1
Basic concepts
Let T be a nonnegative (i.e. T ≥ 0) continuous random variable with distribution function
F (t) = P (T ≤ t)
and survival function
S(t) = P (T > t).
Since T is continuous, it has a density function defined by
f (t) = F ′ (t) = lim
h↓0
1
1
[F (t + h) − F (t)] = lim P(t < T ≤ t + h).
h↓0 h
h
Also note that
f (t) =
d
d
F (t) = [1 − S(t)] = −S ′ (t).
dt
dt
(1)
In survival analysis, we will be working with the hazard rate function
defined as
1
µt = lim P(T ≤ t + h|T > t)
h↓0 h
Claim 1. The hazard rate function can be expressed as
µt =
f (t)
S(t)
µt = −
and
d
log S(t).
dt
Proof. Using the identity P(A|B) = P(A ∩ B)/ P(B), it follows that
1
P(T ≤ t + h|T > t)
h
P(T ≤ t + h, T > t)
= lim
h↓0
h P(T > t)
P(t < T ≤ t + h)
= lim
h↓0
h P(T > t)
F (t + h) − F (t)
= lim
h↓0
hS(t)
f (t)
=
.
S(t)
µt = lim
h↓0
The second claim follows by applying the chain rule and (1):
−
S ′ (t)
f (t)
d
log S(t) = −
=
= µt .
dt
S(t)
S(t)
Claim 2. The survival function can be expressed in terms of the hazard rate
as follows:
Z t
S(t) = exp −
µs ds .
0
1
Proof. Integrating on both sides the second result in Claim 1, we get that
Z t
Z t
d
µs ds = −
log S(s)ds
ds
0
0
= − [log S(t) − log S(0)]
= − log S(t),
where we made use of the fact that S(0) = 1. Solving for S(t) yields the
required result.
The following claim provides an alternative expression for the expected
value of a nonnegative random variable.
Claim 3. For any nonnegative random variable T ,
Z ∞
Z ∞
P (T > u)du =
S(u)du,
E(T ) =
0
0
provided that E(T ) < ∞.
Proof. We first prove the claim if T is continuous and bounded, i.e. 0 ≤ T ≤
ω for some constant limiting age ω < ∞. Using integration by parts and
the fact that f (u) = −S ′ (u),
Z ω
Z ω
ω
E(T ) =
uf (u)du = −[uS(u)]0 +
S(u)du
0
0
Z ω
= −ωS(ω) + 0 × S(0) +
S(u)du.
0
Since S(0) = 1 and S(ω) = 0, the result follows. To prove the claim for
unbounded T requires some technical results from calculus, and is left to
those who are interested (you can come ask me if you are interested, but it
is not for exam purposes).
If T is discrete, the proof follows as discussed in class (and in your
actuarial notes). I will maybe also add it here later.
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Conditional survival times
The above results can be generalised to the case of conditional survival times.
Let Tx denote the future lifetime after time x of an individual who is alive
at age x. The distribution function of Tx is defined by
Fx (t) = P(Tx ≤ t) = P(T ≤ x + t|T > x),
and the survival function by
Sx (t) = P(Tx > t) = P(T > x + t|T > x).
2
Here, T denotes the lifetime of the individual from birth, i.e. T = T0 .
The following relationships hold between the conditional random variable
Tx and the unconditional variable T .
Claim 4. It holds that
Sx (t) =
S(x + t)
S(x)
and
Fx (t) =
F (x + t) − F (x)
.
S(x)
Proof. Notice that
Sx (t) = P(T > x + t|T > x)
P(T > x + t, T > x)
=
P(T > x)
P(T > x + t)
=
P(T > x)
S(x + t)
,
=
S(x)
which proves the first claim. The second claim follows from observing that
S(x + t)
S(x)
S(x) − S(x + t)
=
S(x)
[1 − F (x)] − [1 − F (x + t)]
=
S(x)
F (x + t) − F (x)
=
.
S(x)
Fx (t) = 1 − Sx (t) = 1 −
Notice that the above result also provides an expression for the density
of Tx in terms of the unconditional density f and unconditional survival
function S:
∂
f (x + t)
∂ F (x + t) − F (x)
fx (t) = Fx (t) =
=
.
(2)
∂t
∂t
S(x)
S(x)
Claim 5. The hazard rate function of Tx can be expressed as
ux+t =
fx (t)
Sx (t)
ux+t = −
and
3
∂
log Sx (t).
∂t
Proof. As before, notice that
1
P(T ≤ x + t + h|T > x + t)
h↓0 h
P(T ≤ x + t + h, T > x + t)
= lim
h↓0
h P(T > x + t)
P(x + t < T ≤ x + t + h)
= lim
h↓0
h P(T > x + t)
F (x + t + h) − F (x + t)
= lim
h↓0
hS(x + t)
f (x + t)
=
S(x + t)
fx (t)
=
,
Sx (t)
ux+t = lim
where the last step follows from (2) and Sx (t) = S(x + t)/S(x) proved in
Claim 4.
Finally, since Tx ≥ 0, it follows immediately from Claim 3 that
Z ∞
Z ∞
E(Tx ) =
P (Tx > u)du =
Sx (u)du,
0
0
if we can assume that E(Tx ) < ∞.
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