2. Weighted Log-rank statistics. Y 2 (u) 1. Perspective 1: weighted difference in cumulative hazard functions

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1 2. Weighted Log-rank statistics Weighted Log-rank statistic W = 0 W (u) Y 1(u)Y 2 (u) Y 1 (u) + Y 2 (u) { dn1 (u) Y 1 (u) dn } 2(u) Y 2 (u) How do we view it? 1. Perspective 1: weighted difference in cumulative hazard functions 2. Perspective 2: weighted sum of (ad bc) of the 2 2 tables over time

2 An equivalent form: W = = W (u) Y 0(u)Y 1 (u) 0 Y 0 (u) + Y 1 (u) 0 W (u) { dn1 (u) Y 1 (u) dn 0(u) Y 0 (u) { Y0 (u) Y (u) dn 1(u) Y 1(u) Y (u) dn 0(u) Z i = 0/1 in group 0/1, respectively Y 1 (u) = n Z i Y i (u) } } Y 1 (u) Y (u) = n Z i Y i (u) n Y i (u) = Z(u) Y 0 (u) Y (u) = 1 Z(u)

3 After substitution, W = = = = 0 W (u) { Y0 (u) Y (u) dn 1(u) Y } 1(u) Y (u) dn 0(u) W (u) { [1 Z(u)]dN 1 (u) + [0 Z(u)]dN 0 (u) } 0 n W (u) { Zi Z(u) } dn i (u) 0 n 0 W (u) { Z i Z(u) } dn i (u) Perspective 3: weighted difference on covariates, because Z(u) = n Z i Y i (u) n Y i (u) E[ZI(X u)] Pr(X u) = E(Z X u) = µ Z (u)

4 Story continues: W = = + n n n 0 W (u) { Z i Z(u) } dn i (u) 0 W (u) { Z i Z(u) } [dn i (u) Y i (u)λ(u)du] 0 W (u) { Z i Z(u) } Y i (u)λ(u)du the second sum is zero, because n { Zi Z(u) } Y i (u) = { n Z i n Z i Y i (u) n Y i (u) } Y i (u) = 0

5 Eureka! W = n 0 W (u){ Z i Z(u) } dm i (u) cf. linear regression model y i = β 0 + β 1 x i + e i LS estimation of the second equation w.r.t. β 1 : n x i (y i ŷ i ) = n x i ê i = n (x i x)ê i

6 Asymptotics under H 0 : λ(t Z i ) = λ(t) U n (t) = n 1/2 n t 0 W (u) { Z i Z(u) } dm i (u) weighted Log-rank statistic: W = n 1/2 U n (τ) F t = σ{n i (u), Y i (u), Z i ; i = 1, 2,... n, 0 u t} M i ( ) are F t -martingales H i (u) = n 1/2 W (u) { Z i Z(u) } are F t -predictable U n (t) = n t0 H i (u)dm i (u)

7 Martingale CLT U n (t) = n t0 n 1/2 W (u) { Z i Z(u) } dm i (u) < U n, U n > (t) should be n t [ n 1/2 W (u) { Z i Z(u) }] 2 Yi (u)λ(u)du = t 0 1 n 0 n Assume that W (u) w(u) W (u) 2 { Z i Z(u) } 2 Yi (u)λ(u)du < U n, U n > (t) P t 0 w(u)2 E =α(t) [ {Z µ Z (u)} 2 ] I(X u) λ(u)du

8 < U n,ɛ, U n,ɛ > (t) P 0, because n t 0 [ n 1/2 W (u) { Z i Z(u) }] 2 I { n 1/2 W (u) { Z i Z(u) } ɛ } Yi (u)λ(u)du therefore, U n U var[u(t)] =α(t) = = = t 0 w(u)2 E t 0 w(u)2e t [ {Z µ Z (u)} 2 I(X u) λ(u)du [ {Z µz (u)} 2 I(X u) ] ] EI(X u)λ(u)du EI(X u) 0 w(u)2 var(z X u)ei(x u)λ(u)du

9 Weighted Log-rank statistic: W = n 1/2 U n (τ) n 1/2 W D N (0, α(τ)) Standardized weighted Log-rank test statistic: n 1/2 W α(τ) D N (0, 1) How to estimate α(τ)?

10 We know α(t) equals t 0 w(u)2 var(z X u)ei(x u)λ(u)du λ(t)dt = d Λ(t) = dn(t)/y (t) ÊI(X u) = Y (u)/n var(z X u) = p q, where p = Ê(Z X u) = Z(u) α(τ) is estimated by n 1 t 0 W (u)2 Z(u)[1 Z(u)]dN(u)

11 Standardized weighted Log-rank statistics n 1/2 W α(τ) = n τ0 W (u) { Z i Z(u) } dn i (u) { n τ0 W (u) 2 Z(u)[1 Z(u)]dN i (u)} 1/2 goes to N (0, 1) Reject H 0 when n 1/2 W α(τ) > 1.96 for type-i error of 5%

12 What is weighted Log-rank test statistic anyway? n 1/2 W α(τ) = n τ0 W (u) { Z i Z(u) } dn i (u) { n τ0 W (u) 2 Z(u)[1 Z(u)]dN i (u)} 1/2 if i = 0, then dn i (u) = 0 if i = 1, dn i (t) = 1 at t = X i and 0 elsewhere τ 0 W (u) { Z i Z(u) } dn i (u) = W (X i ){Z i Y 1 (X i )/Y (X i )} = w i {Z i Y i1 /Y i } Numerator is n w i i (Z i Y i1 /Y i ) Denominator is { n w 2 i iy i1 Y i0 /Y 2 i }1/2

13 Numerator is n w i i (Z i Y i1 /Y i ) Denominator is { n w 2 i iy i1 Y i0 /Y 2 i }1/2 2 2 table for ith failure, i = 1 t Z dn(t) = 1 Y (t) dn(t) Y (t) X i Z i = 1 1 Y i1 1 Y i1 Z i = 0 0 Y i0 Y i0 1 Y i 1 Y i X i Z i = 1 0 Y i1 Y i1 Z i = 0 1 Y i0 1 Y i0 1 Y i 1 Y i O i = Z i = 0/1, E i = 1 Y i1 /Y i var(o i ) = 1 (Y i 1) Y i1 Y i0 /[Y i 2 (Y i 1)]

14 Power analysis of weighted Log-rank test statistics 1. type-i error: α = 5% 2. power level 3. alternative hypothesis 4. error bound Under H 0 : λ 0 (t) = λ 1 (t) = λ(t), n 1/2 W N (0, α(τ))

15 Alternative hypothesis H 1 : λ 1 (t) = λ 0 (t)e β n θ(t) log[λ 1 (t Z i )/λ 0 (t)] = β n Z i θ(t) θ(t): take into account of nonproportionality β n : distance between the null and an alternative 1. n 1/2 β n ξ (0, ) 2. local alternatives: β n 0 Given a sample size n, { Power = Pr n 1/2 W / α(τ) > z 1 α/2 H 1 }

16 Aysmptotic distribution of n 1/2 W under H 1 n 1/2 W = n 1/2 n τ0 W (u) { Z i Z(u) } dn i (u) under H 1, E[dN i (u) F u ] = Y i (u)λ i (u)du = Y i (u)λ 0 (u)e β nz i θ(u) du n 1/2 W = n 1/2 +n 1/2 n τ n τ 0 W (u){ Z i Z(u) } dm i (u) 0 W (u){ Z i Z(u) } Y i (u)λ 0 (u)e β nz i θ(u) du apply MCLT

17 Aysmptotic distribution of n 1/2 W under H 1 n 1/2 W = n 1/2 n τ0 W (u) { Z i Z(u) } dn i (u) under H 1, E[dN i (u) F u ] = Y i (u)λ i (u)du = Y i (u)λ 0 (u)e β nz i θ(u) du n 1/2 W = n 1/2 +n 1/2 n τ =Term I + Term II n τ 0 W (u){ Z i Z(u) } dm i (u) 0 W (u){ Z i Z(u) } Y i (u)λ 0 (u)e β nz i θ(u) du

18 Term I: predictable variation τ 0 n 1 n W (u) 2{ Z i Z(u) } 2 Yi (u)λ 0 (u)e β nz i θ(u) du β n 0 e β nz i θ(u) 1 and H 1n H 0 Term I τ 0 w(u) 2 E[(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du Term I asymptotically N (0, τ 0 w(u)2 E H0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du)

19 Term II: n 1/2 n τ 0 W (u){ Z i Z(u) } Y i (u)λ 0 (u)e β nz i θ(u) du Taylor expansion: e β nz i θ(u) = 1 + β n Z i θ(u) + O(β 2 n ) O(β 2 n )/β2 n is bounded Term II = Term IIa + Term IIb + Term IIc Term IIa n 1/2 n τ 0 W (u){ Z i Z(u) } Y i (u)λ 0 (u)du = 0

20 Term IIb = = =ξ n 1/2 n τ τ 0 W (u)n 1 τ 0 W (u)n 1 τ 0 W (u){ Z i Z(u) } Z i Y i (u)β n θ(u)λ 0 (u)du n n { Zi Z(u) } Z i Y i (u) n 1/2 β n θ(u)λ 0 (u)du { Zi Z(u) } 2 Yi (u) n 1/2 β n θ(u)λ 0 (u)du 0 w(u)e H 0 [(Z i µ Z (u)) 2 I(X u)] ξ θ(u)λ 0 (u)du τ 0 w(u)θ(u)e H 0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du

21 Term IIc O(βn 2)/β2 n < M no(β2 n ) = O(nβ2 n ) n 1/2 O(βn) 2 = n 1/2 O(nβn) 2 = o(n 1/2 ) = 0 n 1/2 n τ 0 τ 0 W (u)n 1 { Zi Z(u) } Y i (u)o(β 2 n )λ 0(u)du n { Zi Z(u) } Y i (u)o(n 1/2 )λ 0 (u)du

22 Term II = Term IIa + Term IIb + Term IIc converges to ξ τ 0 w(u)θ(u)e H 0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du Recall on Term I N (0, τ 0 w(u)2 E H0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du) Under H 1n : A(w 2 ) = τ 0 w(u) 2 E H0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du) n 1/2 W N (ξa(θw), A(w))

23 Recap on power calculation of weighted Log-rank n 1/2 W = n 1/2 n τ0 W (u) { Z i Z(u) } dn i (u) Under H 0, n 1/2 W N (0, α(τ)) Alternative hypothesis: H 1n : λ 1 (t) = λ 0 (t)e β n θ(t) A breakdown n 1/2 W = n 1/2 +n 1/2 n τ =Term I + Term II n τ 0 W (u){ Z i Z(u) } dm i (u) 0 W (u){ Z i Z(u) } Y i (u)λ 0 (u)e β nz i θ(u) du

24 Term I: mean zero contribute random variation Term I asymptotically N (0, τ 0 w(u)2 E H0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du)

25 Term II: n 1/2 n τ 0 W (u){ Z i Z(u) } Y i (u)λ 0 (u)e β nz i θ(u) du Taylor expansion: e β nz i θ(u) = 1 + β n Z i θ(u) + O(βn 2) Term II = Term IIa + Term IIb + Term IIc

26 Term IIa is zero: n 1/2 n τ 0 W (u){ Z i Z(u) } Y i (u)λ 0 (u)du = 0 Term IIb converges to ξ τ 0 w(u)θ(u)e H 0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du Term IIc converges to zero

27 Term I converges in distribution to N (0, τ 0 w(u)2 E H0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du Term II converges in probability to ξ τ 0 w(u)θ(u)e H 0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du Let s define: A(w 2 ) = τ 0 w(u)2 E H0 [(Z i µ Z (u)) 2 I(X u)]λ 0 (u)du Under H 1n : n 1/2 W N (ξa(θw), A(w 2 ))

28 Summary on n 1/2 W Under H 0 : n 1/2 W N (0, A(w 2 )) Under H 1n : n 1/2 W N (ξa(θw), A(w 2 )) Binary versus Time-to-event Binary Time-to-event T.S. p 1 p 0 weighted Log-rank H 0 p 1 = p 0 λ 1 (t) = λ 0 (t) Dist n N (0, σ0 2) N (0, A(w2 )) H 1 p 1 = p 0 + d λ 1 (t) = λ 0 (t)e β nθ(u) Dist n N (d, σ1 2) N (ξa(θw), A(w2 ))

29 Power P = Φ ( ξa(θw) A(w 2 ) 1/2 z 1 α/2 ) t Probability density

30 What would affect power? ξa(θw) A(w 2 ) 1/2 increases, power increases ξ: usually predetermined w(u): weight functions How do we choose w to maximize A(θw) A(w 2 ) 1/2

31 Consider A( ): for any constant b A[(w bθ) 2 ] 0 A[(w bθ) 2 ] =A(w 2 2bwθ + b 2 θ 2 ) =A(θ 2 )b 2 2bA(wθ) + A(w 2 ) 0 A(wθ) 2 A(θ 2 )A(w 2 ) 0 equality satisfied only when Cauchy-Schwarz Inequality A(θw) A(w 2 ) 1/2 A(θ2 ) 1/2 w(u) = θ(u).

32 Some examples of optimal w(u) in nonproportional alternatives Additive hazards model (Lin & Ying, 1994, BMKA): λ(t Z) = λ 0 (t) + βz λ 0 (t)e βz λ 0 (t) w(u) = 1 λ 0 (u) Accelerated hazards model (Chen & Wang, 1999, JASA): λ(t Z) = λ 0 (te βz ) λ 0 (t)+λ 0 (t)tβz w(u) = λ 0 (u)u λ 0 (u)

33 Why n 1/2 β n ξ suppose n k β n ξ for some k 0 n 1/2 β n = n 1/2 k n k β n n 1/2 k ξ, we can verify in Term IIb if k > 1/2, n 1/2 β n 0; Term II goes to 0 no power whatsoever if k < 1/2, n 1/2 β n ; Term II goes to always 100% power for any w(u)

34 3. Sample size calculation In practice, we have a fixed β 0 to be detected H 0 : λ 1 (t) = λ 0 (t) H 1 : λ 1 (t) = λ 0 (t)e β 0 θ(u) Standardized weighted Log-rank T S: under H 0 : T S N (0, 1) under H 1 : T S N ( n 1/2 ) β 0 A(θw) A(w 2 ) 1/2, 1

35 Power P = Pr{ T S z 1 α/2 } = 1 β n 1/2 β 0 A(θw) A(w 2 ) 1/2 = z 1 α/2 +z 1 β n = (z α/2 + zβ)2a(w2 ) β 0 A(θw) 2 w = θ = 1 Log-rank for proportional hazards model sample size n = (z α/2 + z β) 2 β 2 0 A(1)

36 what is A(1)? recall on A(1) = 0 E[(Z µ Z (u)) 2 I(X u)]λ 0 (u)du A(1) = π Z (1 π Z ) Pr( = 1) Sample size is then n Pr( = 1) = (z α/2 + z β) 2 β0 2π Z(1 π Z ) Expected # failures/events: E D = n Pr( = 1) HR = e β is hazards ratio 1-to-1 treatment-control assignment E D = 4(z α/2 + z β) 2 (log HR) 2

37 Example: type-i error: 5% power: 90% HR = 2 E D = 42/(log HR) 2 : 88

38 Summary on comparing survival functions Weighted Log-rank statistic perspectives asymptotics power calculation Alternatives Yet to cover stratified Log-rank K-samples staggered entry in sample size calculation