Z-estimators and auxiliary information for strong mixing processes

Abstract

This paper introduces a weighted Z-estimator for moment condition models, assuming auxiliary information on the unknown distribution of the data and under the assumption of weak dependence (strong mixing processes). We model serial dependence through a simple nonparametric blocking device, routinely used in the bootstrap literature. The weights that carry the auxiliary information are computed by means of generalized empirical likelihood. The resulting weighted estimator is shown to be consistent and asymptotically normal. The proposed estimator is computationally simple and shows nice finite sample features when compared to asymptotically equivalent estimators.

Appendix

In what follows we present the proofs of the theorems presented in Sect. 3 and some auxiliary results. In addition, we use the following notation: $\rightarrow _{p}$ and $\rightarrow _{d}$ denote convergence in probability and convergence in distribution; C is a generic positive constant; CS and T denote Cauchy–Schwarz inequality and triangular inequality respectively; $\left\| \cdot \right\|$ is the Euclidean norm of $\cdot$. The CLT is meant to be a central limit theorem for strong mixing sequences (see e.g. Ibragimov and Linnik 1971) and CMT is the continuous mapping theorem.

Proof of Lemma 1

Consider again

$$\begin{aligned} \widehat{R}\left( \lambda \right) =\frac{1}{b}\sum _{i=1}^{b}\rho \left( \lambda ^{\prime }g_{i}\right) . \end{aligned}$$

Since $\rho$ is concave on the real line, $\widehat{R}$ inherits this property, since concavity is closed under linear combinations. Moreover, Assumptions 2–4 match Assumptions (i)–(iii) as in Theorem 2.7 of Newey and McFadden (1994), implying consistency of $\widehat{\lambda }$. The mean value expansion of the first order for the GEL criterion function can be written as

$$\begin{aligned} 0&=\frac{\partial \widehat{R}\big ( \widehat{\lambda }\big )}{\partial \lambda } = \frac{1}{b}\sum _{i=1}^{b}\rho ^{(1)}\left( \dot{\lambda }'g_{i}\right) g_{i} \\&=-\bar{g}+\left( \frac{q}{b}\sum _{i=1}^{b}\rho ^{(1)}\left( \dot{\lambda }'g_{i}\right) g_{i}g_{i}'\right) \frac{\widehat{\lambda }}{q} \end{aligned}$$

where $\bar{g}: =\sum _{i=1}^{b}g_{i}/b$ and $\dot{\lambda }$ is a mean value between zero and $\widehat{\lambda }$. Since $\widehat{\lambda }$ is consistent and $\left\| \dot{\lambda }\right\| \le \left\| \widehat{\lambda }\right\|$, we have that $\rho ^{(1)}\left( \dot{\lambda }'g_{i}\right) =-1+o_{p}\left( 1\right)$. Multiplying both sides of the equation above by $\sqrt{n}$, we obtain

$$\begin{aligned} 0=-\sqrt{n}\bar{g}-\widehat{\varSigma }\sqrt{n}\frac{\widehat{\lambda }}{q}-o_{p}\left( 1\right) \widehat{\varSigma }\sqrt{n}\frac{\widehat{\lambda }}{q} \end{aligned}$$

where $\widehat{\varSigma }=q\sum _{i=1}^{b}g_{i}g_{i}'/b$. Notice that $\widehat{\varSigma }\sqrt{n}\frac{\widehat{\lambda } }{q}=O_{p}\left( 1\right)$; therefore, direct inspection shows that

$$\begin{aligned} \sqrt{n}\frac{\widehat{\lambda }}{q}=-\widehat{\varSigma }^{-1}\sqrt{n}\bar{g} +o_{p}\left( 1\right) . \end{aligned}$$

(7)

The proof is then completed through direct application of CLT as well as Slutsky theorem. $\square$

Proof of Theorem 1

[Consistency of $\widehat{\beta }_{\pi }$] Let us compute a mean value expansion of

$$\begin{aligned} \widehat{\pi }_{i}=\frac{\rho ^{(1)}\left( \widehat{\lambda }'g_{i}\right) }{ \sum _{j}\rho ^{(1)}\left( \widehat{\lambda }'g_{j}\right) } \end{aligned}$$

about $\lambda =0$, where $\widehat{\lambda }$ is a consistent estimator for $\lambda$:

$$\begin{aligned} \widehat{\pi }_{i}= & {} \frac{1}{b}+\frac{1}{b}\left( \frac{\rho ^{(2)}\left( \dot{ \lambda }'g_{i}\right) g_{i}'}{\frac{1}{b}\sum _{j}\rho ^{(1)}\left( \dot{\lambda }'g_{j}\right) }-\frac{\rho ^{(1)}\left( \widehat{ \lambda }'g_{i}\right) \frac{1}{b}\sum _{j}\rho ^{(2)}\left( \dot{ \lambda }'g_{j}\right) g_{j}'}{\left( \frac{1}{b} \sum _{j}\rho ^{(1)}\left( \dot{\lambda }'g_{j}\right) \right) ^{2}} \right) \left( \widehat{\lambda }-0\right) \\= & {} \frac{1}{b}+\frac{1}{b}\left( \frac{\rho ^{(2)}\left( \dot{\lambda }'g_{i}\right) \widehat{\lambda }'g_{i}}{\frac{1}{b}\sum _{j}\rho ^{(1)}\left( \dot{\lambda }'g_{j}\right) }-\frac{\rho ^{(1)}\left( \widehat{ \lambda }'g_{i}\right) \frac{1}{b}\sum _{j}\rho ^{(2)}\left( \dot{ \lambda }'g_{j}\right) \widehat{\lambda }^{\prime }g_{j}}{\left( \frac{1}{ b}\sum _{j}\rho ^{(1)}\left( \dot{\lambda }'g_{j}\right) \right) ^{2}} \right) . \end{aligned}$$

From results of Lemma 1 we obtain

$$\begin{aligned} \widehat{\pi }_{i}=\frac{1}{b}+\frac{1}{b}\left( \widehat{\lambda }' g_{i}+o_{p}\left( 1\right) \right) \end{aligned}$$

(8)

and

$$\begin{aligned} \widehat{\pi }_{i}=\frac{1}{b}\left( 1+o_{p}\left( 1\right) \right) . \end{aligned}$$

(9)

From Bravo (2009) we have $\widehat{h}_{\pi }\left( \beta \right) = \widehat{m}\left( \beta \right) +O_{p}\left( M/n\right)$. Then, by adding and subtracting $\widehat{h}_{\pi }\left( \widehat{\beta }_{\pi }\right)$ and by T, we have

$$\begin{aligned} \left\| m\left( \widehat{\beta }_{\pi }\right) \right\| \le \left\| m\left( \widehat{\beta }_{\pi }\right) -\widehat{h}_{\pi }\left( \widehat{\beta }_{\pi }\right) \right\| +\left\| \widehat{h}_{\pi }\left( \widehat{\beta }_{\pi }\right) \right\| \end{aligned}$$

Moreover, by optimality of $\widehat{\beta }_{\pi }$ and since $m\left( \beta _{0}\right) =0$, and by some results in Bravo (2009) and T, we get

$$\begin{aligned} \left\| m\left( \widehat{\beta }_{\pi }\right) \right\|\le & {} \left\| m\left( \widehat{\beta }_{\pi }\right) -\widehat{m}\left( \widehat{\beta }_{\pi }\right) \right\| +\left( 1+o_{p}\left( 1\right) \right) \left\| \widehat{m}\left( \beta _{0}\right) -m\left( \beta _{0}\right) \right\| +O_{p}\left( \frac{M }{n}\right) \\\le & {} \sup _{\beta \in {\mathcal {B}}}\left\| m\left( \beta \right) -\widehat{m} \left( \beta \right) \right\| +\left( 1+o_{p}\left( 1\right) \right) \sup _{\beta \in {\mathcal {B}}}\left\| \widehat{m}\left( \beta \right) -m\left( \beta \right) \right\| +O_{p}\left( \frac{M}{n}\right) . \end{aligned}$$

Assumption 3 implies that $\sup _{\beta \in {\mathcal {B}}}\left\| m\left( \beta \right) -\widehat{m}\left( \beta \right) \right\| =o_{p}\left( 1\right)$; hence

$$\begin{aligned} \left\| m\left( \widehat{\beta }_{\pi }\right) \right\| \le o_{p}\left( 1\right) . \end{aligned}$$

Since $m\left( \beta \right)$ is bounded away from zero for $\left\| \beta -\beta _{0}\right\| >\delta$ (Assumption 2), it follows that $\widehat{ \beta }_{\pi }\in \left\| \beta -\beta _{0}\right\| <\delta$. As $\delta$ is arbitrary, $\widehat{\beta }_{\pi }{\mathop {\longrightarrow }\limits ^{p}}\beta _{0}.$$\square$

Proof of Theorem 2

[Asymptotic Normality of $\widehat{\beta }_{\pi }$] Let us consider $\sum _{i}\widehat{\pi }_{i}h_{i}\left( \widehat{\beta }_{\pi }\right) =0,$ by replacing the probabilities with the expression in Eq. (8), we have

$$\begin{aligned} 0=\frac{1}{b}\sum _{i}\left( 1+\widehat{\lambda }^{\prime }g_{i}+o_{p}\left( 1\right) \right) h_{i}\left( \widehat{\beta }_{\pi }\right) . \end{aligned}$$

Then, via a mean value expansion of $h_{i}\left( \widehat{\beta }_{\pi }\right)$ about $\beta _{0}$, for $\widehat{\beta }_{\pi }$ being consistent,

$$\begin{aligned} 0= & {} \sum _{i}\left( 1+\widehat{\lambda }^{\prime }g_{i}+o_{p}\left( 1\right) \right) h_{i}\left( \widehat{\beta }_{\pi }\right) \\= & {} \sum _{i}\left( 1+\widehat{\lambda }^{\prime }g_{i}\right) \left( h_{i}\left( \beta _{0}\right) +\frac{\partial h_{i}\left( \dot{\beta } \right) }{\partial \beta }\left( \widehat{\beta }_{\pi }-\beta _{0}\right) \right) +o_{p}\left( 1\right) \widehat{h}\left( \widehat{\beta }_{\pi }\right) \end{aligned}$$

where $\left\| \dot{\beta }-\beta _{0}\right\| \le \left\| \widehat{ \beta }_{\pi }-\beta _{0}\right\|$. Let us define $\widehat{B}\left( \beta \right) =q\sum _{i}h_{i}\left( \beta \right) g_{i}^{\prime }/b$. Then, by appropriate rescaling and Eq. (7)

$$\begin{aligned} 0= & {} \sqrt{n}\widehat{h}\left( \beta _{0}\right) +\widehat{B}\left( \beta _{0}\right) \widehat{\varSigma }^{-1}\sqrt{n}\bar{g} \\&+\left( \frac{1}{b}\sum _{i}\frac{\partial h_{i}\left( \dot{\beta }\right) }{ \partial \beta }+\widehat{\lambda }^{\prime }\sum _{i}g_{i}\frac{\partial h_{i}\left( \dot{\beta } \right) }{\partial \beta }/b\right) \sqrt{n}\left( \widehat{\beta }_{\pi }-\beta _{0}\right) \\&+o_{p}\left( 1\right) \sqrt{n}\widehat{h}\left( \widehat{\beta }_{\pi }\right) \\= & {} A_{1}+A_{2}+A_{3} \end{aligned}$$

where

$$\begin{aligned} A_{1}= & {} \sqrt{n}\widehat{h}\left( \beta _{0}\right) +\widehat{B}\left( \beta _{0}\right) \widehat{\varSigma }^{-1}\sqrt{n}\bar{g},\\ A_{2}= & {} \left( \sum _{i}\partial h_{i}\left( \dot{\beta }\right) /\partial \beta +\widehat{\lambda }^{\prime }\sum _{i}g_{i}\partial h_{i}\left( \beta _{0}\right) /\partial \beta /b\right) \sqrt{n}\left( \widehat{\beta }_{\pi }-\beta _{0}\right) ,\\ A_{3}= & {} o_{p}\left( 1\right) \sqrt{n}\widehat{h}\left( \widehat{\beta }_{\pi }\right) . \end{aligned}$$

From Assumption 3 and results in Bravo (2009) $\sqrt{n}\widehat{h} \left( \beta _{0}\right) {\mathop {\longrightarrow }\limits ^{d}}N\left( 0,S\left( \beta _{0}\right) \right)$ and $\sqrt{n}\bar{g}{\mathop {\longrightarrow }\limits ^{d}}N\left( 0,\varSigma \right)$. Then, after simple calculations, we get $A_{1}{\mathop {\longrightarrow }\limits ^{d}}N\left( 0,W\right)$, where

$$\begin{aligned} W= & {} \left( \begin{array}{cc} I,&-B\left( \beta _{0}\right) \varSigma ^{-1} \end{array} \right) \left( \begin{array}{cc} S\left( \beta _{0}\right) &{} B\left( \beta _{0}\right) \\ B\left( \beta _{0}\right) ^{\prime } &{} \varSigma \end{array} \right) \left( \begin{array}{c} I \\ -\varSigma ^{-1}B\left( \beta _{0}\right) ^{\prime } \end{array} \right) \\= & {} S\left( \beta _{0}\right) -B\left( \beta _{0}\right) \varSigma ^{-1}B\left( \beta _{0}\right) ^{\prime } \end{aligned}$$

Let us now focus attention on $A_{2}$. By results in Bravo (2009) and T

$$\begin{aligned} \left\| \widehat{\lambda }^{\prime }\frac{1}{b}\sum _{i}g_{i}\frac{\partial h_{i}\left( \dot{\beta }\right) }{\partial \beta ^{\prime }}\right\|\le & {} \left\| \widehat{\lambda }\right\| \left\| \frac{1}{n}\sum _{t}f_{t} \frac{\partial m_{t}\left( \dot{\beta }\right) }{\partial \beta ^{\prime }} +O_{p}\left( \frac{q}{n}\right) \right\| \\\le & {} \left\| \widehat{\lambda }\right\| \frac{1}{n}\sum _{t}\sup _{\beta \in {\mathcal {B}}}\left\| f_{t}\frac{\partial m_{t}\left( \beta \right) }{ \partial \beta ^{\prime }}\right\| +o_{p}\left( 1\right) . \end{aligned}$$

Thus,

$$\begin{aligned} \left\| \widehat{\lambda }^{\prime }\frac{1}{b}\sum _{i}g_{i}\frac{\partial h_{i}\left( \dot{\beta }\right) }{\partial \beta }\right\| \le o_{p}\left( 1\right) . \end{aligned}$$

By CMT and Assumption 3 $\sqrt{n}\widehat{h}\left( \widehat{\beta }_{\pi }\right)$ is normally distributed. Thus, its order of magnitude is $O_{p}\left( 1\right)$ and $A_{3}=o_{p}\left( 1\right)$. Finally,

$$\begin{aligned} M\left( \beta _{0}\right) ^{\prime }\sqrt{n}\left( \widehat{\beta }_{\pi }-\beta _{0}\right) =-\left( \begin{array}{cc} I,&-\widehat{B}\left( \beta _{0}\right) \varSigma ^{-1} \end{array} \right) \left( \begin{array}{c} \sqrt{n}\widehat{h}\left( \beta _{0}\right) \\ \sqrt{n}\bar{g} \end{array} \right) +o_{p}\left( 1\right) \end{aligned}$$

and

$$\begin{aligned} \sqrt{n}\left( \widehat{\beta }_{\pi }-\beta _{0}\right) =-\left( M\left( \beta _{0}\right) ^{\prime }\right) ^{-1}\left( \begin{array}{cc} I,&-\widehat{B}\left( \beta _{0}\right) \varSigma ^{-1} \end{array} \right) \left( \begin{array}{c} \sqrt{n}\widehat{h}\left( \beta _{0}\right) \\ \sqrt{n}\bar{g} \end{array} \right) +o_{p}\left( 1\right) \end{aligned}$$

which implies, by CLT applied to $\sqrt{n}\bar{g}$, Assumption 3 and CMT,

$$\begin{aligned} \sqrt{n}\left( \widehat{\beta }_{\pi }-\beta _{0}\right) {\mathop {\longrightarrow }\limits ^{d}}N\left( 0,\left( M\left( \beta _{0}\right) ^{\prime }\right) ^{-1}\left( S\left( \beta _{0}\right) -B\left( \beta _{0}\right) \varSigma ^{-1}B\left( \beta _{0}\right) ^{\prime }\right) \left( M\left( \beta _{0}\right) \right) ^{-1}\right) . \end{aligned}$$

$\square$

Proof of Corollary 1

From results in Lemma 1 and Theorem 1 we have

$$\begin{aligned} \widehat{\mu }_{\pi }\left( z\right)= & {} \frac{1}{b}\sum _{t=1}^{b}1_{M}\left( z_{i}\le z\right) \left( 1+\widehat{\lambda }^{\prime }g_{i}+o_{p}\left( 1\right) \right) \\&\widehat{\mu }_{b}\left( z\right) -\bar{g}^{\prime }\widehat{\varSigma }^{-1}\frac{q}{ \sqrt{n}b}\sum _{t=1}^{b}g_{i}1_{q}\left( z_{i}\le z\right) +o_{p}\left( \frac{1}{\sqrt{n}}\right) \end{aligned}$$

Then, by adding and subtracting $\mu \left( x\right)$ and multiplying both sides by $\sqrt{n}$, we get

$$\begin{aligned} \sqrt{n}\left( \widehat{\mu }_{\pi }\left( z\right) -\mu \left( x\right) \right)&=\sqrt{n}\left( \widehat{\mu }_{b}\left( z\right) -\mu \left( x\right) \right) - \sqrt{n}\bar{g}^{\prime }\widehat{\varSigma }\frac{1}{b}\sum _{t=1}^{b}g_{i}1_{q} \left( z_{i}\le z\right) +o_{p}\left( 1\right) \\&=\left( \begin{array}{cc} 1,&-\widehat{a}^{\prime } \end{array} \right) \left( \begin{array}{c} \sqrt{n}\left( \widehat{\mu }_{b}\left( z\right) -\mu \left( x\right) \right) \\ \sqrt{n}\bar{g} \end{array} \right) \\&{\mathop {\longrightarrow }\limits ^{d}} \, N\left( 0,\sigma ^{2}-a^{\prime }\varSigma ^{-1}a\right) . \end{aligned}$$

The result follows from an application of CLT and Slutsky’s theorem. $\square$