Beran-based approach for single-index models under censoring

Abstract

In this paper we propose a new method for estimating parameters in a single-index model under censoring based on the Beran estimator for the conditional distribution function. This, likelihood-based method is also a useful and simple tool used for bandwidth selection. Additionally, we perform an extensive simulation study comparing this new Beran-based approach with other existing method based on Kaplan–Meier integrals. Finally, we apply both methods to a primary biliary cirrhosis data set and propose a bootstrap test for the parameters.

Appendix: Proof of Theorem 1

Lemma 1

Let $\tilde{F}_{n\theta }(y|\theta '\mathbf {x})$ be the Beran estimator. Under B1-B5, we have

• a) $\nabla _{\theta }\log (1-\tilde{F}_{n\theta }(y|\theta '\mathbf {x}))\mathop {\rightarrow }\limits ^{n\rightarrow \infty } \nabla _{\theta }\log (1- F_{\theta }(y|\theta '\mathbf {x}))$ in probability and consequently

• b) $\nabla _{\theta }\tilde{F}_{n\theta }(y|\theta '\mathbf {x})\mathop {\rightarrow }\limits ^{n\rightarrow \infty } \nabla _{\theta }F_{\theta }(y|\theta '\mathbf {x})$ in probability.

Proof

Recall that

$$\begin{aligned} \tilde{F}_{n\theta }(y|\theta '\mathbf {x})=1-\prod \limits _{i=1}^n\left[ 1-\frac{B_{in}(\theta '\mathbf {x})1_{\{Y_i\le y\}}\delta _i}{\sum _{j=1}^n 1_{\{Y_j\ge Y_i\} } B_{jn}(\theta '\mathbf {x})}\right] , \end{aligned}$$

where

$$\begin{aligned} B_{in}(\theta '\mathbf {x})=\frac{K\left( \frac{\theta '\mathbf {x}-\theta 'X_i}{h_1}\right) }{\sum _{j=1}^n K\left( \frac{\theta '\mathbf {x}-\theta 'X_j}{h_1}\right) }. \end{aligned}$$

Then

$$\begin{aligned}&{\log (1-\tilde{F}_{n\theta }(y|\theta '\mathbf {x}))= \sum _{i=1}^n \log \left( \frac{\sum _{j=1}^n 1_{\{Y_j> Y_i\} } B_{jn}(\theta '\mathbf {x})}{\sum _{j=1}^n 1_{\{Y_j\ge Y_i\} } B_{jn}(\theta '\mathbf {x})}\right) ^{1_{\{Y_i\le y\}}\delta _i}}\\&\quad = \sum _{i=1}^n 1_{\{Y_i\le y\}}\delta _i\left( \log \left( \sum _{j=1}^n 1_{\{Y_j> Y_i\} } B_{jn}(\theta '\mathbf {x})\right) -\log \left( \sum _{j=1}^n 1_{\{Y_j\ge Y_i\} } B_{jn}(\theta '\mathbf {x})\right) \right) \end{aligned}$$

Hence

$$\begin{aligned}&{\nabla _{\theta }\log (1-\tilde{F}_{n\theta }(y|\theta '\mathbf {x})) }\\&\quad = \sum _{i=1}^n 1_{\{Y_i\le y\}}\delta _i\left( \frac{\sum _{j=1}^n 1_{\{Y_j> Y_i\} } \nabla _{\theta }B_{jn}(\theta '\mathbf {x})}{\sum _{j=1}^n 1_{\{Y_j> Y_i\} } B_{jn}(\theta '\mathbf {x})}-\frac{\sum _{j=1}^n 1_{\{Y_j\ge Y_i\} } \nabla _{\theta }B_{jn}(\theta '\mathbf {x})}{\sum _{j=1}^n 1_{\{Y_j\ge Y_i\} } B_{jn}(\theta '\mathbf {x})}\right) \\&\quad = \sum _{i=1}^n 1_{\{Y_i\le y\}}\delta _i \frac{B_{in}(\theta '\mathbf {x})\sum _{j=1}^n 1_{\{Y_j> Y_i\} } \nabla _{\theta }B_{jn}(\theta '\mathbf {x})-\nabla _{\theta }B_{in}(\theta '\mathbf {x} )\sum _{j=1}^n 1_{\{Y_j> Y_i\} } B_{jn}(\theta '\mathbf {x} )}{\biggr (\sum _{j=1}^n 1_{\{Y_j> Y_i\} } B_{jn}(\theta '\mathbf {x} )\biggr )\biggr (\sum _{j=1}^n 1_{\{Y_j\ge Y_i\} } B_{jn}(\theta '\mathbf {x} )\biggr )}. \end{aligned}$$

Moreover,

$$\begin{aligned} \nabla _{\theta }B_{jn}(\theta '\mathbf {x} )&= \frac{\frac{\mathbf {x}-X_i}{nh_1^2} K'\left( \frac{\theta '\mathbf {x}-\theta 'X_i}{h_1}\right) }{\frac{1}{nh_1}\sum _{j=1}^n K\left( \frac{\theta '\mathbf {x} -\theta 'X_j}{h_1}\right) }\\&\quad - \frac{1}{nh_1}K\left( \frac{\theta '\mathbf {x}-\theta 'X_i}{h_1}\right) \frac{\frac{1}{nh_1^2}\sum _{j=1}^n (\mathbf {x}-X_j)K'\left( \frac{\theta '\mathbf {x}-\theta 'X_j}{h_1}\right) }{\left( \frac{1}{nh_1}\sum _{j=1}^n K\left( \frac{\theta '\mathbf {x} -\theta 'X_j}{h_1}\right) \right) ^2}. \end{aligned}$$

Let

$$\begin{aligned} \tilde{H}(y|\theta '\mathbf {x} )=\mathbb {P}(Z\le y, \delta =1|\theta 'X=\theta '\mathbf {x} ) \end{aligned}$$

and

$$\begin{aligned} H(y|\theta '\mathbf {x} )=\mathbb {P}(Z\le y|\theta 'X=\theta '\mathbf {x} ). \end{aligned}$$

Hence if $H$ is continuous, it is easy to show that

$$\begin{aligned} \nabla _{\theta }\log (1-\tilde{F}_{n\theta }(y|\theta '\mathbf {x} ))\rightarrow -\nabla _{\theta }\int \limits _0^y \frac{\tilde{H}(dz|\theta '\mathbf {x} )}{1- H(z|\theta '\mathbf {x} )} \end{aligned}$$

Finally, using (vi) in González-Manteiga and Cadarso-Suárez (1994), it is obvious that

$$\begin{aligned} \nabla _{\theta }\log (1-F_{\theta }(y|\theta '\mathbf {x}))=-\nabla _{\theta }\int \limits _0^y \frac{\tilde{H}(dz|\theta '\mathbf {x} )}{1- H(z|\theta '\mathbf {x} )}. \end{aligned}$$

As to b), since $\tilde{F}_{n\theta }(y|\theta '\mathbf {x} )\rightarrow F_{\theta }(y|\theta '\mathbf {x} )$ and $\nabla _{\theta }\tilde{F}_{n\theta }(y|\theta '\mathbf {x} )=-\nabla _{\theta }\log (1-\tilde{F}_{n\theta }(y|\theta '\mathbf {x} ))(1-\tilde{F}_{n\theta }(y|\theta '\mathbf {x} ))$, the proof is completed.

Lemma 2

Let $\tilde{f}_{\theta }(y|\theta '\mathbf {x} )$ be the estimated density and $\tilde{F}_{\theta }^S(y|\theta '\mathbf {x} )$ the smoothed distribution function estimator defined in (2) and (3). Then, under B1-B5 and when $n\rightarrow \infty$

• a) $\nabla _{\theta }\tilde{f}_{\theta }(y|\theta '\mathbf {x} )_{\theta =\theta _0}\rightarrow \nabla _{\theta }f_{\theta }(y|\theta '\mathbf {x} )_{\theta =\theta _0}$.

• b) $\nabla _{\theta }(1-\tilde{F}_{\theta }^S(y|\theta '\mathbf {x} ))_{\theta =\theta _0}\rightarrow \nabla _{\theta }(1-F_{\theta }(y|\theta '\mathbf {x} ))_{\theta =\theta _0}$.

Proof

Recall,

$$\begin{aligned} \tilde{f}_{\theta }(y|\theta '\mathbf {x} )=\frac{1}{h_2}\sum _{i=1}^n W_{in}(\theta '\mathbf {x} ) K\left( \frac{y-Y_i}{h_2}\right) . \end{aligned}$$

Since $\sum _{i=1}^n B_{in}(\theta '\mathbf {x} )=1$, we have

$$\begin{aligned} W_{in}(\theta '\mathbf {x} )=\frac{\delta _i B_{in}(\theta '\mathbf {x})}{1-G_{n\theta }(Y_i-|\theta '\mathbf {x} )}, \end{aligned}$$

where

$$\begin{aligned} 1-G_{n\theta }(y|\theta '\mathbf {x})=\prod \limits _{j=1}^n\left[ 1-\frac{B_{jn}(\theta '\mathbf {x} )1_{\{Y_j\le y\}}(1-\delta _j)}{\sum _{k=1}^n 1_{\{Y_k\ge Y_j\} } B_{kn}(\theta '\mathbf {x} )}\right] . \end{aligned}$$

Hence

$$\begin{aligned} \nabla _{\theta }W_{in}(\theta '\mathbf {x} )&= \frac{\delta _i \nabla _{\theta }B_{in}(\theta '\mathbf {x})}{1-G_{n\theta }(Y_i-|\theta '\mathbf {x} )}\\&\quad -\frac{\delta _i B_{in}(\theta '\mathbf {x} )}{1-G_{n\theta }(Y_i-|\theta '\mathbf {x})}\nabla _{\theta }(\log (1-G_{n\theta }(Y_i-|\theta '\mathbf {x} )). \end{aligned}$$

Furthermore,

$$\begin{aligned} \nabla _{\theta }\tilde{f}_{\theta }(y|\theta '\mathbf {x})=\frac{1}{h_2}\sum _{i=1}^n \nabla _{\theta } W_{in}(\theta '\mathbf {x}) K\left( \frac{y-Y_i}{h_2}\right) =A_n(y,\theta '\mathbf {x})+B_n(y,\theta '\mathbf {x} ), \end{aligned}$$

where

$$\begin{aligned} A_n(y,\theta '\mathbf {x} )=\frac{1}{h_2}\sum _{i=1}^n \frac{\delta _i \nabla _{\theta }B_{in}(\theta '\mathbf {x})}{1-G_{n\theta }(Y_i-|\theta '\mathbf {x} )} K\left( \frac{y-Y_i}{h_2}\right) \end{aligned}$$

and

$$\begin{aligned} B_n(y,\theta '\mathbf {x} )=-\frac{1}{h_2}\sum _{i=1}^n W_{in}(\theta '\mathbf {x} ) \nabla _{\theta }(\log (1-G_{n\theta }(Y_i-|\theta '\mathbf {x}))K\left( \frac{y-Y_i}{h_2}\right) . \end{aligned}$$

As in the proof of Lemma 1, we have

$$\begin{aligned} \nabla _{\theta }B_{in}(\theta '\mathbf {x} )&= \frac{\frac{\mathbf {x}-X_i}{nh_1^2} K'\left( \frac{\theta '\mathbf {x}-\theta 'X_i}{h_1}\right) }{\frac{1}{nh_1}\sum _{j=1}^n K\left( \frac{\theta '\mathbf {x} -\theta 'X_j}{h_1}\right) } \\&\quad - \frac{1}{nh_1}K\left( \frac{\theta '\mathbf {x}-\theta 'X_i}{h_1}\right) \frac{\frac{1}{nh_1^2}\sum _{j=1}^n (\mathbf {x}-X_j)K'\left( \frac{\theta '\mathbf {x}-\theta 'X_j}{h_1}\right) }{\left( \frac{1}{nh_1}\sum _{j=1}^n K\left( \frac{\theta '\mathbf {x} -\theta 'X_j}{h_1}\right) \right) ^2}. \end{aligned}$$

Hence, it is easy to show that

$$\begin{aligned} A_n(y,\theta '\mathbf {x} )=A_{1n}(y,\theta '\mathbf {x})+A_{2n}(y,\theta '\mathbf {x} )+o_{\mathbb {P}}(1), \end{aligned}$$

where

$$\begin{aligned}&A_{1n}(y,\theta '\mathbf {x} )\\&\quad =\frac{1}{f_{\theta 'X}(\theta '\mathbf {x})}\frac{1}{nh_1^2 h_2}\sum _{i=1}^n \frac{\delta _i }{1\!-\!G_{\theta }(Y_i-|\theta '\mathbf {x} )} (\mathbf {x}\!-\!X_i)K'\left( \frac{\theta '\mathbf {x}\!-\!\theta 'X_i}{h_1}\right) K\left( \frac{y-Y_i}{h_2}\right) \end{aligned}$$

and

$$\begin{aligned}&A_{2n}(y,\theta '\mathbf {x})\\&\quad =\frac{\nabla _{\theta }f_{\theta 'X}(\theta '\mathbf {x})}{f_{\theta 'X}^2(\theta '\mathbf {x} )}\frac{1}{nh_1 h_2}\sum _{i=1}^n \frac{\delta _i }{1-G_{\theta }(Y_i-|\theta '\mathbf {x})}K\left( \frac{\theta '\mathbf {x}-\theta 'X_i}{h_1}\right) K\left( \frac{y-Y_i}{h_2}\right) . \end{aligned}$$

As to $A_{1n}(y,\theta '\mathbf {x} )$, it is a sum of iid random variables, whose variance goes to zero when $n\rightarrow \infty$ and $n h_1^3 h_2\rightarrow \infty$. Moreover, the expectation of $A_{1n}(y,\theta '\mathbf {x} )$ equals

$$\begin{aligned}&E(A_{1n}(y,\theta '\mathbf {x} ))\\&\quad =\frac{1}{f_{\theta 'X}(\theta '\mathbf {x})}\frac{1}{h_1^2 h_2}E\left( \frac{1_{\{Z\le C\}} (\mathbf {x}-X)}{1-G_{\theta }(Z-|\theta '\mathbf {x} )}K'\left( \frac{\theta '\mathbf {x}-\theta 'X}{h_1}\right) K\left( \frac{y-Z}{h_2}\right) \right) . \end{aligned}$$

Setting, $\theta =\theta _0$, under B2, B3 and using Taylor expansion, we obtain

$$\begin{aligned} E(A_{1n}(y,\theta _0'\mathbf {x} ))&= \frac{1}{f_{\theta _0'X}(\theta _0'\mathbf {x} )}\frac{1}{h_1^2 h_2}\int \frac{1-G_{\theta _0}(z-|u)}{1-G_{\theta _0}(z-|\theta _0'\mathbf {x} )} E(\mathbf {x} -X|\theta _0' X=u)\\&K'\left( \frac{\theta _0'\mathbf {x} -u}{h_1}\right) K\left( \frac{y-z}{h_2}\right) f_{\theta _0}(z,u)dzdu\\&= \frac{\frac{d}{dt}\left\{ (\mathbf {x} -E(X|\theta _0' X=t)(1-G_{\theta _0}(y-|t))f_{\theta _0}(y,t)\right\} _{t=\theta _0'\mathbf {x} }}{f_{\theta _0'X}(\theta _0'\mathbf {x} )(1-G_{\theta _0}(y-|\theta _0'\mathbf {x} ))}+o(1)\\&= \frac{1}{f_{\theta _0'X}(\theta _0'\mathbf {x} )}\frac{d}{dt}\left\{ (\mathbf {x} -E(X|\theta _0' X=t)f_{\theta _0}(y,t)\right\} _{t=\theta _0'\mathbf {x} }\\&+\frac{\frac{d}{dt}(1-G_{\theta _0}(y-|t))_{t=\theta _0'\mathbf {x} }}{1-G_{\theta _0}(y-|\theta _0'\mathbf {x} )}(\mathbf {x} \\&-E(X|\theta _0' X=\theta _0'\mathbf {x} )f_{\theta _0}(y|\theta _0'\mathbf {x} )+o(1). \end{aligned}$$

Furthermore, using Lemmas 4 and 5 of Strzalkowska-Kominiak and Cao (2013), we have

$$\begin{aligned} \frac{d}{dt}\left\{ (\mathbf {x} -E(X|\theta _0' X=t)f_{\theta _0}(y,t)\right\} _{t=\theta _0'\mathbf {x}}=\nabla _{\theta }f_{\theta }(y,\theta '\mathbf {x} )_{\theta =\theta _0} \end{aligned}$$

and

$$\begin{aligned} \frac{\frac{d}{dt}(1-G_{\theta _0}(y-|t))_{t=\theta _0'\mathbf {x}}}{1-G_{\theta _0}(y-|\theta _0'\mathbf {x} )}(\mathbf {x} -E(X|\theta _0' X=\theta _0'\mathbf {x}))=\nabla _{\theta }\log (1-G_{\theta }(y-|\theta '\mathbf {x}))_{\theta =\theta _0}. \end{aligned}$$

Recall, that the last equation holds even if the conditional distribution function of $C$ given $\theta _0'X$, $G_{\theta _0}(y|\theta _0'\mathbf {x} )$, does not follow the single-index model assumption.

Finally, we obtain

$$\begin{aligned} A_{1n}(y,\theta _0'\mathbf {x} )\rightarrow \frac{\nabla _{\theta }f_{\theta }(y,\theta '\mathbf {x})_{\theta =\theta _0}}{f_{\theta _0'X}(\theta _0'\mathbf {x})}+\nabla _{\theta }\log (1-G_{\theta }(y-|\theta '\mathbf {x}))_{\theta =\theta _0}f_{\theta _0}(y|\theta _0'\mathbf {x} ). \end{aligned}$$

Similarly, we may show that

$$\begin{aligned} A_{2n}(y,\theta _0'\mathbf {x} )\rightarrow \frac{\nabla _{\theta }f_{\theta 'X}(\theta '\mathbf {x})}{f_{\theta 'X}^2(\theta '\mathbf {x} )}f_{\theta _0}(y,\theta _0'\mathbf {x}). \end{aligned}$$

Hence

$$\begin{aligned} A_n(y,\theta _0'\mathbf {x} )\rightarrow \nabla _{\theta }f_{\theta }(y|\theta '\mathbf {x})_{\theta =\theta _0}+\nabla _{\theta }\log (1-G_{\theta }(y-|\theta '\mathbf {x}))_{\theta =\theta _0}f_{\theta _0}(y|\theta _0'\mathbf {x} ). \end{aligned}$$

As to $B_n(y,\theta '\mathbf {x} )$, by a Taylor expansion and using Lemma 1 for the Beran estimator $G_{n\theta }$, we obtain

$$\begin{aligned} B_n(y,\theta _0'\mathbf x)\rightarrow -\nabla _{\theta }\log (1-G_{\theta }(y-|\theta '\mathbf {x}))_{\theta =\theta _0} f_{\theta _0}(y|\theta _0'\mathbf {x} ). \end{aligned}$$

This completes the proof of a). Finally, since

$$\begin{aligned} \tilde{F}_{\theta }^S(y|\theta '\mathbf {x} )=\sum _{i=1}^n W_{in}(\theta '\mathbf {x} ) {\mathbb {K}}\left( \frac{y-Y_i}{h_2}\right) \end{aligned}$$

the proof of b) is similar.

Lemma 3

Let

$$\begin{aligned} l_n(\theta )=\frac{1}{n}\sum _{i=1}^n \biggr (\delta _i \log f_{\theta }(Y_i|\theta 'X_i)+(1-\delta _i)\log (1- F_{\theta }(Y_i|\theta 'X_i))\biggr ), \end{aligned}$$

be the theoretical log-likelihood function and $\theta _0$ the true parameter. Then

$$\begin{aligned} \nabla _{\theta }\tilde{l}_n(\theta )_{\theta =\theta _0}- \nabla _{\theta } l_n(\theta )_{\theta =\theta _0}\rightarrow 0, \end{aligned}$$

when $n$ goes to infinity.

Proof

We have

$$\begin{aligned} \nabla _{\theta }\tilde{l}_n(\theta )=\frac{1}{n}\sum _{i=1}^n \biggr (\delta _i \frac{\nabla _{\theta }\tilde{f}_{\theta }^{-i}(Y_i|\theta 'X_i)}{\tilde{f}_{\theta }^{-i}(Y_i|\theta 'X_i)}+(1-\delta _i)\frac{\nabla _{\theta }(1-\tilde{F}_{\theta }^{S,-i}(Y_i|\theta 'X_i))}{1-\tilde{F}_{\theta }^{S,-i}(Y_i|\theta 'X_i)}\biggr ). \end{aligned}$$

Lemma 2 complete the proof.

Similarly, we can show that

Lemma 4

Under conditions B1–B5,

$$\begin{aligned} \tilde{l}_n^{[2]}(\theta )\rightarrow l^{[2]}(\theta ), \end{aligned}$$

where $\tilde{l}_n^{[2]}(\theta )$ and $l^{[2]}(\theta )$ denote the Hessian matrices of $\tilde{l}_n(\theta )$ and $E(l_n^{[2]}(\theta ))$, respectively.

Proof of Theorem 1

Using Theorem 1 in Strzalkowska-Kominiak and Cao (2013), it is easy to prove that, under B1,

$$\begin{aligned} \theta _0=\arg \max _{\theta }E(l_n(\theta )). \end{aligned}$$

Hence using a Taylor expansion, we have

$$\begin{aligned} E(\nabla _{\theta } l_n(\theta )_{\theta =\theta _0})=0=\nabla _{\theta } \tilde{l}_n(\theta )_{\theta =\tilde{\theta }_n}=\nabla _{\theta } \tilde{l}_n(\theta )_{\theta =\theta _0}+\hat{l}_n^{[2]}(\tilde{\theta }_n^*)(\tilde{\theta }_n-\theta _0), \end{aligned}$$

where $\hat{\theta }_n^*$ is between $\hat{\theta }_n$ and $\theta _0$. Furthermore,

$$\begin{aligned} \tilde{\theta }_n-\theta _0=-\left[ \tilde{l}_n^{[2]}(\tilde{\theta }_n^*)\right] ^{-1}(\nabla _{\theta } \tilde{l}_n(\theta )_{\theta =\theta _0}-E(\nabla _{\theta } l_n(\theta )_{\theta =\theta _0})). \end{aligned}$$

Finally, using Lemmas 3 and 4 and since $\nabla _{\theta } l_n(\theta )\rightarrow E(\nabla _{\theta } l_n(\theta )_{\theta =\theta _0})$ in probability, the proof is completed.