Top Banner

Click here to load reader

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

ReportDownload

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