$\displaystyle\hat{\Sigma}_{2}=$
$\displaystyle\frac{1}{K}\sum_{k=1}^{K}\mathbb{E}_{n,k}\left[\left(\psi_{2}\left(W,\tilde{\theta},\hat{p}_{k},\hat{\eta}_{1k}\right)+\hat{G}_{2p}\left(D-\hat{p}_{k}\right)+\hat{G}_{2\lambda}\left(T-\hat{\lambda}_{k}\right)\right)^{2}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{K}\sum_{k=1}^{K}\mathbb{E}_{n,k}\left[\left(\frac{1}{\hat{\lambda}_{k}\left(1-\hat{\lambda}_{k}\right)\hat{p}_{k}}\frac{D-\hat{g}_{k}\left(X\right)}{1-\hat{g}_{k}\left(X\right)}\left(\left(T-\hat{\lambda}_{k}\right)Y-\hat{\ell}_{2k}\left(X\right)\right)-\frac{D\tilde{\theta}}{\hat{p}_{k}}+\hat{G}_{2\lambda}\left(T-\hat{\lambda}_{k}\right)\right)^{2}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{K}\sum_{k=1}^{K}\mathbb{E}_{n,k}\left[\bar{\psi}_{2}\left(W,\tilde{\theta},\hat{p}_{k},\hat{\lambda}_{k},\hat{G}_{2\lambda},\hat{\eta}_{2k}\right)^{2}\right],$

where the second inequality follows from
$\hat{G}_{2p}=-\tilde{\theta}/\hat{p}_{k}$ .

Since $K$ is fixed, which is independent of $N$, it suffices to show that

$J_{k}\coloneqq\mid\mathbb{E}_{n,k}\left[\bar{\psi}_{2}\left(W,\tilde{\theta},\hat{p}_{k},\hat{\lambda}_{k},\hat{G}_{2\lambda},\hat{\eta}_{2k}\right)^{2}\right]-E_{P}\left[\bar{\psi}_{2}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)^{2}\right]\mid=o_{P}\left(1\right).$

By the triangle inequality, we have

$J_{k}\leq J_{5,k}+J_{6,k},$

where

$J_{5,k}\coloneqq\mid\mathbb{E}_{n,k}\left[\bar{\psi}_{2}\left(W,\tilde{\theta},\hat{p}_{k},\hat{\lambda}_{k},\hat{G}_{2\lambda},\hat{\eta}_{2k}\right)^{2}\right]-\mathbb{E}_{n,k}\left[\bar{\psi}_{2}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)^{2}\right]\mid,$

$J_{6,k}\coloneqq\mid\mathbb{E}_{n,k}\left[\bar{\psi}_{2}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)^{2}\right]-E_{P}\left[\bar{\psi}_{2}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)^{2}\right]\mid.$

By the same arguments for $I_{4,k}$ in the proof of repeated outcomes and
(A.15), we can show $J_{6,k}=o_{P}\left(1\right)$. Also, by the same arguments
for $I_{3,k}$ in the proof of repeated outcomes, we have

$J_{5,k}^{2}\lesssim S_{N}^{\prime}\times\left(\frac{1}{n}\sum_{i\in
I_{k}}\parallel\bar{\psi_{2}}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)\parallel^{2}+S_{N}^{\prime}\right),$

where

$S_{N}^{\prime}\coloneqq\frac{1}{n}\sum_{i\in
I_{k}}\parallel\bar{\psi_{2}}\left(W,\tilde{\theta},\hat{p}_{k},\hat{\lambda}_{k},\hat{G}_{2\lambda},\hat{\eta}_{2k}\right)-\bar{\psi_{2}}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)\parallel^{2}.$

Since $\frac{1}{n}\sum_{i\in
I_{k}}\parallel\bar{\psi_{2}}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)\parallel^{2}=O_{P}\left(1\right)$, it remains to bound
$S_{N}^{\prime}$. Define
$\bar{\psi}_{20}\coloneqq\bar{\psi_{2}}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)$. By the triangle inequality, we have

$\displaystyle S_{N}^{\prime}\leq$ $\displaystyle\frac{1}{n}\sum_{i\in
I_{k}}\parallel\bar{\psi}_{2}\left(W_{i},\theta_{0},\hat{p}_{k},\hat{\lambda}_{k},G_{2\lambda
0},\hat{\eta}_{2k}\right)-\bar{\psi}_{20}\parallel^{2}$
$\displaystyle+\frac{1}{n}\sum_{i\in
I_{k}}\parallel\partial_{\theta}\bar{\psi}_{2}\left(W_{i},\bar{\theta},\hat{p}_{k},\hat{\lambda}_{k},\hat{G}_{2\lambda},\hat{\eta}_{2k}\right)\left(\tilde{\theta}-\theta_{0}\right)\parallel^{2}$
$\displaystyle+\frac{1}{n}\sum_{i\in
I_{k}}\parallel\partial_{G_{2\lambda}}\bar{\psi}_{2}\left(W_{i},\theta_{0},\hat{p}_{k},\hat{\lambda}_{k},\overline{G_{2\lambda}},\hat{\eta}_{2k}\right)\left(\hat{G}_{2\lambda}-G_{2\lambda
0}\right)\parallel^{2},$

where $\bar{\theta}\in\left(\tilde{\theta},\theta_{0}\right)$ and
$\overline{G_{2\lambda}}\in\left(\widehat{G_{2\lambda}},G_{2\lambda
0}\right)$. Then we have

$\displaystyle S_{N}^{\prime}\leq$ $\displaystyle\frac{1}{n}\sum_{i\in
I_{k}}\parallel\bar{\psi}_{2}\left(W_{i},\theta_{0},\hat{p}_{k},\hat{\lambda}_{k},G_{2\lambda
0},\hat{\eta}_{2k}\right)-\bar{\psi}_{20}\parallel^{2}$
$\displaystyle+\frac{1}{n}\sum_{i\in
I_{k}}\parallel\frac{D_{i}}{\hat{p}_{k}}\left(\tilde{\theta}-\theta_{0}\right)\parallel^{2}+\frac{1}{n}\sum_{i\in
I_{k}}\parallel\left(T_{i}-\hat{\lambda}_{k}\right)\left(\hat{G}_{2\lambda}-G_{2\lambda
0}\right)\parallel^{2}.$

The two terms in the last line are bounded by

$\left(\frac{1}{n}\sum_{i\in
I_{k}}\left(\frac{D_{i}}{\hat{p}_{k}}\right)^{2}\right)\times\parallel\tilde{\theta}-\theta_{0}\parallel^{2}=O_{P}\left(1\right)\times
O_{P}\left(N^{-1}\right)$

and

$\left(\frac{1}{n}\sum_{i\in
I_{k}}\left(T_{i}-\hat{\lambda}_{k}\right)^{2}\right)\times\parallel\hat{G}_{2\lambda}-G_{2\lambda
0}\parallel^{2}=O_{P}\left(1\right)\times\left(o_{P}\left(1\right)\right)^{2}.$

Conditional on the auxiliary sample $I_{k}^{c}$,
$\left(\hat{p}_{k},\hat{\lambda}_{k},\hat{\eta}_{2k}\right)$ can be treated as
fixed. Also, under the event that $\hat{p}_{k}\in\mathcal{P}_{N}$,
$\hat{\lambda}_{k}\in\Lambda_{N}$, and $\hat{\eta}_{2k}\in\mathcal{T}_{N}$, we
have

$\displaystyle
E_{P}\left[\parallel\bar{\psi}_{2}\left(W_{i},\theta_{0},\hat{p}_{k},\hat{\lambda}_{k},G_{2\lambda
0},\hat{\eta}_{2k}\right)-\bar{\psi}_{2}\left(W_{i},\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)\parallel^{2}\mid\left(W_{i}\right)_{i\in
I_{k}^{c}}\right]$
$\displaystyle\leq\sup_{p\in\mathcal{P}_{N},\lambda\in\Lambda_{N},\eta_{2}\in\mathcal{T}_{N}}E_{P}\left[\parallel\bar{\psi}_{2}\left(W_{i},\theta_{0},p,\lambda,G_{2\lambda
0},\eta_{2}\right)-\bar{\psi}_{2}\left(W_{i},\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)\parallel^{2}\right]$
$\displaystyle=\left(\varepsilon_{N}\right)^{2}$

by (A.14). It follows that
$S_{N}^{\prime}=O_{P}\left(N^{-1}+\varepsilon_{N}^{2}\right)+\left(o_{P}\left(1\right)\right)^{2}$
so that

$J_{k}=o_{P}\left(1\right)+O_{P}\left(N^{-1/2}+\varepsilon_{N}\right)+o_{P}\left(1\right)=o_{P}\left(1\right).$

_Step 2._ It remains to show (A.14) and (A.15). Define
$\bar{\psi}_{20}\coloneqq\bar{\psi}_{2}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)$. By the triangle inequality and

$\bar{\psi}_{2}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{2}\right)-\bar{\psi}_{20}=\psi_{2}\left(W,\theta_{0},p_{0},\lambda_{0},\eta_{2}\right)-\psi_{2}\left(W,\theta_{0},p_{0},\lambda_{0},\eta_{20}\right),$

we have

$\displaystyle\parallel\bar{\psi}_{2}\left(W,\theta_{0},p,\lambda,G_{2\lambda
0},\eta_{2}\right)-\bar{\psi}_{20}\parallel_{P,2}\leq$
$\displaystyle\parallel\psi_{2}\left(W,\theta_{0},p_{0},\lambda_{0},\eta_{2}\right)-\psi_{2}\left(W,\theta_{0},p_{0},\lambda_{0},\eta_{20}\right)\parallel_{P,2}$
$\displaystyle+\parallel\partial_{\lambda}\bar{\psi}_{2}\left(W_{i},\theta_{0},p_{0},\bar{\lambda},G_{2\lambda
0},\eta_{2}\right)\parallel_{P,2}\mid\lambda-\lambda_{0}\mid$
$\displaystyle+\parallel\partial_{p}\bar{\psi}_{2}\left(W_{i},\theta_{0},\bar{p},\lambda,G_{2\lambda
0},\eta_{2}\right)\parallel_{P,2}\mid p-p_{0}\mid,$

where $\bar{p}\in\left(p,p_{0}\right)$ and
$\bar{\lambda}\in\left(\lambda,\lambda_{0}\right)$. The term in the second
line is bounded by

$\displaystyle\parallel\partial_{\lambda}\bar{\psi}_{2}\left(W_{i},\theta_{0},p_{0},\bar{\lambda},G_{2\lambda
0},\eta_{2}\right)\parallel_{P,2}\leq$ $\displaystyle\frac{\mid
1-2\bar{\lambda}\mid}{p_{0}\bar{\lambda}^{2}\left(1-\bar{\lambda}\right)^{2}}\parallel\frac{D-g\left(X\right)}{1-g\left(X\right)}\left(\left(T-\bar{\lambda}\right)Y-\ell_{2}\left(X\right)\right)\parallel_{P,2}$
$\displaystyle+\frac{1}{p_{0}\bar{\lambda}\left(1-\bar{\lambda}\right)}\parallel\frac{D-g\left(X\right)}{1-g\left(X\right)}\times
Y\parallel_{P,2}+\mid G_{2\lambda 0}\mid$ $\displaystyle\leq$ $\displaystyle
O\left(1\right)$

by the same arguments in (A.9)-(A.11) and

$\displaystyle\mid G_{2\lambda 0}\mid=$ $\displaystyle\mid
E_{P}\left[-\frac{1-2\lambda_{0}}{\lambda_{0}^{2}\left(1-\lambda_{0}\right)^{2}p_{0}}\frac{D-g_{0}}{1-g_{0}}\left(\left(T-\lambda_{0}\right)Y-\ell_{20}\right)-\frac{Y}{\lambda_{0}\left(1-\lambda_{0}\right)p_{0}}\frac{D-g_{0}}{1-g_{0}}\right]\mid$
$\displaystyle\leq$ $\displaystyle\frac{\mid
1-2\lambda_{0}\mid}{\lambda_{0}^{2}\left(1-\lambda_{0}\right)^{2}p_{0}\kappa}\mid
E_{P}\left[UV_{2}\right]\mid+\frac{1}{\lambda_{0}\left(1-\lambda_{0}\right)p_{0}\kappa}\mid
E_{P}\left[YU\right]\mid$ $\displaystyle\leq$ $\displaystyle\frac{\mid
1-2\lambda_{0}\mid}{\lambda_{0}^{2}\left(1-\lambda_{0}\right)^{2}p_{0}\kappa}C+\frac{1}{\lambda_{0}\left(1-\lambda_{0}\right)p_{0}\kappa}C$
$\displaystyle=$ $\displaystyle O\left(1\right)$

since $\mid E_{P}\left[UV_{2}\right]\mid\leq\parallel
UV_{2}\parallel_{P,4}\leq C$ and $\mid E_{P}\left[YU\right]\mid\leq C$. Also,
we have

$\displaystyle\parallel\partial_{p}\bar{\psi}_{2}\left(W_{i},\theta_{0},\bar{p},\lambda,G_{2\lambda
0},\eta_{2}\right)\parallel_{P,2}\leq$
$\displaystyle\frac{1}{\lambda\left(1-\lambda\right)\bar{p}^{2}}\parallel\frac{D-g\left(X\right)}{1-g\left(X\right)}\left(\left(T-\lambda\right)Y-\ell_{2}\left(X\right)\right)\parallel_{P,2}$
$\displaystyle+\parallel\frac{D\theta_{0}}{\bar{p}^{2}}\parallel_{P,2}$
$\displaystyle\leq$ $\displaystyle O\left(1\right)$

by the same arguments in (A.9)-(A.11) and

$\displaystyle\mid\theta_{0}\mid=$ $\displaystyle\mid
E_{P}\left[\frac{D-g_{0}\left(X\right)}{p_{0}^{\prime}\left(1-g_{0}\left(X\right)\right)}\left(T-\lambda_{0}\right)Y\right]\mid$
$\displaystyle\leq$ $\displaystyle\frac{1}{p_{0}^{\prime}\kappa}\mid
E_{P}\left[\left(T-\lambda_{0}\right)YU\right]\mid$ $\displaystyle=$
$\displaystyle\frac{1}{p_{0}\kappa}\mid
E_{P}\left[\left(\ell_{20}\left(X\right)+V_{2}\right)U\right]\mid$
$\displaystyle=$ $\displaystyle\frac{1}{p_{0}\kappa}\mid
E_{P}\left[UV_{2}\right]\mid$ $\displaystyle\leq$
$\displaystyle\frac{C}{p_{0}\kappa}$

since $\mid E_{P}\left[UV_{2}\right]\mid\leq\parallel
UV_{2}\parallel_{P,4}\leq C$. Together with (A.5), we have

$\displaystyle\parallel\bar{\psi}_{2}\left(W,\theta_{0},p,\lambda,G_{2\lambda
0},\eta_{2}\right)-\bar{\psi}_{20}\parallel_{P,2}\leq$ $\displaystyle
O\left(\varepsilon_{N}\right)+O\left(1\right)O\left(N^{-1/2}\right)+O\left(1\right)O\left(N^{-1/2}\right)$
$\displaystyle=$ $\displaystyle O\left(\varepsilon_{N}\right),$

where I assume that $\varepsilon_{N}$ converges to zero no faster than
$N^{-1/2}$.

For (A.15), we have

$\displaystyle\parallel\bar{\psi}_{2}\left(W,\theta_{0},p_{0},\lambda_{0},G_{2\lambda
0},\eta_{20}\right)\parallel_{P,4}=$
$\displaystyle\parallel\frac{1}{\lambda_{0}\left(1-\lambda_{0}\right)p_{0}}\frac{UV_{2}}{1-g_{0}}-\frac{D\theta_{0}}{p_{0}}+G_{2\lambda
0}\left(T-\lambda_{0}\right)\parallel_{P,4}$ $\displaystyle\leq$
$\displaystyle\frac{1}{\lambda_{0}\left(1-\lambda_{0}\right)p_{0}\kappa}\parallel
UV_{2}\parallel_{P,4}+\frac{1}{p_{0}}\mid\theta_{0}\mid+\mid G_{2\lambda
0}\mid$ $\displaystyle\leq$ $\displaystyle O\left(1\right)$

since $\parallel UV_{2}\parallel_{P,4}\leq C$.

Proof of Theorem 3:

By the same arguments in the proof of Theorem 1, we can have

$\sqrt{N}\left(\tilde{\theta}-\theta_{0}\right)=\frac{1}{\sqrt{N}}\sum_{i=1}^{N}\psi_{1}\left(W_{i},\theta_{0},p_{0},\eta_{10}\right)+G_{1p0}\left(D-p_{0}\right)+O_{P}\left(\varepsilon_{N}+\sqrt{N}\left(\varepsilon_{N}\right)^{2}\right)$

for repeated outcomes and

$\displaystyle\sqrt{N}\left(\tilde{\theta}-\theta_{0}\right)=$
$\displaystyle\frac{1}{\sqrt{N}}\sum_{i=1}^{N}\psi_{2}\left(W_{i},\theta_{0},p_{0},\lambda_{0},\eta_{20}\right)+G_{2p0}\left(D-p_{0}\right)+G_{2\lambda
0}\left(T-\lambda_{0}\right)$
$\displaystyle+O_{P}\left(\varepsilon_{N}+\sqrt{N}\left(\varepsilon_{N}\right)^{2}\right)$

for repeated cross sections. The term $\varepsilon_{N}$ is the rate of
convergence of the kernel estimators $\hat{g}_{kh}$, $\hat{\ell}_{1kh}$, and
$\hat{\ell}_{2kh}$. It remains to show
$\parallel\hat{g}_{kh}-g_{0}\parallel_{P,2}=o_{P}\left(N^{-1/4}\right)$,
$\parallel\hat{\ell}_{1kh}-\ell_{10}\parallel_{P,2}=o_{P}\left(N^{-1/4}\right)$,
and
$\parallel\hat{\ell}_{2kh}-\ell_{20}\parallel_{P,2}=o_{P}\left(N^{-1/4}\right)$.

Here I use the standard result of kernel estimation in Newey McFadden (1994).
Let $\hat{\gamma}_{kh}\left(x\right)$ denote the kernel estimator of
$\gamma_{0}\left(x\right)=f_{0}\left(x\right)E_{P}\left[z\mid x\right]$ using
the auxiliary sample $I_{k}^{c}$, where
$z\in\left\\{1,D,Y\left(1\right)-Y\left(0\right)\mid
D=0,\left(T-\lambda_{0}\right)Y\mid D=0\right\\}$. By Assumption (3.3) and
Lemma 8.10 of Newey McFadden (1994), we have

$\displaystyle\sup_{x\in\mathcal{X}}\mid\hat{\gamma}_{kh}\left(x\right)-\gamma_{0}\left(x\right)\mid=$
$\displaystyle O_{P}\left(\left(\log
N\right)^{1/2}\left(Nh^{d+2s}\right)^{-1/2}+h^{m}\right)$ $\displaystyle=$
$\displaystyle o_{P}\left(N^{-1/4}\right)$

by the conditions on $h$. Let $\hat{f}_{kh}\left(x\right)$ denote
$\hat{\gamma}_{k}\left(x\right)$ with $z=1$ and $\hat{m}_{kh}\left(x\right)$
denote $\hat{\gamma}_{k}\left(x\right)$ with $z=D$. Then

$\hat{g}_{kh}\left(x\right)=\frac{\hat{m}_{kh}\left(x\right)}{\hat{f}_{kh}\left(x\right)}=\frac{\hat{m}_{kh}\left(x\right)/f_{0}\left(x\right)}{\hat{f}_{kh}\left(x\right)/f_{0}\left(x\right)}.$

For the denominator, we have

$\sup_{x\in\mathcal{X}}\mid\frac{\hat{f}_{kh}\left(x\right)}{f_{0}\left(x\right)}-1\mid\leq\frac{\sup_{x\in\mathcal{X}}\mid\hat{f}_{kh}\left(x\right)-f_{0}\left(x\right)\mid}{\inf_{x\in\mathcal{X}}f_{0}\left(x\right)}=o_{P}\left(N^{-1/4}\right)$

given $\inf_{x\in\mathcal{X}}f_{0}\left(x\right)\neq 0$. For the numerator,
let $m_{0}\left(x\right)$ denote $\gamma_{0}\left(x\right)$ with $z=D$, we
have

$\sup_{x\in\mathcal{X}}\mid\frac{\hat{m}_{kh}\left(x\right)}{f_{0}\left(x\right)}-g_{0}\left(x\right)\mid\leq\frac{\sup_{x\in\mathcal{X}}\mid\hat{m}_{kh}\left(x\right)-m_{0}\left(x\right)\mid}{\inf_{x\in\mathcal{X}}f_{0}\left(x\right)}=o_{P}\left(N^{-1/4}\right)$

given $\inf_{x\in\mathcal{X}}f_{0}\left(x\right)\neq 0$. The above two
inequalities imply that uniformly over $x\in\mathcal{X}$,

$\hat{g}_{kh}\left(x\right)=\frac{g_{0}\left(x\right)+o_{P}\left(N^{-1/4}\right)}{1+o_{P}\left(N^{-1/4}\right)}=g_{0}\left(x\right)+o_{P}\left(N^{-1/4}\right).$

That is,
$\sup_{x\in\mathcal{X}}\mid\hat{g}_{kh}\left(x\right)-g_{0}\left(x\right)\mid=o_{P}\left(N^{-1/4}\right)$.
Using the same arguments, one can also show that
$\sup_{x\in\mathcal{X}}\mid\hat{\ell}_{1kh}\left(x\right)-\ell_{0}\left(x\right)\mid=o_{P}\left(N^{-1/4}\right)$
and
$\sup_{x\in\mathcal{X}}\mid\hat{\ell}_{2kh}\left(x\right)-\ell_{0}\left(x\right)\mid=o_{P}\left(N^{-1/4}\right)$.
Since uniform convergence implies $L^{2}$-norm convergence, we complete the
proof.

Proof of Theorem 4:

The proof is the same as the proof in Theorem 2 provided the assumptions in
Theorem 3 hold.

Lemma A.1

(CONDITIONAL CONVERGENCE IMPLIES UNCONDITIONAL)

_Let $\left\\{X_{m}\right\\}$ and $\left\\{Y_{m}\right\\}$ be sequences of
random vectors. (i) If for $\epsilon_{m}\rightarrow 0$,
$\text{Pr}\left(\parallel X_{m}\parallel>\epsilon_{m}\mid
Y_{m}\right)\stackrel{{\scriptstyle p}}{{\rightarrow}}0$, then
$\text{Pr}\left(\parallel X_{m}\parallel>\epsilon_{m}\right)\rightarrow 0$.
This occurs if $E\left[\parallel X_{m}\parallel^{q}/\epsilon_{m}^{q}\mid
Y_{m}\right]\stackrel{{\scriptstyle p}}{{\rightarrow}}0$ for some $q\geq 1$,
by Markov’s inequality. (ii) Let $\left\\{A_{m}\right\\}$ be a sequence of
positive constants. If $\parallel X_{m}\parallel=O_{P}\left(A_{m}\right)$
conditional on $Y_{m}$, namely, that for any $\ell_{m}\rightarrow\infty$,
$\text{Pr}\left(\parallel X_{m}\parallel>\ell_{m}A_{m}\mid
Y_{m}\right)\stackrel{{\scriptstyle p}}{{\rightarrow}}0$, then $\parallel
X_{m}\parallel=O_{P}\left(A_{m}\right)$ unconditionally, namely, that for any
$\ell_{m}\rightarrow\infty$, $\text{Pr}\left(\parallel
X_{m}\parallel>\ell_{m}A_{m}\right)\stackrel{{\scriptstyle}}{{\rightarrow}}0$.
_

PROOF: This lemma is the Lemma 6.1 in Chernozhukov . (2018).

SIMULATION

Figure 3: Repeated outcomes: $N=200$ and $p=300$. Figure 4: Repeated outcomes:
$N=200$ and $p=100$. Figure 5: Repeated outcomes: $N=500$ and $p=300$. Figure
6: Repeated outcomes: $N=500$ and $p=100$. Figure 7: Repeated outcomes:
$N=200$ Figure 8: Repeated outcomes: $N=500$ Figure 9: Repeated cross
sections: $N=200$ and $p=300$. Figure 10: Repeated cross sections: $N=200$ and
$p=100$. Figure 11: Repeated cross sections: $N=500$ and $p=300$. Figure 12:
Repeated cross sections: $N=500$ and $p=100$. Figure 13: Repeated cross
sections: $N=200$ Figure 14: Repeated cross sections: $N=500$ Figure 15:
Multilevel treatment: $N=200$ and $p=300$. Figure 16: Multilevel treatment:
$N=200$ and $p=100$. Figure 17: Multilevel treatment: $N=500$ and $p=300$.
Figure 18: Multilevel treatment: $N=500$ and $p=100$. Figure 19: Multilevel
treatment: $N=200$ Figure 20: Multilevel treatment: $N=500$


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