Dimension reduction for kernel-assisted M-estimators with missing response at random

Abstract

To obtain M-estimators of a response variable when the data are missing at random, we can construct three bias-corrected nonparametric estimating equations based on inverse probability weighting, mean imputation, and augmented inverse probability weighting approaches. However, when the dimension of covariate is not low, the estimation efficiency will be affected due to the curse of dimensionality. To address this issue, we propose a two-stage estimation procedure by using the dimension-reduced kernel estimators in conjunction with bias-corrected estimating equations. We show that the resulting three kernel-assisted estimating equations yield asymptotically equivalent M-estimators that achieve the desirable properties. The finite-sample performance of the proposed estimators for response mean, distribution function and quantile is studied through simulation, and an application to HIV-CD4 data set is also presented.

Appendix

• (C1) The true value $\theta _0$ is the unique root of $n^{-1}\sum _{i=1}^ng_l(Y_i,S_i,\delta _i,\theta )=0$, $n^{-1}\sum _{i=1}^ng_l(Y_i,S_i,\delta _i,\theta )$ is differentiable at $\theta =\theta _0$ for $l=1, 2, 3$ with $\sum _{i=1}^n{\partial g_l(Y_i,S_i,\delta _i,\theta _0)}/{\partial \theta } \ne 0$.

• (C2) The function $\varphi (Y,\theta )$ is monotone and continuous in $\theta$, $E|\varphi (Y,\theta )|< \infty$, $\partial \varphi (Y,\theta )/\partial \theta$ is continuous at $\theta =\theta _0$; $E|\partial \varphi (Y,\theta _0)/\partial \theta |< \infty$, $E\{\varphi ^2(Y,\theta )|S\} < \infty$.

• (C3) The kernel $K(\cdot )$ is bounded and has compact support, and is of order $m \ge 2$, i.e., $\int K(s_1,...,s_{d})ds_1 \cdots ds_{d}=1$, $\int s_j^tK(s_1,...,s_{d})ds_1\cdots ds_{d}=0$, and $\int s_j^mK(s_1,...,s_{d})ds_1\cdots ds_{d}\ne 0$ for any $j =1,...,d$ and $t=1,..., m-1$.

• (C4) The function $\pi (S)$ and the S-density function f(S) have continuous and bounded partial derivatives with respect to S up to order m, and $\pi (S)$ are bounded away from 0 and 1.

• (C5) The function $m_{\varphi }(S, \theta )$ is twice continuously differentiable in the neighborhood of S; has bounded partial derivatives up to order m.

• (C6) As $n\rightarrow \infty$, $nh^{2d}\rightarrow \infty$, $nh^{d}/\log n\rightarrow \infty$, $nh^{2m}\rightarrow 0$, and the estimator $\hat{B}$ obtained by SDR is a root-n consistent estimator of B.

Proof of Theorem 1

For $g_2(Y_i, \hat{S}_i,\delta _i,\theta )$, note that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^ng_2(Y_i,\hat{S}_i,\delta _i,\theta ) =&\frac{1}{n}\sum \limits _{i=1}^n\{\delta _i\varphi (Y_i,\theta ) +(1-\delta _i){m}_{\varphi }(S_i, \theta )\}\\&+\frac{1}{n}\sum \limits _{i=1}^n(1-\delta _i)\{{\hat{m}}_{\varphi }(\hat{S}_i,\theta )-m_{\varphi }(S_i, \theta )\}, \end{aligned}$$

where $S_i=BX_i$ and $\hat{S}_i=\hat{B}X_i$. Define $G(S)=f(S)\pi (S)$ and

$$\begin{aligned} \hat{G}_n(S)=\dfrac{1}{n}\sum \limits _{j=1}^{n}\delta _j{K}_h(S_j-S). \end{aligned}$$

Let $\varDelta _n(\hat{S}_i,S_i)=\hat{G}_n(\hat{S})-G(S_i)$. Then,

$$\begin{aligned} \frac{1}{n}\sum \limits _{i=1}^n(1-\delta _i)\{{\hat{m}}_{\varphi }(\hat{S}_i,\theta )-m_{\varphi }(S_i, \theta )\}=A_{n1}+A_{n2}-A_{n3}, \end{aligned}$$

where

$$\begin{aligned} \begin{array}{llll} A_{n1}=\dfrac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n (1-\delta _i){K}_h(\hat{S}_j-\hat{S}_i)\dfrac{\delta _j\{\varphi (Y_j,\theta )-m_{\varphi }(S_j, \theta )\}}{ G(S_i)},\\ A_{n2}=\dfrac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n(1-\delta _i) {K}_h(\hat{S}_j-\hat{S}_i)\dfrac{\delta _j\{m_{\varphi }(S_j, \theta )-m_{\varphi }(S_i, \theta )\}}{G(S_i)},\\ A_{n3}=\dfrac{1}{n}\sum \limits _{i=1}^n(1-\delta _i) \{{\hat{m}}_{\varphi }(\hat{S}_i,\theta )-m_{\varphi }(S_i, \theta )\} \dfrac{\varDelta _n(\hat{S}_i,S_i)}{G(S_i)}. \end{array} \end{aligned}$$

Using the fact $\delta \varphi (Y, \theta )\perp X|B X$, we can show that

$$\begin{aligned} E[\delta _j\{\varphi (Y_j, \theta )-m_{\varphi }(S_j, \theta )\}|X_j]=0. \end{aligned}$$

As in Wang and Chen (2009), we can prove that

$$\begin{aligned} A_{n1}=\dfrac{1}{n}\sum \limits _{i=1}^n\delta _i\{\pi ^{-1}(S_i)-1\} \{\varphi (Y_i, \theta )-m_{\varphi }(S_i, \theta )\}+o_p(n^{-1/2}), \end{aligned}$$

and $A_{n2}=o_p(n^{-1/2})$. Using the arguments in Andrews (1995) and $\Vert \hat{B}-B\Vert =O_p(n^{-1/2})$, it leads to

$$\begin{aligned} \sup _i\big |\hat{m}_{\varphi }(\hat{S}_i, \theta )-m_{\varphi }(S_i, \theta )\big |=o_p(n^{-1/4}), \end{aligned}$$

such that $A_{n3}=o_p(n^{-1/2})$. Thus, we have

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^ng_2(Y_i,\hat{S}_i,\delta _i,\theta )=&\frac{1}{n}\sum _{i=1}^n\{\delta _i\varphi (Y_i,\theta ) +(1-\delta _i){m}_{\varphi }({S}_i, \theta )\}\\&+\dfrac{1}{n}\sum \limits _{i=1}^n\delta _i\{\pi ^{-1}(S_i)-1\} \{\varphi (Y_i, \theta )-m_{\varphi }(S_i, \theta )\}\\&+o_p(n^{-1/2}). \end{aligned}$$

It leads to

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^ng_2(Y_i,\hat{S}_i,\delta _i,\theta ) \rightarrow E\varphi (Y,\theta ). \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n\{g_2(Y_i,\hat{S}_i,\delta _i,\theta )-E\varphi (Y,\theta )\} \rightarrow N(0,V(\theta )^2). \end{aligned}$$

For $g_1(Y_i, \hat{S}_i,\delta _i,\theta )$, we have

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^ng_1(Y_i,\hat{S}_i,\delta _i,\theta )&=\frac{1}{n}\sum \limits _{i=1}^n\Big [\frac{\delta _i\varphi (Y_i,\theta )}{\pi (S_i)} +\frac{\delta _i\varphi (Y_i,\theta )\{\pi (S_i)- \hat{\pi }(\hat{S}_i)\}}{\pi ^2(S_i)}\Big ]\\&\quad +\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _i\varphi (Y_i,\theta )\{\pi (S_i)- \hat{\pi }(\hat{S}_i)\}^2}{\pi ^2(S_i)\hat{\pi }(\hat{S}_i)}. \end{aligned}$$

Using the similar arguments in Wang (2007), we can prove that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^ng_1(Y_i,\hat{S}_i,\delta _i,\theta ) =\frac{1}{n}\sum \limits _{i=1}^n\Big [\frac{\delta _i\varphi (Y_i,\theta )}{\pi (S_i)}+\big \{1-\frac{\delta _i}{\pi (S_i)}\big \}{m}_{\varphi }(S_i, \theta )\Big ], \end{aligned}$$

which leads to

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^ng_1(Y_i,\hat{S}_i,\delta _i,\theta ) \rightarrow E\varphi (Y,\theta ), \end{aligned}$$

and

$$\begin{aligned} \frac{1}{\sqrt{n}}\sum _{i=1}^n\{g_1(Y_i,\hat{S}_i,\delta _i,\theta )-E\varphi (Y,\theta )\} \rightarrow N(0,V(\theta )^2). \end{aligned}$$

For $g_3(Y_i, \hat{S}_i,\delta _i,\theta )$, it can be seen that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^ng_3(Y_i,\hat{S}_i,\delta _i,\theta )&=\frac{1}{n}\sum \limits _{i=1}^n\Big [\frac{\delta _i\varphi (Y_i,\theta )}{\pi (S_i)}+\big \{1-\frac{\delta _i}{\pi (S_i)}\big \}{m}_{\varphi }(S_i, \theta )\Big ]\\&\quad +\frac{1}{n}\sum \limits _{i=1}^n\Big [\Big \{\frac{\delta _i}{\hat{\pi }(\hat{S}_i)}-\frac{\delta _i}{{\pi }({S}_i)}\Big \}\Big \{\varphi (Y_i,\theta )-{m}_{\varphi }(S_i, \theta )\Big \}\Big ]\\&\quad +\frac{1}{n}\sum \limits _{i=1}^n\Big [\Big \{1-\frac{\delta _i}{\hat{\pi }(\hat{S}_i)}\Big \}\Big \{\hat{m}_{\varphi }(\hat{S}_i, \theta )-{m}_{\varphi }(S_i, \theta )\Big \}\Big ]. \end{aligned}$$

Using the similar arguments for $g_1(Y_i, \hat{S}_i,\delta _i,\theta )$ and $g_2(Y_i, \hat{S}_i,\delta _i,\theta )$, it can be proved that the last two terms on the right side of the above equation are $o_p(1)$. The proof is completed. $\square$

Proof of Theorem 2

By Taylor expansion, there exists ${\theta }_l^*$ between $\hat{\theta }_l$ and $\hat{\theta }_0$, $l=1,2,3,$ such that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^ng_l(Y_i,\hat{S}_i,\delta _i,\hat{\theta }_l) =\frac{1}{n}\sum _{i=1}^ng_l(Y_i,\hat{S}_i,\delta _i,{\theta }_0)+\frac{1}{n}\sum _{i=1}^n \frac{\partial g_l(Y_i,\hat{S}_i,\delta _i,{\theta }_l^*)}{\partial \theta } (\hat{\theta }_l-\hat{\theta }_0). \end{aligned}$$

Since $\sum _{i=1}^ng_l(Y_i,\hat{S}_i,\delta _i,\hat{\theta }_l)=0$, we have

$$\begin{aligned} \sqrt{n}(\hat{\theta }_l-{\theta }_0)=-\sqrt{n}\Big \{\frac{1}{n}\sum _{i=1}^n \frac{\partial g_l(Y_i,\hat{S}_i,\delta _i,{\theta }_l^*)}{\partial \theta }\Big \}^{-1}\frac{1}{n}\sum _{i=1}^ng_l(Y_i,\hat{S}_i,\delta _i,{\theta }_0). \end{aligned}$$

Similar to Theorem 1, as $n \rightarrow \infty$, it can be proved that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n \frac{\partial g_l(Y_i,\hat{S}_i,\delta _i,{\theta }_l^*)}{\partial \theta } \rightarrow E\Big \{\frac{\varphi (Y,\theta _0)}{\partial \theta }\Big \}. \end{aligned}$$

The proof is completed. $\square$

Lemma 1

Assume that $P(\delta = 1|X)>0$ and $P(Y=0|X)=0$. For any given y, it can be verified that

$$\begin{aligned} \mathcal {S}_{\delta I(Y \le y)|X} \subseteq \mathcal {S}_{\delta Y|X}, \end{aligned}$$

where $\mathcal {S}$ denotes for the central subspace (Cook 1994).

Proof of Lemma 1

Suppose that B is a basis of $\mathcal {S}_{\delta Y|X}$, such that $\delta Y \perp X | BX.$ Then, we have $\mathrm{Pr}(\delta Y=0|X)=\mathrm{Pr}(\delta Y=0|BX)$ and $\mathrm{Pr}(\delta Y \le y |X)=\mathrm{Pr}(\delta Y \le y |BX)$. Note $\mathrm{Pr}(\delta Y \le y |X)=\mathrm{Pr}(\delta =1, Y \le y |X)+I(y \ge 0)\mathrm{Pr}(\delta =0|X)$ and $\mathrm{Pr}(\delta Y =0 |X)=\mathrm{Pr}(\delta =1, Y=0 |X)+\mathrm{Pr}(\delta =0|X)=\mathrm{Pr}(\delta =0 |X).$ We have

$$\begin{aligned} \mathrm{Pr}(\delta I(Y \le y)=1|X)&=\mathrm{Pr}(\delta =1, Y \le y |X)\\&=\mathrm{Pr}(\delta Y \le y |X)-I(y \ge 0)\mathrm{Pr}(\delta =0|X)\\&=\mathrm{Pr}(\delta Y \le y |X)-I(y \ge 0)\mathrm{Pr}(\delta Y=0|X)\\&=\mathrm{Pr}(\delta Y \le y |BX)-I(y \ge 0)\mathrm{Pr}(\delta Y=0|BX)\\&=\mathrm{Pr}(\delta I(Y \le y)=1|BX).\\ \mathrm{Pr}(\delta I(Y \le y)=0|X)&=\mathrm{Pr}(\delta =0|X)+\mathrm{Pr}(\delta =1, Y \ge y |X)\\&=\mathrm{Pr}(\delta Y=0|X)+\mathrm{Pr}(\delta Y \ge y |X)-I(y \le 0)\mathrm{Pr}(\delta =0|X)\\&=\mathrm{Pr}(\delta Y \ge y |X)+I(y \ge 0)\mathrm{Pr}(\delta Y=0|X)\\&=\mathrm{Pr}(\delta Y \ge y |BX)+I(y \ge 0)\mathrm{Pr}(\delta Y=0|BX)\\&=\mathrm{Pr}(\delta I(Y \le y)=0|BX). \end{aligned}$$

$\square$