Penalized empirical likelihood for partially linear errors-in-variables panel data models with fixed effects

Abstract

For the partially linear errors-in-variables panel data models with fixed effects, we, in this paper, study asymptotic distributions of a corrected empirical log-likelihood ratio and maximum empirical likelihood estimator of the regression parameter. In addition, we propose penalized empirical likelihood (PEL) and variable selection procedure for the parameter with diverging numbers of parameters. By using an appropriate penalty function, we show that PEL estimators have the oracle property. Also, the PEL ratio for the vector of regression coefficients is defined and its limiting distribution is asymptotically chi-square under the null hypothesis. Moreover, empirical log-likelihood ratio for the nonparametric part is also investigated. Monte Carlo simulations are conducted to illustrate the finite sample performance of the proposed estimators.

Appendix: Proofs of the main results

We use Frobenius norm of a matrix A, defined as $||A||=\{tr(A^{\tau }A)\}^{1/2}$. Before we give the details of the proofs, we present some regularity conditions.

• (B1) The random vector $Z_{it}$ has a continuous density function $f(\cdot )$ with a bounded support $\mathcal {Z}$. $0<inf_{z\in \mathcal {Z}}f(\cdot )\le sup_{z\in \mathcal {Z}}f(\cdot )<\infty$.

• (B2) The functions $E(X_{it}|Z_{it}=z)$ and $g(\cdot )$ have two bounded and continuous derivatives on $\mathcal {Z}$.

• (B3) The kernel K(v) is a symmetric probability density function with a continuous derivative on its compact support $\mathcal {Z}$.

• (B4) $(\mu _i,W_{it}, Z_{it}, \varepsilon _{it}),~ i=1,\ldots ,n,~t=1,\ldots ,T$ are i.i.d. $E(\varepsilon |W,Z,\mu )=0$ almost surely. Furthermore, for some integer $k\ge 4,$ $E(||W\varepsilon ||^k)\le \infty ,$ $E(||W||^k)\le \infty ,$ $E(|\varepsilon |^k)\le \infty .$

• (B5) $E|\breve{X}_{it}|^{2+\delta }<\infty$,  $\Sigma =E[\breve{X}_{it}\breve{X}_{it}^{\tau }]$ is non-singular, where $\breve{X}_{it}=X_{it}-E(X_{it}|Z_{it}).$

• (B6) The bandwidth h satisfies $h\rightarrow 0$, $Nh^8\rightarrow 0$ and $Nh^2/(logN)^2\rightarrow \infty$ as $n\rightarrow \infty$.

• (B7) $\Sigma _1$ and $\Sigma _2$ are positive definite matrices with all eigenvalues being uniformly bounded away from zero and infinity.

• (B8) Let $\varpi _1=\sum _{t=1}^T\frac{T-1}{T}(X_{it}-E(X_{it}|U_{it}))(\varepsilon _{it}-\nu _{it}\beta _0)$, $\varpi _2=\sum _{t=1}^T\frac{T-1}{T}\nu _{it}\varepsilon _{it}$, $\varpi _3=\sum _{t=1}^T\frac{T-1}{T}(\nu _{it}\nu _{it}^{\tau }-\Sigma _{\nu })\beta _0$ and $\varpi _{sj}, j=1,\ldots ,p$ be the j-th component of $\varpi _{s}.$ For k of condition (B4), there is a positive constant c such that as $n\rightarrow \infty$, $E(||\varpi _{s}/\sqrt{p}||^k)\le c, s=1,2,3$.

• (B9) The $p_{\lambda }(\cdot )$ satisfy $\mathop {max}\limits _{j\in \mathcal {B}}p_{\lambda }'(|\beta _{j0}|)=o((np)^{-1/2})$ and $\mathop {max}\limits _{j\in \mathcal {B}}p_{\lambda }''(|\beta _{j0}|)=o(p^{-1/2})$.

Note that the obove conditions are assumed to hold uniformly in $z\in \mathcal {Z}$. Conditions (B1)–(B9) while look a bit lengthy, are actually quite mild and can be easily satisfied. (B1)–(B2) are standard in the literature on local linear/polynomial estimation. B5 implies $E(\varepsilon _{it}|X_i,Z_i,\mu _i)=E(\varepsilon _{it}|X_{it},Z_{it},\mu _{it})=0.$ (B1)–(B5) can be founded in Su and Ullah 2006. (B6) and (B7) have been used in Zhou et al. (2010).

For the convenience and simplicity, let $\vartheta _k=\int z^kK(z)dz$, $c_N=\{log(1/h)/(Nh)\}^{1/2}+h^2$ and $\widetilde{M}_{\widetilde{D}}=\widetilde{D}(\widetilde{D}^{\tau }\widetilde{D})^{-1}\widetilde{D}^{\tau }$

$$\begin{aligned} a_n= & {} \max _{1\le j\le p}\{p_{\lambda }'(\beta _{j0})|,\beta _{j0}\ne 0\},~~~~b_n=\max _{1\le j\le p}\{p_{\lambda }''(\beta _{j0})|,\beta _{j0}\ne 0\},\\ B_n= & {} \{\beta : ||\beta -\beta _0||\ge d_n\},~~~d_n=n^{-1/3-\delta }+a_n,~~~0<\delta <1/6 \end{aligned}$$

Lemma A.1

Suppose that Assumptions (B1)–(B6) hold. Then

$$\begin{aligned}&\displaystyle G^{\tau }(z,h)W_h(z)G(z,h)=Nf(z)\times \begin{pmatrix}1&{}\quad 0\\ 0&{}\quad v_2\end{pmatrix}\{1+O_p(c_N)\},\\&\displaystyle G^{\tau }(z,h)W_h(z)X=Nf(z)E(X|Z)\times (1,0)^{\tau }\{1+O_p(c_N)\}, \end{aligned}$$

Proof

Note that

$$\begin{aligned}&G^{\tau }(z,h)W_h(z)G(z,h)\\&\quad =\left( \begin{array}{cc} \sum _{i=1}^n\sum _{t=1}^TK_h(Z_{it}-z)&{}\quad \sum _{i=1}^n\sum _{t=1}^T(\frac{Z_{it}-z}{h})K_h(Z_{it}-z)\\ \sum _{i=1}^n\sum _{t=1}^T(\frac{Z_{it}-z}{h})K_h(Z_{it}-z) &{}\quad \sum _{i=1}^n\sum _{t=1}^T(\frac{Z_{it}-z}{h})^2K_h(Z_{it}-z) \end{array}\right) . \end{aligned}$$

Each element of the above matrix is in the form of a kernel regression. Similar to the proof of Lemma A.2 in Fan and Huang (2005), we can derive the desired result. $\square$

Lemma A.2

Suppose that Assumptions (B1)–(B6) hold, we have

$$\begin{aligned} E|g(Z_{it})-\sum _{k=1}^n\sum _{l=1}^TS_{kl}g(Z_{kl})|^2=O(h^4). \end{aligned}$$

(A.1)

Proof

Similar to the proof of Lemma 5.1 in He et al. (2017). $\square$

Lemma A.3

Suppose that Assumptions (B1)–(B6) hold, we have

$$\begin{aligned} \frac{1}{N}\widetilde{{W}}^{\tau }H\widetilde{{W}}{\mathop {\rightarrow }\limits ^{d}}\frac{T-1}{T}(\Sigma _2+\Sigma _{\nu }), \end{aligned}$$

where $\Sigma _2=E\{[{X_{11}}-E({X_{11}}|Z_{11})]^{\tau }[{X_{11}}-E({X_{11}}|Z_{11})]\}.$

Proof

By Lemma A.1, we can obtain

$$\begin{aligned}{}[I_q~~0^\tau _q]^{-1}(G^{\tau }(z,h)W_h(z)G(z,h))^{-1}G^{\tau }(z,h)W_h(z)X=E(X|Z)+O_p(c_N). \end{aligned}$$

Then we have

$$\begin{aligned} \widetilde{{X}}= & {} [{X}_{11}-E(X_{11}|Z_{11}),\ldots ,{X}_{1T}-E(X_{1T}|Z_{1T}),\ldots ,{X}_{nT}-E(X_{nT}|Z_{nT})]^{\tau }\\&+\,O_p(c_N), \end{aligned}$$

and

$$\begin{aligned} \widetilde{{W}}=\widetilde{{X}}+\nu +O_p(c_N)\triangleq A+O_p(c_N). \end{aligned}$$

By the law of large numbers, we have

$$\begin{aligned} \frac{1}{N}\widetilde{{W}}^{\tau }\widetilde{{W}}= & {} \frac{1}{N}\sum _{i=1}^n\sum _{t=1}^T\big \{[{X}_{it}-E(X_{it}|Z_{it})]^{\tau }[{X}_{it}-E(X_{it}|Z_{it})]+\nu _{it}^{\tau }\nu _{it}\big \}\nonumber \\&+\,O_p(c_N) {\mathop {\rightarrow }\limits ^{p}}\Sigma _2+\Sigma _{\nu }. \end{aligned}$$

(A.2)

Hence, to prove the lemma, we consider the limit of $N^{-1}\widetilde{{W}}^{\tau }\widetilde{M}_{\widetilde{D}}\widetilde{{W}}$. It is easy to show that $N^{-1}\widetilde{{W}}^{\tau }\widetilde{M}_{\widetilde{D}}\widetilde{{W}}=N^{-1}A^{\tau }\widetilde{M}_{\widetilde{D}}A+O_p(c_N)$. Let $(\widetilde{M}_{\widetilde{D}})_{e_{kl}e_{it}}\triangleq m_{e_{kl}e_{it}}$ and $(A)_{it}\triangleq a_{it}=\widetilde{{W}}_{it}$, where $e_{kl}=(k-1)T+l$. Then

$$\begin{aligned} \frac{1}{N}A^{\tau }\widetilde{M}_{\widetilde{D}}A&=\frac{1}{N}\sum _{i=1}^n\sum _{t=1}^T\sum _{k=1}^n\sum _{l=1}^T a_{kl}m_{e_{kl}e_{it}}a_{it}=\frac{1}{N}\sum _{i=1}^n\sum _{t=1}^Ta_{it}m_{e_{it}e_{it}}a_{it}\\&\quad +\,\frac{1}{N}\sum _{e_{kl}\ne e_{it}}\sum _{k=1}^n\sum _{l=1}^Ta_{kl}m_{e_{kl}e_{it}}a_{it} \triangleq I_1+I_2. \end{aligned}$$

For the term $I_2$, we have

$$\begin{aligned} EI_2^2=\frac{1}{N^2}E\left[ \sum _{e_{kl}\ne e_{it}}\sum _{k=1}^n\sum _{l=1}^T\sum _{e_{rs}\ne e_{uv}}\sum _{r=1}^n\sum _{s=1}^Ta_{kl}m_{e_{kl}e_{it}}a_{it}a_{rs}m_{e_{rs}e_{uv}}a_{uv}\right] . \end{aligned}$$

Note that $(X_{11},Z_{11}),\ldots ,(X_{nT},Z_{nT})$ are i.i.d. and $E(a_{it}|Z_{it})=0$, when $e_{kl}\ne e_{rs}$ and $e_{it}\ne e_{uv}$, we have

$$\begin{aligned}&E(a_{kl}m_{e_{kl}e_{it}}a_{it}a_{rs}m_{e_{rs}e_{uv}}a_{uv})\\&\quad =E[m_{e_{kl}e_{it}}m_{e_{rs}e_{uv}}E(a_{kl}a_{it}a_{rs}a_{uv}|Z_{kl},Z_{it},Z_{rs},Z_{uv})]\\&\quad =E[m_{e_{kl}e_{it}}m_{e_{rs}e_{uv}}E(a_{it}a_{rs}a_{uv}|Z_{it},Z_{rs},Z_{uv})E(a_{kl}|Z_{kl})]=0. \end{aligned}$$

Using the same argument and $m_{e_{kl}e_{it}}=m_{e_{it}e_{kl}}$, we have

$$\begin{aligned} EI_2^2=\frac{1}{N^2}\sum _{e_{kl}\ne e_{it}}\sum _{k=1}^n\sum _{l=1}^TE(a_{kl}m_{e_{kl}e_{it}}a_{it})^2+\frac{1}{N^2}\sum _{e_{kl}\ne e_{it}}\sum _{k=1}^n\sum _{l=1}^TE(m_{e_{kl}e_{it}}^2a_{kl}a_{it}a_{it}a_{kl}). \end{aligned}$$

By Conditions (B3), we obtain

$$\begin{aligned} EI_2^2\le \frac{2c}{N^2}\sum _{e_{kl}\ne e_{it}}E(m_{e_{kl}e_{it}})^2 \le \frac{1}{N^2}tr(\widetilde{M}^2)\le \frac{2c}{N}, \end{aligned}$$

where c is a constant. Hence

$$\begin{aligned} EI_2=o_p(1). \end{aligned}$$

(A.3)

Note that $I_1$ can be decomposed as

$$\begin{aligned} I_1=\frac{1}{N}\sum _{i=1}^n\sum _{t=1}^Tm_{e_{it}e_{it}} [a_{it}a_{it}-E(a_{it}a_{it})]+\frac{1}{N}\sum _{i=1}^n\sum _{t=1}^Tm_{e_{it}e_{it}}E(a_{it}a_{it})\triangleq \Pi _1+\Pi _2. \end{aligned}$$

By the definition of S, it is easy to show that

$$\begin{aligned} S=(S_{11},\ldots ,S_{1T},S_{21},\ldots ,S_{nT})^{\tau }[I+diag\{O_p(c_n)\}], \end{aligned}$$

where

$$\begin{aligned} S_{it}=\left( \frac{K_h(Z_{11}-Z_{it})}{Nf(Z_{it})},\ldots ,\frac{K_h(Z_{1T}-Z_{it})}{Nf(Z_{it})},\ldots ,\frac{K_h(Z_{nT}-Z_{it})}{Nf(Z_{it})}\right) ^{\tau }. \end{aligned}$$

Let $D_1$ is the first column vector of D, thus we have

$$\begin{aligned} D_1^{\tau }(I-S^{\tau })(I-S)D_1= & {} \left\{ T-2\sum _{t=1}^T \left[ \sum _{l=1}^T \frac{K_h(Z_{1l}-Z_{1t})}{Nf(Z_{1t})}\right] \right. \\&\left. +\sum _{i=1}^n\sum _{t=1}^T\left[ \sum _{l=1}^T \frac{K_h(Z_{1l}-Z_{it})}{Nf(Z_{it})}\right] ^2\right\} \{1+O_p(c_N)\}. \end{aligned}$$

Because

$$\begin{aligned} \sum _{t=1}^T \left[ \sum _{l=1}^T \frac{K_h(Z_{1l}-Z_{1t})}{Nf(Z_{1t})}\right] \{1+O_p(c_N)\}=O_p\left( \frac{1}{Nh}\right) , \end{aligned}$$

and

$$\begin{aligned} \sum _{i=1}^n\sum _{t=1}^T\left[ \sum _{l=1}^T \frac{K_h(Z_{1l}-Z_{it})}{Nf(Z_{it})}\right] ^2\{1+O_p(c_N)\}=O_p\left( \frac{1}{Nh}\right) , \end{aligned}$$

we have

$$\begin{aligned} D_1^{\tau }(I-S^{\tau })(I-S)D_1 =T\left[ 1+O_p\left( \frac{1}{Nh}\right) \right] . \end{aligned}$$

Consider the projection matrix, for $i=1,\ldots ,T$, we obtain

$$\begin{aligned} (\widetilde{M}_{{\widetilde{D}}_1})_{ii}= & {} (I-S)D_1[D_1^{\tau }(I-S^{\tau })(I-S)D_1]^{-1}D_1^{\tau }(I-S^{\tau })\\= & {} \frac{1}{T}\left[ 1+O_p\left( \frac{1}{Nh}\right) \right] \left[ 1-\sum _{l=1}^T \frac{K_h(Z_{1l}-Z_{it})}{Nf(Z_{it})}\{1+O_p(c_N)\}\right] ^2\\= & {} \frac{1}{T}+O_p\left( \frac{1}{Nh}\right) . \end{aligned}$$

Because ${\widetilde{D}}_1$ is the first column vector of ${\widetilde{D}}$. It is easy to show that $\widetilde{M}_{\widetilde{D}}\widetilde{M}_{{\widetilde{D}}_1}=\widetilde{M}_{{\widetilde{D}}_1}\widetilde{M}_{\widetilde{D}}=\widetilde{M}_{{\widetilde{D}}_1}$. Hence, $\widetilde{M}_{\widetilde{D}}-\widetilde{M}_{{\widetilde{D}}_1}$ is also a projection matrix. Thus $\widetilde{M}_{\widetilde{D}}-\widetilde{M}_{\widetilde{D}_1}=(\widetilde{M}_{\widetilde{D}}-\widetilde{M}_{{\widetilde{D}}_1})^2\ge 0$. We obtain $(\widetilde{M}_{\widetilde{D}})_{ii}\ge (\widetilde{M}_{{\widetilde{D}}_1})_{ii}=\frac{1}{T}+O_p(\frac{1}{Nh}),~i=1,\ldots ,T$. By a similar argument, we can show that $(\widetilde{M}_{\widetilde{D}})_{ii}\ge \frac{1}{T}+O_p(\frac{1}{Nh}),i=T+1,\ldots ,N$. Thus, we have

$$\begin{aligned} 1\ge m_{e_{it}e_{it}}\ge \frac{1}{T}+O_p\left( \frac{1}{Nh}\right) , \end{aligned}$$

then, it is easy to show that

$$\begin{aligned} tr(\widetilde{M}_{\widetilde{D}})=\sum _{i=1}^{n}\sum _{t=1}^{T} m_{e_{it}e_{it}}\ge \frac{N}{T}+O_p\left( \frac{1}{Nh}\right) . \end{aligned}$$

Hence, we have

$$\begin{aligned} \Pi _2=\frac{1}{N}\sum _{i=1}^n\sum _{t=1}^Tm_{e_{it}e_{it}}E(a_{it}a_{it})=\frac{1}{T}(\Sigma +\Sigma _{\eta })+O_p\left( \frac{1}{Nh}\right) . \end{aligned}$$

(A.4)

By (A.2), it is easy to show that

$$\begin{aligned} \frac{1}{N}\sum _{i=1}^n\sum _{t=1}^T(m_{e_{it}e_{it}}-T^{-1})=O_p\left( \frac{1}{Nh^2}\right) , \end{aligned}$$

By the law of large numbers, $\Pi _1$ is bounded as

$$\begin{aligned} \Pi _1&=\frac{1}{N}\sum _{i=1}^n\sum _{t=1}^T(m_{e_{it}e_{it}}-T^{-1})(a_{it}a_{it}-E(a_{it}a_{it}))\nonumber \\&\quad +\frac{1}{NT}\sum _{i=1}^n\sum _{t=1}^T(a_{it}a_{it}-E(a_{it}a_{it}))\nonumber \\&\le \frac{1}{N}\left[ \sum _{i=1}^n\sum _{t=1}^T(m_{e_{it}e_{it}}-T^{-1})^2\right] ^{1/2}+o_p(1)=o_p(1). \end{aligned}$$

(A.5)

By (A.4) and (A.5), we have

$$\begin{aligned} I_1=\frac{1}{T}(\Sigma _2+\Sigma _{\eta })+o_p(1). \end{aligned}$$

(A.6)

By (A.2), (A.3) and (A.6), the lemma holds. $\square$

Lemma A.4

Under the conditions of Theorem 2.1, if $\beta$ is the true value of the parameter, we have

$$\begin{aligned} \sum _{i=1}^n\sum _{t=1}^T \nu _{it}H\widetilde{g}_{it}= & {} o_p(N^{1/2}), \end{aligned}$$

(A.7)

$$\begin{aligned} \sum _{i=1}^n\sum _{t=1}^T \varepsilon _{it}H\widetilde{g}_{it}= & {} o_p(N^{1/2}), \end{aligned}$$

(A.8)

Proof

Since the proof of (A.8) is similar of (A.7), we prove only (A.7)here. Let $\zeta _N=N^{1/2}/log(N)$,

$$\begin{aligned} P\left( |\sum _{i=1}^n\sum _{t=1}^T \nu _{it}H\widetilde{g}_{it}|>\zeta _N\right)\le & {} P\left( |\sum _{i=1}^n\sum _{t=1}^T \nu _{it}H\widetilde{g}_{it}|>\zeta _N, max_{i,t}|\widetilde{g}_{it}|\le ch^4\right) \nonumber \\&+P(max_{i,t}|\widetilde{g}_{it}|\ge ch^4) \end{aligned}$$

(A.9)

The second term is $o_p(1)$ by Lemma A.2. For the first term, let $R_{it}$ be the event that $|\widetilde{g}_{it}|\le ch^4$. Then

$$\begin{aligned}&P\left( |\sum _{i=1}^n\sum _{t=1}^T \nu _{it}H\widetilde{g}_{it}|>\zeta _N, \{I(R_{it})=1,\forall i,t\}\right) \\&\quad \le \zeta _N^{-2}\sum _{i=1}^n\sum _{t=1}^T E[\nu _{it}H\widetilde{g}_{it}\{I(R_{it})=1\}]^2\\&\qquad +\,\,\zeta _N^{-2}\sum _{i\ne k}^n\sum _{t\ne s}^TE[\nu _{it}H\widetilde{g}_{it}\nu _{ks}\Omega \widetilde{g}_{ks}\{I(R_{ks})=1\}]. \end{aligned}$$

Since $\widetilde{g}_{it}\{I(R_{it})=1\}\le ch^4$ is independent of $\nu _{it}$, the first term is $O\{N\zeta _N^{-2}c^2h^8\}=o(1)$. The second term is easily seen to equal zero. $\square$

Lemma A.5

Under the conditions of Theorem 2.1, if $\beta _0$ is the true value of the parameter, we have

$$\begin{aligned}&\displaystyle \mathop {max}\limits _{1\le i\le n}||\Gamma _i(\beta _0)||=o_p(\sqrt{n/p}), \end{aligned}$$

(A.10)

$$\begin{aligned}&\displaystyle \frac{\{N^{-1/2}\sum _{i=1}^n\Gamma _i^{\tau }(\beta _0)\}\Sigma _1^{-1}\{N^{-1/2}\sum _{i=1}^n\Gamma _i(\beta _0)\}-p}{\sqrt{2p}}{\mathop {\rightarrow }\limits ^{d}}N(0,1). \end{aligned}$$

(A.11)

Proof

From the definition of $\Gamma _{i}(\beta )$ by (2.7), and a simple calculation, yields

$$\begin{aligned} \frac{1}{\sqrt{N}}\sum _{i=1}^n\Gamma _i(\beta _0)= & {} \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\big \{ \widetilde{X}_{it}H\widetilde{g}_{it}-\widetilde{X}_{it}H \widetilde{\nu }_{it}\beta _0+\widetilde{X}_{it}H\widetilde{\varepsilon }_{it}\\&+\,\widetilde{\nu }_{it}H\widetilde{g}_{it}-\widetilde{\nu }_{it}H \widetilde{\nu }_{it}^{\tau }\beta _0+\widetilde{\nu }_{it}H\widetilde{\varepsilon }_{it}+\Sigma _{\nu }\beta _0\big \}. \end{aligned}$$

By Lemma A.1, we have $S\varepsilon =O_p(c_N)$. Similar to the proof of Lemma A.3 and under Assumption (B7), we have $\frac{1}{\sqrt{N}}\widetilde{{X}}^{\tau }HS\varepsilon =O(\sqrt{N}c_N^2)=o_p(1)$. Therefore

$$\begin{aligned} \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\widetilde{X}_{it}H\widetilde{\varepsilon }_{it}=\frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T \frac{T-1}{T}(X_{it}-E(X_{it}|Z_{it}))\varepsilon _{it} \end{aligned}$$

(A.12)

Similar to the proofs of (A.12), we can derive that

$$\begin{aligned} \displaystyle \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\widetilde{X}_{it}H \widetilde{\nu }_{it}= & {} \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\frac{T-1}{T}(X_{it}-E(X_{it}|U_{it}))\nu _{it},\\ \displaystyle \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\widetilde{\nu }_{it}H \widetilde{\nu }_{it}^{\tau }= & {} \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\frac{T-1}{T}\nu _{it}\nu _{it},\\ \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\widetilde{\nu }_{it}H\widetilde{\varepsilon }_{it}= & {} \frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\frac{T-1}{T}\nu _{it}\varepsilon _{it}, \end{aligned}$$

which combining with Lemma A.4, it is easy to obtain

$$\begin{aligned}&\frac{1}{\sqrt{N}}\sum _{i=1}^n\Gamma _i(\beta _0)\nonumber \\&\quad =\frac{1}{\sqrt{N}}\sum _{i=1}^n\sum _{t=1}^T\frac{T-1}{T}\big \{[(X_{it}-E(X_{it}|U_{it}))(\varepsilon _{it}-\nu _{it}\beta _0)]+\nu _{it}\varepsilon _{it}\nonumber \\&\qquad +(\Sigma _{\nu }-\nu _{it} \nu _{it}^{\tau })\beta _0]\big \}+o_p(1)\nonumber \\&\quad =\frac{1}{\sqrt{N}}\sum _{i=1}^nG_{i}(\beta _0)+o_p(1). \end{aligned}$$

(A.13)

Therefore, we have

$$\begin{aligned} \Gamma _i(\beta _0)&=\frac{T-1}{T}\sum _{t=1}^T\big \{[(X_{it}-E(X_{it}|U_{it}))(\varepsilon _{it}-\nu _{it}\beta _0)]+\nu _{it}\varepsilon _{it}\nonumber \\&\quad +(\Sigma _{\nu }-\nu _{it} \nu _{it}^{\tau })\beta _0]\big \}+o_p(1)\nonumber \\&=\sum _{t=1}^T\left\{ \frac{T-1}{T}(X_{it}-E(X_{it}|U_{it}))(\varepsilon _{it}-\nu _{it}\beta _0)+\frac{T-1}{T}\nu _{it}\varepsilon _{it}\right. \nonumber \\&\quad \left. +\,\frac{T-1}{T}(\nu _{it}\nu _{it}^{\tau }-\Sigma _{\nu })\beta _0\right\} +o_p(1)=\varpi _1+\varpi _2+\varpi _3+o_p(1). \end{aligned}$$

(A.14)

Let $\varpi _s^{*}=\mathop {max}\limits _{1\le i\le n}||\varpi _{si}||,~s=1,2,3$, and $\{\varpi _{si}, i=1,\ldots , n \}$ is a sequence of independent random variables with common distribution. for any $\varepsilon \ge 0$, then

$$\begin{aligned} P\{\varpi _1^{*}\ge (p)^{1/2}n^{1/k}\epsilon \}\le & {} \sum _{i=1}^nP\{||\varpi _{1i}||\ge (p)^{1/2}n^{1/k}\epsilon \}\\\le & {} \frac{1}{np^{k/2}\epsilon ^k}\sum _{i=1}^nE||\varpi _{1i}||^k\\\le & {} \frac{1}{\epsilon ^k}E||\varpi _{11}/p^{1/2}||^k. \end{aligned}$$

From conditions (B4) and (B7) and Cauchy-Schwarz inequality yields that $\varpi _1^{*}=o_p(\sqrt{p}n^{1/k}).$ By the condition $p=o(n^{(k-2)/(2k)}$ in Theorem 2.1, it is easy to check that $\varpi _1^{*}=o_p(\sqrt{n/p}n^{2-k/(2k)}p)=o_p(\sqrt{n/p})$. Similar to the proof, we obtain $\varpi _2^{*}=o_p(\sqrt{n/p})$ and $\varpi _3^{*}=o_p(\sqrt{n/p})$. Then $\mathop {max}\nolimits _{1\le i\le n}||\Gamma _i(\beta _0)||=o_p(\sqrt{n/p}).$

By applying the martingale central limit theorem as give in Hall and Heyde (1980) and (A.13), it is easy to obtain (A.11). The proof of Lemma A.5 is thus completed. $\square$

Lemma A.6

Under the conditions of Theorem 2.1. Denote$D_n=\{\beta :||\beta -\beta _0||\le ca_n\}$ Then $||\gamma (\beta )||=O_p(a_n)$, for $\beta \in D_n$.

Proof

For $\beta \in D_n$, let$\gamma (\beta )=\rho \theta$, where$\rho \ge 0, \theta \in R^p$, and $||\theta ||=1$. Set

$$\begin{aligned} J(\beta )=\frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta )\Gamma _i^{\tau }(\beta ).~\bar{\Gamma }(\beta )=\frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta ), ~\Gamma ^{*}(\beta )=\mathop {max}\limits _{1\le i\le n}||\Gamma _i(\beta )||. \end{aligned}$$

From (2.7), we can obtain

$$\begin{aligned} 0= & {} \frac{1}{n}\sum _{i=1}^n\frac{\Gamma _i(\beta )}{1+\gamma ^\tau \Gamma _i(\beta )}= \frac{1}{n}\sum _{i=1}^n\theta ^{\tau }\Gamma _i(\beta )-\rho \frac{1}{n}\sum _{i=1}^n\frac{(\theta ^{\tau }\Gamma _i(\beta ))^2}{1+\rho \theta ^{\tau }\Gamma _i(\beta )}\\\le & {} \theta ^{\tau }\bar{\Gamma }(\beta )-\frac{\rho }{1+\rho \Gamma ^{*}}\theta ^{\tau }J(\beta )\theta . \end{aligned}$$

Then

$$\begin{aligned} \rho \left[ \theta ^{\tau }J(\beta )\theta -\mathop {max}\limits _{1\le i\le n}||\Gamma _i(\beta )||n^{-1}\left| \sum _{i=1}^n\theta ^{\tau }\Gamma _i(\beta )\right| \right] \le |\theta ^{\tau }\bar{\Gamma }(\beta )|. \end{aligned}$$

Observe that

$$\begin{aligned} \Gamma ^{*}(\beta )\le \Gamma ^{*}(\beta _0)+|\mathop {max}\limits _{1\le i\le n}\left\| \frac{1}{N}\sum _{i=1}^n\sum _{t=1}^T\widetilde{W}_{it}H\widetilde{W}_{it}(\beta -\beta _0)\right\| \end{aligned}$$

Let $\mathcal {X}_{it}=\widetilde{W}_{it}H\widetilde{W}_{it}$ According to Condition (B.7) and Minkowski inequality, we have

$$\begin{aligned} Var(||\mathcal {X}_{it}||^{r/2})\le & {} E(||\mathcal {X}_{it}||^r)=E\left[ \left( \sum _{j=1}^p\mathcal {X}_{itj}^2\right) ^{r/2}\right] \\\le & {} \left\{ \sum _{j=1}^pE[\mathcal {X}_{itj}^r]^{2/r}\right\} ^{r/2}=O(p^{r/2}). \end{aligned}$$

Then we obtain that

$$\begin{aligned}&\mathop {max}\limits _{1\le i\le n}||\mathcal {X}_{it}||\\&\quad \le \Big [\{Var(||\mathcal {X}_{it}||^{r/2})\}^{1/2}\mathop {max}\limits _{1\le i\le n}\left\{ \frac{||\mathcal {X}_{it}||^{r/2})-E||\mathcal {X}_{it}||^{r/2})}{\{Var(||\mathcal {X}_{it}||^{r/2})\}^{1/2}}\right. +E||\mathcal {X}_{it}||^{r/2}\Big ]^{r/2}\\&\quad \le o_p(\sqrt{p}n^{1/r})+O_p(\sqrt{p})=o_p(\sqrt{p}n^{1/r}). \end{aligned}$$

which combining with (A.11)

$$\begin{aligned} \Gamma ^{*}(\beta )=o_p(n/p) \end{aligned}$$

(A.15)

For $\bar{\Gamma }(\beta )$, it is easy to see that

$$\begin{aligned} \bar{\Gamma }(\beta )=\bar{\Gamma }(\beta _0)+\frac{1}{N}\sum _{i=1}^n\sum _{t=1}^T\widetilde{W}_{it}H\widetilde{W}_{it}(\beta -\beta _0) \end{aligned}$$

Similar to the proofs of (A.10) in Fan et al. (2016), we obtain

$$\begin{aligned} \theta ^{\tau }\bar{\Gamma }(\beta )=O_p(a_n) \end{aligned}$$

(A.16)

Therefore, it follows from (A.15) and (A.16), we have $\mathop {max}\nolimits _{1\le i\le n}||\Gamma _i(\beta )||=n^{-1}|\sum _{i=1}^n$.

From (2.7), similar to the proof (A.11) in Fan et al. (2016) and Lemma B.4 in Li et al. (2012), we can derive $tr[(J(\beta _0)-\Sigma _1)^2]=O_p(p^2(c_n^4+1/n))$ which means that all the eigenvalues of $J(\beta _0)$ converge to those of $\Sigma _1$ at the rate of $O_p(p^2(c_n^4+1/n))$. Therefore, by Lemma A.2, (2.7), (A.16), together with Condition (B7), we have

$$\begin{aligned} J(\beta )&=\frac{1}{n}\sum _{i=1}^n\sum _{t=1}^T\big \{\widetilde{W}_{it}H(\widetilde{Y}_{it}-\widetilde{W}_{it}\beta _0)-(T-1)\Sigma _{\nu }\beta _0 -\widetilde{W}_{it}H\widetilde{W}_{it}^{\tau }(\beta -\beta _0)\nonumber \\&\quad +(T-1)\Sigma _{\nu }(\beta -\beta _0)\big \}^{\oplus 2}\nonumber \\&=J(\beta _0)+\frac{1}{n}\sum _{i=1}^n\sum _{t=1}^T\big \{\widetilde{W}_{it}H\widetilde{W}_{it}^{\tau }(\beta -\beta _0)+(T-1)\Sigma _{\nu }(\beta -\beta _0)\big \}^{\oplus 2} \nonumber \\&\quad -\frac{2}{n}\sum _{i=1}^n\sum _{t=1}^T\big \{\widetilde{W}_{it}H(\widetilde{Y}_{it}-\widetilde{W}_{it}\beta _0)-(T-1)\Sigma _{\nu }\beta _0\big \}\big \{\widetilde{W}_{it}H\widetilde{W}_{it}^{\tau }(\beta -\beta _0)\nonumber \\&\quad +(T-1)\Sigma _{\nu }(\beta -\beta _0)\big \} \nonumber \\&= \Sigma _1+O_p(p^2(c_n^4+1/n)). \end{aligned}$$

(A.17)

we can obtain $\theta ^{\tau }J(\beta )\theta =\theta ^{\tau }\Sigma _1\theta {\mathop {\rightarrow }\limits ^{P}}c.$ Therefore, we obtain $\rho \le c|\theta ^{\tau }\bar{\Gamma }(\beta )|=O_p(a_n),$ then $||\gamma (\beta )||=O_p(a_n)$. $\square$

Lemma A.7

Under the conditions of Theorem 2.1. as $n\rightarrow \infty$, with probability tending to 1, $R_{n}(\beta )$ has a minimum in $D_n$.

Proof

For $\beta \in D_n$,

$$\begin{aligned} H_{1n}(\beta ,\gamma )=\frac{1}{n}\sum _{i=1}^n\frac{\Gamma _i(\beta )}{1+\gamma ^\tau \Gamma _i(\beta )}=0 \end{aligned}$$

According to Lemma A.6, we have $\gamma ^\tau \Gamma _i(\beta )=o_p(1)$. Apply Taylor expansion to $H_{1n}(\beta ,\gamma )$, we obtain $\bar{\Gamma }(\beta )-J(\beta )\gamma +\delta _n=0,$ where $\bar{\Gamma }(\beta )=\frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta )$, $\delta _n=\frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta )(\gamma ^\tau \Gamma _i(\beta ))^2/[1+\zeta _i]^3$ for some $|\zeta _i|\le |\gamma ^\tau \Gamma _i(\beta )|$. We have $\gamma =J(\beta )^{-1}\bar{\Gamma }(\beta )+J(\beta )^{-1}\delta _n$. Substituting $\gamma$ into (2.6), it is easy to see that

$$\begin{aligned} 2R_{n}(\beta )=n\bar{\Gamma }(\beta )^{\tau }J(\beta )^{-1}\bar{\Gamma }(\beta )-n\delta _n^{\tau }J(\beta )^{-1}\delta _n+\frac{2}{3}\sum _{i=1}^n(\gamma ^\tau \Gamma _i(\beta ))^3(1+\zeta _i)^{-4} \end{aligned}$$

(A.18)

For $\beta \in \partial D_n$, where $\partial D_n$ denotes the boundary of $D_n$, we write $\beta =\beta _0+ca_n\phi$ where $\phi$ is a unit vector, we have a decomposition as $2R_{n}(\beta )=\Pi _0+\Pi _1+\Pi _2$, where $\Pi _0=n\bar{\Gamma }(\beta _0)^{\tau }\Sigma _1^{-1}\bar{\Gamma }(\beta _0)$, $\Pi _1=n(\bar{\Gamma }(\beta )-\bar{\Gamma }(\beta _0))^{\tau }J(\beta )^{-1}(\bar{\Gamma }(\beta )-\bar{\Gamma }(\beta _0))$, $\Pi _2=n[\bar{\Gamma }(\beta _0)^{\tau }(J(\beta )^{-1}-\Sigma _1^{-1})\bar{\Gamma }(\beta _0)+2\bar{\Gamma }(\beta _0)^{\tau }J(\beta )^{-1}(\bar{\Gamma }(\beta )-\bar{\Gamma }(\beta _0)]-n\delta _n^{\tau }J(\beta )^{-1}\delta _n+\frac{2}{3}\sum _{i=1}^n(\gamma ^\tau \Gamma _i(\beta ))^3(1+\zeta _i)^{-4}$ As $n\rightarrow \infty$, we see that

$$\begin{aligned} \Pi _1= & {} n\left\{ \left[ \frac{1}{n}\widetilde{{W}}^{\tau }H\widetilde{{W}}-(T-1) \Sigma _{\nu }\right] (\beta -\beta _0)\right\} ^{\tau }J(\beta )^{-1}\left\{ \left[ \frac{1}{n}\widetilde{{W}}^{\tau }H\widetilde{{W}}\right. \right. \\&\left. \left. -\,(T-1)\Sigma _{\nu }\right] (\beta -\beta _0)\right\} \\= & {} c^2na_n^2\phi ^{\tau }\Sigma _0\Sigma _1^{-1}\Sigma _0\phi \{1+o_p(1)\}=O_p(na_n^2), \end{aligned}$$

$\Pi _2/\Pi _1{\mathop {\rightarrow }\limits ^{P}}0$ and $2R_{n}(\beta _0)-\Pi _0=o_p(1)$. This implies that for any c given, as $n\rightarrow \infty$, $Pr\{2[R_{n}(\beta )- R_{n}(\beta _0)]\ge c\}\rightarrow 1$. In addition, note that for n large,

$$\begin{aligned} \mathcal {L}_{n}(\beta )- \mathcal {L}_{n}(\beta _0)= & {} R_{n}(\beta )- R_{n}(\beta _0)+n\sum _{j=1}^p\{p_{\lambda }(|\beta _j|)-p_{\lambda }(|\beta _{j0}|)\\\ge & {} R_{n}(\beta )- R_{n}(\beta _0)+n\sum _{j\in \mathcal {B}}\{p_{\lambda }(|\beta _j|)-p_{\lambda }(|\beta _{j0}|)\ge R_{n}(\beta )- R_{n}(\beta _0), \end{aligned}$$

where the last inequality holds due to Conditions (B9) and the unbiased property of the SCAD penalty so that $j\in \mathcal {B}$, $p_{\lambda }(|\beta _j|)=p_{\lambda }(|\beta _{j0}|)$ when n is large. Hence, $Pr\{\mathcal {L}_{n}(\beta )\ge \mathcal {L}_{n}(\beta _0)\}\rightarrow 1$ for $\beta \in \partial D_n$, which establishes Lemma A.7. $\square$

Proof of Theorem 2.1

Let $U_i=\gamma ^\tau \Gamma _i(\beta _0)$. Apply Taylor expansion to (2.10), we have

$$\begin{aligned} 0= \frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta _0)\left( 1-U_i+\frac{U_i^2}{1+U_i}\right) =\bar{\Gamma }(\beta _0)-J(\beta _0)\gamma +\delta _n, \end{aligned}$$

(A.19)

where $\delta _n=\frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta _0)U_i^2-\frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta _0)\frac{U_i^3}{1+U_i}$.

From (A.11) and Lemma A.6, we have

$$\begin{aligned} \mathop {max}\limits _{1\le i\le n}|U_i|\le ||\gamma (\beta )||\mathop {max}\limits _{1\le i\le n}||\Gamma _i(\beta _0)||=O_p(p/n^{1/2-1/r}). \end{aligned}$$

Similar to the proof of (A.19) in Li et al. (2012), we can get $||\delta _n ||=o_p(p^{5/2}n^{-1}(n^{-1/2}+c_n^2))+o_p(p^2n^{-1}c_n).$ From (A.19), we obtain that $\gamma =J(\beta _0)^{-1}\bar{\Gamma }(\beta _0)+J(\beta _0)^{-1}\delta _n$. Taylor expansion implies $\ln (1+U_i)=U_i-U_i^2/2+U_i^3/3(1+\varsigma _i)^4,$ for some $\varsigma _i$ such that $|\varsigma _i|\le |U_i|$. Therefore, combining (A.16) and some elementary calculation, we have

$$\begin{aligned} 2R_{n}(\beta _0)&=2\sum _{i=1}^n\ln \{1+U_i\}=n\bar{\Gamma }^{\tau }(\beta _0)J(\beta _0)^{-1}\bar{\Gamma }(\beta _0)-n\delta _n^{\tau }J(\beta _0)^{-1}\delta _n\nonumber \\&\quad +\frac{2}{3}\mathcal {R}_n\{1+o_p(1)\}\nonumber \\&=n\bar{\Gamma }^{\tau }(\beta _0)\Sigma _1^{-1}\bar{\Gamma }(\beta _0)+n\bar{\Gamma }^{\tau }(\beta _0)(J(\beta _0)^{-1}-\Sigma _1^{-1})\bar{\Gamma }(\beta _0)-n\delta _n^{\tau }J(\beta _0)^{-1}\delta _n^{\tau }\nonumber \\&\quad +\frac{2}{3}\mathcal {R}_n\{1+o_p(1)\} \end{aligned}$$

(A.20)

where $\mathcal {R}_n=\sum _{i=1}^n[\gamma ^\tau \Gamma _i(\beta _0)]^3$, By using the proving method of (A.22) and Lemma B.6 in Li et al. (2012), we can easily derive $n\delta _n^{\tau }J(\beta _0)^{-1}\delta _n=o_p(\sqrt{p})$ and $n\bar{\Gamma }^{\tau }(\beta _0)(J(\beta _0)^{-1}-\Sigma _1^{-1})\bar{\Gamma }(\beta _0)=o_p(\sqrt{p})$. The proof of Theorem 2.1 is concluded from the above results together with (A.11). $\square$

Proof of Theorem 2.2

Let $H_{1n}(\beta ,\gamma )=\frac{1}{n}\sum _{i=1}^n\frac{\Gamma _i(\beta )}{1+\gamma ^\tau \Gamma _i(\beta )}$ and $H_{2n}(\beta ,\gamma )=\frac{1}{n}\sum _{i=1}^n\frac{\Gamma _i(\beta )}{1+\gamma ^\tau \Gamma _i(\beta )}(\frac{\partial \Gamma _i(\beta )}{\partial \beta }^{\tau })^{\tau }\gamma$. Note that $\hat{\beta }$ and $\hat{\gamma }$ satisfy $H_{1n}(\hat{\beta },\hat{\gamma })=0$ and $H_{2n}(\hat{\beta },\hat{\gamma })=0$ Let $\varphi =(\beta ^{\tau },\gamma ^{\tau })^{\tau }$, $\varphi _0=(\beta _0^{\tau },0)^{\tau }$ and $\hat{\varphi }_0=(\hat{\beta }_0^{\tau },\hat{\gamma }^{\tau }_0)^{\tau }$. Then by

$$\begin{aligned} 0=H_{jn}(\hat{\beta },\hat{\gamma })=H_{jn}(\beta _0^{\tau },0)+\frac{\partial \Gamma _i(\beta ,0)}{\partial \beta }(\hat{\beta }-\beta _0)+\frac{\partial \Gamma _i(\beta ,0)}{\partial \gamma }(\hat{\gamma }-0)+\delta _{jn}, \end{aligned}$$

where $\delta _{jn}$ with $\delta _{jn}=\frac{1}{2}(\hat{\varphi }_0-\varphi _0)^{\tau }H_{jn}^{\prime \prime }(\varphi )(\hat{\varphi }_0-\varphi _0)$ for $j=1,2$. Here $H_{jn}^{\prime \prime }(\varphi )$ denotes the Hessian matric of $H_{jn}(\varphi )$. Then

$$\begin{aligned} \left( \begin{array}{cc} \hat{\gamma }\\ \hat{\beta }-\beta _0\\ \end{array}\right) = \left[ \begin{array}{cc} \frac{\partial H_{1n}(\beta ,\gamma )}{\partial \gamma } &{}\frac{\partial H_{1n}(\beta ,\gamma )}{\partial \beta }\\ \frac{\partial H_{2n}(\beta ,\gamma )}{\partial \gamma } &{}\frac{\partial H_{2n}(\beta ,\gamma )}{\partial \beta }\\ \end{array}\right] ^{-1}_{(\beta _0,0)}\left( \begin{array}{cc} H_{1n}(\beta _0,0)+\delta _{1n}\\ H_{2n}(\beta _0,0)+\delta _{2n}\\ \end{array}\right) , \end{aligned}$$

from Lemma A.3, we have

$$\begin{aligned} n^{-1}\sum _{i=1}^n\frac{\partial \Gamma _i(\beta ,0)}{\partial \beta }=\frac{1}{n}\widetilde{{W}}^{\tau }H\widetilde{{W}}-(T-1)\Sigma _{\nu }{\mathop {\rightarrow }\limits ^{d}}(T-1)\Sigma _2\{1+o_p(1)\}=\Sigma _0\{1+o_p(1)\}. \end{aligned}$$

(A.21)

Note that $||\hat{\gamma }(\beta )||=O_p(a_n)$ by Lemma A.6 and $||\hat{\beta }-\beta _0||=O_p(a_n)$ by Lemma A.7. Then using the Cauchy-Schwarz inequality, we find

$$\begin{aligned} ||\delta _{1n}||^2=\sum _{k=1}^p\delta _{1n,k}\le c\sum _{k=1}^p||\hat{\varphi }_0-\varphi _0||^4||H_{jn}^{\prime \prime }(\varphi )||^2=O_p(a_n^4p^3). \end{aligned}$$

(A.22)

and Condition (B9) yields that

$$\begin{aligned} ||\sqrt{N}\Sigma _0^{-1}\delta _{1n}||^2\le N\Sigma _0^{-2}||\delta _{1n}||=O_p(Na_n^4p^3)=o_p(1). \end{aligned}$$

and combining with $H_{1n}(\beta _0,0)=n^{-1}\sum _{i=1}^n\Gamma _i(\beta _0)$ , we have

$$\begin{aligned} \sqrt{n}(\widehat{\beta }-\beta _0 )= & {} \Sigma _0^{-1}\cdot \frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta _0)+o_p(1)\\= & {} \Sigma _0^{-1}\cdot \frac{1}{n}\sum _{i=1}^n\sum _{t=1}^T\frac{T-1}{T}\big \{[(X_{it}-E(X_{it}|U_{it}))(\varepsilon _{it}-\nu _{it}\beta _0)]+\nu _{it}\varepsilon _{it}\\&+\,(\Sigma _{\nu }-\nu ^{\tau }\nu _{11})\beta _0]\big \} +o_p(1)=\Sigma _0^{-1}\cdot \frac{1}{n}\sum _{i=1}^n\Gamma _i(\beta _0)+o_p(1). \end{aligned}$$

Note that

$$\begin{aligned} Cov(\Gamma _i(\beta _0))= & {} (T-1)\big \{E[(X_{11}-E(X_{11}|U_{11}))(\varepsilon _{11}-\nu _{11}\beta _0)]^2 + E[\nu _{11}^{\tau }\varepsilon _{11}]^2\\&+\,E[(\Sigma _{\nu }-\nu ^{\tau }\nu _{11})\beta _0]^2\big \}. \end{aligned}$$

Therefore,

$$\begin{aligned} \mathop {lim}\limits _{n\rightarrow \infty }Cov(\Gamma _i(\beta _0))= & {} (T-1)\big \{E(\varepsilon _{11}-\nu _{11}\beta _0)^2\Sigma +\sigma ^2\Sigma _{\nu }\\&+ \,E[(\nu _{11}\nu _{11}^{\tau }-\Sigma _{\nu })\beta _0]^2\big \}= \Sigma _1. \end{aligned}$$

Invoking the Slutsky theorem and the central limit theorem, we can prove Theorem 2.2. $\square$

Proof of Theorem 2.3

From the Lemma A.7, we note that the minimizer of $\mathcal {L}_{n}(\beta )$ is in $\mathcal {D}_n$. Considering $\beta \in \mathcal {D}_n$, we have that for each of its components

$$\begin{aligned} \frac{1}{n}\frac{\partial \mathcal {L}_{n}(\beta )}{\partial \beta _i}= \frac{1}{n}\sum _{i=1}^n\frac{\gamma \left[ \sum _{t=1}^T\widetilde{W}_{it}H\widetilde{W}_{it}+(T-1)\Sigma _{\nu }\right] }{1+\gamma ^\tau \Gamma _i(\beta )}+p_{\lambda }'(|\beta _j|)sign(\beta _j)=I_j+II_j, \end{aligned}$$

(A.23)

First, $\mathop {max}\nolimits _{j\in \mathcal {B}}|I_j|\le \gamma \Sigma _j(1+o_p(1))=O_p(a_n)$, because $\gamma ^\tau \Gamma _i(\beta )=o_p(1)$,where $\Sigma _j$ denotes the jth column of $\Sigma$. as $\tau (n/p)^{1/2}\rightarrow \infty$. $Pr(\mathop {max}\nolimits _{j\in \mathcal {B}}|I_j|>\tau /2)\rightarrow 0$. it can be seen that $p_{\lambda }'(|\beta _j|)sign(\beta _j)$ dominates the sign of $\frac{\partial \mathcal {L}_{n}(\beta )}{\partial \beta _i}$ asymptotically for all $j\notin \mathcal {B}$, as $n\rightarrow \infty$, for any $j\notin \mathcal {B}$, with probability tending to 1,

$$\begin{aligned} \frac{\partial \mathcal {L}_{n}(\beta )}{\partial \beta _i}>0,~\beta _j\in (0,ca_n); \frac{\partial \mathcal {L}_{n}(\beta )}{\partial \beta _i}<0, ~\beta _j\in (0,-ca_n) \end{aligned}$$

which implies that $\hat{\beta }_j=0$ for all $j\notin \mathcal {B}^c$, with probability tending to 1. Thus part (a) of Theorem 2.3 follows.

Next, we establish part (b), Let $\Psi _1$ and $\Psi _2$ be matrices such that $\Psi _1\beta =\beta _1$ and $\Psi _2\beta =\beta _2$. As we have shown that as $n\rightarrow \infty$, $Pr(\hat{\beta }_2=0)\rightarrow 1$, thus by the Lagange multiplier method, finding the minimizer of $\mathcal {L}_{n}(\beta )$ is asymptotic equivalent to solve the minimization of the following objective function

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\log \{1+\gamma ^\tau \Gamma _i(\beta )\}+n\sum _{j=1}^pp_{\lambda }(|\beta _j|)+v^{\tau }\Psi _2\beta , \end{aligned}$$

(A.24)

where v is $p-s$ dimensional column vector of an other Lagrange multiplier. Define

$$\begin{aligned}&\displaystyle \tilde{Q}_{1n}(\beta ,\gamma ,v)=\frac{1}{n}\sum _{i=1}^n\frac{\Gamma _i(\beta )}{1+\gamma ^\tau \Gamma _i(\beta )},\\&\displaystyle \tilde{Q}_{2n}(\beta ,\gamma ,v)=\frac{1}{n}\sum _{i=1}^n\frac{\gamma }{1+\gamma ^\tau \Gamma _i(\beta )}+b(\beta )+\Psi _2^{\tau }v, \end{aligned}$$

and $\tilde{Q}_{3n}(\beta ,\gamma ,v)=\Psi _2\beta$. where

$$\begin{aligned} b(\beta )=\{p_{\lambda }'(|\beta _1|)sign(\beta _1),p_{\lambda }'(|\beta _2|)sign(\beta _2),\ldots ,p_{\lambda }'(|\beta _p|)sign(\beta _p),0^{\tau }\}^{\tau }. \end{aligned}$$

The minimizer $(\beta ,\gamma ,v)$ of (A.24) satisfies $\tilde{Q}_{in}(\beta ,\gamma ,v)=0, (i=1,2,3)$. Since $||\gamma ||=O_p(a_n)$ for $\beta \in \mathcal {B}$, we can obtain that $||v||=O_p(a_n)$ from $\tilde{Q}_{2n}(\beta ,\gamma ,v)=0$, In order to expand $\tilde{Q}_{in}(\beta ,\gamma ,v)(i=1,2,3)$ around the value $(\beta _0,0,0)$, we first give the following partial derivatives,

$$\begin{aligned}&\displaystyle \frac{\partial \tilde{Q}_{1n}(\beta _0,0,0)}{\partial \gamma }=-J(\beta _0), ~ \frac{\partial \tilde{Q}_{1n}(\beta _0,0,0)}{\partial \beta }=\Sigma (\beta _0),~\frac{\partial \tilde{Q}_{1n}(\beta _0,0,0)}{\partial v}=0,\\&\displaystyle \frac{\partial \tilde{Q}_{2n}(\beta _0,0,0)}{\partial \gamma }=\Sigma (\beta _0),~\frac{\partial \tilde{Q}_{2n}(\beta _0,0,0)}{\partial \beta }=b'(\beta ),~\frac{\partial \tilde{Q}_{2n}(\beta _0,0,0)}{\partial v} =\Psi _2^{\tau },\\&\displaystyle \frac{\partial \tilde{Q}_{3n}(\beta _0,0,0)}{\partial \gamma }=0,~\frac{\partial \tilde{Q}_{3n}(\beta _0,0,0)}{\partial \beta }=\Psi _2,~\frac{\partial \tilde{Q}_{3n}(\beta _0,0,0)}{\partial v} =0. \end{aligned}$$

Then by Taylor expansion, we immediately derive that

$$\begin{aligned} \left( \begin{array}{ccc} \tilde{Q}_{1n}(\beta _0,0,0)\\ 0\\ 0\\ \end{array}\right) = \left( \begin{array}{ccc} -\Sigma _1 &{}\Sigma _0 &{}0\\ \Sigma _0^{\tau }&{}0&{}\Psi _2^{\tau }\\ 0&{}\Psi _2&{}0\\ \end{array}\right) \left( \begin{array}{ccc}\tilde{\gamma }\\ \tilde{\beta }\\ \tilde{v}\\ \end{array}\right) +R_n, \end{aligned}$$

(A.25)

where $\Sigma (\beta _0)=n^{-1}\sum _{i=1}^n\partial \Gamma _i(\beta _0)/\partial \beta$, $R_n=\sum _{l=1}^5R_n^{(l)}$, $R_n^{(1)}=(R_{1n}^{\tau 1},R_{2n}^{\tau 1},0)^{\tau }$, $R_{jn}^{\tau 1}\in R^p$ and the kth component of $R_{jn}^{\tau 1}$ for $j=1,2$ is given by

$$\begin{aligned} R_{jn,k}^{(1)}=\frac{1}{2}(\hat{\vartheta }-\vartheta _0)^{\tau }\frac{\partial ^2\tilde{Q}_{jn,k}(\tilde{\vartheta })}{\partial \vartheta \partial \vartheta ^{\tau }}(\hat{\vartheta }-\vartheta _0), \end{aligned}$$

$\vartheta =(\beta ,\gamma )^{\tau }, \tilde{\vartheta }=(\tilde{\beta },\tilde{\gamma })^{\tau }$ satisfying $||\tilde{\vartheta }-\vartheta _0||\le ||\hat{\vartheta }-\vartheta _0||$. $R_n^{(2)}=(0,b^{\tau }(\beta _0),0)^{\tau }$, $R_n^{(3)}=[0,\{b'(\tilde{\vartheta })(\hat{\vartheta }-\vartheta _0)\},0]^{\tau }$, $R_n^{(4)}=[\{(J(\beta _0)-\Sigma _1))\hat{\gamma }\}^{\tau }+(\Sigma (\beta _0)-\Sigma _0)(\hat{\beta }-\beta )\}^{\tau },0,0]^{\tau }$ and $R_n^{(5)}=[0,\{(\Sigma (\beta _0)-\Sigma _0)\hat{\gamma }\}^{\tau },0]^{\tau }$. Similar to the proof of (A.22), we can get $R_n^{(1)}=o_p(n^{-1/2})$. Given Condition (B8) and (B9), we see that $R_n^{(2)}=o_p(n^{-1/2})$ and $R_n^{(3)}=o_p(n^{-1/2})$. By (A.21) and (A.17) which together with Lemma A.6 yields that $R_n^{(4)}=o_p(n^{-1/2})$ and $R_n^{(5)}=o_p(n^{-1/2})$. Hence, we can get $R_n^{(k)}=o_p(n^{-1/2}),k=1,\ldots ,5$.

Define $K_{11}=-\Sigma _1, K_{12}=[\Sigma _0,~~0]$ and $K_{21}=K_{12}^{\tau }$,

$$\begin{aligned} K_{22}=\left( \begin{array}{ccc}0&{}\quad \Psi _2^{\tau }\\ \Psi _2&{}\quad 0\\ \end{array}\right) ,~K=\left( \begin{array}{ccc}K_{11}&{}\quad K_{12}\\ K_{21}&{}\quad K_{22}\\ \end{array}\right) \end{aligned}$$

and let $\kappa =(\beta ^{\tau },v^{\tau })^{\tau }$. Then by inverted (A.25), we find

$$\begin{aligned} \left( \begin{array}{ccc} \hat{\gamma }\\ \hat{\kappa }-\kappa _0\\ \end{array}\right) =K^{-1}\Big \{ \left( \begin{array}{ccc} -\tilde{Q}_{1n}(\beta _0,0,0)\\ 0\\ \end{array}\right) +R_n\Big \}, \end{aligned}$$

(A.26)

As matrix K is partitioned into four blocks, it can be inverted blockwise as follows

$$\begin{aligned} K^{-1}=\left[ \begin{array}{ccc} K_{11}^{-1}+K_{11}^{-1}K_{12}A^{-1}K_{21}K_{11}^{-1}&{}\quad -K_{11}^{-1}K_{12}A^{-1}\\ -A^{-1}K_{21}K_{11}^{-1}&{}\quad A^{-1}\\ \end{array}\right] , \end{aligned}$$

where $A=K_{22}-K_{21}K_{11}^{-1}K_{12}=\left[ \begin{array}{ccc} \Omega ^{-1} &{}\quad \Psi _2^{\tau }\\ \Psi _2&{}\quad 0\\ \end{array}\right]$ and $\Sigma$ is defined in Theorem 2.2. Thus, we get

$$\begin{aligned} \hat{\kappa }-\kappa _0=A^{-1}K_{21}K_{11}^{-1}\tilde{Q}_{1n}(\beta _0,0,0)+o_p(n^{-1/2}). \end{aligned}$$

Matric A can also be inverted blockwise by using the analytic inversion formula,ie.,

$$\begin{aligned} A^{-1}=\left[ \begin{array}{ccc} \Omega -\Omega \Psi _2^{\tau }(\Psi _2\Omega \Psi _2^{\tau })^{-1}\Psi _2\Omega &{}\quad \Omega \Psi _2^{\tau }(\Psi _2\Omega \Psi _2^{\tau })^{-1}\\ (\Psi _2\Omega \Psi _2^{\tau })^{-1}\Psi _2\Omega &{}\quad -(\Psi _2\Omega \Psi _2^{\tau })^{-1}\\ \end{array}\right] , \end{aligned}$$

Further, we have

$$\begin{aligned} \hat{\beta }-\beta _0=[\Omega -\Omega \Psi _2^{\tau }(\Psi _2\Omega \Psi _2^{\tau })^{-1}\Psi _2\Omega ](\Sigma _0\Sigma _1\bar{\Gamma }(\beta _0)+o_p(n^{-1/2})). \end{aligned}$$

It follows by an expansion of $\hat{\beta }_1$ that

$$\begin{aligned} \hat{\beta }_1-\beta _0=[\Psi _1\Omega -\Psi _1\Omega \Psi _2^{\tau }(\Psi _2\Omega \Psi _2^{\tau })^{-1}\Psi _2\Omega ](\Sigma _0\Sigma _1\bar{\Gamma }(\beta _0)+o_p(n^{-1/2})). \end{aligned}$$

(A.27)

Then similar to the proof of Theorem 2.3 in Fan et al. (2016), we have $n^{1/2}W_n\Omega _p^{-1/2}(\widehat{\beta }_1-\beta _{10}){\mathop {\rightarrow }\limits ^{d}}N(0,G)$, which completes the proof of Theorem 2.3. $\square$

Proof of Theorem 2.4

Let $\hat{\beta }$ be the minimizer (2.13) and $U_i=\hat{\gamma }^{\tau }\Gamma _i(\beta )$. Taylor expansion gives

$$\begin{aligned} \mathcal {L}_{n}(\beta )=\sum _{i=1}^nU_i-\sum _{i=1}^nU_i^2/2+\sum _{i=1}^nU_i^3/3(1+\xi _i)^4+o_p(1), \end{aligned}$$

where $|\xi _i|\rightarrow |U_i|$ and $o_p(1)$ is due to the penalty function. From (A.26), we have $\hat{\gamma }=[\Sigma _1^{-1}+\Sigma _1^{-1}\Sigma _0\{\Omega -\Omega \Psi _2^{\tau }(\Psi _2\Omega \Psi _2^{\tau })^{-1}\Psi _2\Omega \}\Sigma _0^{\tau }\Sigma _1^{-1}][\bar{\Gamma }(\beta _0)+o_p(n^{-1/2})].$

Similar to Tang and Leng (2010), Substituting the expansion of $\hat{\gamma }$ and $\hat{\beta }$ given by (A.24) into $U_i$, we show that

$$\begin{aligned} 2\mathcal {L}_{n}(\hat{\beta })= n\bar{\Gamma }(\beta _0)^{\tau }\Psi _2^{\tau }(\Psi _2\Omega ^{-1}\Psi _2^{\tau })^{-1}\Psi _2\bar{\Gamma }(\beta _0)+o_p(1) \end{aligned}$$

(A.28)

Under the null hypothesis, because $L_nL_n^{\tau }=I_q$, there exists $\tilde{\Psi }_2$ such that $\tilde{\Psi }_2\beta =0$ and $\tilde{\Psi }_2\tilde{\Psi }_2^{\tau }=I_{p-d+q}.$ Now by repeating the proof of Theorem 2.3, we establish that under the null hypothesis, the estimation of $\beta$ can be obtained by minimizing (A.27), where $\Psi _2$ is replaced by $\tilde{\Psi }_2$, we can easily obtain that

$$\begin{aligned} 2\mathcal {L}_{n}(\tilde{\beta })= n\bar{\Gamma }(\beta _0)^{\tau }\tilde{\Psi }_2^{\tau }(\tilde{\Psi }_2\Omega ^{-1}\tilde{\Psi }_2^{\tau })^{-1}\tilde{\Psi }_2\bar{\Gamma }(\beta _0)+o_p(1). \end{aligned}$$

Combining Eqs. (A.28), we have

$$\begin{aligned} \mathcal {L}_{n}= n\bar{\Gamma }(\beta _0)^{\tau }\Omega ^{-1/2}(P_1-P_2)\Omega ^{-1/2}\bar{\Gamma }(\beta _0)+o_p(1). \end{aligned}$$

where

$$\begin{aligned} P_1=\Omega ^{-1/2}\Psi _2^{\tau }(\Psi _2\Omega ^{-1}\Psi _2^{\tau })^{-1}\Psi _2\Omega ^{-1/2}, \end{aligned}$$

and

$$\begin{aligned} P_2=\Omega ^{-1/2}\tilde{\Psi }_2^{\tau }(\tilde{\Psi }_2\Omega ^{-1}\tilde{\Psi }_2^{\tau })^{-1}\tilde{\Psi }_2\Omega ^{-1/2}, \end{aligned}$$

are two idempotent matrices. As the rank of $P_1-P_2$ is q, $P_1-P_2$ can be written as $\Upsilon ^{\tau }\Upsilon$, where $\Upsilon$ is $q\times p$ matrix such that $\Upsilon ^{\tau }\Upsilon =I_q$, further, we see that

$$\begin{aligned} \sqrt{n}\Upsilon \Omega ^{-1/2}\bar{\Gamma }(\beta _0){\mathop {\rightarrow }\limits ^{d}}N(0,I_q). \end{aligned}$$

Then

$$\begin{aligned} n\bar{\Gamma }(\beta _0)^{\tau }\Omega ^{-1/2}(P_1-P_2)\Omega ^{-1/2}\bar{\Gamma }(\beta _0){\mathop {\rightarrow }\limits ^{d}}\chi _q^2. \end{aligned}$$

and the proof of Theorem 2.4 is finished. $\square$

Lemma A.8

Under the conditions of Theorem 2.5. For a given z, if g(z) is the true value of the parameter, then

$$\begin{aligned}&\displaystyle \frac{1}{\sqrt{Nh}}\sum _{i=1}^n\hat{\Xi }_{i}\{g(z)\}-b(z){\mathop {\rightarrow }\limits ^{d}}N(0,R). \end{aligned}$$

(A.29)

$$\begin{aligned}&\displaystyle \frac{1}{Nh}\sum _{i=1}^n\hat{\Xi }_{i}\{g(z)\}\hat{\Xi }_{i}^{\tau }\{g(z)\}{\mathop {\rightarrow }\limits ^{P}}R . \end{aligned}$$

(A.30)

$$\begin{aligned}&\displaystyle \mathop {max}\limits _{1\le i\le n}||\hat{\Xi }_{i}\{g(z)\}||=o_p(\sqrt{Nh}), \phi =O_p(N^{-1/2}). \end{aligned}$$

(A.31)

where $b(z)=\left( \frac{N}{h}\right) ^{1/2}\frac{T-1}{T}E[g(Z_{it})-g(z)]f(z)\int K(z)dz$ and $R=\sigma ^2f(z)\int K^2(z)dz$.

Proof

Observe that

$$\begin{aligned} \frac{1}{\sqrt{Nh}}\sum _{i=1}^n\hat{\Xi }_{i}\{g(z)\}-b(z)=S_1(z)+S_2(z)+S_3(z) \end{aligned}$$

where

$$\begin{aligned}&\displaystyle S_1(z)=\frac{1}{\sqrt{Nh}}\sum _{i=1}^n\sum _{t=1}^T(I_{it}-Q)K_h(Z_{it}-z)\varepsilon _{it},\\&\displaystyle S_2(z)=\frac{1}{\sqrt{Nh}}\sum _{i=1}^n\sum _{t=1}^T(I_{it}-Q)\{[g(Z_{it})-g(z)]K_h(Z_{it}-z)-(h/N)^{1/2}b(z)\},\\&\displaystyle S_3(z)=\frac{1}{\sqrt{Nh}}\sum _{i=1}^n\sum _{t=1}^T(I_{it}-Q)X_{it}^{\tau }(\beta -\hat{\beta })K_h(Z_{it}-z). \end{aligned}$$

It is not difficult to prove $E[S_1(z)]=0$ and $Var[S_1(z)]=R+o(1)$. $S_1(z)$ satisfies the conditions of the Cramer–Wold theorem and the Lindeberg condition. Therefore, we get

$$\begin{aligned} S_1(z){\mathop {\rightarrow }\limits ^{d}}N(0,R). \end{aligned}$$

(A.32)

We can also prove that

$$\begin{aligned} S_2(z)=o_p(1). \end{aligned}$$

(A.33)

Theorems 2.2 and condition (B8) imply that $\beta -\hat{\beta }=O_p(N^{-1/2})$. Therefore, we get $S_3(z)=O_p(h^{1/2})$. This together with (A.32) and (A.33) proves (A.29).

Analogously to the proof of (A.30). We can verify (A.30) easily. As to (A.31), we find

$$\begin{aligned} \mathop {max}\limits _{1\le i\le n}||\hat{\Xi }_{i}\{g(z)\}||\le & {} \mathop {max}\limits _{1\le i\le n}||(I_{it}-Q)K_h(Z_{it}-z)\varepsilon _{it}||\\&+\mathop {max}\limits _{1\le i\le n}||(I_{it}-Q)[g(Z_{it})-g(z)]K_h(Z_{it}-z)||\\&+\mathop {max}\limits _{1\le i\le n}||(I_{it}-Q)X_{it}^{\tau }(\beta -\hat{\beta })K_h(Z_{it}-z)||=J_1+J_2+J_3 \end{aligned}$$

From Markov inequality and conditions (B3) and (B4), one can obtain

$$\begin{aligned} P(J_1\ge \sqrt{Nh})\le (Nh)^{-s}\sum _{i=1}^nE[(I_{it}-Q)\varepsilon _{it}K_h(Z_{it}-z)]^{2s}\le C(Nh)^{1-s}\rightarrow 0 \end{aligned}$$

which implies that $J_1=o_p(\sqrt{Nh})$. Using some arguments similar to those used in the proof of Lemma A.6, we can prove $J_2=o_p(\sqrt{Nh})$ and $J_3=o_p(\sqrt{Nh})$. Therefore we obtain that $\mathop {max}\nolimits _{1\le i\le n}||\hat{\Xi }_{i}\{g(z)\}||=o_p(\sqrt{Nh}).$

Applying (A.30) and the proof in Owen (1990), one can derive that $\phi =O_p(N^{-1/2})$, which completes the proof of Lemma A.8. $\square$

Proof of Theorem 2.5

Invoking some arguments similar to those used in the proof of can be proved Theorems 2.4, we can proof

$$\begin{aligned} 2\mathcal {Q}_{n}(g(z))=\left[ \frac{1}{Nh}\sum _{i=1}^n\hat{\Xi }_{i}^2\{g(z)\}\right] ^{-1}\Big \{\frac{1}{\sqrt{Nh}}\sum _{i=1}^n\hat{\Xi }_{i}\{g(z)\}-b(z)\Big \}^2 \end{aligned}$$

From Lemma A.8, we can prove that $2\mathcal {Q}_{n}(g(z)){\mathop {\rightarrow }\limits ^{d}}\chi _1^2.$ $\square$