HSRP 734:
Advanced Statistical Methods
July 10, 2008
Objectives



Describe the Kaplan-Meier estimated survival
curve
Describe the log-rank test
Use SAS to implement
Kaplan-Meier Estimate of Survival Function S(t)

The Kaplan-Meier estimate of the survival
function is a simple, useful and popular estimate
for the survival function.
This estimate incorporates both censored and
noncensored observations
 Breaks the estimation problem down into small
pieces

Kaplan-Meier Estimate of the
Survival Function S(t)

For grouped survival data,
Sˆ (t )  Estimated Pr( Survive beyong t )

y j Lj
  1 
j : bins 1 thru t 
Nj






Let interval lengths Lj become very small – all of length
L=Dt and let t1, t2, … be times of events (survival times)
Kaplan-Meier Estimate of the
Survival Function S(t)

2 cases to consider in the previous equation

Case 1. No event in a bin (interval)
y j Lj
Nj

 1

0 Lj
Nj
y j Lj
Nj
0
1
Sˆ (t ) does not change — which means that we can ignore
bins with no events
Kaplan-Meier Estimate of the
Survival Function S(t)

Case 2. yj events occur in a bin (interval)
Also: nj persons enter the bin
assume any censored times that occur in the bin occur at
the end of the bin
1
y j Lj
Nj
 1
y j Dt
n j Dt

nj  y j
nj
Kaplan-Meier Estimate of the
Survival Function S(t)

So, as Dt → 0, we get the Kaplan- Meier estimate of the survival
function S(t)
 nj  y j 
ˆ

S (t )   

j :t j  t 
 nj 
Sˆ (0)  1 (by convention )


Also called the “product-limit estimate” of the survival function
S(t)
Note: each conditional probability estimate is obtained from the
observed number at risk for an event and the observed number
of events (nj-yj) / nj
Kaplan-Meier Estimate of Survival
Function S(t)

We begin by






Rank ordering the survival times (including the censored
survival times)
Define each interval as starting at an observed time and
ending just before the next ordered time
Identify the number at risk within each interval
Identify the number of events within each interval
Calculate the probability of surviving within that interval
Calculate the survival function for that interval as the
probability of surviving that interval times the probability of
surviving to the start of that interval
Example - AML
Group
Weeks in remission -- ie,
time
to relapse
Maintenance
chemo (X=1)
9, 13, 13+, 18, 23, 28+, 31,
34, 45+, 48, 161+
5, 5, 8, 8, 12, 16+, 23, 27,
30+, 33, 43, 45
No maintenance
chemo (X=0)
+ indicates a censored time to relapse; e.g., 13+ = more than 13 weeks to relapse
Example – AML

Calculation of Kaplan-Meier estimates:
In the “not maintained on chemotherapy” group:
Sˆ (t )
Time
At risk
Events
tj
nj
yj
n  yj
Sˆ (t j )  Sˆ (t j 1 )  j
nj
0
12
0
1.000
5
12
2
1.000 x ((12-2)/12) = 0.833
8
10
2
0.833 x ((10-2)/10) = 0.666
12
8
1
0.666 x ((8-1)/8) = 0.583
23
6
1
0.583 x ((6-1)/6) = 0.486
27
5
1
0.486 x ((5-1)/5) = 0.389
33
3
1
0.389 x ((3-1)/3) = 0.259
43
2
1
0.259 x ((2-1)/2) = 0.130
45
1
1
0.130 x ((1-1)/1) = 0
Example – AML (cont’d)
In the “maintained on chemotherapy” group:
Sˆ (t )
Time
At risk
Events
tj
nj
yj
n  yj
Sˆ (t j )  Sˆ (t j 1 )  j
nj
0
11
0
1.000
9
11
1
1.000 x ((11-1)/11) = 0.909
13
10
1
0.909 x ((10-1)/10) = 0.818
18
8
1
0.716
23
7
1
0.614
31
5
1
0.491
34
4
1
0.368
48
2
1
0.184
Example – AML (cont’d)
1.0
The “Kaplan-Meier curve” plots the estimated survival function vs. time —
separate curves for each group
0.6
0.8
Maintained=0
Maintained=1
0.0
0.2
0.4
Survival

0
50
100
Time
150
Example – AML (cont’d)

Notes
— Can count the total number of events by
counting the number of steps (times)
— If feasible, picture the censoring times on the
graph as shown above.
Kaplan-Meier Estimate Using SAS
Comments on the Kaplan-Meier
Estimate
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If the event and censoring times are tied, we
assume that the censoring time is slightly larger
than the death time.
If the largest observation is an event, the
Kaplan-Meier estimate is 0.
If the largest observation is censored, the
Kaplan-Meier estimate remains constant forever.
Comments on the Kaplan-Meier
Estimate
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
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If we plot the empirical survival estimates, we
observe a step function. If there are no ties and
no censoring, the step function drops by 1/n.
With every censored observation the size of the
steps increase.
When does the number of intervals equal the
number of deaths in the sample?
When does the number of intervals equal n?
Comments on the Kaplan-Meier
Estimate
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
The Kaplan-Meier is a consistent estimate of the
true S(t). That means that as the sample size
gets large, KM estimate converges to the true
value.
The Kaplan-Meier estimate can be used to
empirically estimate any cumulative distribution
function
Comments on the Kaplan-Meier
Estimate

The step function in K-M curve really looks like this:

If you have a failure at t1 then you want to say
survivorship at t1 should be less than 1.
For small data sets it matters, but for large data sets it
does not matter.

Confidence Interval for S(t) –
Greenwood’s Formula

Greenwood’s formula for the variance of Sˆ (t ) :
 
Vaˆr Sˆ (t )  Sˆ (t ) 2 

j:t j  t
yj
n j (n j  y j )
Using Greenwood’s formula, an approximate 95% CI
for S(t) is
 
Sˆ (t )  1.96 Vaˆr Sˆ (t )

There is a “problem”: the 95% CI is not constrained
to lie within the interval (0,1)
Confidence Interval for S(t) –
Alternative Formula

Based on log(-log(S(t)) which ranges from -∞ to ∞

Find the standard error of above, find the CI of above,
then transform CI to one for S(t)

This CI will lie within the interval [0,1]

This is the default in SAS
Log-rank test for comparing survivor
curves
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Are two survivor curves the same?
Use the times of events: t1, t2, ...
(do not include censoring times)
Treat each event and its “set of persons still at risk” (i.e., risk
set) at each time tj as an independent table
Make a 2×2 table at each tj
Event
No Event
Total
Group A
aj
njA- aj
njA
Group B
cj
njB-cj
njB
Total
dj
nj-dj
nj
Log-rank test for comparing survivor
curves


At each event time t j, under assumption of equal
survival (i.e., SA(t) = SB(t) ), the expected number of
events in Group A out of the total events (dj=aj +cj) is
in proportion to the numbers at risk in group A to the
total at risk at time tj:
Eaj = dj x njA / nj
Differences between aj and Eaj represent evidence
against the null hypothesis of equal survival in the two
groups
Log-rank test for comparing survivor
curves

Use the Cochran Mantel-Haenszel idea of pooling over
events j to get the log-rank chi-squared statistic with
one degree of freedom
2
 


a

Ea
 j
j 

j
 ~ 2
2  
1
ˆ
V
a
r
a

j
j
Vaˆ r a j 
d j (n j  d j )n jA n jB
n 2j (n j  1)
Log-rank test for comparing survivor
curves

Idea summary:
Create a 2x2 table at each uncensored failure time
 The construct of each 2x2 table is based on the
corresponding risk set
 Combine information from all the tables

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The null hypothesis is SA(t) = SB(t) for all time t.
Comparisons across Groups


Extensions of the log-rank test to several groups
require knowledge of matrix algebra. In general,
these tests are well approximated by a chisquared distribution with G-1 degrees of
freedom.
Alternative tests:
Wilcoxon family of tests (including Peto test)
 Likelihood ratio test (SAS)

Comparison between Log-Rank and
Wilcoxon Tests



The log-rank test weights each failure time equally. No
parametric model is assumed for failure times within a stratum.
The Wilcoxon test weights each failure time by a function of the
number at risk. Thus, more weight tends to be given to early
failure times. As in the log-rank test, no parametric model is
assumed for failure times within a stratum.
Between these two tests (Wilcoxon and log-rank tests), the
Wilcoxon test will tend to be better at picking up early
departures from the null hypothesis and the log-rank test will
tend to be more sensitive to departures in the tail.
Comparison with Likelihood Ratio
Test in SAS

The likelihood ratio test employed in SAS
assumes the data within the various strata are
exponentially distributed and censoring in noninformative. Thus, this is a parametric method
that smoothes across the entire curve.