1. INTRODUCTION
Survey data are frequently used in economics and have one prominent characteristic: data heterogeneity across population subgroups. The presence of heterogeneity naturally makes attractive the application of conditional quantile regressions; see Koenker and Bassett (1978, 1982), Buchinsky (1994), Powell (1986), and Newey and Powell (1990), among others. A major concern of modeling survey data is whether the data are poolable. To our knowledge, testing for poolability within the framework of a conditional quantile regression has not yet been studied in the literature, although within the framework of a conditional mean regression it has been investigated by Baltagi, Hidalgo, and Li (1996) and Lavergne (2001), among others. This paper aims to fill in the literature gap by providing a consistent nonparametric test for testing for the equality of unknown conditional quantile curves across population subgroups. The proposed test is a residual-based test derived by recognizing that an orthogonal moment condition only holds under the null hypothesis. Tests based on orthogonal moment(s) have been investigated extensively in other contexts also; related work includes Fan and Li (1996), Zheng (1996, 1998), and many others.
The proposed nonparametric test requires the availability of a consistent nonparametric estimator of the unknown conditional quantile curve. In the literature of nonparametric conditional quantile curve estimation, related work includes the kernel and k-nearest neighbor estimator of Bhattacharya and Gangopadtyay (1990), the spline smoothing estimator of Koenker, Ng, and Portnoy (1994), the weighted Nadaraya–Watson estimator of Hall, Wolff, and Yao (1999), the local linear regression approach of Fan, Hu, and Truong (1994), the double kernel method of Yu and Jones (1998), and many others. Because the proposed test is designed to handle data containing both discrete (with or without natural ordering) and continuous regressors—usually the case of survey data—we extend the estimation method of Fan et al. (1994) to the mixed data case.
We show that the proposed nonparametric test not only is consistent against any fixed alternative but it also has asymptotic power of one against local alternatives converging to the null hypothesis at usual nonparametric rates. Monte Carlo evidence, however, indicates a size-distortion problem as well documented in the literature of nonparametric hypothesis tests in finite samples. In a mean regression context, the technique of wild bootstrap is used to improve the finite-sample performance of a nonparametric test; see Gozalo (1997), Härdle and Mammen (1993), and Li and Wang (1998), for example. The commonly used wild bootstrap, however, fails in a quantile context because the conditional mean of disturbance terms may not be zero for a general quantile regression model. Hence, we propose a modified version of the wild bootstrap procedure in the quantile context. Monte Carlo simulation results indicate that the proposed bootstrap procedure significantly improves the performance of the test in small samples.
The rest of the paper is organized as follows. Section 2 gives the test and its asymptotic results. Section 3 describes the bootstrap procedure and provides theoretical justification. Section 4 presents a Monte Carlo simulation study of the test. Section 5 concludes. All relevant proofs are given in the Appendixes.
2. EQUALITY TEST AND MAIN THEOREMS
Suppose we fit the following nonparametric quantile regression model (at a probability mass θ ∈ (0,1)) to a set of survey data comprising J subgroups with total sample size n:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm001.gif?pub-status=live)
where the θth-conditional quantile of vij given xij is 0 and gθ,j(x) is an unknown function of x, i = 1,2,…, nj; j = 1,2,…, J. Here J (≥2) is a finite number. Denote the sample size of the jth subgroup by nj. We have aj = limn→∞(nj /n), where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm001.gif?pub-status=live)
. Let xij = (x1,ij′,x2,ij′)′ ∈ Rq−p × Rp, where xij is a q × 1 column vector of explanatory variables with p (≥1) continuously distributed variables x2,ij and q − p(≥0) discretely distributed variables x1,ij.
Let Fj(y|x) be the conditional distribution of Y given X = x of the jth subgroup. Then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm002.gif?pub-status=live)
Throughout this paper we assume that Fj(y|x) are absolutely continuous in y for almost all x, which implies that Fj(gθ,j(x)|x) = θ for almost all x. To simplify notation, we suppress the subscript θ of gθ,j(·). The aim of this paper is to test whether gj(·) are the same across subgroups. If they are, the pooled data will be used to estimate the unknown conditional quantile curve to gain more efficiency. Hence, the null and alternative hypotheses under consideration are
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm002.gif?pub-status=live)
Using the pooled data to estimate the unknown θth-conditional quantile curve, g(x), Lemma 1, which follows, shows that g(x) is a weighted average of gj(x) and that g(x) ≡ gj(x) holds for all j only under H0. Hence, the equivalent null and alternative hypotheses can be written as
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm003.gif?pub-status=live)
As in Zheng (1998), we transfer model (1) to the following conditional mean regression model:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm003.gif?pub-status=live)
Let uij = I(yij ≤ g(xij)) − θ. Then under H0, uij = εij; under H1, uij = εij + I(yij ≤ g(xij)) − I(yij ≤ gj(xij)). Hence, E [uij E(uij|xij)] ≥ 0, and the equality holds if and only if H0 is true. To avoid the random denominator problem in nonparametric kernel estimation, we use the density weighted version of E [uij E(uij|xij)] as in Powell, Stock, and Stoker (1989). The leave-one-out estimator of E [uij E(uij|xij) fj(xij)] is
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-95642-mediumThumb-S0266466606060300frm004.jpg?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm004.gif?pub-status=live)
. Also, for any fixed j,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm005.gif?pub-status=live)
where λ = (λ1,…, λq−p)′ are the smoothing parameters for the discrete variables and H = diag(h12,…, hp2) for the continuous variables, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm005.gif?pub-status=live)
. Moreover,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm006.gif?pub-status=live)
defined in Lemma 1 is the nonparametric estimator of g(xij) from the pooled data by using a different set of smoothing parameters
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm007.gif?pub-status=live)
in (7), where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm008.gif?pub-status=live)
.
Let
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm009.gif?pub-status=live)
for any matrix A and x = (x1′,x2′)′ ∈ Rq−p × Rp. We list all the relevant assumptions as follows.
Assumption 1. (xij,yij) are independent and identically distributed (i.i.d.) in the subscript i and independently distributed in the subscript j.
Assumption 2. Let f1,j(x1) be the probability of x1,ij = x1 and f2|1,j(x2|x1) be the conditional density function of x2,ij given x1,ij = x1. Denote fj(x1,x2) = f1,j(x1) f2|1,j(x2|x1). Then they satisfy the following restrictions:
(i) x1,ij can take any value from
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm010.gif?pub-status=live)
, which contains
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm011.gif?pub-status=live)
different points {0,1,…, ct − 1 : ct ≥ 2}t=1q−p. Also,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm012.gif?pub-status=live)
with f1,j(x1) > 0 for any
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm013.gif?pub-status=live)
.
(ii) f2|1,j(x2|x1) > 0, and its first- and second-order (partial) derivatives with respect to x2 are all uniformly bounded.
Assumption 3. The conditional cumulative distribution function (c.d.f.) and probability density function (p.d.f.) of errors vij given xij = x are Fv,j(v|x) and fv,j(v|x), respectively. Also, fv,j(v|x) has uniformly bounded first- and second-order partial derivatives with respect to v and x2. Moreover, fv,j(0|x) > 0 and Fv,j(0|x) = Pr(vij ≤ 0|xij = x) = θ hold for all j on its domain.
Assumption 4. The unknown quantile curves gj(x) satisfy the conditions in Fan et al. (1994). For instance, gj(x) are twice continuously differentiable with respect to x2, and gj(x) and their first- and second-order partial derivatives with respect to x2 are all uniformly bounded.
Assumption 5. The product kernel function is used, K(u1,u2) = K1(u1;λ) × K2(u2), where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm014.gif?pub-status=live)
. (i) k1(u1,s;λs) = 1 if u1,s = 0,λs otherwise, where 0 ≤ λs < 1 for all s; (ii) k2(·) is a nonnegative, symmetric, and bounded function taking values on [−1,1] such that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm015.gif?pub-status=live)
, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm016.gif?pub-status=live)
.
Assumption 6. As n → ∞, the smoothing parameters satisfy the following restrictions: (i) n|H|1/2 → ∞, ∥H∥ → 0, λs → 0 for all s; (ii)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm017.gif?pub-status=live)
; (iii)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm018.gif?pub-status=live)
; and (iv)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm019.gif?pub-status=live)
.
The term K1(u1;λ) in Assumption 5 is similar to what is defined by Racine and Li (2004). One advantage of this smoothing method relative to the conventional estimator with λs = 0 for all s lies in its ability to borrow information from nearby cells.
The construction of Jn in (4) requires consistent nonparametric estimation of g(x). For the continuous case Fan et al. (1994) develop a local linear regression approach to estimate g(x) and give the consistency and the asymptotic normality results of the estimator. Our paper extends their estimation method to the multivariate mixed categorical and continuous cases, and the result is presented in the following lemma.
LEMMA 1. Suppose Assumptions 1–5 and (ii) of Assumption 6 hold. Then we have the following condition.
(i) For any fixed
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm020.gif?pub-status=live)
, there exists a set of weights {λj(x0)}j=1J satisfying
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm021.gif?pub-status=live)
so that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm022.gif?pub-status=live)
.
(ii) Solving the following optimization problem:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm006.gif?pub-status=live)
gives
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm023.gif?pub-status=live)
, where ρθ(u) = u[θ − I(u ≤ 0)] is called the “check function” and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm007.gif?pub-status=live)
and (iii)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm008.gif?pub-status=live)
The discrete variables are excluded from the check function in (6) because the validity of Lemma 1 requires the error term of the Taylor expansion (A.10) in Appendix A to vanish at some appropriate rate.
The following theorem gives the main asymptotic results of the proposed test and Lemmas 6–9 in Appendix A give the proof of the theorem.
THEOREM 2. If Assumptions 1–6 hold, under H0 we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm009.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm024.gif?pub-status=live)
can be consistently estimated by
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm010.gif?pub-status=live)
Under H1 we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm011.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm025.gif?pub-status=live)
. Thus
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm012.gif?pub-status=live)
where Bn = o(n|H|1/4).
To investigate the local power of the test, we consider the following local alternative hypotheses:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm013.gif?pub-status=live)
where dn → 0 as n → ∞, the unknown functions mj(x) are assumed to have continuous second-order partial derivatives with respect to x2, and mj(x) and their first- and second-order partial derivatives are all uniformly bounded.
THEOREM 3. (Local power). If Assumptions 1–6 hold, under local alternatives H1a with dn = n−1/2|H|−1/8, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm014.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm026.gif?pub-status=live)
, and λl(x) are as in Lemma 1.
Theorems 2 and 3 indicate that the proposed test is consistent against any fixed alternative and against local alternatives converging to the null hypothesis at usual nonparametric rates. Let
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm027.gif?pub-status=live)
; then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm028.gif?pub-status=live)
under H0. This is a one-sided test. If Tn is greater than the 95% critical value of the standard normal distribution, then the null hypothesis will be rejected at the significance level of 5%.
3. HOW TO BOOTSTRAP
Bootstrap procedure is usually used to improve the performance of a test in small samples. Härdle and Mammen (1991, 1993) show that the two-point wild bootstrap is valid in the context of nonparametric estimation of conditional mean curves and nonparametric model specification tests. However, the commonly used wild bootstrap fails in a quantile context where the assumption of disturbances having zero conditional mean is not true. Therefore, this paper proposes a modified version of the wild bootstrap procedure for the quantile case.
In our context, we define the estimated residuals as
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm015.gif?pub-status=live)
Let vijb be the bootstrap resamples of
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm029.gif?pub-status=live)
. For each pair of index (i,j), randomly draw with replacement the bootstrap residuals vijb from a two-point distribution
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm030.gif?pub-status=live)
such that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm016.gif?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm017.gif?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm018.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm031.gif?pub-status=live)
denotes a probability measure that puts mass one at x, pij ∈ (0,1). If aij ≤ 0 < bij, (16) gives pij = θ. Given pij = θ, solving (17) and (18) gives
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm032.gif?pub-status=live)
where cij = −dij /(1 + dij) and dij is one of the real roots of the following function:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm019.gif?pub-status=live)
For example, when θ = 0.25, we obtain1
When
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm033.gif?pub-status=live)
, we add a small positive number, δ, to
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm034.gif?pub-status=live)
such that bij > 0 will be obtained. For example, we take δ = 10−8.
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-69057-mediumThumb-S0266466606060300ffm035.jpg?pub-status=live)
We now construct the bootstrap resamples {(xy, yijb)}, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm020.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm036.gif?pub-status=live)
are calculated from the pooled data {(xij,yij)} as described in Lemma 1 with smoothing parameters
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm037.gif?pub-status=live)
. The proposed bootstrap method guarantees that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm038.gif?pub-status=live)
for all i and j, where the probability is taken over the distribution of yijb conditional on the true observations {(xij,yij)}. Lemma 4, which follows, shows that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm039.gif?pub-status=live)
needs to be oversmoothed, as found by Härdle and Mammen (1991) in the conditional mean regression context.
LEMMA 4. Given the assumptions in Lemma 1, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm021.gif?pub-status=live)
Moreover, if
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm040.gif?pub-status=live)
, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm022.gif?pub-status=live)
Given that the bootstrap resamples {(xij,yijb)}, we construct the bootstrap test Tn*. We conduct B bootstrap repetitions to obtain the bootstrap test statistics {Tn,b*}b=1B so that we can construct the (1 − α) quantile tα*. If Tn calculated from the true data is greater than tα*, we reject the null hypothesis of poolability. The validity of the proposed bootstrap procedure requires that the distribution of Tn* should consistently approximate that of Tn derived under the null hypothesis no matter whether the data are or are not generated from the null hypothesis. In what follows, we give the theoretical results and delay the proof to Appendix B.
THEOREM 5. If the assumptions in Lemma 4 hold, then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm023.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm041.gif?pub-status=live)
is the conditional distribution of the bootstrap test Tn* given {(xij,yij)} and d2(G,H), the Mallows distance, is the infimum of E(X − Y)2 over all joint distributions for the pair of random variables X and Y whose marginal distribution functions are G and H, respectively. The functions G and H belong to a functional space Γ2 = {G : ∫x2 dG(x) < ∞}.
Bickel and Freedman (1981) show that convergence in Mallows distance is equivalent to the weak convergence; Pólya's theorem will extend the weak convergence to the strong convergence (Serfling 2002, p. 18).
4. MONTE CARLO SIMULATIONS
In this section we investigate the finite-sample properties of the test by means of Monte Carlo simulations. We consider θ = 0.25, 0.5, and 0.75; n = 200 and 400. The number of Monte Carlo replications is M = 1,000 for each experiment. The data generating processes take the form of
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm024.gif?pub-status=live)
where uij ∼ i.i.d.N(ξθ,1) such that the θth quantile of uij is zero, x1,ij ∈ {0,1,2,3} with natural ordering, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm042.gif?pub-status=live)
for l = 0,1,2,3, for i = 1,2,…, nj, j = 1,2, and n1 = n2. When j = 1, we generate x2,i1 ∼ i.i.d. U [0,1], σ1(x) ≡ 0.5, and d = 0. When j = 2, we generate
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm043.gif?pub-status=live)
, σ2(xi2) = cx2,i2 with c determined by
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm044.gif?pub-status=live)
, and d ∈ {0, 0.5, 1} where d = 0 gives the null hypothesis.
Two pairs of smoothing parameters are used: (λj,hj) for nonparametric estimation from subgroup data and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm045.gif?pub-status=live)
for nonparametric estimation from the pooled data. We set
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm046.gif?pub-status=live)
, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm047.gif?pub-status=live)
are the standard deviations of x2,ij for a given j and for the pooled data, respectively. Assumption 6 implies that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm048.gif?pub-status=live)
; we take β = 0.2 and take α = 0.6 and 0.65 to measure the sensitivity of the test to the choice of bandwidth. Following Racine and Li (2004), we define k1(u;λ) = λs if |u| = s. Here
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm049.gif?pub-status=live)
is the Epanechnikov kernel. We acknowledge the use of the Koenker and d'Orey (1987) algorithm.
Table 1 presents the percentage rejection rates of the test. Generally speaking, the power of the test improves as the sample size increases; the size of the test, however, does not improve much when the sample size increases to 400 from 200, and this size distortion is observed in all cases; a smaller bandwidth (α = 0.65) is good for size and a larger bandwidth (α = 0.6) is good for power; the optimal bandwidth will be different for the test at the different probability masses.
Percentage of rejections using critical values from the standard normal distribution
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-39677-mediumThumb-S0266466606060300tbl001.jpg?pub-status=live)
To improve the performance of the test in small samples, we apply the proposed bootstrap procedure to the case of θ = 0.25 and n = 200 and present the results in Table 2. By Lemma 4, we use
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm050.gif?pub-status=live)
to calculate
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm051.gif?pub-status=live)
, and we replace
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm052.gif?pub-status=live)
defined earlier by
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm053.gif?pub-status=live)
. With 100 bootstrap replications, the results in Table 2 indicate that the performance of the test is improved significantly when the bootstrap critical values are used—smaller size distortions and stronger powers are observed.
Percentage of rejections using bootstrap critical values: θ = 0.25 and n = 200
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-22070-mediumThumb-S0266466606060300tbl002.jpg?pub-status=live)
5. CONCLUSION
This paper proposes a consistent nonparametric equality test for testing whether the unknown conditional quantile functions are the same across subgroups within a survey data framework. To improve the small-sample performance of the test, we propose a modified version of the two-point wild bootstrap in the quantile context. Monte Carlo evidence shows that the performance of the test in small samples is adequate but is improved significantly when the bootstrap critical values are used.
APPENDIX A
Proof of Lemma 1. First, for any given
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm054.gif?pub-status=live)
, we will show that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm055.gif?pub-status=live)
calculated from the pooled data converges to a weighted average of g1(x0),…, gJ(x0). To simplify the proof, we consider the linear constant approach with an objective function
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm025.gif?pub-status=live)
If Assumptions 1–5 hold and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm056.gif?pub-status=live)
for all s, as n → ∞, we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-11408-mediumThumb-S0266466606060300frm026.jpg?pub-status=live)
and similarly,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm057.gif?pub-status=live)
. Hence,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm027.gif?pub-status=live)
Because
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm058.gif?pub-status=live)
is convex in a and Q0(a) is continuous in a, by the convexity lemma of Pollard (1991) and Theorem 2.1 of Newey and McFadden (1994), we have that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm129.gif?pub-status=live)
converges in probability to g(x0), the unique minimizer of Q0(a) and the solution of the following equation:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm028.gif?pub-status=live)
By Assumption 3, taking Taylor expansion at 0 yields
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm029.gif?pub-status=live)
where ζ0,j is between 0 and g(x0) − gj(x0). Substituting (A.5) into (A.4) gives
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm030.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm059.gif?pub-status=live)
for all j and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm060.gif?pub-status=live)
. Assumption 2 and the continuity of vij conditional on xij indicate that g(x0) = gj(x0) for all j if and only if H0 is true.
Next, closely following Fan et al. (1994), we give the proof of the rest of Lemma 1. For the sake of convenience, in what follows, the summations begin with 1 and end with n, and the heterogeneity in the j subscript is suppressed without hurting the essence of the proof of this lemma. Hence, the modified linear quantile regression approach minimizes the following objective function:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm031.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm061.gif?pub-status=live)
. Here we define
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm032.gif?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm033.gif?pub-status=live)
and yi* = yi − g(x0) − [dtri ]x2′ g(x0)(x2,i − x2,0), where [dtri ]x2 g(x0) is the first-order partial derivative of g(x) with respect to x2 evaluated at x0, the subscripts 1 correspond to the discrete variables, and the subscripts 2 refer to the continuous variables. Then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm062.gif?pub-status=live)
minimizes the following function:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-96272-mediumThumb-S0266466606060300ffm063.jpg?pub-status=live)
Our proof will repeatedly use the following three results: (a) Taylor expansion of g(x1,0,x2,i) at x0 = (x1,0,x2,0)′ implies that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm034.gif?pub-status=live)
holds uniformly over
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm064.gif?pub-status=live)
; (b) the law of large numbers gives
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm065.gif?pub-status=live)
, where I(yi* = 0) = 1 if yi* = 0, 0 otherwise; and (c) for any x ≠ 0
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm035.gif?pub-status=live)
As a result, if
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm066.gif?pub-status=live)
as n → ∞, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm067.gif?pub-status=live)
, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm036.gif?pub-status=live)
and ξi = g(x0) + [dtri ]x2′ g(x0)(x2,i − x2,0) − g(x1,i,x2,i); also,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm068.gif?pub-status=live)
with ξi lying between ξi + t and ξi. Similarly, we can show that those conditions are also sufficient for Rn = op(1).
Therefore, by the convexity lemma of Pollard (1991), we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm069.gif?pub-status=live)
holds uniformly. Then simple calculations yield
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm037.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm038.gif?pub-status=live)
It is easy to calculate that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm070.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm071.gif?pub-status=live)
. This will complete the proof of (8). █
LEMMA 6. If Assumptions 1–5 and (i) of Assumption 6 hold, then under H0,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm072.gif?pub-status=live)
as n → ∞.
Proof. Under the null hypothesis, uij = εij. For a given j(1 ≤ j ≤ J), denote
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm073.gif?pub-status=live)
. Let zij = (εij,xij); then Hn,j(zij,zkj) = εijεkj Kij,kj is symmetric and E [Hn,j(zij,zkj)|zij] = 0 for any i ≠ k. The central limit theorem of Hall (1984) for a degenerate U-statistic will be used to derive the asymptotic normality result of Jn1,j. Then the asymptotic normality of
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm074.gif?pub-status=live)
will follow.
For k ≠ i ≠ s, given the assumption, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-04710-mediumThumb-S0266466606060300frm039.jpg?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm040.gif?pub-status=live)
where σj2 = θ2(1 − θ)2R2(K2)E [fj(x1,ij,x2,ij)], λmax = max1≤s≤q−p λs, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm041.gif?pub-status=live)
Hence 0 ≤ {E [Gn,j2(zij,zkj)] + nj−1E [Hn,j4(zij,zkj)]}/{E [Hn,j2(zij,zkj)]}2 → 0 if
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm075.gif?pub-status=live)
, and λmax → 0 as nj → ∞. Then Theorem 1 of Hall (1984) gives
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm076.gif?pub-status=live)
. Immediately, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm077.gif?pub-status=live)
, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm078.gif?pub-status=live)
. █
LEMMA 7. If Assumptions 1–6 hold, then under H0 we have n|H|1/4Jn2 = op(1).
Proof. Under
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm079.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm080.gif?pub-status=live)
. Simple calculation gives
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-30360-mediumThumb-S0266466606060300frm042.jpg?pub-status=live)
where i ≠ k ≠ m ≠ s ≠ t. The computational complexity of (A.18) stems from the dependency between
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm081.gif?pub-status=live)
because
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm082.gif?pub-status=live)
depends on vij. To overcome this problem, we rewrite
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm083.gif?pub-status=live)
by (A.12)–(A.14):
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm084.gif?pub-status=live)
, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm043.gif?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm085.gif?pub-status=live)
is the product kernel with smoothing parameters
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm086.gif?pub-status=live)
. The rest of the proofs follow by recognizing that only Δkj,ij is correlated with εij for all k ≠ i conditional on x. Without hurting the essence of the proof, we also assume that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm087.gif?pub-status=live)
is the leave-one-out estimate of g(xkj) to remove the dependence between
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm088.gif?pub-status=live)
; this is only assumed to shorten the proof.
First, we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-46984-mediumThumb-S0266466606060300frm044.jpg?pub-status=live)
where ξkj lies between
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm089.gif?pub-status=live)
. Second, define
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm090.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm091.gif?pub-status=live)
. The terms
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm092.gif?pub-status=live)
are independent of εij and εsj for all k ≠ s ≠ i conditional on x. Some calculations give
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm093.gif?pub-status=live)
, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm094.gif?pub-status=live)
. Hence E(n|H|1/4Jn2)2 = o(1) under Assumption 6.
Similarly, we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm095.gif?pub-status=live)
if
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm096.gif?pub-status=live)
because
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm097.gif?pub-status=live)
holds for k ≠ i. This will complete the proof of this lemma. █
LEMMA 8. If Assumptions 1–6 hold, then under H0 we have n|H|1/4Jn3 = op(1).
Proof. First we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm045.gif?pub-status=live)
if
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm098.gif?pub-status=live)
as n → ∞. Second, we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-25741-mediumThumb-S0266466606060300frm046.jpg?pub-status=live)
where i ≠ s ≠ m ≠ k ≠ t under Assumptions 1–6, because
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-65187-mediumThumb-S0266466606060300ffm099.jpg?pub-status=live)
Equations (A.21) and (A.22) together give n|H|1/4Jn3 = op(1). █
LEMMA 9. If Assumptions 1–6 hold, then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm100.gif?pub-status=live)
as n → ∞.
Proof. For a given j, let
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm101.gif?pub-status=live)
, a U-statistic with Hn,j(xij,xkj) = Kij,kj2, which is a symmetric function of
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm102.gif?pub-status=live)
. Hence, Lemma 3.1 of Powell et al. (1989) yields
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm103.gif?pub-status=live)
Hence
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm104.gif?pub-status=live)
. █
LEMMA 10. If Assumptions 1–6 hold, then under H1, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm047.gif?pub-status=live)
Proof. Lemma 1 shows that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm105.gif?pub-status=live)
also holds under H1, and it is easy to show that Jn2 and Jn3 are both op(1). Hence, by (4), we only need to show that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-58832-mediumThumb-S0266466606060300frm048.jpg?pub-status=live)
where uij = I(yij ≤ g(xij)) − θ .
Because
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm106.gif?pub-status=live)
, by Lemma 3.1 of Powell et al. (1989), we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-61715-mediumThumb-S0266466606060300frm049.jpg?pub-status=live)
where wij = I(yij ≤ g(xij)) − I(yij ≤ gj(xij)). This gives (A.24) and will complete the proof of this lemma. █
Proof of Theorem 3. Under
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm107.gif?pub-status=live)
as n → ∞. Similar to the proofs in Lemmas 7 and 8, we obtain that n|H|1/4Jn2 and n|H|1/4Jn3 are both op(1). So we only need to show that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm108.gif?pub-status=live)
.
By (4), we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-72333-mediumThumb-S0266466606060300frm050.jpg?pub-status=live)
where Lemma 6 gives
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm109.gif?pub-status=live)
and simple calculations give Bn = op(1).
Let
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm110.gif?pub-status=live)
. We obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-59666-mediumThumb-S0266466606060300ffm111.jpg?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm112.gif?pub-status=live)
. Following the proof of Lemma 8, we have var(Cn) = o(1). Hence, Cn = μ + op(1), which will complete the proof of this theorem. █
APPENDIX B
We have bootstrap resamples {(xij,yijb)}, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm051.gif?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm052.gif?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm053.gif?pub-status=live)
where a1,b2 > 0 and a2,b1 < 0. To simplify the proofs, we assume that the leave-one-out nonparametric estimation of g(·) is applied.
Proof of Lemma 4. The proof of this lemma closely follows that of Lemma 1 in Appendix A, replacing yi by yib, yi* by yib*, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm113.gif?pub-status=live)
by
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm114.gif?pub-status=live)
in the proof of Lemma 1, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm115.gif?pub-status=live)
, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm116.gif?pub-status=live)
.
By (B.1)–(B.3), a2 < 0 < a1, and b1 < 0 < b2, simple calculations give
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-15280-mediumThumb-S0266466606060300frm054.jpg?pub-status=live)
Treating the last equation as a function of t and applying Taylor expansion to this function at t = 0 yields
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm055.gif?pub-status=live)
where c = a2−1 − a1−1 = b1−1 − b2−1. We then obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm117.gif?pub-status=live)
, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm118.gif?pub-status=live)
If Assumptions 1–5 and (ii) of Assumption 6 hold, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300frm056.gif?pub-status=live)
where ξ(x0,θ) = θ if ∂2g(x0)/∂x2∂x2′ is a nonnegative definite matrix, otherwise 1 − θ. It can be shown that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm119.gif?pub-status=live)
has the same order as
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm120.gif?pub-status=live)
in the proof of Lemma 1, and this will complete the proof of (21).
Next, we are going to show that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm121.gif?pub-status=live)
, where e1 is a (p + 1) × 1 vector with the first element being one and zero for the rest. By (B.4), we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-10798-mediumThumb-S0266466606060300ffm122.jpg?pub-status=live)
where w(·) is a uniformly bounded function over its domain. If
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm123.gif?pub-status=live)
, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm124.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm125.gif?pub-status=live)
because it can be shown that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm126.gif?pub-status=live)
based on the results from the proof of Lemma 1. This will complete the proof of this lemma. █
Proof of Theorem 5. The bootstrap test is given by
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary-alt:20170409100816-44163-mediumThumb-S0266466606060300frm057.jpg?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm127.gif?pub-status=live)
, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20170215065628874-0502:S0266466606060300:S0266466606060300ffm128.gif?pub-status=live)
. The proposed bootstrap procedure ensures that Eb(εijb) = 0, Eb(εijb2) = θ(1 − θ), and Eb(εijbεkjb) = 0 for i ≠ k. Following the proofs of Lemma 6, the asymptotic normality of n|H|1/4Jn1b can be easily obtained. Following the proofs in Lemmas 7 and 8, we can show that n|H|1/4Jn2b = op(1) and n|H|1/4Jn3b = op(1), respectively, although the proofs are more complicated because more terms are involved. █