Uniform}({\rm O}(p-1))$. Therefore, the validity of the second generating
procedure is shown.∎

##  Appendix B Proofs

###  B.1 Proof of 2.1

We first show the uniqueness of $R_{n}(\cdot)$. If the uniqueness fails, then
necessarily, there exist two different (non-random) permutations $\sigma_{1}$
and $\sigma_{2}$ such that
$\sum_{i=1}^{n}c(\mathbf{X}_{i},\mathbf{h}_{\sigma_{1}(i)})=\sum_{i=1}^{n}c(\mathbf{X}_{i},\mathbf{h}_{\sigma_{2}(i)})$.
Without loss of generality, suppose $\sigma_{1}(1)\neq\sigma_{2}(1)$. Then

$c(\mathbf{X}_{1},\mathbf{h}_{\sigma_{1}(1)})-c(\mathbf{X}_{1},\mathbf{h}_{\sigma_{2}(1)})=\sum_{i=2}^{n}\left(c(\mathbf{X}_{i},\mathbf{h}_{\sigma_{2}(i)})-c(\mathbf{X}_{i},\mathbf{h}_{\sigma_{1}(i)})\right).$
(B.1)

Define the distance from a point $\mathbf{x}\in\mathbb{R}^{p}$ to a set
$A\subset\mathbb{R}^{p}$ as $d(\mathbf{x},A):=\inf_{\mathbf{y}\in
A}\|\mathbf{x}-\mathbf{y}\|$. Note that
$c(\mathbf{X}_{1},\mathbf{h}_{\sigma_{1}(1)})-c(\mathbf{X}_{1},\mathbf{h}_{\sigma_{2}(1)})=d(\mathbf{X}_{1},\mathcal{G}\mathbf{h}_{\sigma_{1}(1)})^{2}-d(\mathbf{X}_{1},\mathcal{G}\mathbf{h}_{\sigma_{2}(1)})^{2}$.
The following lemma shows that as long as the orbits
$\mathcal{G}\mathbf{h}_{\sigma_{1}(1)}\neq\mathcal{G}\mathbf{h}_{\sigma_{2}(1)}$
are different,

$\\{\mathbf{x}\in\mathbb{R}^{p}:d(\mathbf{x},\mathcal{G}\mathbf{h}_{\sigma_{1}(1)})^{2}-d(\mathbf{x},\mathcal{G}\mathbf{h}_{\sigma_{2}(1)})^{2}=C\\}$

has Lebesgue measure 0 for any constant $C$. This implies (B.1) holds with
probability 0 by considering the conditional distribution of $\mathbf{X}_{1}$
given $\mathbf{X}_{2},\ldots,\mathbf{X}_{n}$. Hence, $\widehat{\sigma}$ and
thus $R_{n}(\cdot)$ are unique.

######  Lemma B.1.

Let $A,B\subset\mathbb{R}^{p}$ be two disjoint compact sets with Lebesgue
measure 0, and $C$ be a constant. Then

$D:=\\{\mathbf{x}\in\mathbb{R}^{p}:d(\mathbf{x},A)^{2}-d(\mathbf{x},B)^{2}=C\\}$

has Lebesgue measure 0.

###### Proof.

It is known that the distance function $f(\mathbf{x}):=d(\mathbf{x},A)$ is
1-Lipschitz and is thus Lebesgue a.e. differentiable (by Rademacher’s
theorem). If $\mathbf{x}_{0}\notin A$ is a differentiable point of $f$ then
its projection onto $A$ must be unique111111To make this uniqueness argument
precise, suppose that $\mathbf{y}_{0}$ (a priori not unique) attains the
minimal distance from $\mathbf{x}_{0}$ to $A$, i.e.,
$f(\mathbf{x}_{0})=\|\mathbf{x}_{0}-\mathbf{y}_{0}\|$. Then $\mathbf{y}_{0}$
also attains the minimal distance from $A$ to any point in the line segment
$[\mathbf{x}_{0},\mathbf{y}_{0}]$. This is because, for any
$\mathbf{y}^{\prime}\in A$, $t\in[0,1]$, and
$\mathbf{z}:=\mathbf{x}_{0}+t(\mathbf{y}_{0}-\mathbf{x}_{0})$, we have by
triangle inequality,
$\|\mathbf{z}-\mathbf{y}^{\prime}\|\geq\|\mathbf{x}_{0}-\mathbf{y}^{\prime}\|-\|\mathbf{x}_{0}-\mathbf{z}\|\geq
f(\mathbf{x}_{0})-\|t(\mathbf{y}_{0}-\mathbf{x}_{0})\|=(1-t)\|\mathbf{y}_{0}-\mathbf{x}_{0}\|=\|\mathbf{z}-\mathbf{y}_{0}\|.$
Therefore, $f(\mathbf{z})=\|\mathbf{z}-\mathbf{y}_{0}\|$; further
$f(\mathbf{z})=f(\mathbf{x}_{0})-(\mathbf{z}-\mathbf{x}_{0})^{\top}\frac{\mathbf{y}_{0}-\mathbf{x}_{0}}{\|\mathbf{y}_{0}-\mathbf{x}_{0}\|}$.
As $f$ is differentiable at $\mathbf{x}_{0}$, for $\mathbf{z}$ close to
$\mathbf{x}_{0}$, we have
$f(\mathbf{z})=f(\mathbf{x}_{0})+(\mathbf{z}-\mathbf{x}_{0})^{\top}\nabla
f(\mathbf{x}_{0})+o(\|\mathbf{z}-\mathbf{x}_{0}\|)$, which further yields
$\nabla
f(\mathbf{x}_{0})=-\frac{\mathbf{y}_{0}-\mathbf{x}_{0}}{\|\mathbf{y}_{0}-\mathbf{x}_{0}\|}$.
Hence, $\mathbf{y}_{0}$ is determined from $\mathbf{x}_{0}$ and the gradient
$\nabla f(\mathbf{x}_{0})$ as
$\mathbf{y}_{0}=\mathbf{x}_{0}-f(\mathbf{x}_{0})\nabla f(\mathbf{x}_{0})$, and
hence $\mathbf{y}_{0}$ unique., otherwise there will be two directions where
$f$ descends the steepest (with rate 1), which is a contradiction to the
differentiability of $f$. Moreover, $\nabla f(\mathbf{x}_{0})$ is a unit
vector on the line that connects $\mathbf{x}_{0}$ and its projection on $A$.

Now, take $\mathbf{x}_{0}\in D$ to be a point that is not in $A$ or $B$, and
is a differentiable point of both $d(\mathbf{x},A)$ and $d(\mathbf{x},B)$.
Then, for $\mathbf{x}$ in a neighborhood of $\mathbf{x}_{0}$, using the
definition of differentiability, $d(\mathbf{x},A)^{2}-d(\mathbf{x},B)^{2}=C$
can be written as:

$2\left[d(\mathbf{x}_{0},A)\nabla
d(\mathbf{x}_{0},A)-d(\mathbf{x}_{0},B)\nabla
d(\mathbf{x}_{0},B)\right]^{\top}(\mathbf{x}-\mathbf{x}_{0})+o(\|\mathbf{x}-\mathbf{x}_{0}\|)=0.$
(B.2)

Let $\mathbf{f}:=2\left[d(\mathbf{x}_{0},A)\nabla
d(\mathbf{x}_{0},A)-d(\mathbf{x}_{0},B)\nabla
d(\mathbf{x}_{0},B)\right]\in\mathbb{R}^{p}$. Then $\mathbf{f}$ is nonzero,
otherwise the projections of $\mathbf{x}_{0}$ onto $A$ and $B$ are the same,
which is a contradiction to $A$, $B$ being disjoint. Hence, (B.2) implies the
set $D$ is locally contained in a hyperplane with arbitrarily small thickness,
i.e., for any $\delta>0$, there exists $\varepsilon_{0}>0$ such that for all
$\varepsilon<\varepsilon_{0}$,

$D\cap B(\mathbf{x}_{0},\varepsilon)\subset\\{\mathbf{x}\in
B(\mathbf{x}_{0},\varepsilon):|\mathbf{f}^{\top}(\mathbf{x}-\mathbf{x}_{0})|\leq\delta\|\mathbf{x}-\mathbf{x}_{0}\|\\}.$

Here
$B(\mathbf{x}_{0},\varepsilon):=\\{\mathbf{x}\in\mathbb{R}^{p}:\|\mathbf{x}-\mathbf{x}_{0}\|<\varepsilon\\}$
is the open ball centered at $\mathbf{x}_{0}$ with radius $\varepsilon>0$.
Hence, with $m$ denoting the Lebesgue measure,

$\displaystyle\limsup_{\varepsilon\to 0}\frac{m(D\cap
B(\mathbf{x}_{0},\varepsilon))}{m(B(\mathbf{x}_{0},\varepsilon))}\leq\limsup_{\varepsilon\to
0}\frac{m(\\{\mathbf{x}\in
B(\mathbf{x}_{0},\varepsilon):|\mathbf{f}^{\top}(\mathbf{x}-\mathbf{x}_{0})|\leq\delta\|\mathbf{x}-\mathbf{x}_{0}\|\\})}{m(B(\mathbf{x}_{0},\varepsilon))}.$

By applying a translation and a rescaling, we observe that

$\frac{m(\\{\mathbf{x}\in
B(\mathbf{x}_{0},\varepsilon):|\mathbf{f}^{\top}(\mathbf{x}-\mathbf{x}_{0})|\leq\delta\|\mathbf{x}-\mathbf{x}_{0}\|\\})}{m(B(\mathbf{x}_{0},\varepsilon))}=\frac{m(\\{\mathbf{x}\in
B(\mathbf{0},1):|\mathbf{f}^{\top}\mathbf{x}|\leq\delta\|\mathbf{x}\|\\})}{m(B(\mathbf{0},1))}$

for all $\mathbf{x}_{0}\in\mathbb{R}^{p}$ and $\varepsilon>0$. Therefore,

$\limsup_{\varepsilon\to 0}\frac{m(D\cap
B(\mathbf{x}_{0},\varepsilon))}{m(B(\mathbf{x}_{0},\varepsilon))}\leq\frac{m(\\{\mathbf{x}\in
B(\mathbf{0},1):|\mathbf{f}^{\top}\mathbf{x}|\leq\delta\|\mathbf{x}\|\\})}{m(B(\mathbf{0},1))}.$

Since $\delta$ is arbitrary (and the set on the right side of the above
display converges to a subset of a hyperplane, as $\delta$ approaches 0), we
further have that

$\lim_{\varepsilon\to 0}\frac{m(D\cap
B(\mathbf{x}_{0},\varepsilon))}{m(B(\mathbf{x}_{0},\varepsilon))}=0.$

But from the Lebesgue differentiation theorem, for Lebesgue
a.e.-$\mathbf{x}_{0}$ in $D$,

$\lim_{\varepsilon\to 0}\frac{m(D\cap
B(\mathbf{x}_{0},\varepsilon))}{m(B(\mathbf{x}_{0},\varepsilon))}=1_{D}(\mathbf{x}_{0})=1.$

Comparing the last two displays we see that $D$ must have Lebesgue measure 0.
∎

Next, we show the uniqueness of $S_{n}(\cdot)$. Since $R_{n}(\mathbf{X}_{1})$
must be one of $\mathbf{h}_{1},\ldots,\mathbf{h}_{n}$, if there are two
elements $Q_{1}\neq Q_{2}\in\mathcal{G}$ that belong to
$\operatorname*{\arg\\!\min}_{Q\in\mathcal{G}}\|Q^{\top}\mathbf{X}_{1}-R_{n}(\mathbf{X}_{1})\|^{2}$,
then necessarily, there exists
$\mathbf{h}\in\\{\mathbf{h}_{1},\ldots,\mathbf{h}_{n}\\}$ such that

$\|Q_{1}^{\top}\mathbf{X}_{1}-\mathbf{h}\|^{2}=\|Q_{2}^{\top}\mathbf{X}_{1}-\mathbf{h}\|^{2},\
{\rm or\ equivalently}\
\langle\mathbf{X}_{1},Q_{1}\mathbf{h}-Q_{2}\mathbf{h}\rangle=0.$

The free group action implies $Q_{1}\mathbf{h}-Q_{2}\mathbf{h}\neq\mathbf{0}$.
Hence the above event happens with probability 0 (as $\mathbf{X}_{1}$ is
absolutely continuous).

Finally, we show below the uniqueness of the signed-rank. We will use the
following lemma.

######  Lemma B.2.

Let $C\subset\mathbb{R}^{p}$ be a compact set. Then for Lebesgue almost every
point, it has a unique nearest point in $C$.

###### Proof.

It is known that the distance to $C$, i.e., $f(\mathbf{x}):=d(\mathbf{x},C)$
is 1-Lipschitz and is thus Lebesgue a.e. differentiable (by Rademacher’s
theorem). For any differentiable point $\mathbf{x}_{0}$ of $f$ that is not in
$C$, its projection onto $C$ must be unique (see the proof of Lemma B.1). When
$\mathbf{x}_{0}\in C$, the projection is $\mathbf{x}_{0}$ itself, and the
uniqueness is trivial. ∎

Since
$S_{n}(\mathbf{X}_{i})=\operatorname*{\arg\\!\min}_{Q\in\mathcal{G}}\|\mathbf{X}_{i}-QR_{n}(\mathbf{X}_{i})\|^{2}$,
$S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i})$ is the point in the orbit of
$R_{n}(\mathbf{X}_{i})$ that is closest to $\mathbf{X}_{i}$. But
$R_{n}(\mathbf{X}_{i})$ must be one of $\mathbf{h}_{1},\ldots,\mathbf{h}_{n}$.
If $S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i})$ is not unique, then there
exists $\mathbf{h}\in\\{\mathbf{h}_{1},\ldots,\mathbf{h}_{n}\\}$ such that the
projection of $\mathbf{X}_{i}$ to $\mathcal{G}\mathbf{h}$ is not unique.
However, by Lemma B.2, for Lebesgue a.e. $\mathbf{x}\in\mathbb{R}^{p}$,
$\mathbf{x}$ has a unique closest point in $\mathcal{G}\mathbf{h}$. Since
$\mathbf{X}_{i}$ is absolutely continuous,
$S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i})$ is almost surely unique. ∎

###  B.2 Proof of 2.2

Since $\mathbf{X}_{1},\ldots,\mathbf{X}_{n}$ are i.i.d., the
$\widehat{\sigma}$ that minimizes the objective function

$\sum_{i=1}^{n}c(\mathbf{X}_{\sigma(i)},\mathbf{h}_{i}),\quad{\rm where}\
c(\mathbf{x},\mathbf{h})=\min_{Q\in\mathcal{G}}\|Q^{\top}\mathbf{x}-\mathbf{h}\|^{2},$

is equally likely to take any of the $n!$ permutations [27]. Hence, the first
conclusion follows.

To show the other two properties, suppose ${\rm H}_{0}$ holds. We consider the
$\sigma$-field $\mathcal{F}$ generated by the $n$ orbits
$\mathcal{G}\mathbf{X}_{1},\ldots,\mathcal{G}\mathbf{X}_{n}$, and the possible
randomness introduced to select $\widehat{\sigma}$ (if $\widehat{\sigma}$ is
not unique). Then given $\mathcal{F}$, $\widehat{\sigma}$ is determined, i.e.,
$\widehat{\sigma}$ is measurable w.r.t. $\mathcal{F}$.

Given $\mathcal{F}$, $\widehat{Q}_{i}$ (see (2.2)) is defined as

$\widehat{Q}_{i}=\operatorname*{\arg\\!\min}_{S_{i}\in\mathcal{G}}\ \lVert
S_{i}^{\top}\mathbf{X}_{\widehat{\sigma}(i)}-\mathbf{h}_{i}\rVert^{2}.$ (B.3)

From Lemma B.3 (see below), when $\mathbf{X}$ has a $\mathcal{G}$-symmetric
distribution, the density $f$ of $\mathbf{X}$ satisfies
$f(\mathbf{x})=f(\mathbf{y})$ whenever $\mathbf{x}$ and $\mathbf{y}$ are on
the orbit of $\mathcal{G}$. Note that given $\mathcal{F}$, only the orbit of
$\mathbf{X}_{\widehat{\sigma}(i)}$ is known, and
$\mathbf{X}_{\widehat{\sigma}(i)}$ is equally likely to be any point in this
orbit due to $\mathcal{G}$-symmetry, i.e., $\mathbf{X}_{\widehat{\sigma}(i)}$
conditionally (on $\mathcal{F}$) follows the uniform distribution over the
$\mathcal{G}$-orbit $\mathcal{G}\mathbf{X}_{\widehat{\sigma}(i)}$. For any
$Q\in\mathcal{G}$, define $\widetilde{Q}_{i}$ as

$\widetilde{Q}_{i}:=\operatorname*{\arg\\!\min}_{S_{i}\in\mathcal{G}}\ \lVert
S_{i}^{\top}Q^{\top}\mathbf{X}_{\widehat{\sigma}(i)}-\mathbf{h}_{i}\rVert^{2}.$
(B.4)

If the above does not have a unique minimizer, choose $\widetilde{Q}_{i}$ from
the distribution ${\rm Uniform}(\mathcal{G})$ restricted to the set of
minimizers of (B.4). Since $\mathbf{X}_{\widehat{\sigma}(i)}$ and
$Q^{\top}\mathbf{X}_{\widehat{\sigma}(i)}$ have the same conditional
distribution given $\mathcal{F}$, we know that $\widetilde{Q}_{i}$ has the
same conditional distribution as $\widehat{Q}_{i}$. On the other hand, if we
view $S_{i}^{\top}Q^{\top}$ in (B.4) as a whole, then (B.4) reduces to the
minimization problem in (B.3), and thus $Q\widetilde{Q}_{i}$ has the same
conditional distribution as $\widehat{Q}_{i}$. Hence,
$Q\widehat{Q}_{i}\overset{d}{=}\widehat{Q}_{i}$ given $\mathcal{F}$. This
implies given $\mathcal{F}$, $\widehat{Q}_{i}$ are i.i.d. following the
uniform distribution over $\mathcal{G}$. Consequently,
$\\{\widehat{Q}_{i}\\}_{i=1}^{n}$ is independent of $\mathcal{F}$ and thus
independent of $\widehat{\sigma}$. Hence, the second and the third conclusions
hold. ∎

###  B.3 Proof of Theorem 3.1

By the independence of $(S_{n}(\mathbf{X}_{1}),\ldots,S_{n}(\mathbf{X}_{n}))$
and $(R_{n}(\mathbf{X}_{1}),\ldots,R_{n}(\mathbf{X}_{n}))$, and the fact that
$(R_{n}(\mathbf{X}_{1}),\ldots,R_{n}(\mathbf{X}_{n}))$ is a permuted version
of $(\mathbf{h}_{1},\ldots,\mathbf{h}_{n})$, $\mathbf{W}_{n}$ has the same
distribution as

$\frac{1}{\sqrt{n}}\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(\mathbf{h}_{i}),$

which has mean $\mathbf{0}$ by 5-$(i)$. We first show the convergence of the
covariance matrix
$\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left[S_{n}(\mathbf{X}_{i})J(\mathbf{h}_{i})J(\mathbf{h}_{i})^{\top}S_{n}(\mathbf{X}_{i})^{\top}\right]$.
Note that
$f(\mathbf{x}):=\mathbb{E}_{S}\left[S\mathbf{x}\mathbf{x}^{\top}S^{\top}\right]$,
where $S\sim{\rm Uniform}(\mathcal{G})$, is continuous in $\mathbf{x}$ (by an
application of the dominated convergence theorem). Then the covariance matrix
can be written as $\mathbb{E}f(J(\mathbf{H}_{n}))$, where
$\mathbf{H}_{n}\sim\frac{1}{n}\sum_{j=1}^{n}\delta_{\mathbf{h}_{j}}$converges
weakly to $\mathbf{H}\sim\nu$.

To show $\mathbb{E}f(J(\mathbf{H}_{n}))\to\mathbb{E}f(J(\mathbf{H}))$ as
$n\to\infty$, by the representation theorem, we may assume that
$\mathbf{H}_{n}\overset{a.s.}{\longrightarrow}\mathbf{H}$ (by 2). Then, as
$J(\cdot)$ is $\nu$-a.e. continuous, by the continuous mapping theorem,
$J(\mathbf{H}_{n})\overset{a.s.}{\longrightarrow}J(\mathbf{H})$. Together with
$\mathbb{E}|J(\mathbf{H}_{n})_{i}J(\mathbf{H}_{n})_{j}|\to\mathbb{E}|J(\mathbf{H})_{i}J(\mathbf{H})_{j}|$
(by 4-$(iii)$), we have
$J(\mathbf{H}_{n})J(\mathbf{H}_{n})^{\top}\overset{L^{1}}{\longrightarrow}J(\mathbf{H})J(\mathbf{H})^{\top}$
by Scheffé’s lemma. Note that
$\|f(\mathbf{x})\|_{F}=\|\mathbb{E}_{S}\left[S\mathbf{x}\mathbf{x}^{\top}S^{\top}\right]\|_{F}\leq\mathbb{E}_{S}[\|S\mathbf{x}\mathbf{x}^{\top}S^{\top}\|_{F}]=\mathbb{E}_{S}[\|\mathbf{x}\mathbf{x}^{\top}\|_{F}]=\|\mathbf{x}\mathbf{x}^{\top}\|_{F}$,
where we have used the convexity of the Frobenius norm and
$\|AS\|_{F}=\|SA\|_{F}=\|A\|_{F}$ for orthogonal $S$ and any matrix $A$.
Hence,
$\|f(J(\mathbf{H}_{n}))\|_{F}\leq\|J(\mathbf{H}_{n})J(\mathbf{H}_{n})^{\top}\|_{F}$.
Together with the fact that $J(\mathbf{H}_{n})J(\mathbf{H}_{n})^{\top}$
converges in $L^{1}$, we have that $\|f(J(\mathbf{H}_{n}))\|_{F}$ is uniformly
integrable. Since
$f(J(\mathbf{H}_{n}))\overset{a.s.}{\longrightarrow}f(J(\mathbf{H}))$, we
further have $\mathbb{E}f(J(\mathbf{H}_{n}))\to\mathbb{E}f(J(\mathbf{H}))$.

Since the covariance converges, we only need to check the Lindeberg’s
condition:

$\displaystyle\quad\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left[\|S_{n}(\mathbf{X}_{i})J(\mathbf{h}_{i})\|^{2}1_{\|S_{n}(\mathbf{X}_{i})J(\mathbf{h}_{i})\|\geq\varepsilon\sqrt{n}}\right]$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}\|J(\mathbf{h}_{i})\|^{2}1_{\|J(\mathbf{h}_{i})\|\geq\varepsilon\sqrt{n}}=\mathbb{E}\left[\|J(\mathbf{H}_{n})\|^{2}1_{\|J(\mathbf{H}_{n})\|\geq\varepsilon\sqrt{n}}\right].$

Note that since $J(\mathbf{H}_{n})$ is bounded in probability,
$\mathbb{P}(\|J(\mathbf{H}_{n})\|\geq\varepsilon\sqrt{n})\to 0$ as
$n\to\infty$, so
$\|J(\mathbf{H}_{n})\|^{2}1_{\|J(\mathbf{H}_{n})\|\geq\varepsilon\sqrt{n}}$
converges in probability to 0. Note that $\|J(\mathbf{H}_{n})\|^{2}$ is
uniformly integrable because $J(\mathbf{H}_{n})J(\mathbf{H}_{n})^{\top}$
converges in $L^{1}$, and hence
$\|J(\mathbf{H}_{n})\|^{2}1_{\|J(\mathbf{H}_{n})\|\geq\varepsilon\sqrt{n}}$ is
uniformly integrable. Thus,
$\mathbb{E}\left[\|J(\mathbf{H}_{n})\|^{2}1_{\|J(\mathbf{H}_{n})\|\geq\varepsilon\sqrt{n}}\right]\to
0$. Hence by Lindeberg-Feller’s CLT, as $n\to\infty$,

$\frac{1}{\sqrt{n}}\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(\mathbf{h}_{i})\overset{d}{\to}N\left(\mathbf{0},\mathbb{E}f(J(\mathbf{H}))\right)\equiv
N\left(\mathbf{0},{\rm Var}(SJ(\mathbf{H}))\right).$

∎

###  B.4 Proof of Theorems 2.1, 2.2, 3.2, and 3.3

######  Definition B.1 ($c$-convexity [101]).

A function $\psi:\mathbb{R}^{p}\to\mathbb{R}\cup\\{+\infty\\}$ is said to be
$c$-convex if it is not identically $+\infty$, and there exists
$\zeta:\mathbb{R}^{p}\to\mathbb{R}\cup\\{\pm\infty\\}$ such that

$\psi(\mathbf{x})=\sup_{\mathbf{y}\in\mathbb{R}^{p}}\left(\zeta(\mathbf{y})-c(\mathbf{x},\mathbf{y})\right),\quad{\rm
for\ all}\ \mathbf{x}\in\mathbb{R}^{p}.$

Its $c$-transform is the function $\psi^{c}$ defined by

$\psi^{c}(\mathbf{y}):=\inf_{\mathbf{x}\in\mathbb{R}^{p}}\left(\psi(\mathbf{x})+c(\mathbf{x},\mathbf{y})\right),\quad{\rm
for\ all}\ \mathbf{y}\in\mathbb{R}^{p}.$ (B.5)

Its $c$-subdifferential is defined as

$\partial_{c}\psi:=\left\\{(\mathbf{x},\mathbf{y})\in\mathbb{R}^{p}\times\mathbb{R}^{p}:\psi^{c}(\mathbf{y})-\psi(\mathbf{x})=c(\mathbf{x},\mathbf{y})\right\\}.$

If $c(\cdot,\cdot)$ is continuous, then a $c$-convex function is automatically
lower semicontinuous (l.s.c.) and Borel measurable [101, Remark 5.5].

We consider the equivalent cost function
$c(\textbf{x},\textbf{y}):=-\max_{Q\in\mathcal{G}}\mathbf{x}^{\top}Q\mathbf{y}$.
Note that with our choice of $c$,
$\psi(\mathbf{x})=\sup_{\mathbf{y},Q}\left(\zeta(\mathbf{y})+\mathbf{x}^{\top}Q\mathbf{y}\right)$
is the supremum of linear functions. This implies that a $c$-convex function
is convex in the usual sense, and is thus Lebesgue a.e. differentiable in its
domain [3]. It is also easy to see that any $c$-convex function $\psi$ is
$\mathcal{G}$-invariant, i.e., $\psi(Q\mathbf{x})=\psi(\mathbf{x})$ for any
$Q\in\mathcal{G}$, and thus $\nabla\psi(\cdot)$ is $\mathcal{G}$-equivariant,
i.e., $\nabla\psi(Q\mathbf{x})=Q\nabla\psi(\mathbf{x})$ whenever $\mathbf{x}$
and $Q\mathbf{x}$ are differentiable points of $\psi$.

######  Lemma B.3.

Let $\mathbf{X}\sim\mu\in\mathcal{P}_{\rm ac}(\mathbb{R}^{p})$, and
$\mathcal{G}$ be a compact subgroup of the orthogonal group ${\rm O}(p)$. Let
$S$ follow the uniform distribution over $\mathcal{G}$, independent of
$\mathbf{X}$. Then $S\mathbf{X}\sim\mu_{S}$ is also Lebesgue absolutely
continuous, and $\mu$ is absolutely continuous w.r.t. $\mu_{S}$.

###### Proof.

Suppose $\mathbf{X}$ has a Lebesgue density $f(\cdot)$. We show in the
following that $S\mathbf{X}$ has a Lebesgue density given by, for
$\mathbf{x}\in\mathbb{R}^{p}$, $g(\mathbf{x})=\mathbb{E}_{S}[f(S\mathbf{x})]$,
where $S\sim{\rm Uniform}(\mathcal{G})$. Let $m$ denote the uniform
distribution over $\mathcal{G}$ (i.e., the Haar measure on $\mathcal{G}$). For
any Borel set $B\subset\mathbb{R}^{p}$,

$\displaystyle\mathbb{P}(S\mathbf{X}\in B)$ $\displaystyle=\int
1_{Q\mathbf{x}\in B}f(\mathbf{x}){\rm d}\mathbf{x}\,{\rm d}m(Q)$
$\displaystyle=\int 1_{\mathbf{y}\in B}f(Q^{-1}\mathbf{y}){\rm
d}\mathbf{y}\,{\rm d}m(Q)\qquad$ $\displaystyle\left({\rm by\ a\ change\ of\
variable\ \mathbf{y}=Q\mathbf{x}}\right)$ $\displaystyle=\int 1_{\mathbf{y}\in
B}f(Q\mathbf{y}){\rm d}\mathbf{y}\,{\rm d}m(Q)\qquad$ $\displaystyle\left({\rm
as\ }S^{-1}\overset{d}{=}S\sim{\rm Uniform}(\mathcal{G})\right)$
$\displaystyle=\int 1_{\mathbf{y}\in B}\mathbb{E}_{S}f(S\mathbf{y}){\rm
d}\mathbf{y}.\qquad$ $\displaystyle\left({\rm by\ Fubini\text{'}s\
theorem}\right)$

Hence, $S\mathbf{X}$ has a Lebesgue density given by
$g(\mathbf{x})=\mathbb{E}_{S}f(S\mathbf{x})$, for
$\mathbf{x}\in\mathbb{R}^{p}$.

Recall that the support of a probability measure $\mu$ on $\mathbb{R}^{p}$ is
the closed set defined as:

${\rm supp}(\mu):=\\{\mathbf{x}\in\mathbb{R}^{p}:\forall{\rm\ open\
neighborhood\ }N_{\mathbf{x}}{\rm\ of\
}\mathbf{x},\,\mu(N_{\mathbf{x}})>0\\}.$

If $\mu$ has a Lebesgue density $f(\cdot)$, then $f(\mathbf{x})>0$ (Lebesgue)
a.e.-$\mathbf{x}$ on ${\rm supp}(\mu)$, and $f(\mathbf{x})=0$ (Lebesgue)
a.e.-$\mathbf{x}$ on $\mathbb{R}^{p}\backslash{\rm supp}(\mu)$.

We will show that every point in the support of $\mathbf{X}$ is in the support
of $S\mathbf{X}$ and thus $\frac{f(\cdot)}{g(\cdot)}$ is well-defined (which
is the Radon-Nikodym derivative of $\mathbf{X}$ w.r.t. $S\mathbf{X}$). This
will establish the absolute continuity of $\mathbf{X}$ w.r.t. $S\mathbf{X}$.

Note that every element $Q\in\mathcal{G}$ is in the support of ${\rm
Uniform}(\mathcal{G})$.121212Otherwise, there exists $Q_{0}\in\mathcal{G}$ and
a neighborhood $U$ of $Q_{0}$ with probability 0. Consider the following open
cover of $\mathcal{G}$: $\\{QU:Q\in\mathcal{G}\\}$. By the compactness of
$\mathcal{G}$, we can find a finite sub-cover $\cup_{i=1}^{k}Q_{i}U$ that
covers $\mathcal{G}$ and has zero probability, yielding a contradiction. In
particular, the identity matrix $I\equiv I_{p}$ is in the support of ${\rm
Uniform}(\mathcal{G})$. Let $\mathbf{x}\in{\rm supp}(\mu)$. For any
neighborhood $N_{\mathbf{x}}$ of $\mathbf{x}$, there exists an open ball
$B(\mathbf{x},\varepsilon)$ contained in $N_{\mathbf{x}}$. By the continuity
of the group action
$\mathcal{G}\times\mathbb{R}^{p}\to\mathbb{R}^{p}:(Q,\mathbf{x})\mapsto
Q\mathbf{x}$, there exists a neighborhood $V$ of $I$ such that for all $Q\in
V$ and $\mathbf{y}\in B(\mathbf{x},\varepsilon/2)$, we have $Q\mathbf{y}\in
N_{\mathbf{x}}$. Therefore, by independence, $\mathbb{P}(S\mathbf{X}\in
N_{\mathbf{x}})\geq\mathbb{P}(S\in V,\mathbf{X}\in
B(\mathbf{x},\varepsilon/2))=\mathbb{P}(S\in V)\mathbb{P}(\mathbf{X}\in
B(\mathbf{x},\varepsilon/2))>0$, where we have used that $I$, $\mathbf{x}$ are
in the supports of ${\rm Uniform}(\mathcal{G})$ and $\mu$, respectively.
Hence, every point in the support of $\mathbf{X}$ is in the support of
$S\mathbf{X}\sim\mu_{S}$.

Note that $g(\cdot)$ is Lebesgue a.e. positive on ${\rm supp}(\mu_{S})$. Thus
for any Borel set $B\subset\mathbb{R}^{p}$,

$\mathbb{P}(\mathbf{X}\in B)=\int_{{\rm supp}(\mu_{S})\cap
B}f(\mathbf{x})\,{\rm d}\mathbf{x}=\int_{{\rm supp}(\mu_{S})\cap
B}\frac{f(\mathbf{x})}{g(\mathbf{x})}g(\mathbf{x})\,{\rm
d}\mathbf{x}=\int_{{\rm supp}(\mu_{S})\cap
B}\frac{f(\mathbf{x})}{g(\mathbf{x})}\,{\rm d}\mu_{S}(\mathbf{x}).$

Thus, $\mathbf{X}$ is absolutely continuous w.r.t. $S\mathbf{X}$ with Radon-
Nikodym derivative $\frac{f(\cdot)}{g(\cdot)}$. ∎

###### Proof of Theorem 2.1.

By C.1, the continuity of $c(\cdot,\cdot)$ implies that there exists
$\pi\in\Pi(\mu,\nu)$ with a $c$-cyclically monotone support. By C.2, for any
$\pi\in\Pi(\mu,\nu)$ with a $c$-cyclically monotone support, there exists a
$c$-convex function $\psi:\mathbb{R}^{p}\to\mathbb{R}\cup\\{+\infty\\}$ such
that the support of $\pi$ is contained in the $c$-subdifferential set

$\partial_{c}\psi:=\left\\{(\mathbf{x},\mathbf{y})\in\mathbb{R}^{p}\times\mathbb{R}^{p}:\psi^{c}(\mathbf{y})-\psi(\mathbf{x})=c(\mathbf{x},\mathbf{y})\right\\}.$

Since $\psi$ is $c$-convex, and $c(\cdot,\cdot)$ is continuous, $\psi$ is a
l.s.c. convex function in the usual sense. Thus, $\psi$ is Lebesgue a.e.
differentiable in its domain (which contains the support of $\mu$). Fix any
pair of $(\mathbf{x},\mathbf{y})$ in the support of $\pi$ and suppose $\psi$
is differentiable at $\mathbf{x}$. Let
$\nabla_{\mathbf{x}}\left(-c(\textbf{x},\textbf{y})\right)$ denote a sub-
gradient of
$g(\textbf{x}):=\max_{Q\in\mathcal{G}}\mathbf{x}^{\top}Q\mathbf{y}$, which
clearly exists (as $g(\cdot)$ is a convex function) and is an element in the
convex hull of $\\{Q\mathbf{y}:Q{\rm\ is\ a\ maximizer\ of\
}\max_{Q\in\mathcal{G}}\mathbf{x}^{\top}Q\mathbf{y}\\}$ (i.e., the gradients
of the active linear functions at $\mathbf{x}$; see e.g., [55, Theorem
4.4.2]). As $\psi(\cdot)$ is differentiable at $\mathbf{x}$, for
$\mathbf{h}\in\mathbb{R}^{p}$ in a neighborhood of
$\mathbf{0}\in\mathbb{R}^{p}$, we have

$\psi(\textbf{x}+\textbf{h})=\psi(\textbf{x})+\nabla\psi(\textbf{x})\cdot\textbf{h}+o(\textbf{h}).$
(B.6)

Now, using the definition of $c$-transform (see (B.5)):

$\displaystyle\psi(\textbf{x}+\textbf{h})$
$\displaystyle\geq\psi^{c}(\textbf{y})-c(\textbf{x}+\textbf{h},\textbf{y})$
$\displaystyle\geq\;\psi^{c}(\textbf{y})-c(\textbf{x},\textbf{y})+\nabla_{\mathbf{x}}\left(-c(\textbf{x},\textbf{y})\right)\cdot\textbf{h}\quad\text{
(by the definition\ of\ sub-gradient of }g(\cdot))$
$\displaystyle=\;\psi(\textbf{x})+\nabla_{\mathbf{x}}\left(-c(\textbf{x},\textbf{y})\right)\cdot\textbf{h}.$

Since $\mathbf{h}$ can take arbitrary directions, comparing the above display
with (B.6), we get

$\nabla\psi(\textbf{x})=\nabla_{\mathbf{x}}\left(-c(\textbf{x},\textbf{y})\right).$
(B.7)

This also implies that the sub-gradient
$\nabla_{\mathbf{x}}\left(-c(\mathbf{x},\mathbf{y})\right)$ is unique, i.e.,
$\\{\widetilde{Q}\mathbf{y}:\widetilde{Q}\in\operatorname*{\arg\\!\max}_{Q\in\mathcal{G}}\mathbf{x}^{\top}Q\mathbf{y}\\}$
consists of only one element. Let
$Q_{0}:=\operatorname*{\arg\\!\max}_{Q\in\mathcal{G}}\mathbf{x}^{\top}Q\mathbf{y}$
be the unique maximizer of $\mathbf{x}^{\top}Q\mathbf{y}$. Then
$\nabla\psi(\textbf{x})=Q_{0}\mathbf{y}$, so that
$\textbf{y}=Q_{0}^{-1}\nabla\psi(\textbf{x})$. Since the orbit
$\\{Q\nabla\psi(\textbf{x}):Q\in\mathcal{G}\\}$ intersects $B$ at one point at
most, we have $\mathbf{y}=q\left(\nabla\psi(\textbf{x})\right)$ (as
$\mathbf{y}\in B$ and $\nu$ is concentrated on $B$).

Thus, for $\mu$-a.e. x, there is only one y such that
$(\textbf{x},\textbf{y})\in\partial_{c}\psi$. So, $\pi$ is given by a
deterministic coupling, i.e., $({\rm identity},q\circ\nabla\psi)\\#\mu$ and
the existence of $R$ is shown. Although $q\circ\nabla\psi$ is only defined
Lebesgue a.e. on the domain of $\psi$, it has a $\mu$-a.e. unique Borel
measurable version (see the proof of Villani, [101, Theorem 5.30]) which we
define to be $R$ (note that $q\circ\nabla\psi$ coincides with $R$ on a Borel
set with $\mu$-measure 1).

Let $S\sim{\rm Uniform}(\mathcal{G})$ independent of everything else. Thus,
$S\stackrel{{\scriptstyle d}}{{=}}SQ$ for any $Q\in\mathcal{G}$. Therefore,
for $\mu$-a.e. $\mathbf{x}$,
$S\nabla\psi(\mathbf{x})\overset{d}{=}SR(\mathbf{x})$ as
$\nabla\psi(\mathbf{x})$ and $R(\mathbf{x})=q(\nabla\psi(\mathbf{x}))$ are on
the same orbit (i.e., $R(\mathbf{x})=Q\nabla\psi(\mathbf{x})$ for some
$Q\in\mathcal{G}$). Therefore,
$S\nabla\psi(\mathbf{X})\overset{d}{=}SR(\mathbf{X})\overset{d}{=}S\mathbf{H}$
(recall $R(\mathbf{X})\sim\mathbf{H}$). Since $\psi$ is a $c$-convex function,
the $\mathcal{G}$-equivariance of $\nabla\psi(\cdot)$ (see the discussion
following Definition B.1) then implies
$\nabla\psi(S\mathbf{X})\overset{d}{=}S\mathbf{H}$. As $\nabla\psi(\cdot)$ —
gradient of a convex function — pushes the symmetrized version $S\mathbf{X}$
to $S\mathbf{H}$, by McCann’s theorem [73] (which does not need any moment
assumptions) $\nabla\psi(\cdot)$ is $\mu_{S}$-a.e. uniquely determined (by the
absolute continuity of $\mu_{S}$). As $\mu$ is absolutely continuous w.r.t.
$\mu_{S}$, $\nabla\psi(\cdot)$ is also $\mu$-a.e. uniquely determined, and the
uniqueness of $R$ is shown.

To show $(iv)$, note that by definition

$S(\mathbf{x},\mathbf{h})=\operatorname*{\arg\\!\min}_{Q\in\mathcal{G}}\|Q^{\top}\mathbf{x}-\mathbf{h}\|^{2}=\operatorname*{\arg\\!\min}_{Q\in\mathcal{G}}\|\mathbf{x}-Q\mathbf{h}\|^{2}=\operatorname*{\arg\\!\max}_{Q\in\mathcal{G}}\mathbf{x}^{\top}Q\mathbf{h},$

and thus, $S(\mathbf{x},R(\mathbf{x}))R(\mathbf{x})$ is a point in the orbit
$\\{QR(\mathbf{x}):Q\in\mathcal{G}\\}$ that is closest to $\mathbf{x}$.
However, as shown in (B.7), for $\mu$-a.e. $\mathbf{x}$,
$\nabla\psi(\mathbf{x})$ is the unique element in
$\\{\widetilde{Q}R(\mathbf{x}):\widetilde{Q}\in\operatorname*{\arg\\!\max}_{Q\in\mathcal{G}}\mathbf{x}^{\top}QR(\mathbf{x})\\}$.
This implies that
$S(\mathbf{X},R(\mathbf{X}))R(\mathbf{X})\overset{a.s.}{=}\nabla\psi(\mathbf{X})$.

Finally, $(v)$ is a direct consequence of Villani, [101, Theorem 5.10]. ∎

At this point, Theorem 2.1 has been proved. In the following, we will develop
some further useful results for Theorems 2.2, 3.2 and 3.3.

######  Lemma B.4 (Continuity and uniqueness of $S(\cdot,\cdot)$ when
$\mathcal{G}$ acts freely on $\mathcal{G}B$).

Under the assumptions of Theorem 2.1, if $\mathcal{G}$ acts freely on
$\mathcal{G}B$, then
$S(\mathbf{x},\mathbf{y}):=\operatorname*{\arg\\!\min}_{Q\in\mathcal{G}}\|Q^{\top}\mathbf{x}-\mathbf{y}\|^{2}$
is unique and continuous for $({\rm identity},{R})\\#\mu$-a.e.
$(\mathbf{x},\mathbf{y})$.

###### Proof.

We first show the uniqueness. Suppose that there are two minimizers
$Q_{1},Q_{2}$ of $\|Q^{\top}\mathbf{x}-\mathbf{y}\|^{2}$. Then $Q_{1},Q_{2}$
are also maximizers of $\mathbf{x}^{\top}Q\mathbf{y}$. Since for $\mu$-a.e.
$\mathbf{x}$ and $\mathbf{y}=R(\mathbf{x})$, from the proof of Theorem 2.1,
$\\{Q\mathbf{y}:Q{\rm\ is\ a\ maximizer\ of\
}\max_{S\in\mathcal{G}}\mathbf{x}^{\top}Q\mathbf{y}\\}$ has only one element
$\nabla\psi(\mathbf{x})$, we must have
$Q_{1}\mathbf{y}=Q_{2}\mathbf{y}=\nabla\psi(\mathbf{x})$. This implies
$Q_{1}^{-1}Q_{2}\mathbf{y}=\mathbf{y}$ and thus $Q_{1}=Q_{2}$, as
$\mathcal{G}$ acts freely on $\mathcal{G}B$. Hence
$\operatorname*{\arg\\!\min}_{Q\in\mathcal{G}}\|Q^{\top}\mathbf{x}-\mathbf{y}\|^{2}$
is unique $({\rm identity},{R})\\#\mu$-a.s.

Next, we show the continuity. Suppose
$(\mathbf{x}_{n},\mathbf{y}_{n})\to(\mathbf{x}_{0},\mathbf{y}_{0})$ as
$n\to\infty$ and $S(\mathbf{x},\mathbf{y})$ is uniquely defined at
$(\mathbf{x}_{0},\mathbf{y}_{0})$. Consider the function
$f_{n}:\mathbb{R}^{p\times p}\to\mathbb{R}$ defined by
$f_{n}(Q)=\|Q^{\top}\mathbf{x}_{n}-\mathbf{y}_{n}\|^{2}$ if $Q\in\mathcal{G}$,
and $f_{n}(Q)=+\infty$ otherwise. Let $f:\mathbb{R}^{p\times p}\to\mathbb{R}$
be the function defined by
$f(Q)=\|Q^{\top}\mathbf{x}_{0}-\mathbf{y}_{0}\|^{2}$ if $Q\in\mathcal{G}$ and
$f(Q)=+\infty$ otherwise. For any $Q\in\mathbb{R}^{p\times p}$, it is easy to
see that for every sequence $\\{Q_{n}\\}_{n\geq 1}\subset\mathbb{R}^{p\times
p}$ such that $\|Q_{n}-Q\|_{F}\to 0$, $\liminf_{n}f_{n}(Q_{n})\geq f(Q)$ (by
using the form of $f_{n}$, which is continuous on $\mathcal{G}$, and the
compactness of $\mathcal{G}$). Moreover, for some sequence $\\{Q_{n}\\}_{n\geq
1}\subset\mathbb{R}^{p\times p}$ such that $\|Q_{n}-Q\|_{F}\to 0$, we also
have $\limsup_{n}f_{n}(Q_{n})\leq f(Q)$. This implies $f_{n}$ epi-converges to
$f$ [89, Proposition 7.2]. Note that $f_{n}$ and $f$ are l.s.c. with bounded
level sets $\\{Q\in\mathbb{R}^{p\times p}\,:\,f_{n}(Q)\leq\alpha\\}$ for all
$\alpha\in\mathbb{R}$. By Rockafellar and Wets, [89, Theorem 7.33], the
minimizer(s) of $f_{n}$ must converge to the minimizer of $f$. Hence, the
continuity of $S(\mathbf{x},\mathbf{y})$ is shown. ∎

The following weak convergence results will be used in the proofs of Theorems
2.2, 3.2 and 3.3.

######  Lemma B.5 (Joint weak convergence of empirical ranks and signed-
ranks).

Suppose Assumptions 1 and 2 hold. Let $\mu_{n}$ denote the empirical
distribution on $\\{\mathbf{X}_{1},\ldots,\mathbf{X}_{n}\\}$ and
$\mathcal{X}_{n}\sim\mu_{n}$. With probability 1, we have
$(\mathcal{X}_{n},R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},R(\mathbf{X}))$
and
$(\mathcal{X}_{n},S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},\nabla\psi(\mathbf{X}))$.

###### Proof.

It is known that $\mu_{n}$ converges weakly to $\mu$ with probability 1. Fix
$\omega\in\Omega$ (the sample space) such that
$\mu_{n}(\omega)\overset{w}{\longrightarrow}\mu$. Consider the sequence of
random measures $({\rm identity},{R}_{n})\\#\mu_{n}(\omega)$, where ${R}_{n}$
is the empirical rank map and the randomness comes from the random draw
according to $\mu_{n}(\omega)$. As ${R}_{n}\\#\mu_{n}=\nu_{n}$ converges
weakly (to $\nu$, by 2) and $\mu_{n}(\omega)$ converges weakly (to $\mu$), we
have that $({\rm identity},{R}_{n})\\#\mu_{n}(\omega)$ is tight. Hence by
Prokhorov’s theorem, every subsequence of $({\rm
identity},{R}_{n})\\#\mu_{n}(\omega)$ has a further subsequence that converges
weakly. We will show that this limit is $({\rm identity},{R})\\#\mu$
regardless of what sub-sequences are chosen. Consequently, since weak
convergence is metrizable, $({\rm identity},{R}_{n})\\#\mu_{n}(\omega)$
converges weakly to the same limit.

Now suppose the convergent sub-sequence is $({\rm
identity},{R}_{n_{k}})\\#\mu_{n_{k}}(\omega)$. We know that $({\rm
identity},{R}_{n_{k}})\\#\mu_{n_{k}}(\omega)$ is supported on a $c$-cyclically
monotone set, following the optimality of ${R}_{n_{k}}$ (see e.g., the proof
of C.1). This, together with the continuity of $c(\cdot,\cdot)$, implies that
its weak limit must also be supported on a $c$-cyclically monotone set (this
is also shown in the proof of C.1). From Theorem 2.1, the unique
$c$-cyclically monotone measure in $\Pi(\mu,\nu)$ is $({\rm
identity},{R})\\#\mu$; this establishes $({\rm
identity},{R}_{n})\\#\mu_{n}(\omega)\overset{w}{\longrightarrow}({\rm
identity},{R})\\#\mu$. Or equivalently, with $\mathcal{X}_{n}\sim\mu_{n}$, we
have
$(\mathcal{X}_{n},R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},R(\mathbf{X}))$.

Next, we use the above weak convergence to show that
$(\mathcal{X}_{n},S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},\nabla\psi(\mathbf{X}))$.
Recall that the signed-rank is the point in the orbit of the rank vector that
is closest to the data point. So by defining $f(\mathbf{x},\mathbf{y}):={\rm
Proj}_{\mathcal{G}\mathbf{y}}(\mathbf{x})$ as the projection of $\mathbf{x}$
onto the orbit of $\mathbf{y}$, we can write
$S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n})=f(\mathcal{X}_{n},R_{n}(\mathcal{X}_{n}))$,
and $\nabla\psi(\mathbf{X})=f(\mathbf{X},R(\mathbf{X}))$. We will show that
$f(\mathbf{x},\mathbf{y})$ is continuous at $({\rm identity},{R})\\#\mu$
a.e.-$(\mathbf{x},\mathbf{y})$. We have shown in the proof of Theorem 2.1 that
the signed-rank $f(\mathbf{x},\mathbf{y})={\rm
Proj}_{\mathcal{G}\mathbf{y}}(\mathbf{x})$ is $({\rm identity},{R})\\#\mu$
a.e.-$(\mathbf{x},\mathbf{y})$ uniquely defined. In case where the nearest
point in $\mathcal{G}\mathbf{y}$ to $\mathbf{x}$ is not unique,
$f(\mathbf{x},\mathbf{y})$ can be chosen as any projection, and we can show
that the continuity still holds.

Let $(\mathbf{x}_{0},\mathbf{y}_{0})$ be a point where the projection ${\rm
Proj}_{\mathcal{G}\mathbf{y}_{0}}(\mathbf{x}_{0})$ is unique. We argue by
contradiction. If $f(\mathbf{x},\mathbf{y})$ is not continuous at
$(\mathbf{x}_{0},\mathbf{y}_{0})$, then there exist $\delta>0$ and a sequence
$(\mathbf{x}_{n},\mathbf{y}_{n})\to(\mathbf{x}_{0},\mathbf{y}_{0})$ such that
$\|{\rm Proj}_{\mathcal{G}\mathbf{y}_{n}}(\mathbf{x}_{n})-{\rm
Proj}_{\mathcal{G}\mathbf{y}_{0}}(\mathbf{x}_{0})\|\geq\delta$ for all $n$.
Define the distance from a point $\mathbf{x}\in\mathbb{R}^{p}$ to a set
$A\subset\mathbb{R}^{p}$ as $d(\mathbf{x},A):=\inf_{\mathbf{y}\in
A}(\mathbf{x},\mathbf{y})$. Let
$d_{0}:=d(\mathbf{x}_{0},\mathcal{G}\mathbf{y}_{0})$, and suppose ${\rm
Proj}_{\mathcal{G}\mathbf{y}_{0}}(\mathbf{x}_{0})=g_{0}\mathbf{y}_{0}$ is the
point in $\mathcal{G}\mathbf{y}_{0}$ that achieves this minimal distance,
i.e., $\|\mathbf{x}_{0}-g_{0}\mathbf{y}_{0}\|=d_{0}$. Observe that
$d(\mathbf{x}_{n},\mathcal{G}\mathbf{y}_{n})\leq
d(\mathbf{x}_{n},\mathbf{x}_{0})+d(\mathbf{x}_{0},\mathcal{G}\mathbf{y}_{n})\leq
d(\mathbf{x}_{n},\mathbf{x}_{0})+d(\mathbf{x}_{0},g_{0}\mathbf{y}_{n})\leq
d(\mathbf{x}_{n},\mathbf{x}_{0})+d(\mathbf{x}_{0},g_{0}\mathbf{y}_{0})+d(g_{0}\mathbf{y}_{0},g_{0}\mathbf{y}_{n})=\|\mathbf{x}_{n}-\mathbf{x}_{0}\|+d_{0}+\|\mathbf{y}_{n}-\mathbf{y}_{0}\|$.
Therefore, $\|{\rm
Proj}_{\mathcal{G}\mathbf{y}_{n}}(\mathbf{x}_{n})-\mathbf{x}_{0}\|\leq\|{\rm
Proj}_{\mathcal{G}\mathbf{y}_{n}}(\mathbf{x}_{n})-\mathbf{x}_{n}\|+\|\mathbf{x}_{n}-\mathbf{x}_{0}\|=d(\mathbf{x}_{n},\mathcal{G}\mathbf{y}_{n})+\|\mathbf{x}_{n}-\mathbf{x}_{0}\|\leq
2\|\mathbf{x}_{n}-\mathbf{x}_{0}\|+d_{0}+\|\mathbf{y}_{n}-\mathbf{y}_{0}\|\to
d_{0}$, as $n\to\infty$. Hence, ${\rm
Proj}_{\mathcal{G}\mathbf{y}_{n}}(\mathbf{x}_{n})$ must have a sub-sequence
that converges to a point in
$\\{\mathbf{x}\in\mathbb{R}^{p}:\|\mathbf{x}-\mathbf{x}_{0}\|\leq
d_{0}\\}\cap\mathcal{G}\mathbf{y}_{0}$, which consists of one single point
${\rm Proj}_{\mathcal{G}\mathbf{y}_{0}}(\mathbf{x}_{0})$, by the uniqueness of
the projection ${\rm Proj}_{\mathcal{G}\mathbf{y}_{0}}(\mathbf{x}_{0})$. This
is a contradiction to $\|{\rm
Proj}_{\mathcal{G}\mathbf{y}_{n}}(\mathbf{x}_{n})-{\rm
Proj}_{\mathcal{G}\mathbf{y}_{0}}(\mathbf{x}_{0})\|\geq\delta>0$ for all $n$.

Now, by continuous mapping applied to
$(\mathbf{x},\mathbf{y})\mapsto(\mathbf{x},f(\mathbf{x},\mathbf{y}))$ and
$(\mathcal{X}_{n},R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},R(\mathbf{X}))$,
we have
$(\mathcal{X}_{n},S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},\nabla\psi(\mathbf{X}))$.
∎

Finally, we present a simple lemma which serves as a convenient tool in
proving Theorem 2.2.

######  Lemma B.6.

Suppose Assumption 2 holds. For a continuous function $f:B\to\mathbb{R}$, if
we have
$\frac{1}{n}\sum_{i=1}^{n}|f(\mathbf{h}_{n,i})|\to\int|f(\mathbf{x})|{\rm
d}\nu(\mathbf{x})$, then $f\\#\nu_{n}$ is uniformly integrable.

###### Proof.

Since $\mathbf{H}_{n}\sim\nu_{n}$ converges weakly to $\nu$, by choosing an
almost sure representative, we may assume
$\mathbf{H}_{n}\overset{a.s.}{\longrightarrow}\mathbf{H}$, and by continuous
mapping $f(\mathbf{H}_{n})\overset{a.s.}{\longrightarrow}f(\mathbf{H})$. Since
the moments also converge, i.e.,
$\mathbb{E}|f(\mathbf{H}_{n})|\to\mathbb{E}|f(\mathbf{H})|$, Scheffé’s lemma
implies $f(\mathbf{H}_{n})$ converges to $f(\mathbf{H})$ in $L^{1}$. This
implies $\\{f(\mathbf{H}_{n})\\}_{n\geq 1}$ is uniformly integrable. ∎

###### Proof of Theorem 2.2.

We first state a continuity property of gradients of convex functions. Note
that the gradient of convex function $\nabla\psi(\cdot)$ is uniquely defined
Lebesgue a.e. In the negligible case where the gradient does not exist at some
point $\mathbf{x}$, we may define $\nabla\psi(\mathbf{x})$ as any subgradient
at $\mathbf{x}$, i.e., an element of the subdifferential set
$\partial\psi(\mathbf{x})$ (see e.g. [73]). A subgradient always exists in the
interior of the domain of a convex function. This notion of
$\nabla\psi(\cdot)$ is continuous in the following sense: if
$\mathbf{x}\in\mathbb{R}^{p}$ is a differentiable point of $\psi(\cdot)$, and
$\mathbf{x}_{n}\to\mathbf{x}$ and
$\mathbf{y}_{n}\in\partial\psi(\mathbf{x}_{n})$, then
$\mathbf{y}_{n}\to\nabla\psi(\mathbf{x})$.

Let us now prove part-1. of Theorem 2.2. From Lemma B.5, we have with
probability 1,
$(\mathcal{X}_{n},R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},R(\mathbf{X}))$
and
$(\mathcal{X}_{n},S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},\nabla\psi(\mathbf{X}))$,
where $\mathcal{X}_{n}\sim\mu_{n}$. As $q(\cdot)$ is continuous,
$R(\cdot)=q(\nabla\psi(\cdot))$ is $\mu$-a.e. continuous. Hence, by the
continuous mapping theorem applied to
$f(\mathbf{x},\mathbf{y})=\rho(R(\mathbf{x}),\mathbf{y})$,
$(\mathcal{X}_{n},R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},R(\mathbf{X}))$
implies
$\rho(R(\mathcal{X}_{n}),R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}\rho(R(\mathbf{X}),R(\mathbf{X}))=0$.
Note that $R_{n}(\mathcal{X}_{n})\sim\nu_{n}$.
$\mathbb{E}\rho(\mathbf{H}_{n},\mathbf{0})\to\mathbb{E}\rho(\mathbf{H},\mathbf{0})$
together with Lemma B.6 implies $\rho(R_{n}(\mathcal{X}_{n}),\mathbf{0})$ is
uniformly integrable. Hence, $\rho(R(\mathbf{X}),\mathbf{0})$, as the weak
limit of $\rho(R_{n}(\mathcal{X}_{n}),\mathbf{0})$, is integrable by Fatou’s
lemma. The strong law of large numbers then implies
$\mathbb{E}\rho(R(\mathcal{X}_{n}),\mathbf{0})$ converges to
$\mathbb{E}\rho(R(\mathbf{X}),\mathbf{0})$. Since
$\rho(R(\mathcal{X}_{n}),\mathbf{0})$ converges in distribution to
$\rho(R(\mathbf{X}),\mathbf{0})$, Scheffé’s lemma again implies
$\rho(R(\mathcal{X}_{n}),\mathbf{0})$ is uniformly integrable. Hence,
$\rho(R(\mathcal{X}_{n}),R_{n}(\mathcal{X}_{n}))\lesssim\rho(R(\mathcal{X}_{n}),\mathbf{0})+\rho(R_{n}(\mathcal{X}_{n}),\mathbf{0})$
is uniformly integrable and its mean converges to the mean of 0, i.e.,

$\frac{1}{n}\sum_{i=1}^{n}\rho(R(\mathbf{X}_{i}),R_{n}(\mathbf{X}_{i}))\to 0.$

Next, we show the convergence of signs. By Lemma B.4 and the continuous
mapping theorem applied to
$f(\mathbf{x},\mathbf{y})=\widetilde{\rho}(S(\mathbf{x},\mathbf{y}),S(\mathbf{x},R(\mathbf{x})))$,
$(\mathcal{X}_{n},R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},R(\mathbf{X}))$
implies
$\widetilde{\rho}(S(\mathcal{X}_{n},R_{n}(\mathcal{X}_{n})),S(\mathcal{X}_{n},R(\mathcal{X}_{n})))\overset{w}{\longrightarrow}0$.
As $\widetilde{\rho}$ is continuous and the signs are bounded, we can apply
the bounded convergence theorem which implies that the expectation also
converges, i.e.,

$\frac{1}{n}\sum_{i=1}^{n}\widetilde{\rho}(S(\mathbf{X}_{i},R_{n}(\mathbf{X}_{i})),S(\mathbf{X}_{i},R(\mathbf{X}_{i})))\to
0\qquad\mbox{as }n\to\infty.$

Since $S(\mathbf{X}_{i},R_{n}(\mathbf{X}_{i}))=S_{n}(\mathbf{X}_{i})$, the
convergence result follows.

Finally, we show the convergence of the empirical signed-ranks. Since, with
probability 1,
$(\mathcal{X}_{n},S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},\nabla\psi(\mathbf{X}))$
(by Lemma B.5), by continuous mapping theorem applied to
$f(\mathbf{x},\mathbf{y})=\rho(\mathbf{y},\nabla\psi(\mathbf{x}))$, we have
$\rho(S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}),\nabla\psi(\mathcal{X}_{n}))\overset{w}{\longrightarrow}0$.
We have shown that $\rho(R_{n}(\mathcal{X}_{n}),\mathbf{0})$ is uniformly
integrable. It follows from
$\rho(Q\mathbf{x},\mathbf{0})\lesssim\rho(\mathbf{x},\mathbf{0})+1$ that
$\rho(S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}),\mathbf{0})$ is also
uniformly integrable. Similarly, since $\nabla\psi(\mathcal{X}_{n})$ and
$R(\mathcal{X}_{n})$ are on a same orbit, the uniform integrability of
$\rho(R(\mathcal{X}_{n}),\mathbf{0})$ implies that
$\rho(\nabla\psi(\mathcal{X}_{n}),\mathbf{0})$ is uniformly integrable. Hence,
$\rho(S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}),\nabla\psi(\mathcal{X}_{n}))$
is uniformly integrable and its mean converges to the mean of 0, i.e.,

$\frac{1}{n}\sum_{i=1}^{n}\rho(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}),\nabla\psi(\mathbf{X}_{i}))\to
0\qquad\mbox{as }n\to\infty.$

∎

###### Proof of the Theorems 3.2 and 3.3.

It suffices to show the following claims:

  1. 1.

$\frac{1}{n}\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})\overset{a.s.}{\longrightarrow}\mathbb{E}[S(\mathbf{X},R(\mathbf{X}))]$;

  2. 2.

$\frac{1}{n}\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))\overset{a.s.}{\longrightarrow}\mathbb{E}[J(\nabla\psi(\mathbf{X}))]$.

From Lemma B.5 we have shown that for a.e. $\omega$, $({\rm
identity},{R}_{n})\\#\mu_{n}(\omega)\overset{w}{\longrightarrow}({\rm
identity},{R})\\#\mu$. From Lemma B.4, $S(\cdot,\cdot)$ is continuous $({\rm
identity},{R})\\#\mu$ a.s. (and for spherical symmetry, $S(\cdot,\cdot,\cdot)$
is continuous $({\rm identity},{R})\\#\mu\otimes P_{\varepsilon}$-a.s. as in
Remark 3.1), by finding an almost sure representative and dominated
convergence, the integration converges:

$\frac{1}{n}\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})=\frac{1}{n}\sum_{i=1}^{n}S(\mathbf{X}_{i},R_{n}(\mathbf{X}_{i}),\varepsilon_{i})\to\mathbb{E}[S(\mathbf{X},R(\mathbf{X}),\varepsilon)].$

To prove part-2. above, let $C\subset\mathbb{R}^{p}$ be the set of points
where $J(\cdot)$ is continuous. Then, $\nu_{S}(C)=1$ by 3. From Theorem 2.1,
as $\nabla\psi(\cdot)$ pushes $\mu_{S}$ to $\nu_{S}$, we have
$\mathbb{P}(\nabla\psi(S\mathbf{X})\in C)=1$. Since $\mu$ is absolutely
continuous w.r.t. $\mu_{S}$ (by Lemma B.3),
$\mathbb{P}(\nabla\psi(\mathbf{X})\in C)=1$. This allows us to apply the
continuous mapping theorem in the following.

From Lemma B.5, for $\mathcal{X}_{n}\sim\mu_{n}(\omega)$, we have
$(\mathcal{X}_{n},S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},\nabla\psi(\mathbf{X}))$.
In particular, by the equivariance and continuity of $J(\cdot)$,
$S_{n}(\mathcal{X}_{n})J(R_{n}(\mathcal{X}_{n}))=J(S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}J(\nabla\psi(\mathbf{X}))$.
Note that
$\|S_{n}(\mathcal{X}_{n})J(R_{n}(\mathcal{X}_{n}))\|=\|J(R_{n}(\mathcal{X}_{n}))\|$
is uniformly integrable as: (i)
$\mathbb{E}\|J(\mathbf{H}_{n})\|\to\mathbb{E}\|J(\mathbf{H})\|$, (ii)
$J(\mathbf{H}_{n})\stackrel{{\scriptstyle
w}}{{\longrightarrow}}J(\mathbf{H})$, and (iii) an application of Scheffé’s
lemma. Hence, we have the convergence of the first moment:

$\displaystyle\frac{1}{n}\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J({R}_{n}(\mathbf{X}_{i}))\to\mathbb{E}[J(\nabla\psi(\mathbf{X}))].$

∎

###  B.5 Proof of Remark 3.2

Under the assumptions of Remark 3.2, $S(X,R(X))={\rm sign}(X)$ and
$R(X)=F_{|X|}(|X|)$, since ${\rm sign}(\cdot)F_{|X|}(|\cdot|)$ transports
$\mu_{S}$ to ${\rm Uniform}(-1,1)$ and appears as the gradient of a convex
function. Hence from Theorem 3.2, the GWSR test is consistent against
$\mathbb{E}\left[{\rm sign}(X)F_{|X|}(|X|)\right]\neq 0$. Suppose $f$ is the
density function of $X$. Then:

$\displaystyle\quad\mathbb{E}\left[{\rm sign}(X)F_{|X|}(|X|)\right]>0$
$\displaystyle\Leftrightarrow\int_{0}^{\infty}\left[\frac{f(x)}{f(x)+f(-x)}-\frac{f(-x)}{f(x)+f(-x)}\right]F_{|X|}(x)\left[f(x)+f(-x)\right]{\rm
d}x>0$
$\displaystyle\Leftrightarrow\int_{0}^{\infty}\left[f(x)-f(-x)\right]F_{|X|}(x){\rm
d}x>0$
$\displaystyle\Leftrightarrow\int_{0}^{\infty}\left[f(x)-f(-x)\right]\int_{0}^{x}\left[f(y)+f(-y)\right]{\rm
d}y{\rm d}x>0$ $\displaystyle\Leftrightarrow\int_{0\leq y\leq
x}\left[f(x)-f(-x)\right]\left[f(y)+f(-y)\right]{\rm d}x{\rm d}y>0$
$\displaystyle\Leftrightarrow\int_{0\leq y\leq
x}f(x)\left[f(y)+f(-y)\right]{\rm d}x{\rm d}y>\int_{0\leq y\leq
x}f(-x)\left[f(y)+f(-y)\right]{\rm d}x{\rm d}y$
$\displaystyle\Leftrightarrow\mathbb{P}(0\leq X_{2}\leq
X_{1})+\mathbb{P}(0\leq-X_{2}\leq X_{1})>\mathbb{P}(0\leq X_{2}\leq-
X_{1})+\mathbb{P}(0\leq-X_{2}\leq-X_{1})$
$\displaystyle\Leftrightarrow\mathbb{P}(-X_{1}\leq X_{2}\leq
X_{1})>\mathbb{P}(X_{1}\leq X_{2}\leq-X_{1})$
$\displaystyle\Leftrightarrow\mathbb{P}(X_{1}+X_{2}\geq 0,X_{2}\leq
X_{1})>\mathbb{P}(X_{1}+X_{2}\leq 0,X_{1}\leq X_{2})$
$\displaystyle\Leftrightarrow\frac{1}{2}\mathbb{P}(X_{1}+X_{2}\geq
0)>\frac{1}{2}\mathbb{P}(X_{1}+X_{2}\leq 0)$
$\displaystyle\Leftrightarrow\mathbb{P}(X_{1}+X_{2}>0)>\frac{1}{2}.$

By replacing $X$ with $-X$, we have:

$\mathbb{E}\left[{\rm
sign}(X)F_{|X|}(|X|)\right]<0\quad\Leftrightarrow\quad\mathbb{P}(X_{1}+X_{2}>0)<\frac{1}{2},$

which concludes the proof. ∎

###  B.6 Proof of Corollary 3.1

By Theorem 3.2, we only need to show that
$\mathbb{E}[\nabla\psi(\mathbf{X})]\neq\mathbf{0}$.

We begin by showing $\nabla\psi(\mathbf{X})$ and
$\nabla\psi(\mathbf{X}-2\mathbf{\Delta})$ are integrable. Note that
$\mathbb{E}[\|\nabla\psi(\mathbf{X})\|]=\mathbb{E}[\|\mathbf{H}\|]$ where
$\mathbf{H}\sim\nu$; hence $\nabla\psi(\mathbf{X})\in L^{1}$. Since
$-I\in\mathcal{G}$ (by Assumption 5-$(ii)$),

$2\mathbf{\Delta}-\mathbf{X}=\mathbf{\Delta}+(\mathbf{\Delta}-\mathbf{X})\overset{d}{=}\mathbf{\Delta}+(\mathbf{X}-\mathbf{\Delta})=\mathbf{X}.$
(B.8)

Since $\nabla\psi$ is equivariant under the group action of $\mathcal{G}$,

$\nabla\psi(\mathbf{X}-2\mathbf{\Delta})=-\nabla\psi(2\mathbf{\Delta}-\mathbf{X})\overset{d}{=}-\nabla\psi(\mathbf{X})\in
L^{1}.$

To show $\mathbb{E}\left[\nabla\psi(\mathbf{X})\right]\neq\mathbf{0}$ we argue
by contradiction. If
$\mathbb{E}\left[\nabla\psi(\mathbf{X})\right]=\mathbf{0}$, then (B.8) would
imply
$\mathbb{E}\left[\nabla\psi(\mathbf{X}-2\mathbf{\Delta})\right]=\mathbf{0}$
and thus:

$\mathbb{E}\left[\mathbf{\Delta}^{\top}\left\\{\nabla\psi(\mathbf{X})-\nabla\psi(\mathbf{X}-2\mathbf{\Delta})\right\\}\right]=0.$

By the monotonicity of gradients of convex functions:
$2\mathbf{\Delta}^{\top}\left\\{\nabla\psi(\mathbf{X})-\nabla\psi(\mathbf{X}-2\mathbf{\Delta})\right\\}\geq
0$. This implies
$\mathbf{\Delta}^{\top}\left\\{\nabla\psi(\mathbf{X})-\nabla\psi(\mathbf{X}-2\mathbf{\Delta})\right\\}=0$
a.s.

Let $A$ be an open set of positive measure w.r.t. $\mu$ where $\psi$ is
strictly convex. Let $\mathbf{x}_{0}\in A$. Consider
$g(t):=\psi(\mathbf{x}_{0}-t\mathbf{\Delta})$, $t\in\mathbb{R}$, which is a
one-dimensional convex function, and is strictly convex in a neighborhood of
0. This implies $g^{\prime}(2)>g^{\prime}(0)$, i.e.,
$\mathbf{\Delta}^{\top}\left\\{\nabla\psi(\mathbf{x}_{0})-\nabla\psi(\mathbf{x}_{0}-2\mathbf{\Delta})\right\\}>0$.
As
$\mathbf{\Delta}^{\top}\left\\{\nabla\psi(\mathbf{x}_{0})-\nabla\psi(\mathbf{x}_{0}-2\mathbf{\Delta})\right\\}\geq
0$ for $\mathbf{x}_{0}\notin A$, we obtain

$\displaystyle\quad\mathbb{E}\left[\mathbf{\Delta}^{\top}\left\\{\nabla\psi(\mathbf{X})-\nabla\psi(\mathbf{X}-2\mathbf{\Delta})\right\\}\right]$
$\displaystyle=\mathbb{E}\left[\mathbf{\Delta}^{\top}\left\\{\nabla\psi(\mathbf{X})-\nabla\psi(\mathbf{X}-2\mathbf{\Delta})\right\\}\mathbf{1}_{A}(\mathbf{X})\right]+\mathbb{E}\left[\mathbf{\Delta}^{\top}\left\\{\nabla\psi(\mathbf{X})-\nabla\psi(\mathbf{X}-2\mathbf{\Delta})\right\\}\mathbf{1}_{A^{c}}(\mathbf{X})\right]>0,$

which yields a contradiction.

∎

###  B.7 Proof of Lemma 3.1

Observe that

$\displaystyle\quad\mathbb{E}\left\|\frac{1}{\sqrt{n}}\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))\right\|^{2}$
(B.9)
$\displaystyle=\frac{1}{n}\mathbb{E}\left\|\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))\right\|^{2}+\frac{1}{n}\mathbb{E}\left\|\sum_{i=1}^{n}S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))\right\|^{2}$
$\displaystyle\quad-\frac{2}{n}\mathbb{E}\left(\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))\right)^{\top}\left(\sum_{i=1}^{n}S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))\right).$

Since
$(S_{n}(\mathbf{X}_{1}),\ldots,S_{n}(\mathbf{X}_{n}))\mathop{\rotatebox[origin={c}]{90.0}{$\models$}}(R_{n}(\mathbf{X}_{1}),\ldots,R_{n}(\mathbf{X}_{n}))$
and follows i.i.d. ${\rm Uniform}(\mathcal{G})$,

$\frac{1}{n}\mathbb{E}\left\|\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))\right\|^{2}=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left\|S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))\right\|^{2}=\mathbb{E}\left\|J(R_{n}(\mathbf{X}_{1}))\right\|^{2},$

as the cross terms all have expectation 0 (by 5-$(i)$).

By the $\mathcal{G}$-symmetry of $\mathbf{X}_{i}$ and the equivariance of
$\nabla\psi$ and $J$ (here $S\sim{\rm Uniform}(\mathcal{G})$),

$\mathbb{E}[S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))]=\mathbb{E}[J(\nabla\psi(\mathbf{X}_{i}))]=\mathbb{E}[J(\nabla\psi(S\mathbf{X}_{i}))]=\mathbb{E}[S]\mathbb{E}[J(\nabla\psi(\mathbf{X}_{i}))]=\mathbf{0}.$

This implies

$\frac{1}{n}\mathbb{E}\left\|\sum_{i=1}^{n}S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))\right\|^{2}=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left\|S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))\right\|^{2}=\mathbb{E}\left\|J(R(\mathbf{X}_{1}))\right\|^{2}.$

To simplify the third term in (B.9), we first show that for $i\neq j$,

$\mathbb{E}\left[\Big{(}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))\Big{)}^{\top}\Big{(}S(\mathbf{X}_{j},R(\mathbf{X}_{j}))J(R(\mathbf{X}_{j}))\Big{)}\right]=0.$

We first condition on the
$\mathbf{X}_{1},\ldots,\mathbf{X}_{j-1},\mathbf{X}_{j+1},\ldots,\mathbf{X}_{n}$
and the orbit of $\mathbf{X}_{j}$, i.e.,
$\\{Q\mathbf{X}_{j}:Q\in\mathcal{G}\\}$. Then
$S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))$ is known, and $\mathbf{X}_{j}$
is still equally likely to take any element in the orbit, due to the
$\mathcal{G}$ symmetry of $\mathbf{X}_{j}$. Hence, by the equivariance of
$\nabla\psi$ and $J(\cdot)$,
$S(\mathbf{X}_{j},R(\mathbf{X}_{j}))J(R(\mathbf{X}_{j}))=J(\nabla\psi(\mathbf{X}_{j}))$
has conditional expectation $\mathbf{0}$ and consequently the unconditional
expectation is $\mathbf{0}$ as well. Hence,

$\displaystyle\quad\frac{1}{n}\mathbb{E}\left(\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))\right)^{\top}\left(\sum_{i=1}^{n}S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))\right)$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\Big{(}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))\Big{)}^{\top}\Big{(}S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))\Big{)}$
$\displaystyle=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left[J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))^{\top}J(\nabla\psi(\mathbf{X}_{i}))\right].$

Recall that from Lemma B.5 if $\mathcal{X}_{n}\sim\mu_{n}(\omega)$, we have
$(\mathcal{X}_{n},S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\overset{w}{\longrightarrow}(\mathbf{X},\nabla\psi(\mathbf{X}))$.
Therefore, by continuous mapping theorem, by 3,

$\displaystyle\quad
J(S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))^{\top}J(\nabla\psi(\mathcal{X}_{n}))+\|J(S_{n}(\mathcal{X}_{n})R_{n}(\mathcal{X}_{n}))\|^{2}+\|J(\nabla\psi(\mathcal{X}_{n}))\|^{2}$
$\displaystyle\overset{w}{\longrightarrow}J(\nabla\psi(\mathbf{X}))^{\top}J(\nabla\psi(\mathbf{X}))+\|J(\nabla\psi(\mathbf{X}))\|^{2}+\|J(\nabla\psi(\mathbf{X}))\|^{2}.$

Since the above random variables are nonnegative, by Fatou’s lemma, the
expectation w.r.t. $\mu_{n}(\omega)$ satisfies:

$\displaystyle\liminf_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\left[J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))^{\top}J(\nabla\psi(\mathbf{X}_{i}))+\|J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))\|^{2}+\|J(\nabla\psi(\mathbf{X}_{i}))\|^{2}\right]$
$\displaystyle\geq\mathbb{E}\left[3\left\|J(\nabla\psi(\mathbf{X}))\right\|^{2}\right]=3\mathbb{E}\left\|J(R(\mathbf{X}))\right\|^{2}.$

Note that the above holds for a.e. $\omega$. Hence we can take the usual
expectation and again by Fatou’s lemma:

$\displaystyle\quad\liminf_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left[J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))^{\top}J(\nabla\psi(\mathbf{X}_{i}))+\|J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))\|^{2}+\|J(\nabla\psi(\mathbf{X}_{i}))\|^{2}\right]$
$\displaystyle\geq\mathbb{E}\left[\liminf_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\left[J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))^{\top}J(\nabla\psi(\mathbf{X}_{i}))+\|J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))\|^{2}+\|J(\nabla\psi(\mathbf{X}_{i}))\|^{2}\right]\right]$
$\displaystyle\geq\mathbb{E}\left[3\mathbb{E}\left\|J(R(\mathbf{X}))\right\|^{2}\right]=3\mathbb{E}\left\|J(R(\mathbf{X}))\right\|^{2}.$

Note that
$\mathbb{E}\left\|J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))\right\|^{2}=\mathbb{E}\left\|J(R_{n}(\mathbf{X}_{1}))\right\|^{2}$
for all $i$, and they converge to
$\mathbb{E}\left\|J(R(\mathbf{X}))\right\|^{2}$ following from the convergence
of the second moments of $J\\#\nu_{n}$ (4-$(ii)$). Therefore,

$\liminf_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left[J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))^{\top}J(\nabla\psi(\mathbf{X}_{i}))\right]\geq\mathbb{E}\left\|J(R(\mathbf{X}))\right\|^{2}.$

Aggregating the previous discussions, we have

$\displaystyle\quad\limsup_{n\to\infty}\mathbb{E}\left\|\frac{1}{\sqrt{n}}\sum_{i=1}^{n}S_{n}(\mathbf{X}_{i})J(R_{n}(\mathbf{X}_{i}))-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}S(\mathbf{X}_{i},R(\mathbf{X}_{i}))J(R(\mathbf{X}_{i}))\right\|^{2}$
$\displaystyle=\limsup_{n\to\infty}\Bigg{[}\mathbb{E}\left\|J(R_{n}(\mathbf{X}_{1}))\right\|^{2}+\mathbb{E}\left\|J(R(\mathbf{X}_{1}))\right\|^{2}-\frac{2}{n}\sum_{i=1}^{n}\mathbb{E}\left[J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))^{\top}J(\nabla\psi(\mathbf{X}_{i}))\right]\Bigg{]}$
$\displaystyle\leq\lim_{n\to\infty}\mathbb{E}\left\|J(R_{n}(\mathbf{X}_{1}))\right\|^{2}+\mathbb{E}\left\|J(R(\mathbf{X}_{1}))\right\|^{2}-2\liminf_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left[J(S_{n}(\mathbf{X}_{i})R_{n}(\mathbf{X}_{i}))^{\top}J(\nabla\psi(\mathbf{X}_{i}))\right]$
$\displaystyle\leq\mathbb{E}\left\|J(R(\mathbf{X}))\right\|^{2}+\mathbb{E}\left\|J(R(\mathbf{X}))\right\|^{2}-2\mathbb{E}\left\|J(R(\mathbf{X}))\right\|^{2}=0,$

which completes the proof. ∎

###  B.8 Proof of Theorem 3.4

Let $Q_{n}$ be the law of $(\mathbf{X}_{1},\ldots,\mathbf{X}_{n})$ under
$\textrm{H}_{1}$, and $P_{n}$ be the law of
$(\mathbf{X}_{1},\ldots,\mathbf{X}_{n})$ under $\textrm{H}_{0}$. Under the
regularity conditions in Section A.1, if
${\bm{\xi}}_{n}\to{\bm{\xi}}\in\mathbb{R}^{p}$, then by Van der Vaart, [99,
Theorem 7.2], under $\textrm{H}_{0}$:

$\log\frac{{\rm d}Q_{n}}{{\rm
d}P_{n}}=\sum_{i=1}^{n}\log\frac{f(\mathbf{X}_{i}-{\bm{\xi}}_{n}/\sqrt{n})}{f(\mathbf{X}_{i})}=-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X}_{i})}{f(\mathbf{X}_{i})}-\frac{1}{2}{\bm{\xi}}^{\top}I_{{\bm{\theta}}_{0}}{\bm{\xi}}+o_{p}(1),$

where $I_{{\bm{\theta}}_{0}}$ is the Fisher information at
${\bm{\theta}}_{0}=\mathbf{0}$.

By Lemma 3.1 and the multivariate central limit theorem:

$\displaystyle\quad\begin{pmatrix}\mathbf{W}_{n}\\\ \log\frac{{\rm
d}Q_{n}}{{\rm d}P_{n}}\\\
\end{pmatrix}=\begin{pmatrix}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}J(\nabla\psi(\mathbf{X}_{i}))\\\
-\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X}_{i})}{f(\mathbf{X}_{i})}-\frac{1}{2}{\bm{\xi}}^{\top}I_{{\bm{\theta}}_{0}}{\bm{\xi}}\\\
\end{pmatrix}+o_{p}(1)$
$\displaystyle\overset{d}{\longrightarrow}N\left(\begin{pmatrix}\mathbf{0}\\\
-\frac{{\bm{\xi}}^{\top}I_{{\bm{\theta}}_{0}}{\bm{\xi}}}{2}\\\
\end{pmatrix},\begin{pmatrix}{\rm
Var}(J(\nabla\psi(\mathbf{X})))&{\bm{\gamma}}\\\
{\bm{\gamma}}^{\top}&{\bm{\xi}}^{\top}I_{{\bm{\theta}}_{0}}{\bm{\xi}}\\\
\end{pmatrix}\right),$

where
${\bm{\gamma}}=-\mathbb{E}_{\textrm{H}_{0}}\left[J(\nabla\psi(\mathbf{X}))\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]$. Now, by Le Cam’s third lemma [99,
Example 6.7], under $\textrm{H}_{1}$,

$\mathbf{W}_{n}\overset{d}{\longrightarrow}N\left({\bm{\gamma}},{\rm
Var}\big{(}J(\nabla\psi(\mathbf{X}))\big{)}\right)=N\left({\bm{\gamma}},\Sigma_{\rm
ERD}\right).$

∎

###  B.9 Proof for Definition 3.2

We first consider a sequence
${\bm{\theta}}_{n}=\frac{{\bm{\xi}}_{n}}{\sqrt{n}}$, where
${\bm{\xi}}_{n}\to{\bm{\xi}}\neq\mathbf{0}$. From Section B.8 we know that
under ${\bm{\theta}}_{n}=\frac{{\bm{\xi}}_{n}}{\sqrt{n}}$,

$\Sigma_{\rm ERD}^{-1/2}\mathbf{W}_{n}\overset{d}{\longrightarrow}\Sigma_{\rm
ERD}^{-1/2}{\bm{\gamma}}+N(\mathbf{0},I_{p}),\quad{\rm
and}\quad\sqrt{n}S_{n}^{-1/2}\overline{\mathbf{X}}_{n}\overset{d}{\longrightarrow}\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}+N(\mathbf{0},I_{p}),$

with
${\bm{\gamma}}\equiv{\bm{\gamma}}({\bm{\xi}})=-\mathbb{E}_{\textrm{H}_{0}}\left[J(\nabla\psi(\mathbf{X}))\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]$.

Hence, the level $\alpha$ GWSR test and Hotelling’s $T^{2}$ test based on $n$
observations have powers converging to:

$\mathbb{P}\left(\left\|\Sigma_{\rm
ERD}^{-1/2}{\bm{\gamma}}+N(\mathbf{0},I_{p})\right\|^{2}\geq\chi_{p}^{2}(\alpha)\right),\quad{\rm
and}\quad\mathbb{P}\left(\left\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}+N(\mathbf{0},I_{p})\right\|^{2}\geq\chi_{p}^{2}(\alpha)\right),$

respectively, where $\chi_{p}^{2}(\alpha)\in\mathbb{R}$ satisfies
$\mathbb{P}\left(\chi_{p}^{2}\geq\chi_{p}^{2}(\alpha)\right)=\alpha$.

If we use $C_{1}n$ observations for the GWSR test with some $C_{1}>0$, then
under
${\bm{\theta}}_{n}=\frac{{\bm{\xi}}_{n}}{\sqrt{n}}=\frac{\sqrt{C_{1}}{\bm{\xi}}_{n}}{\sqrt{C_{1}n}}$,
where $\sqrt{C_{1}}{\bm{\xi}}_{n}\to\sqrt{C_{1}}{\bm{\xi}}$, the power of the
level $\alpha$ GWSR test converges to

$\mathbb{P}\left(\left\|\Sigma_{\rm
ERD}^{-1/2}\sqrt{C_{1}}{\bm{\gamma}}+N(\mathbf{0},I_{p})\right\|^{2}\geq\chi_{p}^{2}(\alpha)\right).$

Note that for a given $\beta\in(\alpha,1)$, there exists a unique $C_{1}>0$
such that

$\mathbb{P}\left(\left\|\Sigma_{\rm
ERD}^{-1/2}\sqrt{C_{1}}{\bm{\gamma}}+N(\mathbf{0},I_{p})\right\|^{2}\geq\chi_{p}^{2}(\alpha)\right)=\beta.$

Similarly, there exists a unique $C_{2}>0$ such that under
${\bm{\theta}}_{n}=\frac{{\bm{\xi}}_{n}}{\sqrt{n}}$, the level $\alpha$
Hotelling’s $T^{2}$ test based on $C_{2}n$ observations has power converging
to

$\mathbb{P}\left(\left\|\Sigma_{\mathbf{X}}^{-1/2}\sqrt{C_{2}}{\bm{\xi}}+N(\mathbf{0},I_{p})\right\|^{2}\geq\chi_{p}^{2}(\alpha)\right)=\beta.$

This implies $\|\Sigma_{\rm
ERD}^{-1/2}\sqrt{C_{1}}{\bm{\gamma}}\|=\|\Sigma_{\mathbf{X}}^{-1/2}\sqrt{C_{2}}{\bm{\xi}}\|$
in order to achieve the same power $\beta$, and

$\lim_{n\to\infty}\frac{N_{2}\left(\alpha,\beta,\frac{{\bm{\xi}}_{n}}{\sqrt{n}}\right)}{N_{1}\left(\alpha,\beta,\frac{{\bm{\xi}}_{n}}{\sqrt{n}}\right)}=\lim_{n\to\infty}\frac{C_{2}n+o(n)}{C_{1}n+o(n)}=\frac{\|\Sigma_{\rm
ERD}^{-1/2}{\bm{\gamma}}\|^{2}}{\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}}.$

Note that for any $\varepsilon_{n}{\bm{\xi}}_{n}\to 0$ such that
$\varepsilon_{n}\to 0$ and ${\bm{\xi}}_{n}\to{\bm{\xi}}\neq 0$, any
subsequence of $\\{\varepsilon_{n}{\bm{\xi}}_{n}\\}_{n\geq 1}$ has a further
subsequence that is contained in a sequence of the form
$\\{\frac{\widetilde{{\bm{\xi}}}_{n}}{\sqrt{n}}\\}_{n\geq 1}$ with
$\widetilde{{\bm{\xi}}}_{n}\to{\bm{\xi}}$. This implies the
$\frac{1}{\sqrt{n}}$ rate is not important, and we also have
$\lim_{n\to\infty}\frac{N_{2}\left(\alpha,\beta,\varepsilon_{n}{\bm{\xi}}_{n}\right)}{N_{1}\left(\alpha,\beta,\varepsilon_{n}{\bm{\xi}}_{n}\right)}=\frac{\|\Sigma_{\rm
ERD}^{-1/2}{\bm{\gamma}}\|^{2}}{\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}}$.

###  B.10 Proof of 3.2

Note that all the score functions $J(\cdot)$ used in this proof satisfy 3, and
hence Theorem 3.4 applies.

Part 1. (Gaussian ERD with identity score function): From Theorem 2.1 we know
that $\nabla\psi$ is the optimal transport map under the square loss that
transports $\mu=\mu_{S}$ to $\nu_{S}=N(\mathbf{0},I)$ under ${\rm H}_{0}$, and
it is also the unique map pushing $\mu$ to $N(\mathbf{0},I)$ that appears as
the gradient of a convex function [73].

In this Gaussian case, $\nabla\psi(\mathbf{x})=\Sigma^{-1/2}\mathbf{x}$, which
maps $N(\mathbf{0},\Sigma)$ to $N(\mathbf{0},I)$, and is the gradient of a
convex function. Let $\mathbf{b}:=\Sigma^{-1/2}{\bm{\xi}}$ and $\mathbf{G}\sim
N(0,I)$. Then:

$\displaystyle\|\Sigma_{\rm
ERD}^{-1/2}{\bm{\gamma}}\|^{2}=\|{\bm{\gamma}}\|^{2}=\left\|\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}$
$\displaystyle=\left\|\mathbb{E}\left[\Sigma^{-1/2}\mathbf{X}\cdot{\bm{\xi}}^{\top}\Sigma^{-1}\mathbf{X}\right]\right\|^{2}$
$\displaystyle=\left\|\mathbb{E}\left[\mathbf{G}\cdot\mathbf{b}^{\top}\mathbf{G}\right]\right\|^{2}$
$\displaystyle=\|\mathbf{b}\|^{2}=\|\Sigma^{-1/2}{\bm{\xi}}\|^{2}.$

This implies ${\rm
ARE}(\mathbf{W}_{n},\overline{\mathbf{X}}_{n};{\bm{\xi}})=1$.

Part 2. (Uniform ERD): In this case, we still have $\nu_{S}=N(\mathbf{0},I)$
so $\nabla\psi(\mathbf{x})=\Sigma^{-1/2}\mathbf{x}$. For ${\rm ERD}={\rm
Uniform}(-1,1)^{p}$, $\Sigma_{\rm ERD}=\frac{1}{3}I$. Using the integration by
parts formula,

$\displaystyle\|\Sigma_{\rm ERD}^{-1/2}{\bm{\gamma}}\|^{2}$
$\displaystyle=3\|{\bm{\gamma}}\|^{2}=3\left\|\mathbb{E}_{\textrm{H}_{0}}\left[J(\nabla\psi(\mathbf{X}))\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}$
$\displaystyle=3\sum_{j=1}^{p}\left(\int
J(\nabla\psi(\mathbf{x}))_{j}\sum_{i=1}^{p}\xi_{i}\frac{\partial}{\partial
x_{i}}f(\mathbf{x}){\rm d}\mathbf{x}\right)^{2}$
$\displaystyle=3\sum_{j=1}^{p}\left(\sum_{i=1}^{p}\mathbb{E}\left[\xi_{i}\frac{\partial}{\partial
X_{i}}J(\nabla\psi(\mathbf{X}))_{j}\right]\right)^{2}$
$\displaystyle=3\sum_{j=1}^{p}\left(\sum_{i=1}^{p}\mathbb{E}\left[\xi_{i}\frac{\partial}{\partial
X_{i}}\left(2\Phi(e_{j}^{\top}\Sigma^{-1/2}\mathbf{X})-1\right)\right]\right)^{2}$
$\displaystyle=3\sum_{j=1}^{p}\left(\sum_{i=1}^{p}\mathbb{E}\left[2\xi_{i}\Sigma^{-1/2}_{ji}\phi(e_{j}^{\top}\Sigma^{-1/2}\mathbf{X})\right]\right)^{2},$

where in the third line, we have used the integration by parts. Here
$\phi(\cdot)$ denotes the density function of the standard normal
distribution. From Deb et al., [25, Eq. (C.32) in Appendix],
$\mathbb{E}\left[\phi(e_{j}^{\top}\Sigma^{-1/2}\mathbf{X})\right]=\frac{1}{2\sqrt{\pi}}$.
Hence, the above equation can be further simplified to

$\displaystyle
3\sum_{j=1}^{p}\left(\sum_{i=1}^{p}\xi_{i}\Sigma^{-1/2}_{ji}\frac{1}{\sqrt{\pi}}\right)^{2}=\frac{3}{\pi}\|\Sigma^{-1/2}{\bm{\xi}}\|^{2}.$

which yields the desired result.

Part 3. (Spherical uniform ERD with identity score function): Write
$\nabla\psi(\mathbf{x})=(r_{1}(\mathbf{x}),\ldots,r_{p}(\mathbf{x}))$. For the
spherical uniform distribution $\Sigma_{\rm ERD}=\frac{1}{3p}I$. Using the
integration by parts formula,

$\displaystyle\|\Sigma_{\rm ERD}^{-1/2}{\bm{\gamma}}\|^{2}$
$\displaystyle=3p\|{\bm{\gamma}}\|^{2}=3p\left\|\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}$
$\displaystyle=3p\sum_{j=1}^{p}\left(\sum_{i=1}^{p}\mathbb{E}\left[\xi_{i}\frac{\partial}{\partial
X_{i}}r_{j}(\mathbf{X})\right]\right)^{2}$
$\displaystyle=3\|\Sigma^{-1/2}{\bm{\xi}}\|^{2}\cdot\frac{1}{p}\left[\frac{1}{2^{p-1}}\cdot\frac{\Gamma(p-0.5)}{(\Gamma(p/2))^{2}}+\frac{\sqrt{2\pi}(p-1)}{2^{p/2}\Gamma(p/2)}\mathbb{E}[|G_{1}|^{p-2}]-\mathcal{C}_{p}\right]^{2},$

where the last equality follows from Deb et al., [25, Eq. (C.44) in Appendix]
with the notations therein. Hence, ${\rm
ARE}(\mathbf{W}_{n},\overline{\mathbf{X}}_{n};{\bm{\xi}})=\kappa_{p}$. ∎

###  B.11 Proof of Theorem 3.5

Part 1. (${\rm ERD}={\rm Uniform}(-1,1)^{p}$ with identity score function): In
this case $\Sigma_{\rm ERD}=\frac{1}{3}I$, and it can be shown that

$\nabla\psi(\mathbf{x})=\left(2F_{1}(x_{1})-1,\ldots,2F_{p}(x_{p})-1\right)^{\top},$

where $F_{i}$ is the cumulative distribution function (cdf) for
$\widetilde{f}_{i}$. To see this, note that the above map pushes $\mu$ to
${\rm Uniform}(-1,1)^{p}$, and is the gradient of the convex function:
$2\sum_{i=1}^{p}\int_{-\infty}^{x_{i}}F_{i}(t_{i}){\rm
d}{t_{i}}-\mathbf{1}^{\top}\mathbf{x}$. Hence,

$\displaystyle\left\|\Sigma_{\rm
ERD}^{-1/2}\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}$
$\displaystyle=3\left\|\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}$
$\displaystyle=3\sum_{i=1}^{p}\left(\mathbb{E}_{\textrm{H}_{0}}\left[(2F_{i}(X_{i})-1)\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right)^{2}$
$\displaystyle=12\sum_{i=1}^{p}\left(\mathbb{E}_{\textrm{H}_{0}}\left[F_{i}(X_{i})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right)^{2}$
$\displaystyle=12\sum_{i=1}^{p}\left(\int
F_{i}(x_{i})\xi_{i}\widetilde{f}_{i}^{\prime}(x_{i}){\rm d}x_{i}\right)^{2}$
$\displaystyle=12\sum_{i=1}^{p}\left(\xi_{i}\int\widetilde{f}_{i}^{2}(x_{i}){\rm
d}x_{i}\right)^{2}.$

Let $\sigma_{i}^{2}$ be the variance of $F_{i}$. Then
$\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}={\bm{\xi}}^{\top}\Sigma_{\mathbf{X}}^{-1}{\bm{\xi}}=\sum_{i=1}^{p}\frac{\xi_{i}^{2}}{\sigma^{2}_{i}}$.

From the derivation of ARE of Wilcoxon’s test w.r.t. the $t$-test in one-
dimensional case [57], we know:

$\sigma_{i}^{2}\left(\int\widetilde{f}_{i}^{2}(x_{i}){\rm
d}x\right)^{2}\geq\frac{9}{125}.$

This implies

${\rm
ARE}(\mathbf{W}_{n},\overline{\mathbf{X}}_{n};{\bm{\xi}})=\frac{\left\|\Sigma_{\rm
ERD}^{-1/2}\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}}{\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}}\geq\frac{108}{125}.$

Part 2. (Gaussian ERD with identity score function): In this case $\Sigma_{\rm
ERD}=I$, and

$\nabla\psi(\mathbf{x})=\left(\Phi^{-1}\circ
F_{1}(x_{1}),\ldots,\Phi^{-1}\circ F_{p}(x_{p})\right).$

To see this, note that the above map pushes $\mu$ to $N(\mathbf{0},I)$, and is
the gradient of the convex function
$\psi(\mathbf{x}):=\sum_{i=1}^{p}\int_{-\infty}^{x_{i}}\Phi^{-1}\circ
F_{i}(t_{i}){\rm d}{t_{i}}$. Hence,

$\displaystyle\left\|\Sigma_{\rm
ERD}^{-1/2}\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}$
$\displaystyle=\left\|\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}$ (B.10)
$\displaystyle=\sum_{i=1}^{p}\left(\mathbb{E}_{\textrm{H}_{0}}\left[\Phi^{-1}\big{(}F_{i}(X_{i})\big{)}\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right)^{2}$
$\displaystyle=\sum_{i=1}^{p}\left(\int\Phi^{-1}\big{(}F_{i}(x_{i})\big{)}\xi_{i}\widetilde{f}_{i}^{\prime}(x_{i})\,{\rm
d}x_{i}\right)^{2}$
$\displaystyle=\sum_{i=1}^{p}\left(\xi_{i}\int\frac{1}{\phi\left(\Phi^{-1}\big{(}F_{i}(x_{i})\big{)}\right)}\widetilde{f}_{i}^{2}(x_{i})\,{\rm
d}x_{i}\right)^{2},$

where the last equality follows from integration by parts, with $\phi$ being
the density of $N(0,1)$. It follows from the optimization problem in one-
dimension [37, Theorem 2.1] that

$\sigma_{i}^{2}\left(\int\frac{1}{\phi\left(\Phi^{-1}\big{(}F_{i}(x_{i})\big{)}\right)}\widetilde{f}_{i}^{2}(x_{i})\,{\rm
d}x_{i}\right)^{2}\geq 1,$ (B.11)

and the minimizing density is that of $N(0,\sigma^{2})$ for any $\sigma>0$.
This implies:

${\rm
ARE}(\mathbf{W}_{n},\overline{\mathbf{X}}_{n};{\bm{\xi}})=\frac{\sum_{i=1}^{p}\left(\xi_{i}\int\frac{1}{\phi\left(\Phi^{-1}\big{(}F_{i}(x_{i})\big{)}\right)}\widetilde{f}_{i}^{2}(x_{i})\,{\rm
d}x_{i}\right)^{2}}{\sum_{i=1}^{p}\frac{\xi_{i}^{2}}{\sigma^{2}_{i}}}\geq 1.$
(B.12)

∎

###  B.12 Proof of Theorem 3.6

We assume the same regularity conditions on $\underline{f}$ as in Deb et al.,
[25, Appendix C.5.3]:

  * •

$\int_{\mathbb{R}^{+}}r^{d+1}\underline{f}(r){\rm d}r<\infty$.

  * •

$\sqrt{\underline{f}}(\cdot)$ admits a weak derivative, which is denoted by
$\left(\sqrt{\underline{f}}\right)^{\prime}(\cdot)$. This means that for all
$\psi:\mathbb{R}^{+}\to\mathbb{R}^{+}$ which is compactly supported and
infinitely differentiable, we have

$\int_{\mathbb{R}^{+}}\sqrt{\underline{f}(r)}\psi^{\prime}(r)\,{\rm
d}r=-\int_{\mathbb{R}^{+}}\left(\sqrt{\underline{f}}\right)^{\prime}(r)\psi(r){\rm
d}r.$

  * •

$\int_{\mathbb{R}^{+}}r^{d-1}\left[\left(\sqrt{\underline{f}}\right)^{\prime}(r)\right]^{2}\,{\rm
d}r<\infty.$

Part 1. (Spherical uniform ERD with identity score): Write
$\overline{\mathbf{X}}:=\Sigma^{-1/2}(\mathbf{X}-{\bm{\theta}})$, which has a
spherically symmetric distribution under the null, and its covariance matrix
is given by $\frac{1}{p}\mathbb{E}\|\overline{\mathbf{X}}\|^{2}I_{p}$. Hence,
$\Sigma_{\mathbf{X}}=\frac{1}{p}\mathbb{E}\|\overline{\mathbf{X}}\|^{2}\cdot\Sigma$
and

$\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}={\bm{\xi}}^{\top}\Sigma_{\mathbf{X}}^{-1}{\bm{\xi}}=\frac{p}{\mathbb{E}\|\overline{\mathbf{X}}\|^{2}}\cdot{\bm{\xi}}^{\top}\Sigma^{-1}{\bm{\xi}}.$

From Deb et al., [25, Eq. (C.55) in Appendix],

$\left\|\Sigma_{\rm
ERD}^{-1/2}\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}=\frac{3}{p}\cdot{\bm{\xi}}^{\top}\Sigma^{-1}{\bm{\xi}}\left(\mathbb{E}\overline{h}\left(\|\overline{\mathbf{X}}\|\right)+(p-1)\mathbb{E}\left[\frac{\overline{H}\left(\|\overline{\mathbf{X}}\|\right)}{\|\overline{\mathbf{X}}\|}\right]\right)^{2},$

where $\overline{H}$ is the distribution function of
$\|\overline{\mathbf{X}}\|$, and $\overline{h}$ is the corresponding density
function. Hence,

$\displaystyle{\rm ARE}(\mathbf{W}_{n},\overline{\mathbf{X}}_{n};{\bm{\xi}})$
$\displaystyle=\frac{\left\|\Sigma_{\rm
ERD}^{-1/2}\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}}{\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}}$
$\displaystyle=\frac{3}{p^{2}}\cdot\mathbb{E}\|\overline{\mathbf{X}}\|^{2}\cdot\left(\mathbb{E}\overline{h}\left(\|\overline{\mathbf{X}}\|\right)+(p-1)\mathbb{E}\left[\frac{\overline{H}\left(\|\overline{\mathbf{X}}\|\right)}{\|\overline{\mathbf{X}}\|}\right]\right)^{2}$
$\displaystyle\geq\frac{81}{500}\cdot\frac{(\sqrt{2p-1}+1)^{5}}{p^{2}(\sqrt{2p-1}+5)},$

where the inequality follows from the optimization problem solved in Deb et
al., [25, Eq. (C.56) in Appendix]. The density that attains the minimum is
also given in Deb et al., [25, Appendix C.5.3].

Part 2. (Gaussian ERD with identity score function): From Deb et al., [25,
Lemma C.2],

$\nabla\psi(\mathbf{X})=\frac{\Sigma^{-1/2}(\mathbf{X}-{\bm{\theta}})}{\|\Sigma^{-1/2}(\mathbf{X}-{\bm{\theta}})\|}\cdot
H_{p}^{-1}\left(\overline{H}\left(\|\Sigma^{-1/2}(\mathbf{X}-{\bm{\theta}})\|\right)\right),$

where $H_{p}$ is the distribution function of a $\sqrt{\chi_{p}^{2}}$
distribution. From Deb et al., [25, Eq. (C.60) in Appendix]:

$\displaystyle{\rm ARE}(\mathbf{W}_{n},\overline{\mathbf{X}}_{n};{\bm{\xi}})$
$\displaystyle=\frac{\left\|\Sigma_{\rm
ERD}^{-1/2}\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}}{\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}}$
$\displaystyle=\frac{\mathbb{E}\|\overline{\mathbf{X}}\|^{2}}{p^{3}}\cdot\left\\{\mathbb{E}\left[\frac{\overline{h}\left(\|\overline{\mathbf{X}}\|\right)}{h_{p}(H_{p}^{-1}(\overline{H}(\|\overline{\mathbf{X}}\|)))}\right]+(p-1)\mathbb{E}\left[\frac{H_{d}^{-1}\circ\overline{H}\left(\|\overline{\mathbf{X}}\|\right)}{\|\overline{\mathbf{X}}\|}\right]\right\\}^{2}$
$\displaystyle\geq 1.$

The equality is attained if and only if the infimum in Deb et al., [25, Eq.
(C.60) in Appendix] is attained, i.e., when $f(\cdot)$ is a multivariate
normal distribution [25, Theorem 3.5]. ∎

###  B.13 Proof of Theorem 3.7

Suppose $\widetilde{f}_{i}(\cdot)$ has variance $\sigma_{i}^{2}$. Then
$\mathbf{W}$ has covariance matrix ${\rm
diag}(\sigma_{1}^{2},\ldots,\sigma_{p}^{2})$, and $\Sigma_{\mathbf{X}}={\rm
Cov}(A\mathbf{W})=A\,{\rm
diag}(\sigma_{1}^{2},\ldots,\sigma_{p}^{2})A^{\top}$,
$\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}={\bm{\xi}}^{\top}A\,{\rm
diag}\left(\frac{1}{\sigma_{1}^{2}},\ldots,\frac{1}{\sigma_{p}^{2}}\right)A^{\top}{\bm{\xi}}$.
Observe that under the null, with $A=(a_{i,j})$, for
$\mathbf{x}=(x_{1},\ldots,x_{p})\in\mathbb{R}^{p}$,

$f(\mathbf{x})=\prod_{i=1}^{p}\widetilde{f}_{i}\left(\sum_{j=1}^{p}a_{j,i}x_{j}\right).$

Hence, with $w_{i}=\sum_{j=1}^{p}a_{j,i}x_{j}$ (or
$\mathbf{w}=A^{\top}\mathbf{x}$),

$\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}=\sum_{i,j=1}^{p}\xi_{i}a_{i,j}\frac{\widetilde{f}_{j}^{\prime}}{\widetilde{f}_{j}}(w_{j}).$

Let $\widetilde{F}_{i}$ be the distribution function corresponding to
$\widetilde{f}_{i}$. Let
$R_{\mathbf{W}}(\mathbf{w})=\left(\Phi^{-1}\circ\widetilde{F}_{1}(w_{1}),\ldots,\Phi^{-1}\circ\widetilde{F}_{p}(w_{p})\right)$.
By Ghosal and Sen, [38, Lemma A.8],
$\nabla\psi(\mathbf{X})=AR_{\mathbf{W}}(A^{\top}\mathbf{X})$. Hence,

$\displaystyle\left\|\Sigma_{\rm
ERD}^{-1/2}\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}$
$\displaystyle=\left\|\mathbb{E}_{\textrm{H}_{0}}\left[AR_{\mathbf{W}}(A^{\top}\mathbf{X})\sum_{i,j=1}^{p}\xi_{i}a_{i,j}\frac{\widetilde{f}_{j}^{\prime}}{\widetilde{f}_{j}}(W_{j})\right]\right\|^{2}$
$\displaystyle=\left\|\mathbb{E}_{\textrm{H}_{0}}\left[R_{\mathbf{W}}(\mathbf{W})\sum_{i,j=1}^{p}\xi_{i}a_{i,j}\frac{\widetilde{f}_{j}^{\prime}}{\widetilde{f}_{j}}(W_{j})\right]\right\|^{2}$
$\displaystyle=\sum_{k=1}^{p}\left(\mathbb{E}_{\textrm{H}_{0}}\left[\Phi^{-1}\circ\widetilde{F}_{k}(W_{k})\sum_{i,j=1}^{p}\xi_{i}a_{i,j}\frac{\widetilde{f}_{j}^{\prime}}{\widetilde{f}_{j}}(W_{j})\right]\right)^{2}.$

Under the null, $W_{j}$ is independent of $W_{k}$ when $j\neq k$ and thus
$\mathbb{E}_{\textrm{H}_{0}}\left[\Phi^{-1}\circ\widetilde{F}_{k}(W_{k})\frac{\widetilde{f}_{j}^{\prime}}{\widetilde{f}_{j}}(W_{j})\right]=0$.
Hence

$\displaystyle\mathbb{E}_{\textrm{H}_{0}}\left[\Phi^{-1}\circ\widetilde{F}_{k}(W_{k})\sum_{i,j=1}^{p}\xi_{i}a_{i,j}\frac{\widetilde{f}_{j}^{\prime}}{\widetilde{f}_{j}}(W_{j})\right]$
$\displaystyle=\sum_{i=1}^{p}\xi_{i}a_{i,k}\cdot\mathbb{E}_{\textrm{H}_{0}}\left[\Phi^{-1}\circ\widetilde{F}_{k}(W_{k})\frac{\widetilde{f}_{k}^{\prime}}{\widetilde{f}_{k}}(W_{k})\right]$
$\displaystyle\geq\sum_{i=1}^{p}\xi_{i}a_{i,k}\cdot\frac{1}{\sigma_{k}}=\frac{(A^{\top}{\bm{\xi}})_{k}}{\sigma_{k}},$

where the inequality follows from (B.10) and (B.11). Hence,

$\displaystyle{\rm ARE}(\mathbf{W}_{n},\overline{\mathbf{X}}_{n};{\bm{\xi}})$
$\displaystyle=\frac{\left\|\Sigma_{\rm
ERD}^{-1/2}\mathbb{E}_{\textrm{H}_{0}}\left[\nabla\psi(\mathbf{X})\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\right\|^{2}}{\|\Sigma_{\mathbf{X}}^{-1/2}{\bm{\xi}}\|^{2}}\geq\frac{\sum_{k=1}^{p}\left(\frac{(A^{\top}{\bm{\xi}})_{k}}{\sigma_{k}}\right)^{2}}{{\bm{\xi}}^{\top}A\,{\rm
diag}\left(\frac{1}{\sigma_{1}^{2}},\ldots,\frac{1}{\sigma_{p}^{2}}\right)A^{\top}{\bm{\xi}}}=1.$

∎

###  B.14 Proof of Theorem 3.9

If the model $(P_{\theta}:{\bm{\theta}}\in\Theta)$ is differentiable in
quadratic mean, then the local experiments converge to a Gaussian limit [99,
Section 15.3]:

$\left(P_{{\bm{\theta}}_{0}+{\bm{\xi}}/\sqrt{n}}^{n}:{\bm{\xi}}\in\mathbb{R}^{p}\right)\to\left(N({\bm{\xi}},I_{{\bm{\theta}}_{0}}^{-1}):{\bm{\xi}}\in\mathbb{R}^{p}\right),$

where $I_{{\bm{\theta}}_{0}}$ is the Fisher information at
${\bm{\theta}}_{0}=\mathbf{0}_{p}$.

1. We consider testing (3.7). The asymptotic representation theorem [99, Theorem 15.1] shows that if we have a sequence of level $\alpha$ tests with power function $\pi_{n}(\cdot)$ such that $\pi_{n}({\bm{\xi}}/\sqrt{n})\to\pi({\bm{\xi}})$ for every ${\bm{\xi}}$ (as $n\to\infty$) and some function $\pi$, then $\pi(\cdot)$ is the power function of a level $\alpha$ test in the Gaussian limit experiment. Now let us consider the Gaussian limit experiment: For testing $\widetilde{{\rm H}}_{0}:{\bm{\xi}}=\mathbf{0}_{p}$ versus $\widetilde{\rm H}_{1}:{\bm{\xi}}\neq\mathbf{0}_{p}$ based on an observation $\mathbf{Y}\sim N({\bm{\xi}},I_{{\bm{\theta}}_{0}}^{-1})$, the level $\alpha$ test that maximizes the minimum power over $\\{{\bm{\xi}}:\|I^{1/2}_{{\bm{\theta}}_{0}}{\bm{\xi}}\|=c\\}$ has the critical region $\\{\mathbf{y}\in\mathbb{R}^{p}:\|I^{1/2}_{{\bm{\theta}}_{0}}\mathbf{y}\|^{2}\geq\chi^{2}_{p}(\alpha)\\}$, for all $c>0$ [99, Problem 15.4]. This implies

$\inf_{{\bm{\xi}}:\|I^{1/2}_{{\bm{\theta}}_{0}}{\bm{\xi}}\|=c}\lim_{n\to\infty}\pi_{n}\left(\frac{{\bm{\xi}}}{\sqrt{n}}\right)\leq\mathbb{P}\left(\|N\big{(}{\bm{\xi}}_{c},I_{p}\big{)}\|^{2}\geq\chi_{p}^{2}(\alpha)\right),$
(B.13)

for any ${\bm{\xi}}_{c}\in\mathbb{R}^{p}$ with $\|{\bm{\xi}}_{c}\|=c$.

Next, we show that by choosing a proper score function $J(\cdot)$, we can
achieve the above optimal power (on the right side of (B.13)). From Theorem
3.4, under ${\rm H}_{1}$,

$\mathbf{W}_{n}\overset{d}{\longrightarrow}N\left({\bm{\gamma}},\Sigma_{\rm
ERD}\right),$

where
${\bm{\gamma}}=-\mathbb{E}_{\textrm{H}_{0}}\left[J(\nabla\psi(\mathbf{X}))\frac{{\bm{\xi}}^{\top}\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right]\in\mathbb{R}^{p}$. In Theorem 2.1, we
show that $\nabla\psi(\mathbf{x})$ is the optimal transport map under square
loss that pushes $P_{{\bm{\theta}}_{0}}$ to $\nu_{S}$. If $\nu_{S}$ has a
Lebesgue density, then McCann’s theorem [73, Remark 16] states that
$\nabla\psi(\cdot)$ admits an almost sure inverse
$\nabla\psi^{*}:\mathbb{R}^{p}\to\mathbb{R}^{p}$ such that
$\nabla\psi^{*}(\nabla\psi(\mathbf{x}))=\mathbf{x}$ ($P_{{\bm{\theta}}_{0}}$
a.e.-$\mathbf{x}$) and $\nabla\psi\circ\nabla\psi^{*}(\mathbf{x})=\mathbf{x}$
($\nu_{S}$ a.e.-$\mathbf{x}$).

With $J(\cdot)=-\frac{\nabla
f(\nabla\psi^{*}(\cdot))}{f(\nabla\psi^{*}(\cdot))}$,
$J(\nabla\psi(\mathbf{x}))=-\frac{\nabla f(\mathbf{x})}{f(\mathbf{x})}$
($P_{{\bm{\theta}}_{0}}$ a.e.-$\mathbf{x}$). Observe that,

$J(Q\nabla\psi(\mathbf{x}))=J(\nabla\psi(Q\mathbf{x}))=-\frac{\nabla
f(Q\mathbf{x})}{f(Q\mathbf{x})},$

for any $Q\in\mathcal{G}$. Since $f$ is $\mathcal{G}$-symmetric,
$f(Q\mathbf{x})=f(\mathbf{x})$ for all $Q\in\mathcal{G}$ and
$\mathbf{x}\in\mathbb{R}^{p}$, which further implies $\nabla
f(Q\mathbf{x})=Q\nabla f(\mathbf{x})$. Hence,
$J(Q\psi(\mathbf{x}))=QJ(\psi(\mathbf{x}))$, showing that $J(\cdot)$ is
$\mathcal{G}$-equivariant ($\nu_{S}$ a.e.). With this choice of $J$,

$\displaystyle{\bm{\gamma}}$
$\displaystyle=\mathbb{E}_{\textrm{H}_{0}}\left[\frac{\nabla
f(\mathbf{X})}{f(\mathbf{X})}\frac{\nabla
f(\mathbf{X})^{\top}}{f(\mathbf{X})}\right]{\bm{\xi}}=I_{{\bm{\theta}}_{0}}{\bm{\xi}},$
$\displaystyle\Sigma_{\rm ERD}$ $\displaystyle={\rm Var}_{\rm
H_{0}}\left(J(\nabla\psi(\mathbf{X}))\right)={\rm Var}_{\rm
H_{0}}\left(\frac{\nabla
f(\mathbf{X})}{f(\mathbf{X})}\right)=I_{{\bm{\theta}}_{0}}.$

Hence, under ${\rm H}_{1}$,
$\mathbf{W}_{n}\overset{d}{\longrightarrow}N\left(I_{{\bm{\theta}}_{0}}{\bm{\xi}},I_{{\bm{\theta}}_{0}}\right)$,
and therefore

$\mathbb{P}_{{\bm{\theta}}_{0}+\frac{{\bm{\xi}}}{\sqrt{n}}}\left(\mathbf{W}_{n}^{\top}\Sigma_{\rm
ERD}^{-1}\mathbf{W}_{n}\geq\chi_{p}^{2}(\alpha)\right)\to\mathbb{P}\left(\|N(I_{{\bm{\theta}}_{0}}^{1/2}{\bm{\xi}},I_{p})\|^{2}\geq\chi_{p}^{2}(\alpha)\right).$

Combining with (B.13), we have

$\inf_{{\bm{\xi}}:\|I^{1/2}_{{\bm{\theta}}_{0}}{\bm{\xi}}\|=c}\lim_{n\to\infty}\pi_{n}\left(\frac{{\bm{\xi}}}{\sqrt{n}}\right)\leq\inf_{{\bm{\xi}}:\|I^{1/2}_{{\bm{\theta}}_{0}}{\bm{\xi}}\|=c}\lim_{n\to\infty}\mathbb{P}_{{\bm{\theta}}_{0}+\frac{{\bm{\xi}}}{\sqrt{n}}}\left(\mathbf{W}_{n}^{\top}\Sigma_{\rm
ERD}^{-1}\mathbf{W}_{n}\geq\chi_{p}^{2}(\alpha)\right),$

which yields the desired result.

2. Recall that $\pi_{n}(\cdot)$ is the power function of a sequence of level $\alpha$ tests for (3.8). By Van der Vaart, [99, Theorem 15.4],

$\limsup_{n\to\infty}\pi_{n}\left({\bm{\theta}}_{0}+\frac{{\bm{\xi}}}{\sqrt{n}}\right)\leq
1-\Phi\left(z_{\alpha}-\frac{{\bm{\xi}}^{\top}\nabla\zeta({\bm{\theta}}_{0})}{\sqrt{\nabla\zeta({\bm{\theta}}_{0})^{\top}I_{{{\bm{\theta}}_{0}}}^{-1}\nabla\zeta({\bm{\theta}}_{0})}}\right),$
(B.14)

where $\Phi(\cdot)$ is the cdf of a standard normal distribution, and
$z_{\alpha}$ satisfies $\Phi(z_{\alpha})=1-\alpha$.

From the proof of part-1.,
$\mathbf{W}_{n}\overset{d}{\longrightarrow}N\left(I_{{\bm{\theta}}_{0}}{\bm{\xi}},I_{{\bm{\theta}}_{0}}\right)$
under ${\bm{\theta}}={{\bm{\theta}}_{0}+\frac{{\bm{\xi}}}{\sqrt{n}}}$, and
therefore

$\nabla\zeta({\bm{\theta}}_{0})^{\top}I_{{\bm{\theta}}_{0}}^{-1}\mathbf{W}_{n}\overset{d}{\longrightarrow}N\big{(}{\bm{\xi}}^{\top}\nabla\zeta({\bm{\theta}}_{0}),\nabla\zeta({\bm{\theta}}_{0})^{\top}I_{{{\bm{\theta}}_{0}}}^{-1}\nabla\zeta({\bm{\theta}}_{0})\big{)}.$

This implies

$\lim_{n\to\infty}\mathbb{P}_{{\bm{\theta}}_{0}+\frac{{\bm{\xi}}}{\sqrt{n}}}\left(\frac{\nabla\zeta({\bm{\theta}}_{0})^{\top}I_{{\bm{\theta}}_{0}}^{-1}\mathbf{W}_{n}}{\sqrt{\nabla\zeta({\bm{\theta}}_{0})^{\top}I_{{{\bm{\theta}}_{0}}}^{-1}\nabla\zeta({\bm{\theta}}_{0})}}\geq
z_{\alpha}\right)=1-\Phi\left(z_{\alpha}-\frac{{\bm{\xi}}^{\top}\nabla\zeta({\bm{\theta}}_{0})}{\sqrt{\nabla\zeta({\bm{\theta}}_{0})^{\top}I_{{{\bm{\theta}}_{0}}}^{-1}\nabla\zeta({\bm{\theta}}_{0})}}\right),$

which attains the upper bound in (B.14).

∎

##  Appendix C Some Auxiliary Results

In the following, we state and prove some auxiliary results. Results of a
similar form can be found in McCann, [73, Lemma 9] and Villani, [101, Theorem
5.10].

######  Proposition C.1.

If $c(\cdot,\cdot)$ is continuous, then there exists $\pi\in\Pi(\mu,\nu)$ with
a $c$-cyclically monotone support (see Definition 2.5).

###### Proof.

Let $\mu_{n}:=\frac{1}{n}\sum_{i=1}^{n}\delta_{\mathbf{x}_{i}}$ and
$\nu_{n}:=\frac{1}{n}\sum_{i=1}^{n}\delta_{\mathbf{y}_{i}}$ be such that
$\mu_{n}\overset{w}{\longrightarrow}\mu$ and
$\nu_{n}\overset{w}{\longrightarrow}\nu$ (for example, we can draw
$\mathbf{x}_{i}$ i.i.d. from $\mu$ and $\mathbf{y}_{i}$ i.i.d. from $\nu$).
For each $n$, solve

$\min_{a_{ij}\geq
0}\sum_{i,j=1}^{n}a_{ij}c(\mathbf{x}_{i},\mathbf{y}_{j}),\qquad{\rm\ s.t.\
}\sum_{j=1}^{n}a_{ij}=1,i=1,\ldots,n,{\rm\ and\
}\sum_{i=1}^{n}a_{ij}=1,j=1,\ldots,n$ (C.1)

to obtain a minimizer $(a_{ij}^{n})$. In fact, this is the assignment problem
[78, 11] and there exists an optimal solution such that
$a_{ij}^{n}\in\\{0,1\\}$ for all $i,j$. Let
$\pi_{n}:=\sum_{i,j=1}^{n}a_{ij}^{n}\delta_{(\mathbf{x}_{i},\mathbf{y}_{j})}$.
Then $\pi_{n}$ is supported on $S$, the set of all couples
$(\mathbf{x}_{i},\mathbf{y}_{j})$ such that $a_{ij}^{n}>0$. To show that
$\pi_{n}\in\Pi(\mu_{n},\nu_{n})$ has a $c$-cyclically monotone support, we
argue by contradiction. Suppose that there exists $N\in\mathbb{N}$ and
$(\mathbf{x}_{i_{1}},\mathbf{y}_{i_{1}}),\ldots,(\mathbf{x}_{i_{N}},\mathbf{y}_{i_{N}})$
in $S$ such that (with $\mathbf{y}_{i_{N+1}}=\mathbf{y}_{i_{1}}$)

$c(\mathbf{x}_{i_{1}},\mathbf{y}_{i_{2}})+c(\mathbf{x}_{i_{2}},\mathbf{y}_{i_{3}})+\ldots+c(\mathbf{x}_{i_{N}},\mathbf{y}_{i_{N+1}})<c(\mathbf{x}_{i_{1}},\mathbf{y}_{i_{1}})+\ldots+c(\mathbf{x}_{i_{N}},\mathbf{y}_{i_{N}}).$

Define

$\widetilde{\pi}_{n}:=\pi_{n}+\varepsilon\sum_{l=1}^{N}\left(\delta_{(\mathbf{x}_{i_{l}},\mathbf{y}_{i_{l+1}})}-\delta_{(\mathbf{x}_{i_{l}},\mathbf{y}_{i_{l}})}\right).$

Suppose
$\widetilde{\pi}_{n}=\sum_{i,j=1}^{n}\widetilde{a}_{ij}^{n}\delta_{(\mathbf{x}_{i},\mathbf{y}_{j})}$.
Then, for a sufficiently small $\varepsilon>0$, $\widetilde{a}_{ij}^{n}\geq
0$, and $(\widetilde{a}_{ij}^{n})$ has a smaller cost than $({a}_{ij}^{n})$ in
(C.1), which yields a contradiction.

Hence, $\pi_{n}\in\Pi(\mu_{n},\nu_{n})$ has a $c$-cyclically monotone support.
The weak convergence of $\mu_{n}$ and $\nu_{n}$ implies that $\\{\mu_{n}\\}$
and $\\{\nu_{n}\\}$ are tight, and thus $\\{\pi_{n}\\}$ must be tight
(Villani, [101, Lemma 4.4]). By Prokhorov’s theorem, there is a subsequence
$\\{\pi_{n_{k}}\\}_{k\geq 1}$ which converges weakly to some probability
measure $\pi$. Since $(\mathbf{x},\mathbf{y})\mapsto\mathbf{x}$ and
$(\mathbf{x},\mathbf{y})\mapsto\mathbf{y}$ are continuous, by continuous
mapping theorem, $\pi$ has marginals $\mu$ and $\nu$. For each $n$, the
$c$-cyclically monotonicity of $\pi_{n}$ implies that for all $N\in\mathbb{N}$
and $\pi_{n}^{\otimes N}$-almost all
$(\mathbf{x}_{1},\mathbf{y}_{1}),\ldots,(\mathbf{x}_{N},\mathbf{y}_{N})$, we
have (with $\mathbf{y}_{N+1}=\mathbf{y}_{1}$)

$\sum_{i=1}^{N}c(\mathbf{x}_{i},\mathbf{y}_{i})\leq\sum_{i=1}^{N}c(\mathbf{x}_{i},\mathbf{y}_{i+1}).$
(C.2)

In other words, $\pi_{n}^{\otimes N}$ is concentrated on the set
$\mathcal{C}(N)$ of all
$((\mathbf{x}_{1},\mathbf{y}_{1}),\ldots,(\mathbf{x}_{N},\mathbf{y}_{N}))\in(\mathbb{R}^{p}\times\mathbb{R}^{p})^{N}$
satisfying (C.2). Since $c(\cdot,\cdot)$ is continuous, $\mathcal{C}(N)$ is a
closed set, so the weak limit $\pi^{\otimes N}$ of $\pi^{\otimes N}_{n_{k}}$
is also concentrated on $\mathcal{C}(N)$ by Portmanteau theorem. Let $\Gamma$
be the support of $\pi$. Then $\Gamma^{N}=({\rm supp}(\pi))^{N}={\rm
supp}(\pi^{\otimes N})\subset\mathcal{C}(N)$. Since this holds true for all
$N$, $\Gamma$ is $c$-cyclically monotone. ∎

######  Proposition C.2.

If $c(\cdot,\cdot)$ is continuous and $\pi\in\Gamma(\mu,\nu)$ has
$c$-cyclically monotone support, then there exists a $c$-convex function
$\psi$ such that $\partial_{c}\psi$ (see Definition B.1) contains the support
of $\pi$.

###### Proof.

Let $\Gamma$ denote the support of $\pi$. Fix
$(\mathbf{x}_{0},\mathbf{y}_{0})\in\Gamma$. Define

$\displaystyle\psi(\mathbf{x})$
$\displaystyle:=\sup_{m\in\mathbb{N}}\sup\big{\\{}[c(\mathbf{x}_{0},\mathbf{y}_{0})-c(\mathbf{x}_{1},\mathbf{y}_{0})]+[c(\mathbf{x}_{1},\mathbf{y}_{1})-c(\mathbf{x}_{2},\mathbf{y}_{1})]$
$\displaystyle\quad+\ldots+[c(\mathbf{x}_{m},\mathbf{y}_{m})-c(\mathbf{x},\mathbf{y}_{m})]:(\mathbf{x}_{1},\mathbf{y}_{1}),\ldots,(\mathbf{x}_{m},\mathbf{y}_{m})\in\Gamma\big{\\}}.$

From the definition, with $m=1$ and
$(\mathbf{x}_{1},\mathbf{y}_{1})=(\mathbf{x}_{0},\mathbf{y}_{0})$, we have
$\psi(\mathbf{x}_{0})\geq 0$. On the other hand, the $c$-cyclical monotonicity
of $\Gamma$ implies that $\psi(\mathbf{x}_{0})\leq 0$, and hence
$\psi(\mathbf{x}_{0})=0$. By renaming $\mathbf{y}_{m}$ as $\mathbf{y}$,

$\displaystyle\psi(\mathbf{x})$
$\displaystyle=\sup_{\mathbf{y}\in\mathbb{R}^{p}}\sup_{m\in\mathbb{N}}\sup_{(\mathbf{x}_{1},\mathbf{y}_{1}),\ldots,(\mathbf{x}_{m-1},\mathbf{y}_{m-1}),\mathbf{x}_{m}}\big{\\{}[c(\mathbf{x}_{0},\mathbf{y}_{0})-c(\mathbf{x}_{1},\mathbf{y}_{0})]+[c(\mathbf{x}_{1},\mathbf{y}_{1})-c(\mathbf{x}_{2},\mathbf{y}_{1})]$
$\displaystyle\quad+\ldots+[c(\mathbf{x}_{m},\mathbf{y})-c(\mathbf{x},\mathbf{y})]:(\mathbf{x}_{1},\mathbf{y}_{1}),\ldots,(\mathbf{x}_{m},\mathbf{y})\in\Gamma\big{\\}}$
$\displaystyle=\sup_{\mathbf{y}\in\mathbb{R}^{p}}[\zeta(\mathbf{y})-c(\mathbf{x},\mathbf{y})],$

where $\zeta(\mathbf{y}):=-\infty$ if there does not exist $\mathbf{x}_{m}$
such that $(\mathbf{x}_{m},\mathbf{y})\in\Gamma$, and otherwise

$\displaystyle\zeta(\mathbf{y})$
$\displaystyle:=\sup_{m\in\mathbb{N}}\sup_{(\mathbf{x}_{1},\mathbf{y}_{1}),\ldots,(\mathbf{x}_{m-1},\mathbf{y}_{m-1}),\mathbf{x}_{m}}\big{\\{}[c(\mathbf{x}_{0},\mathbf{y}_{0})-c(\mathbf{x}_{1},\mathbf{y}_{0})]+[c(\mathbf{x}_{1},\mathbf{y}_{1})-c(\mathbf{x}_{2},\mathbf{y}_{1})]$
$\displaystyle\quad+\ldots+[c(\mathbf{x}_{m},\mathbf{y})-c(\mathbf{x},\mathbf{y})]:(\mathbf{x}_{1},\mathbf{y}_{1}),\ldots,(\mathbf{x}_{m},\mathbf{y})\in\Gamma\big{\\}}.$

Thus $\psi(\cdot)$ is a $c$-convex function. For any
$(\overline{\mathbf{x}},\overline{\mathbf{y}})\in\Gamma$, with
$(\mathbf{x}_{m},\mathbf{y}_{m})=(\overline{\mathbf{x}},\overline{\mathbf{y}})$
in the definition of $\psi$,

$\displaystyle\psi(\mathbf{x})$ $\displaystyle\geq\sup_{m\geq
2}\sup_{(\mathbf{x}_{1},\mathbf{y}_{1}),\ldots,(\mathbf{x}_{m-1},\mathbf{y}_{m-1})\in\Gamma}\big{\\{}[c(\mathbf{x}_{0},\mathbf{y}_{0})-c(\mathbf{x}_{1},\mathbf{y}_{0})]+[c(\mathbf{x}_{1},\mathbf{y}_{1})-c(\mathbf{x}_{2},\mathbf{y}_{1})]$
$\displaystyle\quad+\ldots+[c(\mathbf{x}_{m-1},\mathbf{y}_{m-1})-c(\overline{\mathbf{x}},\mathbf{y}_{m-1})]+[c(\overline{\mathbf{x}},\overline{\mathbf{y}})-c(\mathbf{x},\overline{\mathbf{y}})]\big{\\}}$
$\displaystyle=\psi(\overline{\mathbf{x}})+[c(\overline{\mathbf{x}},\overline{\mathbf{y}})-c(\mathbf{x},\overline{\mathbf{y}})].$

Hence,
$\psi(\mathbf{x})+c(\mathbf{x},\overline{\mathbf{y}})\geq\psi(\overline{\mathbf{x}})+c(\overline{\mathbf{x}},\overline{\mathbf{y}})$.
Taking the infimum over $\mathbf{x}\in\mathbb{R}^{p}$, we deduce that
$\psi^{c}(\overline{\mathbf{y}})\geq\psi(\overline{\mathbf{x}})+c(\overline{\mathbf{x}},\overline{\mathbf{y}})$.
Since the reverse inequality always holds from the definition of $\psi^{c}$
(Definition B.1), we have
$\psi^{c}(\overline{\mathbf{y}})=\psi(\overline{\mathbf{x}})+c(\overline{\mathbf{x}},\overline{\mathbf{y}})$.
This means that
$(\overline{\mathbf{x}},\overline{\mathbf{y}})\in\partial_{c}\psi$, and so
$\Gamma\subset\partial_{c}\psi$. ∎

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