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

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,τ ] |S1(t)− S0(t)|∫ τ

0|S1(t)− S0(t)|dt∫ τ

0{S1(t)− S0(t)}dt

• The null distributions are complex.

3

Logrank Test

• The most popular method is the logrank test

1. 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

• τ1 < τ2 < · · · < τK are distinct failure times

• Yi(τj) = # persons in group i at risk at τj

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

• dij = # of failures in group i at τj

• dj = d0j + d1j total # of failures at τj

5

Two by two table

• At the time τj

observed to at risk

fail at τj at τj

group 0 d0j Y0(τj)− d0j Y0(τj)

group 1 d1j Y1(τj)− d1j Y1(τj)

dj Y (τj)− dj Y (τj)

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Logrank test

• Under the null hypothesis H0 : S1(t) = S0(t), 0 < t < ∞, d1j has the hypergeometric distribution

conditional on the margins {Y0(τj), Y1(τj), dj , Y (τj)− dj}

pr(d1j = d) =

dj

d

Y (τj)− dj

Y1(τj)− d

/ Y (τj)

Y1(τj)

• 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.

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Logrank test

E(d1j |marginals) =(

Y1(τj)Y (τj)

)dj

Var(d1j |marginals) =Y (τj)−Y1(τj)

Y (τj)−1 · Y1(τj)(

dj

Y (τj)

)(1− dj

Y (τj)

)=

Y0(τj)Y1(τj)dj{Y (τj)−dj}Y (τj)2{Y (τj)−1}

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Logrank Test

• Oj = d1j : observed number of failures

• Ej = djY1(τj)Y (τj)

: expected number of failures

• Vj =Y0(τj)Y1(τj)dj(Y (τj)−dj)

Y (τj)2(Y (τj)−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 τ1, . . . , τ5 = 3.1, 8.7, 9, 16.2, 18.7

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Example

τ1 = 3.1

1 5 6

0 6 6

1 11 12

τ2 = 8.7

0 4 4

1 5 6

1 9 10

τ3 = 9

2 2 4

1 4 5

3 6 9

τ4 = 16.2

1 0 1

0 2 2

1 2 3

τ5 = 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)

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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 don’t

contribute to the logrank test.

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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(β) ⇔ h1(t) = h0(t)e

β , β = 0

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

Φ(|β|

√Dπ0(1− π0)− 1.96

),

where D is the expected number of failures and π0 is the proportion of patient in groups 0.

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Targeting the hazard funciton

k∑j=1

(Oj − Ej) =k∑

j=1

(d1j − dj

Y1(τj)

Y (τj)

)

=

k∑j=1

d1j(Y1(τj) + Y0(τj))− (d0j + d1j)Y1(τj)

Y (τj)

=

k∑j=1

Y0(τj)Y1(τj)

Y (τj)

(d1j

Y1(τj)− d0j

Y0(τj)

)

=

∫ ∞

0

Y0(s)Y1(s)

Y0(s) + Y1(s)d{H1(s)− H0(s)

}

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Weighted Logrank test

• Attach weight wj to the two by two table at time τj :

Z =

∑j wj(Oj − Ej)√∑

j w2jVj

15

Generalized Wilcoxon test

• Set wj = Y (τj) :

k∑j=1

wj(Oj − Ej) =k∑

j=1

{d1jY0(τj)− d0jY1(τj)}

=∑i,j

{I(Ui0 ≥ Uj1)δj1 − I(Uj1 ≥ Ui0)δi0}

• The commonly used Wilcoxon test statistics without censoring is∑i,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.

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The Generalized Logrank test

• In general, the test statistics is in the form of

Zw =

∫ τ

0w(s)d

{H1(s)− H0(s)

}σw

• 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|}.

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More than two groups

• H0 : S0(·) = S1(·) = · · · = Sp(·)

• At τj

at risk

fail at τj at τj

Group 0 d0j Y0(τj)− d0j Y0(τj)

Group 1 d1j Y1(τj)− d1j Y1(τj)

Group 2 d2j Y2(τj)− d2j Y2(τj)...

......

...

Group p dpj Yp(τj)− dpj Yp(τj)

dj Y (τj)− dj Y (τj)

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More than two groups

Oj = (d1j , d2j , · · · , dpj)′

Ej = (E1j , E2j , · · · , Epj)′

where Eij = djYi(τj)Y (τj)

.

Vj = (V(j)kl )p×p :

where V(j)kl =

−djYk(τj)Yl(τj)(Y (τj)−dj)Y (τj)2(Y (τj)−1) if k = l

djYk(τj)(Y (τj)−dj)(Y (τj)−Yk(τj))Y (τj)2(Y (τj)−1) if k = l

19

More than two groups

The test statistics:

∑j

(Oj −Ej)

′ ∑j

Vj

−1 ∑j

(Oj −Ej)

∼ χ2p

under 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