Kaplan-Meier methods and
Parametric Regression
methods
Kristin Sainani Ph.D.
http://www.stanford.edu/~kcobb
Stanford University
Department of Health Research and Policy
1
More on Kaplan-Meier estimator of S(t)
(“product-limit estimator” or “KM estimator”)

When there are no censored data, the KM
estimator is simple and intuitive:



When there are censored data, KM provides
estimate of S(t) that takes censoring into
account (see last week’s lecture).


Estimated S(t)= proportion of observations with failure
times > t.
For example, if you are following 10 patients, and 3 of
them die by the end of the first year, then your best
estimate of S(1 year) = 70%.
If the censored observation had actually been a failure:
S(1 year)=4/5*3/4*2/3=2/5=40%
KM estimator is defined only at times when
events occur! (empirically defined)
2
KM (product-limit) estimator,
formally
k distinct event times t1  t j  ...  t k
at each event time t j , there are n j individual s at - risk
d j is the number who have the event at time t j
S (t ) 
dj
 [1  n
j:t j t
]
j
3
KM (product-limit) estimator,
formally
Observed event times
k distinct event times t1  t j  ...  t k
at each event time t j , there are n j individual s at - risk
d j is the number who have the event at time t j
S (t ) 
dj
 [1  n
j:t j t
j
]
The risk set nj at time tj consists of the
originaldj=
sample
minus
all those
Typically
1 person,
unless
datawho
are have
been censored
or had the
event
before tj
grouped
in time intervals
(e.g.,
everyone
rd
who had the event in the 3 month).
d /n =proportion that failed at the event time tj
j j
S(t) represents estimated survival probability at time t:
1- dj/nj=proportion surviving the event time
P(T>t)
Multiply the probability of surviving event time
This formula gives the product-limit estimate of survival at each time an event happens.
t with the probabilities of surviving all the
previous event times.
4
Example 1: time-to-conception
for subfertile women
“Failure” here is a good thing.
38 women (in 1982) were treated for infertility with
laparoscopy and hydrotubation.
All women were followed for up to 2-years to describe
time-to-conception.
The event is conception, and women "survived" until
they conceived.
Example from: BMJ, Dec 1998; 317: 1572 - 1580.
5
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4
7
7
8
8
9
9
9
11
24
24
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
S(t) is estimated at 9 event times.
(step-wise function)
7
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4
7
7
8
8
9
9
9
11
24
24
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4
7
7
8
8
9
9
9
11
24
24
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
6 women conceived in 1st month
(1st menstrual cycle). Therefore,
32/38 “survived” pregnancy-free
past 1 month.
10
Corresponding Kaplan-Meier
Curve
S(t=1) = 32/38 = 84.2%
S(t) represents estimated survival probability: P(T>t)
Here P(T>1).
11
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2.1
3
4
7
7
8
8
9
9
9
11
24
24
Important detail of how the data were coded:
Censoring at t=2 indicates survival PAST the 2nd cycle (i.e., we
know the woman “survived” her 2nd cycle pregnancy-free).
Thus, for calculating KM estimator at 2 months, this person
should still be included in the risk set.
Think of it as 2+ months, e.g., 2.1 months.
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
13
Corresponding Kaplan-Meier
Curve
5 women conceive in 2nd month.
The risk set at event time 2 included 32
women.
Therefore, 27/32=84.4% “survived” event
time 2 pregnancy-free.
S(t=2) = ( 84.2%)*(84.4%)=71.1%
Can get an estimate of the hazard rate
here, h(t=2)= 5/32=15.6%. Given that
you didn’t get pregnant in month 1, you
have an estimated 5/32 chance of
conceiving in the 2nd month.
And estimate of density (marginal
probability of conceiving in month 2):
f(t)=h(t)*S(t)=(.711)*(.156)=11%
14
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2.1
3.1
4
7
7
8
8
9
9
9
11
24
24
Risk set at 3
months includes
26 women
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
16
Corresponding Kaplan-Meier
Curve
3 women conceive in the 3rd month.
The risk set at event time 3 included 26
women.
23/26=88.5% “survived” event time 3
pregnancy-free.
S(t=3) = ( 84.2%)*(84.4%)*(88.5%)=62.8%
17
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3.1
4
7
7
8
8
9
9
9
11
24
24
Risk set at 4
months includes
22 women
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
19
Corresponding Kaplan-Meier
Curve
3 women conceive in the 4th month, and 1
was censored between months 3 and 4.
The risk set at event time 4 included 22
women.
19/22=86.4% “survived” event time 4
pregnancy-free.
S(t=4) = ( 84.2%)*(84.4%)*(88.5%)*(86.4%)=54.2%
Hazard rates (conditional chances of
conceiving, e.g. 100%-84%) look similar
over time.
And estimate of density (marginal
probability of conceiving in month 4):
f(t)=h(t)*S(t)=(.136)* (.542)=7.4%
20
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4.1
7
7
8
8
9
9
9
11
24
24
Risk set at 6
months includes
18 women
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Corresponding Kaplan-Meier
Curve
22
Corresponding Kaplan-Meier
Curve
2 women conceive in the 6th month of the
study, and one was censored between
months 4 and 6.
The risk set at event time 5 included 18
women.
16/18=88.8% “survived” event time 5
pregnancy-free.
S(t=6) = (54.2%)*(88.8%)=42.9%
23
Skipping ahead to the 9th and
final event time (months=16)…
S(t=13)  22%
(“eyeball” approximation)
24
Raw data: Time (months) to conception or censoring in 38 sub-fertile women
after laparoscopy and hydrotubation (1982 study)
Conceived (event)
1
1
1
1
1
1
2
2
2
2
2
3
3
3
4
4
4
6
6
9
9
9
10
13
16
Did not conceive (censored)
2
3
4
7
7
8
8
9
9
9
11
24
24
2 remaining at 16
months (9th event
time)
Data from: Luthra P, Bland JM, Stanton SL. Incidence of pregnancy after laparoscopy
and hydrotubation. BMJ 1982; 284: 1013-1014
Skipping ahead to the 9th and
final event time (months=16)…
S(t=16) =( 22%)*(2/3)=15%
Tail here just represents that the
final 2 women did not conceive
(cannot make many inferences
from the end of a KM curve)!
26
Kaplan-Meier: SAS output
The LIFETEST Procedure
Product-Limit Survival Estimates
time
0.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
2.0000
2.0000
2.0000
2.0000
2.0000
2.0000*
3.0000
3.0000
3.0000
3.0000*
4.0000
4.0000
4.0000
4.0000*
Survival
1.0000
.
.
.
.
.
0.8421
.
.
.
.
0.7105
.
.
.
0.6285
.
.
.
0.5428
.
Failure
0
.
.
.
.
.
0.1579
.
.
.
.
0.2895
.
.
.
0.3715
.
.
.
0.4572
.
Survival
Standard
Error
0
.
.
.
.
.
0.0592
.
.
.
.
0.0736
.
.
.
0.0789
.
.
.
0.0822
.
Number
Failed
Number
Left
0
1
2
3
4
5
6
7
8
9
10
11
11
12
13
14
14
15
16
17
17
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
27
Kaplan-Meier: SAS output
time
6.0000
6.0000
7.0000*
7.0000*
8.0000*
8.0000*
9.0000
9.0000
9.0000
9.0000*
9.0000*
9.0000*
10.0000
11.0000*
13.0000
16.0000
24.0000*
24.0000*
Survival
.
0.4825
.
.
.
.
.
.
0.3619
.
.
.
0.3016
.
0.2262
0.1508
.
.
Failure
.
0.5175
.
.
.
.
.
.
0.6381
.
.
.
0.6984
.
0.7738
0.8492
.
.
Survival
Standard
Error
.
0.0834
.
.
.
.
.
.
0.0869
.
.
.
0.0910
.
0.0944
0.0880
.
.
Number
Failed
Number
Left
18
19
19
19
19
19
20
21
22
22
22
22
23
23
24
25
25
25
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
NOTE: The marked survival times are censored observations.
28
Not so easy to get a plot of the actual hazard function!
In SAS, need a complicated MACRO, and depends on
assumptions…here’s what I get from Paul Allison’s
macro for these data…
At best, you can get the
cumulative hazard function…
t
S (t )  e
  h ( u ) du
0
t

  log S (t )  h(u )du
0
Linear cumulative
hazard function
indicates a constant
hazard.
See lecture 1 if you
want more math!
30
t

 log S (t )  h(u )du
0
Cumulative Hazard Function
If the hazard function is constant, e.g. h(t)=k, then the cumulative hazard function
will be linear (and higher hazards will have steeper slopes):

t
 k du k t
0

If the hazard function is increasing with time, e.g. h(t)=kt, then the cumulative
hazard function will be curved up, for example h(t)=kt gives a quadratic:
t
k t2
k tdu 
2
0


If the hazard function is decreasing over time, e.g. h(t)=k/t, then the
cumulative hazard function should be curved down, for example:
t

0
k
du k log( t )
t
31
Kaplan-Meier: example 2
Researchers randomized 44 patients with chronic active
hepatitis were to receive prednisolone or no treatment
(control), then compared survival curves.
Example from: BMJ 1998;317:468-469 ( 15 August )
32
Survival times (months) of 44 patients with chronic active hepatitis randomised to receive
prednisolone or no treatment.
Prednisolone (n=22)
Control (n=22)
2
2
6
3
12
4
54
7
56 *
10
68
22
89
28
96
29
96
32
125*
37
128*
40
131*
41
140*
54
141*
61
143
63
145*
71
146
127*
148*
140*
162*
146*
168
158*
173*
167*
181*
182*
Data from: BMJ 1998;317:468-469 ( 15 August )
*=censored
Kaplan-Meier: example 2
Are these two curves
different?
Big drops at the end of
the curve indicate few
patients left. E.g., only
2/3 (66%) survived this
drop.
Misleading to the eye—
apparent convergence by end
of study. But this is due to 6
controls who survived fairly
long, and 3 events in the
treatment group when the
sample size was small.
34
Control group:
time
6 controls
made it past
100 months.
0.000
2.000
3.000
4.000
7.000
10.000
22.000
28.000
29.000
32.000
37.000
40.000
41.000
54.000
61.000
63.000
71.000
127.000*
140.000*
146.000*
158.000*
167.000*
182.000*
Survival
1.0000
0.9545
0.9091
0.8636
0.8182
0.7727
0.7273
0.6818
0.6364
0.5909
0.5455
0.5000
0.4545
0.4091
0.3636
0.3182
0.2727
.
.
.
.
.
.
Failure
0
0.0455
0.0909
0.1364
0.1818
0.2273
0.2727
0.3182
0.3636
0.4091
0.4545
0.5000
0.5455
0.5909
0.6364
0.6818
0.7273
.
.
.
.
.
.
Survival
Standard
Error
0
0.0444
0.0613
0.0732
0.0822
0.0893
0.0950
0.0993
0.1026
0.1048
0.1062
0.1066
0.1062
0.1048
0.1026
0.0993
0.0950
.
.
.
.
.
.
Number
Failed
Number
Left
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
16
16
16
16
16
16
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
treated group:
time
5/6 of 54%
rapidly drops
the curve to
45%.
2/3 of 45%
rapidly drops
the curve to
30%.
0.000
2.000
6.000
12.000
54.000
56.000*
68.000
89.000
96.000
96.000
125.000*
128.000*
131.000*
140.000*
141.000*
143.000
145.000*
146.000
148.000*
162.000*
168.000
173.000*
181.000*
Survival
1.0000
0.9545
0.9091
0.8636
0.8182
.
0.7701
0.7219
.
0.6257
.
.
.
.
.
0.5475
.
0.4562
.
.
0.3041
.
.
Failure
0
0.0455
0.0909
0.1364
0.1818
.
0.2299
0.2781
.
0.3743
.
.
.
.
.
0.4525
.
0.5438
.
.
0.6959
.
.
Survival
Standard
Error
0
0.0444
0.0613
0.0732
0.0822
.
0.0904
0.0967
.
0.1051
.
.
.
.
.
0.1175
.
0.1285
.
.
0.1509
.
.
Number
Failed
Number
Left
0
1
2
3
4
4
5
6
7
8
8
8
8
8
8
9
9
10
10
10
11
11
11
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Point-wise confidence intervals
We will not worry about mathematical formula for confidence bands.
The important point is that there is a confidence interval for each
estimate of S(t). (SAS uses Greenwood’s formula.)
37
Log-rank test
Test of Equality over Strata
Test
Log-Rank
Wilcoxon
-2Log(LR)
Chi-Square
4.6599
6.5435
5.4096
DF
1
1
1
Pr >
Chi-Square
0.0309
0.0105
0.0200
Chi-square test (with 1 df) of the (overall)
difference between the two groups.
Groups appear significantly different.
38
Log-rank test
Log-rank test is just a Cochran-Mantel-Haenszel chi-square test!
Anyone remember (know) what this is?
39
CMH test of conditional
independence
K Strata =
unique event
times
k
[

Event
No Event
Group 1
a
b
Group 2
c
d
Nk
(ak  E (ak ))]2
i 1
k
Var(a )
k
i 1
~ 12
E ( ak ) 
(ak  bk ) * (ak  ck )
Nk
Var(ak ) 
(ak  bk ) * (ck  d k ) * (ak  ck ) * (bk  d k )
N k2 ( N k  1)
CMH test of conditional
independence
K Strata =
unique event
times
k
[

Event
No Event
Group 1
a
b
Group 2
c
d
Nk
(ak  E (ak ))]2
i 1
k
Var(a )
k
i 1
~ 12
E ( ak ) 
row1k * col1k
Nk
Var(ak ) 
row1k * row2 k * col1k * col2 k
N k2 ( N k  1)
CMH test of conditional
independence
)
How
do E(you
know
that
 events
 events
observed  expected

Z
standard
deviation
this Var
is a events
chi-square with
1
df?
Z 
k event times
2
1
k
[
 (a
No Event
Group 1
a
b
Group 2
c
d
k event times
k event times
2
Event
 E (ak ))]
2
k
i 1
k
Var(a )
k
~
2
1
Why is this the
expected value in
each stratum?
E ( ak ) 
row1k * col1k
Nk
Var(ak ) 
row1k * row2 k * col1k * col2 k
N k2 ( N k  1)
i 1
Variance is the variance of a
hypergeometric distribution
At
risk=22
Event time 1 (2 months), control group:
time
1st
event at
month
2.
0.000
2.000
3.000
4.000
7.000
10.000
22.000
28.000
29.000
32.000
37.000
40.000
41.000
54.000
61.000
63.000
71.000
127.000*
140.000*
146.000*
158.000*
167.000*
182.000*
Survival
1.0000
0.9545
0.9091
0.8636
0.8182
0.7727
0.7273
0.6818
0.6364
0.5909
0.5455
0.5000
0.4545
0.4091
0.3636
0.3182
0.2727
.
.
.
.
.
.
Failure
0
0.0455
0.0909
0.1364
0.1818
0.2273
0.2727
0.3182
0.3636
0.4091
0.4545
0.5000
0.5455
0.5909
0.6364
0.6818
0.7273
.
.
.
.
.
.
Survival
Standard
Error
0
0.0444
0.0613
0.0732
0.0822
0.0893
0.0950
0.0993
0.1026
0.1048
0.1062
0.1066
0.1062
0.1048
0.1026
0.0993
0.0950
.
.
.
.
.
.
Number
Failed
Number
Left
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
16
16
16
16
16
16
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Event time 1 (2 months), treated group:
time
Survival
Failure
Survival
Standard
Error
At
risk=22
Number
Failed
Number
Left
0
1
2
3
4
4
5
6
7
8
8
8
8
8
8
9
9
10
10
10
11
11
11
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
1st
event at
month
2.
0.000
2.000
6.000
12.000
54.000
56.000*
68.000
89.000
96.000
96.000
125.000*
128.000*
131.000*
140.000*
141.000*
143.000
145.000*
146.000
148.000*
162.000*
168.000
173.000*
181.000*
1.0000
0.9545
0.9091
0.8636
0.8182
.
0.7701
0.7219
.
0.6257
.
.
.
.
.
0.5475
.
0.4562
.
.
0.3041
.
.
0
0.0455
0.0909
0.1364
0.1818
.
0.2299
0.2781
.
0.3743
.
.
.
.
.
0.4525
.
0.5438
.
.
0.6959
.
.
0
0.0444
0.0613
0.0732
0.0822
.
0.0904
0.0967
.
0.1051
.
.
.
.
.
0.1175
.
0.1285
.
.
0.1509
.
.
Stratum 1= event time 1
Event time 1:
1 died from each
group. (22 at risk in
each group)
Event
No Event
treated
1
21
control
1
21
44
a1  1
(22) * (2)
1
44
(22) * (22) * (2) * (42)
Var(a1 ) 
 .244
2
44 (43)
E (a1 ) 
Event time 2 (3 months), control group:
time
Next
event at
month
3.
0.000
2.000
3.000
4.000
7.000
10.000
22.000
28.000
29.000
32.000
37.000
40.000
41.000
54.000
61.000
63.000
71.000
127.000*
140.000*
146.000*
158.000*
167.000*
182.000*
Survival
1.0000
0.9545
0.9091
0.8636
0.8182
0.7727
0.7273
0.6818
0.6364
0.5909
0.5455
0.5000
0.4545
0.4091
0.3636
0.3182
0.2727
.
.
.
.
.
.
Failure
0
0.0455
0.0909
0.1364
0.1818
0.2273
0.2727
0.3182
0.3636
0.4091
0.4545
0.5000
0.5455
0.5909
0.6364
0.6818
0.7273
.
.
.
.
.
.
Survival
Standard
Error
0
0.0444
0.0613
0.0732
0.0822
0.0893
0.0950
0.0993
0.1026
0.1048
0.1062
0.1066
0.1062
0.1048
0.1026
0.0993
0.0950
.
.
.
.
.
.
At
risk=21
Number
Failed
Number
Left
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
16
16
16
16
16
16
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Event time 2 (3 months), treated group:
time
No
events
at 3
months
0.000
2.000
6.000
12.000
54.000
56.000*
68.000
89.000
96.000
96.000
125.000*
128.000*
131.000*
140.000*
141.000*
143.000
145.000*
146.000
148.000*
162.000*
168.000
173.000*
181.000*
Survival
1.0000
0.9545
0.9091
0.8636
0.8182
.
0.7701
0.7219
.
0.6257
.
.
.
.
.
0.5475
.
0.4562
.
.
0.3041
.
.
Failure
0
0.0455
0.0909
0.1364
0.1818
.
0.2299
0.2781
.
0.3743
.
.
.
.
.
0.4525
.
0.5438
.
.
0.6959
.
.
Survival
Standard
Error
0
0.0444
0.0613
0.0732
0.0822
.
0.0904
0.0967
.
0.1051
.
.
.
.
.
0.1175
.
0.1285
.
.
0.1509
.
.
At
risk=21
Number
Failed
Number
Left
0
1
2
3
4
4
5
6
7
8
8
8
8
8
8
9
9
10
10
10
11
11
11
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Stratum 2= event time 2
Event time 2:
Event
No Event
At 3 months, 1 died in
the control group.
treated
0
21
At that time 21 from
each group were at risk
control
1
20
42
a1  0
(1) * (21)
 .5
42
(21) * (21) * (1) * (41)
Var(a1 ) 
 .25
2
42 (41)
E (a1 ) 
Event time 3 (4 months), control group:
time
1 event at
month 4.
0.000
2.000
3.000
4.000
7.000
10.000
22.000
28.000
29.000
32.000
37.000
40.000
41.000
54.000
61.000
63.000
71.000
127.000*
140.000*
146.000*
158.000*
167.000*
182.000*
Survival
1.0000
0.9545
0.9091
0.8636
0.8182
0.7727
0.7273
0.6818
0.6364
0.5909
0.5455
0.5000
0.4545
0.4091
0.3636
0.3182
0.2727
.
.
.
.
.
.
Failure
0
0.0455
0.0909
0.1364
0.1818
0.2273
0.2727
0.3182
0.3636
0.4091
0.4545
0.5000
0.5455
0.5909
0.6364
0.6818
0.7273
.
.
.
.
.
.
Survival
Standard
Error
0
0.0444
0.0613
0.0732
0.0822
0.0893
0.0950
0.0993
0.1026
0.1048
0.1062
0.1066
0.1062
0.1048
0.1026
0.0993
0.0950
.
.
.
.
.
.
Number
Failed
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
16
16
16
16
16
16
At
risk=20
Number
Left
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Event time 3 (4 months), treated group:
time
0.000
2.000
6.000
12.000
54.000
56.000*
68.000
89.000
96.000
96.000
125.000*
128.000*
131.000*
140.000*
141.000*
143.000
145.000*
146.000
148.000*
162.000*
168.000
173.000*
181.000*
Survival
1.0000
0.9545
0.9091
0.8636
0.8182
.
0.7701
0.7219
.
0.6257
.
.
.
.
.
0.5475
.
0.4562
.
.
0.3041
.
.
Failure
0
0.0455
0.0909
0.1364
0.1818
.
0.2299
0.2781
.
0.3743
.
.
.
.
.
0.4525
.
0.5438
.
.
0.6959
.
.
Survival
Standard
Error
0
0.0444
0.0613
0.0732
0.0822
.
0.0904
0.0967
.
0.1051
.
.
.
.
.
0.1175
.
0.1285
.
.
0.1509
.
.
At
risk=21
Number
Failed
Number
Left
0
1
2
3
4
4
5
6
7
8
8
8
8
8
8
9
9
10
10
10
11
11
11
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Stratum 3= event time 3
(4 months)
Event time 3:
At 4 months, 1 died in
the control group.
At that time 21 from
the treated group and
20 from the control
group were at-risk.
Event
No Event
treated
0
21
control
1
19
a1  0
(1) * (21)
 .51
41
(21) * (20) * (1) * ( 40)
Var(a1 ) 
 .25
2
41 (40)
E ( a1 ) 
41
Etc.
22
[

(a k  E (a k ))]2
i 1
22
 Var(a
i 1
k
)
[(1  1)  (0  .5)  (0  .51)  ...............]2

 4.66
.244  .25  .25  .....
Log-rank test, et al.
Wilcoxon is just a version of the
log-rank test that weights
strata by their size (giving more
weight to earlier time points).
More sensitive to differencesTest
at of Equality over Strata
earlier time points.
Test
Log-Rank
Wilcoxon
-2Log(LR)
Chi-Square
4.6599
6.5435
5.4096
Likelihood Ratio test is not ideal here
because it assumes exponential
distribution (constant hazard).
DF
1
1
1
Pr >
Chi-Square
0.0309
0.0105
0.0200
Log-rank test has most power
to test differences that fit the
proportional hazards model—so
works well as a set-up for
subsequent Cox regression.
53
Estimated –log(S(t))
Maybe hazard function
decreases a little then
increases a little? Hard
to say exactly…
54
One more graph from SAS…
log(-log(S(t))=
log(cumulative hazard)
If group plots are parallel,
this indicates that the
proportional hazards
assumption is valid.
Necessary assumption for
calculation of Hazard
Ratios…
55
Uses of Kaplan-Meier



Commonly used to describe survivorship
of study population/s.
Commonly used to compare two study
populations.
Intuitive graphical presentation.
56
Limitations of Kaplan-Meier
•
•
•
Mainly descriptive
Doesn’t control for covariates
Requires categorical predictors
•
•
SAS does let you easily discretize continuous variables
for KM methods, for exploratory purposes.
Can’t accommodate time-dependent variables
57
Parametric Models for the
hazard/survival function

The class of regression models
estimated by PROC LIFEREG is known
as the accelerated failure time models.
58
Recall: two parametric models
Components:
•A baseline hazard function (that may change over time).
•A linear function of a set of k fixed covariates that when
exponentiated (and a few other things) gives the relative risk.
Exponential model assumes fixed baseline hazard that we can estimate.
log hi (t )     1 xi1  ...   k xik
Weibull model models the baseline hazard as a function of time. Two parameters (baseline
hazard and scale) must be estimated to describe the underlying hazard function over time.
log hi (t )     log t   1 xi1  ...   k xik
59
To get Hazard Ratios (relative
risk)…
•Weibull (and thus exponential) are proportional hazards models, so
hazard ratio can be calculated.
•For other parametric models, you cannot calculate hazard ratio
(hazards are not necessarily proportional over time).
Exponentia l Model :
HR  e

Weibull Model :
HR  e

scale
More tricky to get confidence intervals here!
60
What’s a hazard ratio?
Distinction between hazard/rate ratio and
odds ratio/risk ratio:
 Hazard/rate ratio: ratio of incidence
rates
 Odds/risk ratio: ratio of proportions
61
Example 1
Using data from pregnancy study…
Recall: roughly, hazard rates were similar
over time
(implies exponential model should be a
good fit).
62
The LIFEREG Procedure
Analysis of Parameter Estimates
Standard
Parameter
DF Estimate
Error
95% Confidence
Limits
Intercept
1
2.2636
0.2049
1.8621
2.6651
Scale
1
1.0217
0.1638
0.7462
1.3987
Weibull Shape
1
0.9788
0.1569
0.7149
1.3401
Scale of 1.0 makes a Weibull an
exponential, so looks exponential.
ChiSquare Pr > ChiSq
122.08
<.0001
Parametric estimates of survival function based on a
Weibull model (left) and exponential (right).
Compare
to KM:
64
Example 2: 2 groups
Using data from hepatitis trial, I fit
exponential and Weibull models in SAS
using LIFEREG (Weibull is default in
LIFEREG)…
65
-2Log Likelihood = 2*68= 176
The LIFEREG Procedure
Dependent Variable
Log(time)
Right Censored Values
17
Left Censored Values
0
Interval Censored Values
0
Name of Distribution
Exponential
Log Likelihood
-68.03461345
Scale parameter is set to 1,
because it’s exponential.
Analysis of Parameter Estimates
Standard
Parameter
DF Estimate
Error
95% Confidence
Limits
P-value for group very
similar to p-value from logrank test.
ChiSquare Pr > ChiSq
Intercept
1
4.4886
0.2500
3.9986
4.9786
322.37
<.0001
group
1
0.9008
0.3917
0.1332
1.6685
5.29
0.0214
Scale
0
1.0000
0.0000
1.0000
1.0000
Weibull Shape
0
1.0000
0.0000
1.0000
1.0000
Hazard ratio (treated vs. control):
e-0.9008 = .406
Interpretation: median time to death was decreased 60% in
treated group; or, equivalently, mortality rate is 60% lower
in treated group.
-2Log Likelihood = 2*67= 174
Model Information
Comparison of models using Likelihood Ratio test:
Dependent Variable
Log(time)
-2LogLikelihood(simpler model)—2LogLikelihood(more complex) = chisquare with 1 df (1 extra17
parameter estimated for weibull model).
Right Censored Values
=176-174 = 2
Left Censored Values
Interval Censored Values
0
NS
0
No evidence that Weibull model is much better than exponential.
Name of Distribution
Weibull
Log Likelihood
-66.94904552
Scale parameter is greater
than 1, indicating decreasing
hazard with time.
P-value for group very
similar to p-value from logrank test and exponential
model.
Analysis of Parameter Estimates
Standard
Parameter
DF Estimate
Error
95% Confidence
Limits
ChiSquare Pr > ChiSq
Intercept
1
4.4811
0.3169
3.8601
5.1022
200.00
<.0001
group
1
1.0544
0.5096
0.0556
2.0533
4.28
0.0385
Scale
1
1.2673
0.2139
0.9103
1.7643
Weibull Shape
1
0.7891
0.1332
0.5668
1.0985
Shape parameter is just
1/scale parameter!
Hazard ratio (treated vs. control):
e-1.05/1.267 = .43
Parametric estimates of cumulative survival based on
Weibull model (left) and exponential (right), by group.
Compare
to KM:
Compare to Cox regression:
Variable
group
DF
Parameter
Estimate
Standard
Error
Chi-Square
Pr > ChiSq
Hazard
Ratio
1
-0.83230
0.39739
4.3865
0.0362
0.435
95% Hazard Ratio
Confidence Limits
0.200
0.948
69
References
Paul Allison. Survival Analysis Using SAS. SAS Institute Inc., Cary, NC: 2003.
70