CHEN Min-qiong

(School of Economics and Trade,Xinhua College of Sun Yat-Sen University,Guangzhou 510520,China)

(School of Mathematics,Sun Yat-Sen University,Guangzhou 510275,China)

Abstract:In this paper,we investigate the problem of testing the conditional symmetry of a random vector given another random vector.We propose a new test based on the concept of conditional energy distance.The test statistic has the form of a U-statistic with random kernel.By using the theory of U-statistic,we prove that the test statistic is asymptotically normal under the null hypothesis of conditional symmetry and consistent against any conditional asymmetric distribution.

Keywords: conditional symmetry test;conditional energy distance;U-statistic with random kernel;consistent;asymptotical normality

1 Introduction

In many regression models,specially the econometric models,for the purpose of identification,some distributional assumptions are often imposed on the error term.The assumptions are conditional moment restrictions,independence between observations,and conditional symmetry around zero given the independent variables.There were a few semiparametric estimators proposed under conditional symmetry.Manski[1]and Newey[2]estimated regression models under conditional symmetry.Powell[3]and Newey[4]proposed semiparametric estimations for Tobit models under conditional symmetry.

Despite the wide use of the property of conditional symmetry,tests for conditional symmetry were not addressed very much in the literature.The first tests were proposed by Powell[5]for censored regression models and by Newey and Powell[6]for linear regression models via asymmetric least squares estimation.However these tests are unlikely to be consistent against all conditional asymmetric distributions.Zheng[7]proposed a consistent test of conditional symmetry using a kernel method,but the test statistic contains integral term and is hard to implement.Bai and Ng[8]proposed an alternative test for conditional symmetry for time series models.The test relied on the correct specification of both conditional mean and conditional variance.Hyndman and Yao[9]developed a bootstrap test for the symmetry of conditional density functions based on their improved methods for conditional density estimation,but they didn’t discuss the asymptotic properties of the test statistic,so it is not clear whether the test be consistent or not.Su[10]gave a simple consistent nonparametric test of conditional symmetry based on the principle of conditional characteristic functions,and he[11]also gave an unconditional method by transforming the conditional symmetry test problem to a unconditional test one.Both of the test statistics he presented in paper need a given characteristic function of the probability measure for the value space of the conditional variable.

In this paper,we propose a simple test for conditional symmetry based on the concept of conditional energy distance.The test is shown to be asymptotically normal under the null hypothesis of conditional symmetry and consistent against any conditional asymmetric distribution.Our test statistic only contains the Euclidean distances and kernel function,so it is easy to compute.

2 The Test Statistic for Conditional Symmetry

Székely[12]introduced a new concept named energy distance to measure the difference between two independent probability distributions.If X and Y are independent random vectors in Rpwith cumulative distribution functions(cdf)F and G respectively,then the energy distance between the distributions F and G is defined as

where X0is an i.i.d.copy of X and Y0is an i.i.d.copy of Y,E is the expected value,and|.|denotes the Euclidean norm.One can also write ε(F,G)as ε(X,Y),and call it the energy distance of X and Y.Székely[12]proved that for real-valued random variables this distance is exactly twice Harald Cramér’s distance,that is

In higher dimensions,however,the two distances are different because the energy distance is rotation invariant while Cramér’s distance is not.The equality becomes

where φX(t)is X’s characteristic function and φY(t)is Y ’s characteristic function,cp=.Thus ε(F,G)≥ 0 with equality to zero if and only if F=G.This property makes it2possible to use ε(F,G)for testing goodness-of- fit,homogeneity,etc.in a consistent way.We shall draw the consistent test statistic for conditional symmetry from the thought of energy distance.

Let X be a p dimensional random vector in Euclidean space Rp,Z be a r dimensional random vector in Euclidean space Rr.Denote f(x|z)as the conditional density function of X given Z.Consider the hypothesis

where S(Z)denotes the support of the density function of Z.Note that the null hypothesis(2.3)can be expressed equivalently as

Analogous to the concept of energy distance for two independent vectors,we can also define the conditional energy distance between X and?X given Z as follows.

Definition 2.1 The conditional energy distance?(X,?X|Z)between X and ?X with finite first moment given Z is defined as the square root of

where φX|Z(t)is the conditional characteristic function of X given Z.Therefore H0holds if and only if?(X,?X|Z)=0.

Let Wi=(Xi,Zi),i=1,2,···,n be a sample from the distribution of(X,Z)and denote W=(X,Z)={W1,W2,···,Wn}.Then for the specific value of ε2(X,?X|Z)when given Z=z,?(X,?X|Z)can be rewritten as the form of expectation by the following lemma.

Lemma 2.1 ε2(X,?X|Z=z)can be rewritten as the form of

Therefore,X|Z=zD=?X|Z=z for any z if and only if

Proof Given the event Z=z,we consider

According to the equation[12],

we have

Let

where f(Z)is the density function of Z.Consequently,X|ZD=?X|Z if and only if Sa=0.Naturally,we can choose test statistic for H0as

where Kik=K(H?1(Zi?Zk)).

The test statistic Unhas the advantage that it has zero mean under H0and hence it does not have a finite sample bias term.We show the consistent of Unand its asymptotical normality under H0.

Here,we choose the Gaussian kernel

in Rr,where H is a diagonal matrix diag{h,h,···,h}determined by bandwidth h.With the Gaussian kernel,Piωi(Z)/n is known to be consistent under the following regularity conditions.

(C1)

(C2)hr?→ 0 and nhr?→ ∞ as n?→ ∞.This requires h to be chosen appropriately according to n.

(C3)The density function of Z and the conditional density function f(·|z)are twice differentiable and all of the derivatives are bounded.

3 Asymptotical Normality of Ununder Null Hypothesis

Using the theory of U-statistic discussed by Fan and Li[13]and Lee[14],we have the following asymptotical normality.

Theorem 3.1(Weak convergence)Assume that conditions(C1)–(C3)hold and the second moment of X exists,if the conditional density of X given Z is symmetric and if h?→0 and nhr?→∞ as n?→∞,we have,where σ2is given in(3.5).

Proof Let Pn(W1,W2,W3)=(|X1+X2|?|X1?X2|)K13K23.Note that Pn(W1,W2,W3)is not symmetric with respect to W1,W2,W3,so we symmetrize Pnas

then Uncan be expressed as a U-statistic of degree 3 with random kernel,

Denote that

and

We use Lemma B.4 in Fan and Li[13]to obtain the asymptotical distribution of Ununder H0in the following steps.

Step 1 Under H0,EPn(W1,W2,W3)=0.Note that

Step 2 Under H0,E[Pn(W1,W2,W3)|W1]=0.Because

which also implies that E[Pn(W1,W3,W2)|W1]=0.Moreover,note that

implies E[Pn(W3,W2,W1)|W1]=0.By the definition of Pn(W1,W2,W3),we have

where

with

Similarly,we can prove the rest three terms in(3.4)are all Op(h2r),which implies that=EP2(W1,W2,W3)=Op(h2r).Thus==o(n)holds.

Step 4 We need to prove that,when n?→∞,

where

Moreover,

Therefore

Therefore,under the conditions nhr?→ ∞ and hr?→ 0,we obtain that

According to Lemma B.4 in Fan and Li[13],it follows that

where

with

and

Therefore,we finally obtain thatwith

4 Consistency of Un

The following result provides the consistency of Un.

Theorem 4.1(Consistency)Assume that conditions(C1)–(C3)hold and the second moment of X exists,then as n?→∞,we have

Proof We will complete this proof by two steps.

Step 1 Un=E[Un]+op(1).

We follow the notation in(3.1)and(3.2).According to Lee[14],we have

where

Therefore,we get

So Un=E[Un]+op(1)by the Chebyshev’s inequality.

Due to the definition of Pn(W1,W2,W3),it’s easy to verify that

Consider E[(|X1+X2|?|X1?X2|)K13K23]as follows

Thus,we get

Combining the results in Step 1 and Step 2,we can finally obtain that