$\displaystyle\quad+\left.(Y_{1}-X_{1})^{2}\frac{X_{1}}{1+\sigma_{d}^{2}}\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\right)\mathbbm{1}\left(\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})<0\right)\mid\mathbf{X}_{d}\right]$
$\displaystyle\qquad\left.-\ell^{2}(\vartheta_{w,\tau}(\ell)/2)(\log\varphi(X_{1}))^{\prime}\right.|$
$\displaystyle\leq\mathbb{E}\left|-X_{1}\ell^{2}\left(\left(\frac{f_{1}(\mathbf{X}_{d})}{1+\sigma_{d}^{2}}-\frac{f_{2}(\mathbf{X}_{d})}{(1+\sigma_{d}^{2})^{2}}\right)-\frac{\vartheta_{w,\tau}(\ell)}{2}\right)+\frac{\sigma_{d}^{4}X_{1}^{3}}{(1+\sigma_{d}^{2})^{3}}f_{2}(\mathbf{X}_{d})\right|$
$\displaystyle\leq\mathbb{E}\left|-X_{1}\ell^{2}\left(\left(\frac{f_{1}(\mathbf{X}_{d})}{1+\sigma_{d}^{2}}-\frac{f_{2}(\mathbf{X}_{d})}{(1+\sigma_{d}^{2})^{2}}\right)-\frac{\vartheta_{w,\tau}(\ell)}{2}\right)\right|+d^{2\tau}\mathbb{E}\left|\frac{\sigma_{d}^{4}X_{1}^{3}}{(1+\sigma_{d}^{2})^{3}}\right|,$

using the triangle inequality and that $0\leq f_{2}(\mathbf{X}_{d})\leq 1$. As
previously,

$d^{2\tau}\mathbb{E}\left|\frac{\sigma_{d}^{4}X_{1}^{3}}{(1+\sigma_{d}^{2})^{3}}\right|\rightarrow
0.$

We also have that

$\frac{f_{1}(\mathbf{X}_{d})}{1+\sigma_{d}^{2}}-\frac{f_{2}(\mathbf{X}_{d})}{(1+\sigma_{d}^{2})^{2}}=\frac{\mathbb{E}[\mathbbm{1}\left(\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\geq
0\right)\mid\mathbf{X}_{d}]}{(1+\sigma_{d}^{2})^{2}}+\frac{\sigma_{d}^{2}f_{1}(\mathbf{X}_{d})}{(1+\sigma_{d}^{2})^{2}}.$

Using that $0\leq f_{1}(\mathbf{X}_{d})\leq 1$, the triangle inequality and
because

$\frac{\ell^{2}\sigma_{d}^{2}}{(1+\sigma_{d}^{2})^{2}}\mathbb{E}|X_{1}|\rightarrow
0,$

we can now focus on

$\mathbb{E}\left|-X_{1}\ell^{2}\left(\frac{\mathbb{E}\left[\mathbbm{1}\left(\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\geq
0\right)\mid\mathbf{X}_{d}\right]}{(1+\sigma_{d}^{2})^{2}}-\frac{\vartheta_{w,\tau}(\ell)}{2}\right)\right|.$

We have that $Y_{i}^{2}-X_{i}^{2}=(Y_{i}-X_{i})(Y_{i}+X_{i})$, and as
previously, we use that given $\mathbf{X}_{d}$, we can write
$Y_{i}=\frac{X_{i}}{1+\sigma_{d}^{2}}+\sqrt{\frac{\sigma_{d}^{2}}{1+\sigma_{d}^{2}}}U_{i}$,
and consequently,

$\displaystyle Y_{i}-X_{i}$
$\displaystyle=-\frac{\sigma_{d}^{2}}{1+\sigma_{d}^{2}}X_{i}+\frac{1}{\sqrt{1+\sigma_{d}^{2}}}\sigma_{d}U_{i},$
$\displaystyle Y_{i}+X_{i}$
$\displaystyle=\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}X_{i}+\frac{1}{\sqrt{1+\sigma_{d}^{2}}}\sigma_{d}U_{i},$

and

$\displaystyle(Y_{i}-X_{i})(Y_{i}+X_{i})$
$\displaystyle=-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}X_{i}^{2}-\frac{1}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}^{3}X_{i}U_{i}$
$\displaystyle\qquad+\frac{(2+\sigma_{d}^{2})}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}U_{i}X_{i}+\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}U_{i}^{2}.$

We define $S_{d}:=\sum_{i=2}^{d}Y_{i}^{2}-X_{i}^{2}$ and
$W_{d}:=-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}+\frac{(2+\sigma_{d}^{2})}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}\sum_{i=2}^{d}U_{i}X_{i}+\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}$.
If $\tau=1/2$, in $S_{d}$, we notice that
$\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}U_{i}^{2}\rightarrow\ell^{2}$
with probability 1 as a result of the strong law of large numbers. Also,
$\frac{1}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}^{3}\sum_{i=2}^{d}X_{i}U_{i}\rightarrow
0$ with probability 1 for the same reason. Therefore, given $\mathbf{X}_{d}$,
$S_{d}$ follows essentially a normal distribution with mean
$-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sum_{i=2}^{d}\sigma_{d}^{2}X_{i}^{2}+\ell^{2}$
and variance
$\frac{(2+\sigma_{d}^{2})^{2}}{(1+\sigma_{d}^{2})^{3}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}$.
We want to use this and that is why we will prove that $S_{d}$ and $W_{d}$ are
asymptotically equivalent; $W_{d}$ has a conditional normal distribution. We
have an explicit expression for $\mathbb{E}\left[\mathbbm{1}\left(W_{d}\geq
0\right)\mid\mathbf{X}_{d}\right]$ and we can use it.

For the rest of the proof, we consider that $\tau=1/2$. If $\tau>1/2$, we can
use the same strategy as below, but with
$W_{d}:=-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}+\frac{(2+\sigma_{d}^{2})}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}\sum_{i=2}^{d}U_{i}X_{i}$
because
$\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}U_{i}^{2}\rightarrow
0$. In this case, $W_{d}$ has a conditional normal distribution whose mean is
$-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}$
and variance
$\frac{(2+\sigma_{d}^{2})^{2}}{(1+\sigma_{d}^{2})^{3}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}$.
Both converge to 0 with probability 1, but the mean converges quicker than the
standard deviation, implying that the limit of the explicit expression for
$\mathbb{E}\left[\mathbbm{1}\left(W_{d}\geq 0\right)\mid\mathbf{X}_{d}\right]$
is $\Phi(0)=1/2$, which allows to conclude.

Let us now return to the case $\tau=1/2$. Using the triangle inequality,

$\displaystyle\mathbb{E}\left|-X_{1}\ell^{2}\left(\frac{\mathbb{E}\left[\mathbbm{1}\left(S_{d}\geq
0\right)\mid\mathbf{X}_{d}\right]}{(1+\sigma_{d}^{2})^{2}}-\frac{\vartheta_{w,\tau}(\ell)}{2}\right)\right|$
$\displaystyle\leq\mathbb{E}\left|-X_{1}\ell^{2}\left(\frac{\mathbb{E}\left[\mathbbm{1}\left(W_{d}\geq
0\right)\mid\mathbf{X}_{d}\right]}{(1+\sigma_{d}^{2})^{2}}-\frac{\vartheta_{w,\tau}(\ell)}{2}\right)\right|$
(16)
$\displaystyle+\mathbb{E}\left|-\frac{X_{1}\ell^{2}}{(1+\sigma_{d}^{2})^{2}}\left(\mathbb{E}\left[\mathbbm{1}\left(W_{d}\geq
0\right)\mid\mathbf{X}\right]-\mathbb{E}\left[\mathbbm{1}\left(S_{d}\geq
0\right)\mid\mathbf{X}\right]\right)\right|.$ (17)

We now prove that the last expectation converges to 0. Using the
Cauchy–Schwarz inequality and Jensen’s inequality, it is sufficient to prove
that $\mathbb{E}[(\mathbbm{1}\left(W_{d}\geq
0\right)-\mathbbm{1}\left(S_{d}\geq 0\right))^{2}]\rightarrow 0$ given that

$\frac{\ell^{2}}{(1+\sigma_{d}^{2})^{2}}\mathbb{E}[X_{1}^{2}]^{1/2}=\frac{\ell^{2}}{(1+\sigma_{d}^{2})^{2}}\rightarrow\ell^{2}.$

We have that

$\displaystyle\mathbb{E}[(\mathbbm{1}\left(W_{d}\geq
0\right)-\mathbbm{1}\left(S_{d}\geq 0\right))^{2}]$
$\displaystyle=\mathbb{P}(W_{d}\geq 0,S_{d}<0)+\mathbb{P}(W_{d}<0,S_{d}\geq
0)$ $\displaystyle=\mathbb{P}(W_{d}\geq 0,S_{d}<0,|W_{d}-S_{d}|>d^{-1/4})$
$\displaystyle\quad+\mathbb{P}(W_{d}\geq 0,S_{d}<0,|W_{d}-S_{d}|\leq
d^{-1/4})$ $\displaystyle\quad+\mathbb{P}(W_{d}<0,S_{d}\geq
0,|W_{d}-S_{d}|>d^{-1/4})$ $\displaystyle\quad+\mathbb{P}(W_{d}<0,S_{d}\geq
0,|W_{d}-S_{d}|\leq d^{-1/4})$ $\displaystyle\leq
2\mathbb{P}(|W_{d}-S_{d}|>d^{-1/4})$ $\displaystyle\quad+\mathbb{P}(W_{d}\geq
0,S_{d}<0,W_{d}-S_{d}\leq d^{-1/4})$
$\displaystyle\quad+\mathbb{P}(W_{d}<0,S_{d}\geq 0,S_{d}-W_{d}\leq d^{-1/4})$
$\displaystyle\leq
2\mathbb{P}(|W_{d}-S_{d}|>d^{-1/4})+\mathbb{P}(-d^{-1/4}\leq W_{d}\leq
d^{-1/4}).$

Using Markov’s inequality,

$\displaystyle\mathbb{P}(|W_{d}-S_{d}|>d^{-1/4})\leq\frac{\mathbb{E}|W_{d}-S_{d}|}{d^{-1/4}}=\frac{\mathbb{E}\left|\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}-\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}U_{i}^{2}+\frac{1}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}^{3}\sum_{i=2}^{d}X_{i}U_{i}\right|}{d^{-1/4}}$

Also,

$\displaystyle
d^{1/4}\mathbb{E}\left|\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}-\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}U_{i}^{2}\right|$
$\displaystyle\leq
d^{1/4}\mathbb{E}\left[\left(\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}-\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}U_{i}^{2}\right)^{2}\right]^{1/2}$
$\displaystyle=d^{1/4}\hbox{var}\left[\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}U_{i}^{2}\right]^{1/2}$
$\displaystyle=\frac{\ell^{2}d^{1/4}\sqrt{d-1}}{(1+\sigma_{d}^{2})d}\rightarrow
0,$

and

$\displaystyle
d^{1/4}\mathbb{E}\left|\frac{1}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}^{3}\sum_{i=2}^{d}X_{i}U_{i}\right|\leq\frac{\ell^{3}(d-1)d^{1/4}}{(1+\sigma_{d}^{2})^{3/2}d^{3/2}}\frac{2}{\pi}\rightarrow
0.$

To compute $\mathbb{P}(-d^{-1/4}\leq W_{d}\leq d^{-1/4})$, we use that given
$\mathbf{X}_{d}$,

$W_{d}\sim\mathcal{N}\left(-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}+\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)},\frac{(2+\sigma_{d}^{2})^{2}}{(1+\sigma_{d}^{2})^{3}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}\right).$

Therefore,

$\displaystyle\mathbb{P}(-d^{-1/4}\leq W_{d}\leq d^{-1/4})$
$\displaystyle=\mathbb{E}\left[\Phi\left(\frac{d^{-1/4}+\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}-\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}}{\sqrt{\frac{(2+\sigma_{d}^{2})^{2}}{(1+\sigma_{d}^{2})^{3}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}}}\right)\right]$
$\displaystyle\quad-\mathbb{E}\left[\Phi\left(\frac{-d^{-1/4}+\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}-\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}}{\sqrt{\frac{(2+\sigma_{d}^{2})^{2}}{(1+\sigma_{d}^{2})^{3}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}}}\right)\right].$

As $d\rightarrow\infty$,

$\frac{\pm
d^{-1/4}+\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}-\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}}{\sqrt{\frac{(2+\sigma_{d}^{2})^{2}}{(1+\sigma_{d}^{2})^{3}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}}}\longrightarrow\frac{\ell}{2}\quad\text{with
probability 1.}$

So, using Lebesgue’s dominated convergence theorem, we know that
$\mathbb{P}(-d^{-1/4}\leq W_{d}\leq d^{-1/4})\rightarrow 0$.

We return to the other term in (16):

$\displaystyle\mathbb{E}\left|-X_{1}\ell^{2}\left(\frac{\mathbb{E}\left[\mathbbm{1}\left(W_{d}\geq
0\right)\mid\mathbf{X}_{d}\right]}{(1+\sigma_{d}^{2})^{2}}-\frac{\vartheta_{w,\tau}(\ell)}{2}\right)\right|\leq\mathbb{E}[X_{1}^{2}]^{1/2}\ell^{2}\mathbb{E}\left[\left(\frac{\mathbb{E}\left[\mathbbm{1}\left(-W_{d}\leq
0\right)\mid\mathbf{X}_{d}\right]}{(1+\sigma_{d}^{2})^{2}}-\frac{\vartheta_{w,\tau}(\ell)}{2}\right)^{2}\right]^{1/2},$

using the Cauchy–Schwarz inequality. We have that

$\mathbb{E}\left[\mathbbm{1}\left(-W_{d}\leq
0\right)\mid\mathbf{X}_{d}\right]=\Phi\left(\frac{-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}+\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}}{\sqrt{\frac{(2+\sigma_{d}^{2})^{2}}{(1+\sigma_{d}^{2})^{3}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}}}\right)\rightarrow\Phi\left(-\frac{\ell}{2}\right),$

with probability 1. Therefore,

$\mathbb{E}\left[\left(\frac{\mathbb{E}\left[\mathbbm{1}\left(-W_{d}\leq
0\right)\mid\mathbf{X}_{d}\right]}{(1+\sigma_{d}^{2})^{2}}-\frac{\vartheta_{w,\tau}(\ell)}{2}\right)^{2}\right]^{1/2}\rightarrow
0,$

using Lebesgue’s dominated convergence theorem, which concludes this part
given that $\mathbb{E}[X_{1}^{2}]=1$.

There remains to prove that

$\mathbb{E}\left|d^{2\tau}\mathbb{E}\left[h^{\prime\prime}(X_{1})\frac{(Y_{1}-X_{1})^{2}}{2}\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{d})\mid\mathbf{X}_{d}\right]-\ell^{2}\frac{\vartheta_{w,\tau}(\ell)}{2}h^{\prime\prime}(X_{1})\right|\rightarrow
0,$

in (11). We proceed as before with a Taylor expansion around $x_{1}$ of
$\alpha_{\text{ideal}}(\mathbf{x}_{d},\mathbf{y}_{d})$, viewed as function of
$y_{1}$. This time it is less complicated because we write

$\alpha_{\text{ideal}}(\mathbf{x}_{d},\mathbf{y}_{d})=\alpha_{\text{ideal}}(\mathbf{x}_{d},\mathbf{y}_{d}^{*})+\left(\left.\frac{\partial}{\partial
y_{1}}\alpha_{\text{ideal}}(\mathbf{x}_{d},\mathbf{y}_{d})\right|_{y_{1}=w}\right)(y_{1}-x_{1}).$

Using that $M$ can be chosen such that $|h^{\prime\prime}|\leq M$, the
triangle inequality, and that $0\leq\exp(x)\mathbbm{1}(x<0)\leq 1$ (see the
partial derivative $\frac{\partial}{\partial
y_{1}}\alpha_{\text{ideal}}(\mathbf{x}_{d},\mathbf{y}_{d})$ (15)),

$\displaystyle\mathbb{E}\left|d^{2\tau}\mathbb{E}\left[h^{\prime\prime}(X_{1})\frac{(Y_{1}-X_{1})^{2}}{2}\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{d})\mid\mathbf{X}_{d}\right]-\ell^{2}\frac{\vartheta_{w,\tau}(\ell)}{2}h^{\prime\prime}(X_{1})\right|$
$\displaystyle\leq
M\mathbb{E}\left|d^{2\tau}\mathbb{E}\left[\frac{(Y_{1}-X_{1})^{2}}{2}\mid\mathbf{X}_{d}\right]f_{1}(\mathbf{X}_{d})-\ell^{2}\frac{\vartheta_{w,\tau}(\ell)}{2}\right|+\frac{d^{2\tau}}{2(1+\sigma_{d}^{2})}\mathbb{E}[|W||Y_{1}-X_{1}|^{3}].$

From what we have seen before, we know that

$\frac{d^{2\tau}}{2(1+\sigma_{d}^{2})}\mathbb{E}[|W||Y_{1}-X_{1}|^{3}]\rightarrow
0.$

We also know that

$\mathbb{E}\left[\frac{(Y_{1}-X_{1})^{2}}{2}\mid\mathbf{X}_{d}\right]=\mathbb{E}\left[\frac{\sigma_{d}^{4}X_{1}^{2}}{2(1+\sigma_{d}^{2})^{2}}-\frac{\sigma_{d}^{2}X_{1}}{1+\sigma_{d}^{2}}\sqrt{\frac{\sigma_{d}^{2}}{1+\sigma_{d}^{2}}}U_{1}+\frac{\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})}U_{1}^{2}\mid\mathbf{X}_{d}\right].$

Therefore, using the triangle inequality and that $0\leq
f_{1}(\mathbf{X}_{d})\leq 1$,

$\displaystyle\mathbb{E}\left|d^{2\tau}\mathbb{E}\left[\frac{(Y_{1}-X_{1})^{2}}{2}\mid\mathbf{X}_{d}\right]f_{1}(\mathbf{X}_{d})-\ell^{2}\frac{\vartheta_{w,\tau}(\ell)}{2}\right|$
$\displaystyle\leq\ell^{2}\mathbb{E}\left|\frac{1}{2(1+\sigma_{d}^{2})}f_{1}(\mathbf{X}_{d})-\frac{\vartheta_{w,\tau}(\ell)}{2}\right|$
$\displaystyle\quad+d^{2\tau}\mathbb{E}\left[\frac{\sigma_{d}^{4}X_{1}^{2}}{2(1+\sigma_{d}^{2})^{2}}\right].$

We have that

$d^{2\tau}\mathbb{E}\left[\frac{\sigma_{d}^{4}X_{1}^{2}}{2(1+\sigma_{d}^{2})^{2}}\right]=\frac{\ell^{4}}{2d(1+\sigma_{d}^{2})^{2}}\rightarrow
0.$

To conclude the proof, there thus remains to show that

$\mathbb{E}\left|\frac{1}{1+\sigma_{d}^{2}}\,\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\right)\mid\mathbf{X}_{d}\right]-\vartheta_{w,\tau}(\ell)\right|\rightarrow
0.$

We proceed similarly as before when we proved that

$\mathbb{E}\left|\mathbb{E}\left[\mathbbm{1}\left(\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\geq
0\right)\mid\mathbf{X}_{d}\right]-\Phi\left(-\frac{\ell}{2}\right)\right|\longrightarrow
0,$

when $\tau=1/2$.

Using the triangle inequality,

$\displaystyle\mathbb{E}\left|\frac{1}{1+\sigma_{d}^{2}}\,\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\right)\mid\mathbf{X}_{d}\right]-\vartheta_{w,\tau}(\ell)\right|$
$\displaystyle\quad\leq\mathbb{E}\left|\frac{1}{1+\sigma_{d}^{2}}\,\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\right)\mid\mathbf{X}_{d}\right]-\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\right)\mid\mathbf{X}_{d}\right]\right|$
$\displaystyle\qquad+\mathbb{E}\left|\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}\sum_{i=2}^{d}(Y_{i}^{2}-X_{i}^{2})\right)\mid\mathbf{X}_{d}\right]-\vartheta_{w,\tau}(\ell)\right|.$

The first term on the RHS converges to 0 given that $0\leq 1\wedge\exp(x)\leq
1$.

Given $\mathbf{X}_{d}$, we saw that we can write

$\displaystyle(Y_{i}-X_{i})(Y_{i}+X_{i})$
$\displaystyle=-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}X_{i}^{2}-\frac{1}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}^{3}X_{i}U_{i}$
$\displaystyle\qquad+\frac{(2+\sigma_{d}^{2})}{(1+\sigma_{d}^{2})^{3/2}}\sigma
U_{i}X_{i}+\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}U_{i}^{2}.$

We define $S_{d}:=\sum_{i=2}^{d}Y_{i}^{2}-X_{i}^{2}$ and
$W_{d}:=-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}+\frac{(2+\sigma_{d}^{2})}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}\sum_{i=2}^{d}U_{i}X_{i}+\frac{\ell^{2}d}{(1+\sigma_{d}^{2})(d-1)}$.
For the rest of the proof, we consider that $\tau=1/2$. If $\tau>1/2$, we can
use the same strategy as below, but with
$W_{d}:=-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}+\frac{(2+\sigma_{d}^{2})}{(1+\sigma_{d}^{2})^{3/2}}\sigma_{d}\sum_{i=2}^{d}U_{i}X_{i}$
because
$\frac{1}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}U_{i}^{2}\rightarrow
0$. In this case, $W_{d}$ has a conditional normal distribution whose mean is
$-\frac{2+\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}$
and variance
$\frac{(2+\sigma_{d}^{2})^{2}}{(1+\sigma_{d}^{2})^{3}}\sigma_{d}^{2}\sum_{i=2}^{d}X_{i}^{2}$.
Both converge to 0 with probability 1, but the mean converges quicker than the
standard deviation, implying that the limit of the explicit expression for
$\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}W_{d}\right)\mid\mathbf{X}_{d}\right]$
is $2\Phi(0)=1$, which allows to conclude.

Let us return to the case $\tau=1/2$. Using the triangle inequality,

$\displaystyle\mathbb{E}\left[\left|\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}S_{d}\right)\mid\mathbf{X}_{d}\right]-\vartheta_{w,\tau}(\ell)\right|\right]$
$\displaystyle\quad\leq\mathbb{E}\left[\left|\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}S_{d}\right)\mid\mathbf{X}_{d}\right]-\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}W_{d}\right)\mid\mathbf{X}_{d}\right]\right|\right]$
$\displaystyle\qquad+\mathbb{E}\left[\left|\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}W_{d}\right)\mid\mathbf{X}_{d}\right]-\vartheta_{w,\tau}(\ell)\right|\right].$

We now show that each term vanishes. We start with the first one:

$\displaystyle\mathbb{E}\left[\left|\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}S_{d}\right)\mid\mathbf{X}_{d}\right]-\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}W_{d}\right)\mid\mathbf{X}_{d}\right]\right|\right]$
$\displaystyle\quad\leq\mathbb{E}\left[\left|1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}S_{d}\right)-1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}W_{d}\right)\right|\right]$
$\displaystyle\quad\leq\frac{1}{2(1+\sigma_{d}^{2})}\mathbb{E}|S_{d}-W_{d}|,$

using Jensen’s inequality and that the function $1\wedge\exp(x)$ is
1-Lipschitz continuous. It has been proved previously that
$\mathbb{E}|S_{d}-W_{d}|\rightarrow 0$.

Given $\mathbf{X}_{d}$,

$\frac{1}{2(1+\sigma_{d}^{2})}W_{d}\sim\mathcal{N}\left(\mu_{d}:=-\frac{2+\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})^{2}}\sigma_{d}^{2}\sum_{i=1}^{d}X_{i}^{2}+\frac{\ell^{2}d}{2(1+\sigma_{d}^{2})^{2}(d-1)},s_{d}^{2}:=\frac{(2+\sigma_{d}^{2})^{2}}{4(1+\sigma_{d}^{2})^{5}}\sigma_{d}^{2}\sum_{i=1}^{d}X_{i}^{2}\right).$

Therefore,

$\displaystyle\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}W_{d}\right)\mid\mathbf{X}_{d}\right]$
$\displaystyle=\Phi\left(\frac{\mu_{d}}{s_{d}}\right)+\exp\left(\mu_{d}+\frac{s_{d}^{2}}{2}\right)\Phi\left(-s_{d}-\frac{\mu_{d}}{s_{d}}\right)$
$\displaystyle\rightarrow 2\Phi\left(-\frac{\ell}{2}\right),$

with probability 1. Lebesgue’s dominated convergence theorem allows to
establish that

$\mathbb{E}\left[\left|\mathbb{E}\left[1\wedge\exp\left(\frac{1}{2(1+\sigma_{d}^{2})}W_{d}\right)\mid\mathbf{X}_{d}\right]-\vartheta_{w,\tau}(\ell)\right|\right]\rightarrow
0,$

which concludes the proof. ∎

###  A.4 Proof of Theorem 3

###### Proof of Theorem 3.

We first prove that if

$\displaystyle\mathbb{E}\left[d^{2\tau}\left|\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\mathbb{E}[w(\mathbf{X}_{d},\mathbf{Y}_{1})\mid\mathbf{X}_{d}]}\right|\right]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{1}(d)\rightarrow
0,$

and

$\displaystyle\mathbb{E}[d^{2\tau}|\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})-\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{2}(d)\rightarrow
0,$

then $\\{Z_{d,\text{MTM}}(t):t\geq 0\\}$ converges weakly towards the same
Langevin diffusion $\\{Z(t):t\geq 0\\}$ as in Theorem 2, where
$\varrho_{1}(d)$ and $\varrho_{2}(d)$ are explicitly defined below.

We saw in the proof of Theorem 2 that to prove a weak converge towards a
diffusion, it is essentially sufficient to prove that the pseudo-generator of
$\\{Z_{d,\text{MTM}}(t):t\geq 0\\}$ converges towards the generator of the
diffusion in the $1$-norm. We thus first derive the pseudo-generator of
$\\{Z_{d,\text{MTM}}(t):t\geq 0\\}$. It is defined as follows:

$\phi_{d,\text{MTM}}(t):=d^{2\tau}\mathbb{E}[h(Z_{d,\text{MTM}}(t+1/d^{2\tau}))-h(Z_{d,\text{MTM}}(t))\mid\mathcal{F}_{\mathbf{Z}_{d,\text{MTM}}}(t)],$

where $h$ is a test function and $\mathcal{F}_{\mathbf{Z}_{d,\text{MTM}}}(t)$
is the natural filtration associated to $\\{\mathbf{Z}_{d,\text{MTM}}(t):t\geq
0\\}$. The Markov property, the fact that
$\mathbf{Z}_{d,\text{MTM}}(0)\sim\pi_{d}$ and that
$\\{\mathbf{X}_{d,\text{MTM}}(m):m\in\operatorname{\mathbb{N}}\\}$ is time-
homogeneous imply that for any $t$,

$\displaystyle\phi_{d,\text{MTM}}(t)$
$\displaystyle=d^{2\tau}\mathbb{E}[h(Z_{d,\text{MTM}}(t+1/d^{2\tau}))-h(Z_{d,\text{MTM}}(t))\mid\mathbf{Z}_{d,\text{MTM}}(t)]$
$\displaystyle\stackrel{{\scriptstyle\text{dist.}}}{{=}}d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}],$

where the last equality is in distribution, $\mathbf{X}_{d}\sim\pi_{d}$ and
$Y_{J,1}$ is the first coordinate of $\mathbf{Y}_{J}$, a proposal generated by
MTM.

The convergence in the $1$-norm of the pseudo-generator of
$\\{Z_{d,\text{MTM}}(t):t\geq 0\\}$ towards the generator of the diffusion
thus corresponds to

$\displaystyle\mathbb{E}|d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}]-Gh(X_{1})|$
$\displaystyle\quad\leq\mathbb{E}|d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}]-d^{2\tau}\mathbb{E}[(h(Y_{1})-h(X_{1}))\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{d})\mid\mathbf{X}_{d}]|$
$\displaystyle\qquad+\mathbb{E}|d^{2\tau}\mathbb{E}[(h(Y_{1})-h(X_{1}))\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{d})\mid\mathbf{X}_{d}]-Gh(X_{1})|,$

using the triangle inequality, where $Y_{1}$ is the first coordinate of
$\mathbf{Y}_{d}\sim Q_{w,\sigma}(\mathbf{X}_{d},\cdot\,)$. We saw in the proof
of Theorem 2 that

$\mathbb{E}|d^{2\tau}\mathbb{E}[(h(Y_{1})-h(X_{1}))\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{d})\mid\mathbf{X}_{d}]-Gh(X_{1})|\rightarrow
0.$

Using the triangle inequality,

$\displaystyle\mathbb{E}|d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}]-d^{2\tau}\mathbb{E}[(h(Y_{1})-h(X_{1}))\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{d})\mid\mathbf{X}_{d}]|$
$\displaystyle\leq\mathbb{E}|d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}]-d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}]|$
$\displaystyle\quad+\mathbb{E}|d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}]-d^{2\tau}\mathbb{E}[(h(Y_{1})-h(X_{1}))\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{d})\mid\mathbf{X}_{d}]|.$

We now prove that each of the two expectations on the RHS converges to 0 if

$\displaystyle\mathbb{E}\left[d^{2\tau}\left|\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\mathbb{E}[w(\mathbf{X}_{d},\mathbf{Y}_{1})\mid\mathbf{X}_{d}]}\right|\right]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{1}(d)\rightarrow
0,$

and

$\displaystyle\mathbb{E}[d^{2\tau}|\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})-\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{2}(d)\rightarrow
0.$

We start with the second one. Given any realisation $\mathbf{x}_{d}$, we have
an explicit expression for the conditional expressions and use them; we thus
use the notation $\mathbb{E}_{\mathbf{x}_{d}}$. Using Proposition 1,

$\displaystyle|\mathbb{E}_{\mathbf{x}_{d}}[(h(Y_{J,1})-h(x_{1}))\,\alpha_{\text{ideal}}(\mathbf{x}_{d},\mathbf{Y}_{J})]-\mathbb{E}_{\mathbf{x}_{d}}[(h(Y_{1})-h(x_{1}))\,\alpha_{\text{ideal}}(\mathbf{x}_{d},\mathbf{Y}_{d})]|$
$\displaystyle\leq\int|h(y_{1,1})-h(x_{1})|\,\alpha_{\text{ideal}}(\mathbf{x}_{d},\mathbf{y}_{1})\left|\frac{w(\mathbf{x}_{d},\mathbf{y}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{x}_{d},\mathbf{y}_{i})}-\frac{w(\mathbf{x}_{d},\mathbf{y}_{1})}{\int
w(\mathbf{x}_{d},\mathbf{y}_{1})\,q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{1})\,d\mathbf{y}_{1}}\right|\prod_{i=1}^{N_{d}}q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{i})\,d\mathbf{y}_{1:N_{d}}$
$\displaystyle\leq
2M\mathbb{E}_{\mathbf{x}_{d}}\left|\frac{w(\mathbf{x}_{d},\mathbf{Y}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{x}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{x}_{d},\mathbf{Y}_{1})}{\mathbb{E}_{\mathbf{x}_{d}}[w(\mathbf{x}_{d},\mathbf{Y}_{1})]}\right|,$

using Jensen’s inequality and the triangle inequality, along with the fact
that there exists a positive constant $M$ such that $|h|\leq M$, where
$y_{1,1}$ is the first coordinate of $\mathbf{y}_{1}$.

Regarding the first one,

$\displaystyle\mathbb{E}|d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}]-d^{2\tau}\mathbb{E}[(h(Y_{J,1})-h(X_{1}))\,\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})\mid\mathbf{X}_{d}]|$
$\displaystyle\quad\leq
2M\mathbb{E}[d^{2\tau}|\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})-\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|],$

using Jensen’s inequality and the triangle inequality, along with the fact
that $|h|\leq M$.

Therefore, if

$\displaystyle\mathbb{E}\left[d^{2\tau}\left|\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\mathbb{E}[w(\mathbf{X}_{d},\mathbf{Y}_{1})\mid\mathbf{X}_{d}]}\right|\right]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{1}(d)\rightarrow
0,$

and

$\displaystyle\mathbb{E}[d^{2\tau}|\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})-\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{2}(d)\rightarrow
0,$

then $\\{Z_{d,\text{MTM}}(t):t\geq 0\\}$ converges weakly towards
$\\{Z(t):t\geq 0\\}$.

We now prove that each of these two expectations converges to 0. We first
prove that

$\displaystyle\mathbb{E}\left[d^{2\tau}\left|\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\mathbb{E}[w(\mathbf{X}_{d},\mathbf{Y}_{1})\mid\mathbf{X}_{d}]}\right|\right]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{1}(d),$
(18)

and next we prove that

$\displaystyle\mathbb{E}[d^{2\tau}|\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})-\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{2}(d).$
(19)

We have that

$\displaystyle\mathbb{E}\left[d^{2\tau}\left|\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\mathbb{E}[w(\mathbf{X}_{d},\mathbf{Y}_{1})\mid\mathbf{X}_{d}]}\right|\right]$
$\displaystyle\quad=\mathbb{E}\left[\mathbb{E}\left[d^{2\tau}\left|\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\mathbb{E}[w(\mathbf{X}_{d},\mathbf{Y}_{1})\mid\mathbf{X}_{d}]}\right|\mid\mathbf{X}_{d}\right]\right].$

For any realisation $\mathbf{x}_{d}$, we have an explicit expression for the
conditional expectation and therefore write it as follows:

$\displaystyle\mathbb{E}_{\mathbf{x}_{d}}\left[d^{2\tau}\left|\frac{w(\mathbf{x}_{d},\mathbf{Y}_{1})}{\sum_{i=1}^{N_{d}}w(\mathbf{x}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{x}_{d},\mathbf{Y}_{1})}{\mathbb{E}_{\mathbf{x}_{d}}[w(\mathbf{x}_{d},\mathbf{Y}_{1})]}\right|\right]$
$\displaystyle\quad=\mathbb{E}_{\mathbf{x}_{d}}\left[d^{2\tau}\left|\frac{w(\mathbf{x}_{d},\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})}{\sum_{i=1}^{N_{d}}w(\mathbf{x}_{d},\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i})}-\frac{w(\mathbf{x}_{d},\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})}{\mathbb{E}_{\mathbf{x}_{d}}[w(\mathbf{x}_{d},\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})]}\right|\right],$

using that $\mathbf{Y}_{i}\sim q_{\sigma_{d}}(\mathbf{x}_{d},\cdot\,)$ is
equal in distribution to $\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i}$ with
$\mathbf{U}_{i}:=(U_{i,1},\ldots,U_{i,d})$, $U_{i,1},\ldots,U_{i,d}$ being $d$
(conditionally) independent standard normal random variables. We prove the
result for the case
$w(\mathbf{x}_{d},\mathbf{y}_{d})=\pi(\mathbf{y}_{d})/\pi(\mathbf{x}_{d})$;
the case
$w(\mathbf{x}_{d},\mathbf{y}_{d})=\sqrt{\pi(\mathbf{y}_{d})/\pi(\mathbf{x}_{d})}$
is proved similarly.

Using the definition of the globally-balanced weight function and the
Cauchy–Schwarz inequality,

$\displaystyle\mathbb{E}_{\mathbf{x}_{d}}\left[d^{2\tau}\left|\frac{w(\mathbf{x}_{d},\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})}{\sum_{i=1}^{N_{d}}w(\mathbf{x}_{d},\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i})}-\frac{w(\mathbf{x}_{d},\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})}{\mathbb{E}_{\mathbf{x}_{d}}[w(\mathbf{x}_{d},\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})]}\right|\right]$
$\displaystyle\quad=\mathbb{E}_{\mathbf{x}_{d}}\left[d^{2\tau}\left|\frac{\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i})-\mathbb{E}_{\mathbf{x}_{d}}[\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})]\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i})\mathbb{E}_{\mathbf{x}_{d}}[\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})]}\right|\right]$
$\displaystyle\quad\leq
d^{2\tau}\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i})}\right)^{2}\right]^{1/2}\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\frac{\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i})}{\mathbb{E}_{\mathbf{x}_{d}}[\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})]}-1\right)^{2}\right]^{1/2}.$

We analyse these two terms separately. First,

$\displaystyle\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i})}\right)^{2}\right]^{1/2}$
$\displaystyle=\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{ij})^{2}\right)}\right)^{2}\right]^{1/2}$
$\displaystyle\leq\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-2\sigma_{d}^{2}\sum_{j=1}^{d}(U_{1j}+x_{j}/\sigma_{d})^{2}\right)\right]^{1/4}\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{\sigma_{d}^{2}}{2}\sum_{j=1}^{d}(U_{ij}+x_{j}/\sigma_{d})^{2}\right)\right)^{-4}\right]^{1/4}$
$\displaystyle\leq\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-2\sigma_{d}^{2}\sum_{j=1}^{d}(U_{1j}+x_{j}/\sigma_{d})^{2}\right)\right]^{1/4}\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(2\sigma_{d}^{2}\sum_{j=1}^{d}(U_{1j}+x_{j}/\sigma_{d})^{2}\right)\right]^{1/4}$
$\displaystyle=\frac{\exp\left(-\frac{\|\mathbf{x}_{d}\|^{2}}{2(1+4\sigma_{d}^{2})}\right)}{(1+4\sigma_{d}^{2})^{d/8}}\frac{\exp\left(\frac{\|\mathbf{x}_{d}\|^{2}}{2(1-4\sigma_{d}^{2})}\right)}{(1-4\sigma_{d}^{2})^{d/8}}=\frac{\exp\left(\frac{4\sigma_{d}^{2}\|\mathbf{x}_{d}\|^{2}}{1-16\sigma_{d}^{4}}\right)}{(1+4\sigma_{d}^{2})^{d/8}(1-4\sigma_{d}^{2})^{d/8}},$

using the Cauchy–Schwarz inequality, Proposition 3, and the explicit
expression of the moment generating function of a non-central chi-squared
distribution. Note that
$\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(2\sigma_{d}^{2}\sum_{j=1}^{d}(U_{1j}+x_{j}/\sigma_{d})^{2}\right)\right]$
exists for large enough $d$; it more precisely exists when
$4\sigma_{d}^{2}<1$.

We now turn to the other term:

$\displaystyle\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\frac{\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i})}{\mathbb{E}_{\mathbf{x}_{d}}[\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})]}-1\right)^{2}\right]^{1/2}$
$\displaystyle=\frac{1}{N_{d}^{1/2}}\hbox{var}_{\mathbf{x}_{d}}\left[\frac{\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})}{\mathbb{E}_{\mathbf{x}_{d}}[\pi_{d}(\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{1})]}\right]^{1/2}$
$\displaystyle\leq\frac{1}{N_{d}^{1/2}}\frac{\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]^{1/2}}{\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]}$
$\displaystyle=\frac{1}{N_{d}^{1/2}}\frac{(1+\sigma_{d}^{2})^{d/2}}{(1+2\sigma_{d}^{2})^{d/4}}\exp\left(\frac{\sigma_{d}^{2}\|\mathbf{x}_{d}\|^{2}}{2(1+2\sigma_{d}^{2})(1+\sigma_{d}^{2})}\right).$

Putting all that together yields

$\displaystyle\mathbb{E}\left[d^{2\tau}\left|\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}-\frac{w(\mathbf{X}_{d},\mathbf{Y}_{1})}{\mathbb{E}[w(\mathbf{X}_{d},\mathbf{Y}_{1})\mid\mathbf{X}_{d}]}\right|\right]$
$\displaystyle\quad\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\frac{(1+\sigma_{d}^{2})^{d/2}}{(1+2\sigma_{d}^{2})^{d/4}(1-16\sigma_{d}^{4})^{d/8}}\,\mathbb{E}\left[\exp\left(\frac{4\sigma_{d}^{2}\|\mathbf{X}_{d}\|^{2}}{(1-16\sigma_{d}^{4})}\right)\exp\left(\frac{\sigma_{d}^{2}\|\mathbf{X}\|^{2}}{2(1+2\sigma_{d}^{2})(1+\sigma_{d}^{2})}\right)\right]$
$\displaystyle\quad=\frac{d^{2\tau}}{N_{d}^{1/2}}\frac{(1+\sigma_{d}^{2})^{d/2}}{(1+2\sigma_{d}^{2})^{d/4}(1-16\sigma_{d}^{4})^{d/8}}\,\mathbb{E}\left[\exp\left(\frac{\sigma_{d}^{2}(9+24\sigma_{d}^{2})\|\mathbf{X}\|^{2}}{2(1-16\sigma_{d}^{4})(1+2\sigma_{d}^{2})(1+\sigma_{d}^{2})}\right)\right]$
$\displaystyle\quad=\frac{d^{2\tau}}{N_{d}^{1/2}}\frac{(1+\sigma_{d}^{2})^{d/2}}{(1+2\sigma_{d}^{2})^{d/4}(1-16\sigma_{d}^{4})^{d/8}}\left(1-\frac{\sigma_{d}^{2}(9+24\sigma_{d}^{2})}{(1-16\sigma_{d}^{4})(1+2\sigma_{d}^{2})(1+\sigma_{d}^{2})}\right)^{-d/2},$

using the explicit expression of the moment generating function of a chi-
squared distribution. Note that the expectation in the penultimate line exists
for large enough $d$.

When $\tau\leq 1/2$,

$\varrho_{1}(d):=\frac{(1+\sigma_{d}^{2})^{d/2}}{(1+2\sigma_{d}^{2})^{d/4}(1-16\sigma_{d}^{4})^{d/8}}\left(1-\frac{\sigma_{d}^{2}(9+24\sigma_{d}^{2})}{(1-16\sigma_{d}^{4})(1+2\sigma_{d}^{2})(1+\sigma_{d}^{2})}\right)^{-d/2}$

converges to a constant. So the expectation converges to 0 if

$N_{d}=d^{4\tau(1+\rho)},$

with any $\rho>0$.

When
$w(\mathbf{x}_{d},\mathbf{y}_{d})=\sqrt{\pi_{d}(\mathbf{y}_{d})/\pi_{d}(\mathbf{x}_{d})}$,
the terms are different, but the speed is the same. Therefore having
$\tau=1/2$ with $N_{d}=d^{4\tau(1+\rho)}$ also makes the expectation vanish,
but we can also use $\tau=1/6$ with $N_{d}=(1+\nu)^{d}$ to make the
expectation vanish, $\nu$ being any positive constant.

There remains to prove the bound in (19) i.e.

$\displaystyle\mathbb{E}[d^{2\tau}|\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})-\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|]\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\,\varrho_{2}(d).$

We first use that the function $1\wedge x$ is 1-Lipschitz continuous:

$\mathbb{E}[d^{2\tau}|\alpha(\mathbf{X}_{d},\mathbf{Y}_{J})-\alpha_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|]\leq\mathbb{E}[d^{2\tau}|r(\mathbf{X}_{d},\mathbf{Y}_{J})-r_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|],$

where

$r(\mathbf{X}_{d},\mathbf{Y}_{J}):=\frac{\pi_{d}(\mathbf{Y}_{J})\,q_{\sigma_{d}}(\mathbf{Y}_{J},\mathbf{X}_{d})\,w(\mathbf{Y}_{J},\mathbf{X}_{d})\bigg{/}\left(\sum_{i=1}^{N_{d}-1}w(\mathbf{Y}_{J},\mathbf{Z}_{i})+w(\mathbf{Y}_{J},\mathbf{X}_{d})\right)}{\pi_{d}(\mathbf{X}_{d})\,q_{\sigma_{d}}(\mathbf{X}_{d},\mathbf{Y}_{J})\,w(\mathbf{X}_{d},\mathbf{Y}_{J})\bigg{/}\left(\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})\right)},$

and

$r_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J}):=\frac{\pi_{d}(\mathbf{Y}_{J})\,Q_{w,\sigma_{d}}(\mathbf{Y}_{J},\mathbf{X}_{d})}{\pi_{d}(\mathbf{X}_{d})\,Q_{w,\sigma_{d}}(\mathbf{X}_{d},\mathbf{Y}_{J})}.$

Recall that

$Q_{w,\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{d}):=\frac{w(\mathbf{x}_{d},\mathbf{y}_{d})\,q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{d})}{\int
w(\mathbf{x}_{d},\mathbf{y}_{d})\,q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{d})\,\mathrm{d}\mathbf{y}_{d}}.$

Using Proposition 1, we can write
$\mathbb{E}[d^{2\tau}|r(\mathbf{X}_{d},\mathbf{Y}_{J})-r_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{J})|]$
as
$\mathbb{E}[d^{2\tau}|r(\mathbf{X}_{d},\mathbf{Y}_{1})-r_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{1})|]$,
where the latter expectation is computed with respect to the following PDF:

$\pi_{d}(\mathbf{x}_{d})\frac{w(\mathbf{x}_{d},\mathbf{y}_{1})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{x}_{d},\mathbf{y}_{i})}\prod_{i=1}^{N_{d}}q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{i})\prod_{i=1}^{N_{d}-1}q_{\sigma_{d}}(\mathbf{y}_{1},\mathbf{z}_{i}).$

This means that the expectation
$\mathbb{E}[d^{2\tau}|r(\mathbf{X}_{d},\mathbf{Y}_{1})-r_{\text{ideal}}(\mathbf{X}_{d},\mathbf{Y}_{1})|]$
can be written as

$\tilde{\mathbb{E}}\left[d^{2\tau}\left|\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}-1}w(\mathbf{Y}_{1},\mathbf{Z}_{i})+w(\mathbf{Y}_{1},\mathbf{X}_{d})}-\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}\frac{\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}\right|\right],$

with respect to a PDF given by

$\pi_{d}(\mathbf{y}_{1})\prod_{i=2}^{N_{d}}q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{i})\prod_{i=1}^{N_{d}-1}q_{\sigma_{d}}(\mathbf{y}_{1},\mathbf{z}_{i})\,q_{\sigma_{d}}(\mathbf{y}_{1},\mathbf{x}_{d}),$

where $\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]$ is a
fonction of the random variable $\mathbf{Y}_{1}$ for which any realisation
$\mathbf{y}_{1}$ is mapped to

$\int
w(\mathbf{y}_{1},\mathbf{x}_{d})\,q_{\sigma_{d}}(\mathbf{y}_{1},\mathbf{x}_{d})\,\mathrm{d}\mathbf{x}_{d};$

$\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]$ is defined
analogously. We noted a change of PDF in the expectation by using the notation
“ $\tilde{\mathbb{E}}$ ”.

Using the triangle inequality,

$\displaystyle\tilde{\mathbb{E}}\left[d^{2\tau}\left|\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}-1}w(\mathbf{Y}_{1},\mathbf{Z}_{i})+w(\mathbf{Y}_{1},\mathbf{X}_{d})}-\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}\frac{\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}\right|\right]$
$\displaystyle\quad\leq\tilde{\mathbb{E}}\left[d^{2\tau}\left|\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}-1}w(\mathbf{Y}_{1},\mathbf{Z}_{i})+w(\mathbf{Y}_{1},\mathbf{X}_{d})}-\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}\right|\right]$
$\displaystyle\qquad+\tilde{\mathbb{E}}\left[d^{2\tau}\left|\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}-\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}\frac{\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}\right|\right].$

The first expectation on the RHS can be seen as an expectation with respect to
the PDF

$\pi_{d}(\mathbf{y}_{1})\prod_{i=1}^{N_{d}-1}q_{\sigma_{d}}(\mathbf{y}_{1},\mathbf{z}_{i})\,q_{\sigma_{d}}(\mathbf{y}_{1},\mathbf{x}_{d});$

so this expectation is equal to that in (18) and it thus converges to 0 under
the same conditions.

For the other one, we have that

$\displaystyle\tilde{\mathbb{E}}\left[d^{2\tau}\left|\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}-\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}\frac{\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}\right|\right]$
$\displaystyle=\tilde{\mathbb{E}}\left[d^{2\tau}\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}\frac{1}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})-\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]\right|\right].$

This expectation can be seen as an expectation with respect to the PDF

$\pi_{d}(\mathbf{y}_{1})\prod_{i=2}^{N_{d}}q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{i})\,q_{\sigma_{d}}(\mathbf{y}_{1},\mathbf{x}_{d})=\pi(\mathbf{y}_{1})\prod_{i=1}^{N_{d}}q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{i}),$

using that $q_{\sigma_{d}}$ is symmetric. We prove the result for the case
$w(\mathbf{x}_{d},\mathbf{y}_{d})=\pi_{d}(\mathbf{y}_{d})/\pi_{d}(\mathbf{x}_{d})$;
the case
$w(\mathbf{x}_{d},\mathbf{y}_{d})=\sqrt{\pi_{d}(\mathbf{y}_{d})/\pi_{d}(\mathbf{x}_{d})}$
is proved similarly. For any realisation $\mathbf{y}_{1}$, we have that

$\displaystyle\frac{w(\mathbf{y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{y}_{1}}[w(\mathbf{y}_{1},\mathbf{X}_{d})]}$
$\displaystyle=\frac{\pi_{d}(\mathbf{X}_{d})}{\frac{1}{(2\pi)^{d/2}}\mathbb{E}_{\mathbf{y}_{1}}\left[\exp\left(-\frac{\sigma_{d}^{2}}{2}\sum_{j=1}^{d}(U_{1j}+y_{1j}/\sigma_{d})^{2}\right)\right]}$
$\displaystyle=\frac{\pi_{d}(\mathbf{X}_{d})}{(2\pi)^{-d/2}(1+\sigma_{d}^{2})^{-d/2}\exp\left(-\frac{1}{2(1+\sigma_{d}^{2})}\sum_{j=1}^{d}y_{1j}^{2}\right)},$

using that $q_{\sigma_{d}}(\mathbf{x}_{d},\cdot\,)$ is a normal distribution,
where, as previously, $U_{11},\ldots,U_{1N_{d}}$ are (conditionally)
independent standard normal random variables. We can thus rewrite the
expectation

$\tilde{\mathbb{E}}\left[d^{2\tau}\frac{w(\mathbf{Y}_{1},\mathbf{X}_{d})}{\mathbb{E}_{\mathbf{Y}_{1}}[w(\mathbf{Y}_{1},\mathbf{X}_{d})]}\frac{1}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})-\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]\right|\right]$

as

$\displaystyle\stackrel{{\scriptstyle\approx}}{{\mathbb{E}}}\left[d^{2\tau}(1+\sigma_{d}^{2})^{d/2}\frac{\exp\left(-\frac{\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})}\,\|\mathbf{Y}_{1}\|^{2}\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})-\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]\right|\right],$

where the expectation is with respect to the PDF
$\pi_{d}(\mathbf{x}_{d})\prod_{i=1}^{N_{d}}q_{\sigma_{d}}(\mathbf{x}_{d},\mathbf{y}_{i})$.
Using the definition of the globally-balanced weight function,

$\displaystyle\stackrel{{\scriptstyle\approx}}{{\mathbb{E}}}\left[d^{2\tau}(1+\sigma_{d}^{2})^{d/2}\frac{\exp\left(-\frac{\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})}\sum_{j=1}^{d}Y_{1j}^{2}\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})}\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}w(\mathbf{X}_{d},\mathbf{Y}_{i})-\mathbb{E}_{\mathbf{X}_{d}}[w(\mathbf{X}_{d},\mathbf{Y}_{1})]\right|\right]$
$\displaystyle\quad=\,\stackrel{{\scriptstyle\approx}}{{\mathbb{E}}}\left[d^{2\tau}(1+\sigma_{d}^{2})^{d/2}\frac{\exp\left(-\frac{\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})}\sum_{j=1}^{d}Y_{1j}^{2}\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{ij}^{2}\right)}\right.$
$\displaystyle\qquad\times\left.\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{ij}^{2}\right)-\mathbb{E}_{\mathbf{X}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{1j}^{2}\right)\right]\right|\right].$

We omit the $\approx$ above $\stackrel{{\scriptstyle\approx}}{{\mathbb{E}}}$
for the rest of the proof to simplify the notation and note that for the rest
of the proof the random variables are such that $\mathbf{X}_{d}\sim\pi_{d}$
and $\mathbf{Y}_{1},\ldots,\mathbf{Y}_{N_{d}}$ are conditionally independent
given $\mathbf{X}_{d}$ with a distribution given by
$\mathbf{Y}_{i}\sim\mathcal{N}(\mathbf{X}_{d},\sigma_{d}^{2}\mathbb{I}_{d})$.

We have that

$\displaystyle\mathbb{E}\left[\frac{\exp\left(-\frac{\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})}\sum_{j=1}^{d}Y_{1j}^{2}\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{ij}^{2}\right)}\right.$
$\displaystyle\qquad\times\left.\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{ij}^{2}\right)-\mathbb{E}_{\mathbf{X}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{1j}^{2}\right)\right]\right|\right]$
$\displaystyle=\mathbb{E}\left[\mathbb{E}\left[\frac{\exp\left(-\frac{\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})}\sum_{j=1}^{d}Y_{1j}^{2}\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{ij}^{2}\right)}\right.\right.$
$\displaystyle\qquad\times\left.\left.\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{ij}^{2}\right)-\mathbb{E}_{\mathbf{X}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{1j}^{2}\right)\right]\right|\mid\mathbf{X}_{d}\right]\right].$

Given any realisation $\mathbf{x}_{d}$, we have an explicit expression for the
conditional expectation and therefore write it as follows:

$\displaystyle\mathbb{E}_{\mathbf{x}_{d}}\left[\frac{\exp\left(-\frac{\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{ij})^{2}\right)}\right.$
$\displaystyle\quad\times\left.\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{ij})^{2}\right)-\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]\right|\right]$
$\displaystyle\quad\leq\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{ij})^{2}\right)\right)^{-2}\right]^{1/2}$
$\displaystyle\qquad\times\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{ij})^{2}\right)-\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]\right)^{2}\right]^{1/2}$

using that $\mathbf{Y}_{i}\sim q_{\sigma_{d}}(\mathbf{x}_{d},\cdot\,)$ is
equal in distribution to $\mathbf{x}_{d}+\sigma_{d}\mathbf{U}_{i}$ and the
Cauchy–Schwarz inequality.

We analyse the two terms on the RHS separately. First,

$\displaystyle\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{ij})^{2}\right)\right)^{-2}\right]^{1/2}$
$\displaystyle\quad\leq\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{4\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]^{1/4}\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{ij})^{2}\right)\right)^{-4}\right]^{1/4}$
$\displaystyle\quad\leq\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{4\sigma_{d}^{2}}{1+\sigma_{d}^{2}}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]^{1/4}\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(2\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]^{1/4}$
$\displaystyle\quad=\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{4\sigma_{d}^{4}}{1+\sigma_{d}^{2}}\sum_{j=1}^{d}(x_{j}/\sigma_{d}+U_{1j})^{2}\right)\right]^{1/4}\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(2\sigma_{d}^{2}\sum_{j=1}^{d}(x_{j}/\sigma_{d}+U_{1j})^{2}\right)\right]^{1/4}$
$\displaystyle\quad=\frac{\exp\left(-\frac{\sigma_{d}^{2}\|\mathbf{x}_{d}\|^{2}}{1+\sigma_{d}^{2}+8\sigma_{d}^{4}}\right)}{\left(1+\frac{8\sigma_{d}^{4}}{1+\sigma_{d}^{2}}\right)^{d/8}}\frac{\exp\left(\frac{\|\mathbf{x}_{d}\|^{2}}{2(1-4\sigma_{d}^{2})}\right)}{\left(1-4\sigma_{d}^{2}\right)^{d/8}}$
$\displaystyle\quad\leq\frac{1}{\left(1+\frac{8\sigma_{d}^{4}}{1+\sigma_{d}^{2}}\right)^{d/8}}\frac{\exp\left(\frac{\|\mathbf{x}_{d}\|^{2}}{2(1-4\sigma_{d}^{2})}\right)}{\left(1-4\sigma_{d}^{2}\right)^{d/8}}$

using the Cauchy–Schwarz inequality, Proposition 3, and the expression for the
moment generating function of a non-central chi-squared distribution. Note
that
$\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(2\sigma_{d}^{2}\sum_{j=1}^{d}(x_{j}/\sigma_{d}+U_{1j})^{2}\right)\right]$
exists for large enough $d$; it more precisely exists when
$4\sigma_{d}^{2}<1$.

Second,

$\displaystyle\mathbb{E}_{\mathbf{x}_{d}}\left[\left(\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{ij})^{2}\right)-\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]\right)^{2}\right]^{1/2}$
$\displaystyle\quad=\frac{1}{N_{d}^{1/2}}\hbox{var}_{\mathbf{x}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}(x_{j}+\sigma_{d}U_{1j})^{2}\right)\right]^{1/2}$
$\displaystyle\quad\leq\frac{1}{N_{d}^{1/2}}\mathbb{E}_{\mathbf{x}_{d}}\left[\exp\left(-\sigma_{d}^{2}\sum_{j=1}^{d}(x_{j}/\sigma_{d}+U_{1j})^{2}\right)\right]^{1/2}$
$\displaystyle\quad=\frac{1}{N_{d}^{1/2}}\frac{\exp\left(\frac{-\|\mathbf{x}_{d}\|^{2}}{2(1+2\sigma_{d}^{2})}\right)}{(1+2\sigma_{d}^{2})^{d/4}}.$

Putting all the results above together yields

$\displaystyle\mathbb{E}\left[d^{2\tau}(1+\sigma_{d}^{2})^{d/2}\frac{\exp\left(-\frac{\sigma_{d}^{2}}{2(1+\sigma_{d}^{2})}\sum_{j=1}^{d}Y_{1j}^{2}\right)}{\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{ij}^{2}\right)}\right.$
$\displaystyle\quad\times\left.\left|\frac{1}{N_{d}}\sum_{i=1}^{N_{d}}\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{ij}^{2}\right)-\mathbb{E}_{\mathbf{X}_{d}}\left[\exp\left(-\frac{1}{2}\sum_{j=1}^{d}Y_{1j}^{2}\right)\right]\right|\right]$
$\displaystyle\quad\leq\frac{d^{2\tau}}{N_{d}^{1/2}}\frac{(1+\sigma_{d}^{2})^{d/2}}{\left(1+\frac{8\sigma_{d}^{4}}{1+\sigma_{d}^{2}}\right)^{d/8}\left(1-4\sigma_{d}^{2}\right)^{d/8}(1+2\sigma_{d}^{2})^{d/4}}\mathbb{E}\left[\exp\left(\frac{3\sigma_{d}^{2}\|\mathbf{X}_{d}\|^{2}}{(1-4\sigma_{d}^{2})(1+2\sigma_{d}^{2})}\right)\right]$
$\displaystyle\quad=\frac{d^{2\tau}}{N_{d}^{1/2}}\frac{(1+\sigma_{d}^{2})^{d/2}}{\left(1+\frac{8\sigma_{d}^{4}}{1+\sigma_{d}^{2}}\right)^{d/8}\left(1-4\sigma_{d}^{2}\right)^{d/8}(1+2\sigma_{d}^{2})^{d/4}\left(1-\frac{6\sigma_{d}^{2}}{(1-4\sigma_{d}^{2})(1+2\sigma_{d}^{2})}\right)^{d/2}}$

using the explicit expression for the moment generating function of a chi-
squared distribution. Note that the expectation in the penultimate line exist
for large enough $d$. The last term, seen as a function of $d$, behaves as
that in the bound (18), and thus converges to 0 under the same conditions.
Note that

$\displaystyle\varrho_{2}(d)$
$\displaystyle:=\frac{(1+\sigma_{d}^{2})^{d/2}}{(1+2\sigma_{d}^{2})^{d/4}(1-16\sigma_{d}^{4})^{d/8}}\left(1-\frac{\sigma_{d}^{2}(9+24\sigma_{d}^{2})}{(1-16\sigma_{d}^{4})(1+2\sigma_{d}^{2})(1+\sigma_{d}^{2})}\right)^{-d/2}$
$\displaystyle\qquad+\frac{(1+\sigma_{d}^{2})^{d/2}}{\left(1+\frac{8\sigma_{d}^{4}}{1+\sigma_{d}^{2}}\right)^{d/8}\left(1-4\sigma_{d}^{2}\right)^{d/8}(1+2\sigma_{d}^{2})^{d/4}\left(1-\frac{6\sigma_{d}^{2}}{(1-4\sigma_{d}^{2})(1+2\sigma_{d}^{2})}\right)^{d/2}}.$

∎