# Heuristic Sensing Schemes for Four-Target Detection in Time-Constrained
Vector Poisson and Gaussian Channels

Muhammad Fahad1, , and Daniel R. Fuhrmann2,
Department of Applied Computing,
Michigan Technological University
Houghton, MI 49931, USA
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

In this work we investigate the different sensing schemes for detection of
four targets as observed through a vector Poisson and Gaussian channels when
the sensing time resource is limited and the source signals can be observed
through a variety of sum combinations during that fixed time. For this purpose
we can maximize the mutual information or the detection probability with
respect to the time allocated to different sum combinations, for a given total
fixed time. It is observed that for both Poisson and Gaussian channels; mutual
information and Bayes risk with $0-1$ cost are not necessarily consistent with
each other. Concavity of mutual information between input and output, for
certain sensing schemes, in Poisson channel and Gaussian channel is shown to
be concave w.r.t given times as linear time constraint is imposed. No optimal
sensing scheme for any of the two channels is investigated in this work.

###### Index Terms:

sensor scheduling, vector Poisson channels, vector Gaussian channels.

##  1 Introduction

In [1], [2] and [3] a two-target detection in vector Poisson and Gaussian
channels is considered. It was observed that prior probability of the targets
have a direct influence in deciding which of the sensing method is better over
the other besides the given available sensing time. As, we study the problem
in higher dimensions (or when targets are greater than two), we are hampered
by the limitations of the deterministic computational methods which fails to
work efficiently, in terms of computational time, due to the curse of the
dimensionality issue. Therefore, to study the problem in higher dimensions;
resorting to some statistical computational method is one way to circumvent
the exponentially rising dimensionality in the objective functions and Monte-
Carlo method is used in this work [4], [5] and [6].

Figure 1: Illustration of sensing paradigm for detection of $4-$long hidden
random vector $X$ from $15-$long observable random vector $Y$ through a vector
Poisson channel under a total time constraint of $T=\sum_{i=1}^{15}T_{i}.$
inline]1) Open draw.io in chrome.
2) Open file from
C:$\backslash$Users$\backslash$mfahad$\backslash$Dropbox$\backslash$Apps$\backslash$drawio$\backslash$Sensing_Paradigm4.html
3) Export to HTML and save in the same folder.
4) Open HTML and print PDF
5) Crop the pdf using acrobats tools: crop
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Figure 2: Illustration of sensing paradigm for detection of $4-$long hidden
random vector $X$ from $15-$long observable random vector $Y$ through a vector
Gaussian channel under a total time constraint. Where $w_{i}(t)$ are
independent white noise processes. Only one of the integrators becomes active
for a time $T_{i}$ such that total time constraint $T=\sum_{i=1}^{15}T_{i}$ is
satisfied by all the integrators. inline]1) Open draw.io in chrome.
2) Open file from
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3) Export to HTML and save in the same folder.
4) Open HTML and print PDF
5) Crop the pdf using acrobats tools: crop
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C:$\backslash$Users$\backslash$mfahad$\backslash$Dropbox$\backslash$PhD_Paper_1

This paper considers an experimental design problem of setting, sub-optimally,
the time-proportions for identifying a four-long binary random vector that is
passed through a vector Poisson and vector Gaussian channels, and then based
on the observation vector; classification of the input vector is performed and
performance of any sensing scheme is then compared. Since, finding the optimal
solution for the problem requires computations to be performed in
(15-dimensional search-space as closed-form solutions don’t exist); we have
instead restricted to a reduced dimensional search-space and studied some
sensing techniques that are sub-optimal.

We start by explaining the problem in the vector Poisson channel [7], [8] and
[9]. The problem is set up such that there is a $4-$long binary input vector
$X=[\allowbreak X_{1},\allowbreak X_{2},\allowbreak X_{3},\allowbreak X_{4}]$
and each $X_{i}$ is a discrete random variable that assumes either of the two
known values: $\lambda_{0}$ or $\lambda_{1}$ with probability $(1-p)$ and $p$,
respectively. All $X_{i}$ are mutually independent and identically
distributed. Conditioned on $X_{i}$, a Poisson process $\mathcal{P}_{i}(t)$ is
initiated in continuous time $t$ [10]. It is known that If we count the
arrivals for time $T_{i}$ from the conditional Poisson process we have another
conditional counting Poisson process whose rate parameter at instant $T_{i}$
is $(T_{i}\cdot x_{i})$. Hence we have, initially, four conditional Poisson
processes: $\mathcal{P}_{1}(T_{1}\cdot x_{1})$; $\mathcal{P}_{2}(T_{2}\cdot
x_{2})$; $\mathcal{P}_{3}(T_{3}\cdot x_{3})$ and $\mathcal{P}_{4}(T_{4}\cdot
x_{4})$ depending on the realization that input vector $X$ assumes. Let
${4\choose 2}$ be the set containing all possible pairs constituted from four
elements of $X$. Summing elements of each of the $6-$pairs we then have
another six conditional Poisson processes:
$\mathcal{P}_{5}(T_{5}\cdot(x_{1}+x_{2}))$;
$\mathcal{P}_{6}(T_{6}\cdot(x_{1}+x_{3}))$;
$\mathcal{P}_{7}(T_{7}\cdot(x_{1}+x_{4}))$;
$\mathcal{P}_{8}(T_{8}\cdot(x_{2}+x_{3}))$;
$\mathcal{P}_{9}(T_{9}\cdot(x_{2}+x_{4}))$ and
$\mathcal{P}_{10}(T_{10}\cdot(x_{3}+x_{4}))$. Considering ${4\choose 3}$ we
have four processes: $\mathcal{P}_{11}(T_{11}\cdot(x_{1}+x_{2}+x_{3}))$;
$\mathcal{P}_{12}(T_{12}\cdot(x_{1}+x_{2}+x_{4}))$;
$\mathcal{P}_{13}(T_{13}\cdot(x_{1}+x_{3}+x_{4}))$ and
$\mathcal{P}_{14}(T_{14}\cdot(x_{2}+x_{3}+x_{4}))$. Summing all the four
components of $X$, we have
$\mathcal{P}_{15}(T_{15}\cdot(x_{1}+x_{2}+x_{3}+x_{4}))$. Hence, there are
$15-$ conditional point processes (in total) that we have to deal with to
extract the maximum possible information or perform the best input signal
detection by setting the counting times from $T_{1}$ to $T_{15}$ in a fixed
given time $\sum_{i=1}^{15}=T$ as illustrated in fig. (2).

The ideal way to address the problem would be to search for a solution in a
$15-\allowbreak$ dimensional search-space by allowing $(T_{1},T_{2},\cdots
T_{15})$ to have fifteen degrees-of-freedom. However, due to the computational
complexity involved in exploring all fifteen dimensions we restrict ourselves
to a reduced dimensional search-space, as said above. Therefore, we have
considered only some special cases of time-configurations. Four different
types of time configurations are studied for each of the channels. We call:
_individual_ sensing when total given time $T$ is equally divided into
$T_{1}=T_{2}=T_{3}=T_{4}=\frac{T}{4}$; _pair-wise_ sensing when
$T_{5}=T_{6}=T_{7}=T_{8}=T_{9}=T_{10}=\frac{T}{6}$; _triplets_ sensing when
$T_{11}=T_{12}=T_{13}=T_{14}=\frac{T}{4}$ and _joint_ sensing when $T_{15}=T$.

The first problem is: does there exist a configuration among these four
configurations which is always performing the best for any given time $T$ and
prior $p$? To answer this we first fixed $p$, and then we consider $T$ as a
free parameter and compute both the mutual information [11], [12] and Bayes
probability of total correct detections [13], for a given set of parameters,
and searched if there exist any instance for which one configuration is the
best for some time and then another configuration becomes the best and so on.
From mutual information perspective: it is computationally observed that when
prior $p<0.5$ then depending on the value of $T$ any of the four schemes can
be better over the others however when $p\geq 0.5$ it is the individual
sensing that works best. However, from the detection perspective this is not
the case as indicated in fig. (3). It is further shown that in each
configuration mutual information is concave in $T$.

The second problem: does there exist a _hybrid_ sensing mechanism that
performs better than any of the above four configurations for fixed time $T$?
A hybrid sensing is one when given time $T$ is divide into any one of the four
sensing configurations and joint sensing according to the proportion:
$(1-\alpha)\cdot T$ and $\alpha\cdot T$ where $0\leq\alpha\leq 1$,
respectively. It turned out that if prior $p\geq 0.5$ then irrespective of
other model parameters; individual sensing is the best among any other
configurations. For $p$ close to zero hybrid sensing is better over any other
as indicated in fig. (4). A concavity of mutual information w.r.t $\alpha$ is
observed, but no proof is given.

For the vector Gaussian model we have a fixed unit covariance matrix and input
$X$ only affects the mean vector. Replace all $T_{i}$ with $\sqrt{T_{i}}$ in
Poisson model; we have the mean vector for Gaussian channel. It is found that
triplet-sensing almost always outperforms any other configuration,
irrespective of model parameters. This is shown in fig. (5) and fig. (6). It
is shown that mutual information is concave in $T$ for any of the four
configurations; further in hybrid sensing the mutual information remains
concave in $\alpha$. However, Bayes probability of total correct detection is
not necessarily consistent with mutual information results.

The paper is organized as follows: Section 2 defines the vector Poisson and
Gaussian channel that we have considered. Section 3 describes the _detection
theoretic_ model of the problem. Section 4 defines the computational setup.
Finally, Section 5 concludes the paper.

Notation: Upper case letters denote random vectors. Realizations of the random
vectors are denoted by lower case letters. A number in subscript is used to
show the component number of the random vector. We use $X_{1}$ and $Y_{1}$ to
represent scalar input and output random variables, respectively. The
superscript $(\cdot)^{\intercal}$ denotes the matrix/vector transpose. $T$ is
a given finite time. $\alpha$ is an arbitrary positive scalar variable. $\Phi$
represents the scaling matrix. $p$ is the prior probability. $f_{X}(x)$
denotes the probability mass function of $X$. $\operatorname{Pois}(U;z)$
denotes the standard Poisson distribution of random variable $U$ with
parameter $z$. We may omit $U$ in some cases. $K-$dimensional multivariate
Gaussian distribution is represented by
$\mathcal{N}(w;\mu,\Sigma)=(2\pi)^{-\frac{k}{2}}\mathtt{det}(\Sigma)^{-\frac{1}{2}}e^{-{\frac{1}{2}}(w-\mu)^{\intercal}\Sigma^{-1}(w-\mu)}$.
$w$ might be omitted for the purpose of brevity.

##  2 Vector Poisson and Gaussian Channels

###  2-A Vector Poisson Channel

We consider the vector Poisson channel model [14]:

$\displaystyle\operatorname{Pois}(Y;\Phi X)$ $\displaystyle=$ $\displaystyle
P_{Y|X}(Y|X)=\prod_{i=1}^{15}P_{Y_{i}|X}(Y_{i}|X)$ (1) $\displaystyle=$
$\displaystyle\>\prod_{i=1}^{15}\operatorname{Pois}(Y_{i};(\Phi X)_{i})$

where $\operatorname{Pois}(U;z)$ denotes the standard Poisson distribution of
random variable $U$ with parameter $z$.

We assume input
$X=(X_{1},X_{2},X_{3},X_{4})\in\\{\lambda_{0},\lambda_{1}\\}^{4}$ such that
$0\leq\lambda_{0}<\lambda_{1}$, each $X_{i}$ is independent and identically
distributed with a pmf: $p_{X_{i}}(x_{i}=\lambda_{0})=1-p$ and
$p_{X_{i}}(x_{i}=\lambda_{1})=p$. $Y=(Y_{1},Y_{2},\cdots
Y_{15})\in\mathbb{Z}_{+}^{15}$ and

$\displaystyle\Phi$ $\displaystyle=\begin{bmatrix}T_{1}&0&0&0\\\
0&T_{2}&0&0\\\ 0&0&T_{3}&0\\\ 0&0&0&T_{4}\\\ T_{5}&T_{5}&0&0\\\
T_{6}&0&T_{6}&0\\\ T_{7}&0&0&T_{7}\\\ 0&T_{8}&T_{8}&0\\\ 0&T_{9}&0&T_{9}\\\
0&0&T_{10}&T_{10}\\\ T_{11}&T_{11}&T_{11}&0\\\ T_{12}&T_{12}&0&T_{12}\\\
T_{13}&0&T_{13}&T_{13}\\\ 0&T_{14}&T_{14}&T_{14}\\\
T_{15}&T_{15}&T_{15}&T_{15}\\\ \end{bmatrix}.$ (2)

The conditional distribution of vector $Y$ given $X$ is a multivariate Poisson
distribution:

$\displaystyle Y\Bigg{|}\Big{(}X=(x_{1}\>x_{2}\>x_{3}\>x_{4})\Big{)}$
$\displaystyle\sim\operatorname{Pois}\Bigg{(}\begin{bmatrix}T_{1}\cdot
x_{1}\\\ T_{2}\cdot x_{2}\\\ T_{3}\cdot x_{3}\\\ T_{4}\cdot x_{4}\\\
T_{5}\cdot(x_{1}+x_{2})\\\ T_{6}\cdot(x_{1}+x_{3})\\\
T_{7}\cdot(x_{1}+x_{4})\\\ T_{8}\cdot(x_{2}+x_{3})\\\
T_{9}\cdot(x_{2}+x_{4})\\\ T_{10}\cdot(x_{3}+x_{4})\\\
T_{11}\cdot(x_{1}+x_{2}+x_{3})\\\ T_{12}\cdot(x_{1}+x_{2}+x_{4})\\\
T_{13}\cdot(x_{1}+x_{3}+x_{4})\\\ T_{14}\cdot(x_{2}+x_{3}+x_{4})\\\
T_{15}\cdot(x_{1}+x_{2}+x_{3}+x_{4})\\\ \end{bmatrix}\Bigg{)},$ (3)

We define mutual information $I(X;Y)$ as

$\displaystyle I(X;Y)=H(Y)-H(Y|X),$ (4)

where $H(Y)$ is an entropy of a finite Poisson mixture model given as

$\displaystyle H(Y)$ $\displaystyle=$
$\displaystyle-\sum_{y_{1}=-\infty}^{\infty}\sum_{y_{2}=-\infty}^{\infty}\cdots\cdots\sum_{y_{15}=-\infty}^{\infty}\>P(Y)\cdot\operatorname{Log}_{2}[P(Y)]$
(5)

where

$\displaystyle P(Y)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{16}\Bigg{(}P_{X_{i}}(X_{i}=x_{1},x_{2},x_{3},x_{4})\cdot\operatorname{Pois}\Big{(}y_{1};{T_{1}}x_{1}\Big{)}\cdot$
(6)
$\displaystyle\>\operatorname{Pois}\Big{(}y_{2};{T_{2}}x_{2}\Big{)}\cdot\operatorname{Pois}\Big{(}y_{3};{T_{3}}x_{3}\Big{)}\cdot\operatorname{Pois}\Big{(}y_{4};{T_{4}}x_{4}\Big{)}\cdot\operatorname{Pois}\Big{(}y_{5};{T_{5}}(x_{1}+x_{2})\Big{)}\cdot$
$\displaystyle\>\operatorname{Pois}\Big{(}y_{6};{T_{6}}(x_{1}+x_{3})\Big{)}\cdot\operatorname{Pois}\Big{(}y_{7};{T_{7}}(x_{1}+x_{4})\Big{)}\cdot\operatorname{Pois}\Big{(}y_{8};{T_{8}}(x_{2}+x_{3})\Big{)}\cdot$
$\displaystyle\>\operatorname{Pois}\Big{(}y_{9};{T_{9}}(x_{2}+x_{4})\Big{)}\cdot\operatorname{Pois}\Big{(}y_{10};{T_{10}}(x_{3}+x_{4})\Big{)}\cdot\operatorname{Pois}\Big{(}y_{11};{T_{11}}(x_{1}+x_{2}+x_{3})\Big{)}\cdot$
$\displaystyle\>\operatorname{Pois}\Big{(}y_{12};{T_{12}}(x_{1}+x_{2}+x_{4})\Big{)}\cdot\operatorname{Pois}\Big{(}y_{13};{T_{13}}(x_{1}+x_{3}+x_{4})\Big{)}\cdot$
$\displaystyle\>\operatorname{Pois}\Big{(}y_{14};{T_{14}}(x_{2}+x_{3}+x_{4})\Big{)}\cdot\operatorname{Pois}\Big{(}y_{15};{T_{15}}(x_{1}+x_{2}+x_{3}+x_{4})\Big{)}\Bigg{)},$

and

$\displaystyle H(Y|X)$ $\displaystyle=$
$\displaystyle-\sum_{y_{1}=-\infty}^{\infty}\sum_{y_{2}=-\infty}^{\infty}\cdots\cdots\sum_{y_{15}=-\infty}^{\infty}\sum_{i=1}^{16}P_{X_{i}}(X_{i})\cdot
P(Y|X_{i})\cdot\operatorname{Log}_{2}[P(Y|X_{i})]$ (7)

where $P(Y|X_{i})=\prod_{j=1}^{15}\operatorname{Pois}(Y_{j};(\Phi X)_{i})$.

###### Theorem 1

$I(X_{1},X_{2},X_{3},X_{4};Y_{1},Y_{2},Y_{3}\cdots Y_{15})$ is symmetric in
variable-groups: $\allowbreak(T_{1},\allowbreak T_{2},\allowbreak
T_{3},\allowbreak T_{4})\allowbreak$;
$\allowbreak(T_{5},T_{6},T_{7},\allowbreak T_{8},T_{9},T_{10})$; and
$(T_{11},T_{12},T_{13},T_{14})$.

###### Proof:

Mutual information $I(X;Y)$ given in (4) is invariant under any permutation of
variables belonging to the same group. That means interchanging the variables
within the same group leaves the expression unchanged. ∎

####  2-A1 Unconstrained objective

For a vector Poisson channel with given prior $p$, $\lambda_{0}$ and
$\lambda_{1}$, which of the following four methods are better over the others
when each expression is a function of $T$ solely,

$\displaystyle
I(X_{1},X_{2},X_{3},X_{4};\overbrace{Y_{1},Y_{2},Y_{3},Y_{4})}^{\rm{4-Singlets}}$
s.t. $\displaystyle T_{1}=T_{2}=T_{3}=T_{4}=\frac{T}{4}$ $\displaystyle
I(X_{1},X_{2},X_{3},X_{4};\overbrace{Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10})}^{\rm{6-Pairs}}\quad$
s.t. $\displaystyle\quad T_{5}=T_{6}=T_{7}=T_{8}=T_{9}=T_{10}=\frac{T}{6}$
$\displaystyle
I(X_{1},X_{2},X_{3},X_{4};\overbrace{Y_{11},Y_{12},Y_{13},Y_{14})}^{\rm{4-Triplets}}\quad$
s.t. $\displaystyle\quad T_{11}=T_{12}=T_{13}=T_{14}=\frac{T}{4}$
$\displaystyle
I(X_{1},X_{2},X_{3},X_{4};\overbrace{Y_{15})}^{\rm{1-Quadruplet}}\quad$ s.t.
$\displaystyle\quad T_{15}=T$

###### Theorem 2

$I(X_{1},X_{2},X_{3},X_{4};Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10})$ s.t.
$T_{5}=T_{6}=T_{7}=T_{8}=T_{9}=T_{10}=\frac{T}{6}$ is concave in $T$.

###### Proof:

$I(X_{1},X_{2},X_{3},X_{4};Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10})=\allowbreak
I((X_{1}+X_{2}),(X_{1}+X_{3}),\allowbreak(X_{1}+X_{4}),\allowbreak(X_{2}+\allowbreak
X_{3}),\allowbreak(X_{2}+\allowbreak
X_{4}),\allowbreak(X_{3}+X_{4});Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10})$.

Consider the R.H.S of the above equation. From [15, p. 1315], for a random
n-tuple vector $X=(X_{1},\cdots X_{n})$ and for $T\geq 0$, let
$Y=(Y_{1},\cdots Y_{n})$ be jointly distributed with $X$ such that given $X$,
the components of $Y$ are independent with
$Y_{i}|X\sim\operatorname{Poiss}(T\cdot X_{i}),1\leq i\leq n.$ Then
$\operatorname{mmle}(T)=E\Big{[}\sum_{i=1}^{n}l(X_{i},E[X_{i}|Y])\Big{]}.$
Since $\frac{\partial}{\partial T}I(X_{i};Y)=\operatorname{mmle}(T).$ Each
$E\Big{[}l(X_{i},E[X_{i}|Y])\Big{]}$ is concave in $T$. Sum of concave
functions result in another Concave function. ∎

###### Corollary 1

Expressions in (LABEL:e21), (LABEL:e23) and (LABEL:e24) are concave in $T$
too.

Note that expressions in (LABEL:e21), (LABEL:e22) and (LABEL:e23), have a
tight upper bound of $H(X_{1},X_{2},X_{3},X_{4})$ as $T\rightarrow\infty$
since the corresponding mappings: from $X$ to $6-$pairs and from $X$ to
$4-$triplets are invertible. Whereas, the expression (LABEL:e24) has a tight
upper bound of $H(\sum_{i=1}^{4}X_{i})$ when $T\rightarrow\infty,$ the mapping
from $(X_{1},X_{2},X_{3},X_{4})\mapsto\sum_{i=1}^{4}X_{i}$ is non-invertible.

####  2-A2 Constraint objective

The second objective is to determine which of the following three
configurations are better over the others for a given prior $p$,
$\lambda_{0}$, $\lambda_{1}$ and given fixed time $T$ i.e.,

$\displaystyle\mathrm{Config-1}:$ $\displaystyle
I(X;\overbrace{Y_{1},Y_{2},Y_{3},Y_{4},Y_{15})}^{\rm{4-Singlets+1-Quadruplet}}\>\text{s.t.}\>T_{1}=T_{2}=T_{3}=T_{4}=\frac{T-\alpha}{4},T_{15}=\alpha.$
(12) $\displaystyle\mathrm{Config-2}:$ $\displaystyle
I(X;\overbrace{Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10},Y_{15})}^{\rm{6-Pairs+1-Quadruplet}}\>\text{s.t.}\>T_{5}=T_{6}=T_{7}=T_{8}=T_{9}=T_{10}=\frac{T-\alpha}{6},$
(13) $\displaystyle\>T_{15}=\alpha.$ $\displaystyle\mathrm{Config-3:}$
$\displaystyle
I(X;\overbrace{Y_{11},Y_{12},Y_{13},Y_{14},Y_{15})}^{\rm{4-Triplets+1-Quadruplet}}\>\text{s.t.}\>T_{11}=T_{12}=T_{13}=T_{14}=\frac{T-\alpha}{4},T_{15}=\alpha.$

where $0\leq\alpha\leq T$.

###  2-B Vector Gaussian Channel

We consider the vector Gaussian channel model as defined in [14] i.e.,
$Y|X\sim\mathcal{N}(\Phi X,\mathrm{I})$. For a scalar Gaussian channel
$Y=\sqrt{T}\cdot X+N$ with $N\sim\mathcal{N}(0,1)$; $I(X;Y)$ is concave in $T$
for arbitrary input signalling [16]. We extend this scalar model to the vector
case. We may also write as

$\displaystyle\begin{bmatrix}Y_{1}\\\ Y_{2}\\\ Y_{3}\\\ Y_{4}\\\ Y_{5}\\\
Y_{6}\\\ Y_{7}\\\ Y_{8}\\\ Y_{9}\\\ Y_{10}\\\ Y_{11}\\\ Y_{12}\\\ Y_{13}\\\
Y_{14}\\\ Y_{15}\\\ \end{bmatrix}$
$\displaystyle=\begin{bmatrix}\sqrt{T_{1}}&0&0&0\\\ 0&\sqrt{T_{2}}&0&0\\\
0&0&\sqrt{T_{3}}&0\\\ 0&0&0&\sqrt{T_{4}}\\\ \sqrt{T_{5}}&\sqrt{T_{5}}&0&0\\\
\sqrt{T_{6}}&0&\sqrt{T_{6}}&0\\\ \sqrt{T_{7}}&0&0&\sqrt{T_{7}}\\\
0&\sqrt{T_{8}}&\sqrt{T_{8}}&0\\\ 0&\sqrt{T_{9}}&0&\sqrt{T_{9}}\\\
0&0&\sqrt{T_{10}}&\sqrt{T_{10}}\\\
\sqrt{T_{11}}&\sqrt{T_{11}}&\sqrt{T_{11}}&0\\\
\sqrt{T_{12}}&\sqrt{T_{12}}&0&\sqrt{T_{12}}\\\
\sqrt{T_{13}}&0&\sqrt{T_{13}}&\sqrt{T_{13}}\\\
0&\sqrt{T_{14}}&\sqrt{T_{14}}&\sqrt{T_{14}}\\\
\sqrt{T_{15}}&\sqrt{T_{15}}&\sqrt{T_{15}}&\sqrt{T_{15}}\\\
\end{bmatrix}\begin{bmatrix}X_{1}\\\ X_{2}\\\ X_{3}\\\ X_{4}\\\
\end{bmatrix}+\begin{bmatrix}N_{1}\\\ N_{2}\\\ N_{3}\\\ N_{4}\\\ N_{5}\\\
N_{6}\\\ N_{7}\\\ N_{8}\\\ N_{9}\\\ N_{10}\\\ N_{11}\\\ N_{12}\\\ N_{13}\\\
N_{14}\\\ N_{15}\\\ \end{bmatrix},$ (15)

where each $X_{i}$ is independent and identical distributed (i.i.d) discrete
random variable with support $\\{\lambda_{0},\lambda_{1}\\}$ such that $p$ is
the probability of occurrence of $\lambda_{1}$ and $(1-p)$ is the occurrence
of $\lambda_{0}$ i.e., probability mass function $f$ of scalar input random
variable $X_{i}$ is

$\displaystyle f_{X_{i}}(x_{i})=\left\\{\begin{array}[]{ll}p&\quad
x=\>\lambda_{1}\\\ (1-p)&\quad x=\>\lambda_{0}\\\ \end{array}\right.$

Noise vector $N$ is a multivariate Gaussian with zero mean and identity
covariance matrix $\mathrm{I}$; and independent of input $X$. The constraint
on the scaling matrix is $\sum_{i=1}^{15}T_{i}=T$. The conditional
distribution of vector $Y$ given $X$ is a multivariate Gaussian:

$\displaystyle Y\Bigg{|}\Big{(}X=(x_{1}\>x_{2}\>x_{3}\>x_{4})\Big{)}$
$\displaystyle\sim\mathcal{N}\Bigg{(}\begin{bmatrix}\sqrt{T_{1}}\cdot x_{1}\\\
\sqrt{T_{2}}\cdot x_{2}\\\ \sqrt{T_{3}}\cdot x_{3}\\\ \sqrt{T_{4}}\cdot
x_{4}\\\ \sqrt{T_{5}}\cdot(x_{1}+x_{2})\\\ \sqrt{T_{6}}\cdot(x_{1}+x_{3})\\\
\sqrt{T_{7}}\cdot(x_{1}+x_{4})\\\ \sqrt{T_{8}}\cdot(x_{2}+x_{3})\\\
\sqrt{T_{9}}\cdot(x_{2}+x_{4})\\\ \sqrt{T_{10}}\cdot(x_{3}+x_{4})\\\
\sqrt{T_{11}}\cdot(x_{1}+x_{2}+x_{3})\\\
\sqrt{T_{12}}\cdot(x_{1}+x_{2}+x_{4})\\\
\sqrt{T_{13}}\cdot(x_{1}+x_{3}+x_{4})\\\
\sqrt{T_{14}}\cdot(x_{2}+x_{3}+x_{4})\\\
\sqrt{T_{15}}\cdot(x_{1}+x_{2}+x_{3}+x_{4})\\\
\end{bmatrix},\mathrm{I}\Bigg{)},$ (17)

where $\mathrm{I}$ is an identity matrix of size $15\times 15$.

We define mutual information $I(X;Y)$ as

$\displaystyle I(X;Y)=H(Y)-H(Y|X),$ (18)

where $H(Y)$ is a differential entropy of a finite Gaussian mixture model
(gmm) given as

$\displaystyle H(Y)$ $\displaystyle=$
$\displaystyle-\int_{y_{1}=-\infty}^{\infty}\int_{y_{2}=-\infty}^{\infty}\cdots\cdots\int_{{y_{15}}=-\infty}^{\infty}\>P(Y)\cdot\operatorname{Log}_{2}[P(Y)]\>\,d{y_{1}}\,d{y_{2}}\,d{y_{3}}\cdots\cdots\,d{y_{15}},$
(19)

where

$\displaystyle P(Y)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{16}\Bigg{(}P_{X_{i}}(X_{i}=x_{1},x_{2},x_{3},x_{4})\cdot\mathcal{N}\Big{(}y_{1};\sqrt{T_{1}}x_{1},1\Big{)}\cdot$
(20)
$\displaystyle\>\mathcal{N}\Big{(}y_{2};\sqrt{T_{2}}x_{2},1\Big{)}\cdot\mathcal{N}\Big{(}y_{3};\sqrt{T_{3}}x_{3},1\Big{)}\cdot\mathcal{N}\Big{(}y_{4};\sqrt{T_{4}}x_{4},1\Big{)}\cdot\mathcal{N}\Big{(}y_{5};\sqrt{T_{5}}(x_{1}+x_{2}),1\Big{)}\cdot$
$\displaystyle\>\mathcal{N}\Big{(}y_{6};\sqrt{T_{6}}(x_{1}+x_{3}),1\Big{)}\cdot\mathcal{N}\Big{(}y_{7};\sqrt{T_{7}}(x_{1}+x_{4}),1\Big{)}\cdot\mathcal{N}\Big{(}y_{8};\sqrt{T_{8}}(x_{2}+x_{3}),1\Big{)}\cdot$
$\displaystyle\>\mathcal{N}\Big{(}y_{9};\sqrt{T_{9}}(x_{2}+x_{4}),1\Big{)}\cdot\mathcal{N}\Big{(}y_{10};\sqrt{T_{10}}(x_{3}+x_{4}),1\Big{)}\cdot\mathcal{N}\Big{(}y_{11};\sqrt{T_{11}}(x_{1}+x_{2}+x_{3}),1\Big{)}\cdot$
$\displaystyle\>\mathcal{N}\Big{(}y_{12};\sqrt{T_{12}}(x_{1}+x_{2}+x_{4}),1\Big{)}\cdot\mathcal{N}\Big{(}y_{13};\sqrt{T_{13}}(x_{1}+x_{3}+x_{4}),1\Big{)}\cdot$
$\displaystyle\>\mathcal{N}\Big{(}y_{14};\sqrt{T_{14}}(x_{2}+x_{3}+x_{4}),1\Big{)}\cdot\mathcal{N}\Big{(}y_{15};\sqrt{T_{15}}(x_{1}+x_{2}+x_{3}+x_{4}),1\Big{)}\Bigg{)},$

and

$\displaystyle H(Y|X)$ $\displaystyle=$ $\displaystyle 15\times
0.5\times\operatorname{Log}_{2}[2\pi\>e].$ (21)

The multidimensional integral defined in (19) have no closed-form solution,
and therefore we need to resort to the Monte Carlo method. The following
method is used to numerically evaluated the integral using sampling from a
finite Gaussian mixture.

$\displaystyle H(Y)$ $\displaystyle=$ $\displaystyle
E[-\operatorname{Log}_{2}[P_{Y}(Y)]]$ (22) $\displaystyle\approx$
$\displaystyle\>-\frac{\sum_{i}\operatorname{Log}_{2}[P_{Y}(s_{i})]}{N_{s}},$

Where $P_{Y}(\cdot)$ is the mixture probability distribution of $Y$, $N_{s}$
is the number of MC samples and $s_{i}$ is the $i^{th}$ sample from
multivariate Gaussian mixture distribution.

###### Theorem 3

$I(X_{1},X_{2},X_{3},X_{4};Y_{1},Y_{2},Y_{3}\cdots Y_{15})$ is symmetric in
variable-groups: $\allowbreak(T_{1},\allowbreak T_{2},\allowbreak
T_{3},\allowbreak T_{4})$; $\allowbreak(T_{5},T_{6},T_{7},\allowbreak
T_{8},T_{9},T_{10})$; and $(T_{11},T_{12},T_{13},T_{14})$.

###### Proof:

Mutual information $I(X;Y)$ given in (18) is invariant under any permutation
of variables belonging to the same group. That means interchanging the
variables within the same group leaves the expression unchanged. ∎

####  2-B1 Unconstrained objective

For a vector Gaussian channel with given prior $p$, $\lambda_{0}$ and
$\lambda_{1}$, which of the following four methods are better over the others
when each expression is a function of $T$ solely,

$\displaystyle
I(X_{1},X_{2},X_{3},X_{4};\overbrace{Y_{1},Y_{2},Y_{3},Y_{4}}^{\rm{4-Singlets}})$
s.t. $\displaystyle T_{1}=T_{2}=T_{3}=T_{4}=\frac{T}{4}$ $\displaystyle
I(X_{1},X_{2},X_{3},X_{4};\overbrace{Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10}}^{\rm{6-Pairs}})\quad$
s.t. $\displaystyle\quad T_{5}=T_{6}=T_{7}=T_{8}=T_{9}=T_{10}=\frac{T}{6}$
$\displaystyle
I(X_{1},X_{2},X_{3},X_{4};\overbrace{Y_{11},Y_{12},Y_{13},Y_{14}}^{\rm{4-Triplets}})\quad$
s.t. $\displaystyle\quad T_{11}=T_{12}=T_{13}=T_{14}=\frac{T}{4}$
$\displaystyle
I(X_{1},X_{2},X_{3},X_{4};\overbrace{Y_{15}}^{\rm{1-Quadruplet}})\quad$ s.t.
$\displaystyle\quad T_{15}=T$

###### Theorem 4

$I(X_{1},X_{2},X_{3},X_{4};Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10})$ in
(LABEL:e42) is concave in $T$.

###### Proof:

It is noted in [17, Theorem 5] that mutual information is a concave function
of the squared singular values $(\boldsymbol{\lambda})$ of the precoder matrix
$P$ if the first $m^{{}^{\prime}}$ eigenvectors of the channel covariance
matrix $(\mathbf{R}_{H}=H^{\intercal}\mathbf{R}_{Z}^{-1}H)$ coincide with the
left singular vectors of the precoder $P$ i-e $\mathsf{H}_{\lambda}I(S;Y)\leq
0$ for the signal model $Y=HPS+Z$ where $H\in\mathbb{R}^{n\times p}$ is the
channel, $S$ is the input signaling $S\in\mathbb{R}^{m}$, $P$ is a precoder
matrix $P\in\mathbb{R}^{p\times m}$ and $Z\in\mathbb{R}^{n}$ is Gaussian noise
independent of the input $S$ and has covariance matrix $\mathbf{R}_{Z}$.

For our problem: $H=I,$ $\mathbf{R}^{-1}_{Z}=\Lambda=I,$ $P=\Phi$, $S=X$ and
$Z=N.$ The singular value decomposition of $\Phi=U\Sigma V^{*}$.

By substituting $T_{1}=T_{2}=T_{3}=T_{4}=0$, $T_{5}=T_{6}=\cdots T_{10}=T$ and
$T_{15}=0$ in (15), the squared singular values of $\Phi$ are
$[\lambda_{1},\lambda_{2}\cdots\lambda_{15}]\allowbreak=\allowbreak[6T,2T,2T,2T,0,0,0,\cdots
0]$. This is just the composition with an affine transformation on the domain.
∎

###### Corollary 2

$I(X;Y_{1},Y_{2},Y_{3},Y_{4})$, $I(X;Y_{11},Y_{12},Y_{13},Y_{14})$ and
$I(X;Y_{15})$ are concave in $T$, since squared singular values are
$\allowbreak[T,T,T,T,0,0\cdots 0]$, $\allowbreak[9T,2T,\allowbreak
2T,\allowbreak 2T,\allowbreak 0,0,0,\allowbreak\cdots 0]$ and
$\allowbreak[4T,\allowbreak 0,0,0,\allowbreak\cdots 0]$ respectively.

####  2-B2 Constraint objective

The second objective for the vector Gaussian channel is which of the following
three configurations are better over the others for a given prior $p$,
$\lambda_{0}$, $\lambda_{1}$ and given fixed time $T$ i.e.,

$\displaystyle\mathrm{Config-1}:$ $\displaystyle
I(X;\overbrace{Y_{1},Y_{2},Y_{3},Y_{4},Y_{15})}^{\rm{4-Singlets+1-Quadruplet}}\>\text{s.t.}\>T_{1}=T_{2}=T_{3}=T_{4}=\frac{T-\alpha}{4},T_{15}=\alpha.$
(27) $\displaystyle\mathrm{Config-2}:$ $\displaystyle
I(X;\overbrace{Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10},Y_{15})}^{\rm{6-Pairs+1-Quadruplet}}\>\text{s.t.}\>T_{5}=T_{6}=T_{7}=T_{8}=T_{9}=T_{10}=\frac{T-\alpha}{6},$
(28) $\displaystyle\>T_{15}=\alpha.$ $\displaystyle\mathrm{Config-3:}$
$\displaystyle
I(X;\overbrace{Y_{11},Y_{12},Y_{13},Y_{14},Y_{15})}^{\rm{4-Triplets+1-Quadruplet}}\>\text{s.t.}\>T_{11}=T_{12}=T_{13}=T_{14}=\frac{T-\alpha}{4},T_{15}=\alpha.$
(29)

where $0\leq\alpha\leq T$.

###### Theorem 5

$I(X_{1},X_{2},X_{3},X_{4};Y_{5},Y_{6},Y_{7},Y_{8},Y_{9},Y_{10},Y_{15})\>\text{
s.t. }\>T_{5}=\allowbreak T_{6}\allowbreak=T_{7}\allowbreak=T_{8}=\allowbreak
T_{9}\allowbreak=T_{10}=\allowbreak\frac{T-\alpha}{6},\allowbreak
T_{15}=\alpha,$ where $0\leq\alpha\leq T,$ is concave in $\alpha$.

###### Proof:

We again resort to the [17, Theorem 5].

For our problem: $H=I,$ $\mathbf{R}^{-1}_{Z}=\Lambda=I,$ $P=\Phi$, $S=X$ and
$Z=N.$ The singular value decomposition of $\Phi=U\Sigma V^{*}$.

By substituting $T_{1}=T_{2}=T_{3}=T_{4}=0$, $T_{5}=T_{6}=\cdots
T_{10}=\frac{T-\alpha}{2}$ and $T_{15}=\alpha$ in (15), the squared singular
values of $\Phi$ are
$[\lambda_{1},\lambda_{2}\cdots\lambda_{15}]\allowbreak=\allowbreak[0,0\cdots
0,\frac{T-\alpha}{3},\allowbreak\frac{T-\alpha}{3},\allowbreak\frac{T-\alpha}{3},\allowbreak
T+3\alpha]$. This is just the composition with an affine transformation on the
domain. Concavity remains preserved under affine transformation [18, page
79-86]. ∎

###### Corollary 3

$I(X;Y_{1},Y_{2},Y_{3},Y_{4},Y_{15})\>\text{ s.t. }\>T_{1}=\allowbreak
T_{2}=\allowbreak T_{3}=\allowbreak
T_{4}=\allowbreak\frac{T-\alpha}{4},T_{15}=\alpha$ and $I(X;Y_{11},\allowbreak
Y_{12},Y_{13},\allowbreak Y_{14},Y_{15})\>\text{ s.t.
}\>T_{11}=T_{12}=T_{13}=T_{14}=\frac{T-\alpha}{4},T_{15}=\alpha$ are concave
in $\alpha$, since squared singular values are
$[\lambda_{1},\lambda_{2}\cdots\lambda_{15}]\allowbreak=\allowbreak[0,0\cdots
0,\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T+15\alpha}{4}]$
and
$[\lambda_{1},\lambda_{2}\cdots\lambda_{15}]\allowbreak=\allowbreak[0,0\cdots
0,\frac{T-\alpha}{4},\allowbreak\frac{T-\alpha}{4},\allowbreak\frac{T-\alpha}{4},\allowbreak\frac{9T+7\alpha}{4}]$,
respectively.

Therefore, the constraint objectives in (27), (28) and (29) are all concave in
$\alpha$.

##  3 Detection Theoretic Description

###  3-A Bayes criterion

In terms of Bayes detection we may consider the problem as deciding among the
$16-$ hypotheses $(\mathcal{H}_{i},1\leq i\leq 16)$ for a fixed time-
proportions. Considering the prior probability of each hypothesis as $\pi_{i}$
such that $\sum_{i=1}^{16}\pi_{i}=1$. Let $C_{il}$ is the cost of deciding
$\mathcal{D}_{i}$ when $\mathcal{H}_{l}$ is correct, then the average cost is
$r=\sum_{i=1}^{16}\sum_{l=1}^{16}\pi_{l}C_{il}P_{il}$, where $P_{il}$ is the
probability of deciding $\mathcal{D}_{i}$ when $\mathcal{H}_{l}$ is true.

For Gaussian problem with fixed sensing-time proportions;
$\mathcal{H}_{i}:\mathcal{N}_{15}(\mu_{i},\mathbb{I}),$ with prior
$\pi_{i},\quad 1\leq i\leq 16.$ Where $\mathcal{N}_{15}(\mu_{i},\mathbb{I})$
is a $15-$dimensional multivariate normal distribution with fixed covariance
unit-matrix and $15-$component random mean vector $\mu_{i}$.
$P_{il}=\int_{y\in\mathcal{R}_{i}}P_{l}(y_{1},y_{2}\cdots
y_{15}|\mathcal{H}_{i})\partial y.$ We only consider the MAP criterion where
cost is

$\displaystyle C_{il}=\left\\{\begin{array}[]{ll}0&\quad i=l\\\ 1&\quad i\neq
l\\\ \end{array}\right.$

This simplifies the detection rule to deciding:

$\displaystyle\mathcal{D}_{i}:\quad\text{if}\quad\pi_{i}p_{i}(y_{1},y_{2},\cdots
y_{15}|\mathcal{H}_{i})\allowbreak\geq\pi_{n}p_{n}(y_{1},y_{2},\cdots
y_{15}|\mathcal{H}_{n})\quad\text{for all}\quad n\neq i;$ (31)

and for any fixed time-proportions $(T_{1},T_{2},\cdots T_{15})$ under
consideration.

For Poisson problem with fixed sensing-times:
$\mathcal{H}_{i}:\operatorname{Poiss}_{15}(\mu_{i}),$ with prior
$\pi_{i},\quad 1\leq i\leq 16.$ Where
$\operatorname{Pois}_{15}(\mu_{i})=\prod_{i=1}^{15}\operatorname{Pois}((\Phi
X)_{i})$ is a $15-$dimensional multivariate Poisson distribution with $15-$
component random mean vector $\mu_{i}$.
$P_{il}=\sum_{y\in\mathcal{R}_{i}}P_{l}(y_{1},y_{2}\cdots
y_{15}|\mathcal{H}_{i})$. Thus we are interested in minimizing the Bayes risk
$r$(under both constrained and unconstrained objectives defined above) for any
given structure in time-proportions $(T_{1},T_{2}\cdots T_{15})$ i-e

$\displaystyle\underset{(T_{1},T_{2},\cdots
T_{15})}{\text{min}}\>r\quad\text{s.t.}\>\>\sum_{i=1}^{15}T_{i}=T.$ (32)

Equivalently, we may write

$\displaystyle\underset{(T_{1},T_{2},\cdots
T_{15})}{\text{max}}\>P_{d}\quad\text{s.t.}\>\>\sum_{i=1}^{15}T_{i}=T$ (33)

where $P_{d}$ is probability of total correct detections, $P_{d}=1-r.$

###### Conjecture 1

The optimal solution of finding the best time-proportions in a given fixed
time $T$, both under information theoretic and detection theoretic metrics,
has a specific structure:
$(T_{1}:=a,T_{2}:=a,T_{3}:=a,T_{4}:=a,T_{5}:=b,T_{6}:=b,T_{7}:=b,T_{8}:=b,T_{9}:=b,T_{10}:=b,T_{11}:=c,T_{12}:=c,T_{13}:=c,T_{14}:=c,T_{15}:=d)$.
Where $0\leq a,b,c,d\leq T$ and $4\cdot a+6\cdot b+4\cdot c+d=T$.

##  4 Computational setup

To compute the mutual information expressions for the Poisson channel given in
(LABEL:e21)-(LABEL:e24) and for the Gaussian channel given in
(LABEL:e41)-(LABEL:e44), we have utilized a Monte Carlo method. For any of the
time settings, under any sensing scheme, we first generate $10^{6}$ samples
from the respective Poisson mixture pmf (or Gaussian mixture pdf). These
samples are generated in a manner that based on the prior of each component
Poisson multivariate (or component Gaussian multivariate), we took the same
percent of samples from that component. Further, as in each component the
random variables are mutually independent, this simplifies the samples’
generation from any component. After the samples are generated from any
component, for fixed time-proportions $(T_{1},T_{2},\cdots T_{15})$ and given
model parameters ($\lambda_{0},\lambda_{1},T,p$), we calculated the $P(Y)$ as
given in (6) and (20). From the computational-time point-of-view, this
calculation of $P(Y)$ is most time-consuming than any other step and this is
due to the calculation of sixteen $15-$dimensional multivariate components
involved in mixture distribution functions. $\operatorname{Log}_{2}$ is then
taken of the $10^{6}$ points of $P(Y)$ before taking the average as given in
(22). Once we calculated the $H(Y)$, then comes the conditional entropy
$H(Y|X)$. For the conditional Gaussian entropy the expression is simple as
given in (21). For the conditional Poisson entropy we first truncate the
conditional Poisson pmfs of each variable $Y_{i}$ to a sufficiently large
value which is calculated as
$y_{i_{max}}=2\cdot\operatorname{PoissCDF}^{-1}(1-1.110223024625157\cdot
10^{-16},\lambda)$, where $y_{i_{max}}$ is the truncation point and
$\operatorname{PoissCDF}^{-1}(m,\lambda)$ is the inverse Poisson cumulative
distribution function (cdf) with parameter $\lambda$ and at point $m$. After
the truncation; a finite summation for the individual variable $y_{i}$ can be
calculated easily i-e
$H(Y_{i}|\lambda)=-\sum_{j=0}^{y_{i_{max}}}\operatorname{Log}_{2}[\operatorname{Poiss}(j;\lambda)]\cdot\operatorname{Poiss}(j;\lambda)$.
Conditional entropy for Poisson channel can then be readily calculated from
(7). For the MAP detection, we use the same samples, for posterior
probabilities of each of the $15-$hypothesis, that are previously used for the
calculation of mutual information. We had generated the samples from each of
the component with the same proportion as defined by the prior of that
component and then calculated the joint probability of that sample point
$Y_{s}$ with each hypothesis. Deciding in favor of the hypothesis for which
the maximum of the joint probability $P(Y_{s},\mathcal{H}_{i})$ happens among
$16$ such probabilities; as given in (31). The computed results for mutual
information and Bayes probability of total correct detections are shown in
fig. (3), (4), (5), and (6).

(a)

(b)

(c)

(d)

(e)

(f)

Figure 3: Poisson channel: (Left) $I(X;Y)$ vs. $T$, (right) $P_{d}$ vs. $T$
for varying _prior_ probabilities $p$.
inline]FourTarget_Poisson_UnConstraint.m
Newpoiss_nn.m

(a)

(b)

(c)

(d)

(e)

(f)

Figure 4: Poisson channel :
$\rm{Config-1:}\Big{(}\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},0,0,0,0,0,0,0,0,0,0,\alpha\Big{)}$;
$\rm{Config-2:}\Big{(}0,0,0,0,\frac{T-\alpha}{6},\frac{T-\alpha}{6},\frac{T-\alpha}{6},\frac{T-\alpha}{6},\frac{T-\alpha}{6},\frac{T-\alpha}{6},0,0,0,0,\alpha\Big{)}$
and
$\rm{Config-3:}\Big{(}0,0,0,0,0,0,0,0,0,0,\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},\alpha\Big{)}$
where $0\leq\alpha\leq T$ and time constraint $\sum_{i=1}^{15}T_{i}=T$ for
$\lambda_{0}=2$, $\lambda_{1}=20$, and varying _prior_ probability $p$.
inline]FourTarget_Poisson_Constraint.m
Newpoiss_nn.m

(a)

(b)

(c)

(d)

(e)

(f)

Figure 5: Gaussian channel: (Left) $I(X;Y)$ vs. $T$, (right) $P_{d}$ vs. $T$
for varying _prior_ probabilities $p$.
inline]FourTarget_Gaussian_UnConstraint.m
Newgauss_nn.m

(a)

(b)

(c)

(d)

(e)

(f)

Figure 6: Gaussian channel
:$\rm{Config-1:}\Big{(}\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},0,0,0,0,0,0,0,0,0,0,\alpha\Big{)}$;
$\rm{Config-2:}\Big{(}0,0,0,0,\frac{T-\alpha}{6},\frac{T-\alpha}{6},\frac{T-\alpha}{6},\frac{T-\alpha}{6},\frac{T-\alpha}{6},\frac{T-\alpha}{6},0,0,0,0,\alpha\Big{)}$
and
$\rm{Config-3:}\Big{(}0,0,0,0,0,0,0,0,0,0,\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},\frac{T-\alpha}{4},\alpha\Big{)}$
where $0\leq\alpha\leq T$ and time constraint $\sum_{i=1}^{15}T_{i}=T$ for
$\lambda_{0}=5$, $\lambda_{1}=10$, and varying _prior_ probability $p$.
inline]FourTarget_Gaussian_Constraint.m
Newgauss_nn.m

##  5 Conclusion

In this work, a sensor scheduling problem for four target detection in a
vector Poisson and Gaussian channel was considered using metrics of mutual
information $I$ and Bayes risk with $0-1$ cost.

First, four sensing schemes: individual-sensing; pairs-sensing; triplets-
sensing; and joint-sensing were considered with the total given time $T$ being
variable. It was shown that mutual information between input and output is
concave w.r.t given time, (and irrespective of any other model parameters) for
either of the two channels. It is further noted that for the Poisson channel;
individual sensing is the best among the four strategies if prior $p\geq 0.5$
from $I$ perspective. However, in the equivalent Bayesian risk minimization
problem, neither the concavity of Bayesian probability, $P_{d},$ of total
correct detections w.r.t time is observed nor individual sensing is always
found to be the best among others. Whereas for the Gaussian channel it is the
triplets-sensing scheme that almost outperform any other sensing-scheme and
Bayesian $P_{d}$ is not consistent with the $I$ computational results.

Secondly, in another constrained configuration: where total time $T$ is always
held fixed while linearly distributed between joint sensing: and individual
sensing; pairs-sensing and triplets-sensing. From computations; concavity of
$I$ is observed w.r.t time shifting parameter $\alpha$. For the Poisson
problem from the $I$ perspective; it is again the individual-sensing that
outperforms any other configuration for $p\geq 0.5$. This is not very much
consistent from the Bayes detection perspective, however. Pair-wise sensing is
more beneficial than individual sensing for prior close to zero. In the
Gaussian channel it is the triplets-sensing scheme that is the best among any
other scheme and this is evident from both $I$ and $P_{d}$ metrics. It is
shown that $I$ is concave in $\alpha$.

The authors are interested in knowing why time-divisions:
$T_{1}=T_{2}=T_{3}=T_{4}$, and $T_{5}=T_{6}=T_{7}=T_{8}=T_{9}=T_{10}$ and
$T_{11}=T_{12}=T_{13}=T_{14}$ are better than being not equal in respective
groups?

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