$\displaystyle\quad\frac{T_{1}(\lambda(C(\eta_{n};\boldsymbol{\delta}_{n})))}{{{\operatorname{{\mathbb{V}ar}}[\lambda(C(\eta_{n};\boldsymbol{\delta}_{n}))]}}},\frac{T_{2}(\lambda(C(\eta_{n};\boldsymbol{\delta}_{n})))}{{{\operatorname{{\mathbb{V}ar}}[\lambda(C(\eta_{n};\boldsymbol{\delta}_{n}))]}}}\leq
C\frac{\left(\frac{(\log
n)^{6+\frac{4}{d}}}{n^{3+\frac{4}{d}}}\right)^{\frac{1}{2}}}{n^{-1-\frac{2}{d}}{{(\log
n)^{-2-\frac{2}{d}}}}}$ $\displaystyle\leq$ $\displaystyle C\frac{(\log
n)^{5+\frac{4}{d}}}{n^{1/2}},$ (4.38)
$\displaystyle\frac{T_{3}(\lambda(C(\eta_{n};\boldsymbol{\delta}_{n})))}{{{\operatorname{{\mathbb{V}ar}}[\lambda(C(\eta_{n};\boldsymbol{\delta}_{n}))]}}^{3/2}}\leq
C\frac{\frac{(\log
n)^{3+\frac{3}{d}}}{n^{2+\frac{3}{d}}}}{\left(n^{-1-\frac{2}{d}}{{(\log
n)^{-2-\frac{2}{d}}}}\right)^{\frac{3}{2}}}$ $\displaystyle\leq$
$\displaystyle C\frac{(\log n)^{6+6/d}}{n^{1/2}}.$

We now apply (A.2) from Theorem A.2 to
$F(\cdot)=\lambda(C(\cdot;\boldsymbol{\delta}_{n}))-\mathbb{E}(\lambda(C(\cdot;\boldsymbol{\delta}_{n})))$,
and substitute the bounds (4.37) and (4.38) to obtain that

III $\displaystyle\leq$ $\displaystyle C\left[{{\frac{(\log
n)^{5+\frac{4}{d}}}{n^{1/2}}}}+{{\frac{(\log
n)^{5+\frac{4}{d}}}{n^{1/2}}}}+{{\frac{(\log
n)^{6+\frac{6}{d}}}{n^{1/2}}}}\right]$ $\displaystyle\leq$ $\displaystyle
C\,\frac{(\log n)^{6+6/d}}{n^{1/2}}.$

Finally, combining the bounds for I, II, and III, we have that

(4.39) $\displaystyle
d_{W}\left(\frac{\delta_{V}(n)-\mathbb{E}[\delta_{V}(n)]}{\sqrt{\operatorname{{\mathbb{V}ar}}[\delta_{V}(n)]}},Z\right)$
$\displaystyle\leq C\left(\frac{(\log
n)^{(1+1/d)}}{n^{\epsilon+\frac{1}{2}+\frac{1}{d}}}+\frac{(\log
n)^{(1+1/d)}}{n^{\frac{\epsilon}{2}}}+\frac{(\log
n)^{6+6/d}}{n^{1/2}}\right).$

Thus, (2.11) holds for the Poisson process.

Case II. The source set is $\beta_{n}$. Let $\beta_{n}$ be a binomial process
on $bD$ satisfying (2.7), and let
$\delta_{V}(n)=\delta_{V}(P(\beta_{n};\boldsymbol{\delta}_{n}),D)$. Once
again, by the triangle inequality, we may write

$d_{W}\left(\frac{\delta_{V}(n)-\mathbb{E}[\delta_{V}(n)]}{\sqrt{\operatorname{{\mathbb{V}ar}}[\delta_{V}(n)]}},Z\right)\leq\text{I}_{B}+\text{II}_{B}+\text{III}_{B},$

where

$\displaystyle\text{I}_{B}$ $\displaystyle:=$ $\displaystyle
d_{W}\left(\frac{\delta_{V}(n)-\mathbb{E}[\delta_{V}(n)]}{\sqrt{\operatorname{{\mathbb{V}ar}}[\delta_{V}(n)]}},\frac{\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))-\mathbb{E}[\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))]}{\sqrt{\operatorname{{\mathbb{V}ar}}[\delta_{V}(n)]}}\right),$
$\displaystyle\text{II}_{B}$ $\displaystyle:=$ $\displaystyle
d_{W}\left(\left(\frac{\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))-\mathbb{E}[\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))]}{\sqrt{{\operatorname{{\mathbb{V}ar}}[\delta_{V}(n)]}}}\right),\frac{\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))-\mathbb{E}[\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))]}{\sqrt{\operatorname{{\mathbb{V}ar}}[\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))]}}\right),$
$\displaystyle\text{III}_{B}$ $\displaystyle:=$ $\displaystyle
d_{W}\left(\frac{\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))-\mathbb{E}[\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))]}{\sqrt{\operatorname{{\mathbb{V}ar}}[\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}))]}},Z\right).$

Owing to the assumed lower bound for
$\operatorname{{\mathbb{V}ar}}[\delta_{V}(n)]$, the bound for IB is the same
as that for I, namely, for some $C>0$,

$\displaystyle\text{I}_{B}$ $\displaystyle\leq$ $\displaystyle C\frac{(\log
n)^{(1+1/d)}}{n^{\epsilon+\frac{1}{2}+\frac{1}{d}}}.$

Next, since (4.34) holds for $\beta_{n}$ as well, the assumed lower bound for
$\operatorname{{\mathbb{V}ar}}[\delta_{V}(n)]$ yields the same bound for
$\text{II}_{B}$ as that for II, i.e., for some $C>0$,

$\text{II}_{B}\leq C\frac{(\log n)^{(1+1/d)}}{n^{\frac{\epsilon}{2}}}.$

To obtain a bound on $\text{III}_{B}$, we use Theorem A.3. First, note that

(4.40)
$|D_{j}\lambda(C({\mathbf{Z}};\boldsymbol{\delta}_{n}))|\leq\lambda(C(W^{j};\boldsymbol{\delta}_{n}(W^{j})))\leq
C\delta_{n}^{d+1},\quad{Z\in Rec({\mathbf{W}})},\ j=1,2.$

From the inclusion-exclusion principle, we have that

(4.41) $D_{1,2}(\lambda(C({\mathbf{W}};\boldsymbol{\delta}_{n})))\neq 0\
\text{implies that}\ C(W^{1};\boldsymbol{\delta}_{n}(W^{1}))\cap
C(W^{2};\boldsymbol{\delta}_{n}(W^{2}))\neq\emptyset.$

Thus, since ${\mathbf{Y}}\overset{d}{=}{\mathbf{W}}$ and (4.40) holds, we have
that

(4.42) $\displaystyle
T^{\prime}_{1}\Big{(}\lambda(C({\mathbf{W}};\boldsymbol{\delta}_{n}))\Big{)}$
$\displaystyle\leq$ $\displaystyle
C\delta_{n}^{4(d+1)}\mathbb{P}\Big{(}C(W^{1};\|g\|_{\infty}\delta_{n})\cap
C(W^{2};\|g\|_{\infty}\delta_{n})\neq\emptyset\Big{)}$ $\displaystyle\leq$
$\displaystyle
C\delta_{n}^{4(d+1)}\mathbb{E}[\sigma_{f}(N(W^{1};\|g\|_{\infty}\delta_{n}))]\leq
C\delta_{n}^{5d+4},$

where, in the final step, we have used Proposition 3.16 to estimate
$\sigma\left(N(W^{1};\|g\|_{\infty}\delta_{n})\right)$. Invoking (4.41) again,
we have the following upper bound for
$T^{\prime}_{2}\Big{(}\lambda(C({\mathbf{W}};\boldsymbol{\delta}_{n}))\Big{)}$:

$\displaystyle C\delta_{n}^{4(d+1)}\sup_{{\mathbf{Y}},{\mathbf{Z}}\in
Rec({\mathbf{W}})}\\!\\!\mathbb{P}\Big{(}C(Y^{1};\boldsymbol{\delta}_{n}(Y^{1}))\cap
C(Y^{2};\boldsymbol{\delta}_{n}(Y^{2}))\neq\emptyset,$ (4.43)
$\displaystyle\qquad\qquad\qquad\qquad\qquad
C(Z^{1};\boldsymbol{\delta}_{n}(Z^{1}))\cap
C(Z^{3};\boldsymbol{\delta}_{n}(Z^{3}))\neq\emptyset\Big{)}.$

To further bound (4.43), note that by the definition of ${\mathbf{Y}}$ and
${\mathbf{Z}}$, $Y^{1},Y^{2},Z^{3}$ are independent and, either $Z^{1}$ is
independent of all of them, or $Z^{1}=Y^{1}$. Thus,

$\displaystyle\qquad
T^{\prime}_{2}\Big{(}\lambda(C({\mathbf{W}};\boldsymbol{\delta}_{n}))\Big{)}$
$\displaystyle\qquad\leq
C\delta_{n}^{4(d+1)}\max\left\\{\mathbb{P}\Big{(}C(W^{1};\boldsymbol{\delta}_{n}(W^{1}))\cap
C(W^{2};\boldsymbol{\delta}_{n}(W^{2}))\neq\emptyset\Big{)}^{2},\right.$
$\displaystyle\qquad\left.\mathbb{P}\Big{(}C(W^{1};\boldsymbol{\delta}_{n}(W^{1}))\cap
C(W^{2};\boldsymbol{\delta}_{n}(W^{2}))\neq\emptyset,C(W^{1};\boldsymbol{\delta}_{n}(W^{1}))\cap
C(W^{3};\boldsymbol{\delta}_{n}(W^{3}))\neq\emptyset\Big{)}\right\\}$
$\displaystyle\qquad\leq
C\delta_{n}^{4(d+1)}\max\left\\{{\mathbb{E}\left[\sigma_{f}(N(W^{1};\|g\|_{\infty}\delta_{n}))\right]^{2}},\mathbb{E}[\sigma_{f}(N(W^{1};\|g\|_{\infty}\delta_{n}))^{2}]\right\\}$
(4.44) $\displaystyle\qquad\leq C\delta_{n}^{6d+4}.$

Next, it is immediate from (4.40) that

(4.45)
$T^{\prime}_{3}\Big{(}\lambda(C({\mathbf{W}};\boldsymbol{\delta}_{n}))\Big{)}\leq
C\delta_{n}^{4(d+1)}\quad\text{and}\quad
T^{\prime}_{4}\Big{(}\lambda(C({\mathbf{W}};\boldsymbol{\delta}_{n}))\Big{)}\leq
C\delta_{n}^{3(d+1)}.$

Lastly, since (4.34) holds for $\beta_{n}$ as well, the assumed lower bound
for
$\operatorname{{\mathbb{V}ar}}[\delta_{V}(P(\beta_{n};\boldsymbol{\delta}_{n}),D)]$
is also a lower bound for
$\operatorname{{\mathbb{V}ar}}[\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}),D)]$.
Now, we apply (A.4) in Theorem A.3 to
$F(\cdot)=\lambda(C(\cdot;\boldsymbol{\delta}_{n}))-\mathbb{E}(\lambda(C(\cdot;\boldsymbol{\delta}_{n})))$,
and substitute the bounds (4.42), (4.44) and (4.45), and the lower bound for
$\operatorname{{\mathbb{V}ar}}[\lambda(C(\beta_{n};\boldsymbol{\delta}_{n}),D)]$
to obtain that

$\displaystyle\text{III}_{B}$ $\displaystyle\leq
C\,\frac{\sqrt{n^{2}\delta_{n}^{5d+4}}+\sqrt{n^{\frac{3}{2}}\delta_{n}^{6d+4}}+\sqrt{n\delta_{n}^{4(d+1)}}}{n^{-(1+2/d)}{{(\log
n)^{-2(1+1/d)}}}}+\frac{Cn\delta_{n}^{3(d+1)}}{\left(n^{-(1+2/d)}{{(\log
n)^{-2(1+1/d)}}}\right)^{\frac{3}{2}}}$ $\displaystyle\leq C\frac{(\log
n)^{6+\frac{6}{d}}}{\sqrt{n}}.$

Finally, combining the bounds for IB, IIB, and IIIB, we have that

$d_{W}\left(\frac{\delta_{V}(n)-\mathbb{E}[\delta_{V}(n)]}{\sqrt{\operatorname{{\mathbb{V}ar}}[\delta_{V}(n)]}},Z\right)\leq
C\left({{\frac{(\log
n)^{(1+1/d)}}{n^{\epsilon+\frac{1}{2}+\frac{1}{d}}}+\frac{(\log
n)^{(1+1/d)}}{n^{\frac{\epsilon}{2}}}+\frac{(\log
n)^{6+6/d}}{n^{1/2}}}}\right).$

Thus, (2.11) holds for $\beta_{n}$ as well, and the proof is complete. ∎

###  4.4. Proposition 2.6: conditions for coverage

###### Proof of Proposition 2.6.

The strategy of the proof is to use Lemma 3.14 to convert containment
probabilities into converage probabilities. We handle the latter by dividing
$bD$ suitably and using a classical balls into bins approach for coverage
problems.

Proof of $(a)$. Let $\eta_{n}$ be a Poisson process on $bD$. We claim that it
suffices to show the existence of a $c_{D}>0$ such that

(4.46) $\lim_{n\to\infty}\mathbb{P}\Big{(}S(z;\varepsilon)\setminus
S(\eta_{n};c_{D}\delta_{n})\neq\emptyset\Big{)}=1,\qquad\text{{for some}}\
z\in bD,\varepsilon>0.$

To see this, assume (4.46) and suppose $\min_{z\in bD}g(z)<c_{D}$. Then, there
exist $z_{0}\in bD$ and $\varepsilon_{0}>0$ such that $\max_{w\in
S(z_{0};2\varepsilon_{0})}g(w)<c_{D}$. Moreover, since $||g||_{\infty}<\infty$
and $\delta_{n}\rightarrow 0$ as $n\rightarrow\infty$, there is an
$n_{g,\varepsilon}$ large enough so that if $n\geq n_{g,\varepsilon}$, then
$S(z_{0};\varepsilon_{0})\cap S(w;\boldsymbol{\delta}_{n}(w))=\emptyset$
whenever $w\notin S(z_{0};2\varepsilon_{0})$. Now, suppose
$S(z_{0};\varepsilon_{0})\setminus S(\eta_{n};c_{D}\delta_{n})\neq\emptyset$
for some $n\geq n_{g,\varepsilon}$. By the monotonicity of cuts,
$S(\eta_{n}\cap S(z_{0};2\varepsilon_{0});\boldsymbol{\delta}_{n})\subset
S(\eta_{n};c_{D}\delta_{n})$. Thus,

$S(z_{0};\varepsilon_{0})\setminus
S(\eta_{n};\boldsymbol{\delta}_{n})=S(z_{0};\varepsilon_{0})\setminus
S(\eta_{n}\cap
S(z_{0};2\varepsilon_{0});\boldsymbol{\delta}_{n})\neq\emptyset.$

In other words, for any $n\geq n_{g,\varepsilon}$,
$S(z_{0};\varepsilon_{0})\setminus S(\eta_{n};c_{D}\delta_{n})\neq\emptyset$
implies that $S(z_{0};\varepsilon_{0})\setminus
S(\eta_{n};\boldsymbol{\delta}_{n})\neq\emptyset$, which (by Lemma 3.14)
further implies that $P(\eta_{n};\boldsymbol{\delta}_{n})\not\subset D$. Thus,
(4.46) implies $(a)$.

We now establish (4.46). Let $m\in\mathbb{N}$ and
$\\{p^{1},\ldots,p^{m}\\}\subset bD$ be as in Lemma 3.15 corresponding to
$\delta=\delta_{n}$. Note that both $m$ and the $p^{j}$’s change with $n$, but
we do not indicate this in the notation for convenience. Now, given $z\in bD$
and $\varepsilon>0$, set

$\ell_{n}:=\ell_{\delta_{n},\varepsilon,z}=|\\{j:p^{j}\in S(z;\epsilon)\\}|.$

Without loss of generality, we assume that $\\{1,...,\ell_{n}\\}=\\{j:p^{j}\in
S(z;\varepsilon)\\}$. We claim that, in order to prove (4.46), it suffices to
show the existence of a $b=b_{D}>0$ such that for all $z\in bD$ and
$\varepsilon>0$, with high probability there exists a
$j\in\\{1,...\ell_{n}\\}$ such that $\eta_{n}\cap
S(p^{j};b\mathfrak{K}\delta_{n})=\emptyset$, i.e.,

(4.47)
$\lim_{n\to\infty}\mathbb{P}\left(\bigcup_{j=1}^{\ell_{n}}\left\\{\eta_{n}\cap
S(p^{j};b\mathfrak{K}\delta_{n})=\emptyset\right\\}\right)=1.$

To see this, set $z=z_{0}$ and $\varepsilon=\varepsilon_{0}$. Now, suppose
$\eta_{n}\cap S(p^{j};b\mathfrak{K}\delta_{n})=\emptyset$ for some
$j\in\\{1,...,\ell_{n}\\}$, where
$\ell_{n}=\ell_{\delta_{n},\varepsilon_{0},z_{0}}$. Then, by the quasi-
symmetry of the quasimetric $\mathtt{d}$, there is a $c_{D}>0$ depending only
on $\mathfrak{K}$ and $b$ such that
$\min\\{\mathtt{d}(p^{j},W):W\in\eta_{n}\\}>c_{D}\delta_{n}$, i.e.,
$p^{j}\notin S(\eta_{n};c_{D}\delta_{n})$. Thus,
$S(z_{0};\varepsilon_{0})\setminus S(\eta_{n};c_{D}\delta_{n})\neq\emptyset$.
In other words, (4.47) implies (4.46).

Finally, it remains to prove (4.47). Let $z\in bD$ and $\varepsilon>0$. Let
$b\in(0,1)$. Using that the caps $S(p^{j};c\mathfrak{K}\delta_{n})$,
$j=1,...,\ell_{n}$, are pairwise disjoint, the Poisson process is completely
independent, $||f||_{\infty}<\infty$, and (3.12), we have that for large
enough $n$,

(4.48)
$\displaystyle\mathbb{P}\left(\bigcap_{j=1}^{\ell_{n}}\left\\{\eta_{n}\cap
S(p^{j};b\mathfrak{K}\delta_{n})\neq\emptyset\right\\}\right)$
$\displaystyle=$ $\displaystyle\prod_{j=1}^{\ell_{n}}\mathbb{P}(\eta_{n}\cap
S(p^{j};b\mathfrak{K}\delta_{n})\neq\emptyset)$ $\displaystyle=$
$\displaystyle\prod_{j=1}^{\ell_{n}}\left(1-e^{-n\sigma_{f}(S(p^{j};b\mathfrak{K}\delta_{n}))}\right)$
$\displaystyle\leq$
$\displaystyle\prod_{j=1}^{\ell_{n}}\left(1-e^{-nC_{f,D}(b\mathfrak{K}\delta_{n})^{d}}\right)$
$\displaystyle=$
$\displaystyle\left(1-n^{-C_{f,D}(b\mathfrak{K})^{d}}\right)^{\ell_{n}}.$

We assume that $n$ is large enough so that (4.48) holds and
$S(w;\delta_{n})\cap S(z;\varepsilon/2)=\emptyset$ whenever $w\notin
S(z;\varepsilon)$. Recalling that $bD\subset\cup_{j=1}^{m}S(p^{j};\delta_{n})$
and $S(p^{j};\delta_{n})\cap S(z;\varepsilon)=\emptyset$ for
$j=\ell_{n}+1,...,m$, we thus obtain that
$S(z;\varepsilon/2)\subset\cup_{j=1}^{\ell_{n}}S(p^{j};\delta_{n})$. Thus, by
(3.12), there is a $C_{D}>0$ such that
$\sigma(S(z;\varepsilon/2))\leq\ell_{n}\left(C_{D}\delta_{n}^{d}\right)$,
i.e.,

$\ell_{n}\geq C_{D}^{-1}\delta_{n}^{-d}\sigma(S(z;\varepsilon/2))\geq
C_{\varepsilon}\delta_{n}^{-d}=C_{\varepsilon}\left(\frac{n}{\log n}\right),$

where $C_{\varepsilon}:=C_{D}^{-1}\min\\{\sigma(S(z;\varepsilon/2)):{z\in
bD}\\}$. Thus, for sufficiently large $n=n_{\varepsilon}$, we have that
$\ell_{n}\geq n^{\beta}$ for some $\beta<1$. Now, choosing $b<1$ so that
$C_{f,D}b^{d}\mathfrak{K}^{d}<\beta$, we have that the right-hand side of
(4.48) is bounded above by
$(1-n^{-C_{f,D}b^{d}\mathfrak{K}^{d}})^{n^{\beta}}$, which converges to $0$ as
$n\rightarrow\infty$. This proves (4.47) and, therefore, (a) for $\eta_{n}$.

Proof of $(b)$. Let $\eta_{n}$ be a Poisson process on $bD$. Let
$\\{p^{1},...,p^{m}\\}$ as in the proof of $(a)$ above. Recall that $m$
depends on $n$, so we denote it by $m_{n}$ in this proof. We claim that it
suffices to prove that for any $\alpha>0$, there is a $b\geq 1$ so that

(4.49) $\mathbb{P}\left(\bigcup_{j=1}^{m_{n}}\\{\eta_{n}\cap
S(p^{j};b\delta_{n})=\emptyset\\}\right)=\mathcal{O}(n^{-\alpha})\ \text{as}\
n\rightarrow\infty.$

To see this, suppose for some $b\geq 1$, $\eta_{n}\cap
S(p^{j};b\delta_{n})\neq\emptyset$ for all $j=1,...,m_{n}$. Then, by the
quasi-symmetry and the quasi-triangle inequality satisfied by $\mathtt{d}$, we
obtain that for for some $C_{D}>0$, $S(p^{j};\delta_{n})\subset
S(\eta_{n};C_{D}\delta_{n})$ for all $j=1,...,m_{n}$. Now, if $\min_{z\in
bD}g(z)\geq C_{D}$, then for all $j$, $S(p^{j};\delta_{n})\subset
S(\eta_{n};\boldsymbol{\delta}_{n})$. Thus, from Lemma 3.15(a) and 3.14, we
have that $P(\eta_{n};\boldsymbol{\delta}_{n})\subset D$. Thus, (4.49) implies
$(b)$.

It now remains to prove (4.49). So, using the union bound and similar
arguments as in the proof of $(a)$ above, we have that

$\displaystyle\mathbb{P}\left(\bigcup_{j=1}^{m_{n}}\\{\eta_{n}\cap
S(p^{j};b\delta_{n})=\emptyset\\}\right)$ $\displaystyle\leq$
$\displaystyle\sum_{j=1}^{m_{n}}\mathbb{P}\left(\eta_{n}\cap
S(p^{j};b\delta_{n})=\emptyset\right)$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{m_{n}}e^{-n\sigma_{f}(S(p^{j};b\delta_{n}))}\leq
m_{n}\left(n^{-C_{f,D}b^{d}}\right).$

Now, since $S(p^{j};\mathfrak{K}\delta_{n})$’s are mutually disjoint, we have
from (3.12) a $C>0$ such that
$m_{n}C\mathfrak{K}^{d}\delta_{n}^{d}\leq\sum_{j=1}^{m_{n}}\sigma\left(S(p^{j};\mathfrak{K}\delta_{n})\right)\leq\sigma(bD)$.
Modifying $C$ as needed, we get $m_{n}\leq C\delta_{n}^{-d}=Cn/(\log n)\leq
Cn$. Thus,

$\mathbb{P}\left(\bigcup_{j=1}^{m_{n}}\\{\eta_{n}\cap
S(p^{j};b\delta_{n})=\emptyset\\}\right)\leq Cn^{1-C_{f,D}b^{d}}.$

Now, (4.49) follows by choosing $b$ suitably large, thus completing the proof
of $(b)$ for $\eta_{n}$.

Proof of $(c)$. The proofs of $(a)$ and $(b)$ for a binomial process
$\beta_{n}$ are the same as those for $\eta_{n}$ — except in the computation
of the probabilities of $\cap_{j=1}^{\ell_{n}}\\{\beta_{n}\cap
S(p^{j};c\delta_{n})\neq\emptyset\\}$ and $\cup_{j=1}^{m_{n}}\\{\beta_{n}\cap
S(p^{j};c\delta_{n})=\emptyset\\}$.

For the proof of $(a)$ for $\beta_{n}$, we couple $\beta_{n}$ with a suitable
Poisson process to bound $\mathbb{P}(\cap_{j=1}^{\ell_{n}}\\{\beta_{n}\cap
S(p^{j};c\delta_{n})\neq\emptyset\\})$. Let $N^{\prime}_{n}$ be a Poisson
random variable with mean $(1+a)n$ for some $a>0$. Let
$\eta^{\prime}_{n}:=\\{W^{1},\ldots,W^{N^{\prime}_{n}}\\}$ be a Poisson
process with intensity measure $(1+a)nf$. Recall that $W^{i}$’s are i.i.d.
with density $f$ and $\beta_{n}=\\{W^{1},\ldots,W^{n}\\}$. Observe that if
$N^{\prime}_{n}\geq n$, then $\beta_{n}\subset\eta^{\prime}_{n}$. Therefore,

$\displaystyle\mathbb{P}\left(\bigcap_{j=1}^{\ell_{n}}\\{\beta_{n}\cap
S(p^{j};c\delta_{n})\neq\emptyset\\}\right)$
$\displaystyle\leq\mathbb{P}\left(\bigcap_{j=1}^{\ell_{n}}\\{\beta_{n}\cap
S(p^{j};c\delta_{n})\neq\emptyset\\}\cap\\{N^{\prime}_{n}\geq
n\\}\right)+\mathbb{P}(N^{\prime}_{n}<n)$
$\displaystyle\leq\mathbb{P}\left(\bigcap_{j=1}^{\ell_{n}}\\{\eta^{\prime}_{n}\cap
S(p^{j};c\delta_{n})\neq\emptyset\\}\right)+\mathbb{P}(|N^{\prime}_{n}-(1+a)n|\geq
an).$

On the right-hand side above, the first term converges to $0$ as
$n\rightarrow\infty$ due to the bound in (4.48) and the subsequent argument.
The second term converges to $0$ as $n\rightarrow\infty$ by Chebyshev’s
inequality. Thus, we obtain the analogue of (4.47) for $\beta_{n}$. The rest
of the proof of $(a)$ for $\beta_{n}$ can be repeated verbatim.

To prove $(b)$ for $\beta_{n}$, we use the obvious inequality $1-x\leq
e^{-x}$, $x>0$, to obtain that

$\mathbb{P}(\beta_{n}\cap
S(p^{j};b\delta_{n})=\emptyset)=\left(1-\sigma_{f}(S(p^{j};b\delta_{n}))\right)^{n}\leq
e^{-n\sigma_{f}(S(p^{j};b\delta_{n}))}$

for all $j=1,...,m_{n}$. The rest of the proof remains the same as in the case
of $\eta_{n}$. This concludes the proof of $(c)$. ∎

##  Appendix A Background on Poisson and Binomial processes

We now collect some notions and results on Poisson and binomial processes that
are used in the article; see [23, 30] for more details.

Let $D\subset\mathbb{C}^{d}$ be as defined above, and
${\mathcal{B}}:=\mathcal{B}(bD)$ be the Borel $\sigma$-field on $bD$. Let
${\mathbf{N}}$ be the set of finite counting measures on $bD$, and
${\mathcal{N}}$ be the evaluation $\sigma$-field i.e., the $\sigma$-field of
subsets of ${\mathbf{N}}$ generated by the sets of the form
$\\{\varphi\in{\mathbf{N}}:\varphi(B)=k\\}$, with $B\in{\mathcal{B}}$,
$k\in{\mathbb{N}}\cup\\{0\\}$.

######  Definition A.1.

Let $(\Omega,{\mathcal{F}},{\mathbb{P}})$ be a probability space. A point
process $\eta$ on $bD$ is a random element of $({\mathbf{N}},{\mathcal{N}})$,
i.e. $\eta:\Omega\rightarrow{\mathbf{N}}$ is measurable. The intensity measure
of a point process $\eta$ is the measure $\lambda$ defined as
$\lambda(B)={\mathbb{E}}[\eta(B)],B\in{\mathcal{B}}$.

Let $\lambda$ be a finite measure. A point process $\eta$ is said to be a
Poisson process on $bD$ with intensity measure $\lambda$ if

  1. (a)

for all $B\in\mathcal{B}$, the random variable $\eta(B)$ has a Poisson
distribution with mean $\lambda(B)$;

  2. (b)

for all $m\geq 1$ and all pairwise disjoint subsets
$B_{1},B_{2},\ldots,B_{m}\in\mathcal{B}$, the random variables
$\eta(B_{1}),\ldots,\eta(B_{m})$ are independent (complete independence).

An elementary way to construct a Poissson process with a finite intensity
measure $\lambda$ on $bD$ is as follows. Let $W^{n},n\geq 1$, be i.i.d. random
variables with distribution $\frac{\lambda(\cdot)}{\lambda(bD)}$ and $N$ be an
independent Poisson random variable with mean $\lambda(bD)$. Then
$\eta=\\{W^{1},\ldots,W^{N}\\}$ is a Poisson process with intensity measure
$\lambda$. We also represent $\eta$ as a counting measure, i.e.,
$\eta=\sum_{i=1}^{N}\delta_{W^{i}}.$ We always work with this explicit
construction of the Poisson process for convenience.

We recall the so-called difference operators [23, Section 18.1]. Let $z\in
bD,\phi\in{\mathbf{N}}$ and $F:{\mathbf{N}}\to{\mathbb{R}}$ be a measurable
function. The first and second-order difference operators are defined as

$\displaystyle D_{z}F(\phi)$ $\displaystyle:=$ $\displaystyle
F(\phi\cup\\{z\\})-F(\phi),$ $\displaystyle D^{2}_{z,w}F(\phi)$
$\displaystyle:=$ $\displaystyle D_{z}(D_{w}F(\phi))=D_{w}(D_{z}F(\phi))$
$\displaystyle=$ $\displaystyle
F(\phi\cup\\{z,w\\})-F(\phi\cup\\{w\\})-F(\phi\cup\\{z\\})+F(\phi).$

Let $F:{\mathbf{N}}\to{\mathbb{R}}$ be a measurable and square-integrable
function with respect to $\eta$, i.e., $\mathbb{E}[F(\eta)^{2}]<\infty$. From
[23, Exercise 18.8 and Theorem 18.7], we have the following bounds on variance

(A.1)
$\int(\mathbb{E}[D_{z}F(\eta)])^{2}\text{d}\lambda(z)\leq\operatorname{{\mathbb{V}ar}}(F(\eta))\leq\int\mathbb{E}[(D_{z}F(\eta))^{2}]\text{d}\lambda(z).$

The upper bound is the well-known Poincaré inequality. We now state a normal
approximation result in the Wasserstein metric (2.10) for $F(\eta)$.

######  Theorem A.2.

(Second-order Poincaré inequality, [24, Theorem 1.1]) Let
$F:{\mathbf{N}}\to{\mathbb{R}}$ be a measurable function with
$\mathbb{E}[F(\eta)]=0$ and $\mathbb{E}[F(\eta)^{2}]<\infty$. Define

$\displaystyle T_{1}(F)$
$\displaystyle:=\left[\int\left(\mathbb{E}[(D^{2}_{z^{1},z^{2}}F)^{2}(D^{2}_{z^{2},z^{3}}F)^{2}]\right)^{1/2}\left(\mathbb{E}[(D_{z^{1}}F)^{2}(D_{z^{2}}F)^{2}]\right)^{1/2}\text{d}\lambda^{3}(z^{1},z^{2},z^{3})\right]^{1/2},$
$\displaystyle T_{2}(F)$
$\displaystyle:=\left[\int\mathbb{E}[(D^{2}_{z^{1},z^{2}}F)^{2}(D^{2}_{z^{1},z^{3}}F)^{2}]\text{d}\lambda^{3}(z^{1},z^{2},z^{3})\right]^{1/2},$
$\displaystyle T_{3}(F)$
$\displaystyle:=\int\mathbb{E}[|D_{z}F|^{3}]\text{d}\lambda(z),$

where $\eta$ has been omitted in the argument of $F$ for convenience. Then

(A.2)
$d_{W}\left(\frac{F(\eta)}{\operatorname{{\mathbb{V}ar}}[F(\eta)]^{1/2}},Z\right)\leq\frac{2T_{1}(F)+T_{2}(F)}{\operatorname{{\mathbb{V}ar}}[F(\eta)]}+\frac{T_{3}(F)}{\operatorname{{\mathbb{V}ar}}[F(\eta)]^{3/2}}.$

We now state similar results for the Binomial process $\beta_{n}$. Recall that
$\beta_{n}=\\{W^{1},\ldots,W^{n}\\}$ with $n$ i.i.d. points sampled according
to the distribution $f\sigma$ where $f$ is a probability density. Let
$F:{\mathbf{N}}\to{\mathbb{R}}$ be a measurable function. From the Efron–Stein
inequality ([21, Remark 2.5]), one can derive the following standard variance
upper bound for $F(\beta_{n})$ satisfying
$\mathbb{E}[F(\beta_{n})^{2}]<\infty$.

(A.3) $\operatorname{{\mathbb{V}ar}}(F(\beta_{n}))\leq
n\int\mathbb{E}[(D_{z}F(\beta_{n-1}))^{2}]f(z)\text{d}\sigma(z).$

We now state a normal approximation theorem for functionals of binomial
process, after introducing the necessary notation. The theorem is obtained by
combining the bounds in [9, Theorem 2.2] and [21, Theorem 5.1] (see also [21,
Remark 5.4]). Let ${\mathbf{W}}=(W^{1},\ldots,W^{n})$ and let
${\mathbf{W}}^{\prime},{\mathbf{W}}^{\prime\prime}$ be independent copies of
${\mathbf{W}}$. We say that ${\mathbf{Z}}=(Z^{1},\ldots,Z^{n})$ is a
recombination of ${\mathbf{W}}$ if
${Z^{i}}\in\\{W^{i},(W^{\prime})^{i},(W^{\prime\prime})^{i}\\}$ for each
$1\leq i\leq n$. We denote the set of possible recombinations of
${\mathbf{W}}$ as $Rec({\mathbf{W}})$. Let
${\mathbf{W}}^{(i)}=(\ldots,W^{i-1},W^{i+1},\ldots),1\leq i\leq n$, be the
vector obtained by dropping the $i$th co-ordinate from ${\mathbf{W}}$.
Similarly, let ${\mathbf{W}}^{(i,j)}=({\mathbf{W}}^{(i)})^{(j)},1\leq i\neq
j\leq n$, be the vector obtained by dropping the $i$th and $j$th co-ordinates
from ${\mathbf{W}}$. We abuse notation and set
$F(\mathbf{V})=F(\\{V^{1},\ldots,V^{m}\\})$ for any vector
$\mathbf{V}=(V^{1},...,V^{m})$. In other words, we view vectors as point sets
when applying $F$. For $1\leq i\neq j\leq n$, define

$\displaystyle D_{i}F({\mathbf{W}})$
$\displaystyle:=F({\mathbf{W}})-F({\mathbf{W}}^{(i)}),$ $\displaystyle
D_{i,j}F({\mathbf{W}})=D_{j,i}F({\mathbf{W}})$
$\displaystyle:=F({\mathbf{W}})-F({\mathbf{W}}^{(i)})-F({\mathbf{W}}^{(j)})+F({\mathbf{W}}^{(i,j)}).$

######  Theorem A.3.

([42, Lemma 2.3]) Let $F:{\mathbf{N}}\to{\mathbb{R}}$ be a measurable function
with $\mathbb{E}[F(\beta_{n})]=0$ and $\mathbb{E}[F(\beta_{n})^{2}]<\infty$.
Define

$\displaystyle T^{\prime}_{1}(F)$
$\displaystyle:=\sup\left\\{\mathbb{E}\left[\mathbf{1}[D_{1,2}F({\mathbf{Y}})\neq
0]D_{1}f({\mathbf{Z}})^{2}D_{2}f({\mathbf{Z}}^{\prime})^{2}\right]:{\mathbf{Y}},{\mathbf{Z}},{\mathbf{Z}}^{\prime}\in
Rec({\mathbf{W}})\right\\},$ $\displaystyle T^{\prime}_{2}(F)$
$\displaystyle:=\sup\left\\{\mathbb{E}\left[\mathbf{1}[D_{1,2}F({\mathbf{Y}})\neq
0,D_{1,3}F({\mathbf{Y}}^{\prime})\neq
0]D_{2}f({\mathbf{Z}})^{2}D_{2}f({\mathbf{Z}}^{\prime})^{2}\right]:{\mathbf{Y}},{\mathbf{Z}},{\mathbf{Z}}^{\prime}\in
Rec({\mathbf{W}})\right\\},$ $\displaystyle T^{\prime}_{3}(F)$
$\displaystyle:=\mathbb{E}[|D_{1}F({\mathbf{W}})|^{4}],$ $\displaystyle
T^{\prime}_{4}(F)$ $\displaystyle:=\mathbb{E}[|D_{1}F({\mathbf{W}})|^{3}].$

Then, there is a finite constant $C>0$ such that

(A.4)
$d_{W}\left(\frac{F(\beta_{n})}{\operatorname{{\mathbb{V}ar}}[F(\beta_{n})]^{1/2}},Z\right)\leq
C\frac{\sqrt{n^{2}T^{\prime}_{1}(F)}+\sqrt{n^{3/2}T^{\prime}_{2}(F)}+\sqrt{nT^{\prime}_{3}(F)}}{\operatorname{{\mathbb{V}ar}}(F(\beta_{n}))}+\frac{nT^{\prime}_{4}(F)}{\operatorname{{\mathbb{V}ar}}(F(\beta_{n}))^{3/2}}.$

While we have only used bounds in terms of the first and second-order
difference operators, we note that higher-order difference operators may be
defined recursively and are important ingredients in the stochastic analysis
of Poisson and binomial processes; see [23, 30, 21].

##  Appendix B The Möbius geometry of strongly ${\mathbb{C}}$-convex domains

In Section 3, we often use the unitary invariance of strong
${\mathbb{C}}$-convexity and induced polyhedra to simplify certain
computations. We give here the proofs of Lemmas 3.2 and 3.3, which are key to
such simplifications. In fact, we prove a slight generalization of Lemma 3.2
to demonstrate why the class of polyhedra considered in this paper are
invariant under the group of complex affine transformations, but not the full
group of linear fractional transformations. We also elaborate on the curvature
function $\nu_{D}$ that appears as a constant in the geometric estimates. Most
of these results are implicitly present in the literature, but we gather them
here for easy reference.

The transformation law in Part $(b)$ of Lemma 3.2 is a special case of a more
general transformation law, for which we need the following definition.

######  Definition B.1.

A linear fractional transformation (or Möbius transformation) on
$\mathbb{C}^{d}$ is a map of the form

(B.1) $\mathcal{M}_{\beta}:z\mapsto\left(\frac{b_{10}+b_{11}z_{1}+\cdots
b_{1d}z_{d}}{b_{00}+b_{01}z_{1}+\cdots+b_{0d}z_{d}},...,\frac{b_{d0}+b_{d1}z_{1}+\cdots+b_{dd}z_{d}}{b_{00}+b_{01}z_{1}+\cdots+b_{0d}z_{d}}\right),$

where $\beta=(b_{jk})_{0\leq j,k\leq
d}\in\operatorname{GL}(d+1;{\mathbb{C}})$. The function
$(z_{1},...,z_{d})\mapsto b_{00}+b_{01}z_{1}+\cdots+b_{0d}z_{d}$ is denoted by
$\beta_{0}$, and $\ell_{\beta}=\\{z\in\mathbb{C}^{d}:\beta_{0}(z)=0\\}$ is the
set of indeterminacy of $\mathcal{M}_{\beta}$.

Any affine transformation on $\mathbb{C}^{d}$ is a linear fractional
transformation. Lemma 3.2 is thus a corollary of the following result, which
is implicitly present in [22, Proposition 4.1] and [5, Theorem 3].

######  Lemma B.2.

Let $D$ and $\rho$ be as in Definition 3.1.

  * $(a)$

For any defining function $\widetilde{\rho}$ of $D$,
$L_{\widetilde{\rho}}(z,w)=L_{\rho}(z,w)$ for all
$(z,w)\in\mathbb{C}^{d}\times bD$.

  * $(b)$

Suppose $\beta\in\operatorname{GL}(d+1;{\mathbb{C}})$ such that
$\overline{D}\cap\ell_{\beta}=\emptyset$. Then, for
$\rho^{*}=\rho\circ\mathcal{M}_{\beta}^{-1}$,

(B.2)
$L_{\rho}(z,w)=\frac{||\partial\rho^{*}(\mathcal{M}_{\beta}(w))||}{||\partial\rho(w)||}\frac{\beta_{0}(z)}{\beta_{0}(w)}L_{\rho^{*}}(\mathcal{M}_{\beta}(z),\mathcal{M}_{\beta}(w)),\quad
z\in\mathbb{C}^{d}\setminus\ell_{\beta},w\in bD.$

###### Proof.

To prove $(a)$, note that there exists a positive $\mathcal{C}^{1}$ function
$h$ defined on some neighborhood $U$ of $bD$, such that,
$\widetilde{\rho}=h\rho$. Thus, since $\rho$ vanishes on $bD$,
$\partial\widetilde{\rho}(w)=h(w)\partial\rho(w)$ for any $w\in bD$, and the
claim follows.

For $(b)$, first observe that
$\mathcal{M}_{\beta}(w)=(\mathcal{M}^{1}_{\beta}(w),...,\mathcal{M}^{d}_{\beta}(w))=({\beta_{0}(w)})^{-1}\Big{(}{\boldsymbol{b}+w\cdot\widetilde{\beta}^{\operatorname{T}}}\Big{)}$,
and

$(D_{\mathbb{C}}\mathcal{M}_{\beta})(w):=\left(\dfrac{\partial{\mathcal{M}^{j}_{\beta}(w)}}{\partial{z_{k}}}\right)_{1\leq
j,k\leq
d}=\frac{1}{\beta_{0}(w)}\widetilde{\beta}-\frac{1}{\beta_{0}(w)^{2}}\big{(}\boldsymbol{b}^{\operatorname{T}}+\widetilde{\beta}\cdot
w^{\operatorname{T}}\big{)}\cdot(b_{01},...,b_{0d}),$

where $\boldsymbol{b}=(b_{10},...,b_{d0})$ and $\widetilde{\beta}$ is the
principal minor $(b_{jk})_{1\leq j,k\leq d}$. Now, setting
$w^{*}=\mathcal{M}_{\beta}(w)$ and denoting the product of two matrices $A$
and $B$ by $A\cdot B$, we get that

$\displaystyle||\partial\rho(w)||L_{\rho}(z,w)$ $\displaystyle=$
$\displaystyle{\partial(\rho^{*}\circ\mathcal{M}_{\beta})(w)\cdot(w-z)^{\operatorname{T}}}$
$\displaystyle=$
$\displaystyle{\partial\rho^{*}(\mathcal{M}_{\beta}(w))\cdot(D_{\mathbb{C}}\mathcal{M}_{\beta})(w)\cdot(w-z)^{\operatorname{T}}}$
$\displaystyle=$
$\displaystyle{\partial\rho^{*}(w^{*})}\cdot\left(\frac{1}{\beta_{0}(w)}\widetilde{\beta}-\frac{1}{\beta_{0}(w)^{2}}\big{(}\boldsymbol{b}^{\operatorname{T}}+\widetilde{\beta}\cdot
w^{\operatorname{T}}\big{)}\cdot(b_{01},...,b_{0d})\right)\cdot(w-z)^{\operatorname{T}}$
$\displaystyle=$
$\displaystyle{{\partial\rho^{*}(w^{*})}\cdot\left(\frac{1}{\beta_{0}(w)}\widetilde{\beta}\cdot(w^{\operatorname{T}}-z^{\operatorname{T}})-\frac{\beta_{0}(w)-\beta_{0}(z)}{\beta_{0}(w)^{2}}\big{(}\boldsymbol{b}^{\operatorname{T}}+\widetilde{\beta}\cdot
w^{\operatorname{T}}\big{)}\right)}$ $\displaystyle=$
$\displaystyle{\partial\rho^{*}(w^{*})}\cdot\left(-\frac{1}{\beta_{0}(w)}(\boldsymbol{b}^{\operatorname{T}}+\widetilde{\beta}\cdot
z^{\operatorname{T}})+\frac{\beta_{0}(z)}{\beta_{0}(w)^{2}}\big{(}\boldsymbol{b}^{\operatorname{T}}+\widetilde{\beta}\cdot
w^{\operatorname{T}}\big{)}\right)$ $\displaystyle=$
$\displaystyle{||\partial\rho^{*}(w^{*})||}\frac{\beta_{0}(z)}{\beta_{0}(w)}\left<\frac{\partial\rho^{*}(w^{*})}{||\partial\rho^{*}(w^{*})||},\mathcal{M}_{\beta}(w)-\mathcal{M}_{\beta}(z)\right>,$

which is precisely (B.2). ∎

Next, as an analogue of the Gaussian curvature for hypersurfaces in
$\mathbb{R}^{d}$, we consider the determinant of the (self-adjoint) linear
operator $\pi\circ S:\mathcal{H}_{w}M\rightarrow\mathcal{H}_{w}M$, where
$S:\mathcal{T}_{w}M\rightarrow\mathcal{T}_{w}M$ is the shape operator, and
$\pi:\mathcal{T}_{w}M\rightarrow\mathcal{H}_{w}M$ is the orthogonal projection
with respect to the standard Euclidean inner product on $\mathcal{T}_{w}M$
(viewed as a subspace of $\mathbb{R}^{2d}$). A more computationally convenient
definition of this second-order quantity is as follows.

######  Definition B.3.

Let $D\subset\mathbb{C}^{d}$ and $\rho$ be as above. For any $w\in bD$, we
define the complex-restricted curvature of $bD$ at $w$ as

$\nu_{D}(w)=(-1)^{d+1}\frac{\det\mathcal{Q}[\rho](w)}{||\nabla\rho(w)||^{2d+2}},$

where $\mathcal{Q}[\rho]$ is the $(2d+2)\times(2d+2)$-matrix function

$\mathcal{Q}[\rho]=\left(\begin{array}[]{c | c|c|c}0&0&\rho_{k}&0\\\ \hline\cr 0&0&0&\rho_{\,\overline{k}}\\\ \hline\cr\rho_{j}&0&\rho_{j,k}&\rho_{j,\overline{k}}\\\ \hline\cr 0&\rho_{\,\overline{j}}&\rho_{\,\overline{j},k}&\rho_{\,\overline{j},\overline{k}}\end{array}\right)_{1\leq j,k\leq d},$

where $\rho_{j}=\frac{\partial{\rho}}{\partial{z_{j}}}$,
$\rho_{j,\overline{k}}=\frac{\partial^{2}{\rho}}{\partial{z_{j}}\text{
}\partial{\overline{z}_{k}}}$, for $1\leq j,k\leq d$.

It is easy to check that $\nu_{D}$ is independent of the choice of defining
function, a positive function for strongly ${\mathbb{C}}$-convex domains, and
invariant under unitary transformations; see [16]. We now prove Lemma 3.3,
which employs a standard trick in the local study of smooth hypersurfaces; see
[44, Lemma 4.1], and the comments preceding Remark 8 in [2].

Proof of Lemma 3.3. Recall that $A\cdot B$ denotes the product of two matrices
$A$ and $B$. First, we choose an $\varepsilon_{D}>0$ so that
$B_{w}(\varepsilon_{D})\cap bD$ is a graph over the tangent space at $w$ for
all $w\in bD$. Now, after a unitary transformation, we may assume that $w=0$,
$\mathcal{T}_{0}bD=\\{z\in\mathbb{C}^{d}:\operatorname{Im}z_{d}=0\\}$ and
$\hat{\eta}(0)=(0,...,0,-i)$. Then, $B_{0}(\varepsilon_{D})\cap D=\\{z\in
B_{0}(\varepsilon_{D}):\rho(z)<0\\}$, where

$\rho(z)=-y_{d}+\sum_{j,k=1}^{d-1}a_{j,k}z_{j}\overline{z}_{k}+\operatorname{Re}\sum_{j,k=1}^{d-1}b_{j,k}z_{j}z_{k}+\operatorname{Im}\sum_{j=1}^{d-1}c_{j}z_{j}x_{d}+ex_{d}^{2}+O||(z^{\prime},x_{d})||^{3},$

for some $a_{j,k},b_{j,k},c_{j}\in{\mathbb{C}}$, $j,k=1,...,d-1$,
$e\in\mathbb{R}$, such that

  * •

$A=(a_{j,k})$ is Hermitian,

  * •

$B=(b_{j,k})$ is symmetric,

  * •

$|\operatorname{Re}(z^{\prime}\cdot B\cdot
z^{\prime\,{\operatorname{T}}})|<\overline{z}^{\prime}\cdot A\cdot
z^{\prime\,{\operatorname{T}}}$ for all nonzero
$z^{\prime}\in{\mathbb{C}}^{d-1}$.

Now, let $U$ be a unitary matrix in $\operatorname{GL}(d-1;{\mathbb{C}})$ such
that $U^{*}\cdot A\cdot U$ is diagonal, and the entries of the symmetric
matrix $U^{\operatorname{T}}\cdot B\cdot U$ are real. Then, after applying the
unitary transformation
$U_{1}=\left(\begin{smallmatrix}U^{-1}&\boldsymbol{0}_{d-1\times 1}\\\
0&1\end{smallmatrix}\right)$, we obtain that $B_{0}(\varepsilon_{D})\cap
U_{1}(\Omega)=\\{z\in B_{0}(\varepsilon_{D}):\rho_{1}(z)<0\\}$ for

$\rho_{1}(z)=-y_{d}+\sum_{j=1}^{d-1}\alpha_{j}|z_{j}|^{2}+\operatorname{Re}\sum_{j,k=1}^{d-1}\widetilde{b}_{j,k}z_{j}z_{k}+\operatorname{Im}\sum_{j=1}^{d-1}\widetilde{c}_{j}z_{j}u_{d}+eu_{d}^{2}+O||(z^{\prime},x_{d})||^{3},$

where $\alpha_{1},...,\alpha_{d-1}>0$ are the eigenvalues of $A$,
$\widetilde{B}=(\widetilde{b}_{j,k})$ is a real symmetric matrix, and
$\widetilde{c}=(\widetilde{c}_{1},...,\widetilde{c}_{d-1})=(c_{1},...,c_{d-1})\cdot~{}U$.
Finally, let $R$ be a real orthogonal matrix that diagonalizes
$\widetilde{B}$. Letting
$U_{w}=\left(\begin{smallmatrix}R^{-1}&\boldsymbol{0}_{d-1\times 1}\\\
0&1\end{smallmatrix}\right)\cdot U_{1}$, we get that
$B_{0}(\varepsilon_{D})\cap U_{w}(\Omega)=\\{\rho_{2}(z)<0\\}$ for

$\rho_{2}(z)=-y_{d}+\sum_{j,k=1}^{d-1}(\alpha_{j}|z_{j}|^{2}+\beta_{j}(\operatorname{Re}z_{j}^{2}))+\operatorname{Im}\sum_{j=1}^{d-1}\gamma_{j}z_{j}x_{d}+ex_{d}^{2}+O||(z^{\prime},x_{d})||^{3},$

where $\beta_{1},...,\beta_{d-1}$ are the eigenvalues of $\widetilde{B}$, and
$(\gamma_{1},...,\gamma_{d-1})=\widetilde{c}\cdot R$. Since unitary
transformations preserve strong ${\mathbb{C}}$-convexity, we have that
$0\leq\beta_{j}<\alpha_{j}$ for all $j=1,...,d-1$.

Now, by the unitary invariance of $\nu_{D}$,
$\nu_{D}(w)=\nu_{B_{0}(\varepsilon_{D})\cap U_{w}(D)}(0)$, and a direct
computation shows that the latter is equal to
$\frac{1}{16}\prod_{j=1}^{d-1}(\alpha_{j}^{2}-\beta_{j}^{2})$. ∎

##  Appendix C Area and volumes estimates for model domains

In the proof of Theorem 3.10, the first-order terms in the estimates come from
the model domains

$D_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}:=\left\\{(z^{\prime},x_{d}+iy_{d})\in\mathbb{C}^{d}:y_{d}>\sum_{j=1}^{d-1}\alpha_{j}|z_{j}|^{2}+\beta_{j}\operatorname{Re}(z_{j}^{2})\right\\},$

where ${\boldsymbol{\alpha}}=(\alpha_{1},...\alpha_{d-1})$ and
${\boldsymbol{\beta}}=(\beta_{1},...,\beta_{d-1})$ satisfy
$0\leq\beta_{j}<\alpha_{j}$, $j=1,...,d-1$. The computations for these domains
is carried out here. Recall that the dependence of cuts, tubular cuts, caps
and visibility regions on the underlying domain is indicated via a
superscript, as in $C^{D}(w;\delta)$, $S^{D}(w;\delta)$, etc., and the
projection $\pi_{2d-1}(C)$ of $C\subset\mathbb{C}^{d}$ onto
${\mathbb{C}}^{d-1}\times{\mathbb{R}}$ is denoted by $\widetilde{C}$.

######  Theorem C.1.

Let $D_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}$ be as above, and
$\delta>0$. Then, there exists a
$C_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}>0$ such that

(C.1)
$\displaystyle\lambda\big{(}C(0;\delta)\big{)}=\dfrac{h_{d+1}\,\kappa_{2d-2}}{d\,\sqrt{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}}\,\delta^{d+1},$
(C.2)
$\displaystyle\left|\sigma\big{(}S(0;\delta)\big{)}-\dfrac{h_{d}\kappa_{2d-2}}{\sqrt{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}}\,\delta^{d}\right|\leq
C_{\alpha,{\boldsymbol{\beta}}}\>\delta^{d+1},$

where
$v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}=\prod\limits_{j=1}^{d-1}(\alpha^{2}_{j}-\beta_{j}^{2})$,
and $\kappa_{d}$ and $h_{d}$ are as in Conjecture 1.1.

###### Proof.

First, we assume that ${\boldsymbol{\alpha}}=\boldsymbol{1}:=(1,...,1)$, and
denote $D_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}$ by
$D_{\boldsymbol{\beta}}$. We claim that

(C.3) $\displaystyle\lambda\left(C(0;\delta)\right)$ $\displaystyle=$
$\displaystyle\dfrac{h_{d+1}\,\kappa_{2d-2}}{d\,\sqrt{v_{\boldsymbol{1},{\boldsymbol{\beta}}}}}\,\delta^{d+1},$
(C.4) $\displaystyle\lambda_{2d-1}\left(\widetilde{C}(0;\delta)\right)$
$\displaystyle=$
$\displaystyle\dfrac{h_{d}\kappa_{2d-2}}{\sqrt{v_{\boldsymbol{1},{\boldsymbol{\beta}}}}}\,\delta^{d},$

for any $\delta>0$. To prove this, we set
$f_{\boldsymbol{\beta}}(z^{\prime},x_{d})=\sum_{j=1}^{d-1}(|z_{j}|^{2}+\beta_{j}\operatorname{Re}z_{j}^{2})$,
and observe that

$\displaystyle C(0;\delta)$ $\displaystyle=$
$\displaystyle\left\\{(z^{\prime},x_{d}+iy_{d})\in\mathbb{C}^{d}:(z^{\prime},x_{d})\in\widetilde{C}(0;\delta),\
f_{\boldsymbol{\beta}}(z^{\prime},x_{d})<y_{d}<\sqrt{\delta^{2}-x_{d}^{2}}\right\\},$
$\displaystyle\widetilde{C}(0;\delta)$ $\displaystyle=$
$\displaystyle\left\\{(z^{\prime},x_{d})\in{\mathbb{C}}^{d-1}\times\mathbb{R}:f_{\boldsymbol{\beta}}(z^{\prime},x_{d})^{2}+x_{d}^{2}<\delta^{2}\right\\}.$

Using spherical coordinates on each slice
$\left\\{\sum_{j=1}^{d-1}(|z_{j}|^{2}+\beta_{j}\operatorname{Re}z_{j}^{2})=r^{2}\right\\}$,
$r>0$, we have that

$\displaystyle\lambda\left(C(0;\delta)\right)$ $\displaystyle=$
$\displaystyle\int_{\widetilde{C}(0;\delta)}\sqrt{\delta^{2}-x_{d}^{2}}-\sum_{j=1}^{d-1}(|z_{j}|^{2}+\beta_{j}\operatorname{Re}z_{j}^{2})\,d\lambda_{2d-1}(z^{\prime},x_{d})$
$\displaystyle=$
$\displaystyle\frac{(2d-2)\kappa_{2d-2}}{\sqrt{1-{\boldsymbol{\beta}}^{2}}}\int\limits_{-\delta}^{\delta}\int\limits_{0}^{(\delta^{2}-x^{2}_{d})^{\frac{1}{4}}}\left(\sqrt{\delta^{2}-x_{d}^{2}}-r^{2}\right)r^{2d-3}\,dr\,dx_{d}=\dfrac{h_{d+1}\,\kappa_{2d-2}}{d\sqrt{v_{\boldsymbol{1},{\boldsymbol{\beta}}}}}\,\delta^{d+1},$
$\displaystyle\lambda_{2d-1}\left(\widetilde{C}(0;\delta)\right)$
$\displaystyle=$
$\displaystyle\int_{\widetilde{C}(0;\delta)}d\lambda_{2d-1}(z^{\prime},x_{d})=\frac{(2d-2)\kappa_{2d-2}}{\sqrt{1-{\boldsymbol{\beta}}^{2}}}\int\limits_{-\delta}^{\delta}\int\limits_{0}^{(\delta^{2}-x^{2}_{d})^{\frac{1}{4}}}r^{2d-3}\,dr\,dx_{d}=\frac{h_{d}\kappa_{2d-2}}{\sqrt{v_{\boldsymbol{1},{\boldsymbol{\beta}}}}}\delta^{d}.$

Thus, we have (C.1) in the special case of $D_{\boldsymbol{\beta}}$.

To prove (C.2) for $D_{\boldsymbol{\beta}}$, we observe that
$\widetilde{S}(0;\delta)=\widetilde{C}(0;\delta)$, and for
$(z^{\prime},x_{d})\in\widetilde{C}(0;\delta)$,

(C.5) $||\nabla
f_{\boldsymbol{\beta}}(z^{\prime},x_{d})||^{2}=4\sum_{j=1}^{d-1}|z_{j}+\beta_{j}\overline{z}_{j}|^{2}\leq
2\sum_{j=1}^{d-1}(|z_{j}|^{2}+\beta_{j}\operatorname{Re}\overline{z}_{j}^{2})<2\delta.$

Thus, since $S(0;\delta)$ is the graph of $f_{\boldsymbol{\beta}}$ over
$\widetilde{C}(0;\delta)$, we have that

$\displaystyle\sigma\big{(}S(0;\delta)\big{)}=\int_{\widetilde{C}(0;\delta)}\left(1+||\nabla
f_{\boldsymbol{\beta}}(z^{\prime},x_{d})||^{2}\right)^{\frac{1}{2}}\,d\lambda_{2d-1}(z^{\prime},x_{d}),$

and, therefore,

$\left|\sigma\left(S(0;\delta)\right)-\lambda_{2d-1}\left(\widetilde{C}(0;\delta)\right)\right|\leq\delta\lambda_{2d-1}\left(\widetilde{C}(0;\delta)\right)$

for all $\delta>0$. Equation (C.2) now follows from (C.4) in the special case
of $D_{\boldsymbol{\beta}}$.

For the general case, consider the map
$A:(z_{1},...,z_{d})\mapsto(\sqrt{\alpha_{1}}z_{1},...,\sqrt{\alpha_{d-1}}z_{d-1},z_{d})$,
which biholomorphically maps $D_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}$
onto $D_{{\boldsymbol{\beta}}/{\boldsymbol{\alpha}}}$, where
${\boldsymbol{\beta}}/{\boldsymbol{\alpha}}=(\beta_{1}/\alpha_{1},...,\beta_{d-1}/\alpha_{d-1})$.
Now, invoking the transformation law for cuts and caps (see Lemma 3.2), and
noting that $A$ and $\left(A|_{{\mathbb{C}}^{d-1}\times\mathbb{R}}\right)$ are
diagonal transformations, we obtain that for all $\delta>0$,

$\displaystyle\lambda\big{(}C^{D_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}(0;\delta)\big{)}=\left(\prod_{j=1}^{d-1}\alpha_{j}^{-1}\right)\lambda\big{(}C^{D_{{\boldsymbol{\beta}}/{\boldsymbol{\alpha}}}}(0;\delta)\big{)},$
$\displaystyle\left|\sigma(S^{D_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}(0;\delta))-\left(\prod_{j=1}^{d-1}\alpha_{j}^{-1}\right)\lambda_{2d-1}\big{(}S^{D_{{\boldsymbol{\beta}}/{\boldsymbol{\alpha}}}}(0;\delta)\big{)}\right|\leq\delta\left(\prod_{j=1}^{d-1}\alpha_{j}^{-1}\right)\lambda_{2d-1}\big{(}S^{D_{{\boldsymbol{\beta}}/{\boldsymbol{\alpha}}}}(0;\delta)\big{)}.$

∎

######  Theorem C.2.

Let $D_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}$ be as above, $\delta>0$,
and $t\in[0,1)$.Then, there is a
$C_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}>0$ such that for
$\delta<\frac{1}{2\max_{j}\alpha_{j}}$,

$\displaystyle\frac{\kappa_{2d-2}}{\sqrt{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}}h_{d}(t)\delta^{d}\leq\sigma(G_{t\delta}(0;\delta))\leq\frac{\kappa_{2d-2}}{\sqrt{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}}h_{d}(t)\delta^{d}+C_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}\delta^{d+1},$

where
$h_{d}(t)=2\int_{0}^{\sin^{-1}\sqrt{1-t^{2}}}(\cos\theta-t)^{d-1}\cos\theta\,d\theta$.

###### Proof.

Since $G_{t\delta}(0;\delta)\subset bD$, it is the graph of
$f_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}$ over
$\widetilde{G}_{t\delta}(0;\delta)$, which is given by

$\displaystyle\left\\{(z^{\prime},x_{d})\in{\mathbb{C}}^{d-1}\times\mathbb{R}:(t\delta+q_{1}(z^{\prime}))^{2}+\left(x_{d}+q_{2}(z^{\prime})\right)^{2}<\delta^{2}(1+4q_{3}(z^{\prime}))\right\\},$

where
$q_{1}(z^{\prime})~{}=~{}\sum_{j=1}^{d-1}\left(\alpha_{j}|z_{j}|^{2}+\beta_{j}\operatorname{Re}\overline{z}_{j}^{2}\right)$,
$q_{2}(z^{\prime})=2\sum_{j=1}^{d-1}\beta_{j}\operatorname{Im}z_{j}^{2}$, and
$q_{3}(z^{\prime})=\sum_{j=1}^{d-1}|\alpha_{j}z_{j}+\beta_{j}\overline{z}_{j}|^{2}$.
Note that $q_{1},q_{2},q_{3}\geq 0$.

Now, setting $A=\max_{j}\alpha_{j}$ and recalling that $\beta_{j}<\alpha_{j}$,
$j=1,...,d-1$, we have that $q_{3}(w^{\prime})\leq 2Aq_{1}(w^{\prime})$. Thus,
if $z\in G_{t\delta}(0;\delta)$,

$q_{3}(z^{\prime})^{2}\leq 4A^{2}\delta^{2}(1+4q_{3}(z^{\prime})),$

i.e., $q_{3}(z^{\prime})\leq 2A\delta+16A^{2}\delta^{2}$. Thus, for
$\delta<1/(2A)$,
$K_{t\delta,\delta}\subseteq\widetilde{G}_{t\delta}(0;\delta)\subseteq
K_{t\delta,\delta(1+8A\delta)}$, where,

$K_{\varepsilon,\eta}=\left\\{(z^{\prime},x_{d})\in{\mathbb{C}}^{d-1}\times\mathbb{R}:(\varepsilon+q_{1}(z^{\prime}))^{2}+\left(x_{d}+q_{2}(z^{\prime})\right)^{2}<\eta^{2}\right\\},\quad
0<\varepsilon<\eta.$

Consequently, for $\delta<1/(2A)$,

(C.6)
$\int\limits_{K_{t\delta,\delta}}d\lambda_{2d-1}(z^{\prime},x_{d})\leq\int\limits_{\widetilde{G}_{t\delta}(0;\delta)}\sqrt{1+4q_{3}(w^{\prime})}\,d\lambda_{2d-1}(z^{\prime},x_{d})\leq\int\limits_{K_{t\delta,\delta+8A\delta^{2}}}(1+8A\delta)d\lambda_{2d-1}(z^{\prime},x_{d}),$

where the term in the center is $\sigma\big{(}G_{t\delta}(0;\delta)\big{)}$.
It now suffices to compute $\lambda_{2d-1}(K_{\varepsilon,\eta}\big{)}$. Now,
setting
$\xi_{j}=\sqrt{\alpha_{j}+\beta_{j}}\,\operatorname{Re}z_{j}+i\sqrt{\alpha_{j}-\beta_{j}}\,\operatorname{Im}z_{j}$,
$1\leq j\leq d-1$, and
$\tau=x_{d}+2\sum_{j=1}^{d-1}\beta_{j}\operatorname{Im}w_{j}^{2}$, we obtain
that

$\displaystyle\lambda_{2d-1}(K_{\varepsilon,\eta})$ $\displaystyle=$
$\displaystyle\frac{1}{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}\lambda_{2d-1}\left\\{(\xi,\tau)\in{\mathbb{C}}^{d-1}\times\mathbb{R}:(\varepsilon+||\xi||^{2})^{2}+\tau^{2}<\eta^{2}\right\\}$
$\displaystyle=$
$\displaystyle\frac{2\kappa_{2d-2}}{\sqrt{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}}\int\limits_{0}^{\sqrt{\eta^{2}-\varepsilon^{2}}}(\sqrt{\eta^{2}-\tau^{2}}-\varepsilon)^{d-1}\,d\tau=\frac{\kappa_{2d-2}}{\sqrt{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}}\,\eta^{d}h_{d}\left(\frac{\varepsilon}{\eta}\right),$

where
$h_{d}(t)=2\int_{0}^{\sin^{-1}\sqrt{1-t^{2}}}(\cos\theta-t)^{d-1}\cos\theta\,d\theta$.

A few comments about $h_{d}$ are in order. Note that
$h_{1}(t)=2\sqrt{1-t^{2}}$, $t\in[0,1]$. When $d\geq 2$, $h_{d}$ is a
decreasing Lipschitz function on $[0,1]$ with $h_{d}(1)=0$. This follows from
observing that $h^{\prime}_{d}=-(d-1)h_{d-1}(t)$, which is negative and
bounded (in absolute value) on $(0,1)$. Thus, for any $t\in(0,1)$, there is an
$r\in(t(1+8A\delta)^{-1},t)$ such that

$h_{d}\left(\frac{t\delta}{\delta+8A\delta^{2}}\right)=h_{d}(t)-h_{d}^{\prime}(r)\frac{8At\delta}{1+8A\delta}\leq
h_{d}(t)+(d-1)h_{d-1}(0)\frac{8At\delta}{1+8A\delta}.$

Combining this with (C.6) gives that

$\displaystyle\frac{\kappa_{2d-2}}{\sqrt{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}}h_{d}(t)\delta^{d}\leq\sigma\big{(}G_{t\delta}(0;\delta)\big{)}$
$\displaystyle\leq$
$\displaystyle\frac{\kappa_{2d-2}}{\sqrt{v_{{\boldsymbol{\alpha}},{\boldsymbol{\beta}}}}}\delta^{d}(1+8A\delta)^{d}\left(h_{d}(t)+(d-1)h_{d-1}(0)\frac{8At\delta}{1+8A\delta}\right),$

which yields our claim. ∎

## Acknowledgements

SA’s research was partially supported by CPDA from the Indian Statistical
Institute and Infosys Chair Professor position at the Chennai Mathematical
Institute, Chennai. PG was partially supported by the Infosys Young
Investigator Award. DY’s research was partially supported by SERB-MATRICS
Grant MTR/2020/000470 and CPDA from the Indian Statistical Institute. DY is
thankful to Matthias Reitzner for comments on the literature of random
polytopes and to Srikanth Iyer for discussions related to Conjecture 1.1. DY
would also like to thank Joe Yukich for very helpful comments on a first draft
of the article.

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