18.3 Inference with Censored Observations 709
ˆ
H(t)
=−log
¡
ˆ
S(t)
¢
.
Better small-sample performance in estimating the cumulative hazard
could be achieved by the Nelson–Aalen estimator,
˜
H(t)
=
X
t
i
<t
d
i
/r
i
, for t > t
1
,
˜
H(t) =0, for t ≤ t
1
,
with an estimated variance
σ
2
H
(t) =
P
t
i
<t
d
i
/r
2
i
. Using
˜
H(t) and σ
2
H
(t), point-
wise confidence intervals on H(t) can be obtained.
Example 18.6. Catheter Complications in Peritoneal Dialysis. The fol-
lowing example is from Chadha et al. (2000). The authors studied a sample
of 36 pediatric patients undergoing acute peritoneal dialysis through Cook
Catheters. They wished to examine how long these catheters performed prop-
erly. They noted the date of complication (either occlusion, leakage, exit-site
infection, or peritonitis).
Half of the subjects had no complications before the catheter was removed.
Reasons for removal of the catheter in this group of patients were that the
patient recovered (n
=4), the patient died (n =9), or the catheter was changed
to a different type electively (n
=5). If the catheter was removed prior to com-
plications, that represented a censored observation, because they knew that
the catheter remained complication free at least until the time of removal.
Day Censored, ` Fail, d At Risk, r 1−(Failures/At Risk) KM
1 8 2 36 1 − 2/36 = 0.944 0.9444
2 2 2 36
− 8 − 2 = 26 1 − 2/26 = 0.92 0.92 · 0.944 = 0.8718
3 1 2 26
− 2 − 2 = 22 1 − 2/22 = 0.91 0.91 · 0.872 = 0.7925
4 1 1 22
− 1 − 2 = 19 1 − 1/19 = 0.95 0.95 · 0.793 = 0.7508
5 6 3 19
− 1 − 1 = 17 1 − 3/17 = 0.82 0.6183
6 0 2 17
− 6 − 3 = 8 1 − 2/8 = 0.75 0.4637
7 0 1 8
− 0 − 2 = 6 1 − 1/6 = 0.83 0.3865
10 0 2 6
− 0 − 1 = 5 1 − 2/5 = 0.60 0.2319
12 0 2 5
− 0 − 2 = 3 1 − 2/3 = 0.33 0.0773
13 0 1 3
− 0 − 2 = 1 1 − 1/1 = 0.00 0.0000
MATLAB script chada.m finds the Kaplan–Meier estimator and gener-
ates Fig. 18.2. plots
%chada.m
times=[1,1,1,1,1,1,1,1,1,1,2,2,2,2,...
3,3,3,4,4,5,5,5,5,5,5,5,5,5,...
6,6,7,10,10,12,12,13];
censored =[1,1,1,1,1,1,1,1,0,0,1,1,...
0,0,1,0,0,1,0,1,1,1,1,1,...
1,0,0,0,0,0,0,0,0,0,0,0];
% Calculate and plot empirical
% cdf and confidence bounds
[f,x,flo,fup] = ecdf(times,’censoring’,censored);
(1-f)’