A model specification test for the variance function in nonparametric regression

Abstract

The problem of testing for the parametric form of the conditional variance is considered in a fully nonparametric regression model. A test statistic based on a weighted $L_2$-distance between the empirical characteristic functions of residuals constructed under the null hypothesis and under the alternative is proposed and studied theoretically. The null asymptotic distribution of the test statistic is obtained and employed to approximate the critical values. Finite sample properties of the proposed test are numerically investigated in several Monte Carlo experiments. The developed results assume independent data. Their extension to dependent observations is also discussed.

A Proofs

1.1 A.1 Sketch of the proofs of results in Sect. 3

Observe that under Assumptions A.1, A.2 and A.3 for $(X_1,Y_1), \ldots , (X_n,Y_n)$ independent from (1), (see, for example, Hansen 2008),

$$\begin{aligned} \begin{array}{lll} \displaystyle \sup _{x\in R}|\hat{m}(x)-m(x)|=o_P(n^{-1/4}),\\ \displaystyle \sup _{x\in R}|\hat{\sigma }(x)-\sigma (x)|=o_P(n^{-1/4}). \\ \end{array} \end{aligned}$$

(13)

Proof of (7) Let $S_n(\theta )$ and $S(\theta )$ be as defined in (3) and (5), respectively. From (13) and the SLLN, it follows that

$$\begin{aligned} S_n(\theta )=S(\theta )+o_P(1), \quad \forall \theta \in \Theta . \end{aligned}$$

Next we prove that the above convergence holds uniformly in $\theta$. Routine calculations show that the derivative

$$\begin{aligned} \frac{\partial }{\partial \theta }\left\{ S_n(\theta )-S(\theta )\right\} , \end{aligned}$$

is bounded (in probability) $\forall \theta \in \Theta _0$, which implies that the family $\left\{ S_n(\theta )-S(\theta ), \, \theta \in \right. \left. \Theta _0\right\}$ is equicontinuous. By the lemma in Yuan (1997), it follows that

$$\begin{aligned} \inf _{\Vert \theta -\theta _0\Vert>\delta } S_n(\theta )-S_n(\theta _0)=\inf _{\Vert \theta -\theta _0\Vert >\delta } S(\theta )-S(\theta _0)+o_p(1). \end{aligned}$$

Now the result follows from Assumption B.1 and Lemma 1 in Wu (1981). $\square$

Proof of (8) Under the assumptions made, routine calculations show that under $H_0$,

$$\begin{aligned} \sqrt{n} \frac{\partial }{\partial \theta } S_n(\theta _0)= & {} -\frac{2}{\sqrt{n}}\sum _{j=1}^n(\varepsilon ^2_j-1)\sigma ^2(X_j; \theta _0)\dot{\sigma }^2(X_j; \theta _0) +o_P(1), \end{aligned}$$

(14)

$$\begin{aligned} \frac{\partial ^2}{\partial \theta \partial \theta ^{T}} S_n(\theta _0)= & {} 2 \Omega +o_P(1). \end{aligned}$$

(15)

By Taylor expansion,

$$\begin{aligned} \frac{\partial }{\partial \theta } S_n(\hat{\theta })-\frac{\partial }{\partial \theta } S_n(\theta _0)=\frac{\partial ^2}{\partial \theta \partial \theta ^{T}} S_n(\theta _0)(\hat{\theta }-\theta _0)+o(\Vert \hat{\theta }-\theta _0\Vert ). \end{aligned}$$

(16)

Taking into account that $\frac{\partial }{\partial \theta } S_n(\hat{\theta })=0$, the result follows from (14)–(16). $\square$

Proof of Theorem 3

From Lemma 10 (i) in Pardo-Fernández et al. (2015b), it follows that

$$\begin{aligned} \begin{array}{rcl} \displaystyle \sqrt{n}\hat{\varphi }(t) &{} = &{} \displaystyle \sqrt{n}\tilde{\varphi }(t)+\mathrm{i}\frac{t}{\sqrt{n}}\sum _{j=1}^n \exp (\mathrm{i}t\varepsilon _j)\frac{m(X_j)-\hat{m}(X_j)}{\sigma (X_j)}\\ &{} &{} \displaystyle +\, \mathrm{i}\frac{t}{\sqrt{n}}\sum _{j=1}^n \exp (\mathrm{i}t\varepsilon _j)\varepsilon _j\frac{\sigma (X_j)-\hat{\sigma }(X_j)}{\sigma (X_j)}+tR_{11}(t)+t^2R_{12}(t), \end{array} \end{aligned}$$

with

$$\begin{aligned} \tilde{\varphi }(t)=\frac{1}{n}\sum _{j=1}^n\exp (\mathrm{i}t \varepsilon _j) \end{aligned}$$

(17)

and $\sup _t |R_{1k}(t)|=o_P(1)$, $k=1,2$. As for $\hat{\varphi }_0(t)$, since

$$\begin{aligned} \begin{array}{rcl} \displaystyle \hat{\varepsilon }_{0j}-\varepsilon _j &{} = &{} \displaystyle \frac{m(X_j)-\hat{m}(X_j)}{\sigma (X_j)}+ \frac{\{m(X_j)-\hat{m}(X_j) \} \{ \sigma (X_j)-\sigma (X_j;\hat{\theta }) \}}{\sigma (X_j)\sigma (X_j;\hat{\theta })}\\ &{} &{} \displaystyle +\, \frac{ \sigma (X_j)-\sigma (X_j;\hat{\theta }) }{\sigma (X_j)}\varepsilon _j+ \frac{ \{ \sigma (X_j)-\sigma (X_j;\hat{\theta }) \}^2}{\sigma (X_j)\sigma (X_j;\hat{\theta })}\varepsilon _j, \end{array} \end{aligned}$$

we have that

$$\begin{aligned} \begin{array}{rcl} \displaystyle \sqrt{n}\hat{\varphi }_0(t) &{} = &{} \displaystyle \sqrt{n}\tilde{\varphi }(t)+\mathrm{i}\frac{t}{\sqrt{n}}\sum _{j=1}^n \exp (\mathrm{i}t\varepsilon _j)\frac{m(X_j)-\hat{m}(X_j)}{\sigma (X_j)}\\ &{} &{} \displaystyle +\, \mathrm{i}\frac{t}{\sqrt{n}}\sum _{j=1}^n \exp (\mathrm{i}t\varepsilon _j)\varepsilon _j\frac{\sigma (X_j)-{\sigma }(X_j;\hat{\theta })}{\sigma (X_j)}+tR_{13}(t)+t^2R_{14}(t), \end{array} \end{aligned}$$

with $\sup _t |R_{1k}(t)|=o_P(1)$, $k=3,4$. From Lemma 11 in Pardo-Fernández et al. (2015b),

$$\begin{aligned} \mathrm{i}\frac{t}{\sqrt{n}}\sum _{j=1}^n \exp (\mathrm{i}t\varepsilon _j)\varepsilon _j\frac{\sigma (X_j)-\hat{\sigma }(X_j)}{\sigma (X_j)}=-\frac{t}{2}\varphi '(t)\frac{1}{\sqrt{n}}\sum _{j=1}^n (\varepsilon _j^2-1)+R_{15}(t) \end{aligned}$$

with $\Vert R_{15}\Vert _w=o_P(1)$. By Taylor expansion and (8),

$$\begin{aligned} \mathrm{i}\frac{t}{\sqrt{n}}\sum _{j=1}^n \exp (\mathrm{i}t\varepsilon _j)\varepsilon _j\frac{\sigma (X_j)-{\sigma }(X_j;\hat{\theta })}{\sigma (X_j)}=V(t)+R_{16}(t), \end{aligned}$$

with $\Vert R_{16}\Vert _w=o_P(1)$ and

$$\begin{aligned} V(t)=\frac{1}{2}\frac{\mathrm{i}t}{n\sqrt{n}} \sum _{j,k=1}^n \exp (\mathrm{i}t\varepsilon _j)\varepsilon _j\frac{\dot{\sigma }^2(X_j;\theta _0)^T}{\sigma ^2(X_j;\theta _0)} \Omega ^{-1} \dot{\sigma }^2(X_k;\theta _0)\sigma ^2(X_k;\theta _0)(\varepsilon _k^2-1). \end{aligned}$$

Routine calculations show that

$$\begin{aligned} V(t)=-\frac{t}{2}\varphi '(t)\frac{1}{\sqrt{n}}\mu ^T\Omega ^{-1}\sum _{j=1}^n\dot{\sigma }^2(X_j;\theta )\sigma ^2(X_j;\theta )(\varepsilon _j^2-1)+R_{17}(t) \end{aligned}$$

with $\Vert R_{17}\Vert _w=o_P(1)$. Therefore, the result follows. $\square$

Proof of Theorem 8

From Lemma 10 (i) in Pardo-Fernández et al. (2015b), it follows that

$$\begin{aligned} \hat{\varphi }(t)=\tilde{\varphi }(t)+tR(t), \end{aligned}$$

(18)

with $\tilde{\varphi }(t)$ as defined in (17) and

$$\begin{aligned} \sup _t |R(t)|=o_P(1). \end{aligned}$$

(19)

As for $\hat{\varphi }_0(t)$, since

$$\begin{aligned}&\hat{\varepsilon }_{0j}-\varepsilon _{0j} = \frac{m(X_j)-\hat{m}(X_j)}{\sigma (X_j;\hat{\theta })}+\varepsilon _j\frac{\sigma (X_j)}{\sigma (X_j;\hat{\theta })\sigma (X_j; \theta _0)} \left\{ \sigma (X_j;\theta _0)-\sigma (X_j;\hat{\theta })\right\} ,\\&\quad \left| \frac{1}{n}\sum _{j=1}^n\frac{m(X_j)-\hat{m}(X_j)}{\sigma (X_j;\hat{\theta })} \right| \le \frac{1}{\displaystyle \inf _{x\in R, \, \theta \in \Theta _0} \sigma (x;\theta )}\sup _{x\in R}|\hat{m}(x)-m(x)|=o_P(1), \end{aligned}$$

and

$$\begin{aligned}&\left| \frac{1}{n}\sum _{j=1}^n \varepsilon _j\frac{\sigma (X_j)}{\sigma (X_j;\hat{\theta })\sigma (X_j; \theta _0)} \left\{ \sigma (X_j;\theta _0)-\sigma (X_j;\hat{\theta })\right\} \right| \\&\quad \le \left( \frac{1}{n}\sum _{j=1}^n \varepsilon _j^2\right) ^{1/2} \frac{\displaystyle \sup _{x\in R} \sigma (X_j)}{\displaystyle \inf _{x\in R, \, \theta \in \Theta _0} \sigma ^3(x;\theta )} \left( \frac{1}{n}\sum _{j=1}^n \left\{ \sigma ^2(X_j;\theta _0)-\sigma ^2(X_j;\hat{\theta })\right\} ^2\right) ^{1/2}\\&\quad =o_P(1), \end{aligned}$$

by Taylor expansion, we get

$$\begin{aligned} \hat{\varphi }_0(t)=\tilde{\varphi }_0(t)+tR_0(t), \end{aligned}$$

(20)

with

$$\begin{aligned} \tilde{\varphi }_0(t)=\frac{1}{n}\sum _{j=1}^n\exp (\mathrm{i}t \varepsilon _{0j}), \qquad \sup _t |R_0(t)|=o_P(1). \end{aligned}$$

(21)

The result follows from (18)–(21), by taking into account that $\Vert \tilde{\varphi }-{\varphi }\Vert _w=o_P(1)$ and $\Vert \tilde{\varphi }_0-{\varphi }_0\Vert _w=o_P(1)$. $\square$

Proof of Theorem 9

Under $H_{1n}$,

$$\begin{aligned} \sqrt{n}(\hat{\theta }-{\theta }_0)= & {} \displaystyle \Omega ^{-1}\frac{1}{\sqrt{n}}\sum _{j=1}^n(\varepsilon _j^2-1)\sigma ^2(X_j;\theta _0)\dot{\sigma }^2(X_j;\theta _0)\\&\displaystyle +\,\Omega ^{-1}E\{r(X)\dot{\sigma }^2(X;\theta _0)\}+o_P(1). \end{aligned}$$

By applying the results in Yuan (1997), we get that (13) also hold under $H_{1n}$. Now, the result follows by proceeding similarly to the proof of Theorem 3. $\square$

1.2 A.2 Sketch of the proofs of results in Sect. 6

Under Assumptions A.2, A.3, C.1–C.4 and C.6 (see, for example, Hansen 2008),

$$\begin{aligned} \begin{array}{lll} \displaystyle \sup _{x\in R_g}|\hat{m}(x)-m(x)|=o_P(n^{-1/4}),\\ \displaystyle \sup _{x\in R_g}|\hat{\sigma }(x)-\sigma (x)|=o_P(n^{-1/4}). \\ \end{array} \end{aligned}$$

(22)

Proof of Theorem 11

Proceeding as in the proof of Theorem 3, we obtain

$$\begin{aligned} \sqrt{n} \left\{ \hat{\varphi }_g(t)-\hat{\varphi }_{0g}(t)\right\} = V_1(t)+V_2(t)+tR_1(t)+t^2R_2(t), \end{aligned}$$

with $\sup _t |R_{k}(t)|=o_P(1)$, $k=1,2$,

$$\begin{aligned} V_1(t)= & {} \mathrm{i}\frac{t}{\sqrt{n}}\sum _{j=1}^n \exp (\mathrm{i}t\varepsilon _j)\varepsilon _jg(X_j)\frac{\sigma (X_j)-\hat{\sigma }(X_j)}{\sigma (X_j)},\\ V_2(t)= & {} \mathrm{i}\frac{t}{\sqrt{n}}\sum _{j=1}^n \exp (\mathrm{i}t\varepsilon _j)\varepsilon _jg(X_j)\frac{\sigma (X_j)-\sigma (X_j;\hat{\theta })}{\sigma (X_j)}. \end{aligned}$$

From (22),

$$\begin{aligned}&\sup _{x \in R_g}\left| \hat{\sigma }(x)-\sigma (x)-\frac{1}{2nf(x)\sigma (x)}\sum _{j=1}^{n} K_h(X_{j}-x)\left[ \left\{ Y_{j}-m(x)\right\} ^2-\sigma ^2(x)\right] \right| \\&\quad =o_p(n^{-1/2}), \end{aligned}$$

where $K_h(\cdot )=\frac{1}{h}K(\frac{\cdot }{h})$. By using this identity, we get that

$$\begin{aligned} V_1(t)=V_3(t)+tR_3(t), \end{aligned}$$

with $\sup _t |R_{3}(t)|=o_P(1)$ and

$$\begin{aligned} V_3(t)= & {} -\mathrm{i}\frac{t}{2n\sqrt{n}}\sum _{j,k=1}^n \exp (\mathrm{i}t\varepsilon _j)\varepsilon _jg(X_j)\frac{1}{\sigma ^2(X_j)f(X_j)}\\&\times \, K_h(X_{j}-X_k)\left[ \left\{ Y_k-m(X_j)\right\} ^2-\sigma ^2(X_j)\right] . \end{aligned}$$

By applying Hoeffding decomposition and Lemma 2 in Yoshihara (1976), we get that

$$\begin{aligned} V_3(t)=-\frac{t}{2} \varphi '(t)\frac{1}{\sqrt{n}}\sum _{j=1}^ng(X_j)(\varepsilon ^2_j-1)+tR_4(t), \end{aligned}$$

with $\sup _t |R_{4}(t)|=o_P(1)$.

By Taylor expansion and following similar steps to those given for $V_1(t)$, we obtain

$$\begin{aligned} V_2(t)=-\frac{t}{2} \varphi '(t)\frac{1}{\sqrt{n}}\sum _{j=1}^n\mu _g^T l(\varepsilon _j, X_j; \theta _0)+tR_5(t), \end{aligned}$$

with $\sup _t |R_{5}(t)|=o_P(1)$. Putting together all above facts, the result follows. $\square$