54208#1 STATISTICS 542 Intro to Clinical Trials SURVIVAL ANALYSIS.

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54208#1
STATISTICS 542STATISTICS 542Intro to Clinical TrialsIntro to Clinical Trials
SURVIVAL ANALYSISSURVIVAL ANALYSIS
54208#2
Survival Analysis Survival Analysis TerminologyTerminology
• Concerned about time to some event
• Event is often death
• Event may also be, for example
1. Cause specific death2. Nonfatal event or death,
whichever comes firstdeath or hospitalizationdeath or MI
death or tumor recurrence
54208#3
0 1 2 3 4 50
0.25
0.5
0.75
1Group A Group B
YEARS•At 5 years, survival rates the same•Survival experience in Group A appears more favorable, considering 1 year, 2 year, 3 year and 4 year rates together
Survival Rates at Yearly Intervals
54208#4
BetaBlocker Heart Attack TrialBetaBlocker Heart Attack Trial
LIFETABLE CUMULATIVE MORTALITY CURVELIFETABLE CUMULATIVE MORTALITY CURVE
54208#5
Survival AnalysisSurvival AnalysisDiscuss
1. Estimation of survival curves
2. Comparison of survival curves
I. Estimation
• Simple Case– All patients entered at the same time and followed for
the same length of time– Survival curve is estimated at various time points by
(number of deaths)/(number of patients)– As intervals become smaller and number of patients
larger, a "smooth" survival curve may be plotted
• Typical Clinical Trial Setting
54208#6
• Each patient has T years of followup• Time for followup taking place may be different for each patient
T years
T years
T years
T years
T0 2T
4
3
2
1
Time Since Start of Trial (T years)
Su
bje
ct
Staggered Entry
54208#7
•Failure time is time from entry until the time of the event•Censoring means vital status of patient is not known beyond that point
o
*
*
•
AdministrativeCensoring
Failure
CensoringLoss to Followup
Time Since Start of Trial (T years)
T 2T0
4
3
2
1
Subject
54208#8
o
*
•
*
T0
4
3
2
1
Subject
AdministrativeCensoring
CensoringLoss to Followup
Failure
Followup Time (T years)
54208#9
Clinical Trial with Common Termination DateClinical Trial with Common Termination Date
*
*
*
1
2
3
4
5
6
7
8
9
10
11
0 T 2TFollowup Time (T years) Trial
Terminated
Subject
*
••• •
•••••
o
o
•
o o
o
o o
•
54208#10
Years of Cohort
FollowUp Patients I IITotalEntered 100 100 200
1Died 20 25 45
Entered 80 75 1552
Died 20
Survived 60
Reduced Sample Estimate (1)
54208#11
– Suppose we estimate the 1 year survival rate
a. P(1 yr) = 155/200 = .775
b. P(1 yr, cohort I) = 80/100 = .80
c. P(1 yr, cohort II) = 75/100 = .75
– Now estimate 2 year survival
Reduced sample estimate = 60/100 = 0.60
Estimate is based on cohort I only
Loss of information
Reduced Sample Estimate (2)
54208#12
Ref: Berkson & Gage (1950) Proc of Mayo ClinicCutler & Ederer (1958) JCDElveback (1958) JASAKaplan & Meier (1958) JASA
 Note that we can express P(2 yr survival asP(2 yrs) = P(2 yrs survivalsurvived 1st yr)
P(1st yr survival) = (60/80) (155/200) = (0.75) (0.775) = 0.58
• This estimate used all the available data
Actuarial EstimateActuarial Estimate (1)
54208#13
• In general, divide the followup time into a series of intervals
Actuarial EstimateActuarial Estimate (2)
• Let pi = prob of surviving Ii given patient alive at beginning of Ii (i.e. survived through Ii 1)
• Then prob of surviving through tk, P(tk)
iP PP P P P(t tP(Sk
1ik321kk
...))
I1 I2 I3 I4 I5
t0 t1 t2 t3 t4 t5
54208#14
 Define the following
ni= number of subjects alive at beginning of Ii (i.e. at ti1)di= number of deaths during interval Ii
li = number of losses during interval Ii (either administrative or lost to followup)
 We know only that di deaths and losses occurred inInterval Ii
Actuarial EstimateActuarial Estimate (3)
Ii
ti1 ti
i
54208#15
a. All deaths precede all losses
b. All losses precede all deaths
c. Deaths and losses uniform, (1/2 deaths before 1/2 losses)
Actuarial Estimate/CutlerEderer
 Problem is that P(t) is a function of the interval choice.
 For some applications, we have no choice, but if weknow the exact date of deaths and losses, theKaplan‑Meier method is preferred.
ii
iii
n
dn
iP
2/
2/P * i
ii
iii
n
dn
i
ii
n
dn
iP
Estimation of PEstimation of Pii
54208#16
Actuarial Lifetime Method (1)Actuarial Lifetime Method (1)
• Used when exact times of death are not known
• Vital status is known at the end of an interval period (e.g. 6 months or 1 year)
• Assume losses uniform over the interval
54208#17
Lifetable
At Number Number Adjusted Prop Prop. Surv. Up toInterval Risk Died Lost No. At Risk Surviving End of Interval
(ni) (di) (li)
01 50 9 0 50 41/500.82 0.82
12 41 6 1 411/2=40.5 34.5/40.5=0.852 0.852 x 0.82=0.699
23 34 2 4 344/2=32 30/32=0.937 0.937 x 0.699=0.655
34 28 1 5 285/2=25.5 24.5/25.5=0.961 0.961 x 0.655=0.629
45 22 2 3 223/2=20.5 18.5/20.5=0.902 0.902 x 0.629=0.567
2i
iinn
ip
ipII
Actuarial Lifetime Method (2)Actuarial Lifetime Method (2)
54208#18
Actuarial Survival CurveActuarial Survival Curve100
80
60
40
20
0
X ___
X___X___
X___X___
X___
1 2 3 4 5
54208#19
KaplanMeier Estimate (1)KaplanMeier Estimate (1)((JASAJASA, 1958), 1958)
• Assumptions
1. "Exact" time of event is known
Failure = uncensored event
Loss = censored event
2. For a "tie", failure always before loss
3. Divide followup time into intervals such that
a. Each event defines left side of an interval
b. No interval has both deaths & losses
54208#20
KaplanMeier Estimate (2)KaplanMeier Estimate (2)((JASAJASA, 1958), 1958)
• Then
ni = # at risk just prior to death at ti
• Note if interval contains only losses, Pi = 1.0
• Because of this, we may combine intervals with only losses with the previous interval containing
only deaths, for convenience
X———o—o—o——
i
ii
i n
dnP
54208#21
Estimate of S(t) or P(t)Estimate of S(t) or P(t)
Suppose that for N patients, there are K distinct failure (death) times. The KaplanMeier estimate of survival curves becomes P(t)=P (Survival t)
KM or Product Limit Estimate
ti t i = 1,2,…,k
where ni = ni1  li1  di1
li1 = # censored events since death at ti1
di1 = # deaths at ti1
i
ii
n
dni
tP)(
)(ˆ
54208#22
Estimate of S(t) or P(t)Estimate of S(t) or P(t)
• Variance of P(t)
Greenwood’s Formula
tt
dnn
d
itPtPV
i
iii
i
)( )](ˆ[)](ˆ[ 2
54208#23
Example (see Table 142 in FFD)
Suppose we follow 20 patients and observe the event time, either failure (death) or censored (+), as
[0.5, 0.6+), [1.5, 1.5, 2.0+), [3.0, 3.5+, 4.0+), [4.8],
[6.2, 8.5+, 9.0+), [10.5, 12.0+ (7 pts)]
There are 6 distinct failure or death times
0.5, 1.5, 3.0, 4.8, 6.2, 10.5
KM Estimate (1)
54208#24
1. failure at t1 = 0.5 [.5, 1.5)n1 = 20d1 = 1l1 = 1 (i.e. 0.6+)
If t [.5, 1.5), p(t) = p1 = 0.95
V [ P(t1) ] = [.95]2 {1/20(19)} = 0.0024
9520
120
n
dnp
1
1 1
1.
KM Estimate (2)KM Estimate (2)
^
^
54208#25
Data [0.5, 0.6+), [1.5, 1.5, 2.0+), 3.0 etc.
2. failure at t2 = 1.5 n2 = n1  d1  1
[1.5, 3.0) = 20  1  1= 18
d2 = 22 = 1 (i.e. 2.0+)
If t [1.5, 3.0), then P(t) = (0.95)(0.89) = 0.84
V [P(t2)] = [0.84]2 { 1/20(19) + 2/18(182) } = 0.0068
89.18
218
2
22
2
n
dnp
KM Estimate (4)KM Estimate (4)
54208#26
Life Table 14.2KaplanMeier Life Table for 20 Subjects
Followed for One Year
Interval Interval Time
Number of death nj dj j
[.5,1.5) 1 .5 20 1 1 0.95 0.95 0.0024
[1.5,3.0) 2 1.5 18 2 1 0.89 0.84 0.0068
[3.0,4.8) 3 3.0 15 1 2 0.93 0.79 0.0089
[4.8,6.2) 4 4.8 12 1 0 0.92 0.72 0.0114
[6.2,10.5) 5 6.2 11 1 2 0.91 0.66 0.0135
[10.5, ) 6 10.5 8 1 7* 0.88 0.58 0.0164
jp̂ )(p̂
jt )](p̂V[
jt
nj : number of subjects alive at the beginning of the jth intervaldj : number of subjects who died during the jth intervalj : number of subjects who were lost or censored during the jth interval : estimate for pj, the probability of surviving the jth interval
given that the subject has survived the previous intervals : estimated survival curve : variance of
* Censored due to termination of study
jp̂
(t)pj
ˆ
(t)]pV[j
ˆ )(ˆ tP
54208#27
0
0.5
0.6
0.7
0.8
0.9
1.0
2 4 6 8 10 12
*
* *
*
*
*
*
Survival Time t (Months)
Est
imat
ed S
urvi
val C
ure
[P(t
)]Survival Curve
KaplanMeier Estimate
^
o
oo o
o o
oooo
ooo
54208#28
Comparison of Comparison of Two Survival CurvesTwo Survival Curves
• Assume that we now have a treatment group and a control group and we wish to make a comparison between their survival experience
• 20 patients in each group
(all patients censored at 12 months)
Control 0.5, 0.6+, 1.5, 1.5, 2.0+, 3.0, 3.5+, 4.0+,
4.8, 6.2, 8.5+, 9.0+, 10.5, 12+'s
Trt 1.0, 1.6+, 2.4+, 4.2+, 4.5, 5.8+, 7.0+, 11.0+, 12+'S
54208#29
1. t1 = 1.0 n1 = 20 p1 = 20  1 = 0.95
d1 = 1 20
l1 = 3
p(t)= .95
2. t2 = 4.5 n2 = 20  1  3 p2 = 16  1 =0 .94
= 16 16
d2 = 1
0024.0)]t(P̂V[1
89.0)94.0)(95.0()t(P̂ 0530.)](ˆ 2
tPV[
KaplanMeier Estimate for KaplanMeier Estimate for TreatmentTreatment
^
54208#30
KaplanMeier EstimateKaplanMeier Estimate
0
0.5
0.6
0.7
0.8
0.9
1.0
2 4 6 8 10 12
*
* *
*
*
*
Survival Time t (Months)
Est
imat
ed S
urvi
val C
ure
[P(t
)]^
TRT
CONTROL
o
* *
o o
ooooo
oo o
54208#31
Comparison of Comparison of Two Survival CurvesTwo Survival Curves
• Comparison of Point Estimates
– Suppose at some time t* we want to compare PC(t*) for the control and PT(t*) for treatment
– The statistic
has approximately, a normal distribution under H0
– Example:
2/1]*)(ˆ*)(ˆ[
*)(ˆ*)(ˆ
tPVtPV
tPtPZ
CT
CT
)(ˆ)(ˆ 6P vs6P CompareCT
3211290
170
0114000530
7208921
..
.
]..[
../
Z
54208#32
• Comparison of Overall Survival Curve
H0: Pc(t) = PT(t)
A. MantelHaenszel Test
Ref: Mantel & Haenszel (1959) J Natl Cancer InstMantel (1966) Cancer Chemotherapy Reports
 Mantel and Haenszel (1959) showed that a series of 2 x 2
tables could be combined into a summary statistic(Note also: Cochran (1954) Biometrics)
 Mantel (1966) applied this procedure to the comparison of
two survival curves
 Basic idea is to form a 2 x 2 table at each distinct deathtime, determining the number in each group who were
atrisk and number who died
54208#33
Suppose we have K distinct times for a death occurring
ti i = 1,2, .., K. For each death time,
Died At Risk at ti Alive (prior to ti)
Treatment ai bi ai + bi
Control ci di ci + di
ai + ci bi + di Ni
• Consider ai, the observed number ofdeaths in the TRT group, under H0
Comparison of Comparison of Two Survival Curves (1)Two Survival Curves (1)
54208#34
E(ai) = (ai + bi)(ai + ci)/Ni
MantelHaenszel Statistic
1)(NN
)d(c)b(a)db)(c(a )V(a
i
2
i
iiiiiiii
i
21iii χ~)V(a Σ/)]}a(E[a{MH 2
N(0,1) ~ MHZ
Comparison of Comparison of Two Survival Curves(2)Two Survival Curves(2)
54208#35
Table 14.3Table 14.3Comparison of Survival Data for a Control Group and an Comparison of Survival Data for a Control Group and an
Intervention Group Using the MantelHaenszel ProcedureIntervention Group Using the MantelHaenszel ProcedureRank Event Intervention Control Total
Times
j tj aj + bj aj j cj + dj cj j aj + cj bj + dj
1 0.5 200 0 201 1 1 39
2 1.0 201 0 180 0 1 37
3 1.5 190 2 182 1 2 35
4 3.0 170 1 151 2 1 31
5 4.5 161 0 120 0 1 27
6 4.8 150 1 121 0 1 26
7 6.2 140 1 111 2 1 24
8 10.5 130 1 8 1 1 20
aj + bj = number of subjects at risk in the intervention group prior to the death at time t j
cj + cj = number of subjects at risk in the control group prior to the death at time t j aj = number of subjects in the intervention group who died at time t j
cj = number of subjects in the control group who died at time t j
j = number of subjects who were lost or censored between time t j and time tj+1
aj + cj = number of subjects in both groups who died at time tj
bj + dj = number of subjects in both groups who are at risk minus the number who died at time t j
54208#36
• Operationally
1. Rank event times for both groups combined2. For each failure, form the 2 x 2 table
a. Number at risk (ai + bi, ci + di)b. Number of deaths (ai, ci)c. Losses (lTi, lCi)
• Example (See table 143 FFD)  Use previous data set
Trt: 1.0, 1.6+, 2.4+, 4.2+, 4.5, 5.8+, 7.0+, 11.0+, 12.0+'s
Control: 0.5, 0.6+, 1.5, 1.5, 2.0+, 3.0, 3.5+, 4.0+, 4.8, 6.2, 8.5+, 9.0+, 10.5, 12.0+'s
MantelHaenszel TestMantelHaenszel Test
54208#37
1. Ranked Failure Times  Both groups combined
0.5, 1.0, 1.5, 3.0, 4.5, 4.8, 6.2, 10.5 C T C C T C C C
8 distinct times for death (k = 8)
2. At t1 = 0.5 (k = 1) [.5, .6+, 1.0)
T: a1 + b1 = 20 a1 = 0 lT1 = 0 c1 + d1 = 20 c1 = 1 lC1 = 1 1 loss @ .6+
D A R
T 0 20 20
C 1 19 20
1 39 40
E(a1)= 1•20/40 = 0.5
V(a1) = 1•39 • 20 • 20 402 •39
54208#38
3. At t2 = 1.0 (k = 2) [1.0, 1.5)
T: a2 + b2 = (a1 + b1)  a1  lT1 a2 = 1.0= 20  0  0
= 20 lT2 = 0
C. c2 + d2 = (c1 + d1)  c1  lC1 c2 = 0 = 20  1  1
= 18 lC2 = 0
so
D A R
T 1 19 20
C 0 18 18
1 37 38
E(a2)= 1•20 38V(a2) = 1•37 • 20 • 18
382 •37
54208#39
Eight 2x2 Tables Corresponding to the Event TimesEight 2x2 Tables Corresponding to the Event TimesUsed in the MantelHaenszel Statistic in Survival Used in the MantelHaenszel Statistic in Survival
Comparison of Treatment (T) and Control (C) GroupsComparison of Treatment (T) and Control (C) Groups
1. (0.5 mo.)* D† A‡ R§ 5. (4.5 mo.)* D A RT 0 20 20 T 1 15 16C 1 19 20 C 0 12 12
1 39 40 1 27 28
2. (1.0 mo) D A R 6. (4.8 mo.) D A RT 1 19 20 T 0 15 15C 0 18 18 C 1 11 12
1 37 38 1 26 27
3. (1.5 mo.) D A R 7. (6.2 mo.) D A RT 0 19 19 T 0 14 14C 2 16 18 C 1 10 11
2 35 37 1 24 25
4. (3.0 mo.) D A R 8. (10.5 mo.) D A RT 0 17 17 T 0 13 13C 1 14 15 C 1 7 8
1 31 32 1 20 21
* Number in parentheses indicates time, tj, of a death in either group† Number of subjects who died at time tj
‡ Number of subjects who are alive between time tj and time tj+1
§ Number of subjects who were at risk before the death at time tj R=D+A)
54208#40
Compute MH StatisticsCompute MH Statistics
Recall K = 1 K = 2 K = 3t1 = 0.5 t2 = 1.0 t3 = 1.5
D A0 20 201 19 201 39 40
D A1 19 200 18 181 37 38
D A0 19 192 16 182 35 37
a. ai = 2 (only two treatment deaths)
b. E(ai ) = 20(1)/40 + 20(1)/38 + 19(2)/37 + . . .= 4.89
c. V(ai) =
= 2.22d. MH = (2  4.89)2/2.22 = 3.76 or ZMH =
K
i 1
K
i 1
K
i 1
...)(
))()((
)(
))()((
3738
1820371
3940
202039122
941.MH
54208#41
B. Gehan Test (Wilcoxon)Ref: Gehan, Biometrika (1965)
Mantel, Biometrics (1966)Gehan (1965) first proposed a modified Wilcoxon rankstatistic for survival data with censoring. Mantel (1967) showed asimpler computational version of Gehan’s proposed test.
1. Combine all observations XT’s and XC’s into a single sample Y1, Y2, . . ., YNC + NT
2. Define Uij wherei = 1, NC + NT j = 1, NC + NT
1 Yi < Yj and death at Yi Uij = 1 Yi > Yj and death at Yj
0 elsewhere
3. Define Ui
i = 1, … , NC + NT ji
ij
NTNC
1ji
U U
54208#42
Note:
Ui = {number of observed times definitely less than i}{number of observed times definitely greater}
4. Define W = Ui (controls)
5. V[W] = NCNT
Variance due to Mantel
6.
• Example (Table 145 FFD)Using previous data set, rank all observations
Gehan TestGehan Test
NTNC
1i
N(NN(N
U
TCTC
2
i
)) 1
~V(W)
WZ
or
(0,1) N ~V(W)
WZ
2
1
2
2
G
G
54208#43
The Gehan Statistics, Gi involves the scores Ui and is defined as
G = W2/V(W)
where W = Ui (Uis in control group only)
and
iC NN
ii
iCiC
iC UNNNN
NNWV
1
2
1)(
))(()(
54208#44
Example of Gehan Statistics Scores UExample of Gehan Statistics Scores Uii for for
Intervention and Control (C) GroupsIntervention and Control (C) GroupsObservation Ranked Definitely Definitely = Ui
i Observed Time Group Less More
1 0.5 C 0 39 392 (0.6)* C 1 0 13 1.0 I 1 37 364 1.5 C 2 35 335 1.5 C 2 35 336 (1.6) I 4 0 47 (2.0) C 4 0 48 (2.4) I 4 0 49 3.0 C 4 31 27
10 (3.5) C 5 0 511 (4.0) C 5 0 512 (4.2) I 5 0 513 4.5 I 5 27 2214 4.8 C 6 26 2015 (5.8) I 7 0 716 6.2 C 7 24 1717 (7.0) I 8 0 818 (8.5) C 8 0 819 (9.0) C 8 0 820 10.5 C 8 20 1221 (11.0) I 9 0 9
2240 (12.0) 12I, 7C 9 0 9
*Censored observations
54208#45
Thus W = (39) + (1) + (36) + (33) + (4) + . . . .
= 87
and V[W] = (20)(20) {(39)2 +12 + (36)2 + . . . } (40)(39)
= 2314.35
so
• Note MH and Gehan not equal
811352314
87.
.
GZ
Gehan TestGehan Test
54208#46
Cox Proportional Hazards ModelCox Proportional Hazards ModelRef: Cox (1972) Journal of the Royal Statistical Association
• Recall simple exponential
S(t) = et
• More complicated
If (s) = , get simple model
• Adjust for covariates
• Cox PHM
(t,x) =0(t) ex
})(exp{)( dsstS t 0
}),(exp{),( dsxsxtS t 0
54208#47
So
S(t1,X) =
=
=
• Estimate regression coefficients (nonlinear estimation) , SE()
• Examplex1 = 1 Trt
2 Controlx2 = Covariate 1
indicator of treatment effect, adjusted for x2, x3 , . . .
• If no covariates, except for treatment group (x1),PHM = logrank
}exp ds(s)e{ xB'
0t
0
xB'00 e(s)ds ]e t[
xB'e
0(t)][S
,ˆ1
Cox Proportional Hazards ModelCox Proportional Hazards Model
54208#48
Homework ProblemHomework Problem
PatientNumber
Treatmenta Length ofObservationb Statusc
1 D 6 D2 P 16 A3 P 2 D4 D 1 D5 D 1 D6 P 4 A7 D 16 A8 D 3 A9 P 16 A
10 D 8 D11 P 16 A12 P 2 D13 D 4 D14 P 16 A15 P 5 A16 D 1 D17 D 10 A18 P 2 A19 P 2 A20 D 10 D21 P 16 A22 D 1 A23 P 16 A24 D 15 A
1. KaplanMeier2. GehanWilcoxon3. MantelHaenszel
a D = drug; P = placebob In weeksc A = alive; D = dead
Source: P.B. Gregory (1974)
54208#49
Survival Analysis SummarySurvival Analysis Summary• Time to event methodology very useful in
multiple settings• Can estimate time to event probabilities or
survival curves• Methods can compare survival curves
– Can stratify for subgroups– Can adjust for baseline covariates using
regression model
• Need to plan for this in sample size estimation & overall design