt
0
Now for X = Tx , F (t) = P (Tx ≤ t) = t qx , and 1 − F (t) = 1 − t qx = t px . We have by (1)
Z ∞
Z ∞
◦
ex = E[Tx ] =
(1 − F (t)) dt =
t px dt.
0
0
The formula for E(Tx2 ) and V(Tx ) can be similarly onbtained.