Appendix
Proof of Theorem 1
The nonparametric estimator of \(S_0(t|x)\) in (4) can be decomposed as follows:
$$\begin{aligned} \hat{S}_{0,h}(t|x)-S_{0}(t|x)=A_{11}+ A_{21} +A_{12}+ A_{22}, \end{aligned}$$
(19)
where the dominant terms of the i.i.d. representation of \(\hat{S}_{0,h}(t|x)\) derive from
$$\begin{aligned} A_{11} =\frac{\hat{S}_{h}(t|x)-S(t|x)}{p(x)}\quad \text { and } \quad A_{21} =\frac{1-S(t|x)}{p^{2}(x)}(\hat{p}_{h}(x)-p(x)), \end{aligned}$$
(20)
and the remaining terms
$$\begin{aligned} A_{12} =\frac{(\hat{S}_{h}(t|x)-S(t|x))(p(x)-\hat{p}_{h}(x))}{\hat{p}_{h}(x)p(x)}\quad \text { and }\quad A_{22}= \frac{S(t|x)-1}{p^2(x)}\frac{\left( \hat{p} _{h}(x)-p(x)\right) ^{2}}{\hat{p}_{h}(x)} \end{aligned}$$
(21)
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):
$$\begin{aligned} A_{11}=-\frac{S(t|x)}{p(x)} \sum _{i=1}^{n}\tilde{B}_{h,i}(x)\xi (T_{i},\delta _{i},t,x)+O\left( \left( \frac{\ln n}{nh}\right) ^{3/4}\right) \text { a.s.} \end{aligned}$$
(22)
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):
$$\begin{aligned} A_{21}= -\frac{(1-S(t|x))}{p^{2}(x)}(1-p(x))\sum _{i=1}^{n} \tilde{B}_{h,i}(x)\xi (T_{i},\delta _{i},\infty ,x) +O\left( \left( \frac{\ln n}{nh}\right) ^{3/4}\right) \text { a.s.} \end{aligned}$$
(23)
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),
$$\begin{aligned} \hat{p}_{h}(x)-p(x)=O\left( \sqrt{\frac{\ln \ln n}{nh}}+h^{2}\right) \text { a.s.} \end{aligned}$$
(24)
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
$$\begin{aligned} A_{12}=O\left( \left( \frac{\ln n}{nh}\right) ^{3/4}\right) \, \mathrm{a.s.} \end{aligned}$$
(25)
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
$$\begin{aligned} A_{22}=O\left( \left( \frac{\ln n}{nh}\right) ^{3/4}\right) \, \text {a.s.} \end{aligned}$$
(26)
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
$$\begin{aligned} \hat{S}_{0,h}(t|x)-S_{0}(t|x)=C_{1}+C_{2}+O\left( \left( \frac{\ln n}{nh} \right) ^{3/4}\right) \text { a.s.}, \end{aligned}$$
where
$$\begin{aligned} C_{1}= & {} -\frac{S(t|x)}{p(x)}\sum _{i=1}^{n}\tilde{B}_{h,i}(x)\xi (T_{i},\delta _{i},t,x),\\ C_{2}= & {} -\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)}\sum _{i=1}^{n}\tilde{B} _{h,i}(x)\xi (T_{i},\delta _{i},\infty ,x), \end{aligned}$$
with \(\tilde{B}_{h,i}(x)\) in (8) and \(\xi \) in (7). Then, the AMSE of \(\hat{S}_{0,h}(t|x)\) is
$$\begin{aligned} \mathrm{AMSE}(\hat{S}_{0,h}(t|x))=E(C_{1}^{2})+E(C_{2}^{2})+2E(C_{1}\cdot C_{2}). \end{aligned}$$
(27)
We start with the first term of AMSE\((\hat{S}_{0, h} (t|x) )\). Note that
$$\begin{aligned} E(C_{1}^{2})=\mathrm{Var}(C_{1})+(E(C_{1}))^{2}, \end{aligned}$$
(28)
where
$$\begin{aligned} \mathrm{Var}(C_{1}) = \frac{1}{n h^2} \left( \frac{S(t|x)}{p(x)}\right) ^{2} \frac{1}{m^2(x)} \mathrm{Var} \left( K \left( \frac{x - X_1}{h} \right) \xi (T_1, \delta _1, t, x) \right) \end{aligned}$$
(29)
and
$$\begin{aligned}&\mathrm{Var} \left( K \left( \frac{x - X_1}{h} \right) \xi (T_1, \delta _1, t, x) \right) \nonumber \\&\quad = E \left( K^2 \left( \frac{x - X_1}{h} \right) \xi ^2(T_1, \delta _1, t, x) \right) - \left[ E \left( K \left( \frac{x - X_1}{h} \right) \xi (T_1, \delta _1, t,x) \right) \right] ^2.\nonumber \\ \end{aligned}$$
(30)
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
$$\begin{aligned} E\left[ K^{2}\left( \frac{x-X_{1}}{h}\right) \xi ^{2}(T_{1},\delta _{1},t,x) \right] =h\varPhi _{1}(x,t,x)m(x)c_{K}+O(h^{3}). \end{aligned}$$
(31)
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
$$\begin{aligned} \left[ E\left( K\left( \frac{x-X_{1}}{h}\right) \xi (T_{1},\delta _{1},t,x)\right) \right] ^{2}=\left[ \varPhi (x,t,x)m(x)h+O(h^{3})\right] ^{2}=O(h^{6}), \end{aligned}$$
(32)
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
$$\begin{aligned} E(C_{1}^{2})= & {} \frac{1}{4}h^{4}d_{K}^{2}\left[ \frac{S(t|x)}{p(x)m(x)}\left( \varPhi ^{\prime \prime }\left( x,t,x\right) m(x)+2\varPhi ^{\prime }\left( x,t,x\right) m^{\prime }(x)\right) \right] ^{2} \nonumber \\&+\frac{1}{nh}\left( \frac{S(t|x)}{p(x)}\right) ^{2}\frac{1}{ m(x)}\varPhi _{1}(x,t,x)c_{K} +o(h^{4})+O\left( \frac{h}{n} \right) . \end{aligned}$$
(33)
Following the same ideas as those for \(C_{1}\), we obtain for \(C_{2}\) that
$$\begin{aligned} E(C_{2}^{2})= & {} \frac{1}{nh}\left( \frac{(1-S(t|x))(1-p(x))}{p^{2}(x)} \right) ^{2}\frac{1}{m(x)}\varPhi _{1}(x,\infty ,x)c_{K} \nonumber \\&+\frac{1}{4}h^{4}d_{K}^{2}\left[ \frac{(1-S(t|x))(1-p(x))}{p^{2}(x)m\left( x\right) } \right. \nonumber \\&\times \left. \left( \varPhi ^{\prime \prime }\left( x,\infty ,x\right) m(x)+2\varPhi ^{\prime }\left( x,\infty ,x\right) m^{\prime }(x)\right) \right] ^{2}o(h^{4})+O\left( \frac{h}{n}\right) .\nonumber \\ \end{aligned}$$
(34)
We continue studying the third term of AMSE\((\hat{S}_{0,h}(t|x))\) in (27):
$$\begin{aligned} E\left( C_{1}\cdot C_{2}\right) =\frac{(1-p(x))S(t|x)(1-S(t|x)}{p^{3}(x)}\left[ n(n-1)\alpha \beta +n\gamma \right] , \end{aligned}$$
where
$$\begin{aligned} \alpha= & {} E\left[ \tilde{B}_{h1}(x)\xi (T_{1},\delta _{1},t,x)\right] , \\ \beta= & {} E\left[ \tilde{B}_{h1}(x)\xi (T_{1},\delta _{1},\infty ,x)\right] \text {,} \\ \gamma= & {} E\left[ \tilde{B}_{h1}^{2}(x)\xi (T_{1},\delta _{1},t,x)\xi (T_{1},\delta _{1},\infty ,x)\right] . \end{aligned}$$
Using a Taylor expansion and \(\varPhi (x,t,x)=0\) for all \(t\ge 0\), the terms \(\alpha \) and \(\beta \) are
$$\begin{aligned} \alpha= & {} \frac{1}{2}\frac{h^{2}}{n}d_{K}\frac{1}{m(x)}\left( \varPhi ^{\prime \prime }\left( x,t,x\right) m(x)+2\varPhi ^{\prime }\left( x,t,x\right) m^{\prime }(x)\right) +o\left( \frac{h^{2}}{n}\right) , \end{aligned}$$
(35)
$$\begin{aligned} \beta= & {} \frac{1}{2}\frac{h^{2}}{n}d_{K}\frac{1}{m(x)}\left( \varPhi ^{\prime \prime }\left( x,\infty ,x\right) m(x)+2\varPhi ^{\prime }\left( x,\infty ,x\right) m^{\prime }(x)\right) +o\left( \frac{h^{2}}{n}\right) . \end{aligned}$$
(36)
For the term \(\gamma \), it follows that
$$\begin{aligned} \gamma= & {} \frac{1}{n^{2}h^{2}}\frac{1}{m^{2}(x)}\int K^{2}\left( \frac{x-y}{h} \right) \varPhi _{2}(y,t,x)m(y)\mathrm{d}y \nonumber \\= & {} \frac{1}{n^{2}h}\frac{1}{m(x)}\varPhi _{2}(x,t,x)c_{K}+O\left( \frac{h}{ n^{2}}\right) , \end{aligned}$$
(37)
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:
$$\begin{aligned} E\left( C_{1}\cdot C_{2}\right)= & {} \frac{(1-p(x))S(t|x)(1-S(t|x)}{p^{3}(x)}\left[ \frac{1}{4}h^{4}d_{K}^{2}\frac{1}{m^{2}(x)}\right. \nonumber \\&\times \left( \varPhi ^{\prime \prime }\left( x,t,x\right) m(x)+2\varPhi ^{\prime }\left( x,t,x\right) m^{\prime }(x)\right) \nonumber \\&\left. \times \left( \varPhi ^{\prime \prime }\right. \left( x,\infty ,x\right) m(x)+2\varPhi ^{\prime }\left( x,\infty ,x\right) m^{\prime }(x)\right) \nonumber \\&+\left. \frac{1}{nh}\frac{1}{m(x)}\varPhi _{2}(x,t,x)c_{K} \right] +o\left( h^{4}\right) +O\left( \frac{h}{n}\right) . \end{aligned}$$
(38)
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 \(A^{\prime \prime }\):
$$\begin{aligned} A^{\prime \prime }=\int _{0}^{t}E\left[ 1\{v\le T\}|X=y\right] \frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}} =\int _{0}^{t}(1-H(v|y))\frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}}. \end{aligned}$$
Then,
$$\begin{aligned} \varPhi \left( y,t,x\right) =\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|y\right) }{ 1-H(v|x)}-\int _{0}^{t}(1-H(v|y))\frac{\mathrm{d}H^{1}(v|x)}{\left( 1-H(v|x)\right) ^{2}}, \end{aligned}$$
(39)
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 \(t \in [a,b]\),
$$\begin{aligned} \varPhi _{1}\left( x,t,x\right) =\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|x\right) }{\left( 1-H(v|x)\right) ^{2}}. \end{aligned}$$
(40)
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:
$$\begin{aligned} \varPhi _{2}(x,t,x)=\int _{0}^{t}\frac{\mathrm{d}H^{1}\left( v|x\right) }{\left( 1-H(v|x)\right) ^{2}}. \end{aligned}$$
(41)
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
$$\begin{aligned} \sqrt{nh}\sum _{i=1}^{n}\eta _{h}(T_{i},\delta _{i},X_{i},t,x)=-\left( I+II+III+IV\right) , \end{aligned}$$
where
$$\begin{aligned} I= & {} \sqrt{nh}\frac{1}{nh}\frac{S(t|x)}{p(x)m(x)} \\\times & {} \sum _{i=1}^{n}\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t,x)-E\left( K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t,x)\right) \right] , \\ II= & {} \sqrt{nh}\frac{1}{nh}\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)m\left( x\right) } \\\times & {} \sum _{i=1}^{n}\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)-E\left( K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)\right) \right] , \\ III= & {} \sqrt{nh}\frac{1}{nh}\frac{S(t|x)}{p(x)m(x)}\sum _{i=1}^{n}E\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t,x)\right] , \\ IV= & {} \sqrt{nh}\frac{1}{nh}\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)m\left( x\right) }\sum _{i=1}^{n}E\left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)\right] . \end{aligned}$$
The deterministic part b(t, x) comes from \(III+IV\). Recall the function \(\varPhi (y,t,x)\) in (39), since \(\varPhi (x,t,x)=0\), then
$$\begin{aligned}&E\left[ K\left( \frac{x-X}{h}\right) \xi (T,\delta ,t,x)\right] \nonumber \\&\quad = \frac{1}{2} h^{3}d_{K}\left( \varPhi ^{\prime \prime }(x,t,x)m(x) + 2\varPhi ^{\prime }(x,t,x)m^{\prime }(x)\right) +o(h^{3}). \end{aligned}$$
(42)
Therefore,
$$\begin{aligned} III= & {} \sqrt{nh^{5}}\frac{S(t|x)}{p(x)m(x)}\frac{1}{2}d_{K}\left( \varPhi ^{\prime \prime }(x,t,x)m(x)+2\varPhi ^{\prime }(x,t,x)m^{\prime }(x)\right) \left( 1+o\left( 1\right) \right) ,\\ IV= & {} \sqrt{nh^{5}}\frac{(1-p(x))(1-S(t|x))}{p^{2}(x)m\left( x\right) }\frac{ 1}{2}d_{K}\nonumber \\&\quad \times \left( \varPhi ^{\prime \prime }(x,\infty ,x)m(x) + 2\varPhi ^{\prime }(x,\infty ,x)m^{\prime }(x)\right) \left( 1+o(1)\right) . \end{aligned}$$
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:
$$\begin{aligned} I+II=\sum _{i=1}^{n}\left( \gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t)\right) , \end{aligned}$$
where
$$\begin{aligned} \gamma _{i,n}(x,t)= & {} \frac{1}{\sqrt{nh}}\frac{S(t|x)}{p(x)m(x)}\\\times & {} \left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t ,x)-E\left( K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},t ,x)\right) \right] , \\ \varGamma _{i,n}(x,t)= & {} \frac{1}{\sqrt{nh}}\frac{(1-p(x))(1-S(t|x))}{ p^{2}(x)m\left( x\right) } \\\times & {} \left[ K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)-E\left( K\left( \frac{x-X_{i}}{h}\right) \xi (T_{i},\delta _{i},\infty ,x)\right) \right] , \end{aligned}$$
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
$$\begin{aligned} \sigma _{i,n}^{2}\left( x,t\right) =\mathrm{Var}\left( \gamma _{i,n}(x,t)\right) +\mathrm{Var}\left( \varGamma _{i,n}(x,t)\right) +2\mathrm{Cov}\left( \gamma _{i,n}(x,t),\varGamma _{i,n}(x,t)\right) \end{aligned}$$
(43)
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
$$\begin{aligned} Var\left( \gamma _{i,n}(x,t)\right) =\frac{1}{n}\left( \frac{S(t|x)}{p(x)} \right) ^{2}\frac{\varPhi _{1}(x,t,x)}{m\left( x\right) }c_{K}+O\left( \frac{ h^{2}}{n}\right) . \end{aligned}$$
(44)
In a similar way, the second term in (43) is
$$\begin{aligned} Var\left( \varGamma _{i,n}(x,t)\right) =\frac{1}{n}\left( \frac{(1-p(x))(1-S(t|x))}{p^{2}(x)}\right) ^{2}\frac{ \varPhi _{1}(x,\infty ,x)}{m\left( x\right) }c_{K}+O\left( \frac{h^{2}}{n} \right) . \end{aligned}$$
(45)
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
$$\begin{aligned} Cov\left( \gamma _{i,n}(x,t),\varGamma _{i,n}(x,t)\right) {=}\frac{1}{n}\frac{ (1\!-\!p(x))S(t|x)(1\!-\!S(t|x))}{p^{3}(x)m(x) }\varPhi _{2}(x,t,x)c_{K}+O\left( \frac{h}{n}\right) . \end{aligned}$$
(46)
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:
$$\begin{aligned} \frac{1}{\sigma _{n}^{2}\left( x,t\right) }\sum _{i=1}^{n}\int _{\{|\gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t)|>\epsilon \sigma _{n}\left( x,t\right) \}}(\gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t))^{2}dP\rightarrow 0,\forall \epsilon >0. \end{aligned}$$
(47)
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
$$\begin{aligned} \frac{1}{\sigma _{n}^{2}\left( x,t\right) }E\left[ \sum _{i=1}^{n}(\gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t))^{2}I_{i,n}\left( x,t\right) \right] =\frac{1 }{\sigma _{n}^{2}\left( x,t\right) }E\left( \eta _{n}\left( x,t\right) \right) , \end{aligned}$$
with
$$\begin{aligned} \eta _{n}\left( x,t\right) =\sum _{i=1}^{n}(\gamma _{i,n}(x,t)+\varGamma _{i,n}(x,t))^{2}I_{i,n}\left( x,t\right) . \end{aligned}$$
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. \(\square \)
