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Nonparametric latency estimation for mixture cure models


A nonparametric latency estimator for mixture cure models is studied in this paper. An i.i.d. representation is obtained, the asymptotic mean squared error of the latency estimator is found, and its asymptotic normality is proven. A bootstrap bandwidth selection method is introduced and its efficiency is evaluated in a simulation study. The proposed methods are applied to a dataset of colorectal cancer patients in the University Hospital of A Coruña (CHUAC).

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The first author’s research was sponsored by the Spanish FPU (Formación de Profesorado Universitario) Grant from MECD (Ministerio de Educación, Cultura y Deporte) with reference FPU13/01371. All the authors acknowledge partial support by the MINECO (Ministerio de Economía y Competitividad) grant MTM2014-52876-R (EU ERDF support included), the MICINN (Ministerio de Ciencia e Innovación) Grant MTM2011-22392 (EU ERDF support included) and Xunta de Galicia GRC Grant CN2012/130. The authors are grateful to Dr. Sonia Pértega and Dr. Salvador Pita, at the University Hospital of A Coruña, for providing the colorectal cancer data set.

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Correspondence to Ana López-Cheda.



Proof of Theorem 1

The nonparametric estimator of \(S_0(t|x)\) in (4) can be decomposed as follows:

$$\begin{aligned} \hat{S}_{0,h}(t|x)-S_{0}(t|x)=A_{11}+ A_{21} +A_{12}+ A_{22}, \end{aligned}$$

where the dominant terms of the i.i.d. representation of \(\hat{S}_{0,h}(t|x)\) derive from

$$\begin{aligned} A_{11} =\frac{\hat{S}_{h}(t|x)-S(t|x)}{p(x)}\quad \text { and } \quad A_{21} =\frac{1-S(t|x)}{p^{2}(x)}(\hat{p}_{h}(x)-p(x)), \end{aligned}$$

and the remaining terms

$$\begin{aligned} A_{12} =\frac{(\hat{S}_{h}(t|x)-S(t|x))(p(x)-\hat{p}_{h}(x))}{\hat{p}_{h}(x)p(x)}\quad \text { and }\quad A_{22}= \frac{S(t|x)-1}{p^2(x)}\frac{\left( \hat{p} _{h}(x)-p(x)\right) ^{2}}{\hat{p}_{h}(x)} \end{aligned}$$

will be proved to be negligible.

The i.i.d. representation of the term \(A_{11}\) in (20) follows, under assumptions (A1)–(A7), (A11) and (A12), from that of \(\hat{S}_h(t|x)\) in Theorem 2 of Iglesias-Pérez and González-Manteiga (1999):

$$\begin{aligned} A_{11}=-\frac{S(t|x)}{p(x)} \sum _{i=1}^{n}\tilde{B}_{h,i}(x)\xi (T_{i},\delta _{i},t,x)+O\left( \left( \frac{\ln n}{nh}\right) ^{3/4}\right) \text { a.s.} \end{aligned}$$

Under assumptions (A1)–(A12), the dominant terms of the i.i.d. representation of \(A_{21}\) in (20) come from the i.i.d. representation of \(\hat{p}_h(x)\) in Theorem 3 of López-Cheda et al. (2017):

$$\begin{aligned} A_{21}= -\frac{(1-S(t|x))}{p^{2}(x)}(1-p(x))\sum _{i=1}^{n} \tilde{B}_{h,i}(x)\xi (T_{i},\delta _{i},\infty ,x) +O\left( \left( \frac{\ln n}{nh}\right) ^{3/4}\right) \text { a.s.} \end{aligned}$$

We continue by proving the negligibility of \(A_{12}\) in (21). Under assumptions (A3a), (A4), (A5) and (A11), we apply Lemma 5 in Iglesias-Pérez and González-Manteiga (1999) to obtain

$$\begin{aligned} \hat{S}_{h}(t|x)-S(t|x)=O\left( \sqrt{\frac{\ln \ln n}{nh}}+h^{2}\right) \text { a.s.} \end{aligned}$$

and, similarly from Theorem 3.3 in Arcones (1997) and the Strong Law of Large Numbers (SLLN),

$$\begin{aligned} \hat{p}_{h}(x)-p(x)=O\left( \sqrt{\frac{\ln \ln n}{nh}}+h^{2}\right) \text { a.s.} \end{aligned}$$

It is straightforward to check that if the bandwidth satisfies \(h\rightarrow 0\), \(\frac{\ln n}{nh}\rightarrow 0\) and \(\frac{nh^{5}}{\ln n}=O(1)\), with the convergence \(\hat{p}_{h}(x)\rightarrow p(x)\text { a.s.}\) proved in Lemma 7 of López-Cheda et al. (2017), it directly follows that

$$\begin{aligned} A_{12}=O\left( \left( \frac{\ln n}{nh}\right) ^{3/4}\right) \, \mathrm{a.s.} \end{aligned}$$

With respect to \(A_{22}\) in (21), if \(h\rightarrow 0\), \(\frac{\ln n}{nh}\rightarrow 0\) and \(\frac{nh^{5}}{\ln n}=O(1)\), using the almost sure consistency of \(\hat{p}_{h}(x)\), it follows from (24) that

$$\begin{aligned} A_{22}=O\left( \left( \frac{\ln n}{nh}\right) ^{3/4}\right) \, \text {a.s.} \end{aligned}$$

The proof of the theorem follows from the decomposition (19) and the results (22), (23), (25) and (26). \(\square \)

Proof of Theorem 2

From Theorem 1, the latency estimator can be decomposed as

$$\begin{aligned} \hat{S}_{0,h}(t|x)-S_{0}(t|x)=C_{1}+C_{2}+O\left( \left( \frac{\ln n}{nh} \right) ^{3/4}\right) \text { a.s.}, \end{aligned}$$


$$\begin{aligned} C_{1}= & {} -\frac{S(t|x)}{p(x)}\sum _{i=1}^{n}\tilde{B}_{h,i}(x)\xi (T_{i},\delta _{i},t,x),\\ C_{2}= & {} -\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)}\sum _{i=1}^{n}\tilde{B} _{h,i}(x)\xi (T_{i},\delta _{i},\infty ,x), \end{aligned}$$

with \(\tilde{B}_{h,i}(x)\) in (8) and \(\xi \) in (7). Then, the AMSE of \(\hat{S}_{0,h}(t|x)\) is

$$\begin{aligned} \mathrm{AMSE}(\hat{S}_{0,h}(t|x))=E(C_{1}^{2})+E(C_{2}^{2})+2E(C_{1}\cdot C_{2}). \end{aligned}$$

We start with the first term of AMSE\((\hat{S}_{0, h} (t|x) )\). Note that

$$\begin{aligned} E(C_{1}^{2})=\mathrm{Var}(C_{1})+(E(C_{1}))^{2}, \end{aligned}$$


$$\begin{aligned} \mathrm{Var}(C_{1}) = \frac{1}{n h^2} \left( \frac{S(t|x)}{p(x)}\right) ^{2} \frac{1}{m^2(x)} \mathrm{Var} \left( K \left( \frac{x - X_1}{h} \right) \xi (T_1, \delta _1, t, x) \right) \end{aligned}$$


$$\begin{aligned}&\mathrm{Var} \left( K \left( \frac{x - X_1}{h} \right) \xi (T_1, \delta _1, t, x) \right) \nonumber \\&\quad = E \left( K^2 \left( \frac{x - X_1}{h} \right) \xi ^2(T_1, \delta _1, t, x) \right) - \left[ E \left( K \left( \frac{x - X_1}{h} \right) \xi (T_1, \delta _1, t,x) \right) \right] ^2.\nonumber \\ \end{aligned}$$

Let us consider \(\varPhi _{1}(y,t,x)\) defined in (9). From a change of variable and a Taylor expansion, then the first term in (30) is

$$\begin{aligned} E\left[ K^{2}\left( \frac{x-X_{1}}{h}\right) \xi ^{2}(T_{1},\delta _{1},t,x) \right] =h\varPhi _{1}(x,t,x)m(x)c_{K}+O(h^{3}). \end{aligned}$$

For the second term in (30), applying a change of variable, a Taylor expansion, and taking into account the symmetry of K, it follows that

$$\begin{aligned} \left[ E\left( K\left( \frac{x-X_{1}}{h}\right) \xi (T_{1},\delta _{1},t,x)\right) \right] ^{2}=\left[ \varPhi (x,t,x)m(x)h+O(h^{3})\right] ^{2}=O(h^{6}), \end{aligned}$$

where \(\varPhi (y,t,x)=E\left[ \xi (T,\delta ,t,x)|X=y\right] \) and, as will be proved in Lemma 4, \(\varPhi (x,t,x)=0\) for all \(t\ge 0\).

From (29), (30), (31) and (32), then

$$\begin{aligned} \mathrm{Var}(C_{1})=\frac{1}{nh}\left( \frac{S(t|x)}{p(x)}\right) ^{2}\frac{1}{m(x)} \varPhi _{1}(x,t,x)c_{K}+O\left( \frac{h}{n}\right) . \end{aligned}$$

Continuing with the second term in the right-hand side of (28):

$$\begin{aligned} E(C_{1}) =-\frac{1}{h}\frac{S(t|x)}{m(x)p(x)}E\left[ K\left( \frac{x-X_{1}}{h} \right) \xi (T_{1},\delta _{1},t,x)\right] . \end{aligned}$$

Using a Taylor expansion, and \(\varPhi (x,t,x)=0 \; \forall t\ge 0\), then

$$\begin{aligned} E(C_{1})=-\frac{1}{2}h^{2}\frac{S(t|x)}{p(x)m(x)}d_{K}\left( \varPhi ^{\prime \prime }\left( x,t,x\right) m(x)+2\varPhi ^{\prime }\left( x,t,x\right) m^{\prime }(x)\right) +o(h^{2}). \end{aligned}$$

So the first term of \(\mathrm{AMSE}(\hat{S}_{0,h}(t|x))\) in (27) is

$$\begin{aligned} E(C_{1}^{2})= & {} \frac{1}{4}h^{4}d_{K}^{2}\left[ \frac{S(t|x)}{p(x)m(x)}\left( \varPhi ^{\prime \prime }\left( x,t,x\right) m(x)+2\varPhi ^{\prime }\left( x,t,x\right) m^{\prime }(x)\right) \right] ^{2} \nonumber \\&+\frac{1}{nh}\left( \frac{S(t|x)}{p(x)}\right) ^{2}\frac{1}{ m(x)}\varPhi _{1}(x,t,x)c_{K} +o(h^{4})+O\left( \frac{h}{n} \right) . \end{aligned}$$

Following the same ideas as those for \(C_{1}\), we obtain for \(C_{2}\) that

$$\begin{aligned} E(C_{2}^{2})= & {} \frac{1}{nh}\left( \frac{(1-S(t|x))(1-p(x))}{p^{2}(x)} \right) ^{2}\frac{1}{m(x)}\varPhi _{1}(x,\infty ,x)c_{K} \nonumber \\&+\frac{1}{4}h^{4}d_{K}^{2}\left[ \frac{(1-S(t|x))(1-p(x))}{p^{2}(x)m\left( x\right) } \right. \nonumber \\&\times \left. \left( \varPhi ^{\prime \prime }\left( x,\infty ,x\right) m(x)+2\varPhi ^{\prime }\left( x,\infty ,x\right) m^{\prime }(x)\right) \right] ^{2}o(h^{4})+O\left( \frac{h}{n}\right) .\nonumber \\ \end{aligned}$$

We continue studying the third term of AMSE\((\hat{S}_{0,h}(t|x))\) in (27):

$$\begin{aligned} E\left( C_{1}\cdot C_{2}\right) =\frac{(1-p(x))S(t|x)(1-S(t|x)}{p^{3}(x)}\left[ n(n-1)\alpha \beta +n\gamma \right] , \end{aligned}$$


$$\begin{aligned} \alpha= & {} E\left[ \tilde{B}_{h1}(x)\xi (T_{1},\delta _{1},t,x)\right] , \\ \beta= & {} E\left[ \tilde{B}_{h1}(x)\xi (T_{1},\delta _{1},\infty ,x)\right] \text {,} \\ \gamma= & {} E\left[ \tilde{B}_{h1}^{2}(x)\xi (T_{1},\delta _{1},t,x)\xi (T_{1},\delta _{1},\infty ,x)\right] . \end{aligned}$$

Using a Taylor expansion and \(\varPhi (x,t,x)=0\) for all \(t\ge 0\), the terms \(\alpha \) and \(\beta \) are

$$\begin{aligned} \alpha= & {} \frac{1}{2}\frac{h^{2}}{n}d_{K}\frac{1}{m(x)}\left( \varPhi ^{\prime \prime }\left( x,t,x\right) m(x)+2\varPhi ^{\prime }\left( x,t,x\right) m^{\prime }(x)\right) +o\left( \frac{h^{2}}{n}\right) , \end{aligned}$$
$$\begin{aligned} \beta= & {} \frac{1}{2}\frac{h^{2}}{n}d_{K}\frac{1}{m(x)}\left( \varPhi ^{\prime \prime }\left( x,\infty ,x\right) m(x)+2\varPhi ^{\prime }\left( x,\infty ,x\right) m^{\prime }(x)\right) +o\left( \frac{h^{2}}{n}\right) . \end{aligned}$$

For the term \(\gamma \), it follows that

$$\begin{aligned} \gamma= & {} \frac{1}{n^{2}h^{2}}\frac{1}{m^{2}(x)}\int K^{2}\left( \frac{x-y}{h} \right) \varPhi _{2}(y,t,x)m(y)\mathrm{d}y \nonumber \\= & {} \frac{1}{n^{2}h}\frac{1}{m(x)}\varPhi _{2}(x,t,x)c_{K}+O\left( \frac{h}{ n^{2}}\right) , \end{aligned}$$

where \(\varPhi _{2}(y,t,x)=E\left[ \xi (T,\delta ,t,x)\xi (T,\delta ,\infty ,x)|X=y\right] .\) From (35), (36) and (37), the third term of \(AMSE( \hat{S}_{0,h}(t|x))\) in (27) is:

$$\begin{aligned} E\left( C_{1}\cdot C_{2}\right)= & {} \frac{(1-p(x))S(t|x)(1-S(t|x)}{p^{3}(x)}\left[ \frac{1}{4}h^{4}d_{K}^{2}\frac{1}{m^{2}(x)}\right. \nonumber \\&\times \left( \varPhi ^{\prime \prime }\left( x,t,x\right) m(x)+2\varPhi ^{\prime }\left( x,t,x\right) m^{\prime }(x)\right) \nonumber \\&\left. \times \left( \varPhi ^{\prime \prime }\right. \left( x,\infty ,x\right) m(x)+2\varPhi ^{\prime }\left( x,\infty ,x\right) m^{\prime }(x)\right) \nonumber \\&+\left. \frac{1}{nh}\frac{1}{m(x)}\varPhi _{2}(x,t,x)c_{K} \right] +o\left( h^{4}\right) +O\left( \frac{h}{n}\right) . \end{aligned}$$

Compiling (33), (34) and (38), the \(\mathrm{AMSE}(\hat{S} _{0,h}(t|x))\) in (27) is

$$\begin{aligned} \mathrm{AMSE}(\hat{S}_{0,h}(t|x))= & {} \frac{1}{nh}\frac{1}{m(x)}c_{K} \left( \left( \frac{S(t|x)}{p(x)}\right) ^{2}\varPhi _{1}(x,t,x)\right. \nonumber \\&\quad \left. + \left( \frac{(1-S(t|x))(1-p(x))}{p^{2}(x)}\right) ^{2}\varPhi _{1}(x,\infty ,x)\right. \\&\quad +\left. 2\frac{(1-p(x))S(t|x)(1-S(t|x))}{p^{3}(x)}\varPhi _{2}(x,t,x)\right) \\&\quad +\frac{1}{4}h^{4}d_{K}^{2}\frac{1}{m^{2}(x)}\left( \frac{S(t|x)}{p(x)}\left( \varPhi ^{\prime \prime }\left( x,t,x\right) m(x)+ 2 \varPhi ^{\prime }\left( x,t,x\right) m^{\prime }(x)\right) \right. \\&\quad +\,\frac{(1-S(t|x))(1-p(x))}{p^{2}(x)} ( \varPhi ^{\prime \prime }\left( x,\infty ,x\right) m(x)\\&\quad \left. +\,2\varPhi ^{\prime }\left( x,\infty ,x\right) m^{\prime }(x)) \right) ^{2} \nonumber \\&\quad +\, o(h^{4}) +O\left( \frac{h}{n}\right) . \end{aligned}$$

Since, from (40) and (41), in Lemmas 5 and 6 it is proven that

$$\begin{aligned} \varPhi _{1}(x,t,x)=\varPhi _{2}(x,t,x)=\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|x\right) }{\left( 1-H(v|x)\right) ^{2}}, \end{aligned}$$

and considering (10)–(14), the AMSE of \(\hat{S}_{0,h}(t|x)\) is, finally, that in (15).

This completes the proof. \(\square \)

Lemma 4

The term \(\varPhi \left( y,t,x\right) \) in (8) has the following expression:

$$\begin{aligned} \varPhi \left( y,t,x\right) =\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|y\right) }{ 1-H(v|x)}-\int _{0}^{t}(1-H(v|y))\frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}}, \end{aligned}$$

and consequently, \(\varPhi \left( x,t,x\right) =0\) for any \(t\ge 0\).

Proof of Lemma 4

Let us recall \(\varPhi \left( y,t,x\right) =E\left[ \xi (T,\delta ,t,x)|X=y\right] \), then

$$\begin{aligned} \varPhi \left( y,t,x\right)= & {} E\left[ \frac{1\{T\le t,\delta =1\}}{1-H(T|x)}\bigg |X=y\right] -E\left[ \int _{0}^{t}\frac{1\{y \le T\} \mathrm{d}H^{1}(u|x)}{ \left( 1-H(u|x)\right) ^{2}}\bigg |X=y\right] \\= & {} A^{\prime }-A^{\prime \prime }. \end{aligned}$$

We start with \(A^{\prime }\):

$$\begin{aligned} A^{\prime } =E\left[ \frac{1\{T\le t\}}{1-H(T|x)}E\left( \delta |T,X=y\right) \right] =\int \limits _{0}^{t}\frac{q(v,y)\mathrm{d}H(v|y)}{1-H(v|x)}=\int _{0}^{t}\frac{ \mathrm{d}H^{1}\left( v|y\right) }{1-H(v|x)}, \end{aligned}$$

where \(q\left( t,y\right) =E\left( \delta |T=t,X=y\right) \) and \(H_{1}\left( t|y\right) =P\left( T\le t,\delta =1|X=y\right) \).

We continue with \(A^{\prime \prime }\):

$$\begin{aligned} A^{\prime \prime }=\int _{0}^{t}E\left[ 1\{v\le T\}|X=y\right] \frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}} =\int _{0}^{t}(1-H(v|y))\frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}}. \end{aligned}$$


$$\begin{aligned} \varPhi \left( y,t,x\right) =\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|y\right) }{ 1-H(v|x)}-\int _{0}^{t}(1-H(v|y))\frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}}, \end{aligned}$$

and therefore, \(\varPhi \left( x,t,x\right) =0\) for any \(t\ge 0\). \(\square \)

Lemma 5

The term \(\varPhi _1(y,t,x)\) in (9) verifies, for any \(t \in [a,b]\),

$$\begin{aligned} \varPhi _{1}\left( x,t,x\right) =\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|x\right) }{\left( 1-H(v|x)\right) ^{2}}. \end{aligned}$$

Proof of Lemma 5

Note that \(\varPhi _{1}\left( y,t,x\right) =E\left[ \xi ^{2}(T,\delta ,t,x)|X=y\right] \), with \(\xi \) in (7). Then,

$$\begin{aligned} \varPhi _{1}\left( y,t,x\right)= & {} E\left[ \frac{1\{T\le t,\delta =1\}}{\left( 1-H(T|x)\right) ^{2}}\bigg |X=y\right] \\&+E\left[ \int _{0}^{t}\int _{0}^{t}\frac{1 \{ u\le T \} 1\{ v\le T \} }{\left( 1-H(u|x)\right) ^{2}\left( 1-H(v|x)\right) ^{2}}\mathrm{d}H^{1}(u|x)\mathrm{d}H^{1}(v|x)\bigg |X=y\right] \\&-2E\left[ \frac{1\{T\le t,\delta =1\}}{1-H(T|x)}\int _{0}^{t}\frac{1\{u\le T \} \mathrm{d}H^{1}(u|x)}{\left( 1-H(u|x)\right) ^{2}}\bigg |X=y\right] \\= & {} A+B-2C. \end{aligned}$$

The first term in the decomposition of \(\varPhi _{1}\left( y,t,x\right) \) is

$$\begin{aligned} A=\int _{0}^{t}\frac{q\left( v,y\right) }{\left( 1-H(v|x)\right) ^{2}} \mathrm{d}H(v|y)=\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|y\right) }{\left( 1-H(v|x)\right) ^{2}}. \end{aligned}$$

The second term is

$$\begin{aligned} B=\int _{0}^{t}\int _{0}^{t}\frac{1-H\left( \max \left( w,v\right) |y\right) }{\left( 1-H(v|x)\right) ^{2}\left( 1-H(w|x)\right) ^{2}} \mathrm{d}H^{1}(v|x)\mathrm{d}H^{1}(w|x). \end{aligned}$$

Integrating in the supports \(\left\{ (v,w) \in \left[ 0,t\right] \times \left[ 0,t\right] /v\le w\right\} \) and \(\left\{ \left( v,w \right) \in \right. \) \(\left[ 0,t\right] \times \left. \left[ 0,t\right] /w < v\right\} \), the term B is

$$\begin{aligned} B=2\int _{0}^{t}\frac{1}{\left( 1-H(v|x)\right) ^{2}}\left( \int _{v}^{t} \frac{1-H\left( w|y\right) }{\left( 1-H(w|x)\right) ^{2}}\mathrm{d}H^{1}(w|x)\right) \mathrm{d}H^{1}(v|x). \end{aligned}$$

Finally, the third term in the decomposition of \(\varPhi _{1}\left( y,t,x\right) \) is

$$\begin{aligned} C=\int _{0}^{t}\frac{1}{\left( 1-H(u|x)\right) ^{2}}\left( \int _{u}^{t}\frac{ \mathrm{d}H^{1}\left( v|y\right) }{1-H(v|x)}\right) \mathrm{d}H^{1}(u|x). \end{aligned}$$

Note that, for \(y=x\), we have that \(B=2C\). This completes the proof. \(\square \)

Lemma 6

The expression for the term \(\varPhi _{2}(x,t,x)\), for any \(t \in [a,b]\), is the following:

$$\begin{aligned} \varPhi _{2}(x,t,x)=\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|x\right) }{\left( 1-H(v|x)\right) ^{2}}. \end{aligned}$$

Proof of Lemma 6

Recall \(\varPhi _{2}(y,t,x)=E\left[ \xi \left( T,\delta ,t,x\right) \xi (T,\delta ,\infty ,x)|X=y\right] \) with \(\xi \) in (7). Then:

$$\begin{aligned}&\varPhi _{2}(y,t,x) \\&\quad =E\left[ \frac{1\{T\le t,\delta =1\}}{\left( 1-H(T|x)\right) ^{2}}\bigg |X=y\right] \\&\qquad -E\left[ \frac{1\{\delta =1\}}{1-H(T|x)}\int _{0}^{\infty }\frac{1\{u\le T\le t\} }{\left( 1-H(u|x)\right) ^{2}}\mathrm{d}H^{1}(u|x)\bigg |X=y \right] \\&\qquad -E\left[ \frac{1\{\delta =1\}}{1-H(T|x)}\int _{0}^{t}\frac{1\{ v\le T\} }{\left( 1-H(v|x)\right) ^{2}}\mathrm{d}H^{1}(v|x)\bigg |X=y\right] \\&\qquad +E\left[ \int _{0}^{t}\frac{1\{ v\le T \} dH^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}}\int _{0}^{\infty }\frac{1\{ u\le T\} \mathrm{d}H^{1}(u|x)}{\left( 1-H(u|x)\right) ^{2}}\bigg |X=y\right] \\&\quad =A-B-C+D. \end{aligned}$$

Straightforward calculations yield:

$$\begin{aligned} A= & {} \int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|y\right) }{\left( 1-H(v|x)\right) ^{2}},\\ B= & {} \int _{0}^{\infty }\left( \int _{u}^{t}\frac{\mathrm{d}H^{1}\left( v|y\right) }{ 1-H(v|x)}\right) \frac{\mathrm{d}H^{1}(u|x)}{\left( 1-H(u|x)\right) ^{2}},\\ C= & {} \int _{0}^{t}\left( \int _{v}^{\infty }\frac{\mathrm{d}H^{1}\left( u|y\right) }{ 1-H(u|x)}\right) \frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}},\\ D= & {} \int _{0}^{t}\frac{1}{\left( 1-H(v|x)\right) ^{2}}\left( \int _{0}^{\infty } \frac{1-H\left( \max \left( u,v\right) |y\right) }{\left( 1-H(u|x)\right) ^{2}}\mathrm{d}H^{1}(u|x)\right) \mathrm{d}H^{1}(v|x). \end{aligned}$$

Integrating in the supports \(\left\{ \left( u,v \right) \in \left[ 0,\infty \right) \times \left[ 0,t \right] /v\le u\right\} \) and \(\left\{ \left( u,v \right) \in \right. \) \( \left[ 0,\infty \right) \left. \times \left[ 0,t \right] /u< v\right\} =\left\{ \left( u, v \right) \in \left[ 0, t \right] \times \left[ 0, t \right] /u < v \right\} \), the term D is

$$\begin{aligned} D= & {} \int _{0}^{t}\left( \int _{v}^{\infty }\frac{1-H\left( u|y\right) }{ \left( 1-H(u|x)\right) ^{2}}\mathrm{d}H^{1}(u|x)\right) \frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}} \\&+\int _{0}^{\infty }\left( \int _{u}^{t}\frac{1-H\left( v|y\right) }{\left( 1-H(v|x)\right) ^{2}}\mathrm{d}H^{1}(v|x)\right) \frac{\mathrm{d}H^{1}(u|x)}{\left( 1-H(u|x)\right) ^{2}}. \end{aligned}$$

When \(y=x\), then \(D=C+B\), which concludes the proof. \(\square \)

Proof of Theorem 3

Under assumptions (A1)–(A10) and using Theorem 1, \(\sqrt{nh}\left( \hat{S}_{0,h}(t|x)-S_{0}(t|x)\right) \) has the same limit distribution as

$$\begin{aligned} \sqrt{nh}\sum _{i=1}^{n}\eta _{h}(T_{i},\delta _{i},X_{i},t,x)=-\left( I+II+III+IV\right) , \end{aligned}$$


$$\begin{aligned} I= & {} \sqrt{nh}\frac{1}{nh}\frac{S(t|x)}{p(x)m(x)} \\\times & {} \sum _{i=1}^{n}\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t,x)-E\left( K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t,x)\right) \right] , \\ II= & {} \sqrt{nh}\frac{1}{nh}\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)m\left( x\right) } \\\times & {} \sum _{i=1}^{n}\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)-E\left( K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)\right) \right] , \\ III= & {} \sqrt{nh}\frac{1}{nh}\frac{S(t|x)}{p(x)m(x)}\sum _{i=1}^{n}E\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t,x)\right] , \\ IV= & {} \sqrt{nh}\frac{1}{nh}\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)m\left( x\right) }\sum _{i=1}^{n}E\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)\right] . \end{aligned}$$

The deterministic part b(tx) comes from \(III+IV\). Recall the function \(\varPhi (y,t,x)\) in (39), since \(\varPhi (x,t,x)=0\), then

$$\begin{aligned}&E\left[ K\left( \frac{x-X}{h}\right) \xi (T,\delta ,t,x)\right] \nonumber \\&\quad = \frac{1}{2} h^{3}d_{K}\left( \varPhi ^{\prime \prime }(x,t,x)m(x) + 2\varPhi ^{\prime }(x,t,x)m^{\prime }(x)\right) +o(h^{3}). \end{aligned}$$


$$\begin{aligned} III= & {} \sqrt{nh^{5}}\frac{S(t|x)}{p(x)m(x)}\frac{1}{2}d_{K}\left( \varPhi ^{\prime \prime }(x,t,x)m(x)+2\varPhi ^{\prime }(x,t,x)m^{\prime }(x)\right) \left( 1+o\left( 1\right) \right) ,\\ IV= & {} \sqrt{nh^{5}}\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)m\left( x\right) }\frac{ 1}{2}d_{K}\nonumber \\&\quad \times \left( \varPhi ^{\prime \prime }(x,\infty ,x)m(x) + 2\varPhi ^{\prime }(x,\infty ,x)m^{\prime }(x)\right) \left( 1+o(1)\right) . \end{aligned}$$

If \(nh^{5}\rightarrow 0\), then \(III+IV=o\left( 1\right) \) and \(b\left( t,x\right) =0\). On the other hand, if \(nh^{5}\rightarrow C^{5}\) then

$$\begin{aligned} b(t,x)= & {} C^{5/2}\frac{S(t|x)}{p(x)m(x)}\frac{1}{2}d_{K}\left( \varPhi ^{\prime \prime }(x,t,x)m(x)+2\varPhi ^{\prime }(x,t,x)m^{\prime }(x)\right) \\&+\,C^{5/2}\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)m\left( x\right) }\frac{1}{2} d_{K}( \varPhi ^{\prime \prime }(x,\infty ,x)m(x)+\, 2\varPhi ^{\prime }(x,\infty ,x)m^{\prime }(x)). \end{aligned}$$

As for the asymptotic distribution of \(I+II\), it is immediate to prove that:

$$\begin{aligned} I+II=\sum _{i=1}^{n}\left( \gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t)\right) , \end{aligned}$$


$$\begin{aligned} \gamma _{i,n}(x,t)= & {} \frac{1}{\sqrt{nh}}\frac{S(t|x)}{p(x)m(x)}\\\times & {} \left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t ,x)-E\left( K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t ,x)\right) \right] , \\ \varGamma _{i,n}(x,t)= & {} \frac{1}{\sqrt{nh}}\frac{(1-p(x))(1-S(t|x))}{ p^{2}(x)m\left( x\right) } \\\times & {} \left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)-E\left( K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)\right) \right] , \end{aligned}$$

are n independent variables with mean 0. To prove the asymptotic normality of \(I+II\), it is only necessary to show that \(\sigma _{i,n}^{2}\left( x,t\right) =Var\left( \gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t)\right) \) \( <\infty \), \(\sigma _{n}^{2}\left( x,t\right) =\sum _{i=1}^{n}\sigma _{i,n}^{2}\left( x,t\right) \) is positive and that the Lindeberg’s condition is satisfied, so Lindeberg’s theorem for triangular arrays (Theorem 7.2 in Billingsley (1968), p. 42) can be applied to obtain

$$\begin{aligned} \frac{\sum _{i=1}^{n}\left( \gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t)\right) }{ \sigma _{n}\left( x,t\right) }\rightarrow N\left( 0,1\right) , \end{aligned}$$

and consequently,

$$\begin{aligned} \frac{\sqrt{nh}\sum _{i=1}^{n}\eta _{h}(T_{i},\delta _{i},X_{i},t,x)}{\sigma _{n}\left( x,t\right) }\rightarrow N\left( 0,1\right) . \end{aligned}$$

We will start proving that the variance

$$\begin{aligned} \sigma _{i,n}^{2}\left( x,t\right) =\mathrm{Var}\left( \gamma _{i,n}(x,t)\right) +\mathrm{Var}\left( \varGamma _{i,n}(x,t)\right) +2\mathrm{Cov}\left( \gamma _{i,n}(x,t),\varGamma _{i,n}(x,t)\right) \end{aligned}$$

is finite. Note that

$$\begin{aligned} Var\left( \gamma _{i,n}(x,t)\right)= & {} \frac{1}{nh}\left( \frac{S(t|x)}{p(x)m(x)}\right) ^{2}\left\{ E\left[ K^{2}\left( \frac{x-X_{1}}{h}\right) \xi ^{2}(T_{1},\delta _{1},t,x)\right] \right. \\&\left. -E\left[ K\left( \frac{x-X_{1}}{h}\right) \xi (T_{1},\delta _{1},t,x)\right] ^{2}\right\} . \end{aligned}$$

Let us define \(\varPhi _{1}(y,t,x)=E\left[ \xi ^{2}(T,\delta ,t,x)|X=y\right] \), using (42), then the first term in (43) is

$$\begin{aligned} Var\left( \gamma _{i,n}(x,t)\right) =\frac{1}{n}\left( \frac{S(t|x)}{p(x)} \right) ^{2}\frac{\varPhi _{1}(x,t,x)}{m\left( x\right) }c_{K}+O\left( \frac{ h^{2}}{n}\right) . \end{aligned}$$

In a similar way, the second term in (43) is

$$\begin{aligned} Var\left( \varGamma _{i,n}(x,t)\right) =\frac{1}{n}\left( \frac{(1-p(x))(1-S(t|x))}{p^{2}(x)}\right) ^{2}\frac{ \varPhi _{1}(x,\infty ,x)}{m\left( x\right) }c_{K}+O\left( \frac{h^{2}}{n} \right) . \end{aligned}$$

Finally, for the third term in (43),

$$\begin{aligned}&Cov\left( \gamma _{i,n}(x,t),\varGamma _{i,n}(x,t)\right) \nonumber \\&\quad = \frac{1}{nh}\left\{ E\left[ K^{2}\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)\xi (T_{i},\delta _{i},t,x)\right] \right. \\&\qquad \left. -E\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t,x)\right] E\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)\right] \right\} . \end{aligned}$$

Let us consider \(\varPhi _{2}(y,t,x)=E\left[ \xi (T,\delta ,t,x)\xi (T,\delta ,\infty ,x)|X=y\right] \). Applying Taylor expansions, the third term in (43) is

$$\begin{aligned} Cov\left( \gamma _{i,n}(x,t),\varGamma _{i,n}(x,t)\right) {=}\frac{1}{n}\frac{ (1\!-\!p(x))S(t|x)(1\!-\!S(t|x))}{p^{3}(x)m(x) }\varPhi _{2}(x,t,x)c_{K}+O\left( \frac{h}{n}\right) . \end{aligned}$$

The results (44), (45) and (46), together with (40) and (41), lead to

$$\begin{aligned} \sigma _{i,n}^{2}\left( x,t\right) =\frac{c_{K}}{n}\left( V_{1}\left( t,x\right) +V_{2}\left( t,x\right) +2V_{3}\left( t,x\right) \right) +O\left( \frac{h}{n}\right) , \end{aligned}$$

where \(V_{1}\left( t,x\right) \), \(V_{2}\left( t,x\right) \) and \(V_{3}\left( t,x\right) \) are defined in (12), (13) and (14), respectively. As a consequence, \(\sigma _{i,n}^{2}\left( x,t\right) <\infty \). The finiteness of the variance \(\sigma _{n}^{2}\left( x,t\right) \) is also proved, since

$$\begin{aligned} \sigma _{n}^{2}\left( x,t\right) \!=\!\sum _{i=1}^{n}\sigma _{i,n}^{2}\left( x,t\right) \!=\!V_{1}\left( t,x\right) c_{K}+V_{2}\left( t,x\right) c_{K}\!+\!2V_{3}\left( t,x\right) c_{K}+O\left( h\right) <+\infty . \end{aligned}$$

We continue studying Lindeberg’s condition:

$$\begin{aligned} \frac{1}{\sigma _{n}^{2}\left( x,t\right) }\sum _{i=1}^{n}\int _{\{|\gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t)|>\epsilon \sigma _{n}\left( x,t\right) \}}(\gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t))^{2}dP\rightarrow 0,\forall \epsilon >0. \end{aligned}$$

Let us define the indicator function \(I_{i,n}( x,t)=1 \left\{ \left( \gamma _{i,n}(x,t){+}\varGamma _{i,n}(x,t)\right) ^{2}\!>\! \epsilon ^{2}\sigma _{n}^{2}\right. \left. ( x,t) \right\} \). Then (47) can be expressed as

$$\begin{aligned} \frac{1}{\sigma _{n}^{2}\left( x,t\right) }E\left[ \sum _{i=1}^{n}(\gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t))^{2}I_{i,n}\left( x,t\right) \right] =\frac{1 }{\sigma _{n}^{2}\left( x,t\right) }E\left( \eta _{n}\left( x,t\right) \right) , \end{aligned}$$


$$\begin{aligned} \eta _{n}\left( x,t\right) =\sum _{i=1}^{n}(\gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t))^{2}I_{i,n}\left( x,t\right) . \end{aligned}$$

Since \(\frac{1}{nh}\rightarrow 0\), and the functions K and \(\xi \) are bounded, one has:

$$\begin{aligned}&\exists n_{0}\in \mathbb {N}/n\ge n_{0}\Rightarrow I_{i,n}(w)=0,\forall w \text { and }\forall i\in \{1,2,\dots ,n\} \\\Leftrightarrow & {} \exists n_{0}\in \mathbb {N}/n\ge n_{0}\Rightarrow \eta _{n}(w)=0,\forall w . \end{aligned}$$

Since \(\eta _{n}(x,t)\) is bounded, then the previous condition implies that \( \exists n_{0}\in \mathbb {N}/n\ge n_{0}\Rightarrow E(\eta _{n}(x,t))=0\), and then \(\lim _{n\rightarrow \infty }\frac{1}{\sigma _{n}^{2}}E(\eta _{n}(x,t))=0.\) Therefore, Lindeberg’s condition is proved. All these previous arguments lead to the proof of Theorem 3. \(\square \)

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López-Cheda, A., Jácome, M.A. & Cao, R. Nonparametric latency estimation for mixture cure models. TEST 26, 353–376 (2017).

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  • Bandwidth selection
  • Bootstrap
  • Censored data
  • Kernel estimation
  • Survival analysis

Mathematics Subject Classification

  • 62N01 Censored data models (Survival analysis and censored data)
  • 62N02 Estimation (Survival analysis and censored data)
  • 62G08 Nonparametric regression (Nonparametric inference)