# Qubit State Discrimination using Post-measurement Information

Donghoon Ha  Department of Applied Mathematics and Institute of Natural
Sciences, Kyung Hee University, Yongin 17104, Republic of Korea  Jeong San Kim
<EMAIL_ADDRESS>Department of Applied Mathematics and Institute of Natural
Sciences, Kyung Hee University, Yongin 17104, Republic of Korea  Younghun Kwon
<EMAIL_ADDRESS>Department of Applied Physics, Center for Bionano
Intelligence Education and Research, Hanyang University, Ansan 15588, Republic
of Korea

###### Abstract

We consider the optimal discrimination of nonorthogonal qubit states with
post-measurement information and provide an analytic structure of the optimal
measurements. We also show that there is always a null optimal measurement
when post-measurement information is given. Further, in discriminating four
states using post-measurement information, we analytically provide the optimal
probability of correct guessing and show that the uniqueness of optimal
measurement is equivalent to the non-existence of non-null optimal measurement
with post-measurement information.

##  I Introduction

Whereas orthogonal quantum states can be perfectly discriminated in quantum
physics, it is not generally true for nonorthogonal statesref:chef2000 ;
ref:barn20091 ; ref:berg2010 ; ref:bae2015 . For these reasons, various
measurement strategies have been studied for optimal discrimination of
nonorthogonal states, such as _minimum-error discrimination_(ME), unambiguous
discrimination, and maximum-confidence discriminationref:hels1976 ;
ref:ivan1987 ; ref:diek1988 ; ref:pere1988 ; ref:jaeg1995 ; ref:ha2015 ;
ref:crok2006 . ME is a discrimination scheme to minimize the average error
probability without inconclusive results. Although a necessary and sufficient
condition for realizing a minimum-error measurement in general cases is well
knownref:hole1979 ; ref:yuen1975 ; ref:elda20031 ; ref:barn20092 , the general
solution for ME is not yet known except for ME of two states, symmetric
states, and qubit statesref:hels1976 ; ref:hunt2004 ; ref:sams2009 ;
ref:jafa2011 ; ref:ha2013 ; ref:ha2014 ; ref:ban1997 ; ref:chou2003 ;
ref:bae20131 ; ref:bae20132 . In some cases, ME can be performed without the
help of measurement, simply by guessing the state with the greatest prior
probability is preparedref:hunt2003 .
When the post-measurement information about the prepared subensemble is
available, some nonorthogonal states can be perfectly
discriminatedref:akib2019 . However, in general, nonorthogonal qubit states
cannot be perfectly discriminated even with post-measurement information about
the prepared subensemble. Therefore, for the case of qubit state, it is
important to investigate _minimizing the average error probability with post-
measurement information_(MEPI)ref:ball2008 ; ref:gopa2010 ; ref:carm2018 .
Also, it is meaningful since MEPI is known to have a relation with the
incompatibility of measurementsref:carm2018 ; ref:hein2016 ; ref:carm2019 ;
ref:skrz20191 ; ref:uola2019 ; ref:skrz20192 .
MEPI problem can be understood in view of ME problem; a MEPI of quantum state
ensemble consisting of $m$ subensembles with $n_{1},\ldots,n_{m}$ states can
be translated into a ME of quantum state ensemble with $\prod_{b=1}^{m}n_{b}$
states by modifying the states and prior probabilities in the original
ensembleref:gopa2010 . This approach can be useful for characterizing MEPI of
qubit states because useful properties and analytical results for ME of qubit
states are already well knownref:hunt2004 ; ref:ha2013 ; ref:ha2014 ;
ref:bae20131 ; ref:bae20132 .
In this paper, we analyze MEPI of nonorthogonal qubit states and provide an
analytic structure of the optimal measurements based on the analysis of some
ME problem. We first show that a null optimal measurement exists for any MEPI
of qubit states. We also analytically provide a necessary and sufficient
condition that pre-measurement information is strictly more favorable than
post-measurement information when all subensembles have two states. Moreover,
we characterize the optimal measurements by classifying MEPI into the two
cases if it is possible or not without the help of measurement. In particular,
for the case where the ensemble consists of two subensembles with two states
and pre-measurement information is strictly more favorable than post-
measurement information, we analytically provide the optimal probability of
correct guessing. In this case, we further show that the uniqueness of optimal
measurement is equivalent to the non-existence of non-null MEPI measurement.
This paper is organized as follows: In Sect. II, we review and analyze ME of
qubit states. By applying the analysis for ME of qubit states to the ME
problem associated with MEPI problem, we provide our results for MEPI of qubit
states in Sect. III. In Sect. IV, we conclude our results.

##  II Minimum-error Discrimination of Qubit States

In two-level quantum systems (qubit), a state is expressed by a density
operator on two-dimensional complex Hilbert space $\mathcal{H}$ and a
measurement with a finite outcome set $\Omega=\\{1,\ldots,n\\}$ is described
by a _positive operator valued measure_(POVM) $\mathcal{M}$, which is composed
of $n$ positive semidefinite operators $M_{1},\ldots,M_{n}$ satisfying
$\sum_{i\in\Omega}M_{i}=\mathbbm{1}$. Here, $\mathbbm{1}$ is the identity
operator on $\mathcal{H}$. We say that $i\in\Omega$ is null(non-null) if
$M_{i}\in\Omega$ is zero(non-zero). We also say that $\mathcal{M}$ is _null_
if it has at least one null outcome, otherwise _non-null_.
In this section, we consider ME of _qubit_ state ensemble
$\mathcal{E}=\\{\eta_{i},\rho_{i}\\}_{i\in\Omega}$ in which the qubit state
$\rho_{i}$ is prepared with the probability $\eta_{i}$. We specify $\eta_{1}$
as the greatest prior probability to reduce the repetitive representation,
that is,

$\eta_{1}\geqslant\eta_{2},\ldots,\eta_{n}.$ (1)

A qubit state $\rho_{i}$ can be described using the Bloch vector
$\bm{\nu}_{i}$ in the three-dimensional real space $\mathbb{R}^{3}$,

$\begin{array}[]{c}\rho_{i}=\frac{1}{2}(\mathbbm{1}+\bm{\nu}_{i}\cdot\bm{\sigma}),\
i\in\Omega.\end{array}$

where $\bm{\sigma}$ is the Pauli matrices
$(\sigma_{X},\sigma_{Y},\sigma_{Z})$.
Given a qubit state ensemble
$\mathcal{E}=\\{\eta_{i},\rho_{i}\\}_{i\in\Omega}$ and distinct
$i,j\in\Omega$, points in $\mathbb{R}^{3}$, which has $|\eta_{i}-\eta_{j}|$ as
distance difference from $\eta_{i}\bm{\nu}_{i}$ and $\eta_{j}\bm{\nu}_{j}$,
form a hyperboloid of two sheets. The one hyperboloid sheet consists of points
${\bm{v}}\in\mathbb{R}^{3}$ satisfying
$\varphi_{i}({\bm{v}})=\varphi_{j}({\bm{v}})$, where

$\varphi_{i}({\bm{v}})=\eta_{i}+\|\eta_{i}\bm{\nu}_{i}-{\bm{v}}\|,\
i\in\Omega.$ (2)

Here $\|\cdot\|$ is the Euclidean norm. The sheet divides $\mathbb{R}^{3}$
into two sets
$\\{\bm{v}\in\mathbb{R}^{3}:\varphi_{i}({\bm{v}})\geqslant\varphi_{j}({\bm{v}})\\}$
and
$\\{\bm{v}\in\mathbb{R}^{3}:\varphi_{i}({\bm{v}})<\varphi_{j}({\bm{v}})\\}$.
We use the following definitions to express various conditions in ME of
$\mathcal{E}$.

######  Definition 1.

For each $\mathsf{S}\subseteq\Omega$,

$\begin{array}[]{rcl}\mathcal{Z}_{\mathsf{S}}:=\\{{\bm{v}}\in\mathcal{P}_{\mathsf{S}}&:&\varphi_{i}({\bm{v}})=\varphi_{j}({\bm{v}})\
\forall(i,j)\in\mathsf{S}\times\mathsf{S},\\\
&&\varphi_{i}({\bm{v}})\geqslant\varphi_{j}({\bm{v}})\
\forall(i,j)\in\mathsf{S}\times(\Omega-\mathsf{S})\\},\end{array}$ (3)

where $\mathcal{P}_{\mathsf{S}}$ is the relative interior of the convex hull
of $\\{\eta_{i}\bm{\nu}_{i}\\}_{i\in\mathsf{S}}$, that is,

$\begin{array}[]{c}\mathcal{P}_{\mathsf{S}}=\\{\sum_{i\in\mathsf{S}}c_{i}\eta_{i}\bm{\nu}_{i}\,:\,c_{i}>0\,\forall
i\in\mathsf{S},\ \sum_{i\in\mathsf{S}}c_{i}=1\ \\}.\end{array}$ (4)

Also, we denote by $\mathcal{Z}$ the union of all $\mathcal{Z}_{\mathsf{S}}$,
that is,
$\mathcal{Z}=\bigcup_{\mathsf{S}\subseteq\Omega}\mathcal{Z}_{\mathsf{S}}$.

In ME of $\mathcal{E}$, we use a POVM $\mathcal{M}=\\{M_{i}\\}_{i\in\Omega}$
as a measurement such that the prepared state is guessed to be $\rho_{i}$ for
each $i\in\Omega$. Then, the maximal average probability of correctly guessing
the given qubit state is

$p_{\rm guess}=\max_{\mathcal{M}}\sum_{i\in\Omega}\eta_{i}{\rm
Tr}[\rho_{i}M_{i}].$ (5)

ME is the task of finding optimal measurements that provides $p_{\rm guess}$,
called the _guessing probability_. We also note that the guessing probability
in ME cannot be less than the greatest prior probability, that is,

$p_{\rm guess}\geqslant\eta_{1}.$ (6)

In some cases, ME can be performed without the help of measurementref:hunt2003
; the guessing probability can be obtained by taking the state with the
greatest prior probability as the prepared state, that is, $p_{\rm
guess}=\eta_{1}$. Even for that case, nontrivial optimal measurements can
possibly exist. The following proposition provides a necessary and sufficient
condition for $\mathcal{M}$ to be optimal when $p_{\rm guess}=\eta_{1}$. The
proof of Proposition 1 is given in Appendix A.

######  Proposition 1.

For ME of qubit state ensemble $\mathcal{E}$,

  1. (a)

$p_{\rm guess}=\eta_{1}$ if and only if $\mathcal{Z}_{\\{1\\}}$ is not an
empty set $\varnothing$ or, equivalently,

$\epsilon_{i}\geqslant\lambda_{i}\ \forall i\in\Omega,$ (7)

where

$\epsilon_{i}=\eta_{1}-\eta_{i},\
\lambda_{i}=\|\eta_{1}\bm{\nu}_{1}-\eta_{i}\bm{\nu}_{i}\|,\ i\in\Omega.$ (8)

  2. (b)

When $p_{\rm guess}=\eta_{1}$, a POVM $\mathcal{M}$ is optimal if and only if

$\forall i\neq 1,\ M_{i}\ \propto\
\left\\{\begin{array}[]{ccc}\mathbbm{1}+\frac{\eta_{i}\bm{\nu}_{i}-\eta_{1}\bm{\nu}_{1}}{\|\eta_{i}\bm{\nu}_{i}-\eta_{1}\bm{\nu}_{1}\|}\cdot\bm{\sigma}&,&\epsilon_{i}=\lambda_{i},\\\
0&,&\epsilon_{i}>\lambda_{i}.\end{array}\right.$ (9)

When $p_{\rm guess}=\eta_{1}$, if $\epsilon_{i}>\lambda_{i}$ for all $i\neq
1$, the optimal measurement has only $1\in\Omega$ as a non-null outcome, and
$M_{1}$ becomes the identity operator. That is, the optimal measurement in
this case is trivial and unique. However, if $\epsilon_{i}=\lambda_{i}$ for
some $i\neq 1$, the optimal measurement can have a non-null outcome other than
1, which implies a non-trivial optimal measurement. Therefore, when $p_{\rm
guess}=\eta_{1}$, a nontrivial optimal measurement exists if and only if
$\epsilon_{i}=\lambda_{i}$ for some $i\neq 1$.
Now, let us consider the qubit state ensemble $\mathcal{E}$ in which a
measurement is a necessary requirement for ME, that is, $p_{\rm
guess}>\eta_{1}$. The following proposition shows that finding
$\mathsf{S}\subseteq\Omega$ with $\mathcal{Z}_{\mathsf{S}}\neq\varnothing$ and
an element of $\mathcal{Z}$ is directly related to obtaining optimal
measurements, where $\varnothing$ is the empty set. The proof of Proposition 2
is given in Appendix A.

######  Proposition 2.

For ME of qubit state ensemble $\mathcal{E}$,

  1. (a)

$\mathcal{Z}$ is always a single-element set $\\{\bm{v}\\}$ and $p_{\rm
guess}=\varphi_{i}(\bm{v})$ for all $i$ in $\mathsf{S}$ with
$\mathcal{Z}_{\mathsf{S}}\neq\varnothing$.

When $p_{\rm guess}>\eta_{1}$,

  1. (b)

there is an optimal measurement having $\mathsf{S}\subseteq\Omega$ as the set
of all non-null outcomes if and only if
$\mathcal{Z}_{\mathsf{S}}\neq\varnothing$.

  2. (c)

Moreover, a POVM $\mathcal{M}$ having $\mathsf{S}$ as the set of all non-null
outcomes is optimal if and only if

$M_{i}\propto\mathbbm{1}+\frac{\eta_{i}\bm{\nu}_{i}-\bm{v}}{\|\eta_{i}\bm{\nu}_{i}-\bm{v}\|}\cdot\bm{\sigma}\
\ \forall i\in\mathsf{S}.$ (10)

When $p_{\rm guess}>\eta_{1}$, all optimal POVMs are characterized by the
_single_ element of $\mathcal{Z}$. Proposition 2 tells the following three
facts. First, if $\mathcal{Z}_{\mathsf{S}}$ is empty, there is no optimal
measurement having $\mathsf{S}$ as the set of all non-null outcomes. This
implies that all optimal measurements are null(that is, $M_{i}=0$ for some
$i\in\Omega$) if $\mathcal{Z}_{\Omega}$ is empty. Second, if
$\mathcal{Z}_{\mathsf{S}}$ is non-empty and
$\mathcal{Z}\subseteq\mathcal{P}_{\mathsf{S}^{\prime}}$ for some
$\mathsf{S}^{\prime}\subseteq\mathsf{S}$, then
$\mathcal{Z}_{\mathsf{S}^{\prime}}$ is also non-empty. This implies that, if
the _affine_ dimensionref:boyd2004 ; ref:affine  of
$\\{\eta_{i}\bm{\nu}_{i}\\}_{i\in\Omega}$ is $D$, the guessing probability can
be obtained by the detection of $D+1$ qubit states ref:ha2014 . Third, if
$\mathcal{Z}_{\mathsf{S}}$ is non-empty and
$\\{\eta_{i}\bm{\nu}_{i}\\}_{i\in\mathsf{S}}$ forms a simplex with affine
dimension $|\mathsf{S}|-1$, the optimal measurement having $\mathsf{S}$ as the
set of all non-null outcomes is unique.
Condition (10) is a necessary and sufficient condition for a POVM
$\mathcal{M}$ having $\mathsf{S}$ as the set of all non-null outcomes to be
optimal when $p_{\rm guess}>\eta_{1}$. Therefore, when $p_{\rm
guess}>\eta_{1}$, all optimal measurements are obtained in the following three
steps. The first step is to distinguish whether $\mathcal{Z}_{\mathsf{S}}$ is
empty or nonempty for each $\mathsf{S}$ with $|\mathsf{S}|\geqslant 2$. Note
that, for $p_{\rm guess}>\eta_{1}$, there exists no subset
$\mathsf{S}\subseteq\Omega$ with $|\mathsf{S}|=1$ and $\mathcal{Z}_{\sf
S}\neq\varnothing$ because all optimal measurements have more than one non-
null outcome. The second step is to find out what the single element of
$\mathcal{Z}$ is. The final step is to get a POVM $\mathcal{M}$ that satisfies
Condition (10).

##  III Main result: MEPI of Qubit States

In this section, we consider MEPI of _qubit_ state ensemble,

$\mathcal{E}=\bigcup_{b\in\mathsf{B}}\\{\eta_{ib},\rho_{ib}\\}_{i\in\mathsf{A}_{b}},$
(11)

where

$\begin{array}[]{lcll}\mathsf{B}&=&\\{1,2,\ldots,m\\},&m\geqslant 2,\\\
\mathsf{A}_{b}&=&\\{1,2,\ldots,n_{b}\\},&n_{b}\geqslant 2.\end{array}$ (12)

The ensemble $\mathcal{E}$ consists of $m$ subensembles(we use the term
“subensemble” regardless of the normalization of prior probability),

$\mathcal{E}_{b}=\\{\eta_{ib},\rho_{ib}\\}_{i\in\mathsf{A}_{b}},\
b\in\mathsf{B}.$ (13)

Without loss of generality, we assume that

$\eta_{1b}\geqslant\eta_{2b},\ldots,\eta_{n_{b}b}\ \forall b\in\mathsf{B}.$
(14)

The classical information $b\in\mathsf{B}$ of the prepared subensemble is
provided after a measurement is performed. We use a POVM
$\mathcal{M}=\\{M_{\bm{\omega}}\\}_{\bm{\omega}\in\Omega}$ to describe a
measurement, where $\Omega$ is the Cartesian product of
$\mathsf{A}_{1}$,$\mathsf{A}_{2}$,…,$\mathsf{A}_{m}$, that is,

$\Omega=\mathsf{A}_{1}\times\mathsf{A}_{2}\times\cdots\times\mathsf{A}_{m}.$
(15)

Each outcome $\bm{\omega}=(\omega_{1},\ldots,\omega_{m})\in\Omega$ means that
the prepared state is guessed to be $\rho_{\omega_{1}1}$,
$\rho_{\omega_{2}2}$, …, or $\rho_{\omega_{m}m}$ according to post-measurement
information $b=1$, $2$, …, or $m$, respectively.
The maximal average probability of correctly guessing the prepared qubit state
is

$p_{\rm guess}^{\rm
post}=\max_{\mathcal{M}}\sum_{b\in\mathsf{B}}\sum_{i\in\mathsf{A}_{b}}\sum_{\begin{subarray}{c}\bm{\omega}\in\Omega\\\
\omega_{b}=i\end{subarray}}\eta_{ib}{\rm
Tr}[\rho_{ib}M_{\bm{\omega}}]=\max_{\mathcal{M}}\sum_{\bm{\omega}\in\Omega}\tilde{\eta}_{\bm{\omega}}{\rm
Tr}[\tilde{\rho}_{\bm{\omega}}M_{\bm{\omega}}],$ (16)

where $\tilde{\eta}_{\bm{\omega}}$ and $\tilde{\rho}_{\bm{\omega}}$ are
positive numbers and density operators, respectively, such that

$\tilde{\eta}_{\bm{\omega}}=\sum_{b\in\mathsf{B}}\eta_{\omega_{b}b},\
\tilde{\rho}_{\bm{\omega}}=\frac{\sum_{b\in\mathsf{B}}\eta_{\omega_{b}b}\rho_{\omega_{b}b}}{\sum_{b^{\prime}\in\mathsf{B}}\eta_{\omega_{b^{\prime}}b^{\prime}}},\
\bm{\omega}\in\Omega.$ (17)

MEPI of $\mathcal{E}$ is the task of finding optimal measurements that
provides $p_{\rm guess}^{\rm post}$. From Eq. (16), we can see that $p_{\rm
guess}^{\rm post}$ is the guessing probability of qubit state ensemble,

$\tilde{\mathcal{E}}=\\{\tilde{\eta}_{\bm{\omega}},\tilde{\rho}_{\bm{\omega}}\\}_{\bm{\omega}\in\Omega}.$
(18)

where $\\{\tilde{\eta}_{\bm{\omega}}\\}_{\bm{\omega}\in\Omega}$ is not
normalized, that is,
$\sum_{\bm{\omega}\in\Omega}\tilde{\eta}_{\bm{\omega}}>1$. Therefore, MEPI of
$\mathcal{E}$ is equivalent to ME of $\tilde{\mathcal{E}}$; a POVM
$\mathcal{M}$ is optimal for MEPI of $\mathcal{E}$ if and only if it is
optimal for ME of $\tilde{\mathcal{E}}$ref:gopa2010 . To distinguish between
optimal measurements for ME and MEPI, we use ME and MEPI measurements,
respectively. We also note that the assumption in (14) implies that
$\tilde{\eta}_{\bm{1}}$ is the greatest prior probability of
$\tilde{\mathcal{E}}$ and a lower bound of $p_{\rm guess}^{\rm post}$, that
is,

$p_{\rm guess}^{\rm
post}\geqslant\tilde{\eta}_{\bm{1}}\geqslant\tilde{\eta}_{\bm{\omega}}\,\forall\bm{\omega}\in\Omega,$
(19)

where

$\bm{1}=(1,1,\ldots,1).$ (20)

###  III.1 Null MEPI measurement

Similar to ME, we use Bloch representation of qubit states as

$\begin{array}[]{lcll}\rho_{ib}&=&\frac{1}{2}(\mathbbm{1}+\bm{\nu}_{ib}\cdot\bm{\sigma}),&b\in\mathsf{B},i\in\mathsf{A}_{b},\\\\[2.84526pt]
\tilde{\rho}_{\bm{\omega}}&=&\frac{1}{2}(\mathbbm{1}+\tilde{\bm{\nu}}_{\bm{\omega}}\cdot\bm{\sigma}),&\bm{\omega}\in\Omega.\end{array}$
(21)

Then,

$\sum_{b\in\mathsf{B}}\eta_{\omega_{b}b}\bm{\nu}_{\omega_{b}b}=\tilde{\eta}_{\bm{\omega}}\tilde{\bm{\nu}}_{\bm{\omega}}=:\tilde{\bm{\mu}}_{\bm{\omega}}\quad\forall\bm{\omega}\in\Omega.$
(22)

When $m=n_{1}=n_{2}=2$, the affine dimension $D$ of
$\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\Omega}$ is less than
three because

$\begin{array}[]{rcl}\tilde{\bm{\mu}}_{(2,1)}-\tilde{\bm{\mu}}_{(1,1)}=\tilde{\bm{\mu}}_{(2,2)}-\tilde{\bm{\mu}}_{(1,2)}=\eta_{21}\bm{\nu}_{21}-\eta_{11}\bm{\nu}_{11},\\\
\tilde{\bm{\mu}}_{(1,2)}-\tilde{\bm{\mu}}_{(1,1)}=\tilde{\bm{\mu}}_{(2,2)}-\tilde{\bm{\mu}}_{(2,1)}=\eta_{22}\bm{\nu}_{22}-\eta_{12}\bm{\nu}_{12}.\end{array}$
(23)

Thus, in this case, ME of $\tilde{\mathcal{E}}$ is possible without detecting
every stateref:ha2014 , and there exists a MEPI measurement of $\mathcal{E}$
that is null(that is, $M_{\bm{\omega}}=0$ for some $\bm{\omega}\in\Omega$).
Other than $m=n_{1}=n_{2}=2$, the number of all outcomes, that is,
$|\Omega|=\prod_{b\in\mathsf{B}}n_{b}$, is greater than four, and a null MEPI
measurement of $\mathcal{E}$ exists because, for any ME of more than four
qubit states, there is a ME measurement that is nullref:ha2014 ; ref:davi1978
; ref:hunt2004 .

######  Corollary 1.

For any qubit state ensemble $\mathcal{E}$, a null MEPI measurement of
$\mathcal{E}$ exists.

###  III.2 Upper bound of $p_{\rm guess}^{\rm post}$

When the classical information $b\in\mathsf{B}$ of the prepared subensemble is
known prior to perform a measurement, the maximal average probability of
correctly guessing the prepared qubit state, $p_{\rm guess}^{\rm prior}$, can
be obtained by performing ME measurement of $\mathcal{E}_{b}$ according to the
pre-measurement information $b$. In other words,

$p_{\rm guess}^{\rm prior}=\sum_{b\in\mathsf{B}}p_{b}^{\mbox{\rm\tiny ME}},$
(24)

where $p_{b}^{\mbox{\rm\tiny ME}}$ is the guessing probability of
$\mathcal{E}_{b}$, that is,

$p_{b}^{\mbox{\rm\tiny ME}}=\max_{\mathcal{M}_{b}}\sum_{i\in{\sf
A}_{b}}\eta_{ib}{\rm Tr}[\rho_{ib}M_{ib}],\ b\in\mathsf{B}.$ (25)

Here, $\mathcal{M}_{b}$ is a POVM with $n_{b}$ elements $M_{ib}$ indicating
the detection of $\rho_{ib}$.
Guessing the prepared qubit state using a POVM $\mathcal{M}$ and post-
measurement information is equivalent to guessing the prepared state by
performing a POVM $\mathcal{M}_{b}$ consisting of

$M_{ib}=\sum_{\begin{subarray}{c}\bm{\omega}\in\Omega\\\
\omega_{b}=i\end{subarray}}M_{\bm{\omega}},\ i\in\mathsf{A}_{b},$ (26)

according to pre-measurement information $b\in\mathsf{B}$. Given an arbitrary
POVM $\mathcal{M}_{b}$ with $n_{b}$ elements $M_{1b},\ldots,M_{n_{b}b}$ for
each $b\in\mathsf{B}$, $m$ POVMs $\mathcal{M}_{1},\ldots,\mathcal{M}_{m}$ are
called _compatible_ if there is a POVM $\mathcal{M}$ satisfying Eq. (26) for
all $b\in\mathsf{B}$; otherwise, they are called _incompatible_ref:hein2016 .
Thus, $p_{\rm guess}^{\rm post}$ is upper bounded as

$p_{\rm guess}^{\rm post}\leqslant p_{\rm guess}^{\rm prior},$ (27)

where the equality holds if and only if there are $m$ ME measurements of
$\mathcal{E}_{1},\ldots,\mathcal{E}_{m}$ that are compatibleref:carm2018 .
Obviously, a POVM with the identity operator is compatible with any POVM;
therefore, $p_{\rm guess}^{\rm post}=p_{\rm guess}^{\rm prior}$ if
$p_{b}^{\mbox{\rm\tiny ME}}=\eta_{1b}$ for some $b\in\mathsf{B}$. Moreover,
POVMs with the same elements are compatible; thus, $p_{\rm guess}^{\rm
post}=p_{\rm guess}^{\rm prior}$ if there are $m$ optimal POVMs for MEs of
$\mathcal{E}_{1},\ldots,\mathcal{E}_{m}$ that have the same elements. The
following lemma provides a sufficient condition for $p_{\rm guess}^{\rm
post}<p_{\rm guess}^{\rm prior}$.

######  Lemma 1.

Suppose that, for some $b,b^{\prime}\in\mathsf{B}$ with $b\neq b^{\prime}$,
the ME measurements for $\mathcal{E}_{b}$ and $\mathcal{E}_{b^{\prime}}$ are
unique and consist of rank-one elements $\\{M_{ib}\\}_{i\in\mathsf{A}_{b}}$,
$\\{M_{jb^{\prime}}\\}_{j\in\mathsf{A}_{b^{\prime}}}$ such that
$M_{ib}\not\propto M_{jb^{\prime}}$ for all $i\in\mathsf{A}_{b}$ and all
$j\in\mathsf{A}_{b^{\prime}}$. Then, $p_{\rm guess}^{\rm post}<p_{\rm
guess}^{\rm prior}$.

###### Proof.

Assume that $p_{\rm guess}^{\rm post}=p_{\rm guess}^{\rm prior}$ in which Eq.
(26) holds for $b,b^{\prime}\in\mathsf{B}$. Then, ${\rm rank}(M_{ib})={\rm
rank}(M_{jb^{\prime}})=1$ implies that $M_{\bm{\omega}}\propto M_{ib}$ for all
$\bm{\omega}\in\Omega$ with $\omega_{b}=i$ and $M_{\bm{\omega}}\propto
M_{jb^{\prime}}$ for all $\bm{\omega}\in\Omega$ with $\omega_{b^{\prime}}=j$.
Thus, $M_{ib}\not\propto M_{jb^{\prime}}$ $\forall i,j$ means
$M_{ib}=M_{jb^{\prime}}=0$ $\forall i,j$ which contradicts the POVM
completeness. Therefore, $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm prior}$.
∎

In the case of $n_{b}=2$ and $p_{b}^{\mbox{\rm\tiny ME}}>\eta_{1b}$ for all
$b\in\mathsf{B}$, from Helstrom boundref:hels1976  or Proposition 2, the
optimal POVM elements for ME of $\mathcal{E}_{b}$ are uniquely determined as
follows:

$M_{1b}=\frac{1}{2}(\mathbbm{1}+\hat{\bm{\mu}}_{2b}\cdot\bm{\sigma}),\
M_{2b}=\frac{1}{2}(\mathbbm{1}-\hat{\bm{\mu}}_{2b}\cdot\bm{\sigma}),\
b\in\mathsf{B},$ (28)

where

$\hat{\bm{\mu}}_{ib}=\frac{\eta_{ib}\bm{\nu}_{ib}-\eta_{1b}\bm{\nu}_{1b}}{\|\eta_{ib}\bm{\nu}_{ib}-\eta_{1b}\bm{\nu}_{1b}\|},\
b\in\mathsf{B}.$ (29)

Thus, if $\hat{\bm{\mu}_{2b}}\times\hat{\bm{\mu}}_{2b^{\prime}}=\bm{0}$ for
all $b,b^{\prime}\in\mathsf{B}$, then $p_{\rm guess}^{\rm post}=p_{\rm
guess}^{\rm prior}$ because $m$ optimal POVMs for MEs of
$\mathcal{E}_{1},\ldots,\mathcal{E}_{m}$ have the same elements; however, if
$\hat{\bm{\mu}}_{2b}\times\hat{\bm{\mu}}_{2b^{\prime}}\neq\bm{0}$ for some
$b,b^{\prime}\in\mathsf{B}$ with $b\neq b^{\prime}$, then $p_{\rm guess}^{\rm
post}>p_{\rm guess}^{\rm prior}$ from Lemma 1.

######  Corollary 2.

When $n_{b}=2$ for all $b\in\mathsf{B}$, $p_{\rm guess}^{\rm post}<p_{\rm
guess}^{\rm prior}$ if and only if $\epsilon_{2b}<\lambda_{2b}$ for all
$b\in\mathsf{B}$ and
$\hat{\bm{\mu}}_{2b}\times\hat{\bm{\mu}}_{2b^{\prime}}\neq\bm{0}$ for some
$b,b^{\prime}\in\mathsf{B}$ with $b\neq b^{\prime}$.

###  III.3 MEPI for $p_{\rm guess}^{\rm post}=\tilde{\eta}_{\bm{1}}$

From Inequalities (19) and (27), $p_{\rm guess}^{\rm post}$ has the following
upper and lower bounds,

$\tilde{\eta}_{\bm{1}}\leqslant p_{\rm guess}^{\rm post}\leqslant p_{\rm
guess}^{\rm prior}.$ (30)

Thus, if $p_{b}^{\mbox{\rm\tiny ME}}=\eta_{1b}$ for all $b\in\mathsf{B}$, then
$p_{\rm guess}^{\rm post}=\tilde{\eta}_{\bm{1}}$ in which MEPI of
$\mathcal{E}$ is possible without the help of measurement; the prepared state
is guessed to be $\rho_{11}$, $\rho_{12}$, …, or $\rho_{1m}$ according to
post-measurement information $b=1$, $2$, …, or $m$, respectively.
From Proposition 1, we can see that $p_{\rm guess}^{\rm
post}=\tilde{\eta}_{\bm{1}}$ if and only if

$\tilde{\epsilon}_{\bm{\omega}}\geqslant\tilde{\lambda}_{\bm{\omega}}\
\forall\bm{\omega}\in\Omega,$ (31)

where

$\tilde{\epsilon}_{\bm{\omega}}=\tilde{\eta}_{\bm{1}}-\tilde{\eta}_{\bm{\omega}},\
\tilde{\lambda}_{\bm{\omega}}=\|\tilde{\bm{\mu}}_{\bm{1}}-\tilde{\bm{\mu}}_{\bm{\omega}}\|,\
\bm{\omega}\in\Omega.$ (32)

Also, we can see that $p_{b}^{\mbox{\rm\tiny ME}}=\eta_{1b}$ if and only if

$\epsilon_{ib}\geqslant\lambda_{ib}\ \forall i\in\mathsf{A}_{b},$ (33)

where

$\epsilon_{ib}=\eta_{1b}-\eta_{ib},\
\lambda_{ib}=\|\eta_{1b}\bm{\nu}_{1b}-\eta_{ib}\bm{\nu}_{ib}\|,\
b\in\mathsf{B},i\in\mathsf{A}_{b}.$ (34)

The following lemma shows that $p_{\rm guess}^{\rm
post}=\tilde{\eta}_{\bm{1}}$ is equivalent to $p_{b}^{\mbox{\rm\tiny
ME}}=\eta_{1b}$ for all $b\in\mathsf{B}$.

######  Lemma 2.

For MEPI of qubit state ensemble $\mathcal{E}$, $p_{\rm guess}^{\rm
post}=\tilde{\eta}_{\bm{1}}$ if and only if
$\epsilon_{ib}\geqslant\lambda_{ib}$ holds for all $b,i$. When $p_{\rm
guess}^{\rm post}=\tilde{\eta}_{\bm{1}}$, a POVM $\mathcal{M}$ is optimal if
and only if

$\forall\bm{\omega}\neq\bm{1},\ M_{\bm{\omega}}\ \propto\
\left\\{\begin{array}[]{ccl}\mathbbm{1}+\frac{\tilde{\bm{\mu}}_{\bm{\omega}}-\tilde{\bm{\mu}}_{\bm{1}}}{\|\tilde{\bm{\mu}}_{\bm{\omega}}-\tilde{\bm{\mu}}_{\bm{1}}\|}\cdot\bm{\sigma}&,&\tilde{\epsilon}_{\bm{\omega}}=\tilde{\lambda}_{\bm{\omega}},\\\
0&,&\tilde{\epsilon}_{\bm{\omega}}>\tilde{\lambda}_{\bm{\omega}}.\end{array}\right.$
(35)

###### Proof.

From the definitions of $\tilde{\eta}_{\bm{\omega}}$ and
$\tilde{\bm{\mu}}_{\bm{\omega}}$ in (17) and (22), we have

$\tilde{\epsilon}_{\bm{\omega}}=\sum_{b\in\mathsf{B}}\epsilon_{\omega_{b}b},\
\tilde{\lambda}_{\bm{\omega}}\leqslant\sum_{b\in\mathsf{B}}\lambda_{\omega_{b}b}\
\forall\bm{\omega}\in\Omega.$ (36)

Thus, if Inequality (31) holds for all $\bm{\omega}\in\Omega$ or,
equivalently, $p_{\rm guess}^{\rm post}=\tilde{\eta}_{\bm{1}}$, then
Inequality (33) holds for all $b,i$ because
$\tilde{\epsilon}_{\bm{\omega}}=\epsilon_{ib}$ and
$\tilde{\lambda}_{\bm{\omega}}=\lambda_{ib}$ for all $\bm{\omega}\in\Omega$
such that $\omega_{b}=i$ for some $b\in\mathsf{B}$ and $\omega_{b^{\prime}}=1$
for all $b^{\prime}\neq b$. Since the converse has already been proved,
$p_{\rm guess}^{\rm post}=\tilde{\eta}_{\bm{1}}$ if and only if
$\epsilon_{ib}\geqslant\lambda_{ib}$ for all $b,i$. In addition, directly from
Proposition 1, we can see that Condition (35) is a necessary and sufficient
condition for a POVM $\mathcal{M}$ to be optimal when $p_{\rm guess}^{\rm
post}=\tilde{\eta}_{\bf 1}$. ∎

For $p_{\rm guess}^{\rm post}=\tilde{\eta}_{\bm{1}}$ and
$\bm{\omega}\in\Omega$, if
$\tilde{\epsilon}_{\bm{\omega}}=\tilde{\lambda}_{\bm{\omega}}$, then

$\epsilon_{\omega_{b}b}=\lambda_{\omega_{b}b}\ \forall b,\
\tilde{\lambda}_{\bm{\omega}}=\sum_{b\in\mathsf{B}}\lambda_{\omega_{b}b}$ (37)

because

$\sum_{b\in\mathsf{B}}\lambda_{\omega_{b}b}\leqslant\sum_{b\in\mathsf{B}}\epsilon_{\omega_{b}b}=\tilde{\epsilon}_{\bm{\omega}}=\tilde{\lambda}_{\bm{\omega}}\leqslant\sum_{b\in\mathsf{B}}\lambda_{\omega_{b}b},$
(38)

where the first and last inequalities follow from the inequalities in (33) and
(36), respectively. Conversely, if Eq. (37) holds,
$\tilde{\epsilon}_{\bm{\omega}}=\tilde{\lambda}_{\bm{\omega}}$ because

$\tilde{\epsilon}_{\bm{\omega}}=\sum_{b\in\mathsf{B}}\epsilon_{\omega_{b}b}=\sum_{b\in\mathsf{B}}\lambda_{\omega_{b}b}=\tilde{\lambda}_{\bm{\omega}}.$
(39)

Therefore, $\tilde{\epsilon}_{\bm{\omega}}=\tilde{\lambda}_{\bm{\omega}}$ is
equivalent to Eq. (37) when $p_{\rm guess}^{\rm post}=\tilde{\eta}_{\bm{1}}$.
The last condition of Eq. (37) is equivalent to that all unit vectors
$\hat{\bm{\mu}}_{\omega_{b}b}$ with $\omega_{b}\neq 1$ are the same, that is,
for all $b\in\mathsf{B}$ with $\omega_{b}\neq 1$,

$\hat{\bm{\mu}}_{\omega_{b}b}=\frac{\tilde{\bm{\mu}}_{\bm{\omega}}-\tilde{\bm{\mu}}_{\bm{1}}}{\|\tilde{\bm{\mu}}_{\bm{\omega}}-\tilde{\bm{\mu}}_{\bm{1}}\|},$
(40)

where $\hat{\bm{\mu}}_{ib}$ is defined in (29).

######  Corollary 3.

For MEPI of $\mathcal{E}$ with $n_{b}=2$ for all $b\in\mathsf{B}$ and $p_{\rm
guess}^{\rm post}=\tilde{\eta}_{\bm{1}}$, a POVM $\mathcal{M}$ is optimal if
and only if

$\forall\bm{\omega}\neq\bm{1},\
M_{\bm{\omega}}\propto\left\\{\begin{array}[]{ccl}\mathbbm{1}+\hat{\bm{\mu}}\cdot\bm{\sigma}&,&\epsilon_{2b}=\lambda_{2b}\,\forall\omega_{b}\neq
1\ \mbox{\rm and}\\\\[2.84526pt] &&\exists\hat{\bm{\mu}}\in\mathbb{R}^{3}\
\mbox{\rm such that}\
\hat{\bm{\mu}}=\hat{\bm{\mu}}_{2b}\,\forall\omega_{b}\neq 1,\\\\[2.84526pt]
0&,&\mbox{\rm otherwise.}\end{array}\right.$ (41)

###  III.4 MEPI for $p_{\rm guess}^{\rm post}>\tilde{\eta}_{\bm{1}}$

In order to consider the case of $p_{\rm guess}^{\rm
post}>\tilde{\eta}_{\bm{1}}$, we redefine $\mathcal{Z}_{\mathsf{S}}$ of
Definition 1 suitable for ME of $\tilde{\mathcal{E}}$.

######  Definition 2.

For each $\mathsf{S}\subseteq\Omega$,

$\begin{array}[]{rcl}\mathcal{Z}_{\mathsf{S}}:=\\{\bm{v}\in\mathcal{P}_{\mathsf{S}}&:&\varphi_{\bm{\omega}}(\bm{v})=\varphi_{\bm{\omega}^{\prime}}(\bm{v})\
\forall(\bm{\omega},\bm{\omega}^{\prime})\in\mathsf{S}\times\mathsf{S},\\\
&&\varphi_{\bm{\omega}}(\bm{v})\geqslant\varphi_{\bm{\omega}^{\prime}}(\bm{v})\
\forall(\bm{\omega},\bm{\omega}^{\prime})\in\mathsf{S}\times(\Omega-\mathsf{S})\\},\end{array}$
(42)

where

$\begin{array}[]{ccl}\varphi_{\bm{\omega}}({\bm{v}})&=&\tilde{\eta}_{\bm{\omega}}+\|\tilde{\bm{\mu}}_{\bm{\omega}}-{\bm{v}}\|,\
\bm{\omega}\in\Omega,\\\ &&\\\
\mathcal{P}_{\mathsf{S}}&=&\\{\sum_{\bm{\omega}\in\mathsf{S}}c_{\bm{\omega}}\tilde{\bm{\mu}}_{\bm{\omega}}\,:\,c_{\bm{\omega}}>0\
\forall\bm{\omega}\in\mathsf{S},\
\sum_{\bm{\omega}\in\mathsf{S}}c_{\bm{\omega}}=1\\}.\end{array}$ (43)

We also use $\mathcal{Z}$ to denote the union of all
$\mathcal{Z}_{\mathsf{S}}$, that is,
$\mathcal{Z}=\bigcup_{\mathsf{S}\subseteq\Omega}\mathcal{Z}_{\mathsf{S}}$.

From Propositions 2, we can obtain the following lemma, showing that all MEPI
measurements of $\mathcal{E}$ are characterized by the _single_ element of
$\mathcal{Z}$ when $p_{\rm guess}^{\rm post}>\tilde{\eta}_{\bm{1}}$.

######  Lemma 3.

For MEPI of qubit state ensemble $\mathcal{E}$, $\mathcal{Z}$ is always a
single-element set $\\{\bm{v}\\}$. If $\mathcal{Z}_{\mathsf{S}}$ is nonempty,

$p_{\rm guess}^{\rm post}=\varphi_{\bm{\omega}}(\bm{v})\ \
\forall\bm{\omega}\in\mathsf{S}.$ (44)

When $p_{\rm guess}^{\rm post}>\tilde{\eta}_{\bm{1}}$, there is an optimal
measurement having $\mathsf{S}\subseteq\Omega$ as the set of all non-null
outcomes if and only if $\mathcal{Z}_{\mathsf{S}}$ is nonempty. Moreover, a
POVM $\mathcal{M}$ having $\mathsf{S}$ as the set of all non-null outcomes is
optimal if and only if

$M_{\bm{\omega}}\propto\mathbbm{1}+\frac{\tilde{\bm{\mu}}_{\bm{\omega}}-{\bm{v}}}{\|\tilde{\bm{\mu}}_{\bm{\omega}}-{\bm{v}}\|}\cdot\bm{\sigma}\
\ \forall\bm{\omega}\in\mathsf{S}.$ (45)

Figure 1: In the case of $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm
post}<p_{\rm guess}^{\rm prior}$, four vectors
$\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\Omega}$ in
$\mathbb{R}^{3}$ forms a parallelogram with nonempty interior
$\mathcal{P}_{\Omega}$(yellow in (a)). For
$\mathcal{Z}\not\subseteq\mathcal{P}_{\Omega}$, the single element of
$\mathcal{Z}$ is in one of two edges $\mathcal{P}_{\\{(1,1),(1,2)\\}}$ and
$\mathcal{P}_{\\{(1,1),(2,1)\\}}$(red in (b)), but it cannot be in one of two
edges $\mathcal{P}_{\\{(1,2),(2,2)\\}}$ and
$\mathcal{P}_{\\{(2,1),(2,2)\\}}$(blue in (b)).

For $p_{\rm guess}^{\rm post}>\tilde{\eta}_{\bf 1}$, there is no subset
$\mathsf{S}\subseteq\Omega$ with $|\mathsf{S}|=1$ and $\mathcal{Z}_{\sf
S}\neq\varnothing$ because all MEPI measurements have more than one non-null
outcome. Therefore, $\mathcal{Z}\not\subseteq\mathcal{P}_{\sf S}$ for all
${\sf S}\subseteq\Omega$ with $|\mathsf{S}|=1$.
For example, let us consider the case of $m=n_{1}=n_{2}=2$ and $p_{\rm
guess}^{\rm post}<p_{\rm guess}^{\rm prior}$. From Corollary 2, Lemma 2, and
Eq. (23), $p_{\rm guess}^{\rm post}>\tilde{\eta}_{\bf 1}$ and the polygon
formed by $\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\Omega}$ is a
_parallelogram_ with nonempty interior $\mathcal{P}_{\Omega}$ illustrated in
Fig. 1. The interior angle between two unit vector $\hat{\bm{\mu}}_{21}$ and
$\hat{\bm{\mu}}_{22}$ defined in (29) is $\Phi\in(0,\pi)$ satisfying

$\hat{\bm{\mu}}_{21}\cdot\hat{\bm{\mu}}_{22}=\cos\Phi.$ (46)

Thus, we can classify this MEPI into two cases if the single element of
$\mathcal{Z}$ is in $\mathcal{P}_{\Omega}$ or not.
Before analyzing two cases $\mathcal{Z}\not\subseteq\mathcal{P}_{\Omega}$ and
$\mathcal{Z}\subseteq\mathcal{P}_{\Omega}$, respectively, let us consider the
uniqueness of MEPI measurement. When $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm
post}<p_{\rm guess}^{\rm prior}$, Lemma 3 implies that, for each
$\mathsf{S}\subsetneq\Omega$ with $|\mathsf{S}|\geqslant 2$, the MEPI
measurement having $\mathsf{S}$ as the set of all non-null outcomes is unique
because $\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\mathsf{S}}$
forms a simplex with affine dimension $|\mathsf{S}|-1$. Since all convex
combination of two different MEPI measurements are also MEPI measurements, if
a non-null MEPI measurement does not exist(that is,
$\mathcal{Z}_{\Omega}=\varnothing$), there cannot be more than one
$\mathsf{S}\subseteq\Omega$ satisfying
$\mathcal{Z}_{\mathsf{S}}\neq\varnothing$; thus, the MEPI measurement is
unique. However, if a non-null MEPI measurements exists(that is,
$\mathcal{Z}_{\Omega}\neq\varnothing$), the MEPI measurement is not unique
because a null MEPI measurement also exists from Corollary 1.

######  Corollary 4.

When $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm
prior}$, the MEPI measurement is unique if and only if a non-null MEPI
measurement does not exist or, equivalently,
$\mathcal{Z}_{\Omega}=\varnothing$.

Let us now analyze MEPI of two cases,
$\mathcal{Z}\not\subseteq\mathcal{P}_{\Omega}$ and
$\mathcal{Z}\subseteq\mathcal{P}_{\Omega}$. If
$\mathcal{Z}\not\subseteq\mathcal{P}_{\Omega}$, then
$\mathcal{Z}\subseteq\mathcal{P}_{\mathsf{S}}$ for only one of the following
four $\mathsf{S}$’s:

$\\{(1,1),(1,2)\\},\,\\{(1,1),(2,1)\\},\,\\{(1,2),(2,2)\\},\,\\{(2,1),(2,2)\\}.$
(47)

In other words, the single element of $\mathcal{Z}$ is in one of the edges
illustrated in Fig. 1. Thus, $\mathcal{Z}_{\mathsf{S}}$ is nonempty for only
one $\mathsf{S}$ in Eq. (47). The following theorem provides a necessary and
sufficient condition that $\mathcal{Z}_{\mathsf{S}}$ is nonempty for one
$\mathsf{S}$ in Eq. (47) when $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm
post}<p_{\rm guess}^{\rm prior}$. The proof of Theorem 1 is given in Appendix
B.

######  Theorem 1.

For MEPI of $\mathcal{E}$ with $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm
post}<p_{\rm guess}^{\rm prior}$, the followings are true:

$\begin{array}[]{l}\mathcal{Z}_{\\{(1,1),(1,2)\\}}\neq\varnothing\
\Leftrightarrow\
\frac{\epsilon_{21}+\lambda_{21}\cos\Phi}{\lambda_{21}+\epsilon_{21}}\geqslant\frac{\lambda_{21}-\epsilon_{21}}{\lambda_{22}-\epsilon_{22}},\frac{\epsilon_{21}-\lambda_{21}\cos\Phi}{\lambda_{21}-\epsilon_{21}}\geqslant\frac{\lambda_{21}+\epsilon_{21}}{\lambda_{22}+\epsilon_{22}},\\\\[4.2679pt]
\mathcal{Z}_{\\{(1,1),(2,1)\\}}\neq\varnothing\ \Leftrightarrow\
\frac{\epsilon_{22}+\lambda_{22}\cos\Phi}{\lambda_{22}+\epsilon_{22}}\geqslant\frac{\lambda_{22}-\epsilon_{22}}{\lambda_{21}-\epsilon_{21}},\frac{\epsilon_{22}-\lambda_{22}\cos\Phi}{\lambda_{22}-\epsilon_{22}}\geqslant\frac{\lambda_{22}+\epsilon_{22}}{\lambda_{21}+\epsilon_{21}},\\\\[4.2679pt]
\mathcal{Z}_{\\{(2,1),(2,2)\\}}=\mathcal{Z}_{\\{(1,2),(2,2)\\}}=\varnothing,\end{array}$
(48)

where $\epsilon_{ib}$ and $\lambda_{ib}$ are defined in (34), and

$p_{\rm guess}^{\rm
post}=\left\\{\begin{array}[]{clc}\eta_{11}+\frac{1}{2}(\eta_{12}+\eta_{22}+\lambda_{22})&,&\mathcal{Z}_{\\{(1,1),(1,2)\\}}\neq\varnothing,\\\\[2.84526pt]
\eta_{12}+\frac{1}{2}(\eta_{11}+\eta_{21}+\lambda_{21})&,&\mathcal{Z}_{\\{(1,1),(2,1)\\}}\neq\varnothing.\\\
\end{array}\right.$ (49)

When $\mathcal{Z}_{\\{(1,1),(1,2)\\}}$ is nonempty, $p_{\rm guess}^{\rm
prior}-p_{\rm guess}^{\rm post}=\frac{1}{2}(\lambda_{21}-\epsilon_{21})$ which
is a finite gap between the guessing probability and the greatest prior
probability of $\mathcal{E}_{1}$. In this case, $p_{\rm guess}^{\rm post}$ can
be obtained by ME of $\mathcal{E}_{2}$. More explicitly, we perform the ME
measurement of $\mathcal{E}_{2}$ before obtaining the classical information
$b$ about the prepared subensemble. If $b=1$, we ignore the result and guess
the prepared state as $\rho_{11}$, whereas, if $b=2$, we accept the
measurement result as it is. Then, the average probability of correctly
guessing the prepared state becomes $p_{\rm guess}^{\rm post}$. Similarly,
MEPI of $\mathcal{E}$ with $\mathcal{Z}_{\\{(1,1),(2,1)\\}}\neq\varnothing$
can be achieved with ME of $\mathcal{E}_{1}$.

Figure 2: When $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm post}<p_{\rm
guess}^{\rm prior}$, if $\mathcal{Z}\subseteq\mathcal{P}_{\Omega}$, the single
element of $\mathcal{Z}$ exists in one of the interiors of two line segments,
$\mathcal{P}_{\\{(1,1),(2,2)\\}}$(red in (a)) and
$\mathcal{P}_{\\{(1,2),(2,1)\\}}$(red in (b)), or in one of the interiors of
four triangles, $\mathcal{P}_{\\{(1,1),(1,2),(2,2)\\}}$,
$\mathcal{P}_{\\{(1,2),(2,1),(2,2)\\}}$(green in (a)),
$\mathcal{P}_{\\{(1,1),(1,2),(2,1)\\}}$(green in (b)), and
$\mathcal{P}_{\\{(1,2),(2,1),(2,2)\\}}$(blue in (b)), but it cannot be an
element of $\mathcal{Z}_{\\{(1,2),(2,1),(2,2)\\}}$.

If $\mathcal{Z}\subseteq\mathcal{P}_{\Omega}$ equivalent to

$\mathcal{Z}_{\\{(1,1),(1,2)\\}}=\mathcal{Z}_{\\{(1,1),(2,1)\\}}=\varnothing$
(50)

from Theorem 1, then $\mathcal{Z}\subseteq\mathcal{P}_{\mathsf{S}}$ for some
$\mathsf{S}$ in the following seven $\mathsf{S}$’s:

$\begin{array}[]{r}\Omega,\\{(1,1),(2,2)\\},\,\\{(1,1),(1,2),(2,2)\\},\,\\{(1,1),(2,1),(2,2)\\},\\\
\\{(1,2),(2,1)\\},\,\\{(1,1),(1,2),(2,1)\\},\,\\{(1,2),(2,1),(2,2)\\}.\end{array}$
(51)

In other words, the single element of $\mathcal{Z}$ exists in one of the
interiors of line segments or triangles illustrated in Fig. 2. Thus,
$\mathcal{Z}_{\mathsf{S}}$ is nonempty for some $\mathsf{S}$ in (51). The
following theorem provides a necessary and sufficient condition that
$\mathcal{Z}_{\mathsf{S}}\neq\varnothing$ for one $\mathsf{S}$ in (51) when
$m=n_{1}=n_{2}=2$, $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm prior}$, and
$\mathcal{Z}\subseteq\mathcal{P}_{\Omega}$. The proof of Theorem 2 is given in
Appendix B.

######  Theorem 2.

For MEPI of $\mathcal{E}$ with $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm
post}<p_{\rm guess}^{\rm prior}$, the followings are true.

  1. (a)

We have

$\begin{array}[]{lclll}\mathcal{Z}_{\\{(1,1),(2,2)\\}}\neq\varnothing&\
\Leftrightarrow&\alpha\geqslant|\beta_{+}|/\gamma_{+},\\\\[3.69885pt]
\mathcal{Z}_{\\{(1,2),(2,1)\\}}\neq\varnothing&\
\Leftrightarrow&\alpha\leqslant-|\beta_{-}|/\gamma_{-},\end{array}$ (52)

and

$p_{\rm guess}^{\rm
post}=\left\\{\begin{array}[]{cclc}\frac{1}{2}(1+\gamma_{+})&,&\mathcal{Z}_{\\{(1,1),(2,2)\\}}\neq\varnothing,\\\\[2.84526pt]
\frac{1}{2}(1+\gamma_{-})&,&\mathcal{Z}_{\\{(1,2),(2,1)\\}}\neq\varnothing,\\\\[2.84526pt]
\end{array}\right.$ (53)

where

$\displaystyle\alpha$ $\displaystyle=$
$\displaystyle\lambda_{21}\lambda_{22}\cos\Phi-\epsilon_{21}\epsilon_{22},$
$\displaystyle\beta_{\pm}$ $\displaystyle=$
$\displaystyle\epsilon_{21}\lambda_{22}^{2}\mp\epsilon_{22}\lambda_{21}^{2}\pm(\epsilon_{21}\mp\epsilon_{22})\lambda_{21}\lambda_{22}\cos\Phi,$
$\displaystyle\gamma_{\pm}$ $\displaystyle=$
$\displaystyle\sqrt{\lambda_{21}^{2}+\lambda_{22}^{2}\pm
2\lambda_{21}\lambda_{22}\cos\Phi}.$ (54)

Here, $\epsilon_{ib}$ and $\lambda_{ib}$ are defined in (34).

  2. (b)

We have

$\begin{array}[]{lclll}\mathcal{Z}_{\\{(1,1),(1,2),(2,1)\\}}\neq\varnothing&\
\Leftrightarrow&\left\\{\begin{array}[]{l}\frac{\epsilon_{21}+\lambda_{21}\cos\Phi}{\lambda_{21}+\epsilon_{21}}<\frac{\lambda_{21}-\epsilon_{21}}{\lambda_{22}-\epsilon_{22}},\\\\[4.2679pt]
\frac{\epsilon_{22}+\lambda_{22}\cos\Phi}{\lambda_{22}+\epsilon_{22}}<\frac{\lambda_{22}-\epsilon_{22}}{\lambda_{21}-\epsilon_{21}},\\\\[4.2679pt]
0\geqslant\alpha>-\beta_{-}/\gamma_{-},\end{array}\right.\\\\[22.76219pt]
\mathcal{Z}_{\\{(1,1),(1,2),(2,2)\\}}\neq\varnothing&\
\Leftrightarrow&\left\\{\begin{array}[]{l}\frac{\epsilon_{21}-\lambda_{21}\cos\Phi}{\lambda_{21}-\epsilon_{21}}<\frac{\lambda_{21}+\epsilon_{21}}{\lambda_{22}+\epsilon_{22}},\\\\[4.2679pt]
\frac{\epsilon_{22}+\lambda_{22}\cos\Phi}{\lambda_{22}+\epsilon_{22}}>-\frac{\lambda_{22}-\epsilon_{22}}{\lambda_{21}-\epsilon_{21}},\\\\[4.2679pt]
0\leqslant\alpha<-\beta_{+}/\gamma_{+},\end{array}\right.\\\\[22.76219pt]
\mathcal{Z}_{\\{(1,1),(2,1),(2,2)\\}}\neq\varnothing&\
\Leftrightarrow&\left\\{\begin{array}[]{l}\frac{\epsilon_{22}-\lambda_{22}\cos\Phi}{\lambda_{22}-\epsilon_{22}}<\frac{\lambda_{22}+\epsilon_{22}}{\lambda_{21}+\epsilon_{21}},\\\\[4.2679pt]
\frac{\epsilon_{21}+\lambda_{21}\cos\Phi}{\lambda_{21}+\epsilon_{21}}>-\frac{\lambda_{21}-\epsilon_{21}}{\lambda_{22}-\epsilon_{22}},\\\\[4.2679pt]
0\leqslant\alpha<\beta_{+}/\gamma_{+},\end{array}\right.\\\\[22.76219pt]
\mathcal{Z}_{\\{(1,2),(2,1),(2,2)\\}}=\varnothing,&&\end{array}$ (55)

and

$p_{\rm guess}^{\rm
post}=\left\\{\begin{array}[]{cclc}\tilde{\eta}_{(1,1)}+\frac{\lambda_{21}^{2}-\epsilon_{21}^{2}}{2[\lambda_{21}\cos(\Theta_{-}-\Xi_{-})+\epsilon_{21}]}&,&\mathcal{Z}_{\\{(1,1),(1,2),(2,1)\\}}\neq\varnothing,\\\\[2.84526pt]
\tilde{\eta}_{(1,2)}-\frac{\lambda_{21}^{2}-\epsilon_{21}^{2}}{2[\lambda_{21}\cos(\Theta_{+}+\Xi_{+})-\epsilon_{21}]}&,&\mathcal{Z}_{\\{(1,1),(1,2),(2,2)\\}}\neq\varnothing,\\\\[2.84526pt]
\tilde{\eta}_{(2,1)}+\frac{\lambda_{21}^{2}-\epsilon_{21}^{2}}{2[\lambda_{21}\cos(\Theta_{+}-\Xi_{+})-\epsilon_{21}]}&,&\mathcal{Z}_{\\{(1,1),(2,1),(2,2)\\}}\neq\varnothing,\end{array}\right.$
(56)

where

$\begin{array}[]{l}\Theta_{\pm}=\arccos\frac{\epsilon_{22}(\lambda_{21}^{2}-\epsilon_{21}^{2})\pm\epsilon_{21}(\lambda_{22}^{2}-\epsilon_{22}^{2})}{\sqrt{\lambda_{21}^{2}(\lambda_{22}^{2}-\epsilon_{22}^{2})^{2}+\lambda_{22}^{2}(\lambda_{21}^{2}-\epsilon_{21}^{2})^{2}\pm
2\lambda_{21}\lambda_{22}(\lambda_{21}^{2}-\epsilon_{21}^{2})(\lambda_{22}^{2}-\epsilon_{22}^{2})\cos\Phi}},\\\\[5.69054pt]
\Xi_{\pm}=\arccos\frac{\lambda_{21}(\lambda_{22}^{2}-\epsilon_{22}^{2})\pm\lambda_{22}(\lambda_{21}^{2}-\epsilon_{21}^{2})\cos\Phi}{\sqrt{\lambda_{21}^{2}(\lambda_{22}^{2}-\epsilon_{22}^{2})^{2}+\lambda_{22}^{2}(\lambda_{21}^{2}-\epsilon_{21}^{2})^{2}\pm
2\lambda_{21}\lambda_{22}(\lambda_{21}^{2}-\epsilon_{21}^{2})(\lambda_{22}^{2}-\epsilon_{22}^{2})\cos\Phi}}.\end{array}$
(57)

  3. (c)

When $\mathcal{Z}\subseteq\mathcal{P}_{\Omega}$, a non-null MEPI measurement
exists if and only if $\alpha$ in Eq. ((a)) is zero.

Finally, we mention the location of the single element of $\mathcal{Z}$
characterizing all MEPI measurements for $p_{\rm guess}^{\rm
post}>\tilde{\eta}_{\bm{1}}$. When $\bm{z}\in\mathcal{Z}_{\mathsf{S}}$ for
some $\mathsf{S}\subseteq\Omega$, the definition of $\mathcal{Z}_{\mathsf{S}}$
in Eq. (42) implies that two points $\bm{z}$ and
$\tilde{\bm{\mu}}_{\bm{\omega}}$ are separated by $p_{\rm guess}^{\rm
post}-\tilde{\eta}_{\bm{\omega}}$ for each $\bm{\omega}\in\mathsf{S}$. Thus,
the location of $\bm{z}$ is easily determined by the value of $p_{\rm
guess}^{\rm post}$ if the affine dimension of
$\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\mathsf{S}}$ is one(that
is, $\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\mathsf{S}}$ form a
line segment). On the other hand, if the affine dimension of
$\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\mathsf{S}}$ is two or
three(that is,
$\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\mathsf{S}}$ form a
polygon or polyhedron), it is difficult to determine the location of $\bm{z}$
only with the value of $p_{\rm guess}^{\rm post}$, and more information about
$\bm{z}$ is required. For example, $\Theta_{\pm}$ and $\Xi_{\pm}$ in Eq. (57)
are used to express the specific angles that determine $\bm{z}$ when
$m=n_{1}=n_{2}=2$, $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm prior}$, and
$\mathcal{Z}_{\mathsf{S}}\neq\varnothing$ for some $\mathsf{S}\subseteq\Omega$
with $|\mathsf{S}|=3$. The details are given in Appendix B.

##  IV Conclusion

We have analyzed MEPI of qubit state ensemble consisting of $m$ subensembles
with $n_{1},\ldots,n_{m}$ states by considering the modified problem, that is,
ME of $\prod_{b=1}^{m}n_{b}$ qubit states, and have characterized all optimal
measurements. For the case where MEPI is impossible without the help of
measurement, we have provided an analytic structure to characterize optimal
measurements with fixed non-null outcomes. We have also shown that a null
optimal measurement always exists for any MEPI of qubit states. In addition,
we have analytically provided a necessary and sufficient condition that pre-
measurement information is strictly more favorable than post-measurement
information(that is, $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm prior}$)
when $n_{b}=2$ for all $b=1,\ldots,m$. For the case of $m=n_{1}=n_{2}=2$ and
$p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm prior}$, we have analytically
provided the optimal probability of correct guessing. In this case, we have
also shown that the uniqueness of optimal measurement is equivalent to the
non-existence of non-null MEPI measurement. Whereas the existing results of
MEPI only deal with some special cases of state ensembles consisting of
equiprobable states or symmetric statesref:ball2008 ; ref:gopa2010 ;
ref:carm2018 , our results consider the general case of qubit state ensembles
having four states and provide the analytic solutions.
Our results are closely related to the incompatibility of
measurementsref:carm2018 ; ref:hein2016 ; ref:carm2019 ; ref:skrz20191 ;
ref:uola2019 ; ref:skrz20192 . Recently, it has been found that the gap
between the two optimal probabilities of correct guessing with pre- and post-
measurement information, $p_{\rm guess}^{\rm prior}$ and $p_{\rm guess}^{\rm
post}$, is a witness of the incompatibility of measurementsref:carm2019 .
Moreover, the incompatibility robustnessref:skrz20191 ; ref:uola2019 ;
ref:skrz20192  of a given set of measurements
$\\{\mathcal{M}_{b}\\}_{b\in\mathsf{B}}$ can be written by $p_{\rm guess}^{\rm
post}$ as

$I_{\rm
R}(\\{\mathcal{M}_{b}\\}_{b\in\mathsf{B}})=\max_{\mathcal{E}}\frac{p_{\rm
corr}^{\rm prior}(\mathcal{E},\\{\mathcal{M}_{b}\\}_{b\in\mathsf{B}})}{p_{\rm
guess}^{\rm post}(\mathcal{E})}-1,$ (58)

where the maximization is over all qubit state ensemble $\mathcal{E}$ such as
Eq. (11) and $p_{\rm corr}^{\rm
prior}(\mathcal{E},\\{\mathcal{M}_{b}\\}_{b\in\mathsf{B}})$ is the average
probability of correct guessing that can be obtained by performing
$\mathcal{M}_{b}$ according to the pre-measurement information $b$. As quantum
incompatibility is related with quantum resource theoryref:chit2019  which is
one of the most general frame work in quantum information processing, it is an
interesting future work to obtain an analytic form of Eq. (58).

## Acknowledgements

We would like to thank Professor Joonwoo Bae for his contribution to the early
stage of this work. We are grateful to Jaehee Shin and Jihwan Kim for useful
comments. This work is supported by the Basic Science Research Program through
the National Research Foundation of Korea(NRF) funded by the Ministry of
Education, Science and Technology (NRF2018R1D1A1B07049420) and Institute of
Information & communications Technology Planning & Evaluation(IITP) grant
funded by the Korea government(MSIT) (No.2020001343, Artificial Intelligence
Convergence Research Center(Hanyang University ERICA). D.H. and J.S.K.
acknowledge support from the National Research Foundation of Korea(NRF) grant
funded by the Korea government(Ministry of Science and
ICT)(NRF2020M3E4A1080088).

##  Appendix A Proofs of Propositions

In the context of convex optimizationref:boyd2004 , ME of $\mathcal{E}$ is a
convex optimization problem that maximizes $\sum_{i\in\Omega}\eta_{i}{\rm
Tr}[\rho_{i}M_{i}]$ subject to the following POVM constraints:

$\begin{array}[]{ll}M_{i}\succeq 0\,\forall i\in\Omega,&(\mbox{positive-
semidefiniteness})\\\\[2.84526pt]
\sum_{i\in\Omega}M_{i}=\mathbbm{1}.&(\mbox{completeness})\end{array}$ (59)

The Lagrange dual problem is to minimize ${\rm Tr}K$ subject to

$\begin{array}[]{ll}W_{i}\succeq 0\,\forall i\in\Omega,&(\mbox{positive-
semidefiniteness})\\\\[2.84526pt] K=\eta_{i}\rho_{i}+W_{i}\,\forall
i\in\Omega,&(\mbox{Lagrangian stability})\end{array}$ (60)

where $W_{i}$ and $K$ are Lagrange multipliers of $M_{i}\succeq 0$ and
$\sum_{i\in\Omega}M_{i}=\mathbbm{1}$, respectivelyref:elda20031 ; ref:bae20132
.
As the primal and dual problems have the same optimal valueref:elda20031 ,
primal and dual feasible variables are optimal if and only if they satisfy

${\rm Tr}[M_{i}W_{i}]=0\,\forall i\in\Omega.\ (\mbox{complementary
slackness})$ (61)

Therefore, ME of $\mathcal{E}$ is equivalent to finding a set of primal and
dual variables,
$\\{M_{i}\\}_{i\in\Omega}\cup\\{W_{i}\\}_{i\in\Omega}\cup\\{K\\}$, that
satisfies the so-called _Karush–Kuhn–Tucker_(KKT) condition consisting of
Conditions (59), (60), and (61)ref:bae20132 .
The primal and dual variables can be expressed as

$M_{i}=p_{i}(\mathbbm{1}+\bm{u}_{i}\cdot\bm{\sigma}),\
W_{i}=\frac{r_{i}}{2}(\mathbbm{1}+\bm{w}_{i}\cdot\bm{\sigma}),\
K=\frac{1}{2}(s\mathbbm{1}+{\bm{v}}\cdot\bm{\sigma}).$ (62)

using $p_{i},r_{i},s\in\mathbb{R}$ and
$\bm{u}_{i},\bm{w}_{i},{\bm{v}}\in\mathbb{R}^{3}$. From the expression of
(62), KKT condition can be expressed as follows:

$\begin{array}[]{lll}\mbox{\bf(P0)}\ p_{i}\geqslant 0,\ \
\|\bm{u}_{i}\|\leqslant 1\ \ \forall i\in\Omega,\ \,\mbox{\bf(P1)}\
\sum_{i\in\Omega}p_{i}=1,&\mbox{\bf(P2)}\
\sum_{i\in\Omega}p_{i}{\bm{u}}_{i}={\bm{0}},\\\ &&\\\ \mbox{\bf(D0)}\
r_{i}\geqslant 0,\ \ \|\bm{w}_{i}\|\leqslant 1\ \ \forall i\in\Omega,\
\mbox{\bf(D1)}\ s=\eta_{i}+r_{i}\ \forall i\in\Omega,&\mbox{\bf(D2)}\
{\bm{v}}=\eta_{i}{\bm{\nu}}_{i}+r_{i}{\bm{w}}_{i}\ \forall i\in\Omega,\\\
&&\\\ \mbox{\bf(C0)}\ p_{i}r_{i}=0\ \ \mbox{or}\ \
{\bm{u}}_{i}\cdot{\bm{w}}_{i}=-1\ \ \forall i\in\Omega.&\end{array}$ (63)

Now, we prove Propositions 1 and 2 by using KKT condition (63) with
$\\{p_{i},{\bm{u}}_{i}\\}_{i\in\Omega}\cup\\{r_{i},{\bm{w}}_{i}\\}_{i\in\Omega}\cup\\{s,{\bm{v}}\\}$
as variables. We use the superscript ⋆ to express the optimal variables. Note
that $s^{\star}$, ${\bm{v}}^{\star}$, and $r_{i}^{\star}$ are always unique,
whereas $p_{i}^{\star}$, $\bm{u}_{i}^{\star}$, and $\bm{w}_{i}^{\star}$ may
notref:bae20132 . We also note that $s^{\star}=p_{\rm guess}$.

### Proof of Proposition 1

###### Proof of (a).

For the proof of Proposition 1(a)($\Rightarrow$), assume $p_{\rm
guess}=\eta_{1}$. From (D1) and (D2) of (63), this assumption implies
$\eta_{1}>\eta_{i}$ for all $i\neq 1$ because $\eta_{j}=\eta_{1}$ means
$\bm{\nu}_{j}=\bm{\nu}_{1}$. Moreover,

$r_{i}^{\star}=\eta_{1}-\eta_{i},\,\
\bm{w}_{i}^{\star}=\frac{\eta_{1}\bm{\nu}_{1}-\eta_{i}\bm{\nu}_{i}}{\eta_{1}-\eta_{i}}\
\ \forall i\neq 1.$ (64)

Thus,

$\begin{array}[]{rcl}\epsilon_{i}=\eta_{1}-\eta_{i}\geqslant\|\bm{w}_{i}^{\star}\|(\eta_{1}-\eta_{i})=\|\eta_{1}\bm{\nu}_{1}-\eta_{i}\bm{\nu}_{i}\|=\lambda_{i}\
\forall i\in\Omega,\end{array}$ (65)

where the inequality is due to (D0) of (63).
For the proof of Proposition 1(a)($\Leftarrow$), suppose that
$\epsilon_{i}\geqslant\lambda_{i}$ for all $i\in\Omega$. Since
$\eta_{j}=\eta_{1}$ means $\bm{\nu}_{j}=\bm{\nu}_{1}$ by
$\epsilon_{j}\geqslant\lambda_{j}$, it follows that $\eta_{1}>\eta_{i}$ for
all $i\neq 1$. Thus, $p_{\rm guess}=\eta_{1}$ because (63) holds for

$\begin{array}[]{lll}p_{i}=\delta_{1i},&\bm{u}_{i}=\bm{0},\\\\[2.84526pt]
r_{i}=\eta_{1}-\eta_{i},&\bm{w}_{i}=\Bigg{\\{}\begin{array}[]{ccl}\bm{0}&\mbox{for}&i=1,\\\
\frac{\eta_{1}\bm{\nu}_{1}-\eta_{i}\bm{\nu}_{i}}{\eta_{1}-\eta_{i}}&\mbox{for}&i\neq
1,\end{array}\\\\[2.84526pt]
s=\eta_{1},&\bm{v}=\eta_{1}\bm{\nu}_{1},\end{array}$ (66)

where $\delta_{ij}$ is the Kronecker delta. This completes our proof of
Proposition 1(a). We also note that $\epsilon_{i}\geqslant\lambda_{i}$ for all
$i\in\Omega$ is equivalent to $\mathcal{Z}_{\\{1\\}}\neq\varnothing$ since
$\mathcal{P}_{\\{1\\}}$ has only one element $\eta_{1}\bm{\nu}_{1}$. ∎

###### Proof of (b).

For the proof of Proposition 1(b)($\Rightarrow$), we assume that $p_{\rm
guess}=\eta_{1}$ and $\mathcal{M}$ is an optimal POVM expressed by
$\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}$. Then, there is
$\\{r_{i},\bm{w}_{i}\\}_{i\in\Omega}\cup\\{s,{\bm{v}}\\}$ such that (63) hold
for
$\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}\cup\\{r_{i},\bm{w}_{i}\\}_{i\in\Omega}\cup\\{s,{\bm{v}}\\}$.
Let us consider a non-null outcome $k\neq 1$. Since $s=p_{\rm
guess}=\eta_{1}$, it follows from (D1) and (D2) of (63) that

$r_{k}=\eta_{1}-\eta_{k},\
r_{k}\bm{w}_{k}=\eta_{1}\bm{\nu}_{1}-\eta_{k}\bm{\nu}_{k}.$ (67)

$r_{k}$ is nonzero because $\eta_{k}=\eta_{1}$ means
${\bm{\nu}}_{k}={\bm{\nu}}_{1}$ by $\epsilon_{k}\geqslant\lambda_{k}$,
therefore,

$\bm{w}_{k}=\frac{\eta_{1}\bm{\nu}_{1}-\eta_{k}\bm{\nu}_{k}}{\eta_{1}-\eta_{k}}.$
(68)

Since $p_{k}$ and $r_{k}$ are nonzero, it follows from (C0) of (63) that
$\bm{u}_{k}$ and $\bm{w}_{k}$ are unit vectors satisfying
$\bm{u}_{k}\cdot\bm{w}_{k}=-1$. Thus,

$\epsilon_{k}=\epsilon_{k}\|{\bm{w}}_{k}\|=\lambda_{k}$ (69)

and

$\bm{u}_{k}=\frac{\eta_{k}\bm{\nu}_{k}-\eta_{1}\bm{\nu}_{1}}{\|\eta_{k}\bm{\nu}_{k}-\eta_{1}\bm{\nu}_{1}\|},$
(70)

that is, Eq. (9) holds for $i=k$. We also note that Eq. (9) is obviously
satisfied for every null outcome. Therefore, $\mathcal{M}$ satisfies Eq. (9).
For the proof of Proposition 1(b)($\Leftarrow$), suppose that $p_{\rm
guess}=\eta_{1}$ and $\mathcal{M}$ is a POVM satisfying Eq. (9). Let
$\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}$ be a set of primal variables expressing
$\mathcal{M}$. Also, let
$\\{r_{i},\bm{w}_{i}\\}_{i\in\Omega}\cup\\{s,\bm{v}\\}$ be a set of dual
variables such that

$\begin{array}[]{ll}r_{i}=\eta_{1}-\eta_{i},&\bm{w}_{i}=\Bigg{\\{}\begin{array}[]{ccl}\bm{0}&\mbox{for}&i=1,\\\
\frac{\eta_{1}\bm{\nu}_{1}-\eta_{i}\bm{\nu}_{i}}{\eta_{1}-\eta_{i}}&\mbox{for}&i\neq
1,\end{array}\\\ s=\eta_{1},&\bm{v}=\eta_{1}\bm{v}_{1}.\end{array}$ (71)

Then, (63) holds for
$\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}\cup\\{r_{i},\bm{w}_{i}\\}_{i\in\Omega}\cup\\{s,\bm{v}\\}$.
Therefore, $\mathcal{M}$ is optimal. ∎

### Proof of Proposition 2

###### Proof of (a), (b)($\Leftarrow$), and (c)($\Leftarrow$).

Let us assume that $\mathsf{S}$ is a subset of $\Omega$ with
$\bm{v}\in\mathcal{Z}_{\mathsf{S}}$. The definition of
$\mathcal{Z}_{\mathsf{S}}$ implies that there exists a real number $s$
fulfilling $\varphi_{i}(\bm{v})=s$ for all $i\in\mathsf{S}$. We will show that
$s=s^{\star}$ and ${\bm{v}}={\bm{v}}^{\star}$. Since $s^{\star}=p_{\rm guess}$
and $\bm{v}^{\star}$ is unique, if $s=s^{\star}$ and
${\bm{v}}={\bm{v}}^{\star}$, then $p_{\rm guess}=\varphi_{i}(\bm{v})$ for all
$i\in\mathsf{S}$ and $\mathcal{Z}_{\mathsf{S}}=\\{\bm{v}^{\star}\\}$ which
implies that $\mathcal{Z}$ is a single-element set having only one element
$\bm{v}$, that is, $\mathcal{Z}=\\{\bm{v}\\}$.
First, let us consider the case of $\eta_{k}=s$ for some $k\in\Omega$. Since
$s\geqslant\varphi_{i}(\bm{v})$ for all $i\in\Omega$ from the definition of
$\mathcal{Z}_{\mathsf{S}}$ and $\varphi_{j}(\bm{v})=s$ for all
$j\in\mathsf{S}$, we have

$\|\eta_{k}\bm{\nu}_{k}-\bm{v}\|=\varphi_{k}(\bm{v})-\eta_{k}\leqslant
s-\eta_{k}=0,$ (72)

which means $\bm{v}=\eta_{k}\bm{\nu}_{k}$. Thus, $k=1$ because

$\eta_{k}=s\geqslant\varphi_{i}(\eta_{k}\bm{\nu}_{k})=\eta_{i}+\|\eta_{i}\bm{\nu}_{i}-\eta_{k}\bm{\nu}_{k}\|>\eta_{i}\
\forall i\neq k.$ (73)

(63) holds for the primal and dual variables satisfying Eq. (66); therefore,
$s=s^{\star}$ and ${\bm{v}}={\bm{v}}^{\star}$.
Now, let us consider the case of $\eta_{i}\neq s$ for all $i\in\Omega$. In
this case, $\eta_{i}\bm{\nu}_{i}\neq\bm{v}$ for all $i\in\mathsf{S}$ because
otherwise

$s=\varphi_{j}(\bm{v})=\eta_{j}+\|\eta_{j}\bm{\nu}_{j}-\bm{v}\|=\eta_{j}$ (74)

for some $j\in\mathsf{S}$ with $\eta_{j}\bm{\nu}_{j}=\bm{v}$. Since
${\bm{v}}\in\mathcal{P}_{\mathsf{S}}$, there exists
$\\{c_{i}\\}_{i\in\Omega}\subseteq\mathbb{R}$ such that

$c_{i}>0\ \forall i\in\mathsf{S},\ \ c_{i}=0\ \forall i\not\in\mathsf{S},\ \
\sum_{i\in\mathsf{S}}c_{i}=1,\ \
\sum_{i\in\mathsf{S}}c_{i}(\eta_{i}\bm{\nu}_{i})={\bm{v}}.$ (75)

Thus, $s=s^{\star}$ and ${\bm{v}}={\bm{v}}^{\star}$ because (63) holds for
$\\{s,{\bm{v}}\\}$ along with
$\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}\cup\\{r_{i},\bm{w}_{i}\\}_{i\in\Omega}$
satisfying

$\begin{array}[]{rclrcl}p_{i}&=&\frac{c_{i}\|\eta_{i}\bm{\nu}_{i}-{\bm{v}}\|}{\sum_{j\in\mathsf{S}}c_{j}\|\eta_{j}\bm{\nu}_{j}-{\bm{v}}\|},&\quad
r_{i}&=&s-\eta_{i},\\\
\bm{u}_{i}&=&\left\\{\begin{array}[]{ccl}\frac{\eta_{i}\bm{\nu}_{i}-{\bm{v}}}{\|\eta_{i}\bm{\nu}_{i}-{\bm{v}}\|}&,&i\in\mathsf{S},\\\
\bm{0}&,&i\not\in\mathsf{S},\end{array}\right.&\bm{w}_{i}&=&\left\\{\begin{array}[]{ccl}\frac{{\bm{v}}-\eta_{i}\bm{\nu}_{i}}{\|{\bm{v}}-\eta_{i}\bm{\nu}_{i}\|}&,&i\in\mathsf{S},\\\
\frac{{\bm{v}}-\eta_{i}\bm{\nu}_{i}}{s-\eta_{i}}&,&i\not\in\mathsf{S}.\end{array}\right.\end{array}$
(76)

This completes our proof of Proposition 2(a).
Moreover, the optimal POVM corresponding to
$\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}$ satisfying Eq. (76) has $\mathsf{S}$ as
the set of all non-null outcomes. Thus, Proposition 2(b)($\Leftarrow$) is
proved. We also note that every POVM $\mathcal{M}$ fulfilling Condition (10)
with $M_{i}=0$ for all $i\notin\mathsf{S}$ is optimal because (63) holds for
$\\{s,\bm{v}\\}$ along with $\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}$ expressing
$\mathcal{M}$ and $\\{r_{i},\bm{w}_{i}\\}_{i\in\Omega}$ satisfying (76);
therefore, Proposition 2(c)($\Leftarrow$) is also proved. ∎

###### Proof of (b)($\Rightarrow$) and (c)($\Rightarrow$).

Suppose that $p_{\rm guess}>\eta_{1}$ and $\mathcal{M}$ is an optimal POVM for
ME of $\mathcal{E}$ having $\mathsf{S}$ as the set of all non-null outcomes.
We also assume that $\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}$ is a set of primal
variables expressing $\mathcal{M}$. Then, there exists a set of dual variables
$\\{r_{i},\bm{w}_{i}\\}_{i\in\Omega}\cup\\{s,{\bm{v}}\\}$ such that (63) holds
for
$\\{p_{i},\bm{u}_{i}\\}_{i\in\Omega}\cup\\{r_{i},\bm{w}_{i}\\}_{i\in\Omega}\cup\\{s,{\bm{v}}\\}$.
Since $p_{\rm guess}>\eta_{1}$, it follows from (D1) of (63) that $r_{i}$ is
nonzero for all $i\in\Omega$. Furthermore, (C0) of (63) implies
$\|\bm{u}_{j}\|=\|\bm{w}_{j}\|=1$ and $\bm{u}_{j}\cdot\bm{w}_{j}=-1$ for each
$j\in\mathsf{S}$ because both $p_{j}$ and $r_{j}$ are nonzero. Thus, from (D2)
of (63), we have

$\begin{array}[]{ll}r_{j}=\|\eta_{j}\bm{\nu}_{j}-{\bm{v}}\|,\
\bm{u}_{j}=\frac{\eta_{j}\bm{\nu}_{j}-{\bm{v}}}{\|\eta_{j}\bm{\nu}_{j}-{\bm{v}}\|}\
\forall j\in\mathsf{S}.\end{array}$ (77)

The right equality in (77) is equivalent to Condition (10); therefore,
Proposition 2(c)($\Rightarrow$) is proved.
Now, we show $\mathcal{Z}_{\mathsf{S}}\neq\varnothing$ to prove Proposition
2(b)($\Rightarrow$). From (D0) and (D2) of (63), we can see that

$r_{i}\geqslant\|r_{i}{\bm{w}}_{i}\|=\|\eta_{i}\bm{\nu}_{i}-{\bm{v}}\|\
\forall i\in\Omega.$ (78)

Thus,

$\begin{array}[]{rcl}\varphi_{i}({\bm{v}})&=&q_{i}+\|\eta_{i}\bm{\nu}_{i}-{\bm{v}}\|=q_{i}+r_{i}=q_{j}+r_{j}\\\
&\geqslant&q_{j}+\|\eta_{j}\bm{\nu}_{j}-{\bm{v}}\|=\varphi_{j}({\bm{v}})\
\forall i\in\mathsf{S},\,j\in\Omega,\end{array}$ (79)

where the second equality is due to Eq. (77) and the third equality is from
(D1) of (63). Since the inequality in (79) for $j\in\mathsf{S}$ becomes an
equality by Eq. (77), $\mathcal{Z}_{\mathsf{S}}$ is a nonempty set with
${\bm{v}}$. Note that ${\bm{v}}\in\mathcal{P}_{\mathsf{S}}$ because

$\sum_{i\in\mathsf{S}}c_{i}(\eta_{i}\bm{\nu}_{i})=\sum_{i\in\mathsf{S}}c_{i}(\bm{v}-r_{i}\bm{w}_{i})=\bm{v}-\frac{\sum_{i\in\mathsf{S}}(p_{i}\bm{w}_{i})}{\sum_{j\in\mathsf{S}}(p_{j}/r_{j})}=\bm{v}+\frac{\sum_{i\in\mathsf{S}}(p_{i}\bm{u}_{i})}{\sum_{j\in\mathsf{S}}(p_{j}/r_{j})}={\bm{v}},$
(80)

where

$c_{i}=\frac{(p_{i}/r_{i})}{\sum_{j\in\mathsf{S}}(p_{j}/r_{j})}.$ (81)

The first equality of (80) is due to (D2) of (63), the second equality is from
$\sum_{i\in\mathsf{S}}c_{i}=1$, and the third and last equality follow from
(C0) and (P2), respectively. Therefore, Proposition 2(b)($\Rightarrow$) is
proved. ∎

##  Appendix B Proofs of Theorems

In this section, we prove Theorems 1 and 2 using
$\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$ and
$\mathcal{Y}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$($\mathsf{S}^{\prime}\subseteq\mathsf{S}\subseteq\Omega$)
defined as

$\begin{array}[]{rcl}\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}:=\\{\bm{v}\in\mathcal{P}_{\mathsf{S}^{\prime}}&:&\varphi_{\bm{\omega}}(\bm{v})=\varphi_{\bm{\omega}^{\prime}}(\bm{v})\
\forall(\bm{\omega},\bm{\omega}^{\prime})\in\mathsf{S}^{\prime}\times\mathsf{S}^{\prime},\\\
&&\varphi_{\bm{\omega}}(\bm{v})\geqslant\varphi_{\bm{\omega}^{\prime}}(\bm{v})\
\forall(\bm{\omega},\bm{\omega}^{\prime})\in\mathsf{S}^{\prime}\times(\mathsf{S}-\mathsf{S}^{\prime})\\},\\\\[5.69054pt]
\mathcal{Y}_{\mathsf{S}^{\prime}}^{\mathsf{S}}:=\\{{\bm{v}}\in\mathbb{R}^{3}&:&\varphi_{\bm{\omega}}(\bm{v})\geqslant\varphi_{\bm{\omega}^{\prime}}(\bm{v})\
\forall(\bm{\omega},\bm{\omega}^{\prime})\in\mathsf{S}^{\prime}\times(\Omega-\mathsf{S})\\},\end{array}$
(82)

where we denote

$\mathcal{X}_{\mathsf{S}}:=\mathcal{X}_{\mathsf{S}}^{\mathsf{S}},\
\mathcal{Y}_{\mathsf{S}}:=\mathcal{Y}_{\mathsf{S}}^{\mathsf{S}}.$ (83)

Note that
$\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}\cap\mathcal{Y}_{\mathsf{S}^{\prime}}^{\mathsf{S}}=\mathcal{Z}_{\mathsf{S}^{\prime}}$
for any $\mathsf{S}^{\prime}\subseteq\mathsf{S}\subseteq\Omega$. We also note
that
$\bigcup_{\mathsf{S}^{\prime}\subseteq\mathsf{S}}\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$
is a single-element set for any $\mathsf{S}\subseteq\Omega$ in terms of ME of
$\\{\tilde{\eta}_{\bm{\omega}},\tilde{\rho}_{\bm{\omega}}\\}_{\bm{\omega}\in\mathsf{S}}$;
thus, at least one of $\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$ and
$\mathcal{X}_{\mathsf{S}^{\prime\prime}}^{\mathsf{S}}$ is empty for any
$\mathsf{S}^{\prime},\mathsf{S}^{\prime\prime}\subseteq\mathsf{S}$ with
$\mathcal{P}_{\mathsf{S}^{\prime}}\cap\mathcal{P}_{\mathsf{S}^{\prime\prime}}=\varnothing$.
In particular, we use the following two lemmas when proving Theorem 2.

######  Lemma 4.

When $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm
prior}$, if $\mathsf{S}$ is a proper subset of $\Omega$ with
$\mathcal{X}_{\mathsf{S}}\neq\varnothing$ and $\mathsf{S}^{\prime}$ is a
proper subset of $\mathsf{S}$ satisfying
$\mathsf{S}-\mathsf{S}^{\prime}=\\{\bm{\omega}\\}$ and
$\mathcal{X}_{\mathsf{S}^{\prime}}=\\{\bm{v}\\}$, then

$\begin{array}[]{lclcl}\varphi_{\bm{\omega}^{\prime}}(\bm{v})<\varphi_{\bm{\omega}}(\bm{v})\
\forall\bm{\omega}^{\prime}\in\mathsf{S}^{\prime}.\end{array}$ (84)

###### Proof.

$\mathcal{P}_{\mathsf{S}}$ and $\mathcal{P}_{\mathsf{S}^{\prime}}$ are
disjoint for any $\mathsf{S^{\prime}}\subsetneq\mathsf{S}\subsetneq\Omega$
because $\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\Omega}$ forms a
parallelogram with nonempty interior when $m=n_{1}=n_{2}=2$ and $p_{\rm
guess}^{\rm post}<p_{\rm guess}^{\rm prior}$. Thus, if
$\mathcal{X}_{\mathsf{S}}$ is nonempty for some $\mathsf{S}\subsetneq\Omega$,
then $\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$ is empty for all
$\mathsf{S}^{\prime}\subsetneq\mathsf{S}$. Now, we assume that $\mathsf{S}$ is
a proper subset of $\Omega$ with $\mathcal{X}_{\mathsf{S}}\neq\varnothing$ and
$\mathsf{S}^{\prime}$ is a proper subset of $\mathsf{S}$ satisfying
$\mathsf{S}-\mathsf{S}^{\prime}=\\{\bm{\omega}\\}$ and
$\mathcal{X}_{\mathsf{S}^{\prime}}=\\{\bm{v}\\}$. If
$\varphi_{\bm{\omega}^{\prime}}(\bm{v})\geqslant\varphi_{\bm{\omega}}(\bm{v})$
for some $\bm{\omega}^{\prime}\in\mathsf{S}^{\prime}$, then
$\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$ becomes a nonempty set with
$\bm{v}$ as a element. Thus,
$\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}=\varnothing$ implies (84),
which completes our proof. ∎

######  Lemma 5.

When $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm
prior}$, $\mathcal{X}_{\mathsf{S}}$ is nonempty for
$\mathsf{S}\subsetneq\Omega$ if
$\mathcal{X}_{\mathsf{S}^{\prime}}\neq\varnothing$ and
$\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}=\varnothing$ for all
$\mathsf{S}^{\prime}\subsetneq\mathsf{S}$ with
$|\mathsf{S}-\mathsf{S}^{\prime}|=1$.

###### Proof.

For each
$\mathsf{S}^{\prime\prime}\subsetneq\mathsf{S}^{\prime}\subsetneq\mathsf{S}$
with $|\mathsf{S}-\mathsf{S}^{\prime}|=1$,
$\mathcal{X}_{\mathsf{S}^{\prime\prime}}^{\mathsf{S}^{\prime}}$ is empty
because
$\bigcup_{\mathsf{S}^{\prime\prime}\subseteq\mathsf{S}^{\prime}}\mathcal{X}_{\mathsf{S}^{\prime\prime}}^{\mathsf{S}^{\prime}}$
is a single-element set and $\mathcal{X}_{\mathsf{S}^{\prime}}$ is a nonempty
set. Note that $\mathcal{P}_{\mathsf{S}^{\prime}}$ and
$\mathcal{P}_{\mathsf{S}^{\prime\prime}}$ are disjoint for any
$\mathsf{S}^{\prime\prime}\subsetneq\mathsf{S}^{\prime}\subsetneq\mathsf{S}$
since $\\{\tilde{\bm{\mu}}_{\bm{\omega}}\\}_{\bm{\omega}\in\Omega}$ forms a
parallelogram with nonempty interior when $m=n_{1}=n_{2}=2$ and $p_{\rm
guess}^{\rm post}<p_{\rm guess}^{\rm prior}$. For
$\mathsf{S}^{\prime\prime}\subsetneq\mathsf{S}^{\prime}\subsetneq\mathsf{S}$,
emptiness of $\mathcal{X}_{\mathsf{S}^{\prime\prime}}^{\mathsf{S}^{\prime}}$
implies emptiness of $\mathcal{X}_{\mathsf{S}^{\prime\prime}}^{\mathsf{S}}$ as
$\mathcal{X}_{\mathsf{S}^{\prime\prime}}^{\mathsf{S}}\subseteq\mathcal{X}_{\mathsf{S}^{\prime\prime}}^{\mathsf{S}^{\prime}}$;
therefore, $\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$ is empty for all
$\mathsf{S}^{\prime}\subsetneq\mathsf{S}$. Since
$\bigcup_{\mathsf{S}^{\prime}\subseteq\mathsf{S}}\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$
is a single-element set, $\mathcal{X}_{\mathsf{S}}$ is nonempty if
$\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$ is empty for all
$\mathsf{S}^{\prime}\subsetneq\mathsf{S}$. Thus, $\mathcal{X}_{\mathsf{S}}$ is
nonempty. ∎

### Proof of Theorem 1

###### Proof.

Suppose that $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm post}<p_{\rm
guess}^{\rm prior}$. We can easily verify that

$\begin{array}[]{lcl}\mathcal{X}_{\\{(1,1),(1,2)\\}}=\\{{\bm{v}}_{11}\\},\ \
\varphi_{(1,1)}(\bm{v}_{11})+\varphi_{(1,2)}(\bm{v}_{11})=2\eta_{11}+\eta_{12}+\eta_{22}+\lambda_{22},\\\\[4.2679pt]
\mathcal{X}_{\\{(1,1),(2,1)\\}}=\\{{\bm{v}}_{12}\\},\ \
\varphi_{(1,1)}(\bm{v}_{12})+\varphi_{(2,1)}(\bm{v}_{12})=2\eta_{12}+\eta_{11}+\eta_{21}+\lambda_{21},\\\\[4.2679pt]
\mathcal{X}_{\\{(2,1),(2,2)\\}}=\\{{\bm{v}}_{21}\\},\ \
\varphi_{(2,1)}(\bm{v}_{21})+\varphi_{(2,2)}(\bm{v}_{21})=2\eta_{21}+\eta_{12}+\eta_{22}+\lambda_{22},\\\\[4.2679pt]
\mathcal{X}_{\\{(1,2),(2,2)\\}}=\\{{\bm{v}}_{22}\\},\ \
\varphi_{(1,2)}(\bm{v}_{22})+\varphi_{(2,2)}(\bm{v}_{22})=2\eta_{22}+\eta_{11}+\eta_{21}+\lambda_{21},\end{array}$
(85)

where

$\begin{array}[]{ll}{\bm{v}}_{11}=\Big{(}\frac{\lambda_{22}+\epsilon_{22}}{2\lambda_{22}}\Big{)}\tilde{\bm{\mu}}_{(1,1)}+\Big{(}\frac{\lambda_{22}-\epsilon_{22}}{2\lambda_{22}}\Big{)}\tilde{\bm{\mu}}_{(1,2)},&{\bm{v}}_{12}=\Big{(}\frac{\lambda_{21}+\epsilon_{21}}{2\lambda_{21}}\Big{)}\tilde{\bm{\mu}}_{(1,1)}+\Big{(}\frac{\lambda_{21}-\epsilon_{21}}{2\lambda_{21}}\Big{)}\tilde{\bm{\mu}}_{(2,1)},\\\\[4.2679pt]
{\bm{v}}_{21}=\Big{(}\frac{\lambda_{22}+\epsilon_{22}}{2\lambda_{22}}\Big{)}\tilde{\bm{\mu}}_{(2,1)}+\Big{(}\frac{\lambda_{22}-\epsilon_{22}}{2\lambda_{22}}\Big{)}\tilde{\bm{\mu}}_{(2,2)},&{\bm{v}}_{22}=\Big{(}\frac{\lambda_{21}+\epsilon_{21}}{2\lambda_{21}}\Big{)}\tilde{\bm{\mu}}_{(1,2)}+\Big{(}\frac{\lambda_{21}-\epsilon_{21}}{2\lambda_{21}}\Big{)}\tilde{\bm{\mu}}_{(2,2)}.\end{array}$
(86)

Since
$\mathcal{X}_{\mathsf{S}}\cap\mathcal{Y}_{\mathsf{S}}=\mathcal{Z}_{\mathsf{S}}$
for all $\mathsf{S}\subseteq\Omega$, it follows from Eq. (85) that

$\begin{array}[]{lclcl}\mathcal{Z}_{\\{(1,1),(1,2)\\}}\neq\varnothing&\
\Leftrightarrow&{\bm{v}}_{11}\in\mathcal{Y}_{\\{(1,1),(1,2)\\}},\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,1),(2,1)\\}}\neq\varnothing&\
\Leftrightarrow&{\bm{v}}_{12}\in\mathcal{Y}_{\\{(1,1),(2,1)\\}},\\\\[2.84526pt]
\mathcal{Z}_{\\{(2,1),(2,2)\\}}\neq\varnothing&\
\Leftrightarrow&{\bm{v}}_{21}\in\mathcal{Y}_{\\{(2,1),(2,2)\\}},\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,2),(2,2)\\}}\neq\varnothing&\
\Leftrightarrow&{\bm{v}}_{22}\in\mathcal{Y}_{\\{(1,2),(2,2)\\}},\end{array}$
(87)

and

$p_{\rm guess}^{\rm
post}=\left\\{\begin{array}[]{cc}\eta_{11}+\frac{1}{2}(\eta_{12}+\eta_{22}+\lambda_{22}),&\mathcal{Z}_{\\{(1,1),(1,2)\\}}\neq\varnothing,\\\\[2.84526pt]
\eta_{12}+\frac{1}{2}(\eta_{11}+\eta_{21}+\lambda_{21}),&\mathcal{Z}_{\\{(1,1),(2,1)\\}}\neq\varnothing,\\\\[2.84526pt]
\eta_{21}+\frac{1}{2}(\eta_{12}+\eta_{22}+\lambda_{22}),&\mathcal{Z}_{\\{(2,1),(2,2)\\}}\neq\varnothing,\\\\[2.84526pt]
\eta_{22}+\frac{1}{2}(\eta_{11}+\eta_{21}+\lambda_{21}),&\mathcal{Z}_{\\{(1,2),(2,2)\\}}\neq\varnothing.\\\\[2.84526pt]
\end{array}\right.$ (88)

From definition of $\mathcal{Y}_{\mathsf{S}}$, we can see that

$\begin{array}[]{lcr}\bm{v}_{11}\in\mathcal{Y}_{\\{(1,1),(1,2)\\}}&\
\Leftrightarrow&\varphi_{(1,1)}({\bm{v}}_{11})\geqslant\varphi_{(2,1)}({\bm{v}}_{11}),\
\
\varphi_{(1,2)}({\bm{v}}_{11})\geqslant\varphi_{(2,2)}({\bm{v}}_{11}),\\\\[4.2679pt]
\bm{v}_{12}\in\mathcal{Y}_{\\{(1,1),(2,1)\\}}&\
\Leftrightarrow&\varphi_{(1,1)}({\bm{v}}_{12})\geqslant\varphi_{(1,2)}({\bm{v}}_{12}),\
\
\varphi_{(2,1)}({\bm{v}}_{12})\geqslant\varphi_{(2,2)}({\bm{v}}_{12}),\\\\[4.2679pt]
\bm{v}_{21}\in\mathcal{Y}_{\\{(2,1),(2,2)\\}}&\
\Leftrightarrow&\varphi_{(1,1)}({\bm{v}}_{21})\leqslant\varphi_{(2,1)}({\bm{v}}_{21}),\
\
\varphi_{(1,2)}({\bm{v}}_{21})\leqslant\varphi_{(2,2)}({\bm{v}}_{21}),\\\\[4.2679pt]
\bm{v}_{22}\in\mathcal{Y}_{\\{(1,2),(2,2)\\}}&\
\Leftrightarrow&\varphi_{(1,1)}({\bm{v}}_{22})\leqslant\varphi_{(1,2)}({\bm{v}}_{22}),\
\
\varphi_{(2,1)}({\bm{v}}_{22})\leqslant\varphi_{(2,2)}({\bm{v}}_{22}).\\\\[4.2679pt]
\end{array}$ (89)

It is straightforward to verify that

$\begin{array}[]{rcl}\varphi_{(1,1)}({\bm{v}}_{11})-\varphi_{(2,1)}({\bm{v}}_{11})&=&\frac{(\epsilon_{21}+\lambda_{21}\cos\Phi)(\lambda_{22}-\epsilon_{22})-(\lambda_{21}^{2}-\epsilon_{21}^{2})}{\epsilon_{21}+\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}_{11}\|+\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}_{11}\|},\\\\[4.2679pt]
\varphi_{(1,2)}({\bm{v}}_{11})-\varphi_{(2,2)}({\bm{v}}_{11})&=&\frac{(\epsilon_{21}-\lambda_{21}\cos\Phi)(\lambda_{22}+\epsilon_{22})-(\lambda_{21}^{2}-\epsilon_{21}^{2})}{\epsilon_{21}+\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}_{11}\|+\|\tilde{\bm{\mu}}_{(2,2)}-\bm{v}_{11}\|},\\\\[8.53581pt]
\varphi_{(1,1)}({\bm{v}}_{12})-\varphi_{(1,2)}({\bm{v}}_{12})&=&\frac{(\epsilon_{22}+\lambda_{22}\cos\Phi)(\lambda_{21}-\epsilon_{21})-(\lambda_{22}^{2}-\epsilon_{22}^{2})}{\epsilon_{22}+\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}_{12}\|+\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}_{12}\|},\\\\[4.2679pt]
\varphi_{(2,1)}({\bm{v}}_{12})-\varphi_{(2,2)}({\bm{v}}_{12})&=&\frac{(\epsilon_{22}-\lambda_{22}\cos\Phi)(\lambda_{21}+\epsilon_{21})-(\lambda_{22}^{2}-\epsilon_{22}^{2})}{\epsilon_{22}+\|\tilde{\bm{\mu}}_{(2,1)}-\bm{v}_{12}\|+\|\tilde{\bm{\mu}}_{(2,2)}-\bm{v}_{12}\|},\\\\[8.53581pt]
\varphi_{(1,1)}({\bm{v}}_{21})-\varphi_{(2,1)}({\bm{v}}_{21})&=&\frac{(\epsilon_{21}+\lambda_{21}\cos\Phi)(\lambda_{22}-\epsilon_{22})+(\lambda_{21}^{2}-\epsilon_{21}^{2})}{\|\tilde{\bm{\mu}}_{(2,1)}-\bm{v}_{21}\|+\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}_{21}\|-\epsilon_{21}},\\\\[4.2679pt]
\varphi_{(1,2)}({\bm{v}}_{21})-\varphi_{(2,2)}({\bm{v}}_{21})&=&\frac{(\epsilon_{21}-\lambda_{21}\cos\Phi)(\lambda_{22}+\epsilon_{22})+(\lambda_{21}^{2}-\epsilon_{21}^{2})}{\|\tilde{\bm{\mu}}_{(2,2)}-\bm{v}_{21}\|+\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}_{21}\|-\epsilon_{21}},\\\\[8.53581pt]
\varphi_{(1,1)}({\bm{v}}_{22})-\varphi_{(1,2)}({\bm{v}}_{22})&=&\frac{(\epsilon_{22}+\lambda_{22}\cos\Phi)(\lambda_{21}-\epsilon_{21})+(\lambda_{22}^{2}-\epsilon_{22}^{2})}{\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}_{22}\|+\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}_{22}\|-\epsilon_{22}},\\\\[4.2679pt]
\varphi_{(2,1)}({\bm{v}}_{22})-\varphi_{(2,2)}({\bm{v}}_{22})&=&\frac{(\epsilon_{22}-\lambda_{22}\cos\Phi)(\lambda_{21}+\epsilon_{21})+(\lambda_{22}^{2}-\epsilon_{22}^{2})}{\|\tilde{\bm{\mu}}_{(2,2)}-\bm{v}_{22}\|+\|\tilde{\bm{\mu}}_{(2,1)}-\bm{v}_{22}\|-\epsilon_{22}}.\end{array}$
(90)

Therefore,

$\displaystyle\mathcal{Z}_{\\{(1,1),(1,2)\\}}\neq\varnothing$
$\displaystyle\Leftrightarrow$
$\frac{\epsilon_{21}+\lambda_{21}\cos\Phi}{\lambda_{21}+\epsilon_{21}}\geqslant\frac{\lambda_{21}-\epsilon_{21}}{\lambda_{22}-\epsilon_{22}},\quad\frac{\epsilon_{21}-\lambda_{21}\cos\Phi}{\lambda_{21}-\epsilon_{21}}\geqslant\frac{\lambda_{21}+\epsilon_{21}}{\lambda_{22}+\epsilon_{22}},$
(91a) $\displaystyle\mathcal{Z}_{\\{(1,1),(2,1)\\}}\neq\varnothing$
$\displaystyle\Leftrightarrow$
$\displaystyle\mbox{$\frac{\epsilon_{22}+\lambda_{22}\cos\Phi}{\lambda_{22}+\epsilon_{22}}\geqslant\frac{\lambda_{22}-\epsilon_{22}}{\lambda_{21}-\epsilon_{21}},\quad\frac{\epsilon_{22}-\lambda_{22}\cos\Phi}{\lambda_{22}-\epsilon_{22}}\geqslant\frac{\lambda_{22}+\epsilon_{22}}{\lambda_{21}+\epsilon_{21}}$},$
(91b) $\displaystyle\mathcal{Z}_{\\{(2,1),(2,2)\\}}\neq\varnothing$
$\displaystyle\Leftrightarrow$
$\frac{\epsilon_{21}+\lambda_{21}\cos\Phi}{\lambda_{21}+\epsilon_{21}}\leqslant-\frac{\lambda_{21}-\epsilon_{21}}{\lambda_{22}-\epsilon_{22}},\,\frac{\epsilon_{21}-\lambda_{21}\cos\Phi}{\lambda_{21}-\epsilon_{21}}\leqslant-\frac{\lambda_{21}+\epsilon_{21}}{\lambda_{22}+\epsilon_{22}},$
(91c) $\displaystyle\mathcal{Z}_{\\{(1,2),(2,2)\\}}\neq\varnothing$
$\displaystyle\Leftrightarrow$
$\frac{\epsilon_{22}+\lambda_{22}\cos\Phi}{\lambda_{22}+\epsilon_{22}}\leqslant-\frac{\lambda_{22}-\epsilon_{22}}{\lambda_{21}-\epsilon_{21}},\,\frac{\epsilon_{22}-\lambda_{22}\cos\Phi}{\lambda_{22}-\epsilon_{22}}\leqslant-\frac{\lambda_{22}+\epsilon_{22}}{\lambda_{21}+\epsilon_{21}}.$
(91d)

Moreover, at least one of the two left-hand sides of Inequalities (91c) is
positive, but the two right-hand sides are both negative; thus,
$\mathcal{Z}_{\\{(2,1),(2,2)\\}}$ is always empty. For similar reasons,
$\mathcal{Z}_{\\{(1,2),(2,2)\\}}$ is always empty. ∎

### Proof of Theorem 2

###### Proof of (a).

Suppose $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm
prior}$. Then, we can easily verify that

$\begin{array}[]{lclcl}\mathcal{X}_{\\{(1,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\mathcal{X}_{\\{(1,1),(2,2)\\}}=\\{{\bm{v}}_{+}\\},\
\
\varphi_{(1,1)}(\bm{v}_{+})+\varphi_{(2,2)}(\bm{v}_{+})=1+\gamma_{+},\\\\[2.84526pt]
\mathcal{X}_{\\{(1,2),(2,1)\\}}\neq\varnothing&\Rightarrow&\mathcal{X}_{\\{(1,2),(2,1)\\}}=\\{{\bm{v}}_{-}\\},\
\
\varphi_{(1,2)}(\bm{v}_{-})+\varphi_{(2,1)}(\bm{v}_{-})=1+\gamma_{-},\end{array}$
(92)

where

$\begin{array}[]{rclcrcl}{\bm{v}}_{+}&=&\Big{(}\frac{\gamma_{+}+\epsilon_{21}+\epsilon_{22}}{2\gamma_{+}}\Big{)}\tilde{\bm{\mu}}_{(1,1)}+\Big{(}\frac{\gamma_{+}-\epsilon_{21}-\epsilon_{22}}{2\gamma_{+}}\Big{)}\tilde{\bm{\mu}}_{(2,2)},\\\\[4.2679pt]
{\bm{v}}_{-}&=&\Big{(}\frac{\gamma_{-}+\epsilon_{21}-\epsilon_{22}}{2\gamma_{-}}\Big{)}\tilde{\bm{\mu}}_{(1,2)}+\Big{(}\frac{\gamma_{-}-\epsilon_{21}+\epsilon_{22}}{2\gamma_{-}}\Big{)}\tilde{\bm{\mu}}_{(2,1)}.\end{array}$
(93)

Since
$\mathcal{X}_{\mathsf{S}}\cap\mathcal{Y}_{\mathsf{S}}=\mathcal{Z}_{\mathsf{S}}$
for all $\mathsf{S}\subseteq\Omega$, Eq. (92) implies

$\begin{array}[]{lclcl}\mathcal{Z}_{\\{(1,1),(2,2)\\}}\neq\varnothing&\Rightarrow&{\bm{v}}_{+}\in\mathcal{Y}_{\\{(1,1),(2,2)\\}},\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,2),(2,1)\\}}\neq\varnothing&\Rightarrow&{\bm{v}}_{-}\in\mathcal{Y}_{\\{(1,2),(2,1)\\}},\end{array}$
(94)

as well as the validity of Eq. (53).
From definition of $\mathcal{Y}_{\mathsf{S}}$, we have

$\begin{array}[]{rcl}\bm{v}_{+}\in\mathcal{Y}_{\\{(1,1),(2,2)\\}}&\Leftrightarrow&\varphi_{(1,1)}({\bm{v}}_{+})\geqslant\varphi_{(1,2)}({\bm{v}}_{+}),\
\varphi_{(1,1)}({\bm{v}}_{+})\geqslant\varphi_{(2,1)}({\bm{v}}_{+}),\\\\[4.2679pt]
\bm{v}_{-}\in\mathcal{Y}_{\\{(1,2),(2,1)\\}}&\Leftrightarrow&\varphi_{(1,1)}({\bm{v}}_{-})\leqslant\varphi_{(1,2)}({\bm{v}}_{-}),\
\varphi_{(2,2)}({\bm{v}}_{-})\leqslant\varphi_{(1,2)}({\bm{v}}_{-}).\end{array}$
(95)

It is straightforward to show that

$\begin{array}[]{rcl}\varphi_{(1,1)}({\bm{v}}_{+})-\varphi_{(1,2)}({\bm{v}}_{+})&=&\frac{\gamma_{+}\alpha-\beta_{+}}{\gamma_{+}(\epsilon_{21}+\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}_{+}\|+\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}_{+}\|)},\\\\[4.2679pt]
\varphi_{(1,1)}({\bm{v}}_{+})-\varphi_{(2,1)}({\bm{v}}_{+})&=&\frac{\gamma_{+}\alpha+\beta_{+}}{\gamma_{+}(\epsilon_{21}+\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}_{+}\|+\|\tilde{\bm{\mu}}_{(2,1)}-\bm{v}_{+}\|)},\\\\[4.2679pt]
\varphi_{(1,1)}({\bm{v}}_{-})-\varphi_{(1,2)}({\bm{v}}_{-})&=&\frac{\gamma_{-}\alpha+\beta_{-}}{\gamma_{-}(\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}_{-}\|+\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}_{-}\|-\epsilon_{22})},\\\\[4.2679pt]
\varphi_{(1,1)}({\bm{v}}_{-})-\varphi_{(2,1)}({\bm{v}}_{-})&=&\frac{\gamma_{-}\alpha-\beta_{-}}{\gamma_{-}(\epsilon_{21}+\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}_{-}\|+\|\tilde{\bm{\mu}}_{(2,2)}-\bm{v}_{-}\|)}.\end{array}$
(96)

By applying Eq. (96) to (95) and (95) to (94), we can show that
(52)($\Rightarrow$) is true.
Now, we prove that (52)($\Leftarrow$) is true. Because non-negativity and non-
positivity of $\alpha$ imply
$\gamma_{+}^{2}\geqslant(\epsilon_{21}+\epsilon_{22})^{2}$ and
$\gamma_{-}^{2}\geqslant(\epsilon_{21}-\epsilon_{22})^{2}$, respectively,

$\begin{array}[]{lclcl}\alpha\geqslant
0&\Rightarrow&\mathcal{X}_{\\{(1,1),(2,2)\\}}=\\{\bm{v}_{+}\\},\\\\[4.2679pt]
\alpha\leqslant
0&\Rightarrow&\mathcal{X}_{\\{(1,2),(2,1)\\}}=\\{\bm{v}_{-}\\}.\end{array}$
(97)

Thus, both $\mathcal{X}_{\\{(1,1),(2,2)\\}}$ and
$\mathcal{Y}_{\\{(1,1),(2,2)\\}}$ have $\bm{v}_{+}$ by Eq. (96) and (97) if
$\alpha\geqslant|\beta_{+}|/\gamma_{+}$. Similarly, both
$\mathcal{X}_{\\{(1,2),(2,1)\\}}$ and $\mathcal{Y}_{\\{(1,2),(2,1)\\}}$ have
$\bm{v}_{-}$ if $\alpha\leqslant-|\beta_{-}|/\gamma_{-}$. Therefore,
(52)($\Leftarrow$) is also proved. ∎

###### Proof of (b).

Assume $m=n_{1}=n_{2}=2$ and $p_{\rm guess}^{\rm post}<p_{\rm guess}^{\rm
prior}$. We can see from Lemma 4 and Eq. (85) that

$\begin{array}[]{ccc}\mathcal{X}_{\\{(1,1),(1,2),(2,1)\\}}\neq\varnothing&\Rightarrow&\varphi_{(1,1)}({\bm{v}}_{11})<\varphi_{(2,1)}({\bm{v}}_{11}),\
\varphi_{(1,1)}({\bm{v}}_{12})<\varphi_{(1,2)}({\bm{v}}_{12}),\\\\[4.2679pt]
\mathcal{X}_{\\{(1,1),(1,2),(2,2)\\}}\neq\varnothing&\Rightarrow&\varphi_{(1,2)}({\bm{v}}_{11})<\varphi_{(2,2)}({\bm{v}}_{11}),\
\varphi_{(1,2)}({\bm{v}}_{22})<\varphi_{(1,1)}({\bm{v}}_{22}),\\\\[4.2679pt]
\mathcal{X}_{\\{(1,1),(2,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\varphi_{(2,1)}({\bm{v}}_{12})<\varphi_{(2,2)}({\bm{v}}_{12}),\
\varphi_{(2,1)}({\bm{v}}_{21})<\varphi_{(1,1)}({\bm{v}}_{21}),\\\\[4.2679pt]
\mathcal{X}_{\\{(1,2),(2,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\varphi_{(2,2)}({\bm{v}}_{21})<\varphi_{(1,2)}({\bm{v}}_{21}),\
\varphi_{(2,2)}({\bm{v}}_{22})<\varphi_{(2,1)}({\bm{v}}_{22}).\end{array}$
(98)

By applying Eq. (90) to (98), we have

$\begin{array}[]{ccl}\mathcal{X}_{\\{(1,1),(1,2),(2,1)\\}}\neq\varnothing&\Rightarrow&\frac{\epsilon_{21}+\lambda_{21}\cos\Phi}{\lambda_{21}+\epsilon_{21}}<\frac{\lambda_{21}-\epsilon_{21}}{\lambda_{22}-\epsilon_{22}},\quad\frac{\epsilon_{22}+\lambda_{22}\cos\Phi}{\lambda_{22}+\epsilon_{22}}<\frac{\lambda_{22}-\epsilon_{22}}{\lambda_{21}-\epsilon_{21}},\\\\[2.84526pt]
\mathcal{X}_{\\{(1,1),(1,2),(2,2)\\}}\neq\varnothing&\Rightarrow&\frac{\epsilon_{21}-\lambda_{21}\cos\Phi}{\lambda_{21}-\epsilon_{21}}<\frac{\lambda_{21}+\epsilon_{21}}{\lambda_{22}+\epsilon_{22}},\quad\frac{\epsilon_{22}+\lambda_{22}\cos\Phi}{\lambda_{22}+\epsilon_{22}}>-\frac{\lambda_{22}-\epsilon_{22}}{\lambda_{21}-\epsilon_{21}},\\\\[2.84526pt]
\mathcal{X}_{\\{(1,1),(2,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\frac{\epsilon_{22}-\lambda_{22}\cos\Phi}{\lambda_{22}-\epsilon_{22}}<\frac{\lambda_{22}+\epsilon_{22}}{\lambda_{21}+\epsilon_{21}},\quad\frac{\epsilon_{21}+\lambda_{21}\cos\Phi}{\lambda_{21}+\epsilon_{21}}>-\frac{\lambda_{21}-\epsilon_{21}}{\lambda_{22}-\epsilon_{22}},\\\\[2.84526pt]
\mathcal{X}_{\\{(1,2),(2,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\frac{\epsilon_{21}-\lambda_{21}\cos\Phi}{\lambda_{21}-\epsilon_{21}}>-\frac{\lambda_{21}+\epsilon_{21}}{\lambda_{22}+\epsilon_{22}},\
\frac{\epsilon_{22}-\lambda_{22}\cos\Phi}{\lambda_{22}-\epsilon_{22}}>-\frac{\lambda_{22}+\epsilon_{22}}{\lambda_{21}+\epsilon_{21}}.\end{array}$
(99)

Since the inequalities given in (99) can also be expressed as

$\begin{array}[]{lcl}\frac{\epsilon_{22}\mp\lambda_{22}\cos\Phi}{\lambda_{22}\mp\epsilon_{22}}<\frac{\lambda_{22}\pm\epsilon_{22}}{\lambda_{21}\pm\epsilon_{21}}&\Leftrightarrow&\Theta_{\pm}>\Xi_{\pm},\\\\[2.84526pt]
\frac{\epsilon_{21}\mp\lambda_{21}\cos\Phi}{\lambda_{21}\mp\epsilon_{21}}<\frac{\lambda_{21}\pm\epsilon_{21}}{\lambda_{22}\pm\epsilon_{22}}&\Leftrightarrow&\mp\Theta_{\pm}<\Phi\mp\Xi_{\pm}<\pi\mp\pi\pm\Theta_{\pm},\\\\[2.84526pt]
\frac{\epsilon_{22}\pm\lambda_{22}\cos\Phi}{\lambda_{22}\pm\epsilon_{22}}>-\frac{\lambda_{22}\mp\epsilon_{22}}{\lambda_{21}\mp\epsilon_{21}}&\Leftrightarrow&\Theta_{\pm}<\pi-\Xi_{\pm},\\\\[2.84526pt]
\frac{\epsilon_{21}\pm\lambda_{21}\cos\Phi}{\lambda_{21}\pm\epsilon_{21}}>-\frac{\lambda_{21}\mp\epsilon_{21}}{\lambda_{22}\mp\epsilon_{22}}&\Leftrightarrow&\mp\pi\pm\Theta_{\pm}<\Phi\mp\Xi_{\pm}<\pi\mp\Theta_{\pm},\end{array}$
(100)

a lengthy calculation show that

$\begin{array}[]{ccccc}\bm{v}\in\mathcal{X}_{\\{(1,1),(1,2),(2,1)\\}}&\Rightarrow&\left\\{\begin{array}[]{ccc}\angle\bm{v}\tilde{\bm{\mu}}_{(1,1)}\tilde{\bm{\mu}}_{(2,1)}&=&\Theta_{-}-\Xi_{-},\\\\[4.2679pt]
\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}\|&=&\frac{\lambda_{21}^{2}-\epsilon_{21}^{2}}{2[\lambda_{21}\cos(\Theta_{-}-\Xi_{-})+\epsilon_{21}]},\\\\[4.2679pt]
\varphi_{(1,1)}({\bm{v}})-\varphi_{(2,2)}({\bm{v}})&=&\frac{-2\alpha}{\|\tilde{\mu}_{(1,1)}-\bm{v}\|+\|\tilde{\mu}_{(2,2)}-\bm{v}\|+\epsilon_{21}+\epsilon_{22}},\end{array}\right.\\\\[22.76219pt]
\bm{v}\in\mathcal{X}_{\\{(1,1),(1,2),(2,2)\\}}&\Rightarrow&\left\\{\begin{array}[]{ccc}\angle\bm{v}\tilde{\bm{\mu}}_{(1,2)}\tilde{\bm{\mu}}_{(2,2)}&=&\pi-\Theta_{+}-\Xi_{+},\\\\[4.2679pt]
\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}\|&=&\frac{\lambda_{21}^{2}-\epsilon_{21}^{2}}{2[\lambda_{21}\cos(\pi-\Theta_{+}-\Xi_{+})+\epsilon_{21}]},\\\\[4.2679pt]
\varphi_{(1,2)}({\bm{v}})-\varphi_{(2,1)}({\bm{v}})&=&\frac{+2\alpha}{\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}\|+\|\tilde{\bm{\mu}}_{(2,1)}-\bm{v}\|+\epsilon_{21}-\epsilon_{22}},\end{array}\right.\\\\[22.76219pt]
\bm{v}\in\mathcal{X}_{\\{(1,1),(2,1),(2,2)\\}}&\Rightarrow&\left\\{\begin{array}[]{ccc}\angle\bm{v}\tilde{\bm{\mu}}_{(2,1)}\tilde{\bm{\mu}}_{(1,1)}&=&\Theta_{+}-\Xi_{+},\\\\[4.2679pt]
\|\tilde{\bm{\mu}}_{(2,1)}-\bm{v}\|&=&\frac{\lambda_{21}^{2}-\epsilon_{21}^{2}}{2[\lambda_{21}\cos(\Theta_{+}-\Xi_{+})-\epsilon_{21}]},\\\\[4.2679pt]
\varphi_{(2,1)}({\bm{v}})-\varphi_{(1,2)}({\bm{v}})&=&\frac{+2\alpha}{\|\tilde{\bm{\mu}}_{(2,1)}-\bm{v}\|+\|\tilde{\bm{\mu}}_{(1,2)}-\bm{v}\|-\epsilon_{21}+\epsilon_{22}},\end{array}\right.\\\\[22.76219pt]
\bm{v}\in\mathcal{X}_{\\{(1,2),(2,1),(2,2)\\}}&\Rightarrow&\left\\{\begin{array}[]{ccc}\angle\bm{v}\tilde{\bm{\mu}}_{(2,2)}\tilde{\bm{\mu}}_{(1,2)}&=&\pi-\Theta_{-}-\Xi_{-},\\\\[4.2679pt]
\|\tilde{\bm{\mu}}_{(2,2)}-\bm{v}\|&=&\frac{\lambda_{21}^{2}-\epsilon_{21}^{2}}{2[\lambda_{21}\cos(\pi-\Theta_{-}-\Xi_{-})-\epsilon_{21}]},\\\\[4.2679pt]
\varphi_{(2,2)}({\bm{v}})-\varphi_{(1,1)}({\bm{v}})&=&\frac{-2\alpha}{\|\tilde{\bm{\mu}}_{(2,2)}-\bm{v}\|+\|\tilde{\bm{\mu}}_{(1,1)}-\bm{v}\|-\epsilon_{21}-\epsilon_{22}}.\end{array}\right.\end{array}$
(101)

Here,
$\angle\bm{v}\tilde{\bm{\mu}}_{\bm{\omega}}\tilde{\bm{\mu}}_{\bm{\omega}^{\prime}}$
is the internal angle between two line segments formed by
$\\{\bm{v},\tilde{\bm{\mu}}_{\bm{\omega}}\\}$ and
$\\{\tilde{\bm{\mu}}_{\bm{\omega}},\tilde{\bm{\mu}}_{\bm{\omega}^{\prime}}\\}$,
respectively.
As
$\mathcal{X}_{\mathsf{S}}\cap\mathcal{Y}_{\mathsf{S}}=\mathcal{Z}_{\mathsf{S}}$
for all $\mathsf{S}\subseteq\Omega$, Eq. (56) is true from Lemma 3 and Eq.
(101). Moreover, Eq. (101) also implies

$\begin{array}[]{ccc}\mathcal{Z}_{\\{(1,1),(1,2),(2,1)\\}}\neq\varnothing&\Leftrightarrow&\mathcal{X}_{\\{(1,1),(1,2),(2,1)\\}}\neq\varnothing,\
\alpha\leqslant 0,\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,1),(1,2),(2,2)\\}}\neq\varnothing&\Leftrightarrow&\mathcal{X}_{\\{(1,1),(1,2),(2,2)\\}}\neq\varnothing,\
\alpha\geqslant 0,\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,1),(2,1),(2,2)\\}}\neq\varnothing&\Leftrightarrow&\mathcal{X}_{\\{(1,1),(2,1),(2,2)\\}}\neq\varnothing,\
\alpha\geqslant 0,\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,2),(2,1),(2,2)\\}}\neq\varnothing&\Leftrightarrow&\mathcal{X}_{\\{(1,2),(2,1),(2,2)\\}}\neq\varnothing,\
\alpha\leqslant 0,\end{array}$ (102)

Now, (97) and (102) lead us to

$\begin{array}[]{ccc}\mathcal{Z}_{\\{(1,1),(1,2),(2,1)\\}}\neq\varnothing\
\mbox{or}\
\mathcal{Z}_{\\{(1,2),(2,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\mathcal{X}_{\\{(1,2),(2,1)\\}}=\\{\bm{v}_{-}\\},\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,1),(1,2),(2,2)\\}}\neq\varnothing\ \mbox{or}\
\mathcal{Z}_{\\{(1,1),(2,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\mathcal{X}_{\\{(1,2),(2,1)\\}}=\\{\bm{v}_{+}\\}.\end{array}$
(103)

Thus, we can show from Eq. (96), (103), and Lemma 4 that

$\begin{array}[]{rclcccc}\mathcal{Z}_{\\{(1,1),(1,2),(2,1)\\}}\neq\varnothing&\Rightarrow&\alpha>-\beta_{-}/\gamma_{-},\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,1),(1,2),(2,2)\\}}\neq\varnothing&\Rightarrow&\alpha<-\beta_{+}/\gamma_{+},\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,1),(2,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\alpha<\beta_{+}/\gamma_{+},\\\\[2.84526pt]
\mathcal{Z}_{\\{(1,2),(2,1),(2,2)\\}}\neq\varnothing&\Rightarrow&\alpha>\beta_{-}/\gamma_{-}.\end{array}$
(104)

Up to this point, (55)($\Rightarrow$) for
$\mathsf{S}\neq\\{(1,2),(2,1),(2,2)\\}$ is proved by (99), (102), and (104).
Since $\beta_{-}/\gamma_{-}<\alpha\leqslant 0$ is a necessary condition for
$\mathcal{Z}_{\\{(1,2),(2,1),(2,2)\\}}\neq\varnothing$ from (102) and (104),
$\mathcal{Z}_{\\{(1,2),(2,1),(2,2)\\}}\neq\varnothing$ requires non-positivity
of $\alpha$ and $\beta_{-}$. However, both $\alpha$ and $\beta_{-}$ cannot be
non-positive because

$\alpha(\epsilon_{21}+\epsilon_{22})+\beta_{-}=\epsilon_{21}(\lambda_{22}^{2}-\epsilon_{22}^{2})+\epsilon_{22}(\lambda_{21}^{2}-\epsilon_{21}^{2})>0.$
(105)

Thus, $\mathcal{Z}_{\\{(1,2),(2,1),(2,2)\\}}$ is always empty.
Now, for the proof of (55)($\Leftarrow$), let $\mathsf{S}$ be a proper subset
of $\Omega$ satisfying the necessary condition of (55). Then,
$\mathcal{X}_{\mathsf{S}^{\prime}}$ is nonempty for all
$\mathsf{S}^{\prime}\subsetneq\mathsf{S}$ with $|\mathsf{S}^{\prime}|=2$ from
(85) and (97). Thus, we can see from Lemma 5 and (102) that
$\mathcal{Z}_{\mathsf{S}}$ is nonempty if
$\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}$ is empty for all
$\mathsf{S}^{\prime}\subsetneq\mathsf{S}$ with $|\mathsf{S}^{\prime}|=2$.
Using (90) and (96), we can easily show that
$\mathcal{X}_{\mathsf{S}^{\prime}}^{\mathsf{S}}=\varnothing$ for all
$\mathsf{S}^{\prime}\subsetneq\mathsf{S}$ with $|\mathsf{S}^{\prime}|=2$.
Therefore, $\mathcal{Z}_{\mathsf{S}}\neq\varnothing$. That is,
(55)($\Leftarrow$) is proved. ∎

###### Proof of (c).

Let us assume $\mathcal{Z}\subseteq\mathcal{P}_{\Omega}$. Since a null MEPI
measurement exists from Corollary 1, $\mathcal{Z}_{\mathsf{S}}\neq\varnothing$
for some $\mathsf{S}$ in (51) with $\mathsf{S}\neq\Omega$. When $\bm{v}$ is
the single element in $\mathcal{Z}_{\mathsf{S}}$, we can see from the
definition of $\mathcal{Z}_{\Omega}$ that
$\mathcal{Z}_{\Omega}\neq\varnothing$ if and only if
$\varphi_{\bm{\omega}}(\bm{v})=\varphi_{\bm{\omega}^{\prime}}(\bm{v})$ for all
$(\bm{\omega},\bm{\omega}^{\prime})\in\mathsf{S}\times(\Omega-\mathsf{S})$.
Thus, when $|\mathsf{S}|=3$, we can show from (101) that
$\mathcal{Z}_{\Omega}\neq\varnothing$ is equivalent to $\alpha=0$.
When $|\mathsf{S}|=2$, we can verify from Eqs. (92) and (96) that
$\mathcal{Z}_{\Omega}\neq\varnothing$ is equivalent to $\alpha=0$. Note that
$\alpha=0$ means $\beta_{+}=0$ when $\mathsf{S}=\\{(1,1),(2,2)\\}$ because
$\alpha\geqslant|\beta_{+}|/\gamma_{+}$ is a necessary condition for
$\mathcal{Z}_{\\{(1,1),(2,2)\\}}\neq\varnothing$. We also note that $\alpha=0$
implies $\beta_{-}=0$ when $\mathsf{S}=\\{(1,2),(2,1)\\}$ because
$\alpha\leqslant-|\beta_{-}|/\gamma_{-}$ is necessary for
$\mathcal{Z}_{\\{(1,2),(2,1)\\}}\neq\varnothing$, therefore $\alpha\neq 0$ and
$\mathcal{Z}_{\Omega}=\varnothing$ when $\mathsf{S}=\\{(1,2),(2,1)\\}$ since
both $\alpha$ and $\beta_{-}$ cannot be zero by Inequality (105). Thus,
$\alpha=0$ is a necessary and sufficient condition for
$\mathcal{Z}_{\Omega}\neq\varnothing$ when
$\mathcal{Z}\subseteq\mathcal{P}_{\Omega}$. This completes our proof of
Theorem 2(c). ∎

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