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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

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Chapter 12

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-Up Patients I II Total

Entered 100 100 200

1

Died 20 25 45

Entered 80 75 155

2

Died 20

Survived 60


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 Number Number Adjusted Prop Prop. Surv. Up to

Interval 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


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

titi = 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 at t1 = 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+), 3.0 etc.

2. failure at t2 = 1.5 n2 = 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

Interval Interval Time

Number of death nj djlj

[.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

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)

    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

    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 = 1.0 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

^


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:


  • 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


  • 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


    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

    Rank Event Intervention Control Total

    Times

    j tj aj + bj ajlj cj + dj cjlj aj + cj bj + dj

    1 0.5 20 0 0 20 1 1 1 39

    2 1.0 20 1 0 18 0 0 1 37

    3 1.5 19 0 2 18 2 1 2 35

    4 3.0 17 0 1 15 1 2 1 31

    5 4.5 16 1 0 12 0 0 1 27

    6 4.8 15 0 1 12 1 0 1 26

    7 6.2 14 0 1 11 1 2 1 24

    8 10.5 13 0 1 8 1 1 20

    • 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 Intervention Group Using the Mantel-Haenszel Procedure

    • 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. Intervention Group Using the Mantel-Haenszel ProcedureRanked 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


    E(a Intervention Group Using the Mantel-Haenszel Procedure2)= 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

    = 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


    Eight 2x2 Tables Corresponding to the Event Times Intervention Group Using the Mantel-Haenszel ProcedureUsed 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 0 20 20 T 1 15 16

    C 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 R

    T 1 19 20 T 0 15 15

    C 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 R

    T 0 19 19 T 0 14 14

    C 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 R

    T 0 17 17 T 0 13 13

    C 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 Intervention Group Using the Mantel-Haenszel Procedure

    Recall K = 1 K = 2 K = 3

    t1 = 0.5 t2 = 1.0 t3 = 1.5

    D A

    0 20 20

    1 19 20

    1 39 40

    D A

    1 19 20

    0 18 18

    1 37 38

    D A

    0 19 19

    2 16 18

    2 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) =

    = 2.22

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

    or ZMH =


    B. Gehan Test (Wilcoxon) Intervention Group Using the Mantel-Haenszel Procedure

    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 Uijwhere 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


    Gehan Test Intervention Group Using the Mantel-Haenszel Procedure

    • 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, G Intervention Group Using the Mantel-Haenszel Procedurei 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 U Intervention Group Using the Mantel-Haenszel Procedurei for Intervention and Control (C) Groups

    Observation Ranked Definitely Definitely = Ui

    i Observed Time Group Less More

    1 0.5 C 0 39 -39

    2 (0.6)* C 1 0 1

    3 1.0 I 1 37 -36

    4 1.5 C 2 35 -33

    5 1.5 C 2 35 -33

    6 (1.6) I 4 0 4

    7 (2.0) C 4 0 4

    8 (2.4) I 4 0 4

    9 3.0 C 4 31 -27

    10 (3.5) C 5 0 5

    11 (4.0) C 5 0 5

    12 (4.2) I 5 0 5

    13 4.5 I 5 27 -22

    14 4.8 C 6 26 -20

    15 (5.8) I 7 0 7

    16 6.2 C 7 24 -17

    17 (7.0) I 8 0 8

    18 (8.5) C 8 0 8

    19 (9.0) C 8 0 8

    20 10.5 C 8 20 -12

    21 (11.0) I 9 0 9

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

    *Censored observations


    Gehan Test Intervention Group Using the Mantel-Haenszel Procedure

    • 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


    Cox Proportional Hazards Model Intervention Group Using the Mantel-Haenszel Procedure

    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 Intervention Group Using the Mantel-Haenszel Procedure

    • So

    • 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


    Survival Analysis Summary Intervention Group Using the Mantel-Haenszel Procedure

    • 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


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