Appendix A. Technical Lemmas
The first two lemmas below concern properties of the \(\xi _{b,i}\)’s and their sum \(W_b\).
Lemma A.1
(Bound on expectation of \(\xi _{b, i}\)) Let \(\xi _{b, i} = \xi _i I(|\xi _i| \le 1) + 1 I(\xi _i >1) - 1 I(\xi _i <-1)\) with \({\mathbb {E}}[\xi _i] = 0\). Then
$$\begin{aligned} \left| {\mathbb {E}}[ \xi _{b, i}]\right| \le {\mathbb {E}}[ \xi _i^2I(|\xi _i| > 1) ] \le {\mathbb {E}}[ \xi _i^2 ] \end{aligned}$$
Proof of Lemma A.1
$$\begin{aligned} \left| {\mathbb {E}}[ \xi _{b, i}]\right|&= \left| {\mathbb {E}}[ (\xi _i - 1) I(\xi _i> 1) + (\xi _i + 1) I(\xi _i < - 1) ]\right| \\&\le {\mathbb {E}}[ (|\xi _i| - 1) I(|\xi _i|> 1)] \le {\mathbb {E}}[ |\xi _i | I(|\xi _i|> 1)] \le {\mathbb {E}}[ |\xi _i |^2 I(|\xi _i|> 1)] \le {\mathbb {E}}[ \xi _i^2 ]. \end{aligned}$$
\(\square \)
Lemma A.2
(Bennett’s inequality for a sum of censored random variables) Let \(\xi _1, \dots , \xi _n\) be independent random variables with \({\mathbb {E}}[\xi _i] = 0\) for all \(i = 1, \dots , n\) and \(\sum _{i=1}^n {\mathbb {E}}[\xi _i^2] \le 1\), and define \(\xi _{b, i} = \xi _i I(|\xi _i| \le 1) + 1 I(\xi _i >1) - 1 I(\xi _i <-1)\). For any \(t >0\) and \(W_b = \sum _{i=1}^n \xi _{b, i}\), we have
$$\begin{aligned} {\mathbb {E}}[e^{t W_b}] \le \exp \left( e^{2t}/4 - 1/4 + t/2\right) \end{aligned}$$
Proof of Lemma A.2
Note that, by Lemma A.1,
$$\begin{aligned} {\mathbb {E}}[e^{t W_b}] = {\mathbb {E}}[e^{t (W_b- {\mathbb {E}}[W_b])}] e^{t {\mathbb {E}}[W_b]} \le {\mathbb {E}}[e^{t \sum _{i=1}^n(\xi _{b,i} - {\mathbb {E}}[\xi _{b, i}])}] e^t. \end{aligned}$$
Moreover, by the standard Bennett’s inequality [3, Lemma 8.1],
$$\begin{aligned} {\mathbb {E}}[e^{t \sum _{i=1}^n(\xi _{b,i} - {\mathbb {E}}[\xi _{b, i}])}] \le \exp \left( 4^{-1} (e^{2t} - 1 - 2t)\right) . \end{aligned}$$
\(\square \)
The next lemmas concern properties of the solution to the Stein equation, \(f_x\) in (2.3). It is customary to define its derivative at x as \(f_x'(x) \equiv xf_x(x) +{\bar{\Phi }}(x)\) so the Stein equation (2.4) is valid for all w. Moreover, we define
$$\begin{aligned} g_x(w) = (wf_x(w))' = f_x(w)+ wf_x'(w). \end{aligned}$$
(A.1)
Precisely,
$$\begin{aligned} f_x' (w)= & {} {\left\{ \begin{array}{ll} \left( \sqrt{2 \pi }w e^{w^2/2} \Phi (w) + 1\right) {\bar{\Phi }}(x)&{} \text { for }\quad w \le x\\ \left( \sqrt{2 \pi }w e^{w^2/2} {\bar{\Phi }}(w) - 1\right) \Phi (x) &{} \text { for }\quad w > x \end{array}\right. }; \end{aligned}$$
(A.2)
$$\begin{aligned} g_x(w)= & {} {\left\{ \begin{array}{ll} \sqrt{2 \pi }{\bar{\Phi }}(x) \left( (1 + w^2) e^{w^2/2} \Phi (w) + \frac{w}{\sqrt{2\pi }}\right) &{} \text { for }\quad w \le x\\ \sqrt{2\pi }\Phi (x) \left( (1 + w^2) e^{w^2/2} {\bar{\Phi }}(w) - \frac{w}{\sqrt{2 \pi }} \right) &{}\text { for }\quad w > x \end{array}\right. }. \end{aligned}$$
(A.3)
Lemma A.3
(Uniform bounds for \(f_x\)) For \(f_x\) and \(f_x'\), the following bounds are true:
$$\begin{aligned} |f_x'(w)| \le 1, \quad \quad 0 < f_x(w) \le 0.63 \quad \text { and }\quad 0 \le g_x(w) \quad \text { for all } \quad w, x \in {\mathbb {R}}. \end{aligned}$$
Moreover, for any \(x \in [0, 1]\), \(g_x(w) \le 2.3\) for all \(w \in {\mathbb {R}}\).
Lemma A.4
(Nonuniform bounds for \(f_x\) when \(x \ge 1\)) For \(x\ge 1\), the following are true:
$$\begin{aligned} f_x (w) \le {\left\{ \begin{array}{ll} 1.7 e^{-x} &{} \quad \textrm{for }\quad w \le x - 1\\ 1/x &{} \quad \textrm{for }\quad x - 1< w \le x \\ 1/w &{} \quad \textrm{for }\quad x < w \end{array}\right. } \end{aligned}$$
(A.4)
and
$$\begin{aligned} | f_x'(w)| \le {\left\{ \begin{array}{ll} e^{1/2 -x } &{} \quad \textrm{for }\quad w \le x - 1\\ 1 &{} \quad \textrm{for }\quad x - 1 < w \le x \\ (1 + x^2)^{-1} &{} \quad \textrm{for }\quad w > x \end{array}\right. }. \end{aligned}$$
(A.5)
Moreover, \(g_x(w) \ge 0\) for all \(w \in {\mathbb {R}}\),
$$\begin{aligned} g_x(w) \le {\left\{ \begin{array}{ll} 1.6 \;{\bar{\Phi }}(x) &{} \quad \textrm{for }\quad w \le 0\\ 1/w &{} \quad \textrm{for }\quad w > x \end{array}\right. }, \end{aligned}$$
(A.6)
and \(g_x(w)\) is increasing for \(0 \le w \le x\) with
$$\begin{aligned} g_x(x - 1) \le x e^{1/2 -x} \;\ \text { and } \;\ g_x(x) \le x+ 2. \end{aligned}$$
We remark that the nonuniform bounds in Lemma A.4 refine the ones previously collected in Shao et al. [9, Lemma 5.3]; as an aside, a property analogous to (A.5) has been incorrectly stated in Shao et al. [9] without the absolute signs \(|\cdot |\) around \(f_x'(w)\). The proofs below repeatedly use the well-known inequality [3, p.16 & 38]
$$\begin{aligned} \frac{w e^{-w^2/2}}{ (1 + w^2)\sqrt{2\pi }}\le {\bar{\Phi }}(w)\le \min \left( \frac{1}{2}, \frac{1}{w \sqrt{2 \pi }}\right) e^{-w^2/2} \text { for } w > 0. \end{aligned}$$
(A.7)
Proof of Lemma A.3
The bounds for \(f_x\) and \(f_x'\), and that \(g_x(w) \ge 0\), are well known; see Chen et al. [3, Lemma 2.3]. We will show that \(g_x\) in (A.3) is less than 2.3 when \(x \in [0,1]\). Using (A.7), for \(w > x\), we have
$$\begin{aligned} g_x(w)\le & {} \sqrt{2\pi }\Phi (x) \left( (1 + w^2) e^{w^2/2} {\bar{\Phi }}(w) - \frac{w}{\sqrt{2 \pi }} \right) \\\le & {} \sqrt{2\pi }\Phi (x) \left( \frac{1}{2} + \frac{w}{\sqrt{2\pi }}- \frac{w}{\sqrt{2 \pi }}\right) \le \frac{\sqrt{2\pi } \Phi (x)}{2} \le 2. \end{aligned}$$
For \(0 \le w \le x\),
$$\begin{aligned} g_x(w)&= \sqrt{2 \pi }{\bar{\Phi }}(x) \left( (1 + w^2) e^{w^2/2} \Phi (w) + \frac{w}{\sqrt{2\pi }}\right) \\&\le \sqrt{2 \pi }{\bar{\Phi }}(x) \left( (1 + x^2 ) e^{x^2/2} \Phi (x) + \frac{x}{\sqrt{2 \pi }} \right) \\&\le \left\{ \left( \frac{\sqrt{2\pi }}{2} + x\right) \Phi (x)+ \frac{e^{-x^2/2}}{\sqrt{2\pi }}\right\} \vee \left( \sqrt{2\pi }{\bar{\Phi }}(0)\cdot \Phi (0) \right) \\&\le \left\{ \left( \frac{\sqrt{2 \pi }}{2} + 1\right) \Phi (1) + \frac{1}{\sqrt{2\pi }} \right\} \vee 0.63 \le 2.3. \end{aligned}$$
For \(w < 0\),
$$\begin{aligned}{} & {} \sqrt{2 \pi }{\bar{\Phi }}(x) \left( (1 + w^2) e^{w^2/2} \Phi (w) + \frac{w}{\sqrt{2\pi }}\right) \le \sqrt{2 \pi }{\bar{\Phi }}(x) \left( \frac{1}{2} + \frac{|w|}{\sqrt{2\pi }}- \frac{|w|}{\sqrt{2 \pi }} \right) \le 1.26. \end{aligned}$$
\(\square \)
Proof of Lemma A.4
Proof of (A.4) by investigating (2.3): When \(w\le 0\), by (A.7), \(x^2 \ge 2x - 1\), and the symmetry of \(\phi (\cdot )\), we have that
$$\begin{aligned} f_x(w) \le e^{w^2/2} \Phi (w) \frac{e^{-x^2/2}}{x} \le \frac{e^{-x^2/2}}{2x} \le \frac{e^{-x +1/2}}{2} \le 0.9 e^{-x}. \end{aligned}$$
When \(0 < w \le x -1\), by (A.7), we have
$$\begin{aligned} f_x (w) \le e^{(x-1)^2/2} \Phi (w) \frac{e^{-x^2/2}}{x} = \Phi (w) e^{-x + 1/2} \le 1.7 e^{-x}. \end{aligned}$$
When \(x - 1 < w \le x\), by (A.7), we have
$$\begin{aligned} f_x (w)\le \frac{e^{(w^2 - x^2)/2} \Phi (w)}{x} \le \frac{1}{x}. \end{aligned}$$
When \(w > x\), by (A.7), we have
$$\begin{aligned} f_x (w)\le \frac{\Phi (x)}{w} \le \frac{1}{w}. \end{aligned}$$
Proof of (A.5) by investigating (A.2): When \(w \le 0\), by the symmetry of \(\phi (\cdot )\), (A.7) and \(x^2 \ge 2x - 1\), we have
$$\begin{aligned} 0 = 0 \cdot {\bar{\Phi }}(x) \le f_x'(w) \le \left( \frac{1}{ 1 + w^2} \right) \frac{e^{-x +1/2}}{\sqrt{2 \pi }} \le 0.4 e^{1/2-x}. \end{aligned}$$
When \( 0 < w \le x - 1\), by (A.7) and \(x^2 \ge 2x - 1\),
$$\begin{aligned} 0 \le f_x'(w) \le \left( \sqrt{2\pi } (x-1) e^{\frac{(x-1)^2}{2}} +1\right) \frac{e^{-x^2/2}}{x\sqrt{2 \pi }} \le \left( \frac{x-1}{x} + \frac{1}{ x\sqrt{2 \pi }}\right) e^{ 1/2 - x} \le e^{1/2-x}, \end{aligned}$$
as \( \left( \frac{x-1}{x} + \frac{1}{ x\sqrt{2 \pi }}\right) \) is increasing as a function in x on \([1, \infty )\). When \(x- 1 < w \le x\), by (A.7) we have
$$\begin{aligned} 0 \le f_x'(w) = \Phi (w)\underbrace{\sqrt{2 \pi } w e^{w^2/2}{\bar{\Phi }}(x) }_{\le 1} + {\bar{\Phi }}(x) \le 1. \end{aligned}$$
When \( w > x\), since \(\sqrt{2\pi }w e^{w^2/2} {\bar{\Phi }}(w) \le 1\) by (A.7), hence \(f'_x(w) \le 0\). Moreover, by applying (A.7) again, we have
$$\begin{aligned} \frac{-1}{x^2 +1}\le \left( \frac{w^2}{w^2 +1} - 1\right) \Phi (x)\le f'_x(w) \le 0. \end{aligned}$$
Proof of (A.6) by investigating (A.3): When \(w < 0\), by the symmetry of \(\phi \) and (A.7),
$$\begin{aligned} 0 = \sqrt{2\pi }{\bar{\Phi }}(x) \cdot 0 \le g_x(w) \le \left( \min \left( \frac{1 + w^2}{ |w|}, \frac{ (1 + w^2)\sqrt{2 \pi }}{2} \right) + w \right) {\bar{\Phi }}(x) \le 1.6 {\bar{\Phi }}(x), \end{aligned}$$
where the last inequality uses the facts that \(\frac{ (1 +w^2)\sqrt{2 \pi }}{2} +w \le 1.6\) for \(w \in [-1, 0]\) and that \(\frac{1 + w^2}{|w|} + w = 1/|w|^2 \le 1\) for \(w < -1\). When \(w > x\), by (A.7),
$$\begin{aligned} 0 \le \sqrt{2\pi } \Phi (x) \cdot 0\le g_x(w) \le \Phi (x) \left( \frac{1 + w^2}{w} - w \right) = \frac{ \Phi (x) }{w} \le 1/w. \end{aligned}$$
When \(0 \le w \le x\), it is easy to see that \(g_x(w)\) is non-negative and increasing in w. Moreover, from (A.7) and \(x^2 \ge 2x - 1\),
$$\begin{aligned} g_x(x-1)&= \sqrt{2\pi } {\bar{\Phi }}(x) \left( (2 + x^2 - 2x) e^{x^2/2 - x +1/2} \Phi (x-1) + \frac{x-1}{\sqrt{2 \pi }} \right) \\&\le \frac{(2 + x^2 - 2x)}{x} e^{1/2 - x} \Phi (x-1) + \frac{x-1}{x \sqrt{2 \pi }} e^{-x^2/2}\\&\le \frac{(4 + 2x^2 - 4x)}{2x} e^{1/2 - x} + \frac{x-1}{2x} e^{1/2 - x}\\&\le \left( x - \frac{3}{2} + \frac{3}{2x}\right) e^{1/2 - x} \le x e^{1/2 -x}. \end{aligned}$$
Lastly, by (A.7), it is easy to see that
$$\begin{aligned} g_x(x)&= \sqrt{2 \pi }{\bar{\Phi }}(x) \left( (1 + x^2) e^{x^2/2} \Phi (x) + \frac{x}{\sqrt{2\pi }}\right) \\&\le \frac{1 +x^2}{x} \Phi (x) + \frac{e^{-x^2/2}}{\sqrt{2 \pi }} \le \left( \frac{1}{x} + x\right) + \frac{1}{2} \le x+ 2 \end{aligned}$$
\(\square \)
Lemma A.5
(Bound on expectation of \(f_x'(W_b^{(i)} +t)\)) Let \(x \ge 1\), \(t \in {\mathbb {R}}\), and \(W_b^{(i)}\) be as defined in Sect. 1 under the assumptions (1.2). Then there exists an absolute constant \(C >0\) such that
$$\begin{aligned} \left| {\mathbb {E}}[f_x'(W_b^{(i)}+ t)]\right| \le C (e^{ -x } + e^{- x +t}). \end{aligned}$$
Proof of Lemma A.5
From (A.5) in Lemma A.4, we have
$$\begin{aligned} |{\mathbb {E}}[f_x'(W_b^{(i)} + t)]|&\le e^{1/2 -x } + {\mathbb {E}}[I(W_b^{(i)} + t > x - 1)] \\&\le e^{1/2 -x } + e^{1 - x +t}{\mathbb {E}}[e^{W_b^{(i)} }] \end{aligned}$$
and then apply the Bennett inequality in Lemma A.2.\(\square \)
Appendix B. Exponential Randomized Concentration Inequality for a Sum of Censored Variables
Lemma B.1
(Exponential randomized concentration inequality for a sum of censored random variables) Let \(\xi _1, \dots , \xi _n\) be independent random variables with mean zero and finite second moments, and for each \(i =1, \dots , n\), define
$$\begin{aligned} \xi _{b, i} = \xi _i I(|\xi _i| \le 1) + 1 I(\xi _i >1) - 1 I(\xi _i <-1), \end{aligned}$$
an upper-and-lower censored version of \(\xi _i\); moreover, let \(W =\sum _{i=1}^n \xi _i \) and \(W_b = \sum _{i=1}^n \xi _{b, i}\) be their corresponding sums, and \(\Delta _1\) and \(\Delta _2\) be two random variables on the same probability space. Assume there exists \(c_1> c_2 >0\) and \(\delta \in (0, 1/2)\) such that
$$\begin{aligned} \sum _{i=1}^n {\mathbb {E}}[\xi _i^2] \le c_1 \end{aligned}$$
and
$$\begin{aligned} \sum _{i=1}^n {\mathbb {E}}[|\xi _i| \min (\delta , |\xi _i|/2 )] \ge c_2. \end{aligned}$$
Then for any \(\lambda \ge 0\), it is true that
$$\begin{aligned}&{\mathbb {E}}[e^{\lambda W_b} I(\Delta _1 \le W_b \le \Delta _2)] \\&\quad \le \left( {\mathbb {E}}\left[ e^{2\lambda W_b}\right] \right) ^{1/2}\exp \left( - \frac{c_2^2}{16 c_1 \delta ^2}\right) \\&\qquad + \frac{2 e^{\lambda \delta }}{c_2} \Biggl \{ \sum _{i =1}^n {\mathbb {E}} [ |\xi _{b, i}|e^{\lambda W_b^{(i)} } (|\Delta _1 - \Delta _1^{(i)}| + |\Delta _2 - \Delta _2^{(i)}|)] \\&\qquad + {\mathbb {E}}[|W_b|e^{\lambda W_b }(|\Delta _2 - \Delta _1| + 2 \delta )]\\&\qquad + \sum _{i=1}^n \Big |{\mathbb {E}}[ \xi _{b, i}]\Big | {\mathbb {E}}[e^{\lambda W_b^{(i)} }(|\Delta _2^{(i)} - \Delta _1^{(i)}| + 2 \delta )]\Biggr \} , \end{aligned}$$
where \(\Delta _1^{(i)}\) and \(\Delta _2^{(i)}\) are any random variables on the same probability space such that \(\xi _i\) and \((\Delta _1^{(i)}, \Delta _2^{(i)}, W^{(i)}, W_b^{(i)})\) are independent, where \(W^{(i)} = W - \xi _i\) and \(W_b^{(i)} = W_b - \xi _{b, i}\).
In particular, by defining \(\beta _2 \equiv \sum _{i=1}^n {\mathbb {E}}[\xi _i^2 I(|\xi _i|>1)]\) and \(\beta _3 \equiv \sum _{i=1}^n {\mathbb {E}}[\xi _i^3 I(|\xi _i|\le 1)]\), if \(\sum _{i=1}^n {\mathbb {E}}[\xi _i^2] = 1\) and \(\beta _2 + \beta _3 \le 1/2\), one can take
$$\begin{aligned} \delta = \frac{\beta _2 + \beta _3}{4}, \quad c_1 = 1 \text { and } \quad c_2 = \frac{1}{4} \end{aligned}$$
(B.1)
to satisfy the conditions of the inequality.
Proof of Lemma B.1
It suffices to show the lemma under the assumptions that
$$\begin{aligned} \Delta _1 \le \Delta _2 \text { and } \Delta _1^{(i)} \le \Delta _2^{(i)}. \end{aligned}$$
(B.2)
If (B.2) is not true, we can let \(\Delta _1^* = \min (\Delta _1, \Delta _2)\), \(\Delta _2^* = \max (\Delta _1, \Delta _2)\), \({\Delta _1^*}^{(i)} = \min (\Delta _1^{(i)}, \Delta _2^{(i)})\), \({\Delta _2^*}^{(i)} = \max (\Delta _1^{(i)}, \Delta _2^{(i)})\). Then the assumptions in (B.2) can be seen to be not forgoing any generality by noting that \(|\Delta _2^* - \Delta _1^*| = |\Delta _2 - \Delta _1|\) (also \(|{\Delta _2^*}^{(i)} - {\Delta _1^*}^{(i)}| = |\Delta _2^{(i)} - \Delta _1^{(i)}|\)),
$$\begin{aligned} {\mathbb {E}}[e^{\lambda W_b} I(\Delta _1 \le W_b \le \Delta _2 )] \le {\mathbb {E}}[e^{\lambda W_b} I(\Delta _1^* \le W_b \le \Delta _2^* )] \end{aligned}$$
and
$$\begin{aligned} |\Delta _1^* - {\Delta _1^*}^{(i)}| + |\Delta _2^* - {\Delta _2^*}^{(i)}| \le |\Delta _1 - \Delta _1^{(i)}| + |\Delta _2 - \Delta _2^{(i)}|, \end{aligned}$$
(B.3)
where (B.3) is true by the following fact: If we have real numbers \(x_1 \le x_2\) and \(y_1 \le y_2\), it must be that
$$\begin{aligned} |x_1 - y_1| + |x_2 - y_2| \le |x_1 - y_2| + |x_2 - y_1|. \end{aligned}$$
(B.4)
Without loss of generality, one can assume \(x_1 \le y_1\) and simply prove (B.4) by case considerations:
-
(i)
If \(x_1 \le x_2 \le y_1 \le y_2\), then
$$\begin{aligned} |x_1 - y_1| + |x_2 - y_2|&= y_1 - x_1 + y_2 - x_2 \\&= y_2 - x_1 + y_1 - x_2 = |x_1 - y_2| + |x_2 - y_1|. \end{aligned}$$
-
(ii)
If \(x_1 \le y_1 \le x_2 \le y_2\), , then
$$\begin{aligned} |x_1 - y_1| + |x_2 - y_2|&= y_1 - x_1 + y_2 - x_2 \\&\le y_2 - x_1 \le |x_1 - y_2| + |x_2 - y_1|. \end{aligned}$$
-
(iii)
If \(x_1 \le y_1 \le y_2 \le x_2\), , then
$$\begin{aligned} |x_1 - y_1| + |x_2 - y_2|&= \underbrace{y_1 - x_1}_{\le y_2 - x_1} + \underbrace{x_2 - y_2}_{\le x_2 - y_1} \le |x_1 - y_2| + |x_2 - y_1|. \end{aligned}$$
More generally, a fact like (B.4) can be proved by the rearrangement inequality [11, p.78], but the details are omitted here.
Under the working assumptions in (B.2), for \(a< b\), we define the function
$$\begin{aligned} f_{a, b}(w) = {\left\{ \begin{array}{ll} 0 &{} \quad \text { for } w \le a- \delta \\ e^{\lambda w} (w - a + \delta ) &{} \quad \text { for } a -\delta < w \le b + \delta \\ e^{\lambda w} (b - a + 2\delta ) &{} \quad \text { for }w > b + \delta \\ \end{array}\right. }, \end{aligned}$$
which has the property
$$\begin{aligned} |f_{a, b}(w) - f_{a_1, b_1} (w)| \le e^{\lambda w} (|a - a_1| + |b - b_1|) \text { for all } w, \;\ a< b \text { and }a_1 < b_1,\nonumber \\ \end{aligned}$$
(B.5)
as well as
$$\begin{aligned} f_{a,b}'(w) \ge 0 \text { almost surely. } \end{aligned}$$
Moreover, we have
$$\begin{aligned} I_1 + I_2 = {\mathbb {E}}[W_b f_{\Delta _1, \Delta _2} (W_b )] - \sum _{i=1}^n {\mathbb {E}}[ \xi _{b, i}] {\mathbb {E}}[f_{\Delta _1^{(i)}, \Delta _2^{(i)}} ( W_b^{(i)} ) )] \end{aligned}$$
(B.6)
where
$$\begin{aligned} I_1&\equiv \sum _{i=1}^n {\mathbb {E}}\left[ \xi _{b, i}\left( f_{\Delta _1, \Delta _2}(W_b ) - f_{\Delta _1, \Delta _2} (W_b^{(i)}) \right) \right] \text { and }\\ I_2&\equiv \sum _{i=1}^n {\mathbb {E}} \left[ \xi _{b, i} \left( f_{\Delta _1, \Delta _2} ( W_b^{(i)} ) -f_{\Delta _1^{(i)}, \Delta _2^{(i)}} ( W_b^{(i)} ) \right) \right] . \end{aligned}$$
Given the property in (B.5), we have
$$\begin{aligned} |I_2| \le \sum _{i =1}^n {\mathbb {E}} \Big [ |\xi _{b, i}|e^{\lambda W_b^{(i)} } \Big (|\Delta _1 - \Delta _1^{(i)}| + |\Delta _2 - \Delta _2^{(i)}|\Big )\Big ]. \end{aligned}$$
(B.7)
Now we estimate \(I_1\), by first rewriting it as
$$\begin{aligned} I_1&= \sum _{i=1}^n{\mathbb {E}}\left[ \xi _{b, i} \Big (f_{\Delta _1, \Delta _2}(W_b ) - f_{\Delta _1, \Delta _2}(W_b^{(i)} ) \Big ) \right] \\&= \sum _{i=1}^n {\mathbb {E}}\left[ \xi _{b, i} \int ^0_{-\xi _{b, i} } f_{\Delta _1, \Delta _2}'(W_b + t) \hbox {d}t \right] = \sum _{i=1}^n {\mathbb {E}}\left[ \int _{-\infty }^\infty f_{\Delta _1, \Delta _2}'(W_b + t) {\hat{K}}_i(t) \hbox {d}t \right] , \end{aligned}$$
where
$$\begin{aligned} {\hat{K}}_i(t) \equiv \xi _{b, i}\{I(- \xi _{b, i} \le t \le 0) - I(0 < t \le - \xi _{b, i} )\}. \end{aligned}$$
Note that \(\xi _{b, i}\) and \(I(-\xi _{b, i} \le t \le 0) - I(0 < t \le - \xi _{b, i} )\) have the same sign, and it is also true that \(0 \le {\tilde{K}}_i(t) \le {\hat{K}}_i(t)\) where
$$\begin{aligned} {\tilde{K}}_i(t) = \xi _{b, i} \{I(- \xi _{b, i}/2 \le t \le 0) - I(0 < t \le - \xi _{b, i} /2)\} \end{aligned}$$
By the fact that \(f_{\Delta _1, \Delta _2}'(w) \ge e^{\lambda w}\ge 0\) for all \(w \in (\Delta _1 - \delta , \Delta _2 + \delta )\), one can lower bound \(I_1\) as
$$\begin{aligned} I_1&\ge \sum _{i=1}^n {\mathbb {E}}\left[ \int _{-\infty }^\infty f_{\Delta _1, \Delta _2}'(W_b + t) {\tilde{K}}_i(t) \hbox {d}t \right] \\&\ge \sum _{i=1}^n {\mathbb {E}}\left[ \int _{|t|\le \delta }I(\Delta _1 \le W_b \le \Delta _2) f_{\Delta _1, \Delta _2}'(W_b + t) {\tilde{K}}_i(t) \hbox {d}t \right] \\&\ge \sum _{i=1}^n {\mathbb {E}}\left[ I(\Delta _1 \le W_b \le \Delta _2) e^{\lambda (W_b - \delta )} |\xi _{b, i}| \min (\delta , |\xi _{b, i}|/2)\right] \\&= {\mathbb {E}}\left[ I(\Delta _1 \le W_b \le \Delta _2) e^{\lambda (W_b - \delta )} \left( \sum _{i=1}^n \eta _i\right) \right] , \end{aligned}$$
where \(\eta _i = |\xi _i| \min (\delta , |\xi _i|/2)\), noting that given \(\delta < 1/2\), \(\min (\delta , |\xi _i|/2) = \min (\delta , |\xi _{b, i}|/2)\) due to the censoring definition of \(\xi _{b, i}\). Hence, continuing, we can further lower bound \(I_1\) as
$$\begin{aligned} I_1&\ge (c_2/2) {\mathbb {E}}\left[ e^{\lambda (W_b - \delta )} I(\Delta _1 \le W_b\le \Delta _2) I\left( \sum _{i=1}^n \eta _i \ge c_2/2\right) \right] \nonumber \\&\ge \frac{c_2}{2 e^{\lambda \delta }} \left\{ {\mathbb {E}}\left[ e^{\lambda W_b } I(\Delta _1 \le W_b \le \Delta _2) \right] - {\mathbb {E}}\left[ e^{\lambda W_b } I\left( \sum _{i=1}^n \eta _i< c_2/2\right) \right] \right\} \nonumber \\&\ge \frac{c_2}{2 e^{\lambda \delta }}\left\{ {\mathbb {E}}\left[ e^{\lambda W_b } I(\Delta _1 \le W_b \le \Delta _2)\right] - \sqrt{{\mathbb {E}}\left[ e^{2\lambda W_b }\right] P\left( \sum _{i=1}^n \eta _i < c_2/2\right) }\right\} \nonumber \\&\ge \frac{c_2}{2 e^{\lambda \delta }}\left\{ {\mathbb {E}}[ e^{\lambda W_b } I(\Delta _1 \le W_b \le \Delta _2)] -\left( {\mathbb {E}}\left[ e^{2\lambda W_b}\right] \right) ^{1/2}\exp \left( - \frac{c_2^2}{16 c_1 \delta ^2}\right) \right\} , \end{aligned}$$
(B.8)
where the last inequality comes from the sub-Gaussian lower tail bound for sum of non-negative random variables [14, Theorem 2.19],
$$\begin{aligned} P\left( \sum _{i=1}^n \eta _i < c_2/2\right) \le \exp \left( - \frac{(c_2/2)^2}{2 \sum _{i=1}^n {\mathbb {E}}[\eta _i^2]}\right) \le \exp \left( - \frac{c_2^2}{8 c_1 \delta ^2}\right) . \end{aligned}$$
Clearly, since \(|f_{\Delta _1, \Delta _2}(w)| \le e^{\lambda w} (\Delta _2 - \Delta _1 + 2 \delta )\), we have, from (B.6),
$$\begin{aligned} I_1 + I_2\le & {} {\mathbb {E}}[|W_b|e^{\lambda W_b }(|\Delta _2 - \Delta _1| + 2 \delta )]\nonumber \\{} & {} + \sum _{i=1}^n \Big |{\mathbb {E}}[ \xi _{b, i}]\Big | {\mathbb {E}}[e^{\lambda W_b^{(i)} }(|\Delta _2^{(i)} - \Delta _1^{(i)}| + 2 \delta )] \end{aligned}$$
(B.9)
Combining (B.7), (B.8) and (B.9), the proof is done.
If \(\sum _{i=1}^n {\mathbb {E}}[\xi _i^2] = 1\) and \(\beta _2 + \beta _3 \le 1/2\), one can apparently take \(c_1 = 1\). The parameter choices of \(c_2\) and \(\delta \) in (B.1) can be justified as follows: Using the fact that [3, p.259]
$$\begin{aligned} \min (x, y) \ge y - \frac{y^2}{4 x}\text { for } x > 0 \text { and } y \ge 0, \end{aligned}$$
by taking \(\delta = (\beta _2 + \beta _3)/4\), we have
$$\begin{aligned} \sum _{i=1}^n {\mathbb {E}}[|\xi _i| \min (\delta , |\xi _i|/2)]&\ge \sum _{i=1}^n {\mathbb {E}}[ |\xi _i | I(|\xi _i| \le 1)\min (\delta , |\xi _i|/2)]\\&\ge \sum _{i=1}^n \Bigg [ \frac{{\mathbb {E}}[\xi _i^2 I(|\xi _i| \le 1)]}{2} - \frac{{\mathbb {E}}[|\xi _i |^3 I(|\xi _i| \le 1)]}{16 \delta } \Bigg ] = \frac{1 - \beta _2}{2} - \frac{\beta _3}{16 \delta }\\&= \frac{1}{2} - \frac{ 8 \delta \beta _2+ \beta _3}{16 \delta } \underbrace{\ge }_{\delta \le 1/8} \frac{1}{2} - \frac{\beta _2 + \beta _3}{16 \delta } = \frac{1}{4}. \end{aligned}$$
\(\square \)
Appendix C. Proof of Theorem 2.1
This section presents the proof of Theorem 2.1. The approach is similar to that of Shao et al. [9, Theorem 3.1], but there are quite a number of differences stemming from correcting the numerous gaps in the latter. It suffices to consider \(x \ge 0\), or else we can consider \(- T_{SN}\) insteadFootnote 2. Moreover, without loss of generality, we can assume
$$\begin{aligned} \beta _2 + \beta _3 < 1/2; \end{aligned}$$
(C.1)
otherwise, it must be true that \(|P(T_{SN} \le x) - \Phi (x)| \le 2 (\beta _2 + \beta _3)\). Since
$$\begin{aligned} 1 + s/2 - s^2/2 \le (1 + s)^{1/2} \le 1 + s/2 \text { for all } s \ge -1, \end{aligned}$$
we have the two inclusions
$$\begin{aligned}{} & {} \{T_{SN}>x\} \subset \{W_n + D_{1n} - x D_{2n}/2 > x\} \cup \{ x + x(D_{2n} - D_{2n}^2)/2 \\{} & {} \quad < W_n + D_{1n} \le x + x D_{2n}/2 \} \end{aligned}$$
and
$$\begin{aligned} \{T_{SN}> x\} \supset \{W_n + D_{1n} - x D_{2n}/2 > x\}. \end{aligned}$$
Hence, it suffices to establish the bounds
$$\begin{aligned} P(x + x(D_{2n} - D_{2n}^2)/2\le & {} W_n + D_{1n} \le x + x D_{2n}/2 ) \le \sum _{j=1}^2 P(|D_{jn}| > 1/2)\nonumber \\{} & {} + C\Bigg \{\beta _2 + \beta _3 + {\mathbb {E}}\Big [(1 + e^{W_b}) {\bar{D}}_{2n}^2 \Big ] \nonumber \\{} & {} + \sum _{j=1}^2 \sum _{i=1}^n\Vert \xi _{b, i}e^{W_b^{(i)}/2} ( {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)}) \Vert _1 \Bigg \} \end{aligned}$$
(C.2)
and
$$\begin{aligned}{} & {} |P(W_n + D_{1n} - x D_{2n}/2 \le x) - \Phi (x)| \le \sum _{j=1}^2P(|D_{jn}| > 1/2) \nonumber \\{} & {} \quad + C \Bigg \{\beta _2 + \beta _3 + \Vert {\bar{D}}_{1n}\Vert _2 + {\mathbb {E}}\Big [(1 +e^{W_b}) {\bar{D}}_{2n}^2\Big ]\nonumber \\{} & {} \quad + \Big |x {\mathbb {E}}[{\bar{D}}_{2n} f_x(W_b)]\Big | \nonumber \\{} & {} \quad + \sum _{j=1}^2 \sum _{i=1}^n \bigg ( {\mathbb {E}}[\xi ^2_{b, i}] \Big \Vert (1+ e^{W_b^{(i)}})({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} )\Big \Vert _1\nonumber \\{} & {} \quad + \Big \Vert \xi _{b, i}( 1+ e^{W_b^{(i)}/2}) ( {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)}) \Big \Vert _1\bigg ) \Bigg \} \end{aligned}$$
(C.3)
separately. Before starting to prove them, we introduce the following notation:
$$\begin{aligned} {\bar{\Delta }}_{1n,x} = \frac{x({\bar{D}}_{2n} - {\bar{D}}_{2n}^2)}{2} - {\bar{D}}_{1n} \text { and } {\bar{\Delta }}_{2n, x} = \frac{x{\bar{D}}_{2n}}{2} - {\bar{D}}_{1n}. \end{aligned}$$
1.1 Proof of (C.2)
We further introduce
$$\begin{aligned} {\bar{\Delta }}_{1n,x}^{(i)} = \frac{x({\bar{D}}_{2n}^{(i)} - ({\bar{D}}_{2n}^{(i)})^2)}{2} - {\bar{D}}_{1n}^{(i)} \text { and } {\bar{\Delta }}_{2n, x}^{(i)} = \frac{x{\bar{D}}_{2n}^{(i)}}{2} - {\bar{D}}_{1n}^{(i)}. \end{aligned}$$
Noting that
$$\begin{aligned}{} & {} P\Big (x + x(D_{2n} - D_{2n}^2)/2 \le W_n + D_{1n} \le x + x D_{2n}/2 \Big ) \nonumber \\{} & {} \quad \le P_0 + \sum _{j=1}^2P(|D_{jn}|> 1/2) + \beta _2, \end{aligned}$$
(C.4)
where
$$\begin{aligned} P_0 = P(x + {\bar{\Delta }}_{1n,x} \le W_b \le x+ {\bar{\Delta }}_{2n, x} ), \end{aligned}$$
it suffices to bound \(P_0\). Since \({\bar{D}}_{2n} - {\bar{D}}_{2n}^2 \ge - 3/4\) and hence \( \frac{1}{2} (x + {\bar{\Delta }}_{1n,x}) \ge \frac{1}{2}( \frac{5x}{8} - \frac{1}{2}) > \frac{x}{4} - \frac{1}{4} \), in light of (C.1), applying Lemma B.1 with the parameters in (B.1) and \(\lambda = 1/2\) implies that
$$\begin{aligned} e^{x/4- 1/4} P_0&\le {\mathbb {E}}[e^{W_b/2} I(x + {\bar{\Delta }}_{1n,x} \le W_b \le x + {\bar{\Delta }}_{2n,x})] \nonumber \\&\le \left( {\mathbb {E}}\left[ e^{ W_b}\right] \right) ^{1/2}\exp \left( - \frac{1}{16(\beta _2 + \beta _3)^2}\right) \nonumber \\&\quad + 8 e^{(\beta _2 + \beta _3)/8}\Biggl \{ \sum _{i =1}^n {\mathbb {E}} \Big [ |\xi _{b, i}|e^{W_b^{(i)} /2} \Big (|{\bar{\Delta }}_{1n,x}- {\bar{\Delta }}_{1n,x}^{(i)}| + |{\bar{\Delta }}_{2n,x}-{\bar{\Delta }}_{2n,x}^{(i)}|\Big )\Big ] \nonumber \\&\quad + {\mathbb {E}}\left[ |W_b|e^{W_b/2 }\left( |{\bar{\Delta }}_{2n,x}- {\bar{\Delta }}_{1n,x}| + \frac{\beta _2 + \beta _3}{2}\right) \right] \nonumber \\&\quad + \sum _{i=1}^n \Big |{\mathbb {E}}[ \xi _{b, i}]\Big | {\mathbb {E}}\left[ e^{ W_b^{(i)} /2 }\left( |{\bar{\Delta }}_{2n,x}^{(i)} - {\bar{\Delta }}_{1n,x}^{(i)}| + \frac{\beta _2 + \beta _3}{2}\right) \right] \Biggr \} \end{aligned}$$
(C.5)
We will bound different terms on the right-hand side of (C.5). First,
$$\begin{aligned} {\mathbb {E}}[e^{W_b} ] \le \exp ( e^2/4 + 1/4) \text { by Lemma }\text {A.2} \end{aligned}$$
(C.6)
and
$$\begin{aligned} \exp \left( \frac{-1}{16 (\beta _2 + \beta _3)^2} \right)&\le C (\beta _2 + \beta _3). \end{aligned}$$
(C.7)
Since \({\bar{D}}_{2n}^2 - ({\bar{D}}_{2n}^{(i)})^2 = ({\bar{D}}_{2n} - {\bar{D}}_{2n}^{(i)})({\bar{D}}_{2n} + {\bar{D}}_{2n}^{(i)})\),
$$\begin{aligned}&{\mathbb {E}} [ |\xi _{b, i}|e^{W_b^{(i)} /2} (|{\bar{\Delta }}_{1n,x}- {\bar{\Delta }}_{1n,x}^{(i)}| + |{\bar{\Delta }}_{2n,x}-{\bar{\Delta }}_{2n,x}^{(i)}|)] \nonumber \\&\quad \le C {\mathbb {E}} [ |\xi _{b, i}|e^{W_b^{(i)} /2} (|{\bar{D}}_{1n} -{\bar{D}}_{1n}^{(i)} | + x|{\bar{D}}_{2n} -{\bar{D}}_{2n}^{(i)}| ) ] . \end{aligned}$$
(C.8)
Moreover, since \( \frac{|W_b|}{2} \le e^{|W_b|/2} \le e^{W_b/2} + e^{-W_b/2} \), by Lemma A.2,
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ |W_b|e^{W_b/2 }\left( |{\bar{\Delta }}_{2n,x}- {\bar{\Delta }}_{1n,x}| + \frac{\beta _2 + \beta _3}{2}\right) \right] \nonumber \\{} & {} \quad \le C_1 x {\mathbb {E}}[(1 + e^{W_b}) {\bar{D}}_{2n}^2 ]+ C_2 (\beta _2 + \beta _3). \end{aligned}$$
(C.9)
Lastly, by Lemma A.1, Bennett’s inequality (Lemma A.2) and (C.1), we have
$$\begin{aligned}&\sum _{i=1}^n\Big |{\mathbb {E}}[ \xi _{b, i}]\Big | {\mathbb {E}}\left[ e^{ W_b^{(i)} /2 }\left( |{\bar{\Delta }}_{2n,x}^{(i)} - {\bar{\Delta }}_{1n,x}^{(i)}| + \frac{\beta _2 + \beta _3}{2}\right) \right] \nonumber \\&\quad \le C \sum _{i=1}^n \Big |{\mathbb {E}}[ \xi _{b, i}]\Big | \underbrace{\Bigg ( x{\mathbb {E}}[ e^{W_b^{(i)}/2} ({\bar{D}}_{2n}^{(i)})^2 ] + \beta _2 + \beta _3\Bigg )}_{\le \quad C(1 +x)} \le C (1+x)\beta _2. \end{aligned}$$
(C.10)
Collecting (C.4)–(C.10), we get (C.2).
1.2 Proof of (C.3)
For this part, as a proof device, we let \(X_1^*, \dots , X_n^*\) be independent copies of \(X_1, \dots , X_n\) and in analogy to (1.4), we introduce
$$\begin{aligned} D_{1n, i^*}= & {} D_{1n}(X_1, \dots , X_{i-1}, X_i^*, X_{i+1}, \dots , X_n) \text { and }\\ D_{2n, i^*}= & {} D_{2n}(X_1, \dots , X_{i-1}, X_i^*, X_{i+1}, \dots , X_n),\\ {\bar{D}}_{1n, i^*}= & {} D_{1n, i^*}I\Bigg (|D_{1n, i^*}| \le \frac{1}{2} \Bigg ) + \frac{1}{2}I\Bigg (D_{1n, i^*}> \frac{1}{2}\Bigg ) - \frac{1}{2} I\Bigg (D_{1n, i^*}<- \frac{1}{2}\Bigg )\text { and }\\ {\bar{D}}_{2n, i^*}= & {} D_{2n, i^*} I\Bigg (| D_{2n, i^*} | \le \frac{1}{2} \Bigg ) + \frac{1}{2}I\Bigg ( D_{2n, i^*} > \frac{1}{2}\Bigg ) - \frac{1}{2} I\Bigg ( D_{2n, i^*} <- \frac{1}{2}\Bigg ), \end{aligned}$$
as well as
$$\begin{aligned} {\bar{\Delta }}_{2n, x, i^*} = \frac{x {\bar{D}}_{2n, i^*} }{2} - {\bar{D}}_{1n, i^*}, \end{aligned}$$
which are correspondingly versions of \(D_{1n}\), \(D_{2n}\), \({\bar{D}}_{1n}\), \({\bar{D}}_{2n}\) and \({\bar{\Delta }}_{2n,x}\) with \(X_i^*\) replacing \(X_i\) as input. For any pair \(1 \le i , i' \le n\) and \(j \in \{1, 2\}\), we also define
$$\begin{aligned} D_{jn, i^*}^{(i')} \equiv {\left\{ \begin{array}{ll} D^{(i')} ( X_1, \dots , X_{i - 1}, X_i^*, X_{i+1}, \dots , X_{i'-1}, X_{i'+1}, \dots , X_n ) &{} \text {if } i < i' \\ D^{(i')} ( X_1, \dots , X_{i'-1}, X_{i'+1}, \dots , X_{i - 1}, X_i^*, X_{i+1}, \dots , X_n ) &{} \text {if } i > i' \\ D^{(i')} ( X_1, \dots , X_{i - 1}, X_{i+1}, \dots , X_n ) &{} \text {if } i = i' \\ \end{array}\right. }, \end{aligned}$$
i.e., \(D_{jn, i^*}^{(i')}\) is a version of the “leave-one-out” \(D_{jn}^{(i')}\) with \(X_i^*\) replacing \(X_i\) as input, and its censored version
$$\begin{aligned} {\bar{D}}_{jn, i^*}^{(i')} \equiv D_{jn, i^*}^{(i')} I\Bigg (|D_{jn, i^*}^{(i')}| \le \frac{1}{2} \Bigg ) + \frac{1}{2}I\Bigg (D_{jn, i^*}^{(i')}> \frac{1}{2}\Bigg ) - \frac{1}{2} I\Bigg (D_{jn, i^*}^{(i')} <- \frac{1}{2}\Bigg ). \end{aligned}$$
It suffices to bound \(|P(W_b - {\bar{\Delta }}_{2n, x} \le x) - \Phi (x)|\) since
$$\begin{aligned} |P(W_n - \Delta _{2n, x} \le x) - \Phi (x)|\le & {} |P(W_b - {\bar{\Delta }}_{2n, x} \le x) \nonumber \\{} & {} - \Phi (x)| + \beta _2 + \sum _{j=1}^2P(|D_{jn}| > 1/2). \end{aligned}$$
(C.11)
First, define the K function
$$\begin{aligned} k_{b, i} (t) = {\mathbb {E}}[\xi _{b, i}\{ I(0 \le t \le \xi _{b, i}) - I(\xi _{b, i} \le t < 0)\}], \end{aligned}$$
which has the properties
$$\begin{aligned} \int _{-\infty }^\infty k_{b, i}(t) \hbox {d}t= & {} \int _{-1}^1 k_{b, i}(t) \hbox {d}t = {\mathbb {E}}[\xi _{b, i}^2] = \Vert \xi _{b, i}\Vert _2^2 \quad \text { and }\nonumber \\ \int _{-\infty }^\infty |t| k_{b, i}(t) \hbox {d}t= & {} \int _{-1}^1 |t| k_{b, i}(t) \hbox {d}t = \frac{{\mathbb {E}}[|\xi _{b, i}|^3]}{2} = \frac{\Vert \xi _{b, i}\Vert _3^3}{2}. \end{aligned}$$
(C.12)
Since
$$\begin{aligned}{} & {} {\mathbb {E}}\Big [\int _{-1}^1 f_x'( W_b^{(i)} - {\bar{\Delta }}_{2n, x, i^*} + t) k_{b, i}(t) \hbox {d}t\Big ] \\{} & {} \quad = {\mathbb {E}}\Big [\xi _{b, i} \{f_x(W_b - {\bar{\Delta }}_{2n, x, i^*} ) - f_x(W_b^{(i)} - {\bar{\Delta }}_{2n, x, i^*} ) \}\Big ] \end{aligned}$$
by independence and the fundamental theorem of calculus, from the Stein equation (2.4), one can then write
$$\begin{aligned}&P(W_b - {\bar{\Delta }}_{2n, x} \le x) - \Phi (x) \\&\quad = {\mathbb {E}}[ f_x'(W_b - {\bar{\Delta }}_{2n, x} )] - {\mathbb {E}}[W_b f_x(W_b - {\bar{\Delta }}_{2n, x})] \\&\qquad + {\mathbb {E}}\Big [ {\bar{\Delta }}_{2n, x} \Big (f_x(W_b - {\bar{\Delta }}_{2n, x}) - f_x(W_b)\Big )\Big ] + {\mathbb {E}}[{\bar{\Delta }}_{2n, x} f_x(W_b) ]\\&\quad = \underbrace{\sum _{i=1}^n {\mathbb {E}}\Big [\int _{-1}^1 \{f_x'(W_b - {\bar{\Delta }}_{2n, x}) - f_x'(W_b^{(i)} - {\bar{\Delta }}_{2n, x, i^*} + t)\} k_{b, i}(t) \hbox {d}t \Big ] }_{R_1}\\&\qquad + \underbrace{\sum _{i=1}^n {\mathbb {E}}[ (\xi _i^2- 1)I(|\xi _i|> 1)] {\mathbb {E}}[f_x'(W_b - {\bar{\Delta }}_{2n,x})] - \sum _{i=1}^n{\mathbb {E}}[\xi _{b, i} f_x(W_b^{(i)} - {\bar{\Delta }}_{2n, x, i^*} )] + {\mathbb {E}}[ {\bar{\Delta }}_{2n, x} f_x (W_b)] }_{R_2}\\&\qquad + \underbrace{\Bigg \{ - \sum _{i=1}^n {\mathbb {E}}\Bigg [\xi _{b, i} \Big \{ f_x(W_b - {\bar{\Delta }}_{2n, x}) - f_x(W_b - {\bar{\Delta }}_{2n, x, i^*} ) \Big \}\Bigg ]\Bigg \}}_{R_3}\\&\qquad + \underbrace{{\mathbb {E}}\Big [ {\bar{\Delta }}_{2n, x} \int _0^{- {\bar{\Delta }}_{2n, x}}f_x'(W_b + t) \hbox {d}t \Big ]}_{R_4}\\&\quad = R_1 + R_2 + R_3 + R_4. \end{aligned}$$
To finish the proof, we will establish the following bounds for \(R_1, R_2, R_3, R_4\):
$$\begin{aligned} |R_1|\le & {} C \Bigg \{ \beta _2 + \beta _3 + \sum _{j=1}^2 \sum _{i=1}^n \bigg ( {\mathbb {E}}[\xi ^2_{b, i}] \Big \Vert (1+ e^{W_b^{(i)}})({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} )\Big \Vert _1\nonumber \\{} & {} + \Big \Vert \xi _{b, i}e^{W_b^{(i)}/2} ( {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)}) \Big \Vert _1\bigg ) \Bigg \} \end{aligned}$$
(C.13)
$$\begin{aligned} |R_2|\le & {} 1.63\beta _2 + 0.63 \Vert {\bar{D}}_{1n}\Vert _2 + \Big |\frac{x}{2} {\mathbb {E}}[{\bar{D}}_{2n} f_x(W_b)]\Big | , \end{aligned}$$
(C.14)
$$\begin{aligned} |R_3|\le & {} C \sum _{j=1}^2 \sum _{i=1}^n \Vert \xi _{b, i} (1 + e^{W_b^{(i)}/2})({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)})\Vert _1, \end{aligned}$$
(C.15)
$$\begin{aligned} |R_4|\le & {} C\Big ( \Vert {\bar{D}}_{1n}\Vert _2 + {\mathbb {E}}[( 1+e^{W_b}) {\bar{D}}_{2n}^2]\Big ). \end{aligned}$$
(C.16)
Then (C.13)–(C.16) together with (C.11) conclude (C.3).
1.2.1 Bound for \(R_1\)
Let \(g_x(w) = (w f_x(w))'\) as defined in (A.1). By the Stein equation (2.4) and defining \(\eta _1 = t - {\bar{\Delta }}_{2n, x, i^*}\) and \(\eta _2 = \xi _{b, i} - {\bar{\Delta }}_{2n, x}\), we can write
$$\begin{aligned} R_1 = R_{11} + R_{12}, \end{aligned}$$
where
$$\begin{aligned} R_{11}&= \sum _{i=1}^n \int _{-1}^1 {\mathbb {E}}\Big [ \int _{t - {\bar{\Delta }}_{2n, x, i^*}}^{\xi _{b, i} - {\bar{\Delta }}_{2n, x}} g_x(W_b^{(i)} + u) du\Big ] k_{b, i} (t) \hbox {d}t\\&= \underbrace{\sum _{i=1}^n \int _{-1}^1 {\mathbb {E}}\Bigg [ \int g_x(W_b^{(i)} + u) I( \eta _1 \le u \le \eta _2) du\Bigg ] k_{b, i} (t) \hbox {d}t}_{R_{11.1}} \\&\quad - \underbrace{\sum _{i=1}^n \int _{-1}^1 {\mathbb {E}}\Bigg [ \int g_x(W_b^{(i)} + u) I(\eta _2\le u \le \eta _1) du\Bigg ] k_{b, i} (t) \hbox {d}t}_{R_{11.2}} \end{aligned}$$
and
$$\begin{aligned} R_{12}&= \sum _{i=1}^n \int _{-1}^1 \{P(W_b - {\bar{\Delta }}_{2n, x} \le x) - P(W_b^{(i)} - {\bar{\Delta }}_{2n, x, i^*} + t \le x)\} k_{b, i}(t) \hbox {d}t. \end{aligned}$$
For \(0 \le x <1\), since \(|g_x| \le 2.3\) (Lemma A.3), using the properties in (C.12), we have
$$\begin{aligned} |R_{11}|&\le C \sum _{i=1}^n \int _{-1}^1 \Big (|t| + \Vert \xi _{b, i}\Vert _1 +\sum _{j=1}^2 \Vert {\bar{D}}_{jn} - {\bar{D}}_{jn, i^*}\Vert _1\Big ) k_{b, i}(t) \hbox {d}t \nonumber \\&\le C\left( \sum _{i=1}^n \Vert \xi _{b, i}\Vert _3^3 + \sum _{i=1}^n \Vert \xi _{b, i}\Vert _2^2 \Vert \xi _{b, i}\Vert _1 + \sum _{j=1}^2 \sum _{i=1}^n \Vert \xi _{b, i}\Vert _2^2 \Vert {\bar{D}}_{jn} - {\bar{D}}_{jn, i^*}\Vert _1\right) \nonumber \\&\le C\left( \beta _2 + \beta _3 + \sum _{j=1}^2 \sum _{i=1}^n \Vert \xi _{b, i}\Vert _2^2 \Vert {\bar{D}}_{jn} - {\bar{D}}_{jn, i^*}\Vert _1\right) \text { for } 0 \le x < 1, \end{aligned}$$
(C.17)
where we have used \(\Vert \xi _{b,i}\Vert _1 \le \Vert \xi _{b,i}\Vert _2 \le \Vert \xi _{b,i}\Vert _3\) and
$$\begin{aligned} \Vert \xi _{b, i}\Vert _3^3&= {\mathbb {E}}[|\xi _i^3| I(|\xi _i| \le 1)] + {\mathbb {E}}[I(|\xi _i|>1)] \nonumber \\&\le {\mathbb {E}}[|\xi _i^3| I(|\xi _i| \le 1)] + {\mathbb {E}}[\xi _i^2 I(|\xi _i|>1)] \end{aligned}$$
(C.18)
in the last inequality.
For \(x \ge 1\), we first bound the integrand of \(R_{11.1}\). Using the identity
$$\begin{aligned} 1&= I(W_b^{(i)} + u \le x - 1) + I(x -1< W_b^{(i)} + u, u \le 3x/4 )\\&\quad + I(x -1< W_b^{(i)} + u, u> 3x/4)\\&\le I(W_b^{(i)} + u \le x - 1) + I(x -1< W_b^{(i)} + u, W_b^{(i)} +1> x/4)\\&\quad + (x -1 < W_b^{(i)} + u, u > 3x/4) \end{aligned}$$
and the bounds for \(g_x(\cdot )\) in Lemma A.4, in light of \(| {\bar{\Delta }}_{2n, x}| \le \frac{x |{\bar{D}}_{2n}|}{2} + |{\bar{D}}_{1n}| \le \frac{1}{2} + \frac{x}{4}\) and \(1.6 {\bar{\Phi }}(x) \le x e^{1/2-x}\),
$$\begin{aligned}&\Bigg |{\mathbb {E}}\Big [\int g_x(W_b^{(i)} + u) I(\eta _1 \le u \le \eta _2) du\Big ]\Bigg |\\&\quad \le x e^{1/2 - x} \Vert \eta _2 - \eta _1\Vert _1 + (x + 2) \Big \{ \Vert I(W_b^{(i)} +1> x/4) (\eta _2 - \eta _1) \Vert _1 \\&\qquad + \Vert I(\eta _2 > 3x/4) (\eta _2 - \eta _1)\Vert _1 \Big \}\\&\quad \le x e^{1/2 - x} \Vert \eta _2 - \eta _1\Vert _1 + \frac{x+2}{e^{x/4 -1}} \Vert e^{W_b^{(i)}}(\eta _2 - \eta _1)\Vert _1 + \frac{x+2}{e^{3x/4}} \Vert e^{\xi _{b, i} - {\bar{\Delta }}_{2n, x}} (\eta _2 - \eta _1) \Vert _1\\&\quad \le \Bigg (x e^{1/2 - x} + \frac{e^{3/2} (x+2)}{e^{x/2}} \Bigg )\Vert \eta _2 - \eta _1\Vert _1 + \frac{x+2}{e^{x/4 -1}} \Vert e^{W_b^{(i)}}(\eta _2 - \eta _1)\Vert _1 \\&\quad \le \frac{C(x+2)}{e^{x/4}}\Bigg \{ |t| + \Vert {\bar{\Delta }}_{2n, x, i^*} - {\bar{\Delta }}_{2n, x} + \xi _{b, i}\Vert _1 + \Vert e^{W_b^{(i)}} ({\bar{\Delta }}_{2n, x, i^*} - {\bar{\Delta }}_{2n, x} + \xi _{b, i})\Vert _1 \Bigg \}\\ \end{aligned}$$
where we have used the Bennett’s inequality (Lemma A.2) via \(\Vert e^{W_b^{(i)}} t\Vert _1 \le C |t|\). Continuing,
$$\begin{aligned}&\Bigg |{\mathbb {E}}\Big [\int g_x(W_b^{(i)} + u) I(\eta _1 \le u \le \eta _2) du\Big ]\Bigg |\nonumber \\&\quad \le \frac{C(x+2)}{e^{x/4}}\Bigg \{ |t| + \Vert x ({\bar{D}}_{2n, i^*}- {\bar{D}}_{2n}) - ({\bar{D}}_{1n, i^*}- {\bar{D}}_{1n}) + \xi _{b, i}\Vert _1\nonumber \\&\qquad + \Vert e^{W_b^{(i)}} [ x ({\bar{D}}_{2n, i^*}- {\bar{D}}_{2n}) - ({\bar{D}}_{1n, i^*}- {\bar{D}}_{1n})+ \xi _{b, i}]\Vert _1 \Bigg \}\nonumber \\&\quad \le C \Bigg \{|t| + (1 + \Vert e^{W_b^{(i)}}\Vert _2)\Vert \xi _{b, i}\Vert _2 +\sum _{j=1}^2 \Vert (1 + e^{W_b^{(i)}}) ({\bar{D}}_{jn, i^*}- {\bar{D}}_{jn})\Vert _1 \Bigg \} \nonumber \\&\quad \le C \Bigg \{|t| + \Vert \xi _{b, i}\Vert _2 +\sum _{j=1}^2 \Vert (1 + e^{W_b^{(i)}}) ({\bar{D}}_{jn, i^*}- {\bar{D}}_{jn})\Vert _1 \Bigg \}, \end{aligned}$$
(C.19)
where the last inequality uses Bennett’s inequality (Lemma A.2 giving \(\Vert e^{W_b^{(i)}}\Vert _2 \le C\)). By a completely analogous argument, we also have the bound
$$\begin{aligned}{} & {} \Bigg |{\mathbb {E}}\Big [\int g_x(W_b^{(i)} + u) I(\eta _2 \le u \le \eta _1) du\Big ]\Bigg | \nonumber \\{} & {} \quad \le C \Bigg \{|t| + \Vert \xi _{b, i}\Vert _2 +\sum _{j=1}^2 \Vert (1 + e^{W_b^{(i)}}) ({\bar{D}}_{jn, i^*}- {\bar{D}}_{jn})\Vert _1 \Bigg \} . \end{aligned}$$
(C.20)
for the integrand of \(R_{11.2}\), for \(x \ge 1\). Combining (C.19) and (C.20), as well as the integral and moment properties in (C.12) and (C.18), via integrating over t, we have
$$\begin{aligned} |R_{11}|&\le C \Bigg \{\beta _2 + \beta _3 + \sum _{i=1}^n \Vert \xi _{b, i}\Vert _2^2 \Bigg ( \Vert \xi _{b, i}\Vert _2 +\sum _{j=1}^2 \Vert (1 + e^{W_b^{(i)}}) ({\bar{D}}_{jn, i^*}- {\bar{D}}_{jn})\Vert _1 \Bigg ) \Bigg \} \nonumber \\&\le C \Bigg \{\beta _2 + \beta _3 + \sum _{j=1}^2\sum _{i=1}^n \Vert \xi _{b, i}\Vert _2^2 \Big \Vert (1 + e^{W_b^{(i)}}) ({\bar{D}}_{jn, i^*}- {\bar{D}}_{jn})\Big \Vert _1\Bigg \} \text { for } x \ge 1 , \end{aligned}$$
(C.21)
where the last inequality also uses \(\Vert \xi _{b, i}\Vert _2^3 \le \Vert \xi _{b, i}\Vert ^3_3\) and (C.18). Combining (C.21) with the bound for \(x \in [0, 1)\) in (C.17), we get, for all \(x \ge 0\),
$$\begin{aligned} |R_{11}|&\le C \Bigg \{\beta _2 + \beta _3 + \sum _{j=1}^2\sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2] \big \Vert (1 + e^{W_b^{(i)}}) ({\bar{D}}_{jn} - {\bar{D}}_{jn, i^*} )\big \Vert _1\Bigg \} \nonumber \\&= C \Bigg \{\beta _2 + \beta _3 + \sum _{j=1}^2\sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2] \big \Vert (1 + e^{W_b^{(i)}}) ({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} + {\bar{D}}_{jn}^{(i)} - {\bar{D}}_{jn, i^*} )\big \Vert _1\Bigg \} \nonumber \\&\le C \Bigg \{\beta _2 + \beta _3 + \sum _{j=1}^2\sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2] \big \Vert (1 + e^{W_b^{(i)}}) ({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} )\big \Vert _1\Bigg \} \end{aligned}$$
(C.22)
where in the last inequality, we have used the fact that \((W_b^{(i)}, {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)}) =_d (W_b^{(i)}, {\bar{D}}_{jn, i^*} - {\bar{D}}_{jn}^{(i)})\) .
For \(R_{12}\), its integrand for a given i is bounded by
$$\begin{aligned}{} & {} P(x + {\bar{\Delta }}_{2n, x} \le W_b \le x - t + {\bar{\Delta }}_{2n, x, i^*} + \xi _{b, i})\nonumber \\{} & {} \quad + P( x - t+ {\bar{\Delta }}_{2n, x, i^*} +\xi _{b, i} \le W_b \le x + {\bar{\Delta }}_{2n, x}) \end{aligned}$$
(C.23)
Since
$$\begin{aligned} (x + {\bar{\Delta }}_{2n, x}) \wedge (x - t + {\bar{\Delta }}_{2n, x, i^*}+\xi _{b, i}) \ge (3x)/4- 5/2 \quad \text { for } \quad |t| \le 1, \end{aligned}$$
and \({\mathbb {E}}[e^{W_b}] \le C\) by Bennett’s inequality (Lemma A.2), by defining
$$\begin{aligned} {\bar{\Delta }}_{2n, x, i^*}^{(i')} \equiv \frac{x {\bar{D}}_{2n, i^*}^{(i')} }{2} - {\bar{D}}_{1n, i^*}^{(i')} \text { for } 1 \le i' \le n, \end{aligned}$$
we can apply the randomized concentration inequality (Lemma B.1) with the parameters in (B.1) and \(\lambda = 1/2\) to bound (C.23) by
$$\begin{aligned}&C e^{-3x/8} \Bigg \{\beta _2 + \beta _3 \nonumber \\&\qquad + \sum _{i' = 1}^n {\mathbb {E}}\Big [ |\xi _{b, i'}|e^{W_b^{(i')}/2} \Big ( |{\bar{\Delta }}_{2n, x} - {\bar{\Delta }}_{2n, x}^{(i')}| +|{\bar{\Delta }}_{2n, x, i^*} - {\bar{\Delta }}_{2n, x, i^*}^{(i')}| + I(i' = i)|\xi _{b, i}| \Big )\Big ] \nonumber \\&\qquad + {\mathbb {E}}\Big [ \underbrace{ |W_b|e^{W_b/2} }_{\le 2 (1 + e^{W_b})}\Big ( |{\bar{\Delta }}_{2n, x} - {\bar{\Delta }}_{2n, x, i^*}| + |\xi _{b, i}| + |t| + \beta _2 + \beta _3 \Big )\Big ] \nonumber \\&\qquad + \sum _{i'=1}^n \Big |{\mathbb {E}}[ \xi _{b, i'}]\Big | {\mathbb {E}}\Big [ e^{W_b^{(i')}/2} \underbrace{\Big (|t| +|\xi _{b, i}|I(i' \ne i) + |{\bar{\Delta }}_{2n, x}^{(i')} - {\bar{\Delta }}_{2n, x, i^*}^{(i')}| + \beta _2 + \beta _3\Big )}_{\le C(1 +x)} \Big ] \Bigg \}\nonumber \\&\quad \le C \Bigg \{\beta _2 + \beta _3 + {\mathbb {E}}[|\xi _{b, i}|^2 e^{ W_b^{(i)}/2}] \nonumber \\&\qquad + \sum _{j=1}^2 \sum _{i' = 1}^n {\mathbb {E}}\bigg [ |\xi _{b, i'}|e^{W_b^{(i')}/2} \Big ( | {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i')}| +|{\bar{D}}_{jn, i^*} - {\bar{D}}_{jn, i^*}^{(i')}| \Big )\bigg ] \nonumber \\&\qquad + {\mathbb {E}}\Big [ (1 + e^{W_b})\Big ( \sum _{j=1}^2 | {\bar{D}}_{jn} -{\bar{D}}_{jn, i^*}| + |\xi _{b, i}| + |t| + \beta _2 + \beta _3 \Big )\Big ] + \sum _{i' = 1}^n \Big |{\mathbb {E}}[ \xi _{b, i'}]\Big | {\mathbb {E}}\Big [ e^{W_b^{(i')}/2} \Big ] \Bigg \}\nonumber \\&\quad \le C \Bigg \{\beta _2 + \beta _3 + {\mathbb {E}}[|\xi _{b, i}|^2] + \sum _{j=1}^2 \sum _{i' = 1}^n {\mathbb {E}}\bigg [ |\xi _{b, i'}|e^{W_b^{( i')}/2} \Big ( | {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i')}| +|{\bar{D}}_{jn, i^*} - {\bar{D}}_{jn, i^*}^{(i')}| \Big )\bigg ]\nonumber \\&\qquad + \sum _{j=1}^2 \Vert (1+ e^{W_b})({\bar{D}}_{jn} - {\bar{D}}_{jn, i^*})\Vert _1 + \Vert \xi _{b, i}\Vert _2 + |t| \Bigg \}; \end{aligned}$$
(C.24)
in (C.24), we have used that \(\sum _{i' =1}^n |{\mathbb {E}}[ \xi _{b, i'}]| \le \beta _2\) by Lemma A.1 and
$$\begin{aligned} \max ( \Vert e^{W_b}\Vert _2, \Vert e^{W_b}\Vert _1, {\mathbb {E}}[e^{W_b^{(i')}/2}], {\mathbb {E}}[e^{W_b^{(i)}/2}] )\le C \end{aligned}$$
by Bennett’s inequality (Lemma A.2). Since (C.24) bounds (C.23) which bounds the integrand of \(R_{12}\), on integration with respect to t which has the properties in (C.12), we get
$$\begin{aligned} |R_{12}|\le & {} C \Bigg \{ \beta _2 + \beta _3 + \sum _{j=1}^2 \Bigg [ \sum _{i=1}^n{\mathbb {E}}[\xi _{b, i}^2] \Big \Vert (1+ e^{W_b^{(i)}})({\bar{D}}_{jn} - {\bar{D}}_{jn, i^*})\Big \Vert _1 \nonumber \\{} & {} + \sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2]\sum _{ i' = 1}^n {\mathbb {E}}\Big [ |\xi _{b, i'}|e^{W_b^{( i')}/2} \big ( | {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i')}| +|{\bar{D}}_{jn, i^*} - {\bar{D}}_{jn, i^*}^{(i')}| \big )\Big ] \Bigg ] \Bigg \}\qquad \quad \end{aligned}$$
(C.25)
where we have used \(\sum _{i=1}^n \Vert \xi _{b, i}\Vert _2^4 \le \sum _{i=1}^n \Vert \xi _{b, i}\Vert _2 \Vert \xi _{b, i}\Vert _2^2 \le \sum _{i=1}^n {\mathbb {E}}[|\xi _{b, i}|^3] \le \beta _2 + \beta _3\) by (C.18). From (C.25), by defining
$$\begin{aligned} W_b^{(i, i')} \equiv {\left\{ \begin{array}{ll} W_b - \xi _{b, i} - \xi _{b, i'} &{} \text { if } i' \ne i \\ W_b - \xi _{b, i} &{} \text { if } i' = i \end{array}\right. }, \end{aligned}$$
with \(e^{W_b^{(i')}/2} \le e^{1/2} e^{W_b^{(i, i')}/2} \), we further get
$$\begin{aligned} |R_{12}|&\le C \Bigg \{ \beta _2 + \beta _3 + \sum _{j=1}^2 \Bigg [ \sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2] \Vert (1+ e^{W_b^{(i)}})({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} + {\bar{D}}_{jn}^{(i)} - {\bar{D}}_{jn, i^*})\Vert _1 \nonumber \\&\quad + \sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2]\sum _{i' = 1}^n {\mathbb {E}}\Big [ |\xi _{b, i'}|e^{W_b^{(i, i')}/2} \big ( | {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i')}| +|{\bar{D}}_{jn, i^*} - {\bar{D}}_{jn, i^*}^{(i')}| \big )\Big ]\Bigg ] \Bigg \} \nonumber \\&\le C \Bigg \{ \beta _2 + \beta _3 + \sum _{j=1}^2 \Bigg [ \sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2] \Vert (1+ e^{W_b^{(i)}})({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} )\Vert _1 \nonumber \\&\quad +\sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2] \sum _{ i' = 1}^n {\mathbb {E}}\Big [ |\xi _{b, i'}|e^{W_b^{(i, i')}/2} | {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i')}| \Big ]\Bigg ] \Bigg \}, \end{aligned}$$
(C.26)
where we have used that
$$\begin{aligned} (e^{W_b^{(i)}}, {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)})= & {} {}_d (e^{W_b^{(i)}}, {\bar{D}}_{jn, i^*} - {\bar{D}}_{jn}^{(i)}) \text { and }\\ (|\xi _{b, i'}|e^{W_b^{(i, i')}/2}, {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i')})= & {} {}_d (|\xi _{b, i'}|e^{W_b^{(i, i')}/2}, {\bar{D}}_{jn, i^*} - {\bar{D}}_{jn, i^*}^{(i')}) \end{aligned}$$
to arrive at (C.26). Lastly, (C.26) can be further simplified as
$$\begin{aligned} |R_{12}|\le & {} C \Bigg \{ \beta _2 + \beta _3 + \sum _{j=1}^2 \sum _{i=1}^n \bigg ( {\mathbb {E}}[\xi ^2_{b, i}] \Big \Vert (1+ e^{W_b^{(i)}})({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} )\Big \Vert _1 \nonumber \\{} & {} + {\mathbb {E}}\Big [ |\xi _{b, i}|e^{W_b^{(i)}/2} | {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)}| \Big ]\bigg ) \Bigg \} \end{aligned}$$
(C.27)
using \(e^{W_b^{(i, i')}/2} \le e^{(W_b^{(i')} +1)/2}\) and \(\sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^2] \le \sum _{i=1}^n {\mathbb {E}}[\xi _i^2] = 1\) by (1.2). Combining (C.22) and (C.27) gives (C.13).
1.2.2 Bound for \(R_2\)
Since \(|f_x'| \le 1\) by Lemma A.3,
$$\begin{aligned}{} & {} |\sum _{i=1}^n {\mathbb {E}}[ (\xi _i^2- 1)I(|\xi _i|> 1)] {\mathbb {E}}[f_x'(W_b - {\bar{\Delta }}_{2n,x})] |\nonumber \\{} & {} \quad \le \sum _{i=1}^n {\mathbb {E}}[\xi _i^2 I(|\xi | >1)] \le \beta _2. \end{aligned}$$
(C.28)
Moreover, by independence, Lemma A.1 and that \(|f_x| \le 0.63\) from Lemma A.3,
$$\begin{aligned}{} & {} \bigg |\sum _{i=1}^n{\mathbb {E}}[\xi _{b, i} f(W_b^{(i)} - {\bar{\Delta }}_{2n, x, i^*})] \bigg |= \bigg | \sum _{i=1}^n{\mathbb {E}}[\xi _{b, i}]{\mathbb {E}}[ f(W_b^{(i)} - {\bar{\Delta }}_{2n, x, i^*})] \bigg |\\{} & {} \quad \le 0.63 \sum _{i=1} ^n |{\mathbb {E}}[\xi _{b, i} ]| \le 0.63 \sum _{i=1}^n {\mathbb {E}}[\xi _i^2 I(|\xi _i| >1)] = 0.63 \beta _2. \end{aligned}$$
Lastly, by \(|f_x| \le 0.63\) and the definition of \({\bar{\Delta }}_{2n, x}\),
$$\begin{aligned} |{\mathbb {E}}[ {\bar{\Delta }}_{2n, x} f_x (W_b)] | \le 0.63 \Vert {\bar{D}}_{1n}\Vert _2 + \Big |\frac{x}{2} {\mathbb {E}}[{\bar{D}}_{2n} f_x(W_b)]\Big | \end{aligned}$$
Hence, we established (C.14).
1.2.3 Bound for \(R_3\)
By mean value theorem, given \(|f'_x| \le 1\) (Lemma A.3),
$$\begin{aligned} |f_x(W_b - {\bar{\Delta }}_{2n, x}) - f_x(W_b - {\bar{\Delta }}_{2n, x, i^*})|&\le C | {\bar{\Delta }}_{2n, x} - {\bar{\Delta }}_{2n, x, i^*}| \\&\le C (|{\bar{D}}_{1n} - {\bar{D}}_{1n, i^*}| + x|{\bar{D}}_{2n} - {\bar{D}}_{2n, i^*}|). \end{aligned}$$
Hence,
$$\begin{aligned} |R_3|\le & {} C \sum _{j=1}^2 \sum _{i=1}^n \Vert \xi _{b, i} ({\bar{D}}_{jn} - {\bar{D}}_{jn, i^*})\Vert _1 \nonumber \\= & {} C \sum _{j=1}^2 \sum _{i=1}^n \Vert \xi _{b, i} ({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} + {\bar{D}}_{jn}^{(i)} - {\bar{D}}_{jn, i^*})\Vert _1 \text { for }0 \le x \le 1. \end{aligned}$$
(C.29)
For \( x >1\), given \(|{\bar{\Delta }}_{2n, x}|\vee |{\bar{\Delta }}_{2n, x, i^*}| \le \frac{1}{2} + \frac{x}{4}\), by (A.5) in Lemma A.4 and \(|f'_x| \le 1\) (Lemma A.3),
$$\begin{aligned}&|f_x(W_b - {\bar{\Delta }}_{2n, x}) - f_x(W_b - {\bar{\Delta }}_{2n, x, i^*})| \\&\quad \le |f_x(W_b - {\bar{\Delta }}_{2n, x}) - f_x(W_b - {\bar{\Delta }}_{2n, x, i^*})| \Big [ I(W_b \le 3x/4 - 3/2) + I(W_b> 3x/4 - 3/2)\Big ]\\&\quad \le C \Big (e^{1/2-x} +I(W_b > 3x/4 - 3/2) \Big )\Big (|{\bar{D}}_{1n} - {\bar{D}}_{1n, i^*}| + x|{\bar{D}}_{2n} - {\bar{D}}_{2n, i^*}|\Big )\\&\quad \le C\Big (e^{-x} + e^{-3x/8} e^{W_b/2} \Big ) \Big (|{\bar{D}}_{1n} - {\bar{D}}_{1n, i^*}| + x|{\bar{D}}_{2n} - {\bar{D}}_{2n, i^*}|\Big ) \\&\quad \le C\Big (e^{-x} + e^{-3x/8} e^{W_b^{(i)}/2} \Big ) \Big (|{\bar{D}}_{1n} - {\bar{D}}_{1n, i^*}| + x|{\bar{D}}_{2n} - {\bar{D}}_{2n, i^*}|\Big ), \end{aligned}$$
where we have used \(e^{W_b/2} \le e^{1/2} e^{W_b^{(i)}/2}\) in the last inequality. Hence,
$$\begin{aligned} |R_3| \le C \sum _{j=1}^2 \sum _{i=1}^n \Vert \xi _{b, i} (1 + e^{W_b^{(i)}/2})({\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)} + {\bar{D}}_{jn}^{(i)}- {\bar{D}}_{jn, i^*})\Vert _1 \text { for } x >1\nonumber \\ \end{aligned}$$
(C.30)
Because \((\xi _{b, i}, W_b^{(i)}, {\bar{D}}_{jn} - {\bar{D}}_{jn}^{(i)}) =_d (\xi _{b, i}, W_b^{(i)}, {\bar{D}}_{jn, i^*} - {\bar{D}}_{jn}^{(i)})\), (C.29) and (C.30) establish (C.15).
1.2.4 Bound for \(R_4\)
Using that \(|f_x'|\le 1\) in Lemma A.3, for \(0 \le x \le 1\),
$$\begin{aligned} {\mathbb {E}}\Big [ {\bar{\Delta }}_{2n, x} \int _0^{- {\bar{\Delta }}_{2n, x}}f_x'(W_b + t) \hbox {d}t \Big ] \le C {\bar{\Delta }}_{2n, x}^2 \le C(\Vert {\bar{D}}_{1n}\Vert _2^2 + \Vert {\bar{D}}_{2n}\Vert _2^2) \le C(\Vert {\bar{D}}_{1n}\Vert _2 + \Vert {\bar{D}}_{2n}\Vert _2^2). \end{aligned}$$
For \(x > 1\), using (A.5) in Lemma A.4 and that \(|f_x'| \le 1\) in Lemma A.3, given \(|{\bar{\Delta }}_{2n, x}| \le \frac{1}{2} + \frac{x}{4}\)
$$\begin{aligned}&{\mathbb {E}}\Big [ {\bar{\Delta }}_{2n, x} \int _0^{- {\bar{\Delta }}_{2n, x}}f_x'(W_b + t) \hbox {d}t \Big ]\\&\quad \le e^{1/2-x} {\mathbb {E}}[{\bar{\Delta }}_{2n, x}^2] + {\mathbb {E}}[I(W_b \ge 3x/4 - 3/2) {\bar{\Delta }}_{2n, x}^2]\\&\quad \le C( e^{-x} {\mathbb {E}}[{\bar{\Delta }}_{2n, x}^2] + e^{-3x/4} {\mathbb {E}}[ e^{W_b} {\bar{\Delta }}_{2n, x}^2 ] )\\&\quad \le C \Bigg \{2 e^{-x} \bigg (\Vert {\bar{D}}_{1n}\Vert _2^2 + \frac{x^2}{4}\Vert {\bar{D}}_{2n}\Vert _2^2\Bigg ) + 2 e^{-3x/4} {\mathbb {E}}\Bigg [e^{W_b} \Bigg ({\bar{D}}_{1n}^2 + \frac{x^2}{4}{\bar{D}}_{2n}^2\Bigg )\Bigg ] \Bigg \}\\&\quad \le C( \Vert {\bar{D}}_{1n}\Vert _2 + {\mathbb {E}}[(1 +e^{W_b} ){\bar{D}}_{2n}^2]), \end{aligned}$$
where we have used \({\mathbb {E}}[e^{W_b}|{\bar{D}}_{1n}|^2]\le {\mathbb {E}}[e^{W_b}|{\bar{D}}_{1n}|] \le \Vert e^{W_b}\Vert _2 \Vert {\bar{D}}_{1n}\Vert _2 \le C \Vert {\bar{D}}_{1n}\Vert _2\) by Lemma A.2 and \(\Vert {\bar{D}}_{1n}\Vert _2^2 \le \Vert {\bar{D}}_{1n}\Vert _2\). This establishes (C.16).
Appendix D. Proof of Theorem 2.3
We first verify (2.8)–(2.10), which will also be used in the proof of Theorem 2.3; (2.10) is immediate from (2.7). We can prove (2.8) with Hölder’s inequality as
$$\begin{aligned} \Vert (1+ e^{W_b^{(i)}})({\bar{D}}_{1n} - {\bar{D}}_{1n}^{(i)} )\Vert _1&\le \Vert 1+ e^{W_b^{(i)}}\Vert _2 \Vert {\bar{D}}_{1n} - {\bar{D}}_{1n}^{(i)} \Vert _2 \\&\le \Big (1 + \exp ( e^{4}/8 - 1/8 + 1/2) \Big )\Big \Vert D_{1n} - D_{1n}^{(i)} \Big \Vert _2, \end{aligned}$$
where we have also used Bennett’s inequality (Lemmas A.2) and (2.6) at the end. Similarly, (2.9) can be proved as
$$\begin{aligned}&\Vert \xi _{b, i} ( 1+e^{W_b^{(i)}/2}) ( {\bar{D}}_{1n} - {\bar{D}}_{1n}^{(i)}) \Vert _1\\&\quad \le \Vert \xi _{b, i}( 1+e^{W_b^{(i)}/2}) \Vert _2 \Vert {\bar{D}}_{1n} - {\bar{D}}_{1n}^{(i)} \Vert _2 \\&\quad = \Vert \xi _{b, i} \Vert _2 \Vert 1+e^{W_b^{(i)}/2} \Vert _2 \Vert {\bar{D}}_{1n} - {\bar{D}}_{1n}^{(i)} \Vert _2\\&\quad \le \Big ( 1+\exp ( e^2/8 - 1/8 + 1/4)\Big ) \Vert \xi _i \Vert _2 \Big \Vert D_{1n} - D_{1n}^{(i)} \Big \Vert _2, \end{aligned}$$
where we have also used the independence of \(e^{W_b^{(i)}}\) and \( \xi _{b, i} \).
Our next task is to bound the other terms in the general bound of Theorem 2.1. Let
$$\begin{aligned} {\bar{\Pi }}_k = \Pi _ k I(|\Pi _k| \le 1) + I(\Pi _k >1) - I(\Pi _k <-1)\text { for } k =1, 2. \end{aligned}$$
Since \(|D_{2n}| \le |\Pi _1| + |\Pi _2|\), and \(|{\bar{D}}_{2n}|\) is precisely \(|D_{2n}|\) as a non-negative random variable upper-censored at 1/2, it must be that \( |{\bar{D}}_{2n}|\le |{\bar{\Pi }}_1| + |{\bar{\Pi }}_2| \), which further implies
$$\begin{aligned} {\bar{D}}_{2n}^2\le 2 ( {\bar{\Pi }}_1^2 + {\bar{\Pi }}_2^2). \end{aligned}$$
(D.1)
From (D.1) and \({\bar{\Pi }}_2^2 \le |{\bar{\Pi }}_2|\), we can get
$$\begin{aligned} {\mathbb {E}}[{\bar{D}}_{2n}^2] \le 2(\Vert \Pi _1\Vert _2^2 + \Vert \Pi _2\Vert _2 ) \end{aligned}$$
(D.2)
On the other hand, define
$$\begin{aligned} D_{2n}^{(i)} = \max \Bigg (-1, \quad \sum _{ 1 \le i' \le n, i' \ne i} (\xi _{b, i'}^2 - {\mathbb {E}}[\xi _{b, i'}^2])+ \Pi _2^{(i)}\Bigg ). \end{aligned}$$
By Property 2.2(i), one can then write
$$\begin{aligned} \Vert (1+ e^{W_b^{(i)}})({\bar{D}}_{2n} - {\bar{D}}_{2n}^{(i)} )\Vert _1&\le \Vert (1+ e^{W_b^{(i)}} )(\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2]) \Vert _1 + \Vert (1+ e^{W_b^{(i)}} )(\Pi _2 -\Pi _2^{(i)}) \Vert _1 \nonumber \\&\le \Vert 1+ e^{W_b^{(i)}} \Vert _3\Vert \xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] \Vert _{3/2} + \Vert 1+ e^{W_b^{(i)}} \Vert _2 \Vert \Pi _2 -\Pi _2^{(i)} \Vert _2 \nonumber \\&\le C\Big ( ({\mathbb {E}}[|\xi _i|^3])^{2/3} + \Vert \Pi _2 -\Pi _2^{(i)} \Vert _2\Big ) \end{aligned}$$
(D.3)
and
$$\begin{aligned}&\Vert \xi _{b, i}( 1+ e^{W_b^{(i)}/2}) ({\bar{D}}_{2n} - {\bar{D}}_{2n}^{(i)} )\Vert _1 \nonumber \\&\quad \le \Vert \xi _{b, i}( 1+e^{W_b^{(i)}/2} )(\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2]) \Vert _1 + \Vert \xi _{b, i}( 1+e^{W_b^{(i)}/2}) (\Pi _2 -\Pi _2^{(i)}) \Vert _1 \nonumber \\&\quad \le \Vert \xi _{b, i}\Vert _3 \Vert 1+e^{W_b^{(i)}}\Vert _3 \Vert \xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] \Vert _{3/2} + \Vert \xi _{b, i} \Vert _2 \Vert 1+e^{W_b^{(i)}}\Vert _2 \Vert \Pi _2 -\Pi _2^{(i)} \Vert _2 \nonumber \\&\quad \le C\Big ( {\mathbb {E}}[|\xi _i|^3] + \Vert \xi _i \Vert _2 \Vert \Pi _2 -\Pi _2^{(i)} \Vert _2\Big ), \end{aligned}$$
(D.4)
where we have applied Bennett’s inequality (Lemma A.2) to both (D.3) and (D.4) at the end. To complete the proof, it suffices to show the bounds
$$\begin{aligned} {\mathbb {E}}[e^{W_b} {\bar{D}}_{2n}^2] \le C \Bigg \{\sum _{i=1}^n \Vert \xi _{b, i}\Vert _3^3 + \Vert \Pi _2\Vert _2 \Bigg \} \end{aligned}$$
(D.5)
and
$$\begin{aligned} \sup _{x \ge 0} |x {\mathbb {E}}[{\bar{D}}_{2n} f_x(W_b)]| \le C \Big ( \Vert \Pi _1\Vert _2^2 + \sum _{i=1}^n \Vert \xi _{b, i}\Vert _3^3 + \Vert \Pi _2\Vert _2 \Big ), \end{aligned}$$
(D.6)
because Theorem 2.3 is then just a corollary of Theorem 2.1 by collecting (2.8)–(2.10), (D.2)–(D.6), as well as the simple facts
$$\begin{aligned} \beta _2 + \beta _3\le & {} \sum _{i=1}^n {\mathbb {E}}[|\xi _i|^3], \qquad {\mathbb {E}}[|\xi _{b, i}|^2] \le \Vert \xi _{b, i}\Vert _2 \le \Vert \xi _i\Vert _2 \le \Vert \xi _i\Vert _3, \\ P(|D_{1n}| > 1/2)\le & {} 2\Vert D_{1n}\Vert _2, \quad \Vert \Pi _1\Vert _2^2 \le \sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^4] \le \sum _{i=1}^n {\mathbb {E}}[|\xi _{b, i}|^3] \le \sum _{i=1}^n {\mathbb {E}}[|\xi _i|^3], \end{aligned}$$
and
$$\begin{aligned} P(|D_{2n}|> 1/2)&\le P(|\Pi _1| + |\Pi _2|> 1/2)\\&\le P(|\Pi _1|>1/4) + P(|\Pi _2| >1/4) \\&\le C(\Vert \Pi _1\Vert ^2_2 + \Vert \Pi _2\Vert _2). \end{aligned}$$
1.1 Proof of (D.5).
First, letting \(W_b^{(i, j)} \equiv W_b - \xi _{b, i} - \xi _{b, j}\) for \(1 \le i \ne j \le n\), we have
$$\begin{aligned}&{\mathbb {E}}[ \Pi _1^2 e^{W_b}] = \sum _{i=1}^n {\mathbb {E}}[ (\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] )^2 e^{\xi _{b, i}}] {\mathbb {E}}[e^{ W_b^{(i)}}] \nonumber \\&\qquad +\sum _{1\le i \ne j \le n} {\mathbb {E}}[(\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] ) e^{ \xi _{b, i}}] {\mathbb {E}}[(\xi _{b, j}^2 - {\mathbb {E}}[\xi _{b, j}^2] ) e^{ \xi _{b, j}}] {\mathbb {E}}[e^{ W_b^{(i,j)}}] \nonumber \\&\quad = \sum _{i=1}^n {\mathbb {E}}[ (\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] )^2 e^{\xi _{b, i}}] {\mathbb {E}}[e^{ W_b^{(i)}}] \nonumber \\&\qquad + \sum _{1\le i \ne j \le n} {\mathbb {E}}[(\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] ) (e^{ \xi _{b, i}} - 1)] {\mathbb {E}}[(\xi _{b, j}^2 - {\mathbb {E}}[\xi _{b, j}^2] ) (e^{ \xi _{b, j}} - 1)] {\mathbb {E}}[e^{ W_b^{(i,j)}}] \nonumber \\&\quad \le C\left( \sum _{i=1}^n {\mathbb {E}}[\xi _{b, i}^4] + \sum _{1 \le i \ne j \le n} {\mathbb {E}}\Big [|\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] || \xi _{b, i}|\Big ] {\mathbb {E}}\Big [|\xi _{b, j}^2 - {\mathbb {E}}[\xi _{b, j}^2] | |\xi _{b, j}|\Big ] {\mathbb {E}}\Big [e^{W_b^{(i,j)}}\Big ] \right) \nonumber \\&\quad \le C \Bigg \{\sum _{i=1}^n \Vert \xi _{b, i}\Vert _3^3 + \sum _{1 \le i \ne j \le n} \Vert \xi _{b, i}\Vert _3^3 \Vert \xi _{b, j}\Vert _2^2 \Bigg \} \le C \sum _{i=1}^n \Vert \xi _{b, i}\Vert _3^3 \end{aligned}$$
(D.7)
by Lemma A.2 that \(|e^s - 1| \le |s| (e^a - 1)/a\) for \(s \le a\) and \(a >0\),
$$\begin{aligned}&{\mathbb {E}}[|\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] || \xi _{b, i}|] \\&\quad \le \Bigg \{(\Vert \xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2]\Vert _{3/2} \Vert \xi _{b, i}\Vert _3)\wedge {\mathbb {E}}[|\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2]|] \Bigg \}\\&\quad \le 2 \Bigg \{ \Vert \xi _{b, i}\Vert _3^3 \wedge \Vert \xi _{b, i}\Vert _2^2 \Bigg \} \text { for any } i = 1, \dots , n, \end{aligned}$$
and \(\sum _{j=1}^n\Vert \xi _{b, i}\Vert _3^3 \Vert \xi _{b, j}\Vert _2^2 \le \Vert \xi _{b, i}\Vert _3^3\). Second, by Lemma A.2,
$$\begin{aligned} {\mathbb {E}}[{\bar{\Pi }}_2^2 e^{W_b}] \le {\mathbb {E}}[{\bar{\Pi }}_2^4]^{1/2} ({\mathbb {E}}[e^{2W_b}])^{1/2} \le C {\mathbb {E}}[\Pi _2^2]^{1/2} = C \Vert \Pi _2\Vert _2 \end{aligned}$$
(D.8)
Combining (D.1), (D.7) and (D.8) gives (D.5).
1.2 Proof of (D.6).
Since \(\sup _{x \ge 0}|xf_x(w)| \le C\) (which uses (A.4) in Lemma A.4 and that \(|f_x| \le 0.63\) in Lemma A.3),
$$\begin{aligned} \sup _{x \ge 0}| x{\mathbb {E}}[(D_{2n} - {\bar{D}}_{2n})f_x(W_b)]|&\le \sup _{x \ge 0} x{\mathbb {E}}[ (|D_{2n}| - 1/2)|f_x(W_b)| I(|D_{2n}|> 1/2)] \\&\le C {\mathbb {E}}[ |D_{2n}| I(|D_{2n}|> 1/2)] \\&\le C \Big ({\mathbb {E}}[|\Pi _1| I(|D_{2n}|> 1/2)] + {\mathbb {E}}[|\Pi _2|] \Big )\\&\le C \Big ( {\mathbb {E}}\Big [|\Pi _1| \Big \{I(|\Pi _1|> 1/4)+I(|\Pi _2| > 1/4) \Big \}\Big ] + {\mathbb {E}}[|\Pi _2|] \Big )\\&\le C\Big ( {\mathbb {E}}[ 4 \Pi _1^2 + 2 |\Pi _1| |\Pi _2|^{1/2}] + {\mathbb {E}}[|\Pi _2|] \Big ) \\&\le C\Big ( {\mathbb {E}}[ 5\Pi _1^2 + |\Pi _2| ] + {\mathbb {E}}[|\Pi _2|] \Big ) \\&\le C\Big (\Vert \Pi _1\Vert _2^2 + \Vert \Pi _2\Vert _2 \Big ), \end{aligned}$$
where we have used that \(I(|\Pi _1| > 1/4) \le 4 |\Pi _1|\), \(I(|\Pi _2| > 1/4) \le 2 |\Pi _2|^{1/2}\) and \(2 |\Pi _1| |\Pi _2|^{1/2} \le |\Pi _1|^2 + |\Pi _2|\). Noting that
$$\begin{aligned} x{\mathbb {E}}[{\bar{D}}_{2n} f_x(W_b)] = x{\mathbb {E}}[({\bar{D}}_{2n} - D_{2n})f_x(W_b)] +x{\mathbb {E}}[ D_{2n}f_x(W_b)], \end{aligned}$$
the above implies
$$\begin{aligned} \sup _{x \ge 0} \Big |x{\mathbb {E}}[{\bar{D}}_{2n} f_x(W_b)] \Big | \le C\Big (\Vert \Pi _1\Vert _2^2 + \Vert \Pi _2\Vert _2 \Big ) + \sup _{x \ge 0} \Big |x{\mathbb {E}}[ D_{2n}f_x(W_b)] \Big |, \end{aligned}$$
(D.9)
so for the rest of this section we focus on bounding \(\sup _{x \ge 0} \Big |x{\mathbb {E}}[ D_{2n}f_x(W_b)] \Big |\). From the form of \(D_{2n}\) in (2.12), by defining \(\Pi = \Pi _1 + \Pi _2\), we have
$$\begin{aligned} x{\mathbb {E}}[D_{2n} f_x(W_b)] = {\mathbb {E}}[x \Pi f_x(W_b)] - {\mathbb {E}}[ xf_x(W_b)I(\Pi < -1) (1 + \Pi )], \end{aligned}$$
so it suffices to establish
$$\begin{aligned}{} & {} \Big |{\mathbb {E}}[x \Pi f_x(W_b)]\Big | \vee \Big |{\mathbb {E}}[ xf_x(W_b)I(\Pi < -1) (1 + \Pi )]\Big | \nonumber \\{} & {} \quad \le C \Bigg ( \sum _{i=1}^n {\mathbb {E}}[|\xi _{b, i}|^3 ] + \Vert \Pi _2\Vert _2 \Bigg ) \text { for all } x\ge 0. \end{aligned}$$
(D.10)
We first bound \(\Big |{\mathbb {E}}[ xf_x(W_b)I(\Pi < -1) (1 + \Pi )]\Big |\). Since
$$\begin{aligned} {\mathbb {E}}[ xf_x(W_b)I(\Pi< -1) (1 + \Pi )]= & {} {\mathbb {E}}[xf_x(W_b)I(\Pi< -1)] \nonumber \\{} & {} + {\mathbb {E}}[xf_x(W_b) \Pi I(\Pi < -1)], \end{aligned}$$
(D.11)
we will bound the two terms on the right hand side separately. As \(x f_x(w)\) is bounded for all \(x \ge 0\) (Lemma A.3 and (A.4) in Lemma A.4), we have
$$\begin{aligned} \Big |{\mathbb {E}}[xf_x(W_b)I(\Pi< -1)]\Big |&\le {\mathbb {E}}\Big [|xf_x(W_b)|I(\Pi< -1)\Big ]\\&\le C \sum _{j=1}^2P(\Pi _j < -1/2) \le C \Big (\Vert \Pi _1\Vert _2^2 + \Vert \Pi _2\Vert _2\Big ) \end{aligned}$$
and
$$\begin{aligned} \Big |{\mathbb {E}}[xf_x(W_b) \Pi I(\Pi< -1)]\Big |&\le C{\mathbb {E}}[|\Pi |I(\Pi< -1)]\\&\le C \Bigg ( {\mathbb {E}}[|\Pi _1| I(\Pi< -1)] + \Vert \Pi _2\Vert _2 \Bigg )\\&\le C \Bigg (\Vert \Pi _1\Vert _2 \sqrt{ \sum _{j=1}^2P(\Pi _j < -1/2)} + \Vert \Pi _2\Vert _2 \Bigg ) \\&\le C \Bigg (\Vert \Pi _1\Vert _2 \sqrt{ \Vert \Pi _1\Vert _2^2 + \Vert \Pi _2\Vert _2 } + \Vert \Pi _2\Vert _2 \Bigg )\\&\le C \Bigg ( \Vert \Pi _1\Vert _2^2 + \Vert \Pi _1\Vert _2\sqrt{ \Vert \Pi _2\Vert _2} + \Vert \Pi _2\Vert _2 \Bigg )\\&\le C \Bigg ( \Vert \Pi _1\Vert _2^2 + \Vert \Pi _2\Vert _2 \Bigg ) , \end{aligned}$$
where the second last inequality uses \(\sqrt{ \Vert \Pi _1\Vert _2^2 + \Vert \Pi _2\Vert _2 } \le \Vert \Pi _1\Vert _2 + \sqrt{\Vert \Pi _2\Vert _2}\) and the last inequality uses that \(2|ab| \le a^2 + b^2\) for any \(a, b \in {\mathbb {R}}\). So the part of (D.10) regarding \(\Big |{\mathbb {E}}[ xf_x(W_b)I(\Pi < -1) (1 + \Pi )]\Big |\) is proved because \(\Vert \Pi _1\Vert _2^2 = \sum _{i=1}^n ({\mathbb {E}}[\xi _{b, i}^4] - ({\mathbb {E}}[\xi _{b, i}^2])^2) \le \sum _{i=1}^n {\mathbb {E}}[|\xi _{b, i}|^3 ]\).
Next we bound \(\Big |{\mathbb {E}}[x \Pi f_x(W_b)]\Big |\), and we will control the two terms on the right-hand side of
$$\begin{aligned} |{\mathbb {E}}[x\Pi f_x(W_b)]| \le x|{\mathbb {E}}[\Pi _1 f_x(W_b)]| + x | {\mathbb {E}}[\Pi _2 f_x(W_b)]|. \end{aligned}$$
(D.12)
For the first term \(x|{\mathbb {E}}[\Pi _1 f_x(W_b)]|\), we write
$$\begin{aligned} \Big | {\mathbb {E}}[\Pi _1 f_x(W_b)]\Big |&= \Bigg | \sum _{i=1}^n {\mathbb {E}}\Big [(\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2] ) (f_x (W_b) - f_x(W_b^{(i)})) \Big ] \Bigg |\nonumber \\&= \Bigg |\sum _{i=1}^n {\mathbb {E}}\Big [(\xi _{b, i}^2 - {\mathbb {E}}[\xi _{b, i}^2]) \int _0^{\xi _{b, i}} {\mathbb {E}}[f_x'(W_b^{(i)} + t)] \hbox {d}t\Big ] \Bigg |\nonumber \\&\le \sum _{i=1}^n {\mathbb {E}}\Big [ (\xi _{b, i}^2 + {\mathbb {E}}[\xi _{b, i}^2]) \int _0^{|\xi _{b, i}|} |{\mathbb {E}}[f_x'(W_b^{(i)} + t)]| \hbox {d}t\Big ] , \end{aligned}$$
(D.13)
where the second equality uses the independence of \(W_b^{(i)} \) and \(\xi _{b, i}\). From (D.13) and Lemma A.5, for any \(x \ge 1\), we have that
$$\begin{aligned} \Big |{\mathbb {E}}[\Pi _1 f_x(W_b)]\Big |&\le C\sum _{i=1}^n {\mathbb {E}}\Big [ (\xi _{b, i}^2 + {\mathbb {E}}[\xi _{b, i}^2]) \int _0^{|\xi _{b, i}|} (e^{ -x } + e^{- x +t})\hbox {d}t\Big ]\\&\le C\sum _{i=1}^n {\mathbb {E}}\Big [( \xi _{b, i}^2 + {\mathbb {E}}[\xi _{b, i}^2] ) \int _0^{|\xi _{b, i}|} (e^{ -x } + e^{- x +1})\hbox {d}t\Big ] \text { ( as }|\xi _{b, i}| \le 1 )\\&\le C e^{-x}\sum _{i=1}^n \bigg ({\mathbb {E}}[|\xi _{b,i}|^3] + {\mathbb {E}}[|\xi _{b,i}|^2] {\mathbb {E}}[|\xi _{b,i}|]\bigg ) \\&\le C e^{-x}\sum _{i=1}^n {\mathbb {E}}[|\xi _{b,i}|^3], \end{aligned}$$
which implies
$$\begin{aligned} \sup _{x \ge 1} x \Big |{\mathbb {E}}[\Pi _1 f_x(W_b)]\Big | \le C \sum _{i=1}^n {\mathbb {E}}[|\xi _{b, i}|^3] . \end{aligned}$$
(D.14)
Moreover, for \(0 \le x < 1\), since \(|f_x'| \le 1\) (Lemma A.3), from (D.13) we get
$$\begin{aligned} \sup _{0 \le x < 1} x \Big | {\mathbb {E}}[\Pi _1 f_x(W_b)]\Big |\le & {} \sum _{i=1}^n \bigg ( {\mathbb {E}}[|\xi _{b, i}|^3] + {\mathbb {E}}[|\xi _{b, i}|^2] {\mathbb {E}}[|\xi _{b, i}|] \bigg ) \nonumber \\\le & {} 2 \sum _{i=1}^n {\mathbb {E}}[|\xi _{b, i}|^3]. \end{aligned}$$
(D.15)
For the term \(x | {\mathbb {E}}[\Pi _2 f_x(W_b)]|\), given that \( \sup _{x \ge 0}|xf_x(w)| \le C \text { for all } w \) (explained at the beginning of Sect. D.2), we have
$$\begin{aligned} \sup _{x \ge 0}x |{\mathbb {E}}[\Pi _2 f_x (W_b)]| \le \sup _{x \ge 0} {\mathbb {E}}[|\Pi _2 | |x f_x(W_b)|] \le C \Vert \Pi _2\Vert _1 \le C \Vert \Pi _2\Vert _2, \end{aligned}$$
(D.16)
Combining (D.12) and (D.14)–(D.16) proves the part of (D.10) regarding \(|{\mathbb {E}}[x\Pi f_x(W_b)]|\).
Appendix E. Proof of Lemma 3.3
In this section, we adopt the following notation: For any natural numbers \(k' \le k\), we denote \([k':k] \equiv \{k', \dots , k\}\) and \([k] \equiv \{1, \dots , k\}\). Moreover, for any natural number \(k \ge 1\), we let
$$\begin{aligned} {\bar{h}}_{k, \{i_1, \dots , i_k\}} \equiv {\bar{h}}_k(X_{i_1}, \dots , X_{i_k}) \end{aligned}$$
with respect to the function \({\bar{h}}_k(\cdot )\) in (3.9). To prove Lemma 3.3, we need the following technical lemmas proven, respectively, in Appendices F.1 and F.2.
Lemma E.1
(Useful kernel bounds) Under assumptions (3.1)–(3.3),
-
(i)
For any \(k \in [m]\),
$$\begin{aligned} {\mathbb {E}}[{\bar{h}}_k^2 ] \le {\mathbb {E}}[h_k^2] \le \frac{k}{m}{\mathbb {E}}[h^2] \end{aligned}$$
-
(ii)
For any \(i \in [n]\),
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ \left( \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-1} \le n\\ i_l \ne i \text { for } l \in [m-1] \end{array}}{\bar{h}}_m(X_i, X_{i_1}, \dots , X_{i_{m-1}}) \right) ^2\right] \\{} & {} \quad \le \frac{2(m-1)^2}{ n(n-m+1)} {n-1 \atopwithdelims ()m-1} {n \atopwithdelims ()m} {\mathbb {E}}[h^2] ; \end{aligned}$$
-
(iii)
For each \(i \in [n]\), consider \(\xi _{b, i}\) defined in (2.1) with \(\xi _i\) defined in (3.8). Given \(k_1, k_2 \in [m]\), for any \(1 \le i_1< \dots < i_{k_1} \le n\) and \(1 \le j_1< \dots < j_{k_2} \le n\), we have
$$\begin{aligned} \Big |{\mathbb {E}}[\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{k_1, \{ i_1, \dots , i_{k_1}\}} {\bar{h}}_{k_2, \{j_1, \dots , j_{k_2}\}}]\Big | \\ \le \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \frac{2d\Vert h\Vert _2}{n} \end{aligned}$$
where
$$\begin{aligned} d = | (\{ i_1, \dots , i_{k_1} \} \cap \{j_1, \dots , j_{k_2}\}) \backslash \{1, 2\}|, \end{aligned}$$
the number of elements in the intersection of \(\{ i_1, \dots , i_{k_1} \}\) and \(\{ j_1, \dots , j_{k_2} \}\) that are not 1 or 2.
-
(iv)
If, in addition to all the conditions in (iii), it is true that \(1 \not \in \{j_1, \dots , j_{k_2}\}\) and \(2 \not \in \{ i_1, \dots , i_{k_1} \} \), then we have the bound
$$\begin{aligned} \Big |{\mathbb {E}}[\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{k_1, \{ i_1, \dots , i_{k_1}\}} {\bar{h}}_{k_2, \{j_1, \dots , j_{k_2}\}}]\Big | \le \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \frac{2d\Vert h\Vert _2}{n^{3/2}} \end{aligned}$$
Lemma E.2
(Counting identities and bounds) Let m, n be non-negative integers such that \(m \le n\).
-
(i)
Suppose \(n_1\) and \(n_2\) are non-negative integers such that \(n_1 + n_2 = n\). Then
$$\begin{aligned} \sum _{k = 0}^m {n_1 \atopwithdelims ()k} {n_2 \atopwithdelims ()m - k} = {n \atopwithdelims ()m}. \end{aligned}$$
-
(ii)
Suppose k is a non-negative integer such that \(k \le m\). Then
$$\begin{aligned} {n \atopwithdelims ()k} {n - k \atopwithdelims ()m - k} = {n \atopwithdelims ()m} {m \atopwithdelims ()k}. \end{aligned}$$
-
(iii)
For positive integers a, b, e such that \(b+ e \le a\), we have
$$\begin{aligned} \left( {\begin{array}{c}a\\ b\end{array}}\right) - \left( {\begin{array}{c}a -e\\ b\end{array}}\right) \le \left( {\begin{array}{c}a\\ b\end{array}}\right) \frac{ b e}{a - b +1}. \end{aligned}$$
In addition to the lemmas above, we will make use of the following enumerative equalities, whenever the binomial coefficients involved are well defined:
$$\begin{aligned} \left( {\begin{array}{c}n-2\\ m-1\end{array}}\right)&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \frac{n-m}{n-1}, \end{aligned}$$
(E.1)
$$\begin{aligned} \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right)&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \frac{m-1}{n-1}, \end{aligned}$$
(E.2)
$$\begin{aligned} \left( {\begin{array}{c}n-3\\ m-2\end{array}}\right)&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \frac{(m-1)(n-m)}{(n-1)(n-2)} \end{aligned}$$
(E.3)
$$\begin{aligned} \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right)&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \frac{(m-1)(m-2)}{(n-1)(n-2)} \quad \text {, and } \end{aligned}$$
(E.4)
$$\begin{aligned} \left( {\begin{array}{c}n-4\\ m-4\end{array}}\right)&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \frac{(m-1)(m-2)(m-3)}{(n-1)(n-2)(n-3)} . \end{aligned}$$
(E.5)
1.1 Proof of Lemma 3.3(i)
We shall further let
$$\begin{aligned} \Pi _{21}\equiv & {} (n^{-1/2} | 0) - \sum _{i=1}^n {\mathbb {E}}[ (\xi ^2_i - 1)I(|\xi _i| > 1) ] \text { and }\nonumber \\ \Pi _{22}\equiv & {} \delta _{2n, b} = \frac{2 (n-1)}{(n-m)} {n-1 \atopwithdelims ()m-1}^{-1} \sum _{i=1}^n \xi _{b, i} \Psi _{n,i}, \end{aligned}$$
(E.6)
so \(\Pi _2 = \Pi _{21} + \Pi _{22}\). It suffices to show these bounds for \(\Pi _{21}\) and \(\Pi _{22}\) in (E.6):
$$\begin{aligned} \Vert \Pi _{21}\Vert _2^2\le & {} C \left( \frac{\Vert g\Vert _3^6}{n} + \frac{1}{n}\right) \le C \frac{\Vert g\Vert _3^6}{n} . \end{aligned}$$
(E.7)
$$\begin{aligned} \Vert \Pi _{22}\Vert _2^2\le & {} C\frac{m^2\Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} \end{aligned}$$
(E.8)
From there, since \(\Vert \Pi _2\Vert _2 \le \Vert \Pi _{21}\Vert _2 + \Vert \Pi _{22}\Vert _2\), Lemma 3.3(i) is proved.
1.1.1 Proof of (E.7)
We first note that
$$\begin{aligned} \sum _{i=1}^n {\mathbb {E}}\Big [ (\xi ^2_i - 1)I(|\xi _i|> 1) \Big ] \le \sum _{i=1}^n {\mathbb {E}}\Big [ \xi ^2_i I(|\xi _i| > 1) \Big ] \le \sum _{i=1}^n{\mathbb {E}}[|\xi _i|^3] = {\mathbb {E}}[|g|^3]/\sqrt{n}, \end{aligned}$$
which gives \((\sum _{i=1}^n {\mathbb {E}}[ (\xi ^2_i - 1)I(|\xi _i| > 1) ])^2 \le ({\mathbb {E}}[|g|^3])^2/n\), and hence (E.7).
1.1.2 Proof of (E.8)
It is trivial for \(m=1\) since \(\Psi _{n,i} = 0\). For \(m\ge 2\), first write
$$\begin{aligned} \Pi _{22}^2 = \frac{4 (n-1)^2}{(n-m)^2n} {n-1 \atopwithdelims ()m-1}^{-2} \left( \sum _{i=1}^n \xi _{b, i} \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-1} \le n\\ i_l \ne i \text { for } l \in [m-1] \end{array}} {\bar{h}}_m(X_i, X_{i_1}, \dots , X_{i_{m-1}}) \right) ^2, \end{aligned}$$
which implies immediately from \(2m < n\) in (3.2) that
$$\begin{aligned} {\mathbb {E}}\left[ \Pi _{22}^2\right] \le \frac{16}{n} {n-1 \atopwithdelims ()m-1}^{-2} {\mathbb {E}}\left[ \left( \sum _{i=1}^n \xi _{b, i} \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-1} \le n\\ i_l \ne i \text { for } l\in [m-1] \end{array}} {\bar{h}}_m(X_i, X_{i_1}, \dots , X_{i_{m-1}}) \right) ^2\right] . \end{aligned}$$
(E.9)
Upon expanding the above expectation,
$$\begin{aligned}&{\mathbb {E}}\left[ \left( \sum _{i=1}^n \xi _{b, i} \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n\\ i_l \ne i \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{i, i_1, \dots , i_{m-1}\}} \right) ^2\right] \nonumber \\&\quad = \sum _{i=1}^n {\mathbb {E}}\left[ \left( \xi _{b, i} \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n\\ i_l \ne i \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{i, i_1, \dots , i_{m-1}\}}\right) ^2\right] \nonumber \\&\qquad + \sum _{ 1\le i \ne j \le n} {\mathbb {E}}\Bigg [ \Bigg (\xi _{b, i} \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n\\ i_l \ne i \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{i, i_1, \dots , i_{m-1}\}}\Bigg )\nonumber \\&\qquad \times \Bigg ( \xi _{b, j} \sum _{\begin{array}{c} 1 \le j_1< \dots< j_{m-1} \le n\\ j_l \ne j \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{j, j_1, \dots , j_{m-1}\}} \Bigg )\Bigg ]\nonumber \\&\quad = n {\mathbb {E}}\left[ \left( \xi _{b, 1} \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-1} \le n\\ i_l \ne 1 \text { for } l = 1, \dots , m-1 \end{array}} {\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}}\right) ^2\right] + \end{aligned}$$
(E.10)
$$\begin{aligned}&n(n-1) {\mathbb {E}}\Bigg [ \Bigg (\xi _{b, 1} \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n\\ i_l \ne 1 \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}}\Bigg ) \nonumber \\ {}&\quad \Bigg ( \xi _{b, 2} \sum _{\begin{array}{c} 1 \le j_1< \dots < j_{m-1} \le n\\ j_l \ne 2 \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{2, j_1, \dots , j_{m-1}\}} \Bigg )\Bigg ]. \end{aligned}$$
(E.11)
We need to control the two expectations in (E.10) and (E.11). We first bound the expectation in (E.10). With the definition in (3.9) and that
$$\begin{aligned}{} & {} {\mathbb {E}}[{\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}} {\bar{h}}_{m, \{1, j_1, \dots , j_{m-1}\}}] \\ {}{} & {} \quad = {\mathbb {E}}[{\bar{h}}_{1, \{1\}}^2] = 0 \text { if } |\{i_1, \dots , i_{m-1}\}\cap \{j_1, \dots , j_{m-1}\}| = 0, \end{aligned}$$
we can write
$$\begin{aligned}&{\mathbb {E}}\left[ \left( \xi _{b, 1} \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n\\ i_l \ne 1 \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}}\right) ^2\right] \nonumber \\&\quad = {\mathbb {E}}\left[ \xi _{b, 1}^2 \sum _{k=0}^{m-1} \left( \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n\\ 1 \le j_1< \dots < j_{m-1} \le n \\ i_l, j_l \ne 1 \text { for } l \in [m-1] \nonumber \\ |\{i_1, \dots , i_{m-1}\}\cap \{j_1, \dots , j_{m-1}\}| = k \end{array}} {\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}}{\bar{h}}_{m, \{1, j_1, \dots , j_{m-1}\}} \right) \right] \nonumber \\&\quad = \sum _{k=1}^{m-1} {n-1 \atopwithdelims ()k} {n -k - 1\atopwithdelims ()m - k - 1} {n - m \atopwithdelims ()m-k-1} {\mathbb {E}}\Bigg [ \xi _{b, 1}^2 {\bar{h}}_{k+1}^2(X_1 \dots , X_{k+1}) \Bigg ]\nonumber \\&\quad \le \sum _{k=1}^{m-1} {n-1 \atopwithdelims ()k} {n -k -1\atopwithdelims ()m - k -1} {n - m \atopwithdelims ()m-k-1} \frac{k+1}{m} {\mathbb {E}}[ h^2 ], \end{aligned}$$
(E.12)
where the last inequality comes from Lemma E.1(i) and that \(\xi _{b, 1}^2 \le 1\). Continuing from (E.12), we can get
$$\begin{aligned}&{\mathbb {E}}\left[ \left( \xi _{b, 1} \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-1} \le n\\ i_l \ne 1 \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}}\right) ^2\right] \nonumber \\&\quad \le \sum _{k=1}^{m-1} {n-1 \atopwithdelims ()k} {n -k -1\atopwithdelims ()m - k -1} {n - m \atopwithdelims ()m-k-1} \frac{k+1}{m} {\mathbb {E}}[ h^2 ] \nonumber \\&\quad = \frac{1}{m}\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \sum _{k=1}^{m-1}\left( {\begin{array}{c}m-1\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-k-1\end{array}}\right) (k+1) {\mathbb {E}}[h^2] \text { by Lemma }\text {E.2}(ii) \nonumber \\&\quad = \frac{m-1}{m}\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \sum _{k=1}^{m-1}\left( {\begin{array}{c}m-2\\ k-1\end{array}}\right) \frac{k+1}{k} \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right) {\mathbb {E}}[h^2] \nonumber \\&\quad \le 2\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \sum _{k=0}^{m-2}\left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) {\mathbb {E}}[h^2] \nonumber \\&\quad =2\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) {\mathbb {E}}[h^2] \text { by Lemma }\text {E.2}(i) \nonumber \\&\quad = 2\frac{m-1}{n-1}\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2{\mathbb {E}}[h^2] \end{aligned}$$
(E.13)
Now we bound the expectation in (E.11). First first expand it as
$$\begin{aligned}&{\mathbb {E}}\Bigg [ \Bigg (\xi _{b, 1} \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n\\ i_l \ne 1 \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}}\Bigg ) \Bigg ( \xi _{b, 2} \sum _{\begin{array}{c} 1 \le j_1< \dots< j_{m-1} \le n\\ j_l \ne 2 \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{2, j_1, \dots , j_{m-1}\}}) \Bigg )\Bigg ] \nonumber \\&\quad = {n-2 \atopwithdelims ()m-1} {n-2-(m -1)\atopwithdelims ()m-1}\Bigg (\underbrace{{\mathbb {E}}\left[ \xi _{b, 1} {\bar{h}}_{m, \{1, \dots , m\}}\right] }_{ = {\mathbb {E}}[{\mathbb {E}}[\xi _{b,1} {\bar{h}}_{m, \{1, \dots , m\}} | X_1]] ={\mathbb {E}}[\xi _{b,1}{\bar{h}}_1 (X_1)] = 0}\Bigg )^2 \nonumber \\&\qquad + 2 \times {n-2 \atopwithdelims ()m-2} {n-2 - (m-2) \atopwithdelims ()m-1}\underbrace{{\mathbb {E}}\Bigg [ \xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, 2, \dots , m\}} {\bar{h}}_{m, \{2, m+1, \dots , 2m-1 \}} \Bigg ] }_{ = {\mathbb {E}}[ {\mathbb {E}}[ \xi _{b, 1} \xi _{b, 2} {\bar{h}}_{2, \{1, 2\}} {\bar{h}}_{1, \{2 \}}| X_1, X_2]]=0 \text { since } {\bar{h}}_{1, \{ 2\}} = 0} \nonumber \\&\qquad + 2 \times \underbrace{\sum _{ \begin{array}{c} 1 \le i_1< \dots< i_{m-2} \le n \\ 1 \le j_1< \dots< j_{m-1} \le n\\ i_l, j_v \ne 1, 2, \text { for } l \in [m-2], v \in [m-1]\\ |\{i_1, \dots , i_{m-2}\}\cap \{j_1, \dots , j_{m-1}\}| \ge 1 \end{array}} {\mathbb {E}}[\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, 2, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{2, j_1, \dots , j_{m-1}\}}] }_{\equiv EA}\nonumber \\&\qquad + \underbrace{ \sum _{ \begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n \\ 1 \le j_1< \dots< j_{m-1} \le n\\ i_l, j_l \ne 1, 2, \text { for } l \in [m-1]\\ |\{i_1, \dots , i_{m-1}\}\cap \{j_1, \dots , j_{m-1}\}| \ge 1 \end{array}}{\mathbb {E}}[\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}} {\bar{h}}_{m, \{2, j_1, \dots , j_{m-1}\}}]}_{\equiv EB} \nonumber \\&\qquad +\underbrace{ \sum _{ \begin{array}{c} 1 \le i_1< \dots< i_{m-2} \le n \\ 1 \le j_1< \dots < j_{m-2} \le n\\ i_l, j_l \ne 1, 2, \text { for } l \in [m-2] \end{array}}{\mathbb {E}}[\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, 2, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{1, 2, j_1, \dots , j_{m-2}\}}]}_{\equiv EC} , \end{aligned}$$
(E.14)
and will then bound each of EA, EB, and EC.
We start with EA, and it suffices to assume \(m \ge 3\), otherwise one cannot expect the two sets \(\{i_1, \dots , i_{m-2}\}\) and \(\{j_1, \dots , j_{m-1}\}\) indexing a given summand
$$\begin{aligned} {\mathbb {E}}[ \xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, 2, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{2, j_1, \dots , j_{m-1} \}}] \end{aligned}$$
of EA to intersect for at least one element. Using the fact that the data \(X_1, \dots , X_n\) are i.i.d., if the two index sets have \(k \in [m-2]\) common elements not in the set \(\{1, 2\}\), one can write the summand as
$$\begin{aligned}{} & {} {\mathbb {E}}[ \xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, 2, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{2, j_1, \dots , j_{m-1} \}}] \\{} & {} \quad = {\mathbb {E}}[\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_m(X_1, X_2, \dots , X_m)\ {\bar{h}}_m(X_2,X_3,\dots , X_{k+2}, X_{m+1},\dots , X_{2m-1-k})]. \end{aligned}$$
From this, we can alternatively write
$$\begin{aligned} EA= & {} \sum _{k=1}^{m-2}\left( {\begin{array}{c}n-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-2-k\\ m-2-k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right) \\{} & {} {\mathbb {E}}\Big [\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_{m, [1:m]}\ {\bar{h}}_{m, [2: (k+2)]\cup [(m+1) : (2m-k-1)]}\Big ]; \end{aligned}$$
from this, we can then form the bound
$$\begin{aligned} |EA|&\le \sum _{k=1}^{m-2}\left( {\begin{array}{c}n-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-2-k\\ m-2-k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right) \\&\quad \Big | {\mathbb {E}}\Big [\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_{m, [1:m]}\ {\bar{h}}_{m, [2: (k+2)]\cup [(m+1) : (2m-k-1)]}\Big ]\Big |\\&= \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \sum _{k=1}^{m-2}\left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right) \\&\quad \Big | {\mathbb {E}}\Big [\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_{m, [1:m]}\ {\bar{h}}_{m, [2: (k+2)]\cup [(m+1) : (2m-k-1)]}\Big ]\Big |\\&\hspace{8cm} \text { by Lemma } \text {E.2}(ii)\\&\le \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \sum _{k=1}^{m-2}\left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right) \bigg \{\frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \frac{2 k\Vert h\Vert _2}{n}\bigg \}\\&\hspace{8cm} \text { by Lemma }\text {E.1} (iii) \\&= \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \bigg \{ \bigg [\left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) - \left( {\begin{array}{c}n-m\\ m-1\end{array}}\right) \bigg ]\frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + (m-2)\left( {\begin{array}{c}n-3\\ m-2\end{array}}\right) \frac{2 \Vert h\Vert _2}{n} \bigg \}, \end{aligned}$$
where the last line comes from the equalities
$$\begin{aligned} \sum _{k=1}^{m-2}\left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right)&= \sum _{k=0}^{m-2}\left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right) - \left( {\begin{array}{c}n-m\\ m-1\end{array}}\right) \\&= \left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) - \left( {\begin{array}{c}n-m\\ m-1\end{array}}\right) \text { by Lemma }\text {E.2}(i) \end{aligned}$$
and
$$\begin{aligned} \sum _{k=1}^{m-2} k \left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right)&= (m-2) \sum _{k=1}^{m-2} \left( {\begin{array}{c}m-3\\ k -1\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-1-k\end{array}}\right) \\&= (m-2) \sum _{k=0}^{m-3} \left( {\begin{array}{c}m-3\\ k \end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \\&= (m-2) \left( {\begin{array}{c}n -3\\ m-2\end{array}}\right) \text { coming from Lemma }\text {E.2}(i) \end{aligned}$$
Continuing, we get
$$\begin{aligned} |EA|&\le \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \bigg \{ \left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) \frac{9.5(m -2)(m-1)\Vert g\Vert _3^2 \Vert h\Vert _3^2}{(n -m)n} + (m-2)\left( {\begin{array}{c}n-3\\ m-2\end{array}}\right) \frac{2 \Vert h\Vert _2}{n} \bigg \} \nonumber \\&\hspace{9cm} \text { by Lemma }\text {E.2}(iii) \nonumber \\&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \bigg \{ \frac{9.5(m-2)(m-1)^2 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{(n-1)^2 n} + \frac{2 (m-1)^2 (m-2)(n-m) \Vert h\Vert _2}{n (n-1)^2 (n-2)}\bigg \} \nonumber \\&\hspace{8cm} \text { by }(\text {E.1}), (\text {E.2})\text { and } (\text {E.3}) \nonumber \\&\le C\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \frac{m^3 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n^3} , \end{aligned}$$
(E.15)
where the last line uses \(2m < n\), and \(1= \sigma _g \le \Vert h\Vert _2 \le \Vert h\Vert _3\).
Now we bound EB. Analogously to EA, we first write
$$\begin{aligned} |EB|&\le \sum _{k=1}^{m-1}\left( {\begin{array}{c}n-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-2-k\\ m-1-k\end{array}}\right) \left( {\begin{array}{c}n-m-1\\ m-1-k\end{array}}\right) \\&\quad \Big |{\mathbb {E}}[\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_m(X_1, X_3, \dots , X_{m+1})\ {\bar{h}}_m(X_2,\underbrace{X_3,\dots , X_{k+2}}_{k\ \text {shared}}, X_{m+2},\dots , X_{2m-k})]\Big |\\&= \left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) \sum _{k=1}^{m-1}\left( {\begin{array}{c}m-1\\ k\end{array}}\right) \left( {\begin{array}{c}n-m-1\\ m-1-k\end{array}}\right) \\&\quad \ \Big |{\mathbb {E}}[\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_m(X_1, X_3, \dots , X_{m+1})\ {\bar{h}}_m(X_2,\underbrace{X_3,\dots , X_{k+2}}_{k\ \text {shared}}, X_{m+2},\dots , X_{2m-k})]\Big |\\&\hspace{8cm} \text { by Lemma }\text {E.2} (ii)\\&\le \left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) \sum _{k=1}^{m-1}\left( {\begin{array}{c}m-1\\ k\end{array}}\right) \left( {\begin{array}{c}n-m-1\\ m-1-k\end{array}}\right) \bigg (\frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \frac{2k\Vert h\Vert _2}{n^{3/2}}\bigg ) \text { by Lemma }\text {E.1}(iv)\\&=\left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) \bigg \{\bigg [\left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) - \left( {\begin{array}{c}n -m - 1\\ m-1\end{array}}\right) \bigg ] \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \left( {\begin{array}{c}n -3\\ m -2\end{array}}\right) \frac{2 (m-1)\Vert h\Vert _2}{n^{3/2}} \bigg \}, \end{aligned}$$
where in the last equality, we have used
$$\begin{aligned} \sum _{k=1}^{m-1}\left( {\begin{array}{c}m-1\\ k\end{array}}\right) \left( {\begin{array}{c}n-m-1\\ m-1-k\end{array}}\right)&= \sum _{k=0}^{m-1}\left( {\begin{array}{c}m-1\\ k\end{array}}\right) \left( {\begin{array}{c}n-m-1\\ m-1-k\end{array}}\right) - \left( {\begin{array}{c}n-m-1\\ m-1\end{array}}\right) \\&= \left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) - \left( {\begin{array}{c}n-m-1\\ m-1\end{array}}\right) \text { by Lemma }\text {E.2}(i) \end{aligned}$$
and
$$\begin{aligned} \sum _{k=1}^{m-1}\left( {\begin{array}{c}m-1\\ k\end{array}}\right) \left( {\begin{array}{c}n-m-1\\ m-1-k\end{array}}\right) k&= (m-1) \sum _{k=1}^{m-1}\left( {\begin{array}{c}m-2\\ k - 1\end{array}}\right) \left( {\begin{array}{c}n-m-1\\ m-1-k\end{array}}\right) \\&= (m-1) \sum _{k=0}^{m-2} \left( {\begin{array}{c}m-2\\ k \end{array}}\right) \left( {\begin{array}{c}n-m-1\\ m-2-k\end{array}}\right) \\&= (m-1) \left( {\begin{array}{c}n -3\\ m -2\end{array}}\right) \text { by Lemma }\text {E.2}(i) \end{aligned}$$
Continuing, we get
$$\begin{aligned} |EB|&\le \left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) \bigg \{ \left( {\begin{array}{c}n-2\\ m-1\end{array}}\right) \frac{(m-1)^2}{n-m}\frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \left( {\begin{array}{c}n -3\\ m -2\end{array}}\right) \frac{2 (m-1)\Vert h\Vert _2}{n^{3/2}} \bigg \} \nonumber \\&\hspace{8cm}\text { by Lemma }\text {E.2}(iii) \nonumber \\&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \bigg \{ \frac{9.5(m-1)^2(n-m) \Vert g\Vert _3^2 \Vert h\Vert _3^2}{ (n-1)^2n} +\frac{2(m-1)^2(n-m)^2\Vert h\Vert _2}{(n-1)^2(n-2)n^{3/2}} \bigg \} \nonumber \\&\hspace{8cm}\text { by }(\text {E.1})\text { and }(\text {E.3}) \nonumber \\&\le C\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \frac{ m^2\Vert g\Vert _3^2 \Vert h\Vert _3^2}{n^2} , \end{aligned}$$
(E.16)
where the last line uses \(2m < n\), and \(1= \sigma _g \le \Vert h\Vert _2 \le \Vert h\Vert _3\).
Lastly, for EC, in an analogous manner as EA and EB, we first write it as
$$\begin{aligned} EC&= \sum _{k=0}^{m-2}\left( {\begin{array}{c}n-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-2-k\\ m-2-k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \\&\quad {\mathbb {E}}[\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_m(X_1, X_2, \dots , X_{m})\\ {}&\quad {\bar{h}}_m(X_1, X_2,\underbrace{X_3,\dots , X_{k+2}}_{k\ \text {shared, empty if }k=0}, X_{m+1},\dots , X_{2m-k-2})]. \end{aligned}$$
Then we can bound
$$\begin{aligned} |EC|&\le \sum _{k=0}^{m-2}\left( {\begin{array}{c}n-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-2-k\\ m-2-k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \\&\quad \Big |{\mathbb {E}}[\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_m(X_1, X_2, \dots , X_{m})\ {\bar{h}}_m(X_1, \dots , X_{k+2}, X_{m+1},\dots , X_{2m-k-2}]\Big |\\&\le \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \sum _{k=0}^{m-2}\left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \\&\quad \Big |{\mathbb {E}}[\xi _{b, 1} \xi _{b, 2}\ {\bar{h}}_m(X_1, X_2, \dots , X_{m})\ {\bar{h}}_m(X_1,\dots , X_{k+2}, X_{m+1},\dots , X_{2m-k-2}]\Big |\\&\hspace{9cm}\text {by Lemma }\text {E.2}(ii) \\&\le \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \sum _{k=0}^{m-2} \left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \bigg (\frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \frac{2k\Vert h\Vert _2}{n}\bigg ) \text { by Lemma }\text {E.1}(iii)\\&= \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \bigg \{ \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \frac{2(m-2)\Vert h\Vert _2}{n}\bigg \}, \end{aligned}$$
where the last equality comes from
$$\begin{aligned} \sum _{k=0}^{m-2} \left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) = \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \text { by Lemma }\text {E.2}(i) \end{aligned}$$
and for \(m \ge 3\),
$$\begin{aligned} \sum _{k=0}^{m-2} \left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) k&= \sum _{k=1}^{m-2} \left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) k \\&= (m-2)\sum _{k=1}^{m-2} \left( {\begin{array}{c}m-3\\ k-1\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \\&= (m-2)\sum _{k=0}^{m-3} \left( {\begin{array}{c}m-3\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-3-k\end{array}}\right) \\&= (m-2) \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \text { by Lemma }\text {E.2}(i). \end{aligned}$$
Continuing, we get by (E.2) and (E.4),
$$\begin{aligned} |EC|&\le \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \bigg \{ \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \frac{2(m-2)\Vert h\Vert _2}{n}\bigg \}\nonumber \\&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \bigg \{ \frac{9.5 (m-1)^2\Vert g\Vert _3^2 \Vert h\Vert _3^2}{n (n-1)^2} +\frac{2(m-1)^2(m-2)^2\Vert h\Vert _2}{n(n-1)^2(n-2)}\bigg \} \nonumber \\&\le C \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \bigg \{ \frac{m^2\Vert g\Vert _3^2 \Vert h\Vert _3^2}{n^3} + \frac{m^4 \Vert h\Vert _2}{n^4}\bigg \} \end{aligned}$$
(E.17)
Substituting (E.15), (E.16), and (E.17) into (E.14), we get that
$$\begin{aligned}{} & {} \Bigg | {\mathbb {E}}\bigg [ \bigg (\xi _{b, 1} \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-1} \le n\\ i_l \ne 1 \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{1, i_1, \dots , i_{m-1}\}}\bigg ) \bigg ( \xi _{b, 2} \sum _{\begin{array}{c} 1 \le j_1< \dots < j_{m-1} \le n\\ j_l \ne 2 \text { for } l \in [m-1] \end{array}} {\bar{h}}_{m, \{2, j_1, \dots , j_{m-1}\}}) \bigg )\bigg ] \Bigg |\nonumber \\{} & {} \quad \le C\left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \frac{m^2\Vert g\Vert _3^2 \Vert h\Vert _3^2}{n^2}, \end{aligned}$$
(E.18)
where we have used that \(2m < n\) and \(1 = \Vert g\Vert _2 \le \Vert h\Vert _2 \le \Vert h\Vert _3\). Finally, collecting (E.9), (E.10), (E.11), (E.13) and (E.18), we obtain (E.8).
1.2 Proof of Lemma 3.3(ii)
Note that
$$\begin{aligned} \delta _{2n, b} - \delta _{2n, b}^{(i)} = A + B, \end{aligned}$$
where
$$\begin{aligned} A = \frac{2 (n-1) }{\sqrt{n}(n-m)} {n-1 \atopwithdelims ()m-1}^{-1} \xi _{b, i} \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-1} \le n \\ i_l \ne i \text { for } l \in [m-1] \end{array}} {\bar{h}}_m (X_i, X_{i_1}, \dots , X_{i_{m-1}}) \end{aligned}$$
and
$$\begin{aligned} B = \frac{2 (n-1) }{\sqrt{n}(n-m)} {n-1 \atopwithdelims ()m-1}^{-1} \sum _{\begin{array}{c} 1 \le j \le n \\ j \ne i \end{array}} \Bigg (\xi _{b, j} \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-2} \le n\\ i_l \ne j, i \text { for } l = 1, \dots , m-2 \end{array}} {\bar{h}}_m (X_j, X_i, X_{i_1}, \dots , X_{i_{m-2}}) \Bigg ). \end{aligned}$$
From (3.26) and (3.28), we first write
$$\begin{aligned} \Vert \Pi _2 - \Pi _2^{(i)} \Vert _2&\le {\mathbb {E}}[ (\xi _i^2 - 1)I(|\xi _i|> 1) ] + \Vert \delta _{2n, b} - \delta _{2n, b}^{(i)}\Vert _2 \nonumber \\&\le \frac{{\mathbb {E}}[g^2]}{n} + \Vert A\Vert _2 + \Vert B\Vert _2, \end{aligned}$$
(E.19)
by Lemma A.1, where
$$\begin{aligned} A = \frac{2 (n-1) }{\sqrt{n}(n-m)} {n-1 \atopwithdelims ()m-1}^{-1} \xi _{b, i} \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-1} \le n \\ i_l \ne i \text { for } l \in [m-1] \end{array}} {\bar{h}}_m (X_i, X_{i_1}, \dots , X_{i_{m-1}}) \end{aligned}$$
and
$$\begin{aligned} B = \frac{2 (n-1) }{\sqrt{n}(n-m)} {n-1 \atopwithdelims ()m-1}^{-1} \sum _{\begin{array}{c} 1 \le j \le n \\ j \ne i \end{array}} \Bigg (\xi _{b, j} \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-2} \le n\\ i_l \ne j, i \text { for } l = 1, \dots , m-2 \end{array}} {\bar{h}}_m (X_j, X_i, X_{i_1}, \dots , X_{i_{m-2}}) \Bigg ). \end{aligned}$$
So we will bound \(\Vert A\Vert _2\) and \(\Vert B\Vert _2\), which is trivial for \(m=1\) as \({\bar{h}}_1(\cdot ) = 0\). For \(m \ge 2\), by Lemma E.1(ii),
$$\begin{aligned} {\mathbb {E}}[A^2]\le & {} \frac{4 (n-1)^2 }{n(n-m)^2} {n-1 \atopwithdelims ()m-1}^{-2} {\mathbb {E}}\Bigg [\Bigg ( \sum _{\begin{array}{c} 1 \le i_1< \dots < i_{m-1} \le n \\ i_l \ne i \text { for } l \in [m-1] \end{array}} {\bar{h}}_m (X_i, X_{i_1}, \dots , X_{i_{m-1}})\Bigg )^2\Bigg ] \nonumber \\\le & {} \frac{8 (n-1)^2 (m-1)^2 {\mathbb {E}}[h^2]}{(n-m)^2n(n-m+1)m } \le C \frac{m{\mathbb {E}}[h^2] }{n^2} \end{aligned}$$
(E.20)
Moreover, for B, we first expand its second moment as
$$\begin{aligned}&{\mathbb {E}}[B^2] \nonumber \\&\quad = \frac{4 (n-1)^2 }{n(n-m)^2} {n-1 \atopwithdelims ()m-1}^{-2} {\mathbb {E}}\Bigg [ \Bigg ( \sum _{\begin{array}{c} 1 \le j \le n \\ j \ne i \end{array}} \Bigg (\xi _{b, j} \sum _{ \begin{array}{c} 1 \le i_1< \dots< i_{m-2} \le n\\ i_l \ne j, i \text { for } l = 1, \dots , m-2 \end{array} } {\bar{h}}_{m, \{j, i, i_1, \dots , i_{m-2}\}} \Bigg )\Bigg )^2\Bigg ] \nonumber \\&\quad = \frac{4 (n-1)^2 }{n(n-m)^2} {n-1 \atopwithdelims ()m-1}^{-2} \nonumber \\&\qquad \times \Bigg \{ (n-1) \underbrace{ \sum _{ \begin{array}{c} 1 \le i_1< \dots< i_{m-2} \le m-2\\ 1 \le j_1< \dots< j_{m-2} \le m-2\\ i_l, j_l \ne 1, 2 \text { for } l \in [m-2] \end{array}} {\mathbb {E}}[\xi _{b, 1}^2 {\bar{h}}_{m, \{1, 2, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{1, 2, j_1, \dots , j_{m-2}\}}] }_{\equiv ED} \nonumber \\&\qquad + (n-1)(n-2) \underbrace{ \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-2} \le n \\ 1 \le j_1< \dots < j_{m-2} \le n \\ i_l \ne 1, 3 \text { for } l \in [m-2]\\ j_l \ne 2, 3 \text { for } l \in [m-2] \end{array}} {\mathbb {E}}[\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, 3, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{2, 3, j_1, \dots , j_{m-2}\}}] }_{\equiv EE} \Bigg \} . \end{aligned}$$
(E.21)
To bound ED, we first note that, by \(|\xi _{b, 1}| \le 1\), Hölder’s inequality and Lemma E.1(i), each of its summand can be bounded as
$$\begin{aligned} \Big |{\mathbb {E}}[\xi _{b, 1}^2 {\bar{h}}_{m, \{1, 2, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{1, 2, j_1, \dots , j_{m-2}\}}]\Big | \le {\mathbb {E}}[h^2] \end{aligned}$$
(E.22)
Then, by considering the number of elements \(k \in [m-2]\) shared by the sets \(\{i_1, \dots , i_{m-2}\}\) and \(\{j_1, \dots , j_{m-2}\}\) indexing each such summand, we have the bound
$$\begin{aligned} |ED|&\le \sum _{k =0}^{m-2} \left( {\begin{array}{c}n-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-2-k\\ m-2-k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) {\mathbb {E}}[h^2] \nonumber \\&= \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) \sum _{k =0}^{m-2} \left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) {\mathbb {E}}[h^2] \text { by Lemma } \text {E.2}(ii) \nonumber \\&= \left( {\begin{array}{c}n-2\\ m-2\end{array}}\right) ^2 {\mathbb {E}}[h^2] \text { by Lemma }\text {E.2}(i) \nonumber \\&= \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \bigg (\frac{m-1}{n-1}\bigg )^2 {\mathbb {E}}[h^2]\text { by }(\text {E.2}) . \end{aligned}$$
(E.23)
To bound EE, we first break it down as
$$\begin{aligned}&EE = \nonumber \\&\quad \underbrace{ \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-2} \le n \\ 1 \le j_1< \dots< j_{m-2} \le n \\ i_l \ne 1, 2, 3 \text { for } l \in [m-2]\\ j_l \ne 1, 2, 3 \text { for } l \in [m-2] \end{array}} {\mathbb {E}}[\xi _{b, 1} \xi _{b, 2}{\bar{h}}_{m, \{1, 3, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{2, 3, j_1, \dots , j_{m-2}\}}}_{\equiv EE_1}\nonumber \\&\qquad +\underbrace{ \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-3} \le n \\ 1 \le j_1< \dots< j_{m-2} \le n \\ i_l \ne 1, 2, 3 \text { for } l \in [m-3]\\ j_l \ne 1, 2, 3 \text { for } l \in [m-2] \end{array}} {\mathbb {E}}[\xi _{b, 1}\xi _{b, 2} {\bar{h}}_{m, \{1, 2, 3, i_1, \dots , i_{m-3}\}} {\bar{h}}_{m, \{2, 3, j_1, \dots , j_{m-2}\}}] }_{\equiv EE_2} \nonumber \\&\qquad + \underbrace{ \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-2} \le n \\ 1 \le j_1< \dots< j_{m-3} \le n \\ i_l \ne 1, 2, 3 \text { for } l \in [m-2]\\ j_l \ne 1, 2, 3 \text { for } l \in [m-3] \end{array}} {\mathbb {E}}[\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, 3, i_1, \dots , i_{m-2}\}} {\bar{h}}_{m, \{1, 2, 3, j_1, \dots , j_{m-3}\}}]}_{\equiv EE_3} \nonumber \\&\qquad + \underbrace{ \sum _{\begin{array}{c} 1 \le i_1< \dots< i_{m-3} \le n \\ 1 \le j_1< \dots < j_{m-3} \le n \\ i_l \ne 1, 2, 3 \text { for } l \in [m-3]\\ j_l \ne 1, 2, 3 \text { for } l \in [m-3] \end{array}} {\mathbb {E}}[\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{m, \{1, 2, 3, i_1, \dots , i_{m-3}\}} {\bar{h}}_{m, \{1, 2, 3, j_1, \dots , j_{m-3}\}} }_{\equiv EE_4}. \end{aligned}$$
(E.24)
Using Lemma E.1(iv), one can then bound \(EE_1\) as
$$\begin{aligned}&|EE_1| \nonumber \\&\quad \le \sum _{k=0}^{m-2} \left( {\begin{array}{c}n-3\\ k\end{array}}\right) \left( {\begin{array}{c}n-3-k\\ m-2-k\end{array}}\right) \left( {\begin{array}{c}n-1 -m\\ m-2-k\end{array}}\right) \bigg ( \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \frac{2d\Vert h\Vert _2}{n^{3/2}}\bigg ) \nonumber \\&\quad \le \sum _{k=0}^{m-2} \left( {\begin{array}{c}n-3\\ k\end{array}}\right) \left( {\begin{array}{c}n-3-k\\ m-2-k\end{array}}\right) \left( {\begin{array}{c}n-1 -m\\ m-2-k\end{array}}\right) \bigg ( \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} + \frac{2\Vert h\Vert _2^2}{n}\bigg )\nonumber \\&\hspace{7cm} \text { by }(\text {3.17})\text { and }d \le m \le n \nonumber \\&\quad \le 11.5 \left( {\begin{array}{c}n-3\\ m-2\end{array}}\right) \sum _{k=0}^{m-2} \left( {\begin{array}{c}m-2\\ k\end{array}}\right) \left( {\begin{array}{c}n-1 -m\\ m-2-k\end{array}}\right) \frac{\Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} \nonumber \\&\hspace{6cm}\text { by Lemma }\text {E.2}(ii)\text { and }\Vert h\Vert _2 \le \Vert h\Vert _3\nonumber \\&\quad = 11.5 \left( {\begin{array}{c}n-3\\ m-2\end{array}}\right) ^2 \frac{\Vert g\Vert _3^2 \Vert h\Vert _3^2}{n} \text { by Lemma }\text {E.1}(i) \nonumber \\&\quad = 11.5 \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \frac{(m-1)^2(n-m)^2\Vert g\Vert _3^2 \Vert h\Vert _3^2}{n(n-1)^2(n-2)^2 } \text { by }(\text {E.3}) . \end{aligned}$$
(E.25)
For \(EE_2\) and \(EE_3\), using Lemma E.1(iii), one can bound them similarly as
$$\begin{aligned}&\max (|EE_2|, |EE_3|) \nonumber \\&\quad \le \sum _{k =0}^{m-3} \left( {\begin{array}{c}n-3\\ k\end{array}}\right) \left( {\begin{array}{c}n-3-k\\ m-3-k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \bigg ( \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n}+ \frac{2(2+k)\Vert h\Vert _2}{n}\bigg )\nonumber \\&\quad = \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \sum _{k =0}^{m-3}\left( {\begin{array}{c}m-3\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \bigg ( \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n}+ \frac{2(2+k)\Vert h\Vert _2}{n}\bigg )\text { by Lemma }\text {E.2}(ii) \nonumber \\&\quad = \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \left( {\begin{array}{c}n-3\\ m-2\end{array}}\right) \frac{9.5\Vert g\Vert _3^2\Vert h\Vert _3^2 + 4 \Vert h\Vert _2}{n} \nonumber \\&\qquad + \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \sum _{k =1}^{m-3} \left( {\begin{array}{c}m-4\\ k-1\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-2-k\end{array}}\right) \frac{2 (m-3)\Vert h\Vert _2 }{n} \text { by Lemma }\text {E.2}(i) \nonumber \\&\quad = \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \bigg \{ \left( {\begin{array}{c}n-3\\ m-2\end{array}}\right) \frac{9.5\Vert g\Vert _3^2\Vert h\Vert _3^2 + 4 \Vert h\Vert _2}{n} + \sum _{k =0}^{m-4} \left( {\begin{array}{c}m-4\\ k\end{array}}\right) \left( {\begin{array}{c}n-m\\ m-3-k\end{array}}\right) \frac{2 (m-3)\Vert h\Vert _2 }{n} \bigg \} \nonumber \\&\quad = \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \bigg \{ \left( {\begin{array}{c}n-3\\ m-2\end{array}}\right) \frac{9.5\Vert g\Vert _3^2\Vert h\Vert _3^2 + 4 \Vert h\Vert _2}{n} + \left( {\begin{array}{c}n-4\\ m-3\end{array}}\right) \frac{2 (m-3)\Vert h\Vert _2 }{n} \bigg \} \text {by Lemma }\text {E.2}(i)\nonumber \\&\quad = \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \bigg \{ \left( {\begin{array}{c}n-3\\ m-2\end{array}}\right) \frac{9.5\Vert g\Vert _3^2\Vert h\Vert _3^2 + 4 \Vert h\Vert _2}{n} + \left( {\begin{array}{c}n-3\\ m-3\end{array}}\right) \frac{2 (m-3)(n-m)\Vert h\Vert _2 }{(n-3)n} \bigg \}\nonumber \\&\quad = \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \bigg \{ \frac{(m-1)^2(m-2)(n-m)(9.5\Vert g\Vert _3^2\Vert h\Vert _3^2 + 4 \Vert h\Vert _2)}{(n-1)^2(n-2)^2 n} \nonumber \\&\qquad + \frac{2(m-1)^2(m-2)^2(m-3)(n-m)\Vert h\Vert _2}{(n-1)^2(n-2)^2(n-3)n} \bigg \} \text { by }(\text {E.3})\text { and }(\text {E.4})\nonumber \\&\quad \le C \left( {\begin{array}{c}n-1\\ m-1\end{array}}\right) ^2 \bigg \{ \frac{m^3 \Vert g\Vert _3^2\Vert h\Vert _3^2 }{n^4} + \frac{ m^5\Vert h\Vert _2}{n^5} \bigg \} \text {by }1 \le \Vert g\Vert _3\text { and }\Vert h\Vert _2 \le \Vert h\Vert _3. \end{aligned}$$
(E.26)
Lastly, for \(EE_4\), using Lemma E.1(iii), one can bound it as
$$\begin{aligned}&|EE_4| \nonumber \\&\quad \le \sum _{k =0}^{m-3} {n -3 \atopwithdelims ()k} {n -3 - k \atopwithdelims ()m - 3 - k} {n -m \atopwithdelims ()m - 3 - k} \bigg ( \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n}+ \frac{2(3+k)\Vert h\Vert _2}{n}\bigg ) \nonumber \\&\quad = {n -3 \atopwithdelims ()m -3} \sum _{k =0}^{m-3} {m-3 \atopwithdelims ()k} {n -m \atopwithdelims ()m - 3 - k} \nonumber \\&\quad \bigg ( \frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n}+ \frac{2(3+k)\Vert h\Vert _2}{n}\bigg ) \text { by Lemma }\text {E.2} (ii) \nonumber \\&\quad = {n -3 \atopwithdelims ()m -3} \bigg \{ {n-3 \atopwithdelims ()m-3}\frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2 + 6 \Vert h\Vert _2}{n} \nonumber \\&\qquad + \frac{2 (m-3)\Vert h\Vert _2}{n} \sum _{k =1}^{m-3} {m-4 \atopwithdelims ()k-1} {n-m \atopwithdelims ()m - 4 - (k-1)} \bigg \}\nonumber \\&\quad = {n -3 \atopwithdelims ()m -3} \bigg \{ {n-3 \atopwithdelims ()m-3}\frac{9.5 \Vert g\Vert _3^2 \Vert h\Vert _3^2 + 6 \Vert h\Vert _2}{n}\nonumber \\&\qquad + \frac{2 (m-3)\Vert h\Vert _2}{n} {n -4 \atopwithdelims ()m - 4} \bigg \} \text { by Lemma }\text {E.2} (i) \nonumber \\&\quad \le C{n -1 \atopwithdelims ()m-1}^2 \bigg \{ \frac{ m^4 \Vert g\Vert _3^2 \Vert h\Vert _3^2 }{n^5 } + \frac{m^6 \Vert h\Vert _2}{n^6} \bigg \} \text {by }(\text {E.4}), (\text {E.5}), \nonumber \\ {}&\qquad 1 \le \Vert g\Vert _3\text { and }\Vert h\Vert _2 \le \Vert h\Vert _3. \end{aligned}$$
(E.27)
Combining (E.24), (E.25), (E.26), (E.27), and \(2m <n\), we get that
$$\begin{aligned} |EE| \le C {n-1 \atopwithdelims ()m-1}^2 \bigg \{ \Vert g\Vert _3^2 \Vert h\Vert _3^2 \bigg (\frac{m^2}{n^3} + \frac{m^3}{n^4} + \frac{m^4}{n^5}\bigg ) + \Vert h\Vert _2 \bigg (\frac{m^5}{n^5} \bigg ) \bigg \}. \end{aligned}$$
(E.28)
Combining (E.21), (E.23), and (E.28), we get
$$\begin{aligned} {\mathbb {E}}[B^2]&\le C \bigg \{\frac{m^2}{n^2} {\mathbb {E}}[h^2] + \bigg [ \Vert g\Vert _3^2 \Vert h\Vert _3^2 \bigg (\frac{m^2}{n^2} + \frac{m^3}{n^3} + \frac{m^4}{n^4}\bigg ) + \Vert h\Vert _2 \bigg (\frac{m^5}{n^4} \bigg ) \bigg ] \bigg \} \nonumber \\&\le C \bigg \{ \frac{m^2 \Vert g\Vert _3^2 \Vert h\Vert _3^2}{n^2} +\frac{m^5 \Vert h\Vert _2 }{n^4} \bigg \}, \end{aligned}$$
(E.29)
where we have used \(2m < n\), as well as \(\Vert h\Vert _2 \le \Vert h\Vert _3\) and \(1 = \Vert g\Vert _2 \le \Vert g\Vert _3\) in the last line. Combining (E.19), (E.20), and (E.29) gives Lemma 3.3(ii).
Appendix F. Proof of Lemmas E.1 and E.2
1.1 Proof of Lemma E.1
The proof for (i) and (ii) can be found in Chen et al. [3, Ch.10, Appendix]. We will focus on proving (iii) and (iv). For any subset \( \{i_1, \dots , i_k\} \subset [n]\), we will denote
$$\begin{aligned} X_{\{i_1, \dots , i_k\}} = \{X_{i_1}, \dots , X_{i_k}\}. \end{aligned}$$
To simplify the notation, we also denote
$$\begin{aligned} I = \{i_1, \dots , i_{k_1}\} \text { and } J = \{j_1, \dots , j_{k_2}\}, \end{aligned}$$
as well as
$$\begin{aligned} h_I = h_{k_1}(X_{i_1}, \dots , X_{i_{k_1}}) \text { and } {\bar{h}}_I = {\bar{h}}_{k_1, \{i_1, \dots , i_{k_1}\} } \end{aligned}$$
and
$$\begin{aligned} h_J = h_{k_2}(X_{j_1}, \dots , X_{j_{k_2}}) \text { and } {\bar{h}}_I = {\bar{h}}_{k_2, \{j_1, \dots , i_{j_2}\} }. \end{aligned}$$
First, it suffices to assume both
$$\begin{aligned} k_1, k_2 \ge 2 \end{aligned}$$
because if any of \(k_1\) and \(k_2\) is equal to 1, then one of \({\bar{h}}_{k_1, \{ i_1, \dots , i_{k_1}\}}\) and \({\bar{h}}_{k_2, \{j_1, \dots , j_{k_2}\}}\) must be equal to zero by the definition in (3.9), so the bound is trivial. Moreover, one can further assume without loss of generality that the index sets I and J are such that
$$\begin{aligned} I \backslash \{1, 2\} = J \backslash \{1, 2\} = [3: (d+2)] \text { if } d >0, \end{aligned}$$
(F.1)
in which case it must be true that \(|I \backslash \{1, 2\}| = |J \backslash \{1, 2\} | = d\). This is because for any I and J, we have
$$\begin{aligned}&{\mathbb {E}}[ \xi _{b, 1} \xi _{b, 2}{\bar{h}}_I {\bar{h}}_J ] \\&\quad = {\mathbb {E}}\Big [{\mathbb {E}}[ \xi _{b, 1} \xi _{b, 2}{\bar{h}}_I {\bar{h}}_J \mid X_{\{1, 2\}\cup (I \cap J)} ] \Big ]\\&\quad = {\mathbb {E}}\Big [\xi _{b, 1} \xi _{b, 2} {\mathbb {E}}[ {\bar{h}}_I {\bar{h}}_J \mid X_{\{1, 2\}\cup (I \cap J)} ] \Big ] \\&\quad = {\mathbb {E}}\Big [\xi _{b, 1} \xi _{b, 2} {\mathbb {E}}[ {\bar{h}}_I \mid X_{\{1, 2\}\cup (I \cap J)} ] {\mathbb {E}}[{\bar{h}}_J \mid X_{\{1, 2\}\cup (I \cap J)}]\Big ]\\&\hspace{1cm} \text { because }I \backslash \Big (\{1, 2\}\cup (I \cap J)\Big )\text { and }J \backslash \Big ( \{1, 2\}\cup (I \cap J)\Big )\text { are disjoint}\\&\quad ={\mathbb {E}}\Big [\xi _{b, 1} \xi _{b, 2} {\mathbb {E}}[ {\bar{h}}_I \mid X_{(I \cap \{1, 2\})\cup (I \cap J)} ] {\mathbb {E}}[{\bar{h}}_J \mid X_{(J\cap \{1, 2\})\cup (I \cap J)}]\Big ] \\&\quad = {\mathbb {E}}\Big [\xi _{b, 1} \xi _{b, 2} {\bar{h}}_{(I \cap \{1, 2\})\cup (I \cap J)} {\bar{h}}_{(J\cap \{1, 2\})\cup (I \cap J)}\Big ]. \end{aligned}$$
Since
$$\begin{aligned} \Big ((I \cap \{1, 2\})\cup (I \cap J) \Big )\backslash \{1, 2\} = (I \cap J) \backslash \{1, 2\} = \Big ((J \cap \{1, 2\})\cup (I \cap J) \Big )\backslash \{1, 2\} \end{aligned}$$
and
$$\begin{aligned} |(I \cap J) \backslash \{1, 2\}| = d \text { by assumption}, \end{aligned}$$
by the i.i.d.’ness of the data \(X_1, \dots , X_n\) it suffices to assume (F.1).
By the definition in (3.9), we perform the expansion
$$\begin{aligned}&{\mathbb {E}}[\xi _{b,1}\xi _{b,2}\ {\bar{h}}_{I}\ {\bar{h}}_{J}]\\&\quad = {\mathbb {E}}\Big [\xi _{b,1}\xi _{b,2}\Big (h_{I} - \sum _{i\in I\cap \{1,2\}}g(X_i)- \sum _{i\in I\setminus \{1,2\}}g(X_i)\Big )\\&\quad \Big (h_{J} -\sum _{j\in J\cap \{1,2\}}g(X_j)- \sum _{j\in J\setminus \{1,2\}}g(X_j)\Big )\Big ]\\&\quad = \underbrace{{\mathbb {E}}[\xi _{b,1}\xi _{b,2}\ h_{I}\ h_{J}]}_{\equiv HH} \\&\qquad - \underbrace{\sum _{i\in I\cap \{1,2\}}{\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_i)\ h_{J}]}_{\equiv GH_{1}} - \underbrace{\sum _{j\in J\cap \{1,2\}}{\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_j)\ h_{I}]}_{\equiv GH_{2}}\\&\qquad - \underbrace{ \sum _{ i \in I \backslash \{1, 2\} }{\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_i)\ h_J]}_{\equiv GH_{3}} - \underbrace{\sum _{ j \in J \backslash \{1, 2\} }{\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_j)\ h_I]}_{\equiv GH_{4}}\\&\qquad + \underbrace{\sum _{i\in I\cap \{1,2\}}\sum _{j\in J\cap \{1,2\}}{\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_i)g(X_j)]}_{\equiv GG_1} + \underbrace{ \sum _{i \in I \backslash \{1, 2\}} \sum _{j \in J \backslash \{1, 2\}} {\mathbb {E}}[\xi _{b,1}\xi _{b,2} g(X_i) g(X_j))]}_{\equiv GG_2}, \end{aligned}$$
recognizing that the last batch of expansion terms
$$\begin{aligned} \sum _{i\in I\cap \{1,2\}} \sum _{j\in J\backslash \{1,2\}}\underbrace{{\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_i)g(X_j)]}_{={\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_i)]{\mathbb {E}}[g(X_j)] = 0} + \sum _{i\in I\backslash \{1,2\}} \sum _{j\in J\cap \{1,2\}}\underbrace{{\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_i)g(X_j)]}_{ {\mathbb {E}}[\xi _{b,1}\xi _{b,2}g(X_j)]{\mathbb {E}}[g(X_i)]=0} \end{aligned}$$
vanish. The remaining terms in each row of the expansion above are bounded as follows:
1.1.1 Bound on HH:
$$\begin{aligned} |HH|&= \Big | {\mathbb {E}}[\xi _{b, 1} \xi _{b, 2}\ h_{I}\ h_{J}]\Big | \le \Big \Vert \xi _{b, 1} \xi _{b,2}\Big \Vert _3 \Big \Vert h_{I}\ h_{J}\Big \Vert _{3/2} \nonumber \\&=\Big ({\mathbb {E}}[|\xi _{b, 1} |^3] {\mathbb {E}}[|\xi _{b, 2} |^3]\Big )^{1/3} \Big ( {\mathbb {E}}\Big [ \Big | h_{I}\Big |^{3/2} \Big | h_{J}\Big |^{3/2}\Big ]\Big )^{2/3} \nonumber \\&\le \Big ({\mathbb {E}}[|\xi _{b, 1} |^3] {\mathbb {E}}[|\xi _{b, 2} |^3]\Big )^{1/3} \left( \Vert |h_{I}|^{3/2}\Vert _2 \Vert |h_{J}|^{3/2}\Vert _2 \right) ^{2/3} \text { by Cauchy's inequality}\nonumber \\&\le n^{-1}\Vert g\Vert _3^2 \Vert h\Vert _3^2, \end{aligned}$$
(F.2)
where the last line comes from (3.10) with \(|I|\vee |J| \le m\).
1.1.2 Bound on \(GH_1 + G H_2\):
$$\begin{aligned}&|GH_1+GH_2| \nonumber \\&\quad \le \sum _{i \in I \cap \{1, 2\}} \Vert \xi _{b,1}\xi _{b,2}g(X_i) \Vert _{3/2} \Vert h_J \Vert _{3} + \sum _{j \in J \cap \{1, 2\}} \Vert \xi _{b,1}\xi _{b,2}g(X_j) \Vert _{3/2} \Vert h_I \Vert _{3} \nonumber \\&\quad = |I \cap \{1, 2\}| \cdot \Vert \xi _{b,1}\xi _{b,2}g(X_1) \Vert _{3/2} \Vert h_J \Vert _{3} + |J \cap \{1, 2\}| \cdot \Vert \xi _{b,1}\xi _{b,2}g(X_1) \Vert _{3/2} \Vert h_I \Vert _{3} \nonumber \\&\quad \le 4 \Vert \xi _{b,1}\xi _{b,2}g(X_1) \Vert _{3/2} \Vert h\Vert _3 \quad \text { by } (\text {3.10}) \nonumber \\&\quad = 4 \Vert \xi _{b, 1} g(X_1)\Vert _{3/2}\Vert \xi _{b, 2}\Vert _{3/2} \Vert h\Vert _3 \quad \text { by independence } \nonumber \\&\quad \le 4 \Big ({\mathbb {E}}[n^{-3/4}|g(X_1)|^3]\Big )^{2/3}\Big ({\mathbb {E}}[n^{-3/4}|g(X_2)|^{3/2}]\Big )^{2/3} \Vert h\Vert _3 \nonumber \\&\quad = 4n^{-1} \Vert g\Vert _3^2 \Vert g\Vert _{3/2} \Vert h\Vert _3 \nonumber \\&\quad \le 4n^{-1}\Vert g\Vert _3^2 \Vert h\Vert _3, \end{aligned}$$
(F.3)
where the last inequality is true because \(\Vert g\Vert _{3/2} \le \Vert g\Vert _2 = \sigma _g = 1\).
1.1.3 General bound on \(GH_3 + G H_4\):
$$\begin{aligned}&|GH_3+GH_4| \nonumber \\&\quad \le \sum _{ i \in I \backslash \{1, 2\} } \Vert \xi _{b,1}\xi _{b,2}g(X_i)\Vert _2 \Vert h_J\Vert _2+ \sum _{ j \in J \backslash \{1, 2\} } \Vert \xi _{b,1}\xi _{b,2}g(X_j)\Vert _2 \Vert h_I\Vert _2 \nonumber \\&\quad = |I \backslash \{1, 2\}| \cdot \Vert \xi _{b,1}\xi _{b,2}g(X_3)\Vert _2 \Vert h_J\Vert _2 + |J \backslash \{1, 2\}| \cdot \Vert \xi _{b,1}\xi _{b,2}g(X_3)\Vert _2 \Vert h_I\Vert _2 \nonumber \\&\quad \le 2d \Vert \xi _{b,1}\xi _{b,2}g(X_3)\Vert _2 \Vert h\Vert _2 \text { by } (\text {3.10}) \text { and } (\text {F.1}) \nonumber \\&\quad \le 2 d\Vert \xi _1\Vert _2 \Vert \xi _2\Vert _2 \Vert g(X_3)\Vert _2 \Vert h\Vert _2 \text { by independence } \nonumber \\&\quad = 2dn^{-1}\Vert h\Vert _2 \text { by }(\text {3.3}). \end{aligned}$$
(F.4)
1.1.4 Special bound on \(GH_3 + G H_4\) under \(1 \not \in J\) and \(2 \not \in I\):
$$\begin{aligned}&|GH_3+GH_4| \nonumber \\&\quad = \bigg |\sum _{ i \in I \backslash \{1, 2\} }{\mathbb {E}}[\xi _{b,1}] {\mathbb {E}}[\xi _{b,2}g(X_i)\ h_J] + \sum _{ j \in J \backslash \{1, 2\} }{\mathbb {E}}[\xi _{b,2} ] {\mathbb {E}}[\xi _{b,1}g(X_j)\ h_I] \bigg | \nonumber \\&\hspace{8cm}\text { by }1 \not \in J\text { and }2 \not \in I \nonumber \\&\quad \le \sum _{ i \in I \backslash \{1, 2\} } \big |{\mathbb {E}}[\xi _{b,1}]\big | \ \Vert \xi _{b,2}g(X_i)\Vert _2 \Vert h\Vert _2 \nonumber \\&\qquad +\sum _{ j \in J \backslash \{1, 2\} } \big |{\mathbb {E}}[\xi _{b,2} ] \big | \ \Vert \xi _{b,1}g(X_j)\Vert _2 \Vert h\Vert _2 \text { by } (\text {3.10}) \nonumber \\&\quad \le 2d \cdot \big |{\mathbb {E}}[\xi _{b,1}]\big | \ \Vert \xi _{b,1}g(X_3)\Vert _2 \Vert h\Vert _2 \text { by }(\text {F.1}) \nonumber \\&\quad \le 2d {\mathbb {E}}[\xi _1^2] \ \Vert \xi _1\Vert _2 \ \Vert g(X_3)\Vert _2 \Vert h\Vert _2 \text { by Lemma }\text {A.1}\text { and independence} \nonumber \\&= 2d n^{-3/2} \Vert h\Vert _2 \text { by }\sigma _g^2 =1\text { in }(\text {3.3}). \end{aligned}$$
(F.5)
1.1.5 Bound on \(GG_1 + GG_2\)
$$\begin{aligned}&|GG_1+GG_2| \\&\quad \le 2 \Big ( {\mathbb {E}}[|\xi _{b,1}g^2(X_1)|] \cdot |{\mathbb {E}}[\xi _{b,2}]| + {\mathbb {E}}[|\xi _{b,1}g(X_1)|]\cdot {\mathbb {E}}[|\xi _{b,2}g(X_2)|] \Big )\\&\qquad + \Big |\sum _{i \in I \backslash \{1, 2\}} \sum _{j \in J \backslash \{1, 2\}} {\mathbb {E}}[\xi _{b,1}\xi _{b,2} g(X_i) g(X_j))] \Big |\\&\quad = 2 \Big ( {\mathbb {E}}[|\xi _{b,1}g^2(X_1)|] \cdot |{\mathbb {E}}[\xi _{b,2}]| + {\mathbb {E}}[|\xi _{b,1}g(X_1)|]\cdot {\mathbb {E}}[|\xi _{b,2}g(X_2)|] \Big ) \\&\qquad + d\cdot |{\mathbb {E}}[\xi _{b, 1}] | \cdot |{\mathbb {E}}[ \xi _{b, 2}]|\cdot {\mathbb {E}}[ g^2(X_3)], \end{aligned}$$
where the last equality uses that
$$\begin{aligned} {\mathbb {E}}[\xi _{b,1}\xi _{b,2} g(X_i) g(X_j))]= & {} {\mathbb {E}}[\xi _{b,1}\xi _{b,2}] {\mathbb {E}}[ g(X_i)] {\mathbb {E}}[ g(X_j))] \\ {}= & {} 0 \text { if } i \ne j \text { and } i, j \not \in \{1, 2\}, \end{aligned}$$
as well as the working assumption in (F.1). Continuing, we get
$$\begin{aligned}&|GG_1+GG_2| \nonumber \\&\quad \le 2 \Big ( {\mathbb {E}}[g^2(X_1)] \cdot |{\mathbb {E}}[\xi _{b,2}]| + n^{-1}{\mathbb {E}}[g^2(X_1)]\cdot {\mathbb {E}}[g^2(X_2)] \Big )\nonumber \\&\qquad + d\cdot |{\mathbb {E}}[\xi _{b, 1}] | \cdot |{\mathbb {E}}[ \xi _{b, 2}]|\cdot {\mathbb {E}}[ g^2(X_3)] \nonumber \\&\quad \le 2 (n^{-1} + n^{-1} ) + d n^{-2} \text { by Lemma }\text {A.1}\text { and } {\mathbb {E}}[g(X_1^2)] = 1 \text { in }(\text {3.3}) \nonumber \\&\quad \le 4n^{-1} + \frac{d}{2m} n^{-1} \text { by }2m < n \nonumber \\&\quad \le 4.5 n^{-1} \text { by }d \le m. \end{aligned}$$
(F.6)
1.1.6 Summary
Recall \(1 = \sigma _g \le \Vert g\Vert _3 \le \Vert h\Vert _3\). Combining (F.2), (F.3), (F.4), (F.6) gives Lemma E.1(iii), and combining (F.2), (F.3), (F.5), (F.6) gives Lemma E.1(iv).
1.2 Proof of Lemma E.2
Statement (i) is the Vandermonde’s identity, which counts the number of ways to choose m balls from \(n_1\) red balls and \(n_2\) green balls, by summing over \(k \in [0:m]\) the number of ways to choose k red balls and \(m-k\) green balls. Statement (ii) counts the number of ways to choose m balls out of a bag of n balls and paint k of the m chosen balls as red, in two different ways. Statement (iii) comes from
$$\begin{aligned} \left( {\begin{array}{c}a\\ b\end{array}}\right) - \left( {\begin{array}{c}a -e\\ b\end{array}}\right)&= \left( {\begin{array}{c}a\\ b\end{array}}\right) \bigg ( 1 - \frac{(a-e) \dots (a -e -b +1)}{a \cdots (a-b+1)} \bigg )\\&= \left( {\begin{array}{c}a\\ b\end{array}}\right) \bigg ( 1 - \prod _{j = a - b+1}^{a} \Big (1 - \frac{e}{j} \Big ) \bigg ) \\&\le \left( {\begin{array}{c}a\\ b\end{array}}\right) \sum _{j = a-b+1}^a \frac{e}{j}\\&\le \left( {\begin{array}{c}a\\ b\end{array}}\right) \frac{ b e}{a - b +1}. \end{aligned}$$