Chapter 12

Survival Analysis

Survival Analysis Terminology

• Concerned about time to some event

• Event is often death

• Event may also be, for example

1. Cause specific death2. Non-fatal event or death,

whichever comes firstdeath or hospitalizationdeath or MI

death or tumor recurrence

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

Beta-Blocker Heart Attack TrialBeta-Blocker Heart Attack Trial

LIFE-TABLE CUMULATIVE MORTALITY CURVELIFE-TABLE CUMULATIVE MORTALITY CURVE

Survival 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

• Each patient has T years of follow-up• Time for follow-up 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

•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 Follow-up

Time Since Start of Trial (T years)

T 2T0

4

3

2

1

Subject

o

*

•

*

T0

4

3

2

1

Subject

AdministrativeCensoring

CensoringLoss to Follow-up

Failure

Follow-up Time (T years)

Clinical Trial with Common Termination DateClinical Trial with Common Termination Date

*

*

*

1

2

3

4

5

6

7

8

9

10

11

0 T 2TFollow-up Time (T years) Trial

Terminated

Subject

*

••• •

•••••

o

o

•

o o

o

o o

•

Years of Cohort

Follow-Up Patients I IITotalEntered 100 100 200

1Died 20 25 45

Entered 80 75 1552

Died 20

Survived 60

Reduced Sample Estimate (1)

– 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)

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 survival|survived 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)

• In general, divide the follow-up 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

1-ik321kk

...))

I1 I2 I3 I4 I5

t0 t1 t2 t3 t4 t5

- Define the following

ni= number of subjects alive at beginning of Ii (i.e. at ti-1)di= number of deaths during interval Ii

li = number of losses during interval Ii (either administrative or lost to follow-up)

- We know only that di deaths and losses occurred inInterval Ii

Actuarial EstimateActuarial Estimate (3)

Ii

ti-1 ti

i

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/Cutler-Ederer

- 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

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

Lifetable

At Number Number Adjusted Prop Prop. Surv. Up toInterval Risk Died Lost No. At Risk Surviving End of Interval

(ni) (di) (li)

0-1 50 9 0 50 41/50-0.82 0.82

1-2 41 6 1 41-1/2=40.5 34.5/40.5=0.852 0.852 x 0.82=0.699

2-3 34 2 4 34-4/2=32 30/32=0.937 0.937 x 0.699=0.655

3-4 28 1 5 28-5/2=25.5 24.5/25.5=0.961 0.961 x 0.655=0.629

4-5 22 2 3 22-3/2=20.5 18.5/20.5=0.902 0.902 x 0.629=0.567

2i

iinn

ip

ipII

Actuarial Lifetime Method (2)

Actuarial Survival CurveActuarial Survival Curve

100

80

60

40

20

0

X ___

X___X___

X___X___

X___

1 2 3 4 5

Kaplan-Meier Estimate (1)(JASA, 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 follow-up time into intervals such that

a. Each event defines left side of an interval

b. No interval has both deaths & losses

Kaplan-Meier Estimate (2)(JASA, 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

Estimate of S(t) or P(t)

Suppose that for N patients, there are K distinct failure (death) times. The Kaplan-Meier estimate of survival curves becomes P(t)=P (Survival t)

K-M or Product Limit Estimate

ti t i = 1,2,…,k

where ni = ni-1 - li-1 - di-1

li-1 = # censored events since death at ti-1

di-1 = # deaths at ti-1

i

ii

n

dni

tP)(

)(ˆ

Estimate of S(t) or P(t)

• Variance of P(t)

Greenwood’s Formula

tt

dnn

d

itPtPV

i

iii

i

)( )](ˆ[)](ˆ[ 2

Example (see Table 14-2 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)

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)

^

^

Data [0.5, 0.6+), [1.5, 1.5, 2.0+), 3.0 etc.

2. failure at t2 = 1.5 n2 = n1 - d1 - l1

[1.5, 3.0) = 20 - 1 - 1= 18

d2 = 2l2 = 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(18-2) } = 0.0068

89.18

218

2

22

2

n

dnp

KM Estimate (4)

Life Table Kaplan-Meier Life Table for 20 Subjects

Followed for One Year

Interval Interval Time

Number of death nj dj lj

[.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 intervallj : 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

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

Kaplan-Meier Estimate^

o

oo o

o o

oooo

ooo

Comparison of Two 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)

Control0.5, 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

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[

Kaplan-Meier Estimate for Treatment

^

Kaplan-Meier EstimateKaplan-Meier 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

Comparison of Two 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

• Comparison of Overall Survival Curve

H0: Pc(t) = PT(t)

A. Mantel-Haenszel 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 a

trisk and number who died

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 Two Survival Curves (1)

E(ai) = (ai + bi)(ai + ci)/Ni

Mantel-Haenszel 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 Two Survival Curves(2)

Comparison of Survival Data for a Control Group and an Intervention Group Using the Mantel-Haenszel Procedure

Rank Event Intervention Control Total

Times

j tj aj + bj aj lj cj + dj cj lj 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

lj = 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

• 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 14-3 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

Mantel-Haenszel TestMantel-Haenszel Test

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

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

Eight 2x2 Tables Corresponding to the Event TimesUsed in the Mantel-Haenszel Statistic in Survival

Comparison 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)

Compute 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

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 where i = 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

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 14-5 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

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)(

))(()(

Example of Gehan Statistics Scores Ui for Intervention and Control (C) Groups

Observation 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

22-40 (12.0) 12I, 7C 9 0 9

*Censored observations

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

Cox Proportional Hazards ModelRef: Cox (1972) Journal of the Royal Statistical Association

• Recall simple exponential

S(t) = e-t

• 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

So

S(t1,X) =

=

=

• Estimate regression coefficients (non-linear 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 Model

Survival 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