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# Survival Analysis: Logrank Test - Stanford University lutian/coursepdf/survweek3.pdf · PDF fileSurvival Analysis: Logrank Test Lu Tian and Richard Olshen Stanford University 1

Jul 11, 2018

## Documents

phungtu

• Survival Analysis: Logrank Test

Lu Tian and Richard Olshen

Stanford University

1

• Two-sample Comparison

Objective: to compare survival functions from two groups.

Requirement: nonparametric, deal with right censoring.

2

• Two-sample comparisons

KM estimators: S1() and S0()

Possible test statistics:

sup[0, ] |S1(t) S0(t)| 0|S1(t) S0(t)|dt

0{S1(t) S0(t)}dt

The null distributions are complex.

3

• Logrank Test

The most popular method is the logrank test1. Adapted from stratified test for 2 by 2 contingency table (Mantel, 1996)

2. Has a nice relationship with the proportional hazards model

3. Targets on the hazard function (not survival function).

4

• Logrank test

1 < 2 < < K are distinct failure times

Yi(j) = # persons in group i at risk at j

Y (j) = Y0(j) + Y1(j), the total # of subjects at risk at j

dij = # of failures in group i at j

dj = d0j + d1j total # of failures at j

5

• Two by two table

At the time jobserved to at risk

fail at j at j

group 0 d0j Y0(j) d0j Y0(j)

group 1 d1j Y1(j) d1j Y1(j)

dj Y (j) dj Y (j)

6

• Logrank test

Under the null hypothesis H0 : S1(t) = S0(t), 0 < t < , d1j has the hypergeometric distributionconditional on the margins {Y0(j), Y1(j), dj , Y (j) dj}

pr(d1j = d) =

djd

Y (j) djY1(j) d

/ Y (j)Y1(j)

The hypergeometric distribution is a discrete probability distribution that describes the probability of d1 successes

in Y1 draws without replacement from a finite population of size Y containing exactly d successes. This is in

contrast to the binomial distribution, which describes the probability of d1 successes in Y1 draws with

replacement.

7

• Logrank test

E(d1j |marginals) =(

Y1(j)Y (j)

)dj

Var(d1j |marginals) = Y (j)Y1(j)Y (j)1 Y1(j)(

djY (j)

)(1 djY (j)

)=

Y0(j)Y1(j)dj{Y (j)dj}Y (j)2{Y (j)1}

8

• Logrank Test

Oj = d1j : observed number of failures

Ej = dj Y1(j)Y (j) : expected number of failures

Vj = Y0(j)Y1(j)dj(Y (j)dj)Y (j)2(Y (j)1) : variance of the observed number of failures

The logrank test statistics

Z =

kj=1(Oj Ej)k

j=1 Vj

N(0, 1) under H0

9

• Example

data:Group 0 : 3.1, 6.8+, 9, 9, 11.3+, 16.2

Group 1 : 8.7, 9, 10.1+, 12.1+, 18.7, 23.1+

k = 5 and 1, . . . , 5 = 3.1, 8.7, 9, 16.2, 18.7

10

• Example

1 = 3.1

1 5 6

0 6 6

1 11 12

2 = 8.7

0 4 4

1 5 6

1 9 10

3 = 9

2 2 4

1 4 5

3 6 9

4 = 16.2

1 0 1

0 2 2

1 2 3

5 = 18.7

0 0 0

1 1 2

1 1 2

Oj = 0 1 1 0 1

Ej = 1/2 6/10 5/9 2/3 1

Oj Ej = 1/2 4/10 4/9 2/3 0

Vj = 1/4 6/25 5/9 2/9 0

Z = .39 (2-sided P = .70)

11

• Logrank test

Symmetric in two groups

Only rank matters

k two by two tables are treated as independent.

If the number of subjects at risk becomes zero in one group, the additional two by two tables dontcontribute to the logrank test.

12

• Logrank test

The power of the logrank test depends on the number of observed failures rather than the sample sizes

Logrank test is most powerful for detecting the alternatives

H1 : S1(t) = S0(t)exp() h1(t) = h0(t)e , = 0

The power of logrank test under the alterative h1(t) = h0(t)e is approximately

(||

D0(1 0) 1.96

),

where D is the expected number of failures and 0 is the proportion of patient in groups 0.

13

• Targeting the hazard funciton

kj=1

(Oj Ej) =k

j=1

(d1j dj

Y1(j)

Y (j)

)

=

kj=1

d1j(Y1(j) + Y0(j)) (d0j + d1j)Y1(j)Y (j)

=

kj=1

Y0(j)Y1(j)

Y (j)

(d1j

Y1(j) d0j

Y0(j)

)

=

0

Y0(s)Y1(s)

Y0(s) + Y1(s)d{H1(s) H0(s)

}

14

• Weighted Logrank test

Attach weight wj to the two by two table at time j :

Z =

j wj(Oj Ej)

j w2jVj

15

• Generalized Wilcoxon test

Set wj = Y (j) :

kj=1

wj(Oj Ej) =k

j=1

{d1jY0(j) d0jY1(j)}

=i,j

{I(Ui0 Uj1)j1 I(Uj1 Ui0)i0}

The commonly used Wilcoxon test statistics without censoring isi,j

{I(Ui0 > Uj1) 1/2} =1

2

i,j

{I(Ui0 Uj1) I(Uj1 Ui0)}

Sensitive to the early differences between two hazard functions.

16

• The Generalized Logrank test

In general, the test statistics is in the form of

Zw =

0w(s)d

{H1(s) H0(s)

}w

The choice of the weight affects the power.

One may construct a test based on several different sets of weights, e.g.,

T = max{|Zw1 |, |Zw2 |, , |ZwL |}.

17

• More than two groups

H0 : S0() = S1() = = Sp()

At j

at risk

fail at j at j

Group 0 d0j Y0(j) d0j Y0(j)Group 1 d1j Y1(j) d1j Y1(j)Group 2 d2j Y2(j) d2j Y2(j)

......

......

Group p dpj Yp(j) dpj Yp(j)dj Y (j) dj Y (j)

18

• More than two groups

Oj = (d1j , d2j , , dpj)

Ej = (E1j , E2j , , Epj)

where Eij = djYi(j)Y (j)

.

Vj = (V(j)kl )pp :

where V(j)kl =

djYk(j)Yl(j)(Y (j)dj)

Y (j)2(Y (j)1) if k = ldjYk(j)(Y (j)dj)(Y (j)Yk(j))

Y (j)2(Y (j)1) if k = l

19

• More than two groups

The test statistics:

j

(Oj Ej)

j

Vj

1 j

(Oj Ej)

2punder the null hypothesis.

20

• More than two groups

Trend test

j c

(Oj Ej)j c

Vjc N(0, 1)

under the null hypothesis. How to choose the vector c to increase the power?

21

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