A Bartlett type correction for Wald test in Cox regression model

Statistics and Probability Letters 78 (2008) 2614–2622

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Statistics and Probability Letters

journal homepage: www.elsevier.com/locate/stapro

A Bartlett type correction for Wald test in Cox regression modelI

Xiao Li a, Yaohua Wu a,∗, Dongsheng Tu ba Department of Statistics and Finance, University of Science and Technology of China, Hefei 230026, Chinab National Cancer Institute of Canada Clinical Trials Group, Queen’s University, Kingston, ON K7L 3N6, Canada

a r t i c l e i n f o

Article history:Received 27 December 2007Received in revised form 13 May 2008Accepted 8 August 2008Available online 28 August 2008

a b s t r a c t

In clinical trials with time to an event endpoint, the Wald test in a Cox regression modelmay be used to assess the difference between two treatments. In this paper, a Bartlett typecorrection to theWald test is derived and applied to the analysis of data from a clinical trialon actuate myelogenous leukemia (AML). Monte Carlo simulations are also performed toassess the actual type I error and power of the proposed Bartlett corrected test.

© 2008 Elsevier B.V. All rights reserved.

1. Introduction

In randomized clinical trials with time to an event (such as disease recurrence or death) as an endpoint, the primaryobjective is usually to compare the difference in survival distributions among treatment groups. When the proportionalhazard assumption is true, Cox proportional hazard model (Cox, 1972) can be used to perform this comparison based onlikelihood ratio,Wald or score tests. Under the null hypothesis that there is no difference between treatment groups, the teststatistics in all of these three tests follow a chi-square distributionwith one degree of freedom (see, for example, Fleming andHarrington (1991)). Gu and Zheng (1993) and Tu et al. (2005) developed Bartlett type correction to likelihood ratio and scoretests respectively. Asmentioned by Tu et al. (2005), the Bartlett correction to theWald test in Coxmodel is, however, still notavailable. SinceWald test is constructed directly from themaximum partial likelihood estimate (MPLE) of the coefficients inthe Cox model and its asymptotic variance, which are used to derive estimate and confidence interval for the hazard ratio,its Bartlett correction is of both theoretical and practical interests. The primary objective of this paper is to derive a Bartletttype correction to the Wald test in a Cox model.There is a rich literature in the area of Bartlett corrections. A review can be found in, for example, Cribari-Neto and

Cordeiro (1996). In this paper, we first establish a two term Edgeworth expansion for the null distribution of Wald teststatistic in the Cox proportional hazard model and then apply the method of Cordeiro and Ferrari (1991) to derive a Bartletttype correctedWald test. The Bartlett corrected test is then evaluated in a simulation study and applied to analyze data froma clinical trial on actuate myelogenous leukemia (AML).

2. Main theoretical results

Let T be the time from the randomization until observation of a specific event, such as death or recurrence of disease, fora subject in a clinical trial and Z a covariate taking values on a bounded subinterval of R, such as an indicator variable for thetreatment this subject received. Denote the hazard function of T given Z as λ(t|Z), which is defined based on the followingrelationship:

I Research partially supported by National Natural Science Foundation of China (Grant No. 10471136), Ph.D. Program Foundation of the Ministry ofEducation of China, Special Foundations of the Chinese Academy of Science and USTC, and Nature Science and Engineering Research Council of Canada.∗ Corresponding author.E-mail addresses: [email protected] (X. Li), [email protected] (Y. Wu), [email protected] (D. Tu).

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X. Li et al. / Statistics and Probability Letters 78 (2008) 2614–2622 2615

P{T ≥ t|Z} = exp{−

∫ t

0λ(y|t)dy

}.

The Cox proportional hazards regression model (Cox, 1972) assumes that λ(t|Z) can be written in the following form:

λ(t|Z) = λ0(t) exp(βZ),

where λ0(t) is an unknown function of t and β ∈ B ⊂ R for an open interval B, which is called the regression coefficient ofcovariate Z .In this paper, we are interested in testing the following hypotheses:

H0 : β = β0 versus H1 : β 6= β0

for some known β0. Due to potential censoring, T may not be observable for every subject. We assume that the observeddata which will be used to test the above hypotheses are in the following form:

Xi = min(Ti, Ci), δi = I(Xi ≤ Ci),

where Ti and Ci are respectively the true failure and censoring times for the ith subject and conditionally independent givenZi, the covariate for the ith subject, i = 1, 2, . . . , n.Let Yi(t) = I(Xi ≥ t) and π(t) = EY1(t) = P{X1 ≥ t}. For a known constant τ such that π(τ) > 0, we can define the

partial log-likelihood function of the observed data by

ln(β) =n∑i=1

∫ τ

0[βZi − log S(0)(β, u)]dNi(u),

where

Ni(t) = I(Xi ≤ t, δi = 1) and S(0)(β, t) = n−1n∑i=1

exp(βZi)Yi(t).

Based on this partial log-likelihood function, β can be estimated by β which satisfies

ln(β) = supβ

ln(β)

or, equivalently,

Un(β) =∂ ln(β)∂β

= 0.

With this maximum partial likelihood estimate, the Wald test rejects the null hypothesis H0 if

Wn = (β − β0)2In(β) > Cα,

where

In(β) =∂2ln(β)∂β2

=

n∑i=1

∫ τ

0

[S(2)(β, u)S(0)(β, u)

−

(S(1)(β, u)S(0)(β, u)

)2]dNi(u)

with

S(j)(β, t) = n−1n∑i=1

Z ji exp(βZi)Yi(t)

for j = 0, 1, 2 and Cα is the upper α percentile of the chi-square distribution with one degree of freedom. Define

Kn = (β − β0)I1/2n (β).

Then,Wn = K 2n . To obtain a Bartlett type correction for theWald test, we first need to develop an asymptotic expression forthe distribution of Kn under H0.

Theorem 1. Uniformly in x,

P{Kn ≤ x|H0} = Φ(x)+ n−1/2[k11 +

k126(x2 − 1)

]φ(x)

+1n[k21x+

124k22(−x3 + 3x)+

k2372(−x5 + 10x3 − 15x)]φ(x)+ o(n−1),

where kij (i = 1, 2; j = 1, 2, 3) are some constants which are defined in the Appendix together with the proof of this theorem.

2616 X. Li et al. / Statistics and Probability Letters 78 (2008) 2614–2622

From this theorem, we can immediately get the following asymptotic expansion for the null distribution ofWn:

Corollary 1.

P{Wn ≤ x|H0} = P{Kn ≤ x1/2|H0} − P{Kn ≤ −x1/2|H0} = P{χ21 ≤ x}

+1n

{2k21x1/2 +

112k22(3x1/2 − x3/2)+

k2336(−x5/2 + 10x3/2 − 15x1/2)

}φ(x1/2)+ o(n−1).

Let χ21 be a central chi-square random variable with 1 degree of freedom and f1(x) its density function. Then from theasymptotic expansion for the distribution function ofWn as given in Corollary 1 and the property of the chi-square densityfunction (Cordeiro and Ferrari, 1991), we can rewrite the above asymptotic expansion into the following form:

P{Wn ≤ x|H0} = P{χ21 ≤ x} +1nf1(x)

3∑j=0

αjxj + o(n−1),

where

α0 = 0, α1 = 2k21 +14k22 −

512k23, α2 = −

112k22 +

518k23, α3 = −

136k23.

From the above expansion and formulae (2) and (3) of Cordeiro and Ferrari (1991), we can prove immediately the followingtheorem:

Theorem 2. Define

W ∗n = Wn

(1+

1n

3∑j=0

αjW j−1n

)= Wn

[1+

1n(α1 + α2Wn + α3W 2n )

].

We have

P{W ∗n > (zα/2)2|H0} = α + o(n−1).

W ∗n is called the Bartlett type corrected Wald test statistic.

Remark. When the baseline hazard function λ0(t) and the censoring distribution are unknown, we may use the samemethod used by Tu et al. (2005) and Gu (1992) to estimate the constants αj (j = 1, 2, 3) to obtain an estimated BartlettcorrectedWald test statistic.With thismethod,α(i)k andΛk(t) are first estimated consistently by respectively their empiricalversions as given by Gu (1992). The estimate for αj can be obtained by plugging in these estimates since it is a function ofα(i)k and Λk(t). Theorem 2 still holds for the estimated Bartlett corrected Wald test statistic since the estimate for α

(i)k and

Λk(t) is root n consistent as shown by Gu (1992).

3. Numerical studies

Weconsider in this section the application of the proposedBartlett type correctedWald test to compare two treatments ina clinical trial and present some results from simulation studies evaluating the actual type I error and power of the proposedtest.In a two arm clinical trial comparing a new treatment with a standard treatment, we can define a covariate zi = 1 if

the patient was assigned a new treatment and zi = 0 if a standard treatment. We assume that the trial is balanced, whichimplies that P{zi = 1} = P{zi = 0} = 1/2. Under the framework of the Cox proportional hazards model, the differencebetween two treatment groups in terms of failure times can be assessed by testing H0 versus H1 defined in Section 2 withβ0 = 0. In this special case of covariate, some algebras can show that the Bartlett correctedWald test statistic can bewrittenas

W ∗n = Wn

[1+

1n(α1 + α2Wn)

],

where

α1 =11Σ2 − 22Σ1 −Σ20 + 7Σ0

4Σ20, α2 =

6Σ20 − 2Σ1 − 3Σ212Σ20

with

Σk =

∫ τ

0Λ0(t)kα0(t)λ0(t)dt =

∫ τ

0Λ0(t)kπ(t)λ0(t)dt.


Table 1Simulation results

n Type 1 error (λ1 = λ0) Power at λ1 = 2λ0Wald Bartlett correction Wald Bartlett correction

L = 1 and B = 9

20 0.0423 0.0480 0.366 0.37830 0.0463 0.0492 0.537 0.54640 0.0457 0.0490 0.659 0.665

L = 5 and B = 5

20 0.0383 0.0453 0.318 0.33030 0.0453 0.0477 0.459 0.47240 0.0495 0.0505 0.589 0.596

L = 9 and B = 1

20 0.0307 0.0443 0.246 0.26830 0.0383 0.0480 0.362 0.38040 0.0440 0.0500 0.476 0.490

When λ0 is unknown,Σk (k = 0, 1, 2) can be estimated by Strawderman (1997)

Σk =1n

n∑i=1

δi[Λ(Xi)]k

where Λ(Xi) is the Nelson–Aalen estimator for the cumulative hazard function based on the data from two treatment groupspooled together. By plugging in these estimates into αj, we can obtain an estimated Bartlett corrected Wald test statistic.This was the approach we used in the data analysis and simulations.We first apply this test to analyze the data from a clinical trial on treatment of maintenance chemotherapy for patients

with actuate myelogenous leukemia (AML) as presented in Table 1.1 of Tableman and Kim (2004). A total of 23 patientsafter remission through treatment by chemotherapy were randomized in this trial to receive maintenance chemotherapy(11 patients) or nothing (12 patients). The endpoint of this trial is the time from randomization until relapse. A total of 18patients relapsed in this dataset. The test statistic from (uncorrected) Wald test in the Cox model was 3.20, correspondingto a p-value of 0.074. The Bartlett correction factors for this dataset were α1 = 0.11 and α2 = 0.019, which led to a Bartlettcorrected Wald test statistic 3.22 and a p-value of 0.073. For this dataset, we may find that the Bartlett corrected Wald testis only slightly different from the uncorrected Wald test.For this reason, we have performed Monte Carlo simulations of 3000 replications to assess the actual type I error and

the powers of the proposed Bartlett type corrected Wald test with nominal level 0.05 for small and moderate sample sizes(n = 20, 30, 40) but with high censoring rates. In the simulations, the true survival time for each patient allocated to thefirst arm was generated from an exponential distribution with parameter λ0 = −log(0.5)/10 and that for a patient in thesecond treatment arm fromanexponential distributionwith parameterλ1 = λ0 (i.e.β0 = 0)orλ1 = 2λ0 (i.e. exp(β0) = 2),depending on whether the type I error or the power was assessed. The censoring time for each patient was generated froma uniform distribution over [B, L + B] with three different sets of L and B: (1) L = 1 and B = 9; (2) L = 5 and B = 5; (3)L = 9 and B = 1, which correspond to respectively censoring rates of 52, 60 and 70%. The results of the simulation studiesare presented in Table 1. From this table, we can find that Bartlett correction improved Wald test, especially when samplesize was small and censoring rate was high.

Appendix

In this Appendix, we provide a proof for Theorem 1 after introducing some notations. Most of these notations appearedin Gu and Zheng (1993) and Tu et al. (2005). Define

ξi,k(t) = Zki exp(βZi)I(Xi ≥ t), α(i)k (t) = E[Z

k exp((i+ 1)β0Z)Y1(t)],

θ(i)k (t) =

α(i)k (t)α0(t)

− 2α(i)k−1(t)α1(t)α20(t)

+ α(k)i−2(t)

α21(t)α30(t)

.

Let θk(t) = θ(0)k (t) and αk(t) = α

(0)k (t). Define also

Λk(t) =∫ t

0

αk(s)α0(s)

λ0(s)ds, Λ(2)1 (t) =

∫ t

0

α21(s)α20

λ0(s)ds,

ηk(t) =α(1)k (t)α0(t)

Λ0(t)−α(1)k−1(t)α0(t)

Λ1(t), η(t) = −(η2(t)− η1(t)

α1(t)α0(t)

).


From these basic notations,we introduce the followingquantitieswhich are related to themoments of theWald test statistic:

σ 2 =

∫ τ

0θ2(t)α0(t)λ0(t)dt, ∆ = −

∫ τ

0

[θ(1)3 (u)Λ0(u)− θ

(0)2 (u)

α1(u)α0(u)

]α0(u)λ0(u)du,

ζ = −

∫ τ

0

[θ(1)3 (u)Λ0(u)− θ

(1)2 (u)

(∫ u

0

α1(s)α0(s)

λ0(s)ds)]α0(u)λ0(u)du, ν =

∫ τ

0θ22 (u)α0(u)λ0(u)du,

µ2 = −

∫ τ

0

[θ4(u)− 2θ3(u)

α1(u)α0(u)

+ θ2(u)α21(u)α20(u)

]α0(u)λ0(u)du,

µ3 = −

∫ τ

0

[η4(u)− 3η3(u)

α1(u)α0(u)

+ 2η2(u)α31(u)α30(u)

]α0(u)λ0(u)du,

µ4 =

∫ τ

0θ2(u)η(u)α0(u)λ0(u)du, µ5 =

∫ τ

0θ(1)2 (u)[Λ2(u)−Λ

(2)1 (u)]α0(u)λ0(u)du,

µ6 =

∫ τ

0[θ(2)4 (u)Λ

20(u)− 2θ

(2)3 (u)Λ1(u)λ0(u)+ θ

(2)2 (u)Λ

21(u)]α0(u)λ0(u)du,

µ7 =

∫ τ

0[θ(1)4 (u)− 2θ

(1)3 (u)Λ1(u)λ(u)+ θ

(1)2 (u)Λ

(2)1 (u)]α0(u)λ0(u)du,

µ8 =

∫ τ

0η2(u)α0(u)λ0(u)du, µ9 =

∫ τ

0θ(1)2 (t)

(∫ t

0η(u)λ0(u)du

)α0(t)λ0(t)dt,

µ = −

∫ τ

0(α(1)4 (t)Λ0(t)− α3(1)(t)Λ1(t))λ0(t)dt +

∫ τ

0

3α1(t)α0(t)

(α(1)3 (t)Λ0(t)− α

(1)2 (t)Λ1(t))λ0(t)dt

+ 3∫ τ

0(θ2(t)−

α21(t)α20(t)

)(α(1)2 (t)Λ0(t)− α

(1)1 (t)Λ1(t))λ0(t)dt

−

∫ τ

0

(3α1(t)α0(t)

θ2(t)−α31(t)α30(t)

)(α(1)1 (t)Λ0(t)− α

(1)0 (t)Λ1(t))λ0(t)dt,

ξ = σ−2(−2µ5 + 2µ7 + ν − σ 4 − µ+

∆′

2

)+ σ−4

[∆2 − 2∆ζ +∆

(∆

2− ζ

)]− γ ,

∆′ = −

∫ τ

0

(α4(t)− 4

α3(t)α1(t)α0(t)

− 3α22(t)α0(t)

+ 12α2(t)α21(t)α20(t)

− 6α41(t)α30(t)

)λ0(t)dt, k3 = −σ−6∆,

k4 = σ−8(µ2 − 4µ3 − 6µ4 − 12µ5 + 12µ7 − 6µ9 + 3ν − 3σ 4 − 2∆′)+ 6σ−10∆(2ζ −∆)− 3σ−10(ζ 2 + 2∆ζ )+ 9σ−12(ζ −∆)2 − 3σ−4

b′ =14σ−6

(−11µ5 − 2µ6 + 11µ7 − 4µ8 − 4µ9 + 4µ− 2∆′ +

92ν −

92σ 4)−12σ−4(ξ + γ )

−12σ−8∆(∆− 2ζ )+

18σ−8(4∆2 − 10∆ζ − ζ 2).

With above notations, we can state and prove some lemmas which are used in the proof of Theorem 1.

Lemma 1.

β − β0 =

n∑i=1

{1nσ 2g(yi)+

[1n2σ 2

ψ0(yi)−1n2σ 4

(g(yi)h(yi)− ζ )+∆

2n2σ 6(g2(yi)− σ 2)

+1n2

(∆(∆− 2ζ )2σ 8

−γ

σ 4+−2µ5 + 2µ7 + ν − σ 4

σ 6

)g(yi)

]}+1n2∑i<j

1σ 2Ψ (yi, yj)

+1n3∑i<j<k

{1σ 2B(yi, yj, yk)−

1σ 6(a(yi)g(yj)g(yk)+ a(yj)g(yi)g(yk)+ a(yk)g(yi)g(yj))

−1σ 4

[h(yi)Ψ (yj, yk)+

∆

σ 4g(yj)g(yk)+ h(yj)(Ψ (yi, yk)

+∆

σ 4g(yi)g(yk))+ h(yk)

(Ψ (yi, yj)+

∆

σ 4g(yi)g(yj)

)]+1σ 4[g(yi)C(yj, yk)+ g(yj)C(yi, yk)+ g(yk)C(yi, yj)]

}+∆/2− ζnσ 4

+ O(n−2),


where

g(yi) =∫ τ

0

(zi −

α1(t)α0(t)

)dMi(t), Mi(t) = Ni(t)−

∫ t

0exp(βZi)I(Yi ≥ s)λ0(s)ds,

ψ0(yi) = ψ0(yi, yi), ψ0(yi, yj) = −∫ τ

0πi(t)dMj(t),

πi(t) =1

α0(t)(ξi,1(t)− α1(t))−

α1(t)α20(t)

(ξi,0(t)− α0(t)),

h(yi) = h1(yi)+ h2(yi), h1(yi) =∫ τ

0θ2(t)dMi(t), h2(yi) =

∫ τ

0ρi(t)λ0(t)dt,

ρi(t) = ξi,2(t)− α2(t)− 2α1(t)α0(t)

(ξi,1(t)− α1(t))+α21(t)α20(t)

(ξi,0(t)− α0(t)),

a(yi) = d(yi)+ b(yi), d(yi) =∫ τ

0

(θ3(t)−

α1(t)α0(t)

θ2(t))dMi(t),

b(yi) =∫ τ

0(ξi,3(t)− α3(t))− 3

α1(t)α0(t)

(ξi,2(t)− α1(t))− 3(θ2(t)−

α21(t)α20(t)

)(ξi,1(t)− α1(t))

+

(3α1(t)α0(t)

(θ2(t)−

α31(t)α30(t)

)(ξi,0(t)− α0(t))

)λ0(t)dt,

Ψ (yi, yj) = ψ(yi, yj)−1σ 2(g(yi)h(yj)+ g(yj)h(yi))+

∆

σ 4g(yi)g(yj), ψ(yi, yj) = ψ0(yi, yj)+ ψ0(yj, yi),

B(yi, yj, yk) =∑i1,i2,i3

B0(yi1 , yi2 , yi3), B0(yi, yj, yk) = −∫ τ

0Πi,j(t)dMk(t),

Πi,j(t) = −ξi,0(t)− α0(t)

α0(t)πj(t), C(yi, yj) = C(yi, yj)+

∆

σ 2Ψ (yi, yj)+

2σ 2h(yi)h(yj),

C(yi, yj) = D(yi, yj)+ D(yj, yi)− E(yi, yj)− E(yj, yi), D(yi, yj) =∫ τ

0Ji,j(t)λ0(t)dt,

Ji,j(t) =α21(t)α30(t)

(ξi,0(t)− α0(t))(ξj,0(t)− α0(t))− 2α1(t)α20(t)

(ξi,0(t)− α0(t))(ξj,1(t)− α1(t))

+1

α0(t)(ξi,1(t)− α1(t))(ξj,1(t)− α1(t)),

E(yi, yj) =∫ τ

0

[ρi(t)α0(t)

− θ2(t)(ξi,0(t)− α0(t))

α0(t)

]dMj(t).

Proof. From the proof of Theorem 1.1 in Gu and Zheng (1993), we have

β − β0 = I−1n (β0)Θ + O(n−2),

where

In(β0) = nσ 2 +n∑i=1

h(yi)−1n

∑i<j

C(yi, yj)+ γ , Θ = Un(β0)+12∆n(β0)Θ

22 +

16n∆′Θ31 ,

Un(β0) =n∑i=1

g(yi)+1n

∑i<j

ψ(yi, yj)+1n

n∑i=1

ψ0(yi)+1n2∑i<j<k

B(yi, yj, yk), ∆n(β0) = n∆−n∑i=1

a(yi),

Θ1 =1nσ 2

n∑i=1

g(yi), Θ2 =1nσ 2

n∑i=1

g(yi)+1n2σ 2

∑i<j

Ψ (yi, yj)+∆/2− ζnσ 4

.

It is easy to show that

I−1n (β0) = I−10,n + O(n

−5/2),

where

I−10,n =1nσ 2−

1n2σ 4

n∑i=1

h(yi)+1n3σ 4

∑i<j

C(yi, yj)−1n2σ 4

γ +Eh2(yi)n2σ 6

+2n3σ 6

∑i<j

h(yi)h(yj).


Therefore, we have

β − β0 = I−10,nΘ + O(n−2).

The lemma can then be proved by some lengthy algebra calculations. �

Lemma 2.

In(β) = nσ 2 +n∑i=1

(h(yi)−

∆

σ 2g(yi)

)+1n

∑i<j

[−C(yi, yj)−

∆

σ 2Ψ (yi, yj)+

1σ 2(g(yi)f (yj)

+ g(yj)f (yi))]+

[γ +

µ−∆′/2σ 2

−∆(∆/2− ζ )

σ 4

]+ O(n−1/2),

where

f (yi) = a(yi)−∆′

2σ 2g(yi).

Proof. From Lemma 4.4 of Gu (1992) and using notations in Lemma 1, we have

β − β0 = O(n−1/2), β − β0 = Θ1 + O(n−1), β − β0 = Θ2 + O(n−3/2).

Using Taylor expansions for In(β), we can obtain

In(β) = In(β0)−∆n(β0)(β − β0)−∆′n(β0)(β − β0)2/2+∆′′n(β)(β − β0)

3.

From the proof of Lemmas 2.2–2.4 in Gu (1992) and noticing that∆′′n(β) = O(n), we have

In(β) = In(β0)− ∆n(β0)Θ2 − n∆′Θ21/2+ O(n−1/2),

where the notations are the same as that in Lemma 1. Some lengthy algebra will lead to the proof of this lemma. �

Proof of Theorem 1. From Lemmas 1 and 2, we have

√n(β − β0) =

n∑i=1

(g1(yi)n1/2

+g ′1(yi)n3/2

)+

∑i<j

g2(yi, yj)n3/2

+

∑i<j<k

g3(yi, yj, yk)n5/2

+∆/2− ζn1/2σ 4

+ O(n−3/2)

1= Tn +

∆/2− ζn1/2σ 4

+ O(n−3/2),

where

g1(yi) = σ−2g(yi),

g ′1(yi) = σ−2ψ0(yi)− σ−4(g(yi)h(yi)− ζ )+12∆σ−6(g2(yi)− σ 2)+

12∆(∆− 2ζ )σ−8g(yi)− γ σ−4g(yi)

+ σ−6(−2µ5 + 2µ7 + ν − σ 4)g(yi),

g2(yi, yj) = σ−2Ψ (yi, yj),

g3(yi, yj, yk) =1σ 2B(yi, yj, yk)−

1σ 6(a(yi)g(yj)g(yk)+ a(yj)g(yi)g(yk)+ a(yk)g(yi)g(yj))

−1σ 4

[h(yi)Ψ (yj, yk)+

∆

σ 4g(yj)g(yk)+ h(yj)

(Ψ (yi, yk)+

∆

σ 4g(yi)g(yk)

)+ h(yk)(Ψ (yi, yj)

+∆

σ 4g(yi)g(yj))

]+1σ 4

[g(yi)C(yj, yk)+ g(yj)C(yi, yk)+ g(yk)C(yi, yj)

],

and

√n(nI−1n (β)− σ

−2) =

n∑i=1

(f1(yi)n1/2+f ′1(yi)n3/2

)+

∑i<j

f2(yi, yj)n3/2

+σ−4ξ

n1/2+ O(n−1),

where

f1(yi) = −σ−4[h(yi)−∆σ−2g(yi)],

f2(yi, yj) = −σ−4[−C(yi, yj)+

∆1

σ 2Ψ (yi, yj)+ σ−2(g(yi)f (yj)+ g(yj)f (yi))

−2σ 2

(h(yi)−

∆

σ 2g(yi)

)(h(yj)−

∆

σ 2g(yj)

)].


Based on these asymptotic expansions and from Theorem A.1 of Strawderman and Wells (1997), we can write TnI1/2n (β) as

an asymptotic U-statistic:

TnI1/2n (β) =

√nTn

√nI−1/2n (β)

=Unν−

η

2ν3n1/2+ o(n−1),

where

ν2 = Eg21 (y1) = σ−2, η = E[g1(y1)f1(y1)] = −σ−6(ζ −∆),

Un =n∑i=1

[g1(yi)n1/2

+χ(yi)n3/2

]+

∑i<j

ω(yi, yj)n3/2

+

∑i<j<k

ϕ(yi, yj, yk)n5/2

,

χ(yi) = g ′1(yi)−σ−4ξ

2ν2g1(yi)+

3η4ν4f1(yi)−

12ν2

(g1(yi)f1(yi)− η)

−12ν2

{E[g1(yj)f2(yi, yj)|yi] − E[f1(yj)g2(yi, yj)|yi]

},

ω(yi, yj) = g2(yi, yj)−12ν2[g1(yi)f1(yj)+ g1(yj)f1(yi)],

ϕ(yi, yj, yk) = g3(yi, yj, yk)−12ν2

(Aijk + Ajik + Akji + Bijk + Bjik + Bkij),

Aijk = f1(yi)g2(yj, yk), Bijk = g1(yi)[f2(yj, yk)−

32f1(yj)f1(yk)

].

Therefore, we have the following asymptotic expansion for Kn = (β − β0)I1/2n (β):

(β − β0)I1/2n (β) = TnI1/2n (β)+∆/2− ζnσ 4

I1/2n (β) =Unν−

η

2ν3n1/2+∆/2− ζn1/2σ 3

+∆/2− ζ2n3/2σ 5

n∑i=1

(h(yi)−

∆

σ 2g(yi))+ o(n−1

)=Unν−

ζ

2n1/2σ 3+ o(n−1),

where

Un =n∑i=1

[g1(yi)n1/2

+χ(yi)n3/2+

∑i<j

ω(yi, yj)n3/2

+

∑i<j<k

ϕ(yi, yj, yk)n5/2

],

χ(yi) = ξ(yi)+∆/2− ζ2σ 6

(h(yi)−∆σ−2g(yi)).

From Theorem A.2 of Strawderman and Wells (1997), we can get the asymptotic expansion for the distribution function ofKn as

P{(β − β0)I1/2n (β) ≤ z} = P

{Unν≤ z +

ζ

2n1/2σ 3

}+ o(n−1)

= Φ(z ′)− n−1/2φ(z ′)P1(z ′)− n−1φ(z ′)P2(z ′)+ o(n−1),

where

z ′ = z +ζ

2n1/2σ 3, P1(z) =

k36ν3

(z2 − 1), P2(z) = b′zν2+k424ν4

(z3 − 3z)+k2372ν6

(z5 − 10z3 + 15z).

Let

δ =ζ

2n1/2σ 3

and note that

−n1/2φ(x+ δ)P1(x+ δ) = −n−1/2φ(x)P1(x)− δn−1/2k36ν3

(−x3 + 3x)φ(x)+ O(n−3/2)

= −n−1/2φ(x)[k36ν3

(x2 − 1)]+ n−1

k3ζ12(x3 − 3x)φ(x)+ O(n−3/2)


and

−n−1φ(x+ δ)P2(x+ δ) = −n−1φ(x)P2(x)+ O(n−3/2)

= n−1φ(x)[−σ 2b′x−

σ 4k424

(x3 − 3x)−σ 6k2372

(x5 − 10x3 + 15x)].

Applying Taylor expansion toΦ(z ′) gives

P{Kn ≤ x|H0} = Φ(x)+ Φ(x)+ n−1/2ζ

2σ 3φ(x)− n−1

ζ 2

8σ 6φ(x)x+ n−1/2φ(x)

k36ν3

(−x2 + 1)

+ n−1φ(x)k3ζ12(x3 − 3x)+ n−1φ(x)

[−σ 2b′x−

σ 4k424

(x3 − 3x)−σ 6k2372

(x5 − 10x3 + 15x)]

= Φ(x)+ n−1/2φ(x)[ζ

2σ 3−σ 3k36(x2 − 1)

]+ n−1φ(x)

[−

(ζ 2

8σ 6+ σ 2b′

)x

+

(k3ζ12−σ 4k424

)(x3 − 3x)−

σ 6k2372

(x5 − 10x3 + 15x)]+ o(n−1)

= Φ(x)+ n−1/2[k11 +

16k12(x2 − 1)

]φ(x)

+ n−1[k21x+

124k22(−x3 + 3x)+

k2372(−x5 + 10x3 − 15x)

]φ(x)+ o(n−1).

Defining

k11 =ζ

2σ 3, k12 = −σ 3k3, k21 = −

(ζ 2

8σ 6+ σ 2b′

), k22 = −2k3ζ + σ 4k4, k23 = σ 6k23,

the proof of Theorem 1 is completed. �

References

Cordeiro, G.M., Ferrari, S.L.P., 1991. A modified score test statistic having chi-squared distribution to order n−1 . Biometrika 78, 573–582.Cox, D.R., 1972. Regression models and life tables (with discussion). Journal of the Royal Statistical Society, Series. B 30, 248–275.Cribari-Neto, F., Cordeiro, G.M., 1996. On Bartlett and Bartlett-type corrections. Econometric Reviews 15, 339–367.Fleming, T.R., Harrington, D.P., 1991. Counting Processes and Survival Analysis. Wiley, New York.Gu, M., 1992. On the Edgeworth expansion and bootstrap approximation for the Cox regression model under random censorship. The Canadian Journal ofStatistics 20, 399–414.

Gu, M., Zheng, Z., 1993. On the Bartlett adjustment for the partial likelihood ratio test in the Cox regression model. Statistica Sinica 3, 543–555.Strawderman, R.L., 1997. An asymptotic analysis of the logrank test. Lifetime Data Analysis 3, 225–249.Strawderman, R.L., Wells, M.T., 1997. Accurate bootstrap confidence limits for the cumulative hazard and survivor functions under random censoring.Journal of the American Statistical Association 92, 1356–1374.

Tableman, M., Kim, J.S., 2004. Survival Analysis Using S. Chapman & Hall, Boca Raton, USA.Tu, D., Chen, J., Shi, P., Wu, Y., 2005. A Bartlett type correction for Rao’s score test in cox regression model. Sankhya 67, 722–735.

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