1. Chapter 12
Survival Analysis

2. 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

3. 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

4. Beta-Blocker Heart Attack Trial
LIFE-TABLE CUMULATIVE MORTALITY CURVE

5. 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

6. Staggered Entry
T years 1
T years 2
Subject
T years 3
T years 4 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

7. • * * Subject o Administrative 1 Censoring 2 Failure 3 Censoring
Loss to Follow-up • * 4 T 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

8. * • * Subject Administrative Censoring 1 o 2 Failure 3 Censoring
Loss to Follow-up 4 * T
Follow-up Time (T years)

9. Clinical Trial with Common Termination Date
Subject 1 o 2 * 3 • 4 • o 5 • * • • 6 7 • • 8 • * 9 • o o 10 • o * 11 • o o T 2T
Follow-up Time (T years)
Trial
Terminated

10. Reduced Sample Estimate (1)
Years of Cohort
Follow-Up Patients I II Total
Entered 1
Died
Entered 2
Died 20
Survived 60

11. 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

12. 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

13. Actuarial Estimate (2)
In general, divide the follow-up time into a series of intervals
I I I I I5
t t t t t t5
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)

14. Actuarial Estimate (3) - Define the followingIi
ti ti
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

15. 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.

16. 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

17. Actuarial Lifetime Method (2)
Lifetable
At Number Number Adjusted Prop Prop. Surv. Up to
Interval Risk Died Lost No. At Risk Surviving End of Interval
(ni) (di) (li) /
/2= /40.5= x 0.82=0.699
/2=32 30/32= x 0.699=0.655
/2= /25.5= x 0.655=0.629
/2= /20.5= x 0.629=0.567

18. Actuarial Survival Curve
100 80 60 40 20
X ___
X___
X___
X___
X___
X___

19. 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

20. 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——

21. 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 = # censored events since death at ti-1
di-1 = # deaths at ti-1

22. Estimate of S(t) or P(t)
Variance of P(t)
Greenwood’s Formula

23. 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, (7 pts)]
There are 6 distinct failure or death times
0.5, 1.5, 3.0, 4.8, 6.2, 10.5

24. KM Estimate (2) 1. failure at t1 = 0.5 [.5, 1.5) n1 = 20 d1 = 1
l1 = 1 (i.e )
If t [.5, 1.5), p(t) = p1 = 0.95
V [ P(t1) ] = [.95]2 {1/20(19)} = ^ ^

25. KM Estimate (4) 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) =
= 18
d2 = 2
l2 = 1 (i.e )
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) } =

26. Kaplan-Meier Life Table for 20 Subjects
Followed for One Year
Interval Interval Time
Number of death nj dj lj
[.5,1.5)
[1.5,3.0)
[3.0,4.8)
[4.8,6.2)
[6.2,10.5)
[10.5, ) *
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

27. Kaplan-Meier Estimate
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 2 4 6 8 10 12
Survival Time t (Months)

28. 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)
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

29. Kaplan-Meier Estimate for Treatment
1. t1 = 1.0 n1 = p1 = = 0.95
d1 =
l1 = 3
p(t) = .95
2. t2 = 4.5 n2 = p2 = =0 .94 =
d2 = 1 ^

30. 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 2 4 6 8 10 12
Survival Time t (Months)

31. 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:

32. Comparison of Overall Survival Curve H0: Pc(t) = PT(t)
A. Mantel-Haenszel Test
Ref: Mantel & Haenszel (1959) J Natl Cancer Inst
Mantel (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 death
time, determining the number in each group who were at
risk and number who died

33. Comparison of Two Survival Curves (1)
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 of
deaths in the TRT group, under H0

34. Comparison of Two Survival Curves(2)
E(ai) = (ai + bi)(ai + ci)/Ni
C Mantel-Haenszel Statistic

35. 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
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

36. 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

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 =
D A R T C
E(a1)= 1•20/40 = 0.5
V(a1) = 1•39 • 20 • 20
402 •39

38. 3. At t2 = 1.0 (k = 2) [1.0, 1.5)
T: a2 + b2 = (a1 + b1) - a1 - lT1 a2 = 1.0 =
= lT2 = 0
C. c2 + d2 = (c1 + d1) - c1 - lC1 c2 = 0 =
= lC2 = 0 so
D A R T C
E(a2)= 1•20 38
V(a2) = 1•37 • 20 • 18
382 •37

39. * Number in parentheses indicates time, tj, of a death in either group
Eight 2x2 Tables Corresponding to the Event Times Used 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 R
T T
C C
2. (1.0 mo) D A R 6. (4.8 mo.) D A R
T T
C C
3. (1.5 mo.) D A R 7. (6.2 mo.) D A R
T T
C C
4. (3.0 mo.) D A R 8. (10.5 mo.) D A R
T T
C C
* 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)

40. Compute MH Statistics Recall K = 1 K = 2 K = 3
t1 = 0.5 t2 = 1.0 t3 = 1.5
D A
D A
D A
a. ai = 2 (only two treatment deaths)
b. E(ai ) = 20(1)/ (1)/ (2)/
= 4.89
c. V(ai) =
= 2.22
d. MH = ( )2/2.22 = 3.76
or ZMH =

41. 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 Uij where i = 1, NC + NT j = 1, NC + NT
-1 Yi < Yj and death at Yi
Uij = Yi > Yj and death at Yj
0 elsewhere
3. Define Ui
i = 1, … , NC + NT

42. 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

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

44. Example of Gehan Statistics Scores Ui for Intervention and Control (C) Groups
Observation Ranked Definitely Definitely = Ui
i Observed Time Group Less More C
2 (0.6)* C I C C
6 (1.6) I
7 (2.0) C
8 (2.4) I C
10 (3.5) C
11 (4.0) C
12 (4.2) I I C
15 (5.8) I C
17 (7.0) I
18 (8.5) C
19 (9.0) C C
21 (11.0) I
(12.0) 12I, 7C
*Censored observations

45. Gehan Test Thus W = (-39) + (1) + (-36) + (-33) + (4) + . . . . = -87
and V[W] = (20)(20) {(-39) (-36) }
(40)(39) = so
Note MH and Gehan not equal

46. 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

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

48. 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