M_{1}(f)\operatorname{\mathds{E}}\left(\left|\Psi^{-1}\bm{R}_{1,+}\right|\mathds{1}_{Y=k-1}+\left|\Psi^{-1}\bm{R}_{1,-}\right|\mathds{1}_{Y=k}\right)$
$\displaystyle\qquad+\sup_{\bm{w}}\left\|\operatorname{Hess}f(\bm{w})\right\|_{\operatorname{H.S.}}\operatorname{\mathds{E}}\left(\left\|\Psi^{-1}\Gamma_{2,+}\right\|_{\operatorname{H.S.}}\mathds{1}_{Y=k-1}+\left\|\Psi^{-1}\Gamma_{2,-}\right\|_{\operatorname{H.S.}}\mathds{1}_{Y=k}\right)$
$\displaystyle\qquad+\frac{1}{2}M_{3}(f)\operatorname{\mathds{E}}\left(\left|(\lambda\Psi)^{-1}\Delta\\!\bm{W}\right|\left|\Delta\\!\bm{W}\right|^{2}\mathds{1}_{Y\in\\{k-1,k\\}}\right).$

Moreover if $\operatorname{\mathds{E}}\left|\bm{W}\right|^{4}<\infty$ then the
last error term on the right hand side can be replaced by

$\frac{M_{4}(f)}{4}\operatorname{\mathds{E}}\left(\left|(\lambda\Psi)^{-1}\Delta\\!\bm{W}\right|\left|\Delta\\!\bm{W}\right|^{3}\mathds{1}_{Y\in\\{k-1,k\\}}\right).$

###### Proof.

Define the symmetric function

$\displaystyle G_{k}(Y,Y^{\prime})$
$\displaystyle:=\mathds{1}_{Y^{\prime}=k,Y=k-1}+\mathds{1}_{Y=k,Y^{\prime}=k-1}$
$\displaystyle=\mathds{1}_{Y^{\prime}=k,\Delta\\!Y=1}+\mathds{1}_{Y=k,\Delta\\!Y=-1}.$

We consider the following mean zero random variable

$\displaystyle\Theta_{f}(\bm{W},Y)=\frac{1}{\lambda}\Delta\\!\bm{W}^{T}\Psi^{-T}\left(\nabla
f(\bm{W})+\nabla f(\bm{W}^{\prime})\right)G_{k}(Y,Y^{\prime}).$

Clearly, we have

$\displaystyle\begin{split}(\nabla f(\bm{W})+\nabla
f(\bm{W}^{\prime}))&-(2\nabla
f(\bm{W})+\operatorname{Hess}f(\bm{W})\Delta\\!\bm{W})\\\ &=\nabla
f(\bm{W}^{\prime})-\nabla
f(\bm{W})-\operatorname{Hess}f(\bm{W})\Delta\\!\bm{W})\\\
&=\int_{0}^{1}(\operatorname{Hess}f(\bm{W}+u\Delta\\!\bm{W})-\operatorname{Hess}f(\bm{W}))du\;\Delta\\!\bm{W}.\end{split}$
(5.3)

Recall that under Assumption IV the pair $(\bm{W},\bm{W}^{\prime})$ satisfies
the following two equalities

$\displaystyle\operatorname{\mathds{E}}\left(\Delta\\!\bm{W}\Delta\\!\bm{W}^{T}\mathds{1}_{\Delta\\!Y=\pm
1}\mid\bm{W},Y\right)=\lambda\left(\Psi\Sigma+\Gamma_{2,\pm}\right),$

for some random matrices $\Gamma_{2,\pm}=\Gamma_{2,\pm}(\bm{W},Y)$. Since
$\Delta\\!\bm{W}\Delta\\!\bm{W}^{T},\Sigma$ are symmetric matrices, Assumption
IV is equivalent to

$\displaystyle\operatorname{\mathds{E}}\left(\Delta\\!\bm{W}\Delta\\!\bm{W}^{T}\mathds{1}_{\Delta\\!Y=\pm
1}\mid\bm{W},Y\right)=\lambda\left(\Sigma\Psi^{T}+\Gamma^{T}_{2,\pm}\right).$

Plugging $(\nabla f(\bm{W})+\nabla f(\bm{W}^{\prime}))$ from equation (5.3)
into $\Theta_{f}$ and using Assumptions III and IV as well as Taylor expansion
we get

$\displaystyle 0$
$\displaystyle=-\operatorname{\mathds{E}}\left(\bm{W}^{T}\nabla
f(\bm{W})\mathds{1}_{Y\in\\{k-1,k\\}}+2(\bm{R}_{1,+}^{T}\mathds{1}_{Y=k-1}+\bm{R}_{1,-}^{T}\mathds{1}_{Y=k})\Psi^{-T}\nabla
f(\bm{W})\right)$ (5.4)
$\displaystyle\qquad+\operatorname{\mathds{E}}\operatorname{Tr}\left((\Sigma\Psi^{T}+\Gamma_{2,+}^{T})\Psi^{-T}\operatorname{Hess}f(\bm{W})\right)\mathds{1}_{Y=k-1}$
(5.5)
$\displaystyle\qquad+\operatorname{\mathds{E}}\operatorname{Tr}\left((\Sigma\Psi^{T}+\Gamma_{2,-}^{T})\Psi^{-T}\operatorname{Hess}f(\bm{W})\right)\mathds{1}_{Y=k}$
(5.6) $\displaystyle\qquad+\textrm{Err},$

where

Err
$\displaystyle=\frac{1}{\lambda}\operatorname{\mathds{E}}\bigl{(}\Delta\\!\bm{W}^{T}\Psi^{-T}\bigl{(}(\operatorname{Hess}f(\bm{W}+U_{1}\Delta
W)-\operatorname{Hess}f(\bm{W}))\Delta\\!\bm{W}\cdot
G_{k}(Y,Y^{\prime})\bigr{)}$ (5.7)

and $U_{1}\sim\textrm{Uniform}[0,1]$ is independent of everything else.
Depending on the moment assumptions one can bound $\left|\textrm{Err}\right|$
in two different ways. If
$\operatorname{\mathds{E}}\left|\bm{W}\right|^{3}<\infty$ then

$\displaystyle\left|\textrm{Err}\right|$
$\displaystyle\leqslant\frac{1}{\lambda}\operatorname{\mathds{E}}\sum_{i,j}^{d}\left|(\Psi^{-1}\Delta\\!\bm{W})_{i}\cdot\Delta\\!\bm{W}_{j}\cdot(\partial_{ij}f(\bm{W}+U_{1}\Delta\\!\bm{W})-\partial_{ij}f(\bm{W}))\right|\cdot
G_{k}(Y,Y^{\prime})$
$\displaystyle\leqslant\frac{1}{2}M_{3}(f)\operatorname{\mathds{E}}\left(\left|(\lambda\Psi)^{-1}\Delta\\!\bm{W}\right|\left|\Delta\\!\bm{W}\right|^{2}\mathds{1}_{Y\in\\{k-1,k\\}}\right).$
(5.8)

On the other hand if $\operatorname{\mathds{E}}\left|\bm{W}\right|^{4}<\infty$
one can use the exchangeability of $(\bm{W},Y)$ and
$(\bm{W}^{\prime},Y^{\prime})$ to derive a better bound. Notice that
$G_{k}(Y,Y^{\prime})$ is a symmetric function and hence remains the same
during such change.

$\displaystyle\operatorname{\mathds{E}}(\Psi^{-1}\Delta\\!\bm{W})_{i}\Delta\\!\bm{W}_{j}\Delta\\!\bm{W}_{\ell}U_{1}\,\partial_{ij\ell}f(\bm{W}+U_{1}U_{2}\Delta\\!\bm{W})G_{k}(Y,Y^{\prime})$
$\displaystyle=-\operatorname{\mathds{E}}(\Psi^{-1}\Delta\\!\bm{W})_{i}\Delta\\!\bm{W}_{j}\Delta\\!\bm{W}_{\ell}U_{1}\,\partial_{ij\ell}f\left(\bm{W}+(1-U_{1}U_{2})\Delta\\!\bm{W}\right)G_{k}(Y,Y^{\prime}),$

where $U_{2}\sim\textrm{Uniform}[0,1]$ and independent of everything else.
This allows us to rewrite the error term (5.7) as the average

Err
$\displaystyle=\operatorname{\mathds{E}}\sum_{i.j,\ell}^{d}((\lambda\Psi)^{-1}\Delta\\!\bm{W})_{i}\Delta\\!\bm{W}_{j}\Delta\bm{W}_{\ell}\cdot
U_{1}$
$\displaystyle\qquad\cdot(\partial_{ij\ell}f(\bm{W}+U_{1}U_{2}\Delta\\!\bm{W})-\partial_{ij\ell}f(\bm{W}+(1-U_{1}U_{2})\Delta\\!\bm{W}))G_{k}(Y,Y^{\prime}).$
(5.9)

Thus, using that
$\operatorname{\mathds{E}}(U_{1}\left|1-2U_{1}U_{2}\right|)=\frac{1}{4}$, we
derive that

$\displaystyle\begin{split}\left|\textrm{Err}\right|&\leqslant
M_{4}(f)\operatorname{\mathds{E}}(U_{1}\left|1-2U_{1}U_{2}\right|)\cdot\operatorname{\mathds{E}}\left(\left|(\lambda\Psi)^{-1}\Delta\\!\bm{W}\right|\left|\Delta\\!\bm{W}\right|^{3}\cdot\mathds{1}_{Y\in\\{k-1,k\\}}\right)\\\
&=\frac{1}{4}M_{4}(f)\operatorname{\mathds{E}}\left(\left|(\lambda\Psi)^{-1}\Delta\\!\bm{W}\right|\left|\Delta\\!\bm{W}\right|^{3}\cdot\mathds{1}_{Y\in\\{k-1,k\\}}\right).\end{split}$
(5.10)

Combining (5.4), (5.5), (5.6), with (5.8) or (5.10), and moving

$\operatorname{\mathds{E}}\left(-\Delta\\!\bm{W}^{T}\nabla
f(\bm{W})+\operatorname{Tr}(\Sigma\operatorname{Hess}f(\bm{W}))\right)\mathds{1}_{Y\in\\{k-1,k\\}}$

to the left-hand side of the equation yields, the desired bound.
$\blacksquare$

###### Proof of Theorem 3.1.

Let

$A_{0}=\sqrt{\operatorname{Tr}(\Sigma)+\operatorname{\mathds{E}}\left(\left|\bm{W}\right|^{2}\,\big{|}\,Y\in\\{k-1,k\\}\right)}.$

If

$\left|\operatorname{\mathds{E}}(\mathscr{S}f(\bm{W})\mid B)\right|\leqslant
A_{1}\cdot
M_{1}(f)+A_{2}\cdot\sup_{w}\left\|\operatorname{Hess}f(w)\right\|_{\operatorname{H.S.}}+A_{4}\cdot\frac{1}{4}M_{4}(f),$

for some event $B$, then using the approximation scheme from Raič [38] and
Fang–Koike [23], we get that

$\displaystyle\begin{split}&\left|\operatorname{\mathds{E}}((h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|\\\ &\qquad\leqslant
M_{1}(h)\left(A_{1}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}\cdot
A_{2}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{3/2}\cdot\sqrt{c_{3}A_{0}A_{4}}\right)\end{split}$
(5.11)

where $c_{3}=(2+8e^{-3/2})/\sqrt{2\pi}<2$. The proof now follows from Lemmas
5.1 and 5.2.

Similarly, for

$\left|\operatorname{\mathds{E}}(\mathscr{S}f(\bm{W})\mid B)\right|\leqslant
A_{1}\cdot
M_{1}(f)+A_{2}\cdot\sup_{w}\left\|\operatorname{Hess}f(w)\right\|_{\operatorname{H.S.}}+A_{3}\cdot
M_{3}(f),$

we get that

$\displaystyle\begin{split}&\left|\operatorname{\mathds{E}}((h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|\leqslant
M_{1}(h)\biggl{(}A_{1}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}\cdot
A_{2}\\\
&\qquad\qquad\qquad+c_{2}\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{2}\cdot
A_{3}\cdot(1+\left|\log{(c_{2}\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{2}\cdot
A_{3}/A_{0})}\right|\biggr{)}\end{split}$ (5.12)

where $c_{2}=4/\sqrt{2\pi e}<1$ and the proof follows similarly. We provide a
proof of (5.11) and (5.12) in Lemma A.1 for completeness. $\blacksquare$

##  6 Change of variables

In this section we provide proofs of Proposition 2.6 and Proposition 3.3. We
treat each of the error terms in a separate lemma.

######  Lemma 6.1.

Under assumptions of Proposition 2.6, recall that we defined the change of
variable

$\displaystyle W^{0}:=X+\lambda\,\psi\,\alpha
XY+\frac{\lambda\,\theta}{2}\left(Y^{2}-\operatorname{\mathds{E}}Y^{2}\right)+\frac{\lambda^{2}(\psi+1)\alpha\theta}{3}(Y^{3}-\operatorname{\mathds{E}}Y^{3}),$

where $\alpha=\frac{a_{+}-a_{-}}{2Q}$ and $\theta=\frac{b_{+}}{Q}$ and
$W=\frac{W^{0}}{\sigma_{W^{0}}}$. Then $W$ satisfies Assumption III with the
error terms given by

$\displaystyle\widetilde{R}_{1,\pm}=\frac{\lambda\theta}{2}\left(1-\frac{\psi}{2}\right)\frac{\overline{Y^{2}}}{\sigma_{W^{0}}}+\frac{1}{\sigma_{W^{0}}}(\widetilde{\varepsilon}_{0,\pm}+\widetilde{\varepsilon}_{1,\pm}+\widetilde{\varepsilon}_{2,\pm}+\widetilde{\varepsilon}_{3,\pm}),$

where

$\displaystyle\widetilde{\varepsilon}_{0,\pm}$
$\displaystyle:=\lambda\psi^{2}\alpha a_{\pm}X(Y\pm
1)-\frac{\lambda\,\alpha\,\psi}{2}XY+R_{1,\pm}\biggl{(}1\pm\lambda\,\psi\,\alpha+\lambda\,\psi\,\alpha
Y\biggr{)},$ $\displaystyle\widetilde{\varepsilon}_{1,\pm}$
$\displaystyle:=\lambda\left(\pm\lambda
b_{\pm}\psi\alpha\mp\frac{a_{\pm}\theta}{2}+(\psi+1)\theta\alpha\left(Q-\frac{1}{3}\lambda
a_{\pm}\right)\right)Y,$ $\displaystyle\widetilde{\varepsilon}_{2,\pm}$
$\displaystyle:=\mp\lambda^{2}\theta(\psi+1)\alpha a_{\pm}Y^{2},$
$\displaystyle\text{and}\quad\widetilde{\varepsilon}_{3,\pm}$
$\displaystyle:=\frac{1}{3}\lambda\theta(\psi+1)\alpha
Q-\lambda^{2}\theta(\psi+1)\alpha
a_{\pm}Y^{3}-\frac{\lambda^{2}\psi(\psi+1)\alpha\theta}{6}\left(Y^{3}-\operatorname{\mathds{E}}Y^{3}\right).$

###### Proof.

With change from $Y$ to $Y^{\prime}$ we have the following change in $W^{0}$

$\displaystyle\Delta W^{0}$ $\displaystyle=\Delta
X+\lambda\psi\alpha\left(\Delta XY+\Delta\\!YX+\Delta
X\Delta\\!Y\right)+\lambda\,\theta\left(Y\Delta\\!Y+\frac{1}{2}\Delta\\!Y^{2}\right)$
$\displaystyle\quad+\lambda^{2}(\psi+1)\alpha\theta\Delta\\!Y\left(Y^{2}+Y\Delta\\!Y+\frac{(\Delta\\!Y)^{2}}{3}\right).$
(6.1)

Using Assumptions II and IIIa, it can be shown that

$\displaystyle M_{1,\pm}\left(W^{0},Y\right)$
$\displaystyle=M_{1,\pm}(X,Y)\left(1\pm\lambda\,\psi\,\alpha+\lambda\,\psi\,\alpha
Y\right)$ $\displaystyle\quad+\lambda\,\biggl{(}\pm\alpha\psi
X+\theta\left(\pm Y+\frac{1}{2}\right)+\lambda(\psi+1)\alpha\theta\left(\pm
Y^{2}+Y\pm\frac{1}{3}\right)\biggr{)}M_{0,\pm}$
$\displaystyle=-\lambda\biggl{[}a_{\pm}\psi X\mp\alpha\psi
QX+b_{\pm}Y+\lambda\psi\alpha b_{\pm}Y^{2}+\varepsilon_{0,\pm}+\left(\mp\theta
Q\right)Y+\varepsilon_{1,\pm}$
$\displaystyle\qquad+\left(\mp\lambda(\psi+1)\alpha\theta Q+\lambda\theta
a_{\pm}\right)Y^{2}+\varepsilon_{2,\pm}-\frac{\theta}{2}Q+\varepsilon_{3,\pm}\biggr{]},$

where

$\displaystyle\varepsilon_{0,\pm}$ $\displaystyle:=\lambda\psi^{2}\alpha
a_{\pm}X(Y\pm
1)+R_{1,\pm}\biggl{(}1\pm\lambda\,\psi\,\alpha+\lambda\,\psi\,\alpha
Y\biggr{)},$ $\displaystyle\varepsilon_{1,\pm}$
$\displaystyle:=\lambda\left(\pm\lambda
b_{\pm}\psi\alpha\mp\frac{a_{\pm}\theta}{2}+(\psi+1)\theta\alpha\left(Q-\frac{1}{3}\lambda
a_{\pm}\right)\right)Y,$ $\displaystyle\varepsilon_{2,\pm}$
$\displaystyle:=\mp\lambda^{2}\theta(\psi+1)\alpha a_{\pm}Y^{2},$
$\displaystyle\text{and}\quad\varepsilon_{3,\pm}$
$\displaystyle:=\frac{1}{3}\lambda\theta(\psi+1)\alpha
Q-\lambda^{2}\theta(\psi+1)\alpha a_{\pm}Y^{3}.$

Notice that $a_{\pm}\mp\alpha
Q=a_{\pm}\mp\frac{a_{+}-a_{-}}{2}=\frac{a_{+}+a_{-}}{2}=\frac{1}{2}$, thus the
coefficient of $X$ is equal to $\frac{1}{2}$ and we can use change of variable
$~{}\eqref{eq:change of var}$ once again to get

$\displaystyle M_{1,\pm}\left(W^{0},Y\right)$
$\displaystyle=-\lambda\biggl{[}\frac{1}{2}\psi
W^{0}-\frac{\lambda\,\alpha\,\psi}{2}XY-\frac{\lambda\,\psi\,\theta}{4}\left(Y^{2}-\operatorname{\mathds{E}}Y^{2}\right)-\frac{\lambda^{2}\psi(\psi+1)\alpha\theta}{6}\left(Y^{3}-\operatorname{\mathds{E}}Y^{3}\right)$
$\displaystyle\quad\quad\quad+\left(b_{\pm}\mp\theta
Q+O(\lambda\theta)\right)Y+\left(\mp\lambda(\psi+1)\alpha\theta
Q+\lambda\theta a_{\pm}+\lambda\psi\alpha
b_{\pm}\right)Y^{2}-\frac{\theta}{2}Q+\sum_{i=0}^{3}\varepsilon_{i}\biggr{]}$
$\displaystyle=-\lambda\biggl{[}\frac{1}{2}W^{0}+\left(b_{\pm}\mp\theta
Q+O(\lambda\theta)\right)Y+\left(\mp\lambda(\psi+1)\alpha\theta
Q+\lambda\theta a_{\pm}+\lambda\psi\alpha
b_{\pm}-\frac{\lambda\,\psi\,\theta}{4}\right)Y^{2}$
$\displaystyle\quad\quad\quad+\frac{\lambda\,\psi\,\theta}{4}\operatorname{\mathds{E}}Y^{2}-\frac{\theta}{2}Q+\widetilde{\varepsilon}_{0,\pm}+\widetilde{\varepsilon}_{1,\pm}+\widetilde{\varepsilon}_{2,\pm}+\widetilde{\varepsilon}_{3,\pm}\biggr{]},$

where $\widetilde{\varepsilon}_{1,\pm}=\varepsilon_{1,\pm}$,
$\widetilde{\varepsilon}_{2,\pm}=\varepsilon_{2,\pm}$,

$\displaystyle\widetilde{\varepsilon}_{0,\pm}:=\varepsilon_{0,\pm}-\frac{\lambda\,\alpha\,\psi}{2}XY\quad\text{and}\quad\widetilde{\varepsilon}_{3,\pm}:=\varepsilon_{3,\pm}-\frac{\lambda^{2}\psi(\psi+1)\alpha\theta}{6}\left(Y^{3}-\operatorname{\mathds{E}}Y^{3}\right).$

The coefficient of $Y$ cancels by Assumption IIIa that says $b_{\pm}\mp\theta
Q=b_{\pm}\mp b_{+}=0$. One can rewrite the coefficient of $Y^{2}$ in the
following way

$\displaystyle\begin{split}\mp\lambda(\psi+1)\alpha\theta Q&+\lambda\theta
a_{\pm}+\lambda\psi\alpha b_{\pm}-\frac{\lambda\,\psi\,\theta}{4}\\\
&=\mp\lambda\psi\alpha\theta Q\mp\lambda\theta\alpha Q+\lambda\theta
a_{\pm}+\lambda\psi\alpha b_{\pm}-\frac{\lambda\,\psi\,\theta}{4}\\\
&=\frac{\lambda\theta}{2}-\frac{\lambda\,\psi\,\theta}{4}=\frac{\lambda\theta}{2}\left(1-\frac{\psi}{2}\right).\end{split}$
(6.2)

The constant term, using $Q=\lambda\operatorname{\mathds{E}}Y^{2}$ from
Assumption II, can be rewritten in the similar fashion

$\displaystyle\frac{\lambda\,\psi\,\theta}{4}\operatorname{\mathds{E}}Y^{2}-\frac{\theta}{2}Q=-\frac{\lambda\theta}{2}\left(1-\frac{\psi}{2}\right)\operatorname{\mathds{E}}Y^{2}.$

Thus we can conclude that

$\displaystyle M_{1,\pm}\left(W^{0},Y\right)$
$\displaystyle=-\lambda\biggl{[}\frac{1}{2}\psi
W^{0}+\frac{\lambda\theta}{2}\left(1-\frac{\psi}{2}\right)\left(Y^{2}-\operatorname{\mathds{E}}Y^{2}\right)+\widetilde{\varepsilon}_{0,\pm}+\widetilde{\varepsilon}_{1,\pm}+\widetilde{\varepsilon}_{2,\pm}+\widetilde{\varepsilon}_{3,\pm}\biggr{]}.$

Scaling both sides of equality by $\sigma_{W^{0}}$ yields the result.
$\blacksquare$

As we discussed in Remark 3.4 the change of variable (3.1) functions very
similarly to its univariate analog (2.4). Thus similar computations to the
ones in the proof of Lemma 6.1 yield the following lemma in the multivariate
case.

######  Lemma 6.2.

Under assumptions of Proposition 3.3, recall that we defined

$\displaystyle\bm{W}^{0}:=\bm{X}+\lambda\mathrm{A}\bm{X}Y+\frac{\lambda\bm{\theta}}{2}\left(Y^{2}-\operatorname{\mathds{E}}Y^{2}\right)+\frac{\lambda^{2}(\mathrm{A}+\alpha)\bm{\theta}}{3}(Y^{3}-\operatorname{\mathds{E}}Y^{3}).$

where, $\mathrm{A}=\frac{\lambda_{+}-\lambda_{-}}{2Q}$,
$\alpha=\frac{a_{+}-a_{-}}{2Q}$, and $\bm{\theta}=\frac{\bm{b}_{+}}{Q}$. Then
the random vector
$\bm{W}:=\left(W^{0}_{i}/\sigma_{W^{0}_{i}}\right)_{1\leqslant i\leqslant d}$
satisfies Assumption III and

$\displaystyle\widetilde{\bm{R}}_{1,\pm}=\Sigma^{-\nicefrac{{1}}{{2}}}_{W^{0}}\frac{\lambda}{2}\left(I-\frac{\Psi}{2}\right)\bm{\theta}\,\overline{Y^{2}}+\Sigma^{-\nicefrac{{1}}{{2}}}_{W^{0}}(\widetilde{\bm{\varepsilon}}_{0,\pm}+\widetilde{\bm{\varepsilon}}_{1,\pm}+\widetilde{\bm{\varepsilon}}_{2,\pm}+\widetilde{\bm{\varepsilon}}_{3,\pm}),$

where

$\displaystyle\widetilde{\bm{\varepsilon}}_{0,\pm}$ $\displaystyle:=\lambda
a_{\pm}\Psi\mathrm{A}\bm{X}(Y\pm
1)-\frac{\lambda\,\Psi\,\mathrm{A}}{2}\bm{X}Y+\biggl{(}1\pm\lambda\,\mathrm{A}+\lambda\,\mathrm{A}Y\biggr{)}\bm{R}_{1,\pm},$
$\displaystyle\widetilde{\bm{\varepsilon}}_{1,\pm}$
$\displaystyle:=\lambda\left(\pm\lambda\mathrm{A}\bm{b}_{\pm}\mp\frac{a_{\pm}\bm{\theta}}{2}+(\mathrm{A}+\alpha)\bm{\theta}\left(Q-\frac{1}{3}\lambda
a_{\pm}\right)\right)Y,$ $\displaystyle\widetilde{\bm{\varepsilon}}_{2,\pm}$
$\displaystyle:=\mp\lambda^{2}a_{\pm}(\mathrm{A}+\alpha)\bm{\theta}Y^{2},$
$\displaystyle\text{and}\quad\widetilde{\bm{\varepsilon}}_{3,\pm}$
$\displaystyle:=\frac{1}{3}\lambda(\mathrm{A}+\alpha)\bm{\theta}Q-\lambda^{2}a_{\pm}(\mathrm{A}+\alpha)\bm{\theta}Y^{3}-\frac{\lambda^{2}\Psi(\mathrm{A}+\alpha)\bm{\theta}}{6}\left(Y^{3}-\operatorname{\mathds{E}}Y^{3}\right).$

###### Proof.

For multivariate case proof is essentially the same as above, but now

$\displaystyle\Delta\\!\bm{W}^{0}$
$\displaystyle=\Delta\bm{X}+\lambda\mathrm{A}(\bm{X}\Delta\\!Y+\Delta\bm{X}Y+\Delta\bm{X}\Delta\\!Y)+\lambda\bm{\theta}\left(Y\Delta\\!Y+\frac{1}{2}\Delta\\!Y^{2}\right)$
$\displaystyle\quad+\lambda^{2}(\mathrm{A}+\alpha)\bm{\theta}\Delta\\!Y\left(Y^{2}+Y\Delta\\!Y+\frac{(\Delta\\!Y)^{2}}{3}\right).$

Notice that $\bm{X}Y$ have the analogous coefficients to the univariate case
and hence functions similarly, _i.e.,_ cancels out with $\Psi_{\pm}\bm{X}$ and
creates $\frac{1}{2}\Psi\bm{X}$ term. The $Y$ terms cancel out in the exact
the same fashion as above. Now we focus our attention on $Y^{2}$ terms and
derive that

$\displaystyle\lambda\left(-\frac{1}{4}\Psi\bm{\theta}+\mathrm{A}\bm{b}_{\pm}\mp(\mathrm{A}+\alpha)\bm{\theta}Q+a_{\pm}\bm{\theta}\right)$
$\displaystyle=\lambda\left(-\frac{1}{4}\Psi\bm{\theta}+\mathrm{A}\left(\bm{b}_{\pm}\mp\bm{\theta}Q\right)+\bm{\theta}(a_{\pm}\pm\alpha
Q)\right)$
$\displaystyle\quad=\left(I-\frac{\Psi}{2}\right)\frac{\lambda\bm{\theta}}{2}Y^{2},$

and notice the the constant term matches it

$\displaystyle-\frac{\bm{\theta}}{2}Q+\lambda\Psi\frac{\bm{\theta}}{4}\operatorname{\mathds{E}}Y^{2}=-\left(I-\frac{\Psi}{2}\right)\frac{\lambda\bm{\theta}}{2}\operatorname{\mathds{E}}Y^{2}.$

Notice that by Assumption IIIa $\bm{b}_{+}=\frac{1}{2}\Psi\bm{b}_{+}$ and
hence

$\left(I-\frac{\Psi}{2}\right)\bm{\theta}=0.$

Therefore we conclude that

$\displaystyle M_{1,\pm}\left(\bm{W}^{0},Y\right)$
$\displaystyle=-\lambda\biggl{[}\frac{\Psi}{2}\bm{W}^{0}+\widetilde{\bm{\varepsilon}}_{0,\pm}+\widetilde{\bm{\varepsilon}}_{1,\pm}+\widetilde{\bm{\varepsilon}}_{2,\pm}+\widetilde{\bm{\varepsilon}}_{3,\pm},\biggr{]}.$

Scaling each coordinate of $\bm{W}^{0}$ appropriately yields the result.
$\blacksquare$

In order to prove Propositions 2.6 and 3.3, it remains to show that the order
of the second-order error terms does not change after the change of variables
as given in (2.4) and (3.1). We present the proof for the univariate case in
the following lemma. The treatment of the multivariate case is completely
analogous.

######  Lemma 6.3.

Under the assumptions of Proposition 2.6 $\left(W,Y\right)$ still satisfy the
Assumption IV. Moreover,

$\displaystyle\left\|\widetilde{R}_{2,\pm}-{R_{2,\pm}}/{\sigma^{2}_{W^{0}}}\right\|_{2p}$
$\displaystyle\lesssim\lambda\psi\left|\alpha\right|\left(\frac{\sigma_{X}}{\sigma_{W^{0}}}\sqrt{\left\|\lambda(1+R_{2,\pm}/\sigma^{2}_{X})\right\|_{2p}\left\|Y\right\|_{2p}}+\frac{\left\|X\right\|_{p}}{\sigma_{W^{0}}}\right)$
$\displaystyle\qquad+\lambda\left|\theta\right|\cdot\frac{\left\|Y\right\|_{p}}{\sigma_{W^{0}}}+\lambda(\psi+1)\left|\alpha\theta\right|\cdot\frac{\lambda\left\|Y^{2}\right\|_{p}}{\sigma_{W^{0}}}.$

###### Proof of Lemma 6.3.

Rewrite the equality in (6) as

$\Delta W^{0}=\Delta X+\delta,$ (6.3)

where

$\displaystyle\delta$ $\displaystyle:=\lambda\psi\alpha\left(\Delta
XY+\Delta\\!YX+\Delta
X\Delta\\!Y\right)+\lambda\,\theta\left(Y\Delta\\!Y+\frac{1}{2}\Delta\\!Y^{2}\right)$
$\displaystyle\quad+\lambda^{2}(\psi+1)\alpha\theta\Delta\\!Y\left(Y^{2}+Y\Delta\\!Y+\frac{(\Delta\\!Y)^{2}}{3}\right).$

Multiplying both sides of (6.3) by $\mathds{1}_{\Delta\\!Y=\pm 1}$, taking
conditional expectation given $(X,Y)$ on both sides, and using triangle
inequality implies that

$\displaystyle\left|\sqrt{\operatorname{\mathds{E}}\left(\left|\Delta
W^{0}\right|^{2}\mathds{1}_{\Delta\\!Y=\pm 1}\mid
X,Y\right)}-\sqrt{\operatorname{\mathds{E}}\left(\left|\Delta
X\right|^{2}\mathds{1}_{\Delta\\!Y=\pm 1}\mid X,Y\right)}\right|$
$\displaystyle\qquad\leqslant\sqrt{\operatorname{\mathds{E}}\left(\left|\delta\right|^{2}\mathds{1}_{\Delta\\!Y=\pm
1}\mid X,Y\right)}$

By Assumption IV we can rewrite it as

$\frac{\lambda\sigma^{2}_{W^{0}}}{2+\frac{R_{2,\pm}}{\sigma^{2}_{W^{0}}}+\frac{\sigma_{X}^{2}}{\sigma^{2}_{W^{0}}}}\cdot\left|\widetilde{R}_{2,\pm}-\frac{R_{2,\pm}}{\sigma^{2}_{W^{0}}}\right|^{2}\leqslant\operatorname{\mathds{E}}\left(\left|\delta\right|^{2}\mathds{1}_{\Delta\\!Y=\pm
1}\mid X,Y\right)$ (6.4)

Assuming
$2+\frac{R_{2,\pm}}{\sigma^{2}_{W^{0}}}+\frac{\sigma_{X}^{2}}{\sigma^{2}_{W^{0}}}\in(1,C)$
for some constant $C$, it remains to derive the following bound

$\displaystyle\left\|\operatorname{\mathds{E}}\left((\delta/\sigma_{W^{0}})^{2}\mathds{1}_{\Delta\\!Y=\pm
1}\mid X,Y\right)\right\|^{\nicefrac{{1}}{{2}}}_{p}$
$\displaystyle\quad\lesssim\lambda\psi\left|\alpha\right|\left(\frac{\sigma_{X}}{\sigma_{W^{0}}}\sqrt{\left\|\lambda(1+R_{2,\pm}/\sigma^{2}_{X})\right\|_{2p}\left\|Y\right\|_{2p}}+\frac{\left\|X\right\|_{p}}{\sigma_{W^{0}}}\right)$
$\displaystyle\qquad\quad+\lambda\left|\theta\right|\cdot\frac{\left\|Y\right\|_{p}}{\sigma_{W^{0}}}+\lambda(\psi+1)\left|\alpha\theta\right|\cdot\frac{\lambda\left\|Y^{2}\right\|_{p}}{\sigma_{W^{0}}}.$

From (6.4) we can see that

$\left\|\widetilde{R}_{2,\pm}-\frac{R_{2,\pm}}{\sigma^{2}_{W^{0}}}\right\|_{2p}\lesssim\lambda^{-\nicefrac{{1}}{{2}}}\left\|\operatorname{\mathds{E}}\left((\delta/\sigma_{W^{0}})^{2}\mathds{1}_{\Delta\\!Y=\pm
1}\mid X,Y\right)\right\|^{\nicefrac{{1}}{{2}}}_{p}.$

This completes the proof. $\blacksquare$

##  7 Closing remarks and further work

Before our work, few results delve into CCLT in general settings. There are
many aspects in which we would like to see our approach extended and improved,
including generalizing the current approach, extending it to other dependency
structures between random variables, and connecting it to concentration
inequalities. In this article, we often utilize subtle, sometimes surprising,
cancellations caused by exchangeability. Hence we believe that there is much
more to understand in this area than we presently know. We discuss our results
and possible future directions and state questions of particular interest in
the remaining of this section.

###  7.1 Change of variable and the assumptions

For our main result, we require random variables to satisfy symmetric
linearity conditions (Assumption III), and the second-moment condition
(Assumption IV). It is important to notice that these assumptions together
with Assumption II yield linearity condition for $(\bm{W},Y/\sigma_{Y})$.
Hence if $R_{i,\pm}$ are small for $i=0,1,2$ the multivariate Stein’s method
for exchangeable pair implies joint convergence to a $d+1$-dimensional
Gaussian vector.

Given the non-symmetric linearity condition, one can make it symmetric by
subtracting the product of random variables with an appropriate coefficient
(the $XY$ term in the change of variable). When working with counting random
variables, it is often the case that the linearity condition is of the form
$\lambda(\psi a_{\pm}W+b_{\pm}Y)$ (Assumption IIIa); as we have seen in
Section 4.2.2 on the sub-pattern count example and in Section 4.2.3 on the
sub-graph count example. However, in such examples, $M_{0,\pm}$ also has a
particular form that we state in the form of the Assumption II. We utilized
this fact to a great extent in the change of variable and the proof of Theorem
2.4 and 3.1, leading to the following question.

######  Question 1.

Is it possible to derive CCLT in the models with $|\Delta\\!Y|=1$ that satisfy
Assumption IIIa but $M_{0,\pm}\approx Q+f_{\pm}(Y)$ for non-linear $f$?

We believe that in such cases, our approach is still applicable, and if the
conditional mean is not known, its approximation in the form of change of
variable would be different, but analogous, from the one presented in (2.4)
and (3.1).

###  7.2 Range of $\Delta\\!Y$

In this article, we focused on the case where $|\Delta\\!Y|\leqslant 1$. When
$Y$ can change by more than $\pm 1$ but
$\operatorname{\mathds{P}}(|\Delta\\!Y|=1)$ is sufficiently large,
conditioning on this event, one can still apply our results similar to the way
how we used the classical methods by conditioning on the event
$\\{\Delta\\!Y=0\\}$ in the Section 4.1. However, in the models where
$\operatorname{\mathds{P}}(|\Delta\\!Y|=1)$ is negligible, it remains open to
extending our approach.

######  Question 2.

Is it possible to derive CCLT with an explicit convergence rate for models
where $\Delta\\!Y$ can take infinitely many values?

For example, consider the number of edges given the number of triangles in the
Erdős–Rényi random graph. Another immediate application would be the extension
of [6, Theorem 4] and [31, Theorem 4.2], for $\delta=1$, to the dependent
settings.

###  7.3 Other types of events than $\\{Y=k\\}$

In this article, we condition on the simplest type of the event $\\{Y=k\\}$
where $Y$ is the sum of indicators, as it already required a considerable
amount of effort and novel techniques. The next step is to extend our result
to CCLT where one conditions on a vector
$\\{(Y_{1},Y_{2},\ldots,Y_{d})=(y_{1},y_{2},\ldots,y_{d})\\}$ sill under the
assumption of joint Gaussian convergence. A natural application would be a
joint CCLT for a sub-graph count given a value for several other sub-graph
counts in the Erdős–Rényi random graph.

######  Question 3.

How does the rate of convergence depend on the dimension of the vector
$\bm{Y}=(Y_{1},Y_{2},\ldots,Y_{d})$?

Another, the more complicated direction, is to relax the assumption of joint
Gaussian convergence and condition on more complicated events such as some
property of the trajectory of a random walk or random environment.

###  7.4 Sufficient statistic

In most of the applications presented in this paper, we condition on the
sufficient statistic. We believe our theorem should reliably work without this
condition, as we demonstrated in Lemma 4.5. However, while writing this paper,
we realized that we do not know of many natural examples satisfying our
assumptions that will also not be sufficient statistics for the parameters of
the model, especially in the univariate case. If one relaxes our condition on
$\Delta\\!Y$ and allows it to range over an infinite set, this will create
many natural examples with conditioning on non-sufficient statistics.

Besides the application presented in Section 4.2.1, where $Y$ was not a
sufficient statistic for the parameter $p$, another application could be a
“noisy” version of a sufficient statistic. For example, suppose we would like
to condition on the number of edges in an inhomogeneous random graph where
each of $N(1-\delta)$ edges is present independently with probability $p$ and
each of the remaining $\delta N$ edges is present with probability
$p\pm\varepsilon$ independently from everything else. One can estimate the
difference between this model and the homogeneous random graph and work with
the latter. However, depending on the $\delta$ and $\varepsilon$, the error
might be significant. Our approach provides an alternative that allows working
directly with inhomogeneous models.

###  7.5 Other approaches

It is of interest to extend other existing approaches of Stein’s method to the
conditional setting, particularly the dependency graph approach in relation to
the subgraph counting problem (See Section 4.4). Since, in this work, we
assume that $(W,Y)$ jointly converges to a Gaussian vector, a natural place to
start would be adopting a multivariate extension of the dependency graph
approach such as in [22] or some other variation of the method to the
conditional setting.

######  Question 4.

Is it possible to get the rate of converges of order $n^{-1}$ in Theorem 4.18
using another approach than the exchangeable pair to match the bound in
Corollary 4.14?

###  7.6 Other distances between distributions

In this article, we use Wasserstein-$1$ distance to quantify the rate of
convergence. Suppose one is interested in bounding other metrics, such as
Kolmogorov-Smirnov distance. Then one has to work with functions with fewer
derivatives. In that case, one can usually approximate those functions by two
or three times differentiable functions to apply our techniques. However, that
would result in an extra loss in the rate of convergence. Thus it remains open
to acquiring optimal bounds on the convergence rates in other distances
directly if at all possible.

##  Appendix A From Smooth functions to Lipschitz functions

######  Lemma A.1.

Let $\bm{W}$ be a $d$-dimensional random vector with mean zero and
$\operatorname{\mathds{E}}\left|\bm{W}\right|^{4}<\infty$, such that for some
event $B$ independent of $\bm{Z}$ with $\operatorname{\mathds{P}}(B)>0$, we
have

$\left|\operatorname{\mathds{E}}(\mathscr{S}f(\bm{W})\mid B)\right|\leqslant
A_{1}\cdot
M_{1}(f)+A_{2}\cdot\sup_{w}\left\|\operatorname{Hess}f(w)\right\|_{\operatorname{H.S.}}+A_{4}\cdot
M_{4}(f)$

for all $f$ with $M_{4}(f)<\infty$. Then, for all Lipschitz functions $h$ with
$M_{1}(h)<\infty$, we have

$\displaystyle\begin{split}&\left|\operatorname{\mathds{E}}((h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|\\\ &\qquad\leqslant
M_{1}(h)\left(A_{1}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}\cdot
A_{2}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{3/2}\cdot\sqrt{4c_{3}A_{0}A_{4}}\right)\end{split}$
(A.1)

where $c_{3}=(2+8e^{-3/2})/\sqrt{2\pi}<2$, $\bm{Z}\sim N_{d}(\bm{0},I_{d})$ is
independent of $\bm{W}$ and

$A_{0}:=\sqrt{\operatorname{\mathds{E}}(\left|\bm{W}\right|^{2}+\operatorname{Tr}(\Sigma)\mid
B)}.$

Similarly, if $\operatorname{\mathds{E}}\left|\bm{W}\right|^{3}<\infty$ and

$\left|\operatorname{\mathds{E}}(\mathscr{S}f(\bm{W})\mid B)\right|\leqslant
A_{1}\cdot
M_{1}(f)+A_{2}\cdot\sup_{w}\left\|\operatorname{Hess}f(w)\right\|_{\operatorname{H.S.}}+A_{3}\cdot
M_{3}(f),$

for all $f$ with $M_{3}(f)<\infty$, we have

$\displaystyle\begin{split}&\left|\operatorname{\mathds{E}}((h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|\leqslant
M_{1}(h)\cdot\biggl{(}A_{1}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}\cdot
A_{2}\\\
&\qquad\qquad+c_{2}\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{2}\cdot
A_{3}\cdot\biggl{(}1+\left|\log\bigl{(}{c_{2}\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{2}\cdot
A_{3}}/{A_{0}}\bigr{)}\right|\biggr{)}\biggr{)}.\end{split}$ (A.2)

where $c_{2}=4/\sqrt{2\pi e}<1.$

###### Proof of Lemma A.1.

We mainly follow the proof of [23, Theorem 1.1] along with estimates from
[38]. Consider a Lipschitz function $h:\mathds{R}^{d}\to\mathds{R}$ and for
all $\alpha\in[0,\pi/2]$, define the function

$h_{\alpha}(\bm{w}):=\operatorname{\mathds{E}}h(\cos\alpha\cdot\bm{w}+\sin\alpha\cdot\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z}),\qquad\bm{w}\in\mathds{R}^{d}$

where $\bm{Z}$ is a $d$-dimensional standard Gaussian vector. Clearly
$h_{0}=h$ and
$h_{\pi/2}\equiv\operatorname{\mathds{E}}h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})$.
It is easy to check that $h_{\alpha}$ is infinitely differentiable. Using [38,
Lemma 4.7], we get that

$\displaystyle M_{r+1}(h_{\alpha})\leqslant
c_{r}\cdot\frac{\cos^{r+1}\alpha}{\sin^{r}\alpha}\cdot
M_{1}(h)\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{r}$

for any non-negative integer $r$ and $\alpha\in(0,\pi/2)$, where $c_{r}$ is as
given in [38, Eqn. 4.9]; in particular

$c_{0}=1,c_{2}=4/\sqrt{2\pi e}<1,c_{3}=(2+8e^{-3/2})/\sqrt{2\pi}<2.$

Moreover, by [12, Lemma 2.2(i)] we get

$\displaystyle\sup_{\bm{w}}\left\|\operatorname{Hess}h_{\alpha}(\bm{w})\right\|_{\operatorname{H.S.}}\leqslant\frac{\cos^{2}\alpha}{\sin\alpha}\cdot
M_{1}(h)\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}.$

Now, using the fact that
$\frac{d}{d\alpha}h_{\alpha}(\bm{w})=\mathscr{S}h_{\alpha}(\bm{w})\
\tan\alpha$, we get that

$\displaystyle\begin{split}&\left|\operatorname{\mathds{E}}(h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|\\\
&=\left|\operatorname{\mathds{E}}(h_{0}(\bm{W})-h_{\pi/2}(\bm{W})\mid
B)\right|\\\
&\leqslant\left|\operatorname{\mathds{E}}(h_{0}(\bm{W})-h_{\varepsilon}(\bm{W}))\mid
B)\right|+\int_{\varepsilon}^{\pi/2}\left|\operatorname{\mathds{E}}(\mathscr{S}h_{\alpha}(\bm{W})\mid
B)\right|\tan\alpha\,d\alpha.\end{split}$ (A.3)

We have for any $\varepsilon\in(0,\pi/2)$

$\displaystyle\left|\operatorname{\mathds{E}}(h_{0}(\bm{W})-h_{\varepsilon}(\bm{W})\mid
B)\right|$ $\displaystyle\leqslant
M_{1}(h)\operatorname{\mathds{E}}(\left|(1-\cos\varepsilon)\cdot\bm{W}+\sin\varepsilon\cdot\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z}\right|\mid
B)$ $\displaystyle\leqslant
M_{1}(h)\sqrt{\operatorname{\mathds{E}}((1-\cos\varepsilon)^{2}\cdot\left|\bm{W}\right|^{2}+\sin^{2}\varepsilon\cdot\left|\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z}\right|^{2}\mid
B)}$ $\displaystyle\leqslant M_{1}(h)\cdot A_{0}\cdot\sin\varepsilon$

where

$A_{0}:=\sqrt{\operatorname{Tr}(\Sigma)+\operatorname{\mathds{E}}(\left|\bm{W}\right|^{2}\mid
B)}.$

Here we used the fact that
$(1-\cos\varepsilon)/\sin\varepsilon=\tan(\varepsilon/2)\leqslant 1$ for
$\varepsilon\in(0,\pi/2)$. Combining with equation (A.3) and integrating, when
$\operatorname{\mathds{E}}\left|\bm{W}\right|^{4}<\infty$, we get that

$\displaystyle\left|\operatorname{\mathds{E}}(h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|$ $\displaystyle\leqslant
M_{1}(h)\cdot\left(A_{0}\cdot\sin\varepsilon+c_{0}\cdot
A_{1}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}\cdot
A_{2}+c_{3}\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{3}\cdot
A_{4}\cdot\frac{1}{\sin\varepsilon}\right)$

for any $\varepsilon\in(0,\pi/2]$. Taking
$\sin^{2}\varepsilon=\min(c_{3}A_{4}\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{3}/A_{0},1)$,
we get

$\displaystyle\left|\operatorname{\mathds{E}}(h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|$ $\displaystyle\leqslant
M_{1}(h)\cdot\left(A_{1}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}\cdot
A_{2}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{3/2}\cdot\sqrt{4c_{3}A_{0}A_{4}}\right).$

Note that, when
$c_{3}A_{4}\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{3}\geqslant
A_{0}$, we can directly use the upper bound

$M_{1}(h)\cdot A_{0}\sin(\pi/2)\leqslant
M_{1}(h)\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{3/2}\cdot\sqrt{4c_{3}A_{0}A_{4}}.$

When, $\operatorname{\mathds{E}}\left|\bm{W}\right|^{3}<\infty$, we similarly
get

$\displaystyle\left|\operatorname{\mathds{E}}(h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|$ $\displaystyle\leqslant
M_{1}(h)\cdot\left(A_{0}\cdot\sin\varepsilon+A_{1}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}\cdot
A_{2}+c_{2}\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{2}\cdot
A_{3}\cdot\log(1/\sin\varepsilon)\right)$

for any $\varepsilon\in(0,\pi/2]$. Here we use the fact that

$\int_{\varepsilon}^{\pi/2}\frac{\cos^{2}\alpha}{\sin\alpha}d\alpha\leqslant\int_{\sin\varepsilon}^{1}\frac{dt}{t}=\log\frac{1}{\sin\varepsilon}.$

Choosing
$\sin\varepsilon=\min(c_{2}\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{2}\cdot
A_{3}/A_{0},1)$, we get the bound

$\displaystyle\left|\operatorname{\mathds{E}}(h(\bm{W})-h(\Sigma^{\nicefrac{{1}}{{2}}}\bm{Z})\mid
B)\right|\leqslant
M_{1}(h)\cdot\biggl{(}A_{1}+\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}\cdot
A_{2}$
$\displaystyle\qquad\qquad\qquad+c_{2}\cdot\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{2}\cdot
A_{3}\cdot\biggl{(}1+\left|\log\bigl{(}{c_{2}\left\|\Sigma^{-1/2}\right\|_{\operatorname{op}}^{2}\cdot
A_{3}}/{A_{0}}\bigr{)}\right|\biggr{)}\biggr{)}.$

This completes the proof. $\blacksquare$

The following lemma shows that under LLT $|p_{k}-p_{k-1}|\ll p_{k}$, and in
particular, $r_{k}=p_{k-1}/p_{k}$ is bounded away from $\infty$, for
$|k|\ll\sigma_{Y}$.

######  Lemma A.2.

Suppose $Y$ is a random variable that take values in $\zeta+\mathds{Z}$ for
some $\zeta\in[0,1)$. Assume $Y$ has mean $0$ and variance $\sigma_{Y}^{2}$
and is such that for all $k\in\zeta+\mathds{Z}$

$\left|\sigma_{Y}p_{k}-\sigma_{Y}\varphi_{\sigma_{Y}^{2}}(k)\right|\leqslant\varepsilon_{Y}<1,$

where $p_{k}:=\operatorname{\mathds{P}}(Y=k)$ and $\varepsilon_{Y}$ is some
function. Then, for $|k|\ll\sigma_{Y}$

$\displaystyle\left|1-p_{k-1}/p_{k}\right|\lesssim(1-\varepsilon_{Y})^{-1}\cdot\max\left\\{\varepsilon_{Y},{k}/{\sigma_{Y}^{2}}\right\\}.$

###### Proof.

By LLT we have that

$\left|\sigma_{Y}p_{k}-\frac{1}{\sqrt{2\pi}}\exp\left(-{k^{2}}/{2\sigma_{Y}^{2}}\right)\right|\leqslant\varepsilon_{Y},$

where $\varepsilon_{Y}=o(1)$ as $n\to\infty$. Thus,

$\displaystyle\left|1-\frac{p_{k-1}}{p_{k}}\right|$
$\displaystyle\leqslant\frac{\left|\exp\left(-{k^{2}}/{2\sigma_{Y}^{2}}\right)-\exp\left(-{(k-1)^{2}}/{2\sigma_{Y}^{2}}\right)\right|}{\sqrt{2\pi}\sigma_{Y}p_{k}}+\frac{2\varepsilon_{Y}}{\sigma_{Y}p_{k}}$
$\displaystyle\lesssim(1-\varepsilon_{Y})^{-1}\cdot\max\left(\varepsilon_{Y},{k}/{\sigma_{Y}^{2}}\right)$

for $|k|\ll\sigma_{Y}$ and the proof is complete. $\blacksquare$

##  Appendix B Computations for Lemma 4.9

Let $\mathcal{H}$ be a graph on $v$ vertices and $m$ edges, and let $H$ be the
centered random variable that counts the number of its copies in $G_{n,p}$.
Let $\bm{s}$ denote a subset of edges in the complete graph $\mathcal{K}_{n}$
on $n$ vertices that form an isomorphic copy of $\mathcal{H}$ and
$\mathcal{S}=\mathcal{S}_{n}(H)$ denote the collection of all such $\bm{s}$’s.
Define $|\mathcal{S}|$ to be the size of the collection $\mathcal{S}$. Thus we
can rewrite the sub-graph count as

$H=\sum_{\bm{s}\in\mathcal{S}}\prod_{e\in\bm{s}}\omega_{e}=\sum_{\bm{s}\in\mathcal{S}}\omega_{\bm{s}},$

It is often more convenient to work with centered edges as it makes as we did
in (4.7) and (4.8). Thus define

$\widetilde{H}:=\sum_{\bm{s}\in\mathcal{S}}\prod_{e\in\bm{s}}\overline{\omega}_{e}=\sum_{\bm{s}\in\mathcal{S}}\widetilde{\omega}_{\bm{s}}.$

######  Lemma B.1.

With the notations as above for fixed $p\in(0,1)$ we have that

$\operatorname{Var}{\widetilde{H}}=O\left(n^{v}\right).$

###### Proof.

Given two $\bm{s}$ and $\bm{s}^{\prime}$ from $\mathcal{S}$. The expectation
$\operatorname{\mathds{E}}\widetilde{\omega}_{\bm{s}}\widetilde{\omega}_{\bm{s}^{\prime}}$
is nonzero if and only if each edge appears twice in the product. Since
$\bm{s}$ and $\bm{s}^{\prime}$ induce isomporphic graphs this yields that
$\bm{s}$ has to be equal to $\bm{s}^{\prime}$. Thus
$V(\bm{s}\cup\bm{s}^{\prime})=v$ and thus the variance of
$\widetilde{H}=O(n^{v})$. $\blacksquare$

In this section, we present explicit computations that we used in the
derivation of the CCLT for the number of wedges given the number of edges in a
random graph in Lemma 4.9. Recall that $G_{n,p}$ is the Erdős–Rényi random
graph and $q:=1-p$. Let $E=\sum_{x<y}\mathds{1}_{x\sim
y}=\sum_{x<y}\omega_{xy}$ be the number of edges in $G_{n,p}$, and
$V=\sum_{x<y,z\neq x,y}\omega_{xy}\omega_{yz}$ be the number of wedges. Define
$X=V-2(n-2)pY-(n-2)\binom{n}{2}p^{2}$ and $Y=E-\operatorname{\mathds{E}}E$.
Finally recall we place bar above the random variable to denote the centered
version of it.

It is straight forward to check that $X$ is uncorrelated with $Y$ and could be
rewritten as

$\displaystyle\sum_{i<j,k\neq
i,j}\overline{\omega}_{ik}\overline{\omega}_{kj}$
$\displaystyle=\sum_{i<j,k\neq i,j}(\omega_{ik}-p)(\omega_{kj}-p)$
$\displaystyle=\sum_{i<j,k\neq
i,j}(\omega_{ik}\omega_{kj}-p\omega_{ik}-p\omega_{jk}+p^{2})$
$\displaystyle=V-2(n-2)pY-(n-2)\binom{n}{2}p^{2}=X.$

In spirit of Lemma B.1, this representation is particularly helpful in
computing the variance of $X$

$\sigma_{X}^{2}=\binom{n}{2}(n-2)p^{2}q^{2}=\frac{1}{2}n(n-1)(n-2)p^{2}(1-p)^{2}.$

To compute the first order terms we also work with centered edges and derive
that

$\displaystyle M_{1,+}(X,Y)$
$\displaystyle=\frac{p}{N}\sum_{i<j}(q-\overline{\omega}_{ij})\left(\sum_{k\neq
i,j}\overline{\omega}_{ik}+\overline{\omega}_{jk}\right)$
$\displaystyle=-\frac{p}{N}\sum_{i\neq j}(\overline{\omega}_{ij}-q)\sum_{k\neq
i\neq j}\overline{\omega}_{ik}$ $\displaystyle=-\frac{p}{N}\sum_{i\neq j\neq
k}\overline{\omega}_{ij}\overline{\omega}_{ik}+\frac{pq}{N}\sum_{i\neq k\neq
j}\overline{\omega}_{ik}=-\frac{2}{N}(pX-pq(n-2)Y).$

Similarly

$\displaystyle M_{1,-}(X,Y)$
$\displaystyle=\frac{q}{N}\sum_{i<j}(\overline{\omega}_{ij}+p)\left(-\sum_{k\neq
i,j}\overline{\omega}_{ik}+\overline{\omega}_{jk}\right)=-\frac{2}{N}(qX+pq(n-2)Y).$

For the second order terms we compute

$\displaystyle M_{2,+}(X,Y)$
$\displaystyle=\frac{p}{N}\sum_{i<j}(q-\overline{\omega}_{ij})\left(\sum_{k\neq
i,j}\overline{\omega}_{ik}+\overline{\omega}_{jk}\right)^{2}$
$\displaystyle=\frac{p}{N}\sum_{i<j}(q-\overline{\omega}_{ij})\left(\overline{d}_{i}+\overline{d}_{j}-2\overline{\omega}_{ij}\right)^{2}$
$\displaystyle=\frac{p}{N}\sum_{i\neq
j}(q-\overline{\omega}_{ij})\left(\overline{d}_{i}^{2}+\overline{d}_{i}\overline{d}_{j}+2\overline{\omega}_{ij}^{2}-4\overline{\omega}_{ij}\overline{d}_{i}\right)$

Letting

$I_{ij}:=\overline{d}_{i}^{2}+\overline{d}_{i}\overline{d}_{j}+2\overline{\omega}_{ij}^{2}-4\overline{\omega}_{ij}\overline{d}_{i},$

we get that

$M_{2,+}(X,Y)=\frac{pq}{N}\sum_{i\neq j}I_{ij}-\frac{p}{N}\sum_{i\neq
j}\overline{\omega}_{ij}I_{ij}.$

The second term counts centered-edge graphs that have at most four distinct
vertices and hence by Lemma B.1

$\operatorname{\mathds{E}}\left|\sum_{i\neq
j}\overline{\omega}_{ij}I_{ij}\right|\lesssim n^{2}.$

Letting $\overline{d}_{i}:=\sum_{j\neq i}\overline{\omega}_{ij}$, the first
summand gives the variance terms required in the Assumption (IV) in the
following way

$\displaystyle\frac{pq}{N}\sum_{i\neq j}I_{ij}$
$\displaystyle=\frac{pq}{N}\sum_{i\neq
j}\left(\overline{d}_{i}^{2}+\overline{d}_{i}\overline{d}_{j}+2\overline{\omega}_{ij}^{2}-4\overline{\omega}_{ij}\overline{d}_{i}\right)$
$\displaystyle=\frac{pq}{N}\left((n-6)\sum_{i}\overline{d}_{i}^{2}+\left(\sum_{i}\overline{d}_{i}\right)^{2}+4\sum_{i<j}\overline{\omega}_{ij}^{2}\right)$
$\displaystyle=\frac{2}{N}\left(\frac{n(n-1)(n-2)}{2}p^{2}q^{2}+(n-6)pqX+R_{2,+}\right),$

where

$R_{2,+}:=pq\left(2\left(E^{2}-Npq\right)+(n-4)\sum_{i<j}(\overline{\omega}_{ij}^{2}-pq)\right).$

with $\operatorname{\mathds{E}}|R_{2,+}|^{2}\lesssim n^{5}.$ $M_{2,-}$ is
treated similarly as one can rewrite in as

$M_{2,-}(X,Y)=\frac{pq}{N}\sum_{i\neq j}I_{ij}+\frac{q}{N}\sum_{i\neq
j}\overline{\omega}_{ij}I_{ij}.$

The third order terms we can bound by

$\displaystyle\lambda^{-1}\operatorname{\mathds{E}}(\left|\Delta
W\right|^{3}\mid W,Y)$
$\displaystyle\leqslant\operatorname{\mathds{E}}(\left|\Delta
W\right|\left|\Delta W\right|^{2}\mid W,Y)$
$\displaystyle\lesssim\lambda^{-1}\frac{n-2}{n^{3/2}}\lambda(\psi+\left|R_{2,+}\right|+\left|R_{2,-}\right|)\lesssim
n^{-\nicefrac{{1}}{{2}}}.$

[Acknowledgments] We thank Persi Diaconis and Jonathon Peterson for their
insightful comments and for pointing out the existing literature. We also
thank two anonymous referees for careful reading, which resulted in the
improved presentation of the article, and the suggestion to adopt Raič’s and
Fang–Koike’s results to derive the Wasserstein distance bound in the
multivariate case. We further thank Felix Christian Clemen, Gleb Chernov,
Kesav Krishnan, Charlie Terlov, and Anush Tserunyan for many enlightening
discussions.

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