&\leq P( F_ \beta (T_n) \in A^{ e_\beta + 3 \delta }) + \frac { \epsilon + C \beta \delta^{-1} \{ B_1 + \beta^2( B_2 + B_3) \}}{ 1 - \epsilon }\qquad(\text{by construction of }g)\\
&\leq P( \tilde{Z}^* \in A^{ 2 e_\beta + 3 \delta}) + \frac {\epsilon + C \beta \delta^{-1} \{ B_1 + \beta^2( B_2 + B_3) \}}{ 1 - \epsilon } \qquad(\text{by} )
\end{align*}

This completes the proof

Taking $\beta = 2 \delta^{-1} \log (p \lor n ) $ in Lemma <ref>, we get

\[\alpha = \beta^2 \delta^2 -1  = 4 \log ^2 ( p \lor n) -1 \geq 2 \log ( p \lor n)\qquad(\text{recall}\;n \geq 3 > e)\]

so that

\[\epsilon \leq 2 \log( p \lor n) / ( p \lor n ) \leq 2 n^{-1} \log n.\]

Then we get from Lemma <ref> that for every $\delta > 0 $,

\[P( | Z  - \tilde{Z} | >  16 \delta ) \leq \delta^{-2}\{ B_1 + \delta^{-2} ( B_2 + B_3)\log(p \lor n )^2\} \log ( p \lor n) + \frac{ \log n }{n}\]

We finally apply this to the discretized empirical process. Construct a tight Gaussian random variable $G_P$ in $ \ell^ \infty ( \mathcal{F}) $ given in assumption (A3), independent of $ X_1, ...X_n$. We note that one can extend $G_P$ to the linear hull of $\mathcal{F}$  in such a way that $G_P$ has linear sample paths.
Let $\{ f_1, ... f_N \}$ be a minimal $ \epsilon || F|| _{P,2}$ -net of $( \mathcal{F}, e_P)$ with $ N = N( \mathcal{F}, e_P, \epsilon \norm{ F}_{P,2}) $. Then for every $f\in \mathcal{F}$, there exists a function $f_j\;(1 \leq j \leq N)$ such that $ e_P( f, f_j) < \epsilon \norm{F}_{ P,2}$. Recall that

\[F_{ \epsilon} = \{ f- g: f, g \in \mathcal{F} , e_P(f,g)< \epsilon || F||_{P,2} \}\]

and define

\begin{equation*}
Z^ \epsilon := \max _{ 1 \leq j \leq N} \mathcal{G}_n f _j,\qquad \tilde{Z}^\epsilon:= \sup G_P f , \tilde{Z^{* \epsilon}} = \max_{1 \leq j \leq N} G_P f_j.
\end{equation*}

Observe that $ \abs{Z - Z^ \epsilon}\leq \norm{\mathcal{G}_n}_{ \mathcal{F}_\epsilon} $ and $ | \tilde{Z}^{ * \epsilon} - \tilde{Z}^ { *}  | \leq || G_P ||_{ \mathcal{F}_\epsilon} $
Under the setup of <cit.>, $\log(N \lor n )= H_n (\epsilon) $.Then for every Borel subset A of $\mathbb{R}$ and $\delta>0$,

\begin{align*}
P(Z^\epsilon \in A)  - P( \tilde{Z}^{* \epsilon} \in A^{16 \delta})
\precsim \delta^{-2} \{ B_1 + \delta^{-2} ( B_2 + B_4 )H_n (\epsilon) + n^{-1} \log n
\end{align*}

Clearly, $B_1 \leq n^{-1/2} E[ \norm{\mathbb{G}_n}_{\mathcal{F}\cdot \mathcal{F}} ] , B_2 \leq n^{-1} \kappa^4$, and $ B_4 \leq n^{-1} P[ F^4 1( F > \delta \sqrt{n} H_n (\epsilon)^{-1})]$. Hence, choosing $\delta >0$ in such a way that

\begin{align*}
C \delta^{-2} n^{-1/2} E[ ||\mathbb{G}_n||_{\mathcal{F}\cdot \mathcal{F}}] H_n ( \epsilon) \leq \frac{ \gamma}{4}\\
C \delta^{-4} n^{-1} \kappa^4 H_n^{7/2}  \epsilon \leq \frac{\gamma}{4}
\end{align*}

that is,

\begin{align*}
\delta \geq C \max \{ \gamma^{-1/2}n ^{-1/4} ( E[ ||\mathbb{G}_n|| _ {\mathcal{F}\cdot \mathcal{F}}])^{1/2} H_n^{1/2}  (\epsilon) , \gamma^{-1/4} n^{-1/4} \kappa H_n ^{7/8} ( \epsilon) \}
\end{align*}

we have

\begin{align*}
P(Z^\epsilon \in A) \leq P( \tilde{Z}^{ * \epsilon} \in A^{16 \delta}) + \frac{\gamma}{2} + \frac{\gamma}{4} \kappa^{-4} P[ F^4 1 ( F > \delta \sqrt{n} H_n ( \epsilon)^{-1}) ] + \frac{C \log n }{n}
\end{align*}

Note that $ \delta \geq c \gamma^{-1/4} n^{1/4} \kappa H_n ^{-1/8} ( \epsilon) $, so that

\begin{equation*}
P[ F^4 1(F > \delta \sqrt{n} H_n (\epsilon)^{-1})] \leq P[ F^4 1( F > c \gamma^{-1/2} n^{1/4} ( E[ || \mathbb{G}_n||_{\mathcal{F} \cdot \mathcal{F}}])^{1/2} H_n^{- 1/3}(\epsilon) ]
\end{equation*}

\begin{align*}
P(Z^\epsilon \in A) &\leq P( \tilde{Z}^{ * \epsilon} \in A^{16 \delta}) + \frac{\gamma}{2} + \frac{\gamma}{4} \kappa^{-4}  P\paren{ F^4 1( F > c \gamma^{-1/2} n^{1/4} ( E[ || \mathbb{G}_n||_{\mathcal{F} \cdot \mathcal{F}}])^{1/2} H_n^{- 1/3}(\epsilon)} + \frac{C \log n }{n} \\
&=: P( \tilde{Z}^{* \epsilon} in A^{16 \delta}) + \frac{\gamma}{2} + \text{(error)}
\end{align*}

By bounding $\norm{\mathbb{G}_n}_{\mathcal{F}_\epsilon} $ and $ \norm{G_P}_{\mathcal{F}_\epsilon}$ by Theorem 5.1 of <cit.> and Borell-Sudakov-Tsirel'son inequality of <cit.> respectively as in the original proof we have

\begin{equation*}
P(Z \in A) \leq P( \tilde{Z}^{*} \in A^{a + b + 16 \delta}) + \gamma + \text{(error)}
\end{equation*}

where a, b is

\begin{align*}
a&: = K(q) \{ \phi_n ( \epsilon) + ( \epsilon ||F||_{P,2} + n^{-1/2} || M||_q ) \gamma^{-1/q} + n^{-1/2} || M|| _2 \gamma^{-2/q} \} \\
b&: = \phi_n (\epsilon) + \epsilon || F||_{P,2} \sqrt{ 2 \log( 4/ \gamma)}
\end{align*}

where $K(q)$ is a constant that depends only on q.
Then, the conclusion follows from Strassen's theorem (Lemma 4.1 of <cit.>).

§.§ Proofs for Section <ref>

The difference between wage and search cost on the right-hand side of equation(<ref>) is the worker's net current income. Then next term on the right-hand side represents the expected capital gain associated with the possible arrival of an outside offer, given that the worker acts optimally by accepting jobs with higher value. The last term reflects the expected
capital loss attributable to job destruction, the difference between the value
of unemployment and the value of employment in the worker’s current
job. Hence, the equation is an arbitrage condition that defines the asset
value of being employed to be that which equates the riskless return on
the asset value of the search while employed option to current net income plus expected capital gains and losses associated with the option. This
relationship is a continuous time equivalent of the well-known Bellman
equation of dynamic programming.
Following the usual fixed point arguments of the contraction map on space of differentiable and real-valued increasing functions as in <cit.>, it can be shown that the value function is the unique fixed point.
By some tedious but straightforward calculations of first order conditions for the interior optimal solutinos, integration by parts, and noting that equation(<ref>) can be rewritten as

\begin{equation}
W(w) = \max_{s \geq 0} \{ \frac{ w - c(s) + \delta U + (\lambda+ \theta \omega) s \int \max [ W(x), W(w) ] dF(x) }{ r + \delta + (\lambda+ \theta \omega)s }
\end{equation}

, it follows that

\begin{align}
\label{eq:jobsearch2}
c^{\prime}[s(w)] &=(\lambda+ \theta \omega) \int_w^{\bar{w}}[W(x)-W(w)] d F(x)=(\lambda+ \theta \omega)\int_w^{\bar{w}} W^{\prime}(x)[1-F(x)] d x \\
&=\lambda \int_w^{\bar{w}} \frac{[1-F(x)] d x}{r+\delta+(\lambda+ \theta \omega) s(x)[1-F(x)]}
\end{align}
where $\bar{w}$ is the upper support of the wage offer distribution. In other words, the optimal search effort function is the unique particular solution to this integral equation. Optimal search effort, $s(w)$, is strictly decreasing and continuous in the wage earned by convexity of the cost of search function.
Consider the same worker when not employed. The value of nonemployment solves the analogous asset-pricing equation

\begin{equation}
\label{eq:jobsearch3}
r U=\max _{s \geq 0}\left(b-c(s)+(\lambda+ \theta \omega) s \int\{\max [W(x), U]-U\} d F(x)\right),
\end{equation}

where $b$ represents income forgone when employed, that is, the unemployment benefit. The worker's reservation wage, $R$, is the solution to

W(R)=U .

Under the assumption that the cost of search effort is the same whether employed or not, a comparison of equations (<ref>), (<ref>), and (<ref>) implies that optimal search effort when unemployed, denoted as $s_0$, equals search effort
when employed at the worker's reservation wage and, consequently, the worker's reservation wage is simply the unemployment compensation, that is,

R=b .

In sum, the overall job duration hazard for any worker employed by an employer paying wage $w$ is

\begin{equation}
\label{eq:jobsearch d def}
d(w)=\delta+(\lambda + \theta_t \omega) s(w)[1-F(w)],
\end{equation}

where $s^{\prime}(w)<0$ and $s(\bar{w})=0$.
, and the overall unemployment duration hazard for the unemployed under the assumptions laid out above is

\begin{equation}
\label{eqLjobsearch d_u def}
d_u = (\lambda + \theta \omega)  s_o [ 1 -F(R) ]
\end{equation}

Under the assumption that an employer pays all workers the same wage and the cost of search is the same for all workers, the function $d(w)$ also represents the employer's separation rate.
To recover the target parameters, when we are interested in the transition from unemployment to employment, then the $d_u$ will be of interest. And the $\theta_t$ will approximately correspond to the treatment effect of job training at time stage $t$on the transition probability, assuming that the model effectively captures the selection into treatment. Then the ATETS will approximately correspond to $\theta_2$ if we set the time stage as short term to t=1 and long-term to t=2. Given the data on the short and the long-term for observational data, we could estimate both $\theta_1$ and $\theta_2$. We omit the remainder(the normalization and likelihood derivation, etc.) since it is identical to <cit.>,

§.§ Proofs for Section <ref>

We note that in the sequel, we omit the dependence on $G=1$ when it does not lead to confusion.

Consider the missingness indicator $Z_x, Z_y$ that is equal to $1$ if and only if the covariate $Y_1(0)$ and the outcome $Y_2(1)$ is observed and $0$ otherwise.

First, by an application of the Bayes rule, and the law of total probability

\begin{align}
&E[ Y_2(0)=1 | Y_1(1)=0, G=1] \\
&= \sum_{j,k} \frac{ P (Y_2 =1 | Y_1 = 0 , Z_y = j, Z_ x = k ) P ( Y_1 = 0 | Z_y =j,Z_ x =k ) P( Z_y = k, Z_x = j) }{ \sum_{ k, j} P( Y_1 = 0 | Z_y = j, Z_x = k ) P (Z_y = j, Z_x = k) }
\end{align}

then by applying <cit.> Corollary 1.1, we obtain the following sharp bounds :

\begin{align}
& P(Y_2 =1 | Y_1 =0, Z_y =1, Z_ x =1 ) P( Y_1 = 0 | Z_y =1, Z_x = 1 )P( Z_y = 1, Z_x =1)\\
&\leq E[ Y_2(0)=1 | Y_1(1)=0, G=1] \\
&\leq P( Y_1 = 0 | Z_y = 0 ,Z_x = 1) P( Z_y = 0 , Z_x =1 ) + P( Z_y = 0, Z _ x = 0) \\
&\qquad+ P( Y_2 =1 | Z_y = 1, Z_ x =0) P  Z_ y =1, Z_x = 0)
\end{align}

As in the brief explanation in the body, what we need to show is that the sharp identification bound for $P( Y_2(0) =1, Y_1(1)=0|G=1)$ is

\begin{align*}
&\biggl[\max(0,P(Y_1=0| W=1, G=0) +P(Y_2=1| W=0, G=1) P(W=0) + P(W=1)- 1) ,\\
&\qquad\min(P(Y_1=0| W=1, G=0) , P(Y_2=1| W=0, G=1) P(W=0) )\biggr]
\end{align*}

Note that by the Frechet-Hoeffding theorem, the sharp identification bound
when both $P( Y_2(0) =1|G=1)$ and $P(  Y_1(1)=0|G=1)$ is identified is

\begin{align}\label{eq:atets bound frechet1}
&[ \max(0, P(Y_2(0)=1|G=1)+P(Y_1(1)=0|G=1) -1),\\
&\qquad \min( P(Y_2(0)=1|G=1), P(Y_1(1)=0|G=1))]
\end{align}

While $P(Y_1(1)=0|G=1)$ is point identified as $P(Y_1 = 0 | W=1,G=0)$ from external validity and experimental internal validity, $P(Y_2(0)=1|G=1)$ is not because the long term outcome
is not available for the experimental data. The sharp bound for this quantity is $[ P(Y_2| W=0, G=1)P(W=0),P(Y_2| W=0, G=1)P(W=0) + P(W=1)]$ by the same argument as in Section <ref>. hence taking this lower bound for the upper bound of <ref>, and the upper bound for the lower bound of <ref>
is the sharp bound. This is because the bound collapses to a point(equals to $0\, a.s.$) for the distribution that takes $P(Y_2=1|W=0,G=1)=0$ , which is admissible under the given assumptions.

Step 1 : Show that $\Delta \dot{ \nu_{it}} ( S_{it} (1) - \Delta \dot{ \nu_{it}} ( S_{it} (0) \geq 0 $

We start from the terminal period $\bar{T}$and recursively show the above inequality.
For time stage $t = \bar{T}$, There is only the continuation value

\begin{align*}
\omega ( Y_{i(t-1)}, e^*( 1, V_{i(t-1)}, A_i || Y_{i(t-1)}), 1,A_i, V_{it} ) - \omega ( Y_{i(t-1)}, e^*( w, V_{i(t-1)}, A_i || Y_{i(t-1)}), 0,A_i, V_{it} )\\
-\Delta \kappa (1, V_{it}, A_i) + \Delta \kappa (0, V_{it}, A_i)
\end{align*}

which is greater than 0
because the first line is greater than 0 by Assumption (i) in the proposition and the second line is greater than 0 by Assumption (ii).
For $t = \bar{T}-1$, w have that

\begin{align*}
&\Delta \dot{ \nu_{it}} ( S_{it} (1) - \Delta \dot{ \nu_{it}} ( S_{it} (0) \geq 0 \\
&= \omega ( Y_{i(t-1)}, e^*( 1, V_{i(t-1)}, A_i || Y_{i(t-1)}), 1,A_i, V_{it} ) -
\omega ( Y_{i(t-1)}, e^*( w, V_{i(t-1)}, A_i || Y_{i(t-1)}), 0,A_i, V_{it} )\\
&-\Delta \kappa (1, V_{it}, A_i) + \Delta \kappa (0, V_{it}, A_i) \\
&+\Delta \gamma (1,V_{it}, A_i) - \Delta \gamma (0,V_{it}, A_i)  \geq 0
\end{align*}

and by the same argument as in $t =\bar{T}$, we have the first two lines greater than zero. Also the third line is also greater than zero because, it is the conditional expectation of a quantity which we know to be greater than zero by the previous argument. Repeating this argument recursively, we can show that for any $t \in \mathcal{T}$, $\Delta \dot{ \nu_{it}} ( S_{it} (1) - \Delta \dot{ \nu_{it}} ( S_{it} (0) \geq 0 $.
Step2 Show that $\Delta \dot{ \nu_{it}} ( S_{it} (1) - \Delta \dot{ \nu_{it}} ( S_{it} (0) \geq 0 \, a.s.  $

$\implies Y_{i t}(1) = 1\{\Delta \dot{ \nu_{it}} ( S_{it} (1)) \geq 0 \} \geq 1\{\Delta \dot{ \nu_{it}} ( S_{it} (0)) \geq 0 \}= Y_{i t}(0) $

For any event such that $ \Delta \dot{ \nu_{it}} ( S_{it} (1)) \geq 0 $, $ Y_{i t}(1) = 1 \geq 1\{\Delta \dot{ \nu_{it}} ( S_{it} (0)) \geq 0 \}= Y_{i t}(0)( \in \{ 0,1\}) $

For any event such that $\Delta \dot{ \nu_{it}} ( S_{it} (1)) < 0 $, by the inequality obtained in Step 1, $\Delta \dot{ \nu_{it}} ( S_{it} (0)) < 0,i.e. Y_{i t}(0) = 0 $,

Hence, with probability 1 the desired inequality holds.

We will be proving our theorem show that stationarity of $Y_{it} (w)$ holds under assumption W(a) and W(b).
To see this, note that
$Y_{it}(0), Y_{it}(1)$ is fully determined by  $(\Delta \dot{\nu}( S_{it}(0)), \Delta \dot{\nu}( S_{it}(1))) $. Under assumption MC and W(b), the latter is only stochastic due to $Y_{i (t-1)} , A_i, (V_{i (t-1)} , V_{it}) $, which is in general fully determined by the distribution of $(A_i , ( V_{ i (1:t) } )$.  First, if assumption W(b) is satisfied, then since $V_{i 1:(t-1)} $ is independent from $V_{it}$, this implies that   $(\Delta \dot{\nu}( S_{it}(0)), \Delta \dot{\nu}( S_{it}(1))) $ is fully determined by teh distribution of $( A_i, V_{it} ) $. Moreover, by Assumption  W(a), $V_{it} $ are stationary, and since $A_i$ is time invariant and hence stationary, we can say that $Y_{it}(0), Y_{it}(1)$ is stationary with order 2.

Note that the collection of selections $(y_t(1), y_t(0) ) , t \in \{1,2 \}  $ , such that    $ (y_t(1), y_t(0) ) $ is a selection of        $ \mathbf{Y_t}$ can be characterized by the family of dominance conditions

\begin{align*}
P( (y_t(1), y_t(0) ) \in K ) \leq P ( \mathbf{ Y(t)}  \cap K \neq \emptyset , \forall K \in \mathcal{K}  t \in \{1, 2 \}
\end{align*}

Finally marginalizing over the unnecessary $y_1(0), y_2(1) $, gives the desired result.

Step 1 : Show that $\Delta \dot{ \nu_{it}} ( S_{it} (w) ,\bar{x}_{it}=x_1)- \Delta \dot{ \nu_{it}} ( S_{it} (w)  ,\bar{x}_{it}=x_2 \geq 0 $

It will follow similarly to Theorem <ref> .
First consider the terminal period $t = \bar{T}$. In this case there is only present value, and by the exclusion restiction and monotonicity,

\begin{align*}
&\omega ( Y_{i(t-1)}, e^*( w, V_{i(t-1)}, A_i || Y_{i(t-1)}), w,A_i, V_{it} ,\bar{X}_{it}=x_1) - \omega ( Y_{i(t-1)}, e^*( w, V_{i(t-1)}, A_i || Y_{i(t-1)}), w,A_i, V_{it} ,\bar{X}_{it}=x_2)\\
-&\Delta \kappa (1, V_{it}, A_i,\bar{X}_{it}=x_1) + \Delta \kappa (0, V_{it}, A_i,\bar{X}_{it}=x_2)
\end{align*}

which is greater than 0
because the first line is  0 by the exclusion restriction  and the second line is greater than 0 by the monotonicity restriction.

For $t = \bar{T}-1$, w have that

\begin{align*}
&\Delta \dot{ \nu_{it}} ( S_{it} (w) ,\bar{x}_{it}=x_1)- \Delta \dot{ \nu_{it}} ( S_{it} (w)  ,\bar{x}_{it}=x_2  \\
&= \omega ( Y_{i(t-1)}, e^*( w, V_{i(t-1)}, A_i || Y_{i(t-1)}), w,A_i, V_{it} ,\bar{X}_{it}=x_1) - \omega ( Y_{i(t-1)}, e^*( w, V_{i(t-1)}, A_i || Y_{i(t-1)}), w,A_i, V_{it} ,\bar{X}_{it}=x_2)\\
-&\Delta \kappa (1, V_{it}, A_i,\bar{X}_{it}=x_1) + \Delta \kappa (0, V_{it}, A_i,\bar{X}_{it}=x_2)\\
&\Delta \gamma(w,V_{it}, A_i ,\bar{x}_{it}=x_1) - \Delta \gamma(w,V_{it}, A_i ,\bar{x}_{it}=x_2)  \geq 0
\end{align*}

and by the same argument as in $t =\bar{T}$, we have the first two lines summed are greater than zero. Also the final line is also greater than zero because, it is the conditional expectation of a quantity which we know to be greater than zero by the previous argument. Repeating this argument recursively, we can show that for any $t \in \mathcal{T}$, $\Delta \dot{ \nu_{it}} ( S_{it} (1) - \Delta \dot{ \nu_{it}} ( S_{it} (0) \geq 0 $.
Step2: Show that  $ \forall x_1  \geq x_2  , Y_{it} (w) | \bar{X}_{it} =x_1   > Y_{it} (w) | \bar{X}_{it} =x_2 $

This goes just like Proposition <ref> and hence will be omitted.

As in Theorem <ref>, the sharp bounds for the iv and monotone iv case have already been extensively discussed in, for example, <cit.>, and hence will be omitted.

§.§ Proofs for Section <ref>

Based on the argument in Section<ref> , by symmetry, it suffices to show the influence function  for
\[E[ Y_2(1) | Y_1(1) , G=1] = E[ E[E [ Y_2 | Y_1 ,W=1,X, G=1] | W=1, X, G=0 ] | G=1]\]

we use $ \partial_t f(t) $ to denote $\frac{ \partial f(t) }{ \partial_t } $ For parameter $\theta $, let $ \theta_t $ be the parameter under a regular parametric sub-model indexed by $t$, that includes the ground-truth model at $t=0$.
Let $V$ be the set of all observed variables. In order to obtain the influence function, we need to find a random variable $\psi $ with mean zero that satisfies,

\begin{equation}
\partial_t \psi_t = E [ \psi S(V) ]
\end{equation}

where $ S(V) = \partial_t log p_t (V) $.
To simplify notation, we assume that all variables are discrete.
First note that

\begin{align}
\partial_t \psi_t = \partial_t \sum_ { y_2, y_1,x} y_2 p_t ( y_2| y_1 , W= 1, X , G= 1) p_t ( y_1 | W =1 , x, G= 0 ) p_t( x | G =1 )   \\
= \label{eq: eif first chain-1} \sum_ { y_2 ,y_1,x} y_2 \partial_t p_t ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 )   \\
+ \sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1)\partial_t p_t ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) \\
+ \sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) \partial_t p_t( x | G =1 )
\end{align}

for the first term in (<ref>),  we have

\begin{align*}
&=\sum_ { y_2 ,y_1,x} y_2 \partial_t p_t ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) \\
&= \sum_ { y_2 ,y_1,x} y_2 p ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) S ( y_2| y_1 , W= 1, X , G= 1) \\
&= \sum_ { y_2 ,y_1,w, x, g}  \frac{ w g }{ p (G = 1) } y_2  \frac{ p( y_2, w) | y_1, x, g}{ p ( W = 1 | y_1(1) , x , g ) }\\
&\qquad\qquad\times p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) \frac{ p ( g, x ) } { p(g =1)} S ( y_2| y_1 , W= 1, X , G= 1) \\
&= \sum_ { y_2 ,y_1,w, x, g}  \frac{ w g }{ p (G = 1) } y_2  \frac{ p( y_2, w | y_1, x, g}{ p ( W = 1 | y_1(1) , x , g ) }   p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 )\\
&\qquad\qquad\times\frac{ p ( g, x ) } { p(g =1)} S ( y_2| y_1 , W, X , G)  \\
&= \sum_ { y_2 ,y_1,w, x, g}  \frac{ w g }{ p (G = 1) } y_2   p( y_2, w | y_1, x, g)  \frac{ P( G =1 | Y_1 , W=1 , X ) }{ P(G=1 | Y_1 , W=1 , X ) } P( W=1 | X , G=1)   \\
&\qquad\qquad\times p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 )
\frac{ p ( g, x ) } { p(G =1)} S ( y_2| y_1 , w, x , g) \qquad( \because\text{Bayes rule})\\
&= E \Square{  \frac{ w g }{ p (G = 1) } y_2     \frac{ P( G =1 | Y_1 , W=1 , X ) }{ P(G=1 | Y_1 , W=1 , X ) } P( W=1 | X , G=1)   S ( y_2| y_1 , w, x , g)}\\
&= E \biggl[  \frac{ w g }{ p (G = 1) } ( y_2 - E [ Y_2 | y_1 , W = 1, X, G =1 ] )     \frac{ P( G =1 | Y_1 , W=1 , X ) }{ P(G=1 | Y_1 , W=1 , X ) }\\
&\qquad\qquad\times P( W=1 | X , G=1)   S ( y_2| y_1 , w, x , g) \biggr]\\
&= E \biggl[  \frac{ w g }{ p (G = 1) } ( y_2 - E [ Y_2 | y_1 , W = 1, X, G =1 ] )     \frac{ P( G =1 | Y_1 , W=1 , X ) }{ P(G=1 | Y_1 , W=1 , X ) } \\
&\qquad\qquad\times P( W=1 | X , G=1)   S ( y_2| y_1 , w, x , g) \biggr]\\
\end{align*}

Note that

\begin{align*}
&E \biggl[  \frac{ w g }{ p (G = 1) } ( y_2 - E [ Y_2 | y_1 , W = 1, X, G =1 ] )\\
&\qquad\times\frac{ P( G =1 | Y_1 , W=1 , X ) }{ P(G=1 | Y_1 , W=1 , X ) } P( W=1 | X , G=1)   S (  y_1 , w, x , g) \biggr] = 0
\end{align*}

\begin{align*}
&\sum_ { y_2 ,y_1,x} y_2 \partial_t p_t ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) \\
&= E \biggl[  \frac{ w g }{ p (G = 1) } ( y_2 - E [ Y_2 | y_1 , W = 1, X, G =1 ] )     \frac{ P( G =1 | Y_1 , W=1 , X ) }{ P(G=1 | Y_1 , W=1 , X ) } P( W=1 | X , G=1)   S ( V) \biggr]
\end{align*}

Likewise for the second term in <ref>

\begin{align*}
&\sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1)\partial_t p_t ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) \\
&=   \sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) S( y_1  | W =1, x, G = 0)  \\
&=   \sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) S( y_1  | W =1, x, G = 0)
\end{align*}

noting that

\begin{align*}
p ( y_1(1) | w = 1, x g = 0 ) &= \frac{ p ( y _1(1) , w = 1, g  = 0 |x) } { p ( G =0, W =1 | X) } = \frac{ w ( 1 -g) p( y_1(1) , w g | x) }{ p  (g =0, w =1 | x) } , \\
p ( x | G= 1) &= \frac{ p(  G =1, X ) } { p(G =1 ) },
\end{align*}

and rearranging terms,

\begin{align*}
&\sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1)\partial_t p_t ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) \\
= &\sum_ { y_2 ,y_1,w,x, g} \frac{ w ( 1 -g ) y_2   p ( y_2| y_1 , W= 1, X , G= 1) p(G =1 | X) }{ p(G=1) p ( G= 0 | X)  p ( W = 1 | X ,G =0) } E [ Y_2 |   Y_1 ,W = 1,X, G =1 ]   S( y_1  | W , x, G )  \\
= &E \Square{   \frac{ W ( 1 -G)   p(G =1 | X) }{ p(G=1) p ( G= 0 | X)  p ( W = 1 | X ,G =0) } E [ Y_2 |   Y_1 ,W = 1, X, G =1 ]   S( y_1  | W , x, G )}\\
= &E \Square{  \frac{  W ( 1 -G)   p(G =1 | X) }{ p(G=1) p ( G= 0 | X)  p ( W = 1 | X ,G =0) } E [ Y_2 |   Y_1 ,W = 1, X, G =1 ]   S( y_1  | W , x, G ) }\\
= &E \biggl[  \frac{  W ( 1 -G)   p(G =1 | X) }{ p(G=1) p ( G= 0 | X)  p ( W = 1 | X ,G =0) } (E [ Y_2 |   Y_1 ,W = 1, X, G =1] \\
&\qquad -E [ E[ Y_2 | Y_1 , W =1 , X, G =1 ] | W=1, X, G= 0 ] )  S( y_1  | W , x, G ) \biggr]
\end{align*}

Note that,

\begin{align*}
&E \biggl[  \frac{  W ( 1 -G)   p(G =1 | X) }{ p(G=1) p ( G= 0 | X)  p ( W = 1 | X ,G =0) } (E [ Y_2 |   Y_1 ,W = 1, X, G =1 ] - \\
&E [ E[ Y_2 | Y_1 , W =1 , X, G =1 ] | W=1, X, G= 0 ]] )  S(  W , x, G ) \biggr] = 0
\end{align*}

\begin{align*}
&\sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1)\partial_t p_t ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) \\
= &E \biggl[  \frac{  W ( 1 -G)   p(G =1 | X) }{ p(G=1) p ( G= 0 | X)  p ( W = 1 | X ,G =0) } (E [ Y_2 |   Y_1 ,W = 1, X, G =1 ] -\\
&E [ E[ Y_2 | Y_1 , W =1 , X, G =1 ] | W=1, X, G= 0 ]] )  S(  V) \biggr]
\end{align*}

for the third term

\begin{align*}
&\sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) \partial_t p_t( x | G =1 )\\
&= \sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) S( x | G= 1) \\
&= \sum_ { y_2 , y_1, x}      \frac{ G } { P(G =1 ) }   y_2  p ( y_2| y_1 , W= 1, X , G= 1)\\
&\qquad\times p ( y_1 | W =1 , x, G= 0 ) p( x | G =1 ) S( x | G= 1) \\
&= \sum_ { y_2 , y_1, x, g}      \frac{ G } { P(G =1 ) }   (E[ E[Y_2| Y_1,  W =1, X, G = 1 ] | W =1 , X, G = 0 ]]  - \\
&\qquad \tau_1 ) p( x , g ) S( x | G)
\end{align*}

note that

\begin{align*}
&= \sum_ { y_2 , y_1, x, g}      \frac{ G } { P(G =1 ) }   (E[ E[Y_2| Y_1,  W =1, X, G = 1 ] | W =1 , X, G = 0 ]]  - \\
&\tau_1 ) p( x , g ) S(  G) = 0
\end{align*}

\begin{align*}
\sum_ { y_2 ,y_1,x} y_2  p ( y_2| y_1 , W= 1, X , G= 1) p ( y_1 | W =1 , x, G= 0 ) \partial_t p_t( x | G =1 )\\
= E\biggl[      \frac{ G } { P(G =1 ) }   (E[ E[Y_2| Y_1,  W =1, X, G = 1 ] | W =1 , X, G = 0 ]  - \tau_1 ) p( x , g ) S(  V) \biggr]
\end{align*}

collectivizing the three results, we have

\begin{align*}
&\partial_t \psi_t = E \biggl[  \frac{ g w }{ p(G=1) } \biggl[ (y_2 - E[ Y_2 | W=1 , Y_1 , X, G =1]  ) \frac{ P( G =1 | Y_1 , W=1 , X ) }{ P(G=1 | Y_1 , W=1 , X ) } P( W=1 |X ,G=1) \\
& \qquad+ E [ E[ Y_2 | W = 1, Y_1 , X, G =1 | W =1, X, G= 0 ] ] - \tau    \biggr]  + \frac{ (1- g) P(G = 1|X)   }{ P(G= 1) P ( G =0 |X) } \\
&\qquad\times\paren{ \frac{ w ( E [ Y _ 2 | Y_1,W =1 , X, G =1 ] - E [ E[ Y_2 | Y_1, W=1, X, G=1 ] | W =1, X, G= 0 ] ] ) }{ p ( W =1| X, G =0 ) } }S(V) \biggr]
\end{align*}

which implies that

\begin{align*}
&\frac{ g w }{ p(G=1) } \biggl[ (y_2 - E[ Y_2 | W=1 , Y_1 , X, G =1]  ) \frac{ P( G =1 | Y_1 , W=1 , X ) }{ P(G=1 | Y_1 , W=1 , X ) } P( W=1 | X , G=1) \\
&+ E [ E[ Y_2 | W= 1,Y_1 , X, G =1 | W =1, X, G= 0 ] ] - \tau    \biggr]  + \\
&\frac{ (1- g) P(G = 1|X)   }{ P(G= 1) P ( G =0 |X) }  \frac{ w ( E [ Y _ 2 | Y_1,W =1 , X, G =1 ] - E [ E[ Y_2 | Y_1, W=1, X, G=1 ] | W =1, X, G= 0 ] ] ) }{ p (W =1| X, G =0 ) }
\end{align*}

We will use <cit.> useful Proposition A.1 as a lemma.
Slightly adjsuting notation from the original[we will use $\bar{w}$ to denote the treatment status w only in this proof], define the notation

\begin{align*}
\psi_0(w)=\psi\left(w, \tau_0, \nu_0, \delta_0, \alpha_0, \eta_0\right)
\end{align*}

\begin{align*}
\psi(w, \tau, \nu, \delta, \alpha, \eta)=\nu(w)+\alpha(w)\{y-\delta(w)\}+\eta(w)\{\delta(w)-\nu(w)\}-\tau
\end{align*}

\begin{align*}
\begin{aligned}
\tau_0 & =\mathbb{E}\left[\nu_0(W)\right] \\
\nu_0(W) & =\int m\left(W_1, w_2, \gamma_0\right) \mathrm{d} \mathbb{P}\left(w_2 \mid W_1\right) \\
\delta_0(W) & =m\left(W_1, W_2, \gamma_0\right)
\end{aligned}
\end{align*}

and $\left(\alpha_0, \eta_0\right)$ are appropriate balancing weights.
The notation for Gateaux differentiation also follows <cit.>

Let $s(w), t(w), u(w), v(w)$ be functions and let $\theta, \zeta \in \mathbb{R}$ be scalars. The Gateaux derivative of $\psi(w, \tau, \nu, \delta, \alpha, \eta)$ with respect to its argument $\nu$ in the direction $s$ is

\begin{align*}
\left\{\partial_\nu \psi(w, \tau, \nu, \delta, \alpha, \eta)\right\}(u)=\left.\frac{\partial}{\partial \theta} \psi(w, \tau, \nu+\theta s, \delta, \alpha, \eta)\right|_{\theta=0}
\end{align*}

The cross derivative of $\psi(w, \tau, \nu, \delta, \alpha, \eta)$ with respect to its arguments $(\nu, \delta)$ in the directions $(s, t)$ is

\begin{align*}
\left\{\partial_{\nu, \delta}^2 \psi(w, \tau, \nu, \delta, \alpha, \eta)\right\}(s, t)=\left.\frac{\partial^2}{\partial \theta \partial \zeta} \psi(w, \tau, \nu+\theta s, \delta+\zeta t, \alpha, \eta)\right|_{\theta=0, \zeta=0} .
\end{align*}

\begin{align*}
\begin{aligned}
\left\{\partial_\nu \psi(w, \tau, \nu, \delta, \alpha, \eta)\right\}(s) & =s(w)-\eta(w) s(w) \\
& =s(w)\{1-\eta(w)\} \\
\left\{\partial_\delta \psi(w, \tau, \nu, \delta, \alpha, \eta)\right\}(t) & =-\alpha(w) t(w)+\eta(w) t(w) \\
& =t(w)\{\eta(w)-\alpha(w)\} \\
\left\{\partial_\alpha \psi(w, \tau, \nu, \delta, \alpha, \eta)\right\}(u) & =u(w)\{y-\delta(w)\} \\
\left\{\partial_\eta \psi(w, \tau, \nu, \delta, \alpha, \eta)\right\}(v) & =v(w)\{\delta(w)-\nu(w)\} \\
\end{aligned}
\end{align*}

For any ( $s, t, u, v)$,

\begin{align*}
\begin{aligned}
\mathbb{E}\left\{\partial_\nu \psi_0(W)\right\}(s) & =\mathbb{E}\left[s(W)\left\{1-\eta_0(W)\right\}\right] \\
\mathbb{E}\left\{\partial_\gamma \psi_0(W)\right\}(t) & =\mathbb{E}\left[t(W)\left\{\eta_0(W)-\alpha_0(W)\right\}\right] \\
\mathbb{E}\left\{\partial_\alpha \psi_0(W)\right\}(u) & =\mathbb{E}\left[u(W)\left\{Y-\delta_0(W)\right\}\right] \\
\mathbb{E}\left\{\partial_\eta \psi_0(W)\right\}(v) & =\mathbb{E}\left[v(W)\left\{\delta_0(W)-\nu_0(W)\right\}\right]
\end{aligned}
\end{align*}

From Lemma <ref>, to show that our moment function satisfies Neyman-orthogonality, we just have to show the four equalities laid out there with

\begin{align*}
s=\nu-\nu_0 ,\\
t=\delta-\delta_0 ,\\
u=\alpha-\alpha_0 ,\\
v=\eta-\eta_0 .
\end{align*}

In the latent confounding model,

\begin{align*}
\begin{aligned}
\gamma_0(y_1,\bar{w}, x, g) & =\mathbb{E}[Y_2 \mid Y_1=y_1,\bar{W}=\bar{w}, X=x, G=g] \\
\mathbb{P}(y_1 \mid \bar{w}, x, g) & =\mathbb{P}(Y_1 = y_1, \mid \bar{W}=\bar{w}, X=x, G=g)
\end{aligned}
\end{align*}

\begin{align*}
\begin{aligned}
\nu_0(W) & =\int \gamma_0(y_1,\bar{w}, X, 1) \mathrm{\bar{w}} \mathbb{P}(y_1 \mid \bar{w}, X, 0) \\
\delta_0(W) & =\gamma_0(Y_1, \bar{W},X, 1) \\
\alpha_0(W) & =\frac{1_{G=1}}{\mathbb{P}(G=1 \mid Y_1, X)} \frac{\mathbb{P}(\bar{w} \mid Y_1, X, G=0) \mathbb{P}(G=0 \mid Y_1, X)}{\mathbb{P}(\bar{w} \mid X, G=0) \mathbb{P}(G=0 \mid X)} \\
\eta_0(W) & =\frac{1_{G=0} 1_{\bar{W}=\bar{w}}}{\mathbb{P}(\bar{w} \mid X, G=0) \mathbb{P}(G=0 \mid X)} .
\end{aligned}
\end{align*}

I will only show how the first Neyman-ortogonality equation,$ \mathbb{E}\left[s(W)\left\{1-\eta_0(W)\right\}\right]$ can be proven, because our proof suggest how the surrogate and the latent unconfounding are essentially the same
Specifically, note that

\begin{align*}
& =\mathbb{E}[s(X)\{1-\mathbb{E}[\frac{1_{G=0} 1_{\bar{W}=\bar{w}}}{\mathbb{P}(\bar{w} \mid X, G=0) \mathbb{P}(G=0 \mid X)} \mid X]\}]
\end{align*}

Then, finally note that,

\begin{align*}
\mathbb{E}\left[\frac{1_{G=0} 1_{\bar{W}=\bar{w}}}{\mathbb{P}(\bar{w} \mid X, G=0) \mathbb{P}(G=0 \mid X)} \mid X\right] & =\frac{\mathbb{E}\left[1_{G=0} 1_{\bar{w}=\bar{w}} \mid X\right]}{\mathbb{P}(\bar{w} \mid X, G=0) \mathbb{P}(G=0 \mid X)} \\
& =\frac{\mathbb{P}(\bar{w}, G=0 \mid X)}{\mathbb{P}(\bar{w} \mid X, G=0) \mathbb{P}(G=0 \mid X)} \\
& =1 .
\end{align*}

proving the result.

Proof. We may assume $\mathcal{M}_4<\infty$ without loss of generality. Throughout the proof, we use the same notation as in the proof of Proposition 4 in <cit.> Also, for two real numbers $a, b$, we write $a \lesssim b$ if there is a universal constant $C>0$ such that $a \leq C b$.
By the proof of Proposition 4 in <cit.>, we have

\begin{aligned}
&\left|\mathrm{P}\left(\widehat{m}_X \in A\right)-\mathrm{P}\left(\widehat{m}_W \in A\right)\right| \leq \frac{1}{\underline{\sigma}}(\sqrt{2 \log M}+2)(\delta+2 \Xi(M, \beta)) \\
&+\left|\mathrm{E}\left[\widetilde{f}_{\beta, \delta, A, \Xi(M, \beta)}(\widetilde{X})\right]-\mathrm{E}\left[\widetilde{f}_{\beta, \delta, A, \Xi(M, \beta)}(\widetilde{W})\right]\right| . \\
\end{aligned}

Take $\beta=\delta^{-1} \log M$ so that $\Xi(M, \beta)=\beta^{-1} \log M=\delta$. Below we prove

\begin{align}
\label{eq:mestkoike1}
\left|\mathrm{E}\left[\widetilde{f}_{\beta, \delta, A, \Xi(M, \beta)}(\widetilde{X})\right]-\mathrm{E}\left[\widetilde{f}_{\beta, \delta, A, \Xi(M, \beta)}(\widetilde{W})\right]\right| \lesssim \frac{\log ^2 M}{\underline{\sigma} \delta} \sqrt{\frac{\mathcal{M}_4}{n}}+\frac{(\log M)^{7 / 2}}{\underline{\sigma} \delta^3} \frac{\mathcal{M}_4}{n}
\end{align}

Combining these two estimates gives

\left|\mathrm{P}\left(\widehat{m}_X \in A\right)-\mathrm{P}\left(\widehat{m}_W \in A\right)\right| \lesssim \frac{\sqrt{\log M}}{\underline{\sigma}} \delta+\frac{\log ^2 M}{\underline{\sigma} \delta} \sqrt{\frac{\mathcal{M}_4}{n}}+\frac{(\log M)^{7 / 2}}{\underline{\sigma} \delta^3} \frac{\mathcal{M}_4}{n} .

Taking $\delta=\left(\mathcal{M}_4(\log M)^3 / n\right)^{1 / 4}$, we obtain the desired result.
Step 1. Write $\widetilde{f}=\widetilde{f}_{\beta, \delta, A, \delta}$ for simplicity. Define the function $\psi: \mathbb{R}^M \rightarrow \mathbb{R}$ as

\psi(x)=\int_0^1 \frac{1}{2 t} \mathrm{E}[\widetilde{f}(\sqrt{t} x+\sqrt{1-t} \widetilde{W})-\widetilde{f}(\widetilde{W})] d t, \quad x \in \mathbb{R}^M .

$\psi$ is a solution to the following Stein equation (cf. Lemma 1 in <cit.>):

\begin{align}
\label{eq:mestkoike2}
\widetilde{f}(x)-\mathrm{E}[\widetilde{f}(\widetilde{W})]=x \cdot \nabla \psi(x)-\sum_{j, k=1}^M \Sigma_X^{j k} \partial_{j k} \psi(x) .
\end{align}

Next, note that $g_\delta^{\prime}(z)=0$ if $z \notin[-\delta, 0]$ by Eq.(19) in [3]. Hence, with $\mu=\mathrm{E}[\bar{X}]$, we have $\partial_j \widetilde{f}(x)=0$ if $h_{\beta, A}(x+\mu)-h_{\beta, A^c}(x+\mu)+\delta \notin[-\delta, 0]$. Since

\begin{aligned}
\max _{m \in A}\left(x_m+\mu_m\right)-\max _{m \in A^c}\left(x_m+\mu_m\right) & \leq h_{\beta, A}(x+\mu)-h_{\beta, A^c}(x+\mu)+\delta \\
& \leq \max _{m \in A}\left(x_m+\mu_m\right)-\max _{m \in A^c}\left(x_m+\mu_m\right)+2 \delta
\end{aligned}

by Eq.(18) in <cit.>, we obtain for all $j_1, \ldots, j_u \in[M]$

\partial_{j_1, \ldots, j_u} \widetilde{f}(x)=\partial_{j_1, \ldots, j_u} \widetilde{f}(x) H^\mu(x ; 3 \delta)

H^y(Y ; x)=1\left\{-x \leq \max _{m \in A}\left(Y_m+y_m\right)-\max _{m \in A^c}\left(Y_m+y_m\right) \leq x\right\} \text {, for all } x \geq 0 \text { and } Y \in \mathbb{R}^M .

One can easily check that

H^y(a Y+b ; x)=H^{(y+b) / a}(Y ; x / a) \text { for any } a>0 \text { and } b \in \mathbb{R}^M .

Hence we obtain for all $j, k \in[M]$

\begin{aligned}
\partial_{j k} \psi(x) &=\frac{1}{2} \int_0^1 \mathrm{E}\left[\partial_{j k} \widetilde{f}(\sqrt{t} x+\sqrt{1-t} \widetilde{W})\right] d t \\
&=\frac{1}{2} \int_0^1 \mathrm{E}\left[H^{y(x, t)}\left(\widetilde{W}, \frac{3 \delta}{\sqrt{1-t}}\right) \partial_{j k} \widetilde{f}(\sqrt{t} x+\sqrt{1-t} \widetilde{W})\right] d t,
\end{aligned}

y(x, t):=\frac{\mu+\sqrt{t} x}{\sqrt{1-t}} .

By Lemma 6 in <cit.>, we have for all $x \in \mathbb{R}^M$

\sum_{j, k=1}^p\left|\partial_{j k} \widetilde{f}(x)\right| \lesssim \delta^{-2} \log M .

Thus we obtain

\begin{align}
\sum_{j, k=1}^p\left|\partial_{j k} \psi(x)\right| & \lesssim \delta^{-2}(\log M) \int_0^1 \mathrm{E}\left[H^{y(x, t)}\left(\widetilde{W}, \frac{3 \delta}{\sqrt{1-t}}\right)\right] d t \\
& \lesssim \underline{\sigma}^{-1} \delta^{-1}(\log M)^{3 / 2} \int_0^1 \frac{1}{\sqrt{1-t}} d t \lesssim \underline{\sigma}^{-1} \delta^{-1}(\log M)^{3 / 2},
\label{eq:mestkoike3}
\end{align}

where the second estimate follows from Lemma 5 in <cit.>. We can similarly prove

\begin{align}
\label{eq:mestkoike4}
\sum_{j, k, l, r=1}^M\left|\partial_{j k l r} \psi(x)\right| \lesssim \underline{\sigma}^{-1} \delta^{-3}(\log M)^{7 / 2} .
\end{align}

Step 2. Set $\xi_i=\widetilde{X}_i / \sqrt{n}$ for $i=1, \ldots, n$, and let $\xi^{\prime}=\left(\xi_1^{\prime}, \ldots, \xi_n^{\prime}\right)$ be an independent copy of $\xi=$ $\left(\xi_1, \ldots, \xi_n\right)$ which is independent of $\widetilde{W}$ as well. Also, let $I$ be a uniform random variable on $\{1, \ldots, n\}$ independent of everything else. Set $\eta_i:=\xi_i^{\prime}-\xi_i$ for $i=1, \ldots, n$ and define $\widetilde{X}^{\prime}:=\widetilde{X}+\eta_I$. It is not difficult to check that $\left(\widetilde{X}, \widetilde{X}^{\prime}\right)$ is an exchangeable pair and satisfies

\begin{align}
\label{eq:mestkoike5}\mathrm{E}\left[\widetilde{X}^{\prime}-\widetilde{X} \mid \widetilde{X}\right] &=-\frac{1}{n} \widetilde{X}, \\
\label{eq:mestkoike6}
\mathrm{E}\left[\left(\widetilde{X}^{\prime}-\widetilde{X}\right)\left(\widetilde{X}^{\prime}-\widetilde{X}\right)^{\top} \mid \widetilde{X}\right] &=\frac{2}{n} \Sigma_X+\frac{1}{n} V,
\end{align}

V=\sum_{i=1}^n\left(\xi_i \xi_i^{\top}-\mathrm{E}\left[\xi_i \xi_i^{\top}\right]\right)

See the proof of e.g., <cit.>. By exchangeability we have, with $D:=\widetilde{X}^{\prime}-\widetilde{X}=\eta_I$,

\begin{aligned}
0 &=\frac{n}{2} \mathrm{E}\left[D \cdot\left(\nabla \psi\left(\tilde{X}^{\prime}\right)+\nabla \psi(\widetilde{X})\right)\right] \\
&=\mathrm{E}\left[\frac{n}{2} D \cdot\left(\nabla \psi\left(\tilde{X}^{\prime}\right)-\nabla \psi(\widetilde{X})\right)+n D \cdot \nabla \psi(\widetilde{X})\right] \\
&=\mathrm{E}\left[\frac{n}{2} \sum_{j, k=1}^M D_j D_k \partial_{j k} \psi(\widetilde{X})+n D \cdot \nabla \psi(\widetilde{X})\right]+R,
\end{aligned}

R=\frac{n}{2} \sum_{j, k, l=1}^M \mathrm{E}\left[D_j D_k D_l U \partial_{j k l} \psi(\tilde{X}+(1-U) D)\right],

and $U$ is a standard uniform random variable independent of everything else. Combining this with (<ref>), (<ref>), (<ref>) and (<ref>), we obtain

\begin{align}
\label{eq:mestkoike7}
|\mathrm{E}[\widetilde{f}(\widetilde{X})]-\mathrm{E}[\widetilde{f}(\widetilde{W})]| \lesssim \underline{\sigma}^{-1} \delta^{-1}(\log M)^{3 / 2} \mathrm{E}\left[\max _{1 \leq j, k \leq M}\left|V_{j k}\right|\right]+|R| .
\end{align}

We have by Nemirovski's inequality [<cit.>, Lemma 14.24] and Jensen's inequality

\begin{align}
\notag \mathrm{E}\left[\max _{1 \leq j, k \leq p}\left|V_{j k}\right|\right] & \lesssim \sqrt{\log M} \mathrm{E}\left[\max _{1 \leq j, k \leq M} \sqrt{\sum_{i=1}^n \mathrm{E}\left[\xi_{i j}^2 \xi_{i k}^2\right]}\right] \\
& \lesssim \sqrt{\log M} \sqrt{\mathrm{E}\left[\max _{1 \leq j \leq M} \sum_{i=1}^n \xi_{i j}^4\right]} \leq \sqrt{\log M} \sqrt{\frac{\mathcal{M}_4}{n}} .
\label{eq:mestkoike8}
\end{align}

Step 3. To estimate $R$, we rewrite it as follows. By exchangeability, we have

\begin{align}
&\mathrm{E}\left[D_j D_k D_l U \partial_{j k l} \psi(\widetilde{X}+(1-U) D)\right] \\
&=-\mathrm{E}\left[D_j D_k D_l U \partial_{j k l} \psi\left(\widetilde{X}^{\prime}-(1-U) D\right)\right] \\
&=-\mathrm{E}\left[D_j D_k D_l U \partial_{j k l} \psi(\widetilde{X}+U D)\right]
\end{align}

Hence we obtain

\begin{align}
R &=\frac{n}{4} \sum_{j, k, l=1}^M \mathrm{E}\left[D_j D_k D_l U\left\{\partial_{j k l} \psi(\tilde{X}+(1-U) D)-\partial_{j k l} \psi(\tilde{X}+U D)\right\}\right] \\
&=\frac{n}{4} \sum_{j, k, l, r-1}^M \mathrm{E}\left[D_j D_k D_l D_r U(1-2 U) \partial_{j k l r} \psi\left(\widetilde{X}+U D+U^{\prime}(1-2 U) D\right)\right],
\end{align}

where $U^{\prime}$ is a standard uniform random variable independent of everything else. Therefore, we obtain by (<ref>)

\begin{align}
\label{eq:mestkoike9}
|R| \lesssim \underline{\sigma}^{-1} \delta^{-3}(\log M)^{7 / 2} \cdot n \mathrm{E}\left[\|D\|_{\infty}^4\right]=\frac{(\log M)^{7 / 2}}{\underline{\sigma} \delta^3} \sum_{i=1}^n \mathrm{E}\left[\left\|\eta_i\right\|_{\infty}^4\right]=\frac{(\log M)^{7 / 2}}{\underline{\sigma} \delta^3} \frac{\mathcal{M}_4}{n} .
\end{align}

Combining (<ref>), (<ref>) and (<ref>), we obtain (<ref>).