Multivariate partially linear regression in the presence of measurement error

Abstract

In this paper, multivariate partially linear model with error in the explanatory variable of nonparametric part, where the response variable is m dimensional, is considered. By modification of local-likelihood method, an estimator of parametric part is driven. Moreover, the asymptotic normality of the generalized least square estimator of the parametric component is investigated when the error distribution function is either ordinarily smooth or super smooth. Applications in the Engel curves are discussed and through Monte Carlo experiments performances of $\hat{\beta }_{n}$ are investigated.

Appendix

The following lemmas are needed to prove Theorem 1. For the proof of the lemmas, the reader is referred to Liang (2000).

Lemma 1

Let $V_{1},\ldots ,V_{n}$ be independent random variables with 0 mean and there is $\delta \ge 0$ for $E\vert V_{ik}\vert ^{2+\delta }\le C<\infty$. Assume $\left\{ a_{ji},i,j=1,\ldots ,n\right\}$ be a sequence of positive numbers such that $\sup _{i,j\le n}\vert a_{ji}\vert \le n^{-p_{1}}$ for some $0<p_{1}<1$ and $\sum _{j=1}^{n}a_{ji}=O\left( n^{p_{2}}\right)$ for $p_{2}\ge \max \left( 0,2/\left( 2+\delta \right) -p_{1}\right)$. Then

$$\begin{aligned} \max _{1\le i\le n} \left| \sum _{j=1}^n a_{ji}V_j\right| =O\left( n^{-s} \log n\right) \quad s=\left( p_1-p_2\right) /2 \; \text {a.s.} \end{aligned}$$

Lemma 2

Suppose that Assumptions 1.1 and 1.3 hold. Then

$$\begin{aligned} \max _{1\le i\le n} \left| G_k\left( T_{i}\right) -\sum _{j=1}^{n}\omega _{n,h}\left( T_{i},T_{j} \right) G_{k}\left( T_{j}\right) \right| =o\left( 1\right) ;\quad k=0,\ldots ,p, \end{aligned}$$

where $G_0\left( .\right) =g\left( .\right)$ and $G_l\left( .\right) =\gamma _l\left( .\right)$ for $l=1,\ldots ,p$.

Lemma 3

Suppose that Assumptions 1.1–1.5 hold. Then

$$\begin{aligned} \lim _{n\rightarrow \infty } n^{-1}\widetilde{\varvec{X}}^\mathrm{T}\widetilde{\varvec{X}}=\mathbf B , \end{aligned}$$

where $\mathbf B$ is given in Assumption 1.1.

Proof of the Theorem

$$\begin{aligned} \varvec{\hat{\beta }}= & {} \left( \widetilde{\varvec{X}}^\mathrm{T}\widetilde{\varvec{X}} \right) ^{-1}\widetilde{\varvec{X}}^\mathrm{T}\widetilde{\varvec{Y}}\\= & {} \left( \widetilde{\varvec{X}}^\mathrm{T}\widetilde{\varvec{X}} \right) ^{-1}\widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \left( \varvec{X\beta }+\varvec{G}+\varvec{\varepsilon } \right) \\= & {} \varvec{\beta }+ \left( \widetilde{\varvec{X}}^\mathrm{T}\widetilde{\varvec{X}} \right) ^{-1}\left[ \widetilde{\varvec{X}}^\mathrm{T} \widetilde{\varvec{G}}+\widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \varvec{\varepsilon } \right] . \end{aligned}$$

Then

$$\begin{aligned} \sqrt{n}\left( \varvec{\hat{\beta }}-\varvec{\beta } \right)= & {} n\left( \widetilde{\varvec{X}}^\mathrm{T}\widetilde{\varvec{X}} \right) ^{-1}\left[ \frac{1}{\sqrt{n}}\widetilde{\varvec{X}}^\mathrm{T} \widetilde{\varvec{G}}+\frac{1}{\sqrt{n}}\widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \varvec{\varepsilon } \right] .\\ \sqrt{n} \text {vec}\left( \varvec{\hat{\beta }}-\varvec{\beta } \right)= & {} n\left[ I_{m}\otimes \left( \widetilde{\varvec{X}}^\mathrm{T}\widetilde{\varvec{X}} \right) ^{-1}\right] \left\{ \frac{1}{\sqrt{n}}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\right] \text {vec}\left( \widetilde{\varvec{G}}\right) \right. \\&+\left. \frac{1}{\sqrt{n}}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{I-\omega }\left( T\right) \right) \right] \text {vec}\left( \varvec{\varepsilon }\right) \right\} \end{aligned}$$

Lemma 3 gives that $n\left[ I_{m}\otimes \left( \widetilde{\varvec{X}}^\mathrm{T}\widetilde{\varvec{X}} \right) ^{-1}\right]$ converges to $\mathbf B ^{-1}$. Hence, we need to show that $\frac{1}{\sqrt{n}}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\right] \text {vec}\left( \widetilde{\varvec{G}}\right)$ and $\frac{1}{\sqrt{n}}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \right] \text {vec}\left( \varvec{\varepsilon }\right)$ converge in probability to zero.

Consider the matrix $\frac{1}{\sqrt{n}}\widetilde{\varvec{X}}^\mathrm{T} \widetilde{\varvec{G}}$. Its krth element is

$$\begin{aligned} \frac{1}{\sqrt{n}}\widetilde{\varvec{x}}_k^\mathrm{T} \widetilde{g}_r=\frac{1}{\sqrt{n}}\left[ \widetilde{\varvec{\gamma }}_k{+}\left( \varvec{ I-\omega }\left( T\right) \right) \mathbf V _{k}\right] ^\mathrm{T}\widetilde{g}_r{=}\frac{1}{\sqrt{n}}\left[ \widetilde{\varvec{\gamma }}_k^\mathrm{T}\widetilde{g}_r{+}{} \mathbf V _{k}^\mathrm{T}\widetilde{g}_r-\left( \varvec{\omega }\left( T\right) \mathbf V _{k}\right) ^\mathrm{T}\widetilde{g}_r\right] . \end{aligned}$$

If we take $\delta =0, V_j=V_k$, $a_{ji}=\widetilde{g}_r$, $1/4 \le p_1 \le 1/3, p_2=1-p_1$ in Lemma 1, then

$$\begin{aligned} \left| \mathbf V _{k}^\mathrm{T}\widetilde{g}_r\right| =O\left( n^{-\left( 2p_1-1\right) /2}\log n\right) . \end{aligned}$$

Using Abel’s inequality $\left| \widetilde{\varvec{\gamma }}_k^\mathrm{T}\widetilde{g}_r \right| \le n \max _{i\le n}\left| \widetilde{g}_r\right| \max _{i\le n}\left| \widetilde{\varvec{\gamma }}_k\right|$. Taking $G_{0}=g_{r}$ and $G_{j}=\varvec{\gamma }_k$ in Lemma 2 one gets

$$\begin{aligned} \max _{i\le n}\left| \widetilde{g}_r\right| =\max _{1\le i\le n} \left| g_r\left( T_{i}\right) -\sum _{j=1}^{n}\omega _{n,h}\left( T_{i},T_{j} \right) g_{r}\left( T_{j}\right) \right| =o\left( 1\right) , \end{aligned}$$

(11)

and

$$\begin{aligned} \max _{i\le n}\left| \widetilde{\varvec{\gamma }}_{k}\right| =\max _{1\le i\le n} \left| \varvec{\gamma }_k\left( T_{i}\right) -\sum _{j=1}^{n}\omega _{n,h}\left( T_{i},T_{j} \right) \gamma _{k}\left( T_{j}\right) \right| =o\left( 1\right) . \end{aligned}$$

(12)

This means

$$\begin{aligned} \left| \widetilde{\varvec{\gamma }}_k^\mathrm{T}\widetilde{g}_r \right| \le o\left( 1\right) . \end{aligned}$$

If we take $\delta =1, V_j=V_{k}$, $a_{ji}=\omega _{n,h}\left( T_{i},T_{j} \right) ,p_1=2/3, p_2=0$ in Lemma 1, we obtain

$$\begin{aligned} \max _{i\le n}\left| \varvec{\omega }\left( T \right) \mathbf V _{k} \right| = O\left( n^{-1/3}\log n\right) . \end{aligned}$$

(13)

Using Abel’s inequality and Eqs. (11) and (13)

$$\begin{aligned} \left| \left( \varvec{\omega }\left( T \right) \mathbf V _{k}\right) ^\mathrm{T}\widetilde{g}_{r}\right| \le n\max _{i\le n}\left| \widetilde{g}_{r}\right| \max _{i\le n}\left| \varvec{\omega }\left( T \right) \mathbf V _{k}\right| = o\left( 1\right) . \end{aligned}$$

All these means that $\frac{1}{\sqrt{n}}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\right] \text {vec}\left( \widetilde{\varvec{G}}\right)$ is $o\left( 1\right)$.

For a given arbitrary fixed vector $\mathbf a \in \mathbb {R}^{pm}/\left\{ 0\right\}$ to show that

$$\begin{aligned} \frac{1}{\sqrt{n}}{} \mathbf a ^\mathrm{T}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \right] \text {vec}\left( \varvec{\varepsilon }\right) \underrightarrow{^{m}} N_{p\times m}\left[ 0, \mathbf a ^\mathrm{T}\left( \varvec{\varSigma }\otimes \mathbf B \right) \mathbf a \right] , \end{aligned}$$

(14)

let $\mathbf c ^\mathrm{T}=\mathbf a ^\mathrm{T}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \right]$. Note that $\mathbf c ^\mathrm{T}=\left( \mathbf c ^\mathrm{T}_1,\ldots ,\mathbf c ^\mathrm{T}_m\right)$, where $\mathbf c ^\mathrm{T}_r=\left( c_{r1},\ldots ,c_{rn}\right)$.

$$\begin{aligned} \mathbf a ^\mathrm{T}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \right] \text {vec}\left( \varvec{\varepsilon }\right) =\mathbf c ^\mathrm{T}\left( \varvec{\varepsilon }\right) =\sum _{r=1}^m\mathbf c ^\mathrm{T}_r\varvec{\varepsilon }_r=\sum _{i=1}^n\mathbf o ^\mathrm{T}_i\mathbf e _i, \end{aligned}$$

where $\mathbf o ^\mathrm{T}_i=\left( c_{1i},\ldots ,c_{mi}\right)$. Using the assumption that the error vectors $\mathbf e _i$ are independent variables with mean vector 0 and matrix of variances and covariances $\varvec{\varSigma }=\sigma _{rs}$ where $r,s=1,\ldots ,m$ we have $E\left( \mathbf o ^\mathrm{T}_i\mathbf e _i\right) =0$ and $Var\left( \mathbf o ^\mathrm{T}_i\mathbf e _i\right) =\mathbf o ^\mathrm{T}_i\varvec{\varSigma }{} \mathbf o _i$.

$$\begin{aligned} Var\left( \sum _{i=1}^n\mathbf o ^\mathrm{T}_i\mathbf e _i\right) =\mathbf a ^\mathrm{T}\left[ \varvec{\varSigma }\otimes \widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \left( \varvec{ I-\omega }\left( T\right) \right) ^\mathrm{T}\widetilde{\varvec{X}}\right] \mathbf a . \end{aligned}$$

Using Lemma 3, $n^{-1}{} \mathbf a ^\mathrm{T}\left[ \varvec{\varSigma }\otimes \widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \left( \varvec{ I-\omega }\left( T\right) \right) ^\mathrm{T}\widetilde{\varvec{X}}\right] \mathbf a \rightarrow \mathbf a ^\mathrm{T}\left[ \varvec{\varSigma }\otimes \varvec{B}\right] \mathbf a$. From Lindeberg condition, we need to show that

$$\begin{aligned} \max _{1\le i\le n}c_{in}^2/\sum _{i=1}^n c_{in}^2\rightarrow 0. \end{aligned}$$

(15)

Hence, if we show that $\max _{1\le i\le n}Var\left( \mathbf o ^\mathrm{T}_i\mathbf e _i\right) =o(n)$, then Eq. (15) will hold. Let $b_r\in \mathbb {R}^{m}$ be the vector where all components are 0 except the rth component, which is 1.

$$\begin{aligned} \max _{1\le i\le n}\left| \mathbf o ^\mathrm{T}_i\varvec{\varSigma }{} \mathbf o _i\right|\le & {} \left\| \varvec{\varSigma }\right\| _2\max _{1\le i\le n}\sum _{r=1}^m c_{ri}^2\le \left\| \varvec{\varSigma }\right\| _2\max _{1\le i\le n}\left( m\max _{1\le r\le m}c_{ri}^2\right) \\= & {} m\left\| \varvec{\varSigma }\right\| _2\max _{1\le r\le m}\left( \max _{1\le i\le n}\left| c_{ri}\right| \right) ^2=m\left\| \varvec{\varSigma }\right\| _2\max _{1\le r\le m}\left\| \left( b^\mathrm{T}_r \otimes \mathbf I _n\right) \mathbf c \right\| _{\infty }^2\\= & {} m\left\| \varvec{\varSigma }\right\| _2\max _{1\le r\le m}\left\| \left( b^\mathrm{T}_r \otimes \mathbf I _n\right) \left[ \mathbf I _m\otimes \left( \varvec{I-\omega }\left( T\right) \right) ^\mathrm{T}\widetilde{\varvec{X}}\right] \mathbf a \right\| _{\infty }^2\\= & {} m\left\| \varvec{\varSigma }\right\| _2\max _{1\le r\le m}\left\| \left( b^\mathrm{T}_r \otimes \left( \varvec{I-\omega }\left( T\right) \right) ^\mathrm{T}\widetilde{\varvec{X}}\right) \mathbf a \right\| _{\infty }^2\\= & {} m\left\| \varvec{\varSigma }\right\| _2\max _{1\le r\le m}\left\| \left( b^\mathrm{T}_r \otimes \left( \varvec{I-\omega }\left( T\right) \right) ^\mathrm{T}\widetilde{\varvec{X}}\right) \mathbf a \right\| _{\infty }^2\\= & {} m\left\| \varvec{\varSigma }\right\| _2\max _{1\le r\le m}\left\| \left( \varvec{I-\omega }\left( T\right) \right) ^\mathrm{T}\widetilde{\varvec{X}}\left( b^\mathrm{T}_r \otimes \mathbf I _p\right) \mathbf a \right\| _{\infty }^2\\= & {} m\left\| \varvec{\varSigma }\right\| _2\max _{1\le r\le m}\left\| \left( \varvec{I-\omega }\left( T\right) \right) ^\mathrm{T}\widetilde{\varvec{X}}k_r\right\| _{\infty }^2, \end{aligned}$$

where $k_r=\left( b^\mathrm{T}_r \otimes \mathbf I _p\right) \mathbf a \in \mathbb {R}^{p}$.

$$\begin{aligned} \left\| \left( \varvec{I-\omega }\left( T\right) \right) ^\mathrm{T}\widetilde{\varvec{X}}k_r\right\| _{\infty }= & {} \left\| \left( \varvec{I-\omega }\left( T\right) \right) ^\mathrm{T}\left[ \left( \varvec{I}-\varvec{\omega }\left( T\right) \right) \varvec{V}+\widetilde{\varvec{\gamma }} \right] k_r\right\| _{\infty }\\= & {} \left\| \left( \varvec{I-\omega }\left( T\right) \right) ^\mathrm{T}\left( \varvec{V}+\widetilde{\varvec{\gamma }}-\varvec{\omega }\left( T\right) \varvec{V} \right) k_r\right\| _{\infty }\\\le & {} \left( 1{+}\left\| \varvec{\omega }\left( T\right) ^\mathrm{T}\right\| _{\infty }\right) \left[ \left\| \varvec{V}k_r\right\| _{\infty }{+}\left\| \widetilde{\varvec{\gamma }}k_r\right\| _{\infty }{+}\left\| \varvec{\omega }\left( T\right) \varvec{V}k_r\right\| _{\infty }\right] . \end{aligned}$$

From Assumption 1.5, $\left\| \varvec{\omega }\left( T\right) ^\mathrm{T}\right\| _{\infty }$ is uniformly bounded. By assumption of theorem, $E\left| \mathbf V _{ik}\right| ^{2+\delta }\le C <\infty$ for $\delta \ge 0$ and all i and k. Using Markov inequality

$$\begin{aligned} P\left( \max _{1\le i\le n}\left| \varvec{V}_{ik}\right| \ge m_n\right) \le \sum _i P\left( \left| \varvec{V}_{ik}\right| \ge m_n\right) \le nCm_n^{-2-\delta } \end{aligned}$$

for any constant $m_n$.

If we take $m_n=n^{1/2}\log n$, then

$$\begin{aligned} P\left( \left\| \varvec{V}_{k}\right\| _\infty > m_n\right) \rightarrow 0. \end{aligned}$$

This means that $\left\| \varvec{V}_{k}\right\| _\infty =o_p(n^{1/2})$ holds.

Using Eqs.  (12) and (13), we get $\left\| \widetilde{\varvec{\gamma }}k_r\right\| _{\infty }=o(1)$ and $\left\| \varvec{\omega }\left( T\right) \varvec{V}k_r\right\| _{\infty }=O\left( n^{-1/3}\log n\right)$, respectively.

Hence, by Cramer–Wold device we have

$$\begin{aligned} n^{-1/2}\left[ I_{m}\otimes \widetilde{\varvec{X}}^\mathrm{T}\left( \varvec{ I-\omega }\left( T\right) \right) \right] \text {vec}\left( \varvec{\varepsilon }\right) \underrightarrow{^{m}} N_{p\times m}\left( 0, \varvec{\varSigma }\otimes \mathbf B \right) . \end{aligned}$$