Download ST5212: Survival Analysis

ST5212: Survival Analysis
2008/2009: Semester II
Tutorial 2
1. A model for lifetimes, with a bathtub-shaped hazard rate, is the exponential power distribution with survival fumction
S(x) = exp{1 − exp[(λx)α ]}.
(a) If α = 0.5, show that the hazard rate has a bathtub shape and find the time at which the
hazard rate changes from decreasing to increasing.
The hazard function is given by
h(x) = αλα xα−1 exp[(λx)α ]
√
√
= 0.5 λx−0.5 exp[ λx] when α = 0.5.
(b) If α = 2, show that the hazard rate of x is monotone increasing.
2. The Gompertz distribution is commonly used by biologists who obelieve that an exponential
hazard rate should occur in nature. The survival function of the Gompertz distribution is given
by
θ
S(x) = exp{ (1 − eαx)}.
α
Suppose that the time to death in months for a mouse exposed to a high dose of radiation
follows a Gompertz distribution with θ = 0.01 and α = 0.25. Find
(a) the probability that a randomly chosen mouse will live at least one year,
The probability is given by
S(12) = exp((1/25)[1 − e0.25×12]) = 0.466.
(b) the probability that a randomly chosen mouse will die within the first six months, and
The probability is given by
1 − S(6) = 1 − exp((1/25)[1 − e0.25×6]) = 0.13.
1
(c) the median time to death.
Solving S(x) = 0.5 to obtain
x = ln(1 + 25 ln(2))/0.25 = 11.63.
3. The time to relapse, in months, for patients on two treatments for lung cancer is compared
using the following log normal regression model:
Y = ln X = 2 + 0.5z + 2W,
where W has a standard normal distribution and Z = 1 if treatment A and 0 if treatment B.
(a) Compare the survival probabilities of the two treatments at 1, 2, and 5 years.
The survival function is given by
S(x) = P (X > x)
= P (ln X > ln x)
= P (2 + 0.5Z + 2W > ln x)
1
= P (W > (ln x − 2 − 0.5Z))
2
1
= 1 − Φ( (ln x − 2 − 0.5Z)),
2
where Φ is the CDF of the standard normal distribution. The time (1, 2, 5) years = (12,
24, 60) months. Thus the survival probabilities for treatment A and B are
Years
1
2
5
A
0.506
0.249
0.055
B
0.314
0.119
0.018
(b) Repeat the calculations if W has a standard logistic distribution. The survival funciton of
the standard logistic distribution is as follows:
S(x) =
1
.
1 + ex
Compare your results with part (a).
The computation using the above survival function yields
2
Years
A
B
1
0.504 0.381
2
0.337 0.235
5
0.169 0.110
4. A model used in the construction of life tables is a piecewise, constant hazard rate model.
Here the time axis is divided into k intervals, [τi−1 , τi ), i = 1, . . . , k, with τ0 = 0 and τk = ∞.
The hazard rate on the ith interval is a constant value, θi ; that is
⎧
θ1
0 ≤ x < τ1
⎪
⎪
⎪
⎪
⎪
τ1 ≤ x < τ2
⎨ θ2
.
..
h(x) =
⎪
⎪
⎪
θk−1 τk−2 ≤ x < τk−1
⎪
⎪
⎩ θ
x ≥ τk−1
k
(a) Find the survival function for this model.
First find the cumulative hazard function H(x). For τi−1 ≤ x < τi ,
H(x) =
=
x
h(t)dt
0
i−1
τj
i−1
x
θj dt +
θi dt
τi−1
τj−1
j=1
=
θj (τj − τj−1 ) + θi (x − τi−1 ).
j=1
Thus,
S(x) = exp[−
i−1
θj (τj − τj−1 ) − θi (x − τi−1 )].
j=1
(b) Find the mean residual-life function.
∞
0
S(t)dt =
k i=1
τi
τ0 = 0, τk = ∞.
S(t)dt,
τi−1
3
Note:
τi
τi
S(t)dt =
exp[−
τi−1
τi−1
i−1
θj (τj − τj−1 ) − θi (t − τi−1 )]dt
j=1
= exp[−
i−1
θj (τj − τj−1 )]
τi
exp[−θi(t − τi−1 )]dt
τi−1
j=1
= exp[−
i−1
1
θj (τj − τj−1 )] (1 − exp[−θi(τi − τi−1 )]).
θi
j=1
Hence
∞
S(t)dt =
0
k
exp[−
i−1
i=1
1
θj (τj − τj−1 )] (1 − exp[−θi(τi − τi−1 )]).
θi
j=1
Similarly, we can derive, for τi−1 ≤ x < τi ,
∞
S(t)dt =
x
k
l=i+1
=
k
l−1
1
exp[−
θj (τj − τj−1 )] (1 − exp[−θl(τl − τl−1 )]) +
θl
j=1
exp[−
τi
S(t)dt
x
l−1
l=i+1
exp[−
1
θj (τj − τj−1 )] (1 − exp[−θl(τl − τl−1 )]) +
θl
j=1
i−1
1
θj (τj − τj−1 )] (exp[−θi(x − τi−1 )] − exp[−θi(τi − τi−1 )]).
θi
j=1
The mean residual life function is then given by
∞
S(t)dt
.
mrlf(x) = x∞
S(t)dt
0
(c) Find the median life.
Solving for each i = 1, . . . , k,
exp(−θi (x − τi−1 ) exp[−
i−1
θj (τj − τj−1 )] = 0.5,
j=1
4
yields
i−1
1
θj (τj − τj−1 )) − log 2 + τi−1 .
xi = − log(
θi
j=1
Then
t0.5 = min{xi : τi−1 ≤ xi < τi }.
5. Given a covariate Z, suppose that the log survival time Y follows a linear model with a
logistic error distribution, that is,
Y = ln X = µ + βZ + σW,
where the pdf of W is given by
f(w) =
ew
, −∞ < w < ∞.
(1 + ew )2
(a) For an individual with covariate Z, find the conditional survival function of the survival
time X, given Z, namely, S(x|Z).
The survival function for W is given by SW (w) = (1 + ew )−1 . By the same reasoning as
in Problem 3,
1
S(x|Z) = SW ( (ln x − µ − βZ)) = [1 + x1/σ e−µ−βZ ]−1 .
σ
(b) The odds that an individual will die prior to time x is expressed by [1 − S(x|Z)]/S(x|Z).
Compute the odds of death prior to time x for this model.
1 − S(x|Z)
= x1/σ e−µ−βZ .
S(x|Z)
(c) Consider two individuals with different covariate values. Show that, for any time x, the
ratio fo their odds of the death is independent of x. The log logistic regression model is
the only model with this property.
It is obvious from (b).
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6. Suppose that the joint survival function of the latent failure times for two competing risks,
X and Y , is
S(x, y) = (1 − x)(1 − y)(1 + 0.5xy), 0 < x < 1, 0 < y < 1.
(a) Find the marginal survival function for X.
The marginal survival function for X is
SX (x) = S(x, 0) = 1 − x.
(b) Find the cumulative incidence functioin F1 .
− ∂S(x,y)
|x=y=t
1
0.5t
∂x
= hy (t).
hx (t) =
=
−
S(t, t))
1 − t 1 + 0.5t2
hT (t) = hx (t) + hy (t) = 2hx (t).
x
t
hx (t)e− 0 hT (u)du dt
Fx(x) =
0
t
0
hT (u)du = 2
Fx(x) =
0
x
[
0
t
hx (u)du = 2
t
[
0
0.5u
1
t2
−
)
]du
=
−2
ln(1−t)+ln(1+
1 − u 1 + 0.5u2
2
t(1 − t)2/2
1−t
−
]dt.
1 + t2 /2 (1 + t2/2)2
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