VMSTA Modern Stochastics: Theory and Applications 2351-6054 2351-6046 2351-6046 VTeXMokslininkų g. 2A, 08412 Vilnius, Lithuania VMSTA03 10.15559/vmsta-2014.1.1.3 Research Article Asymptotic normality of corrected estimator in Cox proportional hazards model with measurement error ChimisovC.kirill.chimisov@gmail.com∗ KukushA.alexander_kukush@univ.kiev.ua Taras Shevchenko National University of Kyiv, Kyiv, Ukraine Corresponding author. 2014 2762014111332 1632014 2952014 562014 © 2014 The Author(s). Published by VTeX2014 Open access article under the CC BY license.

Cox proportional hazards model is considered. In Kukush et al. (2011), Journal of Statistical Research, Vol. 45, No. 2, 77–94 simultaneous estimators λn(·) and βn of baseline hazard rate λ(·) and regression parameter β are studied. The estimators maximize the objective function that corrects the log-likelihood function for measurement errors and censoring. Parameter sets for λ(·) and β are convex compact sets in C[0,τ] and Rk , respectively. In present paper the asymptotic normality for βn and linear functionals of λn(·) is shown. The results are valid as well for a model without measurement errors. A way to compute the estimators is discussed based on the fact that λn(·) is a linear spline.

 Asymptotic normality of estimators classical measurement error Corrected Maximum Likelihood Estimator Cox proportional hazards model estimator of baseline hazards function 62N02 62N01 62J12 Introduction

We deal with Cox proportional hazards model where a lifetime T≥0 has the following intensity function λ(t|X;λ,β)=λ(t)exp(βTX),t≥0. Here we say that positive random variable ξ has intensity function λ˜(·) if λ˜(t)=limh→0+h−1P{t≤ξ<t+h|ξ≥t},t≥0.

In (1.1) covariate X is a random vector distributed in Rk , λ(·)∈Θλ⊂C[0,τ] is the baseline hazard function and β is a parameter from Θβ⊂Rk . We observe only censored value Y:=min{T,C} , where censor C is distributed in [0,τ] . Survival function of C, GC(u)=1−FC(u) , is unknown but we know τ. Censorship indicator Δ:=I{T≤C} is observed as well. X is not observed directly, instead a surrogate data W=X+U is observed, where U has known and finite moment generating function MU(β):=EeβTU . Here E stands for expectation. A couple (T,X) , censor C and measurement error U are stochastically independent. We mention that recently measurement error models become quite popular, e.g., in [9] an autoregressive model with measurement error was studied.

Consider independent copies of the model (Xi,Ti,Ci,Yi,Δi) , i=1,…,n . Based on (Yi,Δi,Wi) , i=1,…,n , we estimate true values of β and λ(·) that we denote by β0 and λ0(·) , respectively. The latter is estimated on [0,τ] only.

There are a lot of papers on estimation of β0 and cumulative hazard Λ(t)=∫0tλ(t)dt . In [1] general ideas are presented based on partial likelihood. Same model but with measurement errors is considered in [4], where, based on Corrected Score method, consistent and asymptotically normal estimators are constructed for regression parameter and cumulative hazard function. Another approach is proposed in [6] where doubly censored data are considered without measurement error. Here cumulative hazard is estimated, and strong consistency and asymptotic normality of maximum likelihood estimators are proven. However, sometimes it is necessary to know the behaviour of baseline hazard function λ(·) itself, not cumulative hazard (see [10]). Our model is presented in [2] and [5] where baseline hazard function is assumed to belong to a parametric space while we consider λ(·) from a compact set of C[0,τ] .

If values of Xi were measured without measurement error, we could use Maximum Likelihood Estimator (MLE) which maximizes the log-likelihood function Q˜n(λ,β):=1n ∑i=1nq(Yi,Δi,Xi;λ,β), where q˜(Y,Δ,X;λ,β)=Δ(logλ(Y)+βTX)−eβTX∫0Yλ(u)du.

Since Xi is contaminated, we have to correct our objective function for measurement error. Due to suggestion of Augustin [2] we construct a new objective function q such that E[q(Yi,Δi,Wi;λ,β)|Yi,Δi,Xi]=q˜(Yi,Δi,Xi;λ,β)a.s.

Then the corrected log-likelihood function is Qn(λ,β):=1n ∑i=1nq(Yi,Δi,Wi;λ,β), where q(Y,Δ,W;λ,β)=Δ(logλ(Y)+βTW)−eβTWMU(β)∫0Yλ(u)du.

As an estimator of true parameters (λ0,β0) , we use a couple (λn,βn) which maximizes (1.2).

Introduce further assumptions.

Θλ={f:[0,τ]→R|f(t)≥a,∀t∈[0,τ],f(0)≤A,|f(t)−f(s)|≤L|t−s|,∀t,s∈[0,τ]} , where a>0,A>a 0,\hspace{2.5pt}A>a$]]> and L>0 0$]]> are fixed constants.

Θβ is a compact and convex set in Rk .

EU=0 and for some ε>0 0$]]>, E[e2D‖U‖]<∞whereD:=maxβ∈Θβ‖β‖+ε.

E[e2D‖X‖]<∞ where D>0 0$]]> is defined in (iii).

τ is right endpoint of the distribution of C, i.e., P{C>τ}=0 \tau \}=0$]]> and for all ε>0 0$]]>, P{C>τ−ε}>0 \tau -\varepsilon \}>0$]]>.

The covariance matrix of random vector X is positive definite.

β0 is an interior point of Θβ .

λ0∈Θλε for some ε>0 0$]]>, where Θλε:={f:[0,τ]→R|f(t)≥a+ε,∀t∈[0,τ],f(0)≤A−ε,|f(t)−f(s)|≤(L−ε)|t−s|,∀t,s∈[0,τ]} .

P{C>0}=1 0\}=1$]]>.

Assumptions (i) to (ix) allow us to consider model without measurement error. One just has to set Ui=0 and MU(β)=1 . All results of the article are valid for this case as well.

In [7] the strong consistency of (λn,βn) is proven and the rate of convergence is presented. Our goal is to provide asymptotic normality for βn and λn . The paper is organised as follows. Section 2 states the main results on the asymptotic normality. Section 3 suggests the procedure for computation of the estimates. Section 4 proves the stochastic boundedness results. Section 5 proves auxiliary results, Section 6 gives the proof of the main result, and Section 7 concludes.

For a sequence of random variables {xn} , notation xn=Op(1) means that {xn} is stochastically bounded. We assume that censor C has pdf fC (this is a technical assumption that can be easily avoided). According to [7], Section 3, conditional density of (Y,Δ) given X at point (λ0,β0) equals f(y,δ|X)=fTδ(y|X)GT1−δ(y|X)fC1−δGCδ(y), where fT is conditional pdf of T given X and GT is conditional survival function: fT(t|X)=Λ(t|X;λ0,β0)exp(−∫0tλ(s|X;λ0,β0)ds),GT(t|X)=exp(−∫0tλ(s|X;λ0,β0)ds).

Let Z be a normed linear space. For a function f:Z→R we denote f(n)(x0) its n-th Fréchet derivative at a point x0∈Z . f(n)(x0) is n-linear form and for h1,…,hn∈Z we denote ⟨f(n)(x0),(h1,…,hn)⟩ the action of f(n)(x0) . If h1=⋯=hn we simply write ⟨f(n)(x0),(h1)n⟩ where it does not cause ambiguity. If a functional F acts on a product space Z1×Z2 then elements of this space are denoted as (h1,h2)∈Z1×Z2 and ⟨F,(h1,h2)⟩ stands for the action of F on (h1,h2) . For x,y∈Z , the following set is called an interval that connects x and y [x,y]={αx+(1−α)y|α∈[0,1]}.

Main result

We make some more notations. Let a(u)=E[Xeβ0TXGT(u|X)],b(u)=E[eβ0TXGT(u|X)],p(u,x)=exp(β0TX)GT(u|X),T(u)=E[XXTp(u,x)]E[p(u,x)]−E[Xp(u,x)]E[XTp(u,x))].

Denote A=E[XXTexp(β0TX)∫0Yλ0(u)du],M=∫0τT(u)K(u)GC(u)du, where K(u)=λ0(u)b(u) . Also introduce a sequence of random vectors ξn:= ∑i=1nζi, with i.i.d. summands ζi=−Δia(Yi)b(Yi)+exp(β0TWi)MU(β0)∫0Yia(u)K(u)du+∂q∂β(Yi,Δi,Wi,β0,λ0).

Let Σβ=4Cov(ζ1) , m(φλ)=∫0τφλ(u)a(u)GC(u)du , Σφ2=4Var[⟨q′(Y,Δ,W,λ0,β0),φ⟩] with φ=(φλ,φβ)∈C[0,τ]×Rk .

Assume conditions (i) to (ix). Then M is invertible and n(βn−β0)→dNk(0,M−1ΣβM−1).

Moreover, for any Lipschitz continuous function f on [0,τ] , n∫0τ(λn−λ0)(u)f(u)GC(u)du→dN(0,σφ2(f)) where σφ2(f)=σφ2 with φ=(φλ,φβ) , φβ=−A−1m(φλ) , and φλ is a unique solution to the Fredholm’s integral equation φλK(u)−aT(u)A−1m(φλ)=f(u).

Let 0<ε<τ . Assume that 1GC is Lipschitz continuous on [0,τ−ε] . Under conditions (i) to (ix), for any Lipschitz continuous function f on [0,τ] with support on [0,τ−ε] , n∫0τ−ε(λn−λ0)(u)f(u)du→dN(0,σφ2(f)) where σφ2(f)=σφ2 with φ=(φλ,φβ) , φβ=−A−1m(φλ) , and φλ is a unique solution to the Fredholm’s integral equation φλK(u)−aT(u)A−1m(φλ)=f(u)GC(u). Here by definition f(τ)GC(τ)=0 .

Note that the corollary immediately follows from the theorem after f is substituted by fGC .

 Computation of estimators

Since Θλ is infinite-dimensional, computation of (λn,βn) is not a parametric problem in general setting. We refer to the ideas of I.J. Schoenberg [11]. We will show that maximum of (1.2) is attained on a linear spline with nodes located at points Yi , i=1,…,n and some other points that can be calculated.

Let i1,…,in∈1,…,n be such a numbering that Yi1≤⋯≤Yin , i.e., (Yi1,…,Yin) is a variational series of (Y1,…,Yn) . Alongside with (λn,βn) we consider (λ‾n,βn) , where λ‾n is the following function. We set λ‾n(Yik)=λ(Yik) , k=1,…,n . For each interval [Yik,Yik+1] , k=1,…,n−1 , perform the next procedure. Draw straight lines Lik1(t)=λ(Yik)+L(Yik−t) and Lik2(t)=λ(Yik+1)+L(t−Yik+1), where L is defined in (i).

Denote Bik the intersection of Lik1(t) and Lik2(t) . Bi0:=0 , Bin:=τ , Yi0:=0 , Yin+1:=τ . We set λ‾n(t)=max{Lik1(t),a}ift∈[Yik,Bik],max{Lik2(t),a}ift∈[Bik,Yik+1].

Note that λn≥λ‾n because λn∈Θλ . Then ∫YikYik+1λn(u)du≥∫YikYik+1λ‾n(u)du.

Thus, one can easily see that Qn(λn,βn)≤Qn(λ‾n,βn) implying λn=λ‾n so that we conclude with the following statement.

Under conditions (i) and (ii), function λn that maximizes Qn is a linear spline constructed in (3.3).

Using maximization in (3.3) makes computation of (λn,βn) inconvenient. Thus, we propose to modify the estimators. As soon as condition (viii) is satisfied and estimator (λn,βn) is strongly consistent, one can induce that eventually λ‾(Bik)>a a$]]>, and thus, eventually there is no need in finding maximum in (3.3). Therefore, instead of (λn,βn) we propose to consider a couple (λˆn,βˆn) with βˆn∈Θβ that maximizes Qn under restrictions:

λˆn(0)≤A .

λˆn(Yik)≥a , k=1,…,n .

λˆn(Yik)+L(Yik−Yik+1)≤λˆn(Yik+1)≤λˆn(Yik)−L(Yik−Yik+1) , k=1,…,n−1 .

λˆn(t):=Lik1(t)ift∈[Yik,Bik],Lik2(t)ift∈[Bik,Yik+1],k=1,…,n−1 .

λˆn(t):=Li02(t)ift∈[0,Yi1],Lin1(t)ift∈[Yin,τ].

Evaluating (λˆn,βˆn) is a parametric problem. We mention that eventually (λˆn,βˆn)=(λn,βn) . We summarise with the next statement.

Assume conditions (i) to (ix). Then estimator (λˆn,βˆn) is strongly consistent and statements of Theorem 1 and Corollary 2 hold true for that estimator.

 Stochastic boundedness of transformed and normalized estimators

Assume (i) to (vi). Then n4‖βn−β0‖=Op(1),n∫0τ(λn(u)−λ0(u))2GC(u)du=Op(1).

The proof is based on the three lemmas. Using integration by parts one can easily prove the following.

For all u∈[0,τ] ∫uτ(fC(y)GT(y|X)+fT(y|X)GC(y))dy=GT(u|X)GC(u)=:G(u|X).

Crucial step of the proof of Theorem 5 is the following.

There exists a closed bounded set A such that μX(A):=P(X∈A)>0 0$]]> and that the identity (vTx−c)IA(x)≡0 , for some v∈Rk , c∈R , implies v=0 and c=0 .

Denote by M the support of μX , so that M is minimal closed set with μX(M)=μX(Rk) . Since μX is not concentrated on a hyperplane due to the condition (vi), there are at least k+1 distinct points m1,…,mk+1 that belong to M and do not lie on a hyperplane. Consider a closed ball B‾(0,r) with radius r>max{‖m1‖,…,‖mk+1‖} \max \{\| m_{1}\| ,\dots ,\| m_{k+1}\| \}$]]>. Now one can take A=M∩B‾(0,r) and make sure that A has all desired properties.  □

Let An(ω) be a collection of assertions (here ω stands for elementary event). We say that {An} hold eventually if for almost all ω there exists Nω such that for all n>Nω N_{\omega }$]]>, An(ω) holds.

Let ηn , ξn be two sequences of random variables, ηn be stochastically bounded, and eventually |ξn|≤|ηn| . Then ξn is stochastically bounded as well.

Step 1. Denote q∞(λ,β)=E[q˜(Y,Δ,W,λ,β)]=E[q˜(Y,Δ,X,λ,β)] . Let us show that (q∞)′ exists for (λ,β)∈B and equals zero at the true point (λ0,β0) , where B is some open set in Rk×C[0,τ] that contains Θβ×Θλ .

Using (iv) one can easily obtain that ∂q∞∂β(λ,β)=E[ΔX−Xexp(βTX)∫0Yλ(u)du],⟨∂q∞∂λ(λ,β),h⟩=E[Δh(Y)λ(Y)−exp(βTX)∫0Yh(u)du], where h∈C[0,τ] . Hence, (q∞)′ exists. According to [7], Section 3 q∞(λ,β)<q∞(λ0,β0) for all (λ,β)≠(λ0,β0) , (λ,β)∈B . Hence, (q∞)′(λ0,β0)=0.

In fact, condition (iv) implies that (q∞)″ and (q∞)‴ exist. Hence, third order Taylor’s formula holds, q∞(λn,βn)−q∞(λ0,β0)=12⟨(q∞)″(λ0,β0),(λn−λ0,βn−β0)2⟩+16⟨(q∞)‴(λ˜n,β˜n),(λn−λ0,βn−β0)3⟩, where (λ˜n,β˜n) belongs to interval [(λn,βn),(λ0,β0)] .

Step 2. We transform (q∞)″ and show that −(q∞)″(λ0,β0) is a positive definite operator. We have ⟨∂2q∞(λ0,β0)∂λ2,(h1,h2)⟩=−E[Δλ02(Y)h1(Y)h2(Y)],∂2q∞∂β2(λ0,β0)=−E[XXTexp(β0TX)∫0Yλ0(u)du],⟨∂2q∞(λ0,β0)∂λ∂β,(hλ,hβ)⟩=−E[(hβTX)exp(β0TX)∫0Yhλ(u)du], where h1,h2,hλ∈C[0,τ] , hβ∈Rk .

We use (1.4) and Lemma 6 for further transformations: ⟨∂2q∞(λ0,β0)∂λ2,(hλ,hλ)⟩=−E[Δλ02(Y)hλ2(Y)]=E(∫0τhλ2(u)λ02(u)fT(u|X)GC(u)du)=E(∫0τhλ2(u)λ02(u)λ0(u)exp(β0TX)exp(−∫0uΛ(s|X;λ0,β0)ds)GC(u)du)=E(∫0τhλ2(u)λ0(u)exp(β0TX)GT(u|X)GC(u)du).

Next, ⟨∂2q∞(λ0,β0)∂β2,(hβ,hβ)⟩=−E[(hβTX)2exp(β0TX)∫0Yλ0(u)du]=−E[(hβTX)2exp(β0TX)(∫0τ(∫0yλ0(u)dufT(y|X)GC(y)+∫0yλ0(u)dufC(y)GT(y|X))dy)]=−E[(hβTX)2exp(β0TX)∫0τλ0(u)∫uτ(fC(y)GT(y|X)+fT(y|X)GC(y))dydu]=−E[(hβTX)2exp(β0TX)∫0τλ0(u)GT(u|X)GC(u)du].

At last, ⟨∂2q∞(λ0,β0)∂λ∂β,(hλ,hβ)⟩=−E[(hβTX)exp(β0TX)∫0Yhλ(u)du]=−E[(hβTX)exp(β0TX)∫0τhλ(u)∫uτ(fC(y)GT(y|X)+fT(y|X)GC(y))dydu]=−E[(hβTX)exp(β0TX)∫0τhλ(u)GT(u|X)GC(u)du].

Hence, from (4.2) to (4.4) it follows that ⟨(q∞)″(λ0,β0),(hλ,hβ)2⟩=−E[exp(β0TX)∫0τ((hβTX)λ0(u)G(u|X)+hλ(u)G(u|x)λ0(u))2du].

Now, condition (vi) implies that −(q∞)″ is positive definite at (λ0,β0) , i.e., ⟨(q∞)″(λ0,β0),(hλ,hβ)2⟩=0⟺(hλ,hβ)=(0,0).

Indeed, if to assume that ⟨(q∞)″(λ0,β0),(hλ,hβ)2⟩=0 and (hλ,hβ)≠(0,0) then (4.5) implies that hβ≠0 and (hβTX)=consta.s. We get a contradiction with (vi).

Step 3. We show that there exist such C>0 0$]]> and δ>0 0$]]> that, whenever max{‖hβ‖2,∫0τhλ2(u)G0(u)Cλ0du}>0 0$]]>, it holds E[−⟨(q˜)″(Y,Δ,X,λ0,β0),(hλ,hβ)2⟩max{‖hβ‖2,∫0τhλ2(u)G0(u)Cλ0du}]≥δ.

Note that G(u|X) is continuous in X. Denote G0(u)=minX∈AG(u|X) , where A is a set from Lemma 7. Note that G0(u)=G(u|X0)>0 0$]]>, for all u∈[0,τ) and some  X0 .

Assume that ‖hβ‖2≥∫0τhλ2(u)G0(u)Cλ0du . Jensen’s inequality and (4.5) yield −⟨(q∞)″(λ0,β0),(hλ,hβ)2⟩≥1τE[IX∈Aexp(β0TX)(∫0τ(hβTX)λ0(u)G0(u)+hλ(u)G0(u)λ0(u)du)2]=1τE[IX∈Aexp(β0TX)((hβTX)∫0τλ0(u)G0(u)du+∫0τhλ(u)G0(u)λ0(u)du)2].

Denote a0=minX∈A1τexp(β0TX),a1=∫0τλ0(u)G0(u)du,Kλ(hβ)=∫0τhλ(u)G0(u)λ0(u)du‖hβ‖.

Inequality (4.7) implies that E[−⟨(q˜)″(Y,Δ,X,λ0,β0),(hλ,hβ)2⟩max{‖hβ‖2,∫0τhλ2(u)G0(u)Cλ0du}]≥a0E[IX∈A((hˆβTX)a1+Kλ)2], where hˆβ=hβ‖hβ‖ . Fix T∈R . Equality E[IX∈A((hˆβTX)a1+T)2]=0 implies that IX∈A((hˆβTX)+Ta1)=consta.s. , which contradicts to the choice of A.

It is easy to see that for a fixed hˆβ , minimum of E[IX∈A((hˆβTX)a1+T)2] is attained at a unique point T=T(hˆβ,A) . Moreover, T(hˆβ,A) is a continuous function of hˆβ . Hence, we have E[−⟨(q˜)″(Y,Δ,X,λ0,β0),(hλ,hβ)2⟩max{‖hβ‖2,∫0τhλ2(u)G0(u)Cλ0du}]≥a0E[IX∈A((hˆβTX)a1+T(hˆβ,A))2]>0. 0.\end{array}\]]]>

Due to ‖hˆβ‖=1 , the right hand side of (4.8) attains its minimum at some point hˆβ0 . Now one can take δ1=a0E[IX∈A((hˆβ0TX)a1+T(hˆβ0,A))2]>0. 0.\]]]>

Consider the second case, where inequality ‖hβ‖2<∫0τhλ2(u)G0(u)Cλ0(u)du holds. Transform right hand side of (4.5): E[exp(β0TX)∫0τ((hβTX)λ0(u)G(u|X)+hλ(u)G(u|x)λ0(u))2du]≥E[IX∈Aexp(β0TX)((hβTX)2∫0τλ0(u)G0(u)du+2∫0τ(hβTX)hλ(u)G0(u)du+∫0τhλ2(u)G0(u)λ0(u)du)].

Denote Φ=∫0τhλ2(u)G0(u)Cλ0(u)du . Hence, the left hand side of (4.6) is transformed to E[−⟨(q˜)″(Y,Δ,X,λ0,β0),(hλ,hβ)2⟩Φ]≥E[IX∈Aexp(β0TX)((h˜βTX)2a2+2Φ∫0τ(hβTX)hλ(u)G0(u)du+C)], where a2=∫0τλ0(u)G(u|X)du,h˜β=hβ∫0τhλ2(u)G0(u)Cλ0(u)du.

Jensen’s inequality implies Φ1/2≥1τC(∫0τ|hλ(u)|G0(u)λ0(u)du).

Since Φ>‖hβ‖ \| h_{\beta }\| $]]>, G0(u)∈[0,1] and λ0 is bounded away from 0, we have |∫0τ(hβTX)hλ(u)G0(u)duΦ|≤‖hβ‖Φ‖X‖|τ1/2∫0τhλ(u)G0(u)du∫0τ|hλ(u)|G0(u)λ0(u)du|C≤C‖X‖D, for some constant D>0 0$]]> which depends only on τ and λ0 . Since ‖h˜β‖<1 , there exist constants K1>0 0$]]>, K2>0 0$]]> that satisfy E[−⟨(q˜)″(Y,Δ,X,λ0,β0),(hλ,hβ)2⟩Φ]≥τa0(−K1−CK2+C).

Choosing C large enough, we get (4.6).

Step 4. Now transform Taylor’s decomposition (4.1): q∞(λn,βn)−q∞(λ0,β0)=E(max{‖hβn‖2,∫0τhλn2(u)G0(u)Cλ0du}[12⟨(q˜)″(Y,Δ,X,λ0,β0),(hλn,hβn)2⟩max{‖hβn‖2,∫0τhλn2(u)G0(u)Cλ0du}+16⟨(q˜)‴(Y,Δ,X,λ˜n,β˜n),(hλn,hβn)3⟩max{‖hβn‖2,∫0τhλn2(u)G0(u)Cλ0du}]), where we denote hλn=λn−λ0 and hβn=βn−β0 . Remember that GT(u|X)∈(0,1] for all X, so that G0(u)≥K3GC(u) for some K3>0 0$]]>. One can see that ∂3q∞∂λ2∂β=0 . Using the same technique as in (4.2)–(4.4) and the assumptions, we get ⟨∂3q∞(λ˜n,β˜n)∂λ3,(h1,h2,h3)⟩=12E[Δλ˜n3(Y)h1(Y)h2(Y)h3(Y)]=12E(∫0τh1(u)h2(u)h3(u)λ˜n2(u)exp(β˜nTX)GT(u|X)GC(u)du)≤K4‖h1‖∫0τh2(u)h3(u)G0(u)du, ⟨∂3q∞(λ˜n,β˜n)∂β3,hβ⟩=−E[(hβTX)3exp(β˜nTX)∫0Yλ˜ndu]≤K5‖hβ‖3, ⟨∂3q∞(λ˜n,β˜n)∂λ∂β2,(hλ,hβ,hβ)⟩=−E[(hβTX)2exp(β˜nTX)∫0Yhλ(u)du]≤K6‖hβ‖2‖hλ‖ where K4 to K6 are positive constants. We note that all constants K3 to K6 depend only on Θ=Θλ×Θβ . Kukush et al. [7] prove strong consistency of the estimator (λn,βn) , that is maxt∈[0,τ]|λn(t)−λ0(t)|→0 and βn→β0 a.s., as n→∞ . One can conclude that limn→∞E[⟨(q˜)‴(Y,Δ,X,λ˜n,β˜n),(hλn,hβn)3⟩max{‖hβn‖2,∫0τhλn2(u)G0(u)Cλ0du}]=0a.s.

Step 5. Set Sn(λ,β)=n(Qn(λ,β)−q∞(λ,β)) . Kukush et al. [7] prove that under assumptions (i) to (vi) Sn(λ,β)n converges in distribution in C(Θ) to a Gaussian measure. Hence, 0≤n(q∞(λ0,β0)−q∞(λn,βn)≤n(Qn(λn,βn)−q∞(λn,βn)−Qn(λ0,β0)+q∞(λ0,β0))≤2nsup(λ,β)∈Θλ×Θβ|Qn(λ,β)−q∞(λ,β)|=Op(1), because q∞(λ,β) and Qn(λ,β) attain their maximums at (λ0,β0) and (λn,βn) , respectively.

Now, (4.9) yields nmax{‖hβn‖2,∫0τhλ2(u)G0(u)Cλ0du}E([12⟨(q˜)″(Y,Δ,X,λ0,β0),(hλn,hβn)2⟩max{‖hβn‖2,∫0τhλ2(u)G0(u)Cλ0du}+16⟨(q˜)‴(Y,Δ,X,λ˜n,β˜n),(hλn,hβn)3⟩max{‖hβn‖2,∫0τhλ2(u)G0(u)Cλ0du}])=n(q∞(λ0,β0)−q∞(λn,βn))=Op(1).

Step 6. Equations (4.6), (4.9) and (4.13) imply that eventually nmax{‖hβn‖2,∫0τhλ2(u)G0(u)Cλ0du}<n(q∞(λ0,β0)−q∞(λn,βn))δ/3.

Lemma 8 proves that nmax{‖hβn‖2,∫0τhλn2(u)G0(u)Cλ0du}=Op(1) . Hence the first equation of Theorem 5 is proved: n‖βn−β0‖2=n‖hβn‖2=Op(1).

Finally, G0(u)≥K3GC(u) . Note that λ0 is bounded away from 0 on [0,τ] . Hence n∫0τhλn2(u)GC(u)du=Op(1).

Thus, Theorem 5 is proved.  □

 Auxiliary results

We use the ideas of [3].

Let θn=(λn,βn) , θ0=(λ0,β0) , Θ=Θλ×Θβ . Denote φ=(φλ,φβ) an admissible shift such that there exists δ>0 0$]]> with θ0±δφ∈Θ . We demand that (vii)–(viii) hold. Note that φ can be a random element and depend on n. However, ‖φ‖  should be bounded from above a.s.

Consider the function f(t)=Qn(θn+t(θ0−θn±δφ)) , 0≤t≤1 . It is well-defined (due to the convexity of Θ) and attains its maximum at point t=0 . Therefore, ⟨Qn′(θn),θ0−θn±δφ⟩≤0 and |⟨Qn′(θn),φ⟩|≤1δ⟨Qn′(θn),Δθn⟩, where Δθn:=θn−θ0 .

Taylor’s expansion at point (λ0,β0) implies |⟨Qn′(θ0),φ⟩+12⟨Qn″(θ0),(Δθn,φ)⟩+16⟨Qn‴(θ˜n),(Δθn2,φ)⟩|≤1δ(⟨Qn′(θ0),Δθn⟩+12⟨Qn″(θ0),Δθn2⟩+16⟨Qn‴(θˆn),Δθn3⟩), for some θˆn and θ˜n from interval [θ0,θn] .

Under conditions (i) to (viii) for every admissible shift φ, one has that n⟨Qn″(θ0),(Δθn,φ)⟩ and n⟨q∞″(θ0),(Δθn,φ)⟩ are stochastically bounded.

Relying on this proposition we will be able to show that n‖βn−β0‖ and n∫0τ(λn−λ0)(u)GC(u)du are stochastically bounded and then prove the asymptotic normality of n⟨Qn″(θ0),(Δθn,φ)⟩ .

Denote Θ−=Θ−Θ . It is clear that it is compact and convex. Before proving the proposition, we show the following.

Under conditions (i) to (viii), nQn′(θ0) and n(Qn″(θ0)−q∞″(θ0)) converge in distribution in C(Θ−) and C(Θ−2) , respectively. Moreover, for all θ=(λ,β)∈Θ , n(∂3Qn∂λ3(θ)−∂3q∞∂λ3(θ)) converges in distribution in C(Θ−3) .

Here only convergence for nQn′(θ0) will be shown, because for n(Qn″(θ0)−q∞″(θ0)) and n(∂3Qn∂λ3(θ)−∂3q∞∂λ3(θ)) the proof is similar. We note that q∞′(θ0)=0 and due to conditions (iii)–(iv) we have E[supβ∈Θβe2βTX‖X‖k]<∞ and E[supβ∈Θβe2βTU‖U‖k]<∞ , for any k∈N .

For (λ,β)∈Θ− let g(λ,β,Y,Δ,W)=⟨q′(Y,Δ,W,λ0,β0),(λ,β)⟩ and ρ((λ1,β1),(λ2,β2))=supu∈[0,τ]|λ1(u)−λ2(u)|+‖β1−β2‖.

(Θ−,ρ) is a compact metric space. We denote by Lip(ρ) a subspace of Lipschitz continuous functions on Θ− with respect to the metric ρ and by ‖·‖ρ the norm induced by ρ, that is for some fixed point (λ∗,β∗)∈Θ− and for all l∈Lip(ρ) we define: ‖l‖ρ:=sup(λ1,β1)≠(λ2,β2)|l(λ1,β1)−l(λ2,β2)|ρ((λ1,β1),(λ2,β2))+l(λ∗,β∗).

We apply Theorem 2 from [12]. It states that nQn′(θ0) converges in distribution in C(Θ−) under the following conditions:

P(g∈Lip(ρ))=1 .

E‖g‖ρ2<∞ .

∫0+H12(Θ−,u)du<∞ , where H is ε-entropy on (Θ−,ρ) , i.e. H(Θ−,u)=log2N(Θ−,u) , where N is a minimal number of balls with diameter not exceeding 2ε that cover Θ− .

Let Θλ−=Θλ−Θλ and Θβ−=Θβ−Θβ , so that Θ−=Θλ−×Θβ− . Consider Θλ− and Θβ− as compact metric spaces with uniform and Euclidean norm, respectively. Then for N(Θ−,2u)≤N(Θβ−,u)N(Θλ−,u) , (3) is equivalent to

∫0+H12(Θλ−,u)du<∞ , and

∫0+H12(Θβ−,u)du<∞ .

 Since Θβ−⊂Rk , we have N(Θβ−,u)<Cuk for some constant C>0 0$]]>, and (3.2) is fulfilled. Note that Θλ− can be considered as a set of Lipschitz continuous functions that map compact connected space [0,τ] into some interval in R . Lemma 1 from [8] implies H(Θλ−,u)≥1+H(Θλ−,4u), so that Θλ− is of “uniform type” (see [8]). According to Theorem 1 from [8] there exists such constant C that H(Θλ−,4Lε)≤CN([0,τ],ε). For the space R1 we have that N([0,τ],u)<C˜1u for some constant C˜ . Hence (3.1) holds.

To verify (1) and (2) note that g(λ,β,Y,Δ,W)=Δλ(Y)λ0(Y)−eβ0TWMU(β0)∫0Yλdu+ΔβTW+βT(MU(β0)W−E(Ueβ0TU))eβ0TWMU2(β0)∫0Yλ0du, and conditions (i)–(ii) imply sup(λ,β)∈Θ−‖g′(λ,β,Y,Δ,W)‖<∞, where g′ is considered as a bilinear operator on C[0,τ]×Rk . Hence, condition (1) is fulfilled. Moreover, there exists such a constant K>0 0$]]> that ‖g(λ,β,Y,Δ,W)‖ρ<K(1+‖W‖+eD‖W‖+‖W‖eD‖W‖) and due to conditions (iii) and (iv), condition (2) is also satisfied. Thus, lemma is proved.  □

Returning to inequality (5.1), because Δθn converges to zero a.s., one can conclude the following.

n⟨Qn′(θ0),φ⟩=Op(1) and ⟨nQn′(θ0),Δθn⟩=op(1) , where op(1) means convergence to zero in probability.

n(∂3Qn∂λ3(θ)−∂3q∞∂λ3(θ)) converges in probability in C(Θ−3) . Inequality (4.10) implies that n⟨∂3q∞∂λ3(θ),(Δθn2,φ)⟩ is stochastically bounded, so is n⟨∂3Qn∂λ3(θ),(Δθn2,φ)⟩ .

n⟨(Qn″(θ0)−q∞″(θ0)),Δθn2⟩ and n⟨(Qn″(θ0)−q∞″(θ0)),(Δθn,φ)⟩ converge to zero in probability. Note that ⟨nQn″(θ0),Δθn2⟩=Op(1) if and only if n⟨q∞″(θ0),Δθn2⟩=Op(1) . The latter equality can be easily derived from Theorem 5, formula (4.1) and convergence (4.13).

To prove the first part of the proposition one has to show that ⟨Qn‴(θˆn),Δθn3⟩=Op(1)n and ⟨Qn‴(θ˜n),(Δθn2,φ)⟩=Op(1)n.

It is clear that (5.3) yields (5.2). After a series of computations one can induce that for some constants C1>0,C2>0 0,\hspace{2.5pt}C_{2}>0$]]> |⟨∂3q(Y,Δ,W,λ,β)∂β3,(hβ)3⟩|≤C1eD‖W‖‖hβ‖3|⟨∂3q(Y,Δ,W,λ,β)∂λ∂β2,(hβ,hβ,hλ)⟩|≤C2eD‖W‖‖hβ‖2‖hλ‖.

Expectations of right hand sides of inequalities in (5.4) are finite. Together with n‖βn−β0‖2=Op(1) and SLLN, this implies that ⟨∂Qn3(θ˜n)∂β3,(Δθn2,φ)⟩ and ⟨∂Qn3(θ˜n)∂β2∂λ,(Δθn2,φ)⟩ are Op(1)n . Noting that ∂Qn3(θ˜n)∂β∂λ2=0 , one can conclude that the first part of the proposition will be proven if one shows that ⟨∂Qn3(θ˜n)∂λ3,(Δθn2,φ)⟩=⟨∂Qn3(λ˜n)∂λ3,((λn−λ0),(λn−λ0),φλ)⟩=1n ∑i=1nΔi(λn−λ0)2(Yi)φλ(Yi)λ˜n3(Yi)=Op(1)n.

From the definition of admissible shifts, φλ belongs to Θ. Since λ˜n is bounded away from zero, there is a constant C such that for the second summand we have |⟨∂Qn3(θ˜n)∂λ3,(Δθn2,φ)⟩|≤C|1n ∑i=1nΔi(λn−λ0)2(Yi)|=Op(1)n, where the last equality holds due to the conclusion (b). Thus, (5.5) holds. This completes the proof of the first part of the proposition. The second part is easily derived from conclusion (c).  □

n‖βn−β0‖=Op(1),n∫0τ(λn−λ0)(u)GC(u)du=Op(1).

Let hβ=βn−β0 , hλ=λn−λ0 . Take some admissible shift φ:=(φλ,φβ) . For this shift one has −⟨q∞″(θ0),(Δθn,φ)⟩=E[(hβTX)(φβTX)exp(β0TX)∫0Yλ0(u)du]+E[(φβTX)exp(β0TX)∫0Yhλ(u)du]+E[Δλ02(Y)hλ(Y)φλ(Y)]+E[(hβTX)exp(β0TX)∫0Yφλ(u)du]=Op(1n).

The idea is to find such φλ that E[(φβTX)exp(β0TX)∫0Yhλ(u)du]+E[Δλ02(Y)hλ(Y)φλ(Y)]=0.

Then after some calculations (using Lemma 6) one can see that (5.7) is equivalent to ∫0τhλφβTa(u)GC(u)du+∫0τhλλ0φλb(u)GC(u)du=0.

One can take φλ(u):=−φβTa(u)b(u)λ0(u) as a solution to (5.8). Since GT(u|X) is differentiable function of u, one can conclude that φλ is an admissible shift for ‖φβ‖ small enough.

Equation (5.6) is now equivalent to ∫0τhβTT(u)φβλ0(u)GC(u)b(u)du=Op(1n).

Using Hölder’s inequality and condition (vi), one can easily see that T(u) is positive definite. Now let h˜β=βn−β0‖βn−β0‖ and take φβ=h˜βC1 , where C1>0 0$]]> such that φ=(φλ,φβ) is an admissible shift. Then (5.6) can be transformed to ‖hβ‖∫0τh˜βTT(u)h˜βλ0(u)GC(u)b(u)du=Op(1n).

Since ‖h˜β‖=1/C1 , left hand side of (5.10) is greater than δ‖hβ‖ for some δ>0 0$]]>. Using Lemma 8 the first part of the corollary is proved.

If now in (5.6) one takes φ=(1C2,0) for large enough C2>0 0$]]>, then (5.6) takes form E[Δλ02(Y)hλ(Y)1C2]+E[(hβTX)exp(β0TX)∫0Y1C2du]=Op(1n).

Due to n‖βn−β0‖=Op(1) , the latter equality implies E[Δλ02(Y)hλ(Y)1C2]=Op(1n) and the second part of the corollary holds.  □

We present the main result of this section.

Under conditions (i) to (ix), for all admissible shifts the following convergence in probability holds n⟨Qn′(θ0),φ⟩+12n⟨q∞″(θ0),(Δθn,φ)⟩→P0. Moreover, if φ is a non-random admissible shift then ⟨Qn′(θ0),φ⟩→dN(0,σφ2) , where σφ2=4Var[⟨q′(Y,Δ,W,λ0,β0),φ⟩] , and n⟨Qn″(θ0),(Δθn,φ)⟩→dN(0,σφ2),n⟨q∞″(θ0),(Δθn,φ)⟩→dN(0,σφ2).

Using Corollary 11 and inequality (5.1), one can repeat the proof of Proposition 9 with a remark that stochastic boundedness should be changed for a convergence to zero in probability. We use (4.2) to (4.4) to show n⟨q∞″(θ0),Δθn2⟩=op(1) . Thus, the convergence (5.11) is proved. The rest of the proof is trivial.  □

 Proof of Theorem <xref rid="j_vmsta03_stat_002">1</xref>

We assume that condition (ix) is satisfied. Thus A is positive definite and, consequently, invertible. Since T(u) is positive definite, M is positive definite as well and therefore, invertible.

Note that due to conditions (vii) and (viii), Theorem 12 is valid for all non-random shifts φ∈Rk×Lip1([0,τ]) , where Lip1([0,τ]) is a class of Lipschitz continuous functions. Take the shift as follows: φ=(φβTa(u)K(u),φβ) , where φβ is a fixed vector of Rk . For this shift one can rewrite ⟨Qn′(θ0),φ⟩ as ⟨Qn′(θ0),φ⟩=φβTξn.

By the CLT applied to ξn one can see that the limit distribution of nξn is in fact Nk(0,14Σβ) . Note that we have already faced with the shift φ in Corollary 11. In particular, (5.9) yields that ⟨q∞″(θ0),(Δθn,φ)⟩ can be rewritten as ∫0τhβTT(u)φβK(u)GC(u)du=hβTMφβ.

Theorem 12 and Cramér-Wold’s theorem yield that hβTM→dNk(0,Σβ). Since M is invertible, the convergence (2.1) is proved.

Now, for a fixed shift φλ take such φβ that E[(hβTX)(φβTX)exp(β0TX)∫0Yλ0(u)du]+E[(hβTX)exp(β0TX)∫0Yφλ(u)du]=hβT(Aφβ+m(φλ))=0.

Hence, φβ=−A−1m(φλ) . From (5.6) it follows that −⟨q∞″(θ0),(Δθn,φ)⟩=E[(φβTX)exp(β0TX)∫0Yhλ(u)du]+E[Δλ02(Y)hλ(Y)φλ(Y)]=∫0τhλ(u)φβTa(u)GC(u)du+∫0τhλ(u)φλK(u)GC(u)du=∫0τhλ(u)(−a(u)Tφβ+φλK(u))GC(u)du.

In view of Theorem 12 and the remark at the beginning of the proof, in order to show the convergence (2.4), one should show that the equation (2.3) has a Lipschitz continuous solution φλ . But if φλ is a solution to (2.3) then φλ(u)=K(u)f(u)+K(u)aT(u)C for some constant C∈Rk and thus, is Lipschitz continuous. After substitution (6.1) in (2.3) we obtain aT(u)[C−∫0τ(f(u)+aT(u)C)A−1a(u)K(u)GC(u)du]=0.

Let S=∫0τf(u)A−1a(u)K(u)GC(u)du and P(u)=E[XXTexp(β0TX)GT(u|X)] . We show that it is possible to choose C so that C−∫0τaT(u)CA−1a(u)K(u)GC(u)du=S.

After transposing both sides and multiplying by A, we have CT(A−∫0τa(u)aT(u)K(u)GC(u)du)=STA.

Transformation of R:=A−∫0τa(u)aT(u)K(u)GC(u)du leads to R=∫0τλ0(u)(P(u)−a(u)aT(u)b(u))GC(u)du.

In the proof of Corollary 11 it was shown that P(u)−a(u)aT(u)b(u)=T(u)b(u) is a positive definite matrix. Therefore, R is positive definite and invertible. Hence, (2.3) has a unique solution and convergence (2.4) holds. This completes the proof.

 Conclusion

Here we studied properties of the Corrected MLE (λn,βn) proposed by Kukush et al. [7] in Cox proportional hazards model with measurement error. Asymptotic normality was obtained for βn and integral functionals of λn . We also present estimator (λˆn,βˆn) that inherits properties of (λn,βn) and transforms the maximization problem to a parametric one.

In future we intend to provide simulations in this model.

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