Let *r* > 2 be an integer. *g*(*x*), *f*(·|*x*) and *F*(·|*x*) are used to denote the density function of *X*,the density and distribution functions of *ε* conditional on *X*_{i} = x, respectively. Let *c* be a positive constant which is independent of *n* and may take a different value in different place. The following conditions will be used in this section.

(C1) {(*Y*_{i}, *X*_{i}): *i* = 1,2,⋯, *n*} are independent and identically distributed random vectors.

(C2) Both *p*(*x*) (the selection probability function) and *g*(*x*) have bounded derivatives up to order *r* almost surely and inf_{x} *p*(*x*) *>* 0:

(C3) *K*(·) is a kernel function of order *r* and is bounded and compactly supported on [– 1, 1]. Furthermore, there exist positive constants C_{1}, C_{2} and *ρ* such that ${C}_{1}{I}_{[||u||\le \rho ]}\le K(u)\le {C}_{2}{I}_{[||u||\le \rho ]}.$

(C4) $P(\parallel X\parallel >{M}_{n})=o({n}^{-1/2}),$ where 0 < *M*_{n} →∞ as *n* → ∞.

(C5) The positive bandwidth parameter *h* satisfies *nh*^{2r} →0 when n →∞.

(C6) The matrices *A* and *B* defined in Theorem 2.2 are both nonsingular. Firstly, some lemmas are introduced to derive the main results.

#### Lemma 6.1

*(see Lemma 2 in [11]) Suppose that Conditions C1 – C6 hold. Then*
$$E\{\hat{p}({X}_{i})-p({X}_{i}){\}}^{2}=O((n{h}^{d}{)}^{-1}{M}_{n}^{d})+O({h}^{2r})+o({n}^{-1/2})$$

*holds uniformly for i* = 1,2,⋯, *n*.

#### Lemma 6.2

*Suppose that Conditions C1 – C6 hold. If β is the true parameter of model (1), then*
$$\begin{array}{l}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{i}(\beta )\stackrel{\mathcal{L}}{\to}N(0,B)\end{array}$$(11)

*and*
$$\begin{array}{l}E(\mathrm{\partial}{Z}_{i}(\beta )/\mathrm{\partial}\beta )=A+o(1),\end{array}$$(12)

*where Z*_{i} (*β*) *is taken to be ${Z}_{iw}^{\ast}(\beta ),{Z}_{iw}(\beta )\phantom{\rule{thickmathspace}{0ex}}\mathit{o}\mathit{r}\phantom{\rule{thickmathspace}{0ex}}{Z}_{iI}(\beta ),A=E\{\pi (X)f(0|X)X{X}^{T}\}$ and $B=\tau (1-\tau )E\{\pi (X)X{X}^{T}\}$* *with π* (*x*) = *1/p*(*x*) *when Z*_{i} (*β*) *is taken to be ${Z}_{iw}^{\ast}(\beta )$ and Z*_{iw}(*β*)*; and π*(*x*) = 1 *when Z*_{i}(*β*) *= Z*_{iI}(*β*).

#### Proof

(a) The case of ${Z}_{i}(\beta )={Z}_{iw}^{\ast}(\beta )$ will be proved firstly for *i* = 1,2,⋯, *n*. Some simple calculation yields
$$\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{iw}^{\ast}(\beta )=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}}{p({X}_{i})}{X}_{i}{\psi}_{i}n({Y}_{i},{X}_{i},\beta )\equiv J.$$

It is easy to obtain *E*(*J*) = 0 and *Cov*(*J) = B*. Then it follows from the central limit theorem that (11) is obtained immediately. In a similar way, we can prove (12).

(b) Now, we prove the case of *Z*_{i}(*β*)*= Z*_{iw} (*β*) *for i* = 1,2,⋯, *n*. Because
$$\begin{array}{ll}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{iw}(\beta )& =\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}}{\hat{p}({X}_{i})}{X}_{i}{\psi}_{i}({Y}_{i},{X}_{i},\beta )\\ & =\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}}{p({X}_{i})}{X}_{i}{\psi}_{i}({Y}_{i},{X}_{i},\beta )+\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}(p({X}_{i})-\hat{p}({X}_{i}))}{\hat{p}({X}_{i})p({X}_{i})}{X}_{i}{\psi}_{i}({Y}_{i},{X}_{i},\beta )\end{array}$$(13)

Similarly to the proof of Theorem 3 in [28], it follows from Conditions C2, C3 and C5 that
$$\begin{array}{l}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}}{\hat{p}({X}_{i})p({X}_{i})}{X}_{i}{\psi}_{i}({Y}_{i},{X}_{i},\beta )={O}_{p}(1)\end{array}$$(14)

Since ${sup}_{x}|\hat{p}(x)-p(x)|={o}_{p}(1),$ (14) indicates
$$\begin{array}{l}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{iw}n=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}}{p({X}_{i})}{X}_{i}{\psi}_{i}({Y}_{i},{X}_{i},\beta )+{o}_{p}(1)=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{iw}^{\ast}(\beta )+{o}_{p}(1).\end{array}$$(15)

Then, (11) is obtained immediately. On the other hand, the proof of (12) is similar to the proof in case (a) and hence is omitted here.

(c) When *Z*_{i}(*β*) = *Z*_{iI}(*β*) *for i* = 1,2,⋯, *n*, direct calculation obtains
$${X}_{i}^{T}\beta -{\hat{Y}}_{i}=\frac{{\delta}_{i}}{\hat{p}({X}_{i})}({X}_{i}^{T}\beta -{Y}_{i})+(1-\frac{{\delta}_{i}}{\hat{p}({X}_{i})}){X}_{i}^{T}(\beta -{\hat{\beta}}_{Q}).$$

Then it is easily shown that $\parallel \frac{1}{n}{\sum}_{i=1}^{n}(1-\frac{{\delta}_{i}}{\hat{p}({X}_{i})}){X}_{i}\parallel ={o}_{p}(1)$ *and*
$${\hat{\beta}}_{Q}-\beta ={A}^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{i}(\beta )+{o}_{p}({n}^{-1/2})={O}_{p}({n}^{-1/2}).$$

Therefore, we have
$$\begin{array}{l}{X}_{i}({I}_{({x}_{i}^{T}\beta -{\hat{Y}}_{i}>0)}-\tau )={X}_{i}({I}_{({x}_{i}^{T}\beta -{Y}_{i}>0)}-\tau )={X}_{i}{\psi}_{i}({Y}_{i},{X}_{i},\beta )\end{array}$$(16)

with which we can prove (11) and (12) by using the similar way in the case (a).□

#### Lemma 6.3

*Suppose that Conditions* C1 – C6 *hold. If β is the true parameter of model (1), then*
$$\begin{array}{l}\frac{1}{n}\sum _{i=1}^{n}{Z}_{i}(\beta ){Z}_{i}^{T}(\beta )\stackrel{\mathcal{P}}{\to}B\end{array}$$(17)

*where Z*_{i}(*β*) *is taken to be ${Z}_{iw}^{\ast}(\beta ),{Z}_{iw}(\beta )\phantom{\rule{thickmathspace}{0ex}}or\phantom{\rule{thickmathspace}{0ex}}{Z}_{iI}(\beta ),\phantom{\rule{thickmathspace}{0ex}}and\phantom{\rule{thickmathspace}{0ex}}B=\tau (1-\tau )E\{\pi (X)X{X}^{T}\}$ with π*(*x*) = 1/*p*(*x*) *when Z*_{i}(*β*) *is taken to be ${Z}_{iw}^{\ast}(\beta )\phantom{\rule{thickmathspace}{0ex}}and\phantom{\rule{thickmathspace}{0ex}}{Z}_{iw}(\beta ),$ and π*(*x*) = 1 *when Z*_{i} (*β*) *= Z*_{iI}(*β*).

#### Proof

(a) When ${Z}_{i}(\beta )={Z}_{iw}^{\ast}(\beta ),$ *for i* = 1,2,⋯, *n*, some simple calculation yields
$$\frac{1}{n}\sum _{i=1}^{n}{Z}_{iw}^{\ast}(\beta ){Z}_{iw}^{\ast T}(\beta )=\frac{1}{n}\sum _{i=1}^{n}{A}_{i1}{A}_{i1}^{T},$$

where ${A}_{i1}={\displaystyle \frac{{\delta}_{i}}{p({X}_{i})}{X}_{i}{\psi}_{i}({Y}_{i},{X}_{i},\beta ).}$ By the law of large numbers, we can derive the result immediately.

(b) When *Z*_{i}(*β*)*= Z*_{iw} (*β*) *for i* = 1,2,⋯,*n*,
$$\begin{array}{ll}\frac{1}{n}\sum _{i=1}^{n}{Z}_{iw}(\beta ){Z}_{iw}^{T}(\beta )& =\frac{1}{n}\sum _{i=1}^{n}{A}_{i1}{A}_{i1}^{T}+\frac{1}{n}\sum _{i=1}^{n}{A}_{i1}{A}_{i2}^{T}+\frac{1}{n}\sum _{i=1}^{n}{A}_{i2}{A}_{i1}^{T}+\frac{1}{n}\sum _{i=1}^{n}{A}_{i2}{A}_{i2}^{T}\\ & ={B}_{1}+{B}_{2}+{B}_{3}+{B}_{4},\end{array}$$(18)

where ${A}_{i2}={\displaystyle \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}(p({X}_{i})-\hat{p}({X}_{i}))}{\hat{p}({X}_{i})p({X}_{i})}{X}_{i}{\psi}_{i}({Y}_{i},{X}_{i},\beta ).}$ By the law of large numbers, we can derive that ${B}_{1}\stackrel{\mathcal{P}}{\to}B.$ Now, ${B}_{2}\stackrel{\mathcal{P}}{\to}0$ will be proved. Let *B*_{2},_{ k}_{s} be the (*k,s*) component of *B*_{2}, A_{ij,r} be the *r*th component of *A*_{ij}, *j* = 1,2. Then we use the Cauchy-Schwarz inequality to get
$$|{B}_{2,ks}|\le {\left(\frac{1}{n}\sum _{i=1}^{n}{A}_{i1,k}^{2}\right)}^{1/2}{\left(\frac{1}{n}\sum _{i=1}^{n}{A}_{i2,r}^{2}\right)}^{1/2}$$

From Lemma 6.1 and 6.2, we can see ${n}^{-1}{\displaystyle \sum _{i=1}^{n}{A}_{i1,k}^{2}={O}_{p}(1)\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thickmathspace}{0ex}}{n}^{-1}\sum _{i=1}^{n}{A}_{i2,r}^{2}={o}_{p}(1).}$ Hence ${B}_{2}\stackrel{\mathcal{P}}{\to}0.$

Using the similar argument, we can prove ${B}_{i}\stackrel{\mathcal{P}}{\to}0$ for *i* = 3, 4.

(c) When *Z*_{i}(*β*) = *Z*_{iI}(*β*) *for i* = 1,2,⋯,*n*, by the (16) and the same methods as that of (a) and (b),

therefore, we can obtain the result.□

#### Lemma 6.4

*Suppose that Conditions* C1 – C6 *hold. If θ is the true parameter of model (1), then*
$$\begin{array}{l}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}({\hat{Y}}_{i}-\theta )\stackrel{\mathcal{L}}{\to}N(0,V),\end{array}$$(19)
$$\begin{array}{l}\frac{1}{n}\sum _{i=1}^{n}({\hat{Y}}_{i}-\theta {)}^{2}\stackrel{\mathcal{P}}{\to}V,\end{array}$$(20)

*and*
$$\begin{array}{l}max|{\hat{Y}}_{i}|={O}_{p}({n}^{1/2}).\end{array}$$(21)

#### Proof

We prove (19) only. (20) and (21) can be proved similarly. It is straightforward to obtain
$$\frac{1}{\sqrt{n}}\sum _{i=1}^{n}({\hat{Y}}_{i}-\theta )={A}_{1}+{A}_{2}+{A}_{3},$$

where ${A}_{1}={\displaystyle \frac{1}{\sqrt{n}}\sum _{i=1}^{n}[\frac{{\delta}_{i}{\epsilon}_{i}}{p({X}_{i})}+\{{X}_{i}^{T}\beta -\theta \}],\phantom{\rule{thinmathspace}{0ex}}{A}_{2}=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}(\frac{1}{\hat{p}({X}_{i})}-\frac{1}{p({X}_{i})}){\delta}_{i}{\epsilon}_{i}}$ and ${A}_{3}={\displaystyle \frac{1}{\sqrt{n}}\sum _{i=1}^{n}(1-\frac{{\delta}_{i}}{\hat{p}({X}_{i})}){X}_{i}^{T}({\hat{\beta}}_{SQ}-\beta ).}$ It follows from the central limit theorem that
$${A}_{1}\stackrel{\mathcal{L}}{\to}N(0,V).$$

To prove (19), we only need to prove that *A*_{2} = *o*_{p}(1) and *A3 = o*_{p}(1)*. A*_{2} will be proved firstly. Direct calculation yields
$${A}_{2}=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}{\epsilon}_{i}\{p({X}_{i})-\hat{p}({X}_{i})\}}{{p}^{2}({X}_{i})}+\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{{\delta}_{i}{\epsilon}_{i}\{p({X}_{i})-\hat{p}({X}_{i}){\}}^{2}}{{p}^{2}({X}_{i})\hat{p}({X}_{i})}={A}_{21}+{A}_{22}.$$

By Lemma 6.1, it is easy to show that *A*_{22} = *o*_{p}(1). Simple calculation yields
$$\begin{array}{ll}{A}_{21}& =\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[\frac{{\delta}_{i}{\epsilon}_{i}}{{p}^{2}({X}_{i})}\sum _{j=1}^{n}{W}_{nj}({X}_{i})\{p({X}_{i})-p({X}_{j})\}\right]\\ & +\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[\frac{{\delta}_{i}{\epsilon}_{i}}{{p}^{2}({X}_{i})}\sum _{j=1}^{n}{W}_{nj}({X}_{i})\{p({X}_{i})-{\delta}_{j})\}\right]+\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[\frac{{\delta}_{i}{\epsilon}_{i}}{p({X}_{i})}\{1-\sum _{j=1}^{n}{W}_{nj}({X}_{i})\}\right]\\ & ={A}_{211}+{A}_{212}+{A}_{213},\end{array}$$(22)

where ${W}_{nj}(x)={\displaystyle \frac{{K}_{h}({X}_{j}-x)}{\sum _{i=1}^{n}{K}_{h}({X}_{i}-x)}.}$ By the Cauchy-Schwarz inequality, we get
$$\begin{array}{ll}E({A}_{211}^{2})& \le \frac{c}{n}\sum _{i=1}^{n}E\left[E\left\{\sum _{j=1}^{n}{W}_{nj}({X}_{i})\{p({X}_{i})-p({X}_{j}{)}^{2}|{X}_{i}\right\}\right]\\ & \le \frac{c}{n}\sum _{i=1}^{n}E\left[E\left\{\sum _{j=1}^{n}{W}_{nj}({X}_{i})||{X}_{i}-{X}_{j}|{|}^{2}|{X}_{i}\right\}\right]\\ & \le c{h}^{2}\to 0\end{array}$$

So we prove *A*_{211} *= o*_{p}(1).

To handle *A*_{212}, write ${S}_{i}^{\prime}={\displaystyle \frac{{\delta}_{i}{\epsilon}_{i}}{{p}^{2}({X}_{i})}\phantom{\rule{thickmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thickmathspace}{0ex}}{S}_{j}^{\u2033}=p({X}_{j})-{\delta}_{j}.}$ We have
$$\begin{array}{ll}E({A}_{212}^{2})& =\frac{c}{n}\sum _{i=1}^{n}\sum _{j=1}^{n}E\{{W}_{nj}({X}_{i}){S}_{i}^{\prime}{S}_{j}^{{}^{\u2033}2}\}\\ & \le \frac{c}{n}\sum _{i=1}^{n}E\left\{\sum _{j=1}^{n}{W}_{nj}^{2}({X}_{i})\right\}\to 0\end{array}$$

Thus, it follows that A_{212} = *o*_{p}(1). It is easy to show that A_{213} = *o*_{p}(1). So A_{21} = *o*_{p}(1) is obtained immediately. In addition, Lemma 6.2 indicates *A*_{3} = *o*_{p}(1). This proves (19) and consequently completes the proof of Lemma 6.4.□

*Proof of Theorem 2.1*. By the Lagrange multiplier method, $\hat{R}(\beta )$ can be represented as
$$\begin{array}{l}\hat{R}(\beta )=2\sum _{i=1}^{n}\mathrm{log}(1+{\lambda}^{T}(\beta ){Z}_{i}(\beta )),\end{array}$$(23)

where *λ*(*β*) is a *d* × 1 vector given as the solution of the equation
$$\begin{array}{l}\sum _{i=1}^{n}\frac{{Z}_{i}(\beta )}{1+{\lambda}^{T}(\beta ){Z}_{i}(\beta ))}=0.\end{array}$$(24)

According to Lemma 6.2 and the arguments in the proof of (2.14) in Owen [16], we can show that
$$\begin{array}{l}\lambda (\beta )=({Z}_{i}(\beta ){Z}_{i}^{T}(\beta ){)}^{-1}\sum _{i=1}^{n}{Z}_{i}(\beta )+{o}_{p}({n}^{-1/2}).\end{array}$$(25)

Applying the Taylor expansion to (23) and invoking Lemma 6.2 and (25), we obtain
$$\begin{array}{l}\hat{R}(\beta )=2\sum _{i=1}^{n}[{\lambda}^{T}(\beta ){Z}_{i}(\beta )-({\lambda}^{T}(\beta ){Z}_{i}(\beta ){)}^{2}/2]+{o}_{p}(1).\end{array}$$(26)

Then it follows from (24) that
$$0=\sum _{i=1}^{n}\frac{{Z}_{i}(\beta )}{1+{\lambda}^{T}(\beta ){Z}_{i}(\beta )}=\sum _{i=1}^{n}{Z}_{i}(\beta )-\sum _{i=1}^{n}{Z}_{i}(\beta ){Z}_{i}^{T}(\beta )\lambda (\beta )+\sum _{i=1}^{n}\frac{{Z}_{i}(\beta )({\lambda}^{T}(\beta ){Z}_{i}(\beta ){)}^{2}}{1+{\lambda}^{T}(\beta ){Z}_{i}(\beta )}.$$

Lemma 6.2 and (25) imply
$$\sum _{i=1}^{n}({\lambda}^{T}(\beta ){Z}_{i}(\beta ){)}^{2}=\sum _{i=1}^{n}{\lambda}^{T}(\beta ){Z}_{i}(\beta )+{o}_{p}(1).$$

Therefore, it is obtained from (26) that
$$\hat{R}(\beta )=(\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{i}^{T}(\beta ))(\frac{1}{n}\sum _{i=1}^{n}{Z}_{i}(\beta ){Z}_{i}^{T}(\beta ){)}^{-1}(\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{i}(\beta ))+{o}_{p}(1).$$

This together with Lemma 6.2 completes the proof of Theorem 2.1.□

*Proof of Theorem 2.2*. First, a Taylor expansion for ${Z}_{i}(\hat{\beta})$ gives
$$\begin{array}{ll}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{i}(\hat{\beta})& =\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{i}(\beta )+\frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{i}^{\prime}(\beta )(\hat{\beta}-\beta )+{o}_{p}({n}^{-1/2})\\ & ={D}_{n}+A(\hat{\beta}-\beta )+{o}_{p}({n}^{-1/2})\end{array}$$(27)

Where ${D}_{n}={\displaystyle \frac{1}{\sqrt{n}}\sum _{i=1}^{n}{Z}_{i}(\beta ).}$ Then, similar to the proofs of (28)-(30) in [22], we have ${\hat{\beta}}_{QEL}-\beta ={o}_{p}({n}^{-1/2})$ by Lemma 6.2. Finally, it follows from Lemma 6.2 and (26), (27) that the conclusion of Theorem 2.2 is obtained directly.□

*Proof of Theorem 3.1*. Similarly to the proof of Theorem 2.1, Theorem 3.1 can be proved by Lemma 6.3. Thus, we omit this proof.□

*Proof of Theorem 3.2*. It follows from (9) and (10) that
$$\sqrt{n}(\hat{\theta}-\theta )=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}({\hat{Y}}_{i}-\theta )+{o}_{p}(1).$$

This together with (19) proves Theorem 3.2.□

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