Appendix
To achieve our conclusions, we assume that regularity conditions to be held (Wang and Wang 2001). These conditions are as follow:
(C1) \(\pi (v,\gamma )\) is bounded and has partial derivatives up to order 2 almost surely.
(C2) The kernel function \(k_{h}(.)\) is continuous and is from order r. It is always at least from order 2.
(C3) The density function of \({\varvec{X}}, f({\varvec{x}})\), exists and has bounded derivatives up to at least order 2.
(C4) The matrix \(\Sigma\) has a finite component and is positive definite.
(C5) \([nh^{2r}+(nh^{2m})^{-1}]\) converges to zero as n goes to infinity. m is the dimension of vector \({\varvec{x}}\).
(C6) \(E(Y^2)\) and \(E({\mathrm{e}}^{2\gamma Y})\) exist and are finite.
(C7) The matrices \(V_A\) and \(V_{{\mathrm{IPW}}}\) which are defined in Theorem 1, are positive definite.
Proof of Lemma 2
We just prove the parts (c) and (d) of Lemma 2. The other parts of Lemma 2 can be concluded in a similar way by taking \(C_n=0\).
1.1 Proof in IPW case
If \(\gamma\) is estimated from Eq. (13), based on the IPW method we can write,
$$\begin{aligned} \sqrt{n}\left( {\hat{\beta }}_{{\mathrm{IPW}}}-\beta \right)&=\left\{ \frac{1}{n}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }g\left( {\varvec{x}}_{i}\right) g^{\mathrm{T}}\left( {\varvec{x}}_{i}\right) \right] \right\} ^{-1}\nonumber \\&\quad \times \left\{ \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }g\left( {\varvec{x}}_{i}\right) \left( y_{i}-g^{\mathrm{T}}\left( {\varvec{x}}_{i}\right) \beta \right) \right] \right\} :=A^{-1}B. \end{aligned}$$
(23)
Now, we can write A as follow:
$$\begin{aligned} A&=\frac{1}{n}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }g\left( {\varvec{x}}_{i}\right) g^{\mathrm{T}}\left( \varvec{{\varvec{x}}}_{i}\right) \right] = \frac{1}{n}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) g^{\mathrm{T}}\left( {\varvec{x}}_{i}\right) \right] \\&\quad +\frac{1}{n}\sum _{i=1}^{n}\left[ \left( \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }-\frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},\gamma \right) }\right) g\left( {\varvec{x}}_{i}\right) g^{\mathrm{T}}\left( {\varvec{x}}_{i}\right) \right] :=A_1+A_2. \end{aligned}$$
For enough large sample size, by \({\hat{\pi }}({\varvec{v}}_i,\gamma )=\pi ({\varvec{v}}_i,\gamma )+o_P(1)\) and regularity condition C1, we can write \(A_1\) as follow:
$$\begin{aligned} A_1=\frac{1}{n}\sum _{i=1}^{n}\frac{\delta _{i}}{\pi ({\varvec{v}}_{i},\gamma )}g({\varvec{x}}_{i})g^{\mathrm{T}}({\varvec{x}}_{i})+o_P(1)=\Sigma +o_P(1). \end{aligned}$$
(24)
\({\hat{\gamma }}\) is a consistent estimator of \(\gamma\). On the other hand, for big n, it is not difficult to prove that \({\hat{\pi }}({\varvec{v}}_i,\gamma )-{\hat{\pi }}({\varvec{v}}_i,{\hat{\gamma }})=o_P(1)\). Therefore,
$$\begin{aligned} A_2=o_P(1). \end{aligned}$$
(25)
Consequently, from (23) and (24), we will have:
$$\begin{aligned} A=A_1+A_2=\Sigma +o_P(1). \end{aligned}$$
(26)
For B, we cn write:
$$\begin{aligned} B&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }g\left( {\varvec{x}}_{i}\right) \left( y_{i}-g^{\mathrm{T}}\left( {\varvec{x}}_{i}\right) \beta \right) \right] \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }g\left( {\varvec{x}}_{i}\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \\&\quad +\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \left( \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }-\frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},\gamma \right) }\right) g\left( {\varvec{x}}_{i}\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \\&:=B_1+B_2. \end{aligned}$$
Now, we rewrite \(B_1\) as follow:
$$\begin{aligned} B_1&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \\&\quad +\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \left( \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},\gamma \right) }-\frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }\right) g\left( {\varvec{x}}_{i}\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \\&:=B_{11}+B_{12}. \end{aligned}$$
We keep \(B_{11}\) without change. For \(B_{12}\) by replacing \({\hat{\pi }}(\cdot )\) and \(\pi (\cdot )\) with their equivalent formula from Eq. (10), we will have:
$$\begin{aligned} B_{12}&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}[\delta _i({\hat{\alpha }}({\varvec{x}}_i,\gamma )-\alpha ({\varvec{x}}_i,\gamma ))g({\varvec{x}}_{i})(\eta _i+C_nG({\varvec{x}}_i))] \nonumber \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}[\delta _i({\hat{\alpha }}({\varvec{x}}_i,\gamma ) \nonumber \\&\quad -\left. \alpha ({\varvec{x}}_i,\gamma ))g({\varvec{x}}_{i})(\eta _i+C_nG({\varvec{x}}_i))\frac{\sum _{j=1}^{n}[(1-\delta _j)-\delta _j{\mathrm{e}}^{\gamma y_{j}}\alpha ({\varvec{x}}_i,\gamma )K_h({\varvec{x}}_i-{\varvec{x}}_j)]}{\sum _{j=1}^{n}[\delta _j{\mathrm{e}}^{\gamma y_{j}}K_h({\varvec{x}}_i-{\varvec{x}}_j)]}\right] , \end{aligned}$$
(27)
where, the last equality follows from the empirical estimator of \({\hat{\alpha }}(\cdot )\) [Eq. (9)] and some simple mathematical operations. For denominator of Eq. (27), we have:
$$\begin{aligned} \frac{1}{n}\sum _{j=1}^{n}\left[ \delta _j{\mathrm{e}}^{\gamma y_j}K_h({\varvec{x}}_i-{\varvec{x}}_j)\right] =\frac{1}{n}\sum _{j=1}^{n}\left[ \delta _j\left( \frac{1}{\pi ({\varvec{v}}_i,\gamma )}-1\right) \alpha ^{-1}({\varvec{x}}_j,\gamma )K_h({\varvec{x}}_i-{\varvec{x}}_j)\right] . \end{aligned}$$
(28)
By the SLN (The expectation has been taken conditionally on \({\varvec{x}}_i\)) and by using some mathematical operations, it is not difficult to prove that
$$\begin{aligned} \frac{1}{n}\sum _{j=1}^{n}\left[ \delta _j\left( \frac{1}{\pi ({\varvec{v}}_i,\gamma )}-1\right) \alpha ^{-1}({\varvec{x}}_j,\gamma )K_h({\varvec{x}}_i-{\varvec{x}}_j)\right] =\alpha ^{-1}({\varvec{x}}_i,\gamma )f({\varvec{x}}_i)E(1-\delta |{\varvec{x}}_i)+o_P(1), \end{aligned}$$
(29)
where \(f({\varvec{x}}_i)\) is the density function of random variable \({\varvec{X}}_i\). Now, by replacing the above statement in Eq. (27), we can write \(B_{12}\) as follow:
$$\begin{aligned} B_{12}&=\frac{1}{n^{\frac{3}{2}}}\sum _{j=1}^{n}\sum _{i=1}^{n}\left[ \frac{\delta _i{\mathrm{e}}^{\gamma y_i}g({\varvec{x}}_i)(\eta _i+C_nG({\varvec{x}}_i))}{\alpha ^{-1}({\varvec{x}}_i,\gamma )f({\varvec{x}}_i)E(1-\delta |{\varvec{x}}_i)}((1-\delta _j)-\delta _j{\mathrm{e}}^{\gamma y_j}\alpha ({\varvec{x}}_i,\gamma )K_h({\varvec{x}}_j-{\varvec{x}}_i))\right] \nonumber \\&\quad +o_P(1)\nonumber \\&=\frac{1}{n^{\frac{3}{2}}}\sum _{j=1}^{n}\sum _{i=1}^{n}\left[ \frac{\delta _iO({\varvec{v}}_i)g({\varvec{x}}_i)(\eta _i+C_nG({\varvec{x}}_i))}{f({\varvec{x}}_i)E(1-\delta |{\varvec{x}}_i)}((1-\delta _j)-\delta _j{\mathrm{e}}^{\gamma y_j}\alpha ({\varvec{x}}_i,\gamma )K_h({\varvec{x}}_j-{\varvec{x}}_i))\right] \nonumber \\&\quad +o_P(1). \end{aligned}$$
(30)
By using SLN for inner summation and by using some mathematical operation, one can conclude that:
$$\begin{aligned} B_{12}=\frac{1}{\sqrt{n}}\sum _{j=1}^{n}\left[ \left( 1-\frac{\delta _j}{\pi ({\varvec{v}}_j,\gamma )}\right) \frac{E(g(X)(\eta +C_nG(X))(1-\delta )|{\varvec{x}}_i)}{E(1-\delta |{\varvec{x}}_i)}\right] +o_P(1). \end{aligned}$$
(31)
Based on the exponential tilting property of Eq. (5), we can write:
$$\begin{aligned} E(g({\varvec{X}})(\eta +C_nG({\varvec{X}}))|{\varvec{x}}_i,\delta =0)&=\frac{E(g({\varvec{X}})(\eta +C_nG({\varvec{X}})){\mathrm{e}}^{\gamma Y}|{\varvec{x}}_i,\delta =1)}{E({\mathrm{e}}^{\gamma Y}|{\varvec{x}}_i,\delta =1)}\nonumber \\&=\frac{E(g({\varvec{X}})(\eta +C_nG({\varvec{X}}))(1-\delta )|{\varvec{x}}_i)}{E((1-\delta )|{\varvec{x}}_i)}. \end{aligned}$$
(32)
Therefore,
$$\begin{aligned} E(g({\varvec{X}})(\eta +C_nG({\varvec{X}}))(1-\delta )|{\varvec{x}}_i)=E((1-\delta )|{\varvec{x}}_i)E(g({\varvec{X}})(\eta +C_nG({\varvec{X}}))|{\varvec{x}}_i,\delta =0). \end{aligned}$$
(33)
Now, by inserting Eq. (33) in Eq. (31), we can conclude that:
$$\begin{aligned} B_{12}=\frac{1}{\sqrt{n}}\sum _{j=1}^{n}\left[ \left( 1-\frac{\delta _j}{\pi ({\varvec{v}}_j,\gamma )}\right) E(g({\varvec{X}})(\eta +C_nG({\varvec{X}}))|{\varvec{x}}_i,\delta =0)\right] +o_P(1). \end{aligned}$$
(34)
For enough big sample size, it is easy to prove that \(E(B_{12})=o_P(1)\) and \(Var(B_{12})=o_P(1)\). Consequently, by Chebyshev’s inequality,
$$\begin{aligned} B_{12}=o_p(1). \end{aligned}$$
(35)
Therefore,
$$\begin{aligned} B_1&=B_{11}+B_{12}=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] +o_P\left( 1\right) \nonumber \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \eta _i\right] +\sqrt{n}C_nE\left( g\left( {\varvec{X}}\right) G\left( {\varvec{X}}\right) \right) +o_P\left( 1\right) . \end{aligned}$$
(36)
For \(B_2\) we have,
$$\begin{aligned} B_2&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \left( \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\right) g\left( {\varvec{x}}_i\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \nonumber \\&=\frac{1}{n}\sum _{i=1}^{n}\left[ \delta _i g\left( {\varvec{x}}_i\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \frac{\partial }{\partial \gamma }\pi ^{-1}\left( {\varvec{x}}_i,\gamma \right) \right] \sqrt{n}\left( {\hat{\gamma }}-\gamma \right) +o_P\left( 1\right) \nonumber \\&=\frac{1}{n}\sum _{i=1}^{n}\left[ \delta _ig\left( {\varvec{x}}_i\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) {\hat{\alpha }}\left( {\varvec{x}}_i,\gamma \right) {\mathrm{e}}^{\gamma y_i}\left( y_i-{\hat{m}}\left( {\varvec{x}}_i,\gamma \right) \right) \right] \sqrt{n}\left( {\hat{\gamma }}-\gamma \right) +o_P\left( 1\right) , \end{aligned}$$
(37)
where the middle equation follows from the mean value theorem and the last equation is concluded by inserting the equivalent function of the derivative function. By \({\hat{m}}({\varvec{x}}_i,\gamma )=m({\varvec{x}}_{i},\gamma )+o_P(1)\) and \({\hat{\alpha }}({\varvec{x}}_i,\gamma ){\mathrm{e}}^{\gamma y_i}=(\frac{1}{{\hat{\pi }}({\varvec{v}}_i,\gamma )}-1)\), we can rewrite \(B_2\) as follow:
$$\begin{aligned} B_2= & {} \frac{1}{n}\sum _{i=1}^{n}\left[ \delta _i\left( \frac{1}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }-1\right) g\left( {\varvec{x}}_i\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \left( y_i-m\left( {\varvec{x}}_{i},\gamma \right) \right) \right] \sqrt{n}\left( {\hat{\gamma }}-\gamma \right) \nonumber \\&+o_P\left( 1\right) , \end{aligned}$$
(38)
Therefore, by the SLN we will have:
$$\begin{aligned} B_2&=E\left( \left( 1-\pi \left( {\varvec{V}},\gamma \right) \right) g\left( {\varvec{X}}\right) \left( \eta -\eta ^*\right) ^2\right) \sqrt{n}\left( {\hat{\gamma }}-\gamma \right) +o_P\left( 1\right) \nonumber \\&=H\sqrt{n}\left( {\hat{\gamma }}-\gamma \right) +o_P\left( 1\right) . \end{aligned}$$
(39)
Now, by using Eqs. (36) and (39), we can write,
$$\begin{aligned} B=B_1+B_2=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi ({\varvec{v}}_{i},\gamma )}g({\varvec{x}}_{i})\eta _i\right] +H\sqrt{n}({\hat{\gamma }}-\gamma )+o_P(1). \end{aligned}$$
(40)
By Lemma 1, we can write the above equation as follow:
$$\begin{aligned} B=B_1+B_2&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \eta _i\right] +\sqrt{n}C_nE\left( g\left( {\varvec{X}}\right) G\left( {\varvec{X}}\right) \right) \nonumber \\&\quad +HM^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ r_i\left( 1-\delta _i\right) \right. \nonumber \\&\quad \left. -\delta _iE\left( r|\delta =0\right) \left( \frac{1}{\pi \left( {\varvec{v}}_i,\gamma \right) }-1\right) \left( y_i-m\left( {\varvec{x}}_{i},\gamma \right) \right) \right] +o_P\left( 1\right) . \end{aligned}$$
(41)
Finally, by replacing Eqs. (26) and (41) in Eq. (23), part (c) of Lemma 2 can be concluded. Also, the part (a) of Lemma 2 obtains by taking \(C_n=0\) in part (c).
1.2 Proof in augmented case
Proof of part (d) of Lemma 2 is very similar to the third part. For this reason, we just review the proof part of (c) of Lemma 2. If the parameters of the general linear model are estimated by the Augmented method, we will have,
$$\begin{aligned} \sqrt{n}\left( {\hat{\beta }}_{{\mathrm{A}}}-\beta \right)&=\left\{ \frac{1}{n}\sum _{i=1}^{n}\left[ g\left( {\varvec{x}}_{i}\right) g^{\mathrm{T}}\left( {\varvec{x}}_{i}\right) \right] \right\} ^{-1}\left\{ \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( v_{i},{\hat{\gamma }}\right) }g\left( {\varvec{x}}_{i}\right) \left( y_{i}-g^{\mathrm{T}}\left( {\varvec{x}}_{i}\right) \beta \right) \right. \right. \nonumber \\&\quad \left. \left. +\left( 1-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\right) g\left( {\varvec{x}}_i\right) \left( {\hat{m}}\left( {\varvec{x}}_i,\gamma \right) -g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \beta \right) \right] \right\} :=C^{-1}D. \end{aligned}$$
(42)
By SLN, C will be,
$$\begin{aligned} C=\Sigma +o_P\left( 1\right) . \end{aligned}$$
(43)
For D, we can write:
$$\begin{aligned} D&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }g\left( {\varvec{x}}_{i}\right) \left( y_{i}-g^{\mathrm{T}}\left( {\varvec{x}}_{i}\right) \beta \right) \right. \\&\quad \left. +\left( 1-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\right) g\left( {\varvec{x}}_i\right) \left( {\hat{m}}\left( {\varvec{x}}_i,\gamma \right) -g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \beta \right) \right] \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},{\hat{\gamma }}\right) }g\left( {\varvec{x}}_{i}\right) \eta _i+\left( 1-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\right) g\left( {\varvec{x}}_i\right) \eta _{i}^{*}+C_nG\left( {\varvec{x}}_i\right) g\left( {\varvec{x}}_i\right) \right] +o_P\left( 1\right) \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \eta _i+\left( 1-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\right) g\left( {\varvec{x}}_i\right) \eta _{i}^{*}+C_nG^{\mathrm{T}}\left( {\varvec{x}}_i\right) g\left( {\varvec{x}}_i\right) \right] \\&\quad +\sum _{i=1}^{n}\left[ \left( \frac{\delta _{i}}{{\hat{\pi }}({\varvec{x}}_{i},\hat{\gamma )}}-\frac{\delta _{i}}{{\hat{\pi }}\left( {\varvec{x}}_{i},\gamma \right) }\right) \left( y_i-m\left( {\varvec{x}}_{i},\gamma \right) \right) \right] +o_P\left( 1\right) \\&:=D_1+D_2+o_P\left( 1\right) , \end{aligned}$$
where the second equality follows from \({\hat{m}}(\cdot )=m(\cdot )+o_P(1)\) and \(\eta ^{*}=E(\eta |{\varvec{x}})\). In a same way to obtaining \(B_1\), we can conclude that
$$\begin{aligned} D_1=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi ({\varvec{x}}_{i},\gamma )}g({\varvec{x}}_{i})\eta _i+\left( 1-\frac{\delta _i}{\pi ({\varvec{v}}_i,\gamma )}\right) g({\varvec{x}}_i)\eta _{i}^{*}\right] +\sqrt{n}C_nE(g({\varvec{X}})G({\varvec{X}}))+o_P(1) \end{aligned}$$
(44)
and in a similar way to obtaining \(B_2\), we can write
$$\begin{aligned} D_2&=\frac{1}{n}\sum _{i=1}^{n}\left[ \delta _i\left( \frac{1}{{\hat{\pi }}({\varvec{v}}_i,\gamma )}-1\right) g({\varvec{x}}_i)(y_i-m({\varvec{x}}_{i},\gamma ))(y_i-m({\varvec{x}}_{i},\gamma ))\right] \sqrt{n}({\hat{\gamma }}-\gamma )+o_P(1)\nonumber \\&=\frac{1}{n}\sum _{i=1}^{n}\left[ \delta _i\left( \frac{1}{{\hat{\pi }}({\varvec{v}}_i,\gamma )}-1\right) g({\varvec{x}}_i)(\eta _i-\eta _{i}^{*})^2\right] \sqrt{n}({\hat{\gamma }}-\gamma )+o_P(1). \end{aligned}$$
(45)
Therefore, by SLN we will have:
$$\begin{aligned} D_2=H\sqrt{n}({\hat{\gamma }}-\gamma )+o_P(1). \end{aligned}$$
(46)
Now, by using Eqs. (44) and (46), we can write,
$$\begin{aligned} D= & {} \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi ({\varvec{v}}_{i},\gamma )}g({\varvec{x}}_{i})\eta _i+\left( 1-\frac{\delta _{i}}{\pi ({\varvec{v}}_{i},\gamma )}\right) g({\varvec{x}}_{i})\eta _{i}^{*}\right] +H\sqrt{n}({\hat{\gamma }}-\gamma ) \nonumber \\&+\sqrt{n}C_nE(g({\varvec{X}})G({\varvec{X}}))+o_P(1). \end{aligned}$$
(47)
Moreover, by using Lemma 1, we can conclude:
$$\begin{aligned} D&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi ({\varvec{v}}_{i},\gamma )}g({\varvec{x}}_{i})\eta _i+\left( 1-\frac{\delta _{i}}{\pi ({\varvec{v}}_{i},\gamma )}\right) g({\varvec{x}}_{i})\eta _{i}^{*}\right] \nonumber \\&\quad +HM^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ r_i(1-\delta _i)-\delta _iE(r|\delta =0)\left( \frac{1}{\pi ({\varvec{v}}_i,\gamma )}-1\right) (y_i-m({\varvec{x}}_{i},\gamma ))\right] \nonumber \\&\quad +\sqrt{n}C_nE(g({\varvec{X}})G({\varvec{X}}))+o_P(1) \end{aligned}$$
(48)
Finally, by replacing Eqs. (43) and (48) in Eq. (42), part (d) of Lemma 2 can be concluded. The part (b) of Lemma 2 comes from taking \(C_n=0\) in part (c).
Proof of Theorem 1
We just prove Theorem 1 under the alternative hypothesis, because the null hypothesis follows by taking \(C_n=0\).
1.1 Empirical properties of test function based on the IPW method
We can rewrite \(T_{{\mathrm{IPW}}}\) as follow:
$$\begin{aligned} T_{{\mathrm{IPW}}}&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\left( y_i-g^{\mathrm{T}}\left( {\varvec{x}}_i\right) {\hat{\beta }}_{{\mathrm{IPW}}}\right) \right] \nonumber \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \nonumber \\&\quad -\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \left( {\hat{\beta }}_{{\mathrm{IPW}}}-\beta \right) \right] \nonumber \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \nonumber \\&\quad -\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \left( {\hat{\beta }}_{{\mathrm{IPW}}}-\beta \right) \right] \nonumber \\&\quad +\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \left( \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\right) \left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] \nonumber \\&\quad -\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \left( \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\right) g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \left( {\hat{\beta }}_{{\mathrm{IPW}}}-\beta \right) \right] \nonumber \\&:=T_{{\mathrm{IPW}},1}+T_{{\mathrm{IPW}},2}+T_{{\mathrm{IPW}},3}+T_{{\mathrm{IPW}},4}. \end{aligned}$$
(49)
In a similar way to obtaining \(B_1\), we can prove that
$$\begin{aligned} T_{{\mathrm{IPW}},1}&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\left( \eta _i+C_nG\left( {\varvec{x}}_i\right) \right) \right] +o_P\left( 1\right) \nonumber \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\eta _i\right] +\sqrt{n}C_nE\left( G\left( {\varvec{X}}\right) \right) +o_P\left( 1\right) , \end{aligned}$$
(50)
and in a similar way to obtaining \(B_{12}\) we can prove that \(T_3=o_P(1)\) and \(T_{{\mathrm{IPW}},4}=o_P(1)\). Also, for \(T_{{\mathrm{IPW}},2}\), we have:
$$\begin{aligned} T_{{\mathrm{IPW}},2}&=-\frac{1}{n}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \right] \sqrt{n}\left( {\hat{\beta }}_{{\mathrm{IPW}}}-\beta \right) \nonumber \\&=-\frac{1}{n}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \right] \sqrt{n}\left( {\hat{\beta }}_{{\mathrm{IPW}}}-\beta \right) \nonumber \\&\quad +\frac{1}{n}\sum _{i=1}^{n}\left[ \left( \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\right) g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \right] \sqrt{n}\left( {\hat{\beta }}_{{\mathrm{IPW}}}-\beta \right) \nonumber \\&=-\frac{1}{n}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \right] \sqrt{n}\left( {\hat{\beta }}_{{\mathrm{IPW}}}-\beta \right) +o_P\left( 1\right) , \end{aligned}$$
(51)
Last equation follows from the fact that \({\hat{\pi }}(\cdot )=\pi (\cdot )+o_P(1)\) and \(E(g(\cdot ))<\infty\). By using the third part of Lemma 2:
$$\begin{aligned} T_{{\mathrm{IPW}},2}&=-\frac{1}{n}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \right] \Sigma ^{-1}\left\{ \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \eta _{i}\right] \right. \nonumber \\&\quad \left. +HM^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ r_i\left( 1-\delta _i\right) -\delta _iE\left( r|\delta =0\right) \left( \frac{1}{\pi \left( {\varvec{v}}_i,\gamma \right) }-1\right) \left( y_i-m\left( {\varvec{x}}_{i},\gamma \right) \right) \right] \right\} \nonumber \\&\quad -\sqrt{n}C_nE\left( g^{\mathrm{T}}\left( X\right) \right) \Sigma ^{-1}E\left( g\left( {\varvec{X}}\right) G\left( {\varvec{X}}\right) \right) +o_{p}\left( 1\right) . \end{aligned}$$
(52)
Therefore, we will have:
$$\begin{aligned} T_{{\mathrm{IPW}}}&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\eta _i\nonumber \\&\quad -\frac{1}{n}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \right] \Sigma ^{-1}\left\{ \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \eta _{i}\right] \right. \nonumber \\&\quad \left. +HM^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ r_i\left( 1-\delta _i\right) -\delta _iE\left( r|\delta =0\right) \left( \frac{1}{\pi \left( {\varvec{v}}_i,\gamma \right) }-1\right) \left( y_i-m\left( {\varvec{x}}_{i},\gamma \right) \right) \right] \right\} \nonumber \\&\quad +\sqrt{n}C_nE\left( G\left( {\varvec{X}}\right) \right) -\sqrt{n}C_nE\left( g^{\mathrm{T}}\left( {\varvec{X}}\right) \right) \Sigma ^{-1}E\left( g\left( {\varvec{X}}\right) G\left( {\varvec{X}}\right) \right) +o_{p}\left( 1\right) . \end{aligned}$$
(53)
Finally, the empirical properties of the test function based on the IPW method are obtained by using the above equation and by applying CLT.
1.2 Empirical properties of test the function based on augmented method
We can rewrite \(T_A\) as follow:
$$\begin{aligned} T_{{\mathrm{A}}}&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\left( y_i-g^{\mathrm{T}}\left( {\varvec{x}}_i\right) {\hat{\beta }}_{{\mathrm{A}}}\right) +\left( 1-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\right) \left( {\hat{m}}\left( {\varvec{x}}_i,\gamma \right) -g^{\mathrm{T}}\left( {\varvec{x}}_i\right) {\hat{\beta }}_{{\mathrm{A}}}\right) \right] \nonumber \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\eta _i+\left( 1-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }\right) \eta _{i}^{*}+C_nG\left( {\varvec{x}}_i\right) \right] \nonumber \\&\quad -\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \left( {\hat{\beta }}_{{\mathrm{A}}}-\beta \right) \right] +o_P\left( 1\right) \nonumber \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\eta _i+\left( 1-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\right) \eta _{i}^{*}+C_nG\left( {\varvec{x}}_i\right) \right] \nonumber \\&\quad -\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \left( {\hat{\beta }}_{{\mathrm{A}}}-\beta \right) \right] \nonumber \\&\quad +\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \left( \frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,{\hat{\gamma }}\right) }-\frac{\delta _i}{{\hat{\pi }}\left( {\varvec{v}}_i,\gamma \right) }\right) \left( y_i-m\left( {\varvec{x}}_{i},\gamma \right) \right) \right] +o_P\left( 1\right) \nonumber \\ :&=T_{A,1}+T_{A,2}+T_{A,3}. \end{aligned}$$
(54)
In similar way to obtaining \(B_1\), we can prove that
$$\begin{aligned} T_{A,1}&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\eta _i+\left( 1-\frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\right) \eta _{i}^{*}+C_nG\left( {\varvec{x}}_i\right) \right] +o_P\left( 1\right) \nonumber \\&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\eta _i+\left( 1-\frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\right) \eta _{i}^{*}\right] +\sqrt{n}C_nE\left( G\left( {\varvec{X}}\right) \right) +o_P\left( 1\right) , \end{aligned}$$
(55)
and in a similar way to obtaining \(B_{12}\), we can prove that \(T_{A,3}=o_P\left( 1\right)\). Also, for \(T_{A,2}\) based on SLN and the forth part of Lemma 2, we will have:
$$\begin{aligned} T_{A,2}&=-\frac{1}{n}\sum _{i=1}^{n}\left[ g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \right] \Sigma ^{-1}\left\{ \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \eta _{i}\right] \right. \nonumber \\&\quad \left. +HM^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ r_i\left( 1-\delta _i\right) -\delta _iE\left( r|\delta =0\right) \left( \frac{1}{\pi \left( {\varvec{v}}_i,\gamma \right) }-1\right) \left( y_i-m\left( {\varvec{x}}_{i},\gamma \right) \right) \right] \right\} \nonumber \\&\quad -\sqrt{n}C_nE\left( g^{\mathrm{T}}\left( {\varvec{X}}\right) \right) \Sigma ^{-1}E\left( g\left( {\varvec{X}}\right) G\left( {\varvec{X}}\right) \right) +o_{p}\left( 1\right) . \end{aligned}$$
(56)
Consequently,
$$\begin{aligned} T_{{\mathrm{A}}}&=\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\eta _i+\left( 1-\frac{\delta _i}{\pi \left( {\varvec{v}}_i,\gamma \right) }\right) \eta _{i}^{*}\right] \nonumber \\&\quad -\frac{1}{n}\sum _{i=1}^{n}\left[ g^{\mathrm{T}}\left( {\varvec{x}}_i\right) \right] \Sigma ^{-1}\left\{ \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ \frac{\delta _{i}}{\pi \left( {\varvec{v}}_{i},\gamma \right) }g\left( {\varvec{x}}_{i}\right) \eta _{i}+\left( 1-\frac{1}{\pi \left( {\varvec{v}}_i,\gamma \right) }\right) \eta _{i}^{*}\right] \right. \end{aligned}$$
(57)
$$\begin{aligned}&\quad \left. +HM^{-1}\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\left[ r_i\left( 1-\delta _i\right) -\delta _iE\left( r|\delta =0\right) \left( \frac{1}{\pi \left( {\varvec{v}}_i,\gamma \right) }-1\right) \left( y_i-m\left( {\varvec{x}}_{i},\gamma \right) \right) \right] \right\} \nonumber \\&\quad +\sqrt{n}C_nE\left( G\left( {\varvec{X}}\right) \right) -\sqrt{n}C_nE\left( g^{\mathrm{T}}\left( {\varvec{X}}\right) \right) \Sigma ^{-1}E\left( g\left( {\varvec{X}}\right) G\left( {\varvec{X}}\right) \right) +o_{p}\left( 1\right) . \end{aligned}$$
(58)
Finally, the part (d) of Theorem 1 follows by the above equation and by applying CLT. Also, the second part of Theorem 1 can be obtained by taking \(C_n=0\) in the last stabilization.
