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Chapter 12

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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 death

2.Non-fatal event or death,

whichever comes first

death or hospitalization

death or MI

death or tumor recurrence

Survival Rates at Yearly Intervals

- 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

Beta-Blocker Heart Attack Trial

LIFE-TABLE CUMULATIVE MORTALITY CURVE

Survival Analysis

Discuss

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

Staggered Entry

T years

1

T years

2

Subject

T years

3

T years

4

0

T

2T

Time Since Start of Trial (T years)

- Each patient has T years of follow-up
- Time for follow-up taking place may be different for each patient

Subject

o

Administrative

Censoring

1

Failure

2

*

•

Censoring

Loss to Follow-up

3

*

4

T

0

2T

Time Since Start of Trial (T years)

- Failure time is time from entry until the time of the event
- Censoring means vital status of patient is not known beyond that point

Subject

Administrative

Censoring

o

1

Failure

2

*

•

3

Censoring

Loss to Follow-up

4

*

T

0

Follow-up Time (T years)

Clinical Trial with Common Termination Date

Subject

o

1

2

*

•

3

•

4

o

•

5

*

•

•

6

•

•

7

•

*

8

•

9

o

o

•

10

o

*

•

11

o

o

0

T

2T

Trial

Terminated

Follow-up Time (T years)

Reduced Sample Estimate (1)

Years of Cohort

Follow-UpPatientsIIITotal

Entered100100200

1

Died202545

Entered8075155

2

Died20

Survived60

Reduced Sample Estimate (2)

- 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

Actuarial Estimate (1)

- Ref:Berkson & Gage (1950) Proc of Mayo Clinic
- Cutler & Ederer (1958) JCD
- Elveback (1958) JASA
- Kaplan & Meier (1958) JASA
- - Note that we can express P(2 yr survival) as
- P(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

I1 I2 I3 I4 I5

t0 t1 t2 t3 t4 t5

Actuarial Estimate (2)

- In general, divide the follow-up time into a series of intervals

- 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)

Ii

ti-1 ti

Actuarial Estimate (3)

-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 in

Interval Ii

Estimation of Pi

- a.All deaths precede all losses
- b.All losses precede all deaths
- 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 we
- know the exact date of deaths and losses, the
- Kaplan‑Meier method is preferred.

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

Actuarial Lifetime Method (2)

Lifetable

At NumberNumber Adjusted Prop Prop. Surv. Up to

Interval Risk Died Lost No. At Risk Surviving End of Interval

(ni) (di) (li)

0-150905041/50-0.820.82

1-2416141-1/2=40.534.5/40.5=0.8520.852 x 0.82=0.699

2-3342434-4/2=3230/32=0.9370.937 x 0.699=0.655

3-4281528-5/2=25.524.5/25.5=0.9610.961 x 0.655=0.629

4-5222322-3/2=20.518.5/20.5=0.9020.902 x 0.629=0.567

Actuarial 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——

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

titi = 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

Estimate of S(t) or P(t)

- Variance of P(t)
Greenwood’s Formula

KM Estimate (1)

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 (2)

1.failure att1 = 0.5 [.5, 1.5)

n1 = 20

d1 = 1

l1 = 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

^

^

KM Estimate (4)

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

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

[1.5, 3.0)= 20 - 1 - 1

= 18

d2= 2

l2= 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

Life Table

Kaplan-Meier Life Table for 20 Subjects

Followed for One Year

IntervalInterval Time

Number of death njdjlj

[.5,1.5)1.520110.950.950.0024

[1.5,3.0)21.518210.890.840.0068

[3.0,4.8)33.015120.930.790.0089

[4.8,6.2)44.812100.920.720.0114

[6.2,10.5)56.211120.910.660.0135

[10.5, )610.581 7*0.880.580.0164

nj:number of subjects alive at the beginning of the jth interval

dj: number of subjects who died during the jth interval

lj : 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

Survival Curve

Kaplan-Meier Estimate

1.0

o

*

0.9

^

*

o

*

0.8

o

o

*

*

Estimated Survival Cure [P(t)]

0.7

o

o

*

o

o

0.6

o

o

*

o

o

o

0.5

0

4

6

8

10

12

2

Survival Time t (Months)

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

Trt1.0, 1.6+, 2.4+, 4.2+, 4.5, 5.8+, 7.0+, 11.0+, 12+'S

Kaplan-Meier Estimate for Treatment

1.t1 = 1.0n1= 20 p1 = 20 - 1 = 0.95

d1= 1 20

l1= 3

p(t) = .95

2.t2 = 4.5n2= 20 - 1 - 3p2 = 16 - 1 =0 .94

= 16 16

d2 = 1

^

Kaplan-Meier Estimate

1.0

o

*

TRT

0.9

*

^

*

o

*

0.8

o

o

*

*

Estimated Survival Cure [P(t)]

0.7

CONTROL

o

o

*

0.6

o

*

o

o

o

o

0.5

0

4

6

8

10

12

2

Survival Time t (Months)

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:

- Comparison of Overall Survival Curve

- -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 death
- time, determining the number in each group who were at
- risk and number who died

Comparison of Two Survival Curves (1)

- Suppose we have K distinct times for a death occurring
- tii = 1,2, .., K. For each death time,
- Died At Risk
- at ti Alive (prior to ti)
- Treatmentaibiai + bi
- Controlcidici + di
- ai + cibi + di Ni
- Consider ai, the observed number of
- deaths in the TRT group, under H0

Comparison of Two Survival Curves(2)

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

CMantel-Haenszel Statistic

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

RankEvent Intervention Control Total

Times

jtjaj + bjajljcj + djcjlj aj + cjbj + dj

10.520002011139

21.020101800137

31.519021821235

43.017011512131

54.516101200127

64.815011210126

76.214011112124

810.5130181120

- aj + bj =number of subjects at risk in the intervention group prior to the death at time tj
- cj + cj =number of subjects at risk in the control group prior to the death at time tj
- aj =number of subjects in the intervention group who died at time tj
- cj =number of subjects in the control group who died at time tj
- lj =number of subjects who were lost or censored between time tj 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 tj

Mantel-Haenszel Test

- Operationally
- 1.Rank event times for both groups combined
- 2.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

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+

DAR

T02020

C11920

13940

E(a1)= 1•20/40 = 0.5

V(a1) = 1•39 • 20 • 20

402 •39

E(a2)= 1•20

38

V(a2) = 1•37 • 20 • 18

382 •37

3.At t2 = 1.0(k = 2) [1.0, 1.5)

T: a2 + b2 = (a1 + b1) - a1 - lT1 a2 = 1.0

= 20 - 0 - 0

= 20lT2 = 0

C. c2 + d2 = (c1 + d1) - c1 - lC1 c2 = 0

= 20 - 1 - 1

= 18 lC2 = 0

so

DAR

T11920

C01818

13738

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.)*DAR

T02020T11516

C11920C01212

1394012728

2.(1.0 mo)DAR6.(4.8 mo.)DAR

T11920T01515

C01818C11112

1373812627

3.(1.5 mo.)DAR7.(6.2 mo.)DAR

T01919T01414

C21618C11011

2353712425

4.(3.0 mo.)DAR8.(10.5 mo.)DAR

T01717T01313

C11415C17 8

1313212021

*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

RecallK = 1K = 2K = 3

t1 = 0.5 t2 = 1.0 t3 = 1.5

DA

02020

11920

13940

DA

11920

01818

13738

DA

01919

21618

23537

a. ai = 2 (only two treatment deaths)

b.E(ai )= 20(1)/40 + 20(1)/38 + 19(2)/37 + . . .

= 4.89

c. V(ai)=

=2.22

d.MH= (2 - 4.89)2/2.22 = 3.76

or ZMH=

B.Gehan Test (Wilcoxon)

Ref: Gehan, Biometrika (1965)

Mantel, Biometrics (1966)

Gehan (1965) first proposed a modified Wilcoxon rank

statistic for survival data with censoring. Mantel (1967) showed a

simpler 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 Uijwherei = 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

Gehan Test

- Note:
- Ui = {number of observed times definitely less than i}
- {number of observed times definitely greater}
- 4.Define W = S Ui (controls)
- 5.V[W] = NCNT
- Variance due to Mantel
- 6.
- Example (Table 14-5 FFD)
- Using previous data set, rank all observations

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

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

Observation RankedDefinitely Definitely =Ui

iObserved TimeGroup Less More

10.5C039-39

2(0.6)*C101

31.0I137-36

41.5C235-33

51.5C235-33

6(1.6)I404

7(2.0)C404

8(2.4)I404

93.0C431-27

10(3.5)C505

11(4.0)C505

12(4.2)I505

134.5I527-22

144.8C626-20

15(5.8)I707

166.2C724-17

17(7.0)I808

18(8.5)C808

19(9.0)C808

2010.5C820-12

21(11.0)I909

22-40(12.0)12I, 7C909

*Censored observations

Gehan Test

- ThusW= (-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

Cox Proportional Hazards Model

Ref:Cox (1972) Journal of the Royal Statistical Association

- Recall simple exponential
S(t) = e-lt

- More complicated
If l(s) = l, get simple model

- Adjust for covariates
- Cox PHM
l(t,x) =l0(t) ebx

Cox Proportional Hazards Model

- So
- S(t1,X) =
- =
- =
- Estimate regression coefficients (non-linear estimation) b, SE(b)
- Example
- x1 =1Trt
- 2Control
- x2 =Covariate 1
- indicator of treatment effect, adjusted for x2, x3 , . . .
- If no covariates, except for treatment group (x1),
- PHM = logrank

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