# Functional Covering of Point Processes

Nirmal V. Shende and Aaron B. Wagner N. V. Shende is with Marvell Technology,
Inc., Santa Clara, CA 95054 (email: nshende@marvell.com). A. B. Wagner is with
the School of Electrical and Computer Engineering, Cornell University, Ithaca,
NY 14853 (email:wagner@cornell.edu). This work was performed when N. V. Shende
was a student at Cornell University. This paper was presented in part at the
IEEE International Symposium on Information Theory, Paris, Jul. 2019 [1]. This
research was supported by the US National Science Foundation under grants
CCF-1956192, CCF-2008266, and CCF-1934985.

###### Abstract

We introduce a new distortion measure for point processes called functional-
covering distortion. It is inspired by intensity theory and is related to both
the covering of point processes and logarithmic-loss distortion. We obtain the
distortion-rate function with feedforward under this distortion measure for a
large class of point processes. For Poisson processes, the rate-distortion
function is obtained under a general condition called constrained functional-
covering distortion, of which both covering and functional-covering are
special cases. Also for Poisson processes, we characterize the rate-distortion
region for a two-encoder CEO problem and show that feedforward does not
enlarge this region.

##  I Introduction

The classical theory of compression [2] focuses on discrete-time, sequential
sources. The theory is thus well-suited to text, audio, speech, genomic data,
and the like. Continuous-time signals are typically handled by reducing to
discrete-time via projection onto a countable basis. Multi-dimensional
extensions enable application to images and video.

Point processes model a distinct data type that appears in diverse domains
such as neuroscience [3, 4, 5, 6, 7, 8], communication networks [9, 10, 11],
imaging [12, 13], blockchains [14, 15, 16, 17], and photonics [18, 19, 20, 21,
22]. Formally, a point process can be viewed as a random counting measure on
some space of interest [23], or if the space is a real line, a random counting
function; we shall adopt the latter view. Informally, it may be viewed as
simply a random collection of points representing epochs in time or points in
space.

Compression of point processes emerges naturally in several of the above
domains. Sub-cranial implants need to communicate the timing of neural firings
to a monitoring station over a wireless link that is low-rate because it must
traverse the skull [24, 25]. In network flow correlation analysis, one cross-
correlates packet timings from different links in the network [11]; this
requires communication of the packet timings from one place to another.
Compressing point process realizations in 2-D (also known as _point clouds_)
arises in computer vision [26, 27, 28], and so on.

Various specialized approaches have been developed for compressing point
processes, and in particular for measuring distortion. One natural approach is
for the compressed representation to be itself a point-process realization. In
this case, the distortion can be the sum of the absolute value of the
differences between the actual and reconstructed epochs, with the constraint
that the two processes must have the same number of points. For the Poisson
point process, Gallager [29] obtained a lower bound on the rate-distortion
function by insisting on the causal reconstruction of the points but allowing
for their reorder. Bedekar [30] determined the rate-distortion function with
the additional constraint of exact orders of epochs in reconstruction. Verdú
[31] allowed the reconstruction to be non-causal. Coleman _et al._ [32]
introduced the queueing distortion function, where the reproduced epochs lead
the actual epochs. Rubin [33] used the $L_{1}$ distance between the counting
functions as a distortion measure. In a more general setting, Koliander _et
al._ [34] gave upper and lower bounds on the rate-distortion function under a
more generic distortion defined between pair of point processes.

Most relevant to the present paper, Lapidoth _et al._ [35] introduced a
_covering_ distortion measure, where the reconstruction of a point process on
$[0,T]$ is a subset $Y$ of $[0,T]$ that must contain all the points, and the
distortion is the Lebesgue measure of the covering set (see also Shen _et al._
[36]).

If we encode the subset $Y$ as an indicator function

$Y_{t}=\begin{cases}1&\text{if $t\in Y$}\\\ 0&\text{otherwise,}\end{cases}$

then $Y_{t}=0$ guarantees that no point occurred at time $t$, while $Y_{t}=1$
indicates that a point may occur at $t$. More generally, $Y_{t}$ could encode
the relative belief that there is a point at $t$. Inspired by this
observation, and the notion of logarithmic-loss distortion [37, 38], we
consider the following formulation. For a realization of a counting (or point)
process $y_{0}^{T}=(y_{t}:t\in[0,T])$ (i.e., $y_{t}$ is integer-valued, non-
decreasing, and has unit jumps) and a non-negative reconstruction
$\hat{y}_{0}^{T}$, we define the _functional-covering distortion_ as

$\displaystyle
d(\hat{y}_{0}^{T},y_{0}^{T})\triangleq\int_{0}^{T}\hat{y}_{t}\,dt-\log(\hat{y}_{t})\,dy_{t}.$
(1)

This is related to the covering distortion measure in the following sense. If
we impose that $\hat{y}_{t}\in\\{0,1\\}$, then (1) reduces to the covering
distortion measure. Yet it is natural to consider the distortion in (1)
without such a restriction, or with a more general set of allowable values for
$\hat{y}_{t}$. In fact, there are advantages to not restricting
$\hat{y}_{0}^{T}$ to the set $\\{0,1\\}$. Consider a remote source setting
where the encoder cannot access the point-process source directly, but instead
observes a thinned version where some of the points in the source point
process are deleted randomly. Then, in case of the covering distortion the
reconstruction can only be the entire interval $[0,T]$ (i.e.
$\hat{y}_{t}=1,t\in[0,T]$). On the other hand, under functional covering
distortion the problem has a nontrivial solution.

The relation functional covering distortion measure to logarithmic-loss is as
follows. If we constrain $\hat{y}_{0}^{T}$ to be bounded, then we can use a
Girsanov-type transformation [39, Chapter VI, Theorems T2-T4] to define a
probability measure on the set of all counting processes using
$\hat{y}_{0}^{T}$, and the distortion can be defined as the expectation of the
negative logarithm of the Radon-Nikodym derivative between this probability
measure and an appropriately chosen reference measure, evaluated at the source
realization, which is equivalent to (1). However, we will allow
$\hat{y}_{0}^{T}$ to be unbounded but integrable
$\mathbb{E}[\int_{0}^{T}\hat{Y}_{t}\,dt]<\infty$.

The relation to intensity theory is as follows. Heuristically, given a random
variable $M$, the _intensity_ of a point process represented by a counting
function $Y_{0}^{T}$ is a non-negative process $\Gamma_{0}^{T}$ such that
$P(Y_{t+\Delta}-Y_{t}=1|M,Y_{0}^{t})\approx\Gamma_{t}\Delta$ (see Definition 2
for the precise statement). From (1), we expect any optimal $\hat{Y}_{0}^{T}$
(in the rate-distortion trade-off sense) to be related to the intensity of
$Y_{0}^{T}$. In fact, we will see in the proof of Theorem 4 that an optimal
reconstruction $\hat{Y}_{0}^{T}$ is the intensity of $Y_{0}^{T}$ given the
encoder’s output.

Beyond the introduction of the functional covering distortion measure and the
accompanying coding theorems, the paper provides a collection of results for
the information-theoretic analysis of point processes, which may be of
independent use. One such contribution is Theorem 1, where we derive the
mutual information between point-processes with intensities and arbitrary
random variables. This is the most general expression available for mutual
informations involving point process with intensities. Theorem 1 subsumes the
existing formulae for mutual informations involving doubly stochastic Poisson
processes [40, 41, 42] and queuing processes [43] as special cases. The other
theorems proved in this paper are: we obtain the rate-distortion trade-off
with feedforward for the functional-covering distortion measure for point
processes which admit intensities (see Theorem 4). For Poisson processes, we
obtain the rate-distortion region when the reconstruction function
$\hat{y}_{0}^{T}$ is constrained to take value in a subset of reals (Theorem
5). The covering distortion in [35, Theorem 1] is a special case of this
constrained functional-covering distortion, hence the rate-distortion function
in [35] can be obtained as the special case of this theorem. We characterize
the rate-distortion region for a two-encoder Poisson CEO problem (see Figure
1) under functional-covering distortion in Theorem 6. To prove the converse of
the CEO problem, we derive a strong data processing inequality for Poisson
processes under superposition (see Theorem 2), which complements the strong
data processing inequality for Poisson processes under thinning due to Wang
[44]. We also provide a self-contained proof of Wang’s theorem in Theorem 3.
The solution to the CEO problem gives the rate-distortion trade-off for remote
Poisson sources as an immediate corollary.

##  II Preliminaries

We will consider a probability space $(\Omega,\mathcal{F},P)$ on which all
stochastic processes considered here are defined. For a finite $T>0$, let
$(\mathcal{F}_{t}:t\in[0,T])$ be an increasing family of $\sigma$-fields with
$\mathcal{F}_{T}\in\mathcal{F}$. We will assume that the given filtration
$(\mathcal{F}_{t}:t\in[0,T])$, $P$, and $\mathcal{F}$ satisfy the “usual
conditions”[39, Chapter III, p. 75]: $\mathcal{F}$ is complete with respect to
$P$, $\mathcal{F}_{t}$ is right continuous, and $\mathcal{F}_{0}$ contains all
the $P$-null sets of $\mathcal{F}_{t}$. Stochastic processes are denoted as
$\hat{Y}_{0}^{T}=\\{\hat{Y}_{t}:0\leq t\leq T\\}$. The process $X_{0}^{T}$ is
said to be _adapted_ to the history $(\mathcal{F}_{t}:t\in[0,T])$ if $X_{t}$
is $\mathcal{F}_{t}$ measurable for all $t\in[0,T]$. The internal history
recorded by the process $X_{0}^{T}$ is denoted by
$\mathcal{F}^{X}_{t}=(\sigma(X_{s}):s\in[0,t])$, where $\sigma(A)$ denotes the
$\sigma$-field generated by $A$.

A process $X_{0}^{T}$ is called $(\mathcal{F}_{t}:t\in[0,T])$-_predictable_ if
$X_{0}$ is $\mathcal{F}_{0}$ measurable and the mapping $(t,\omega)\to
X_{t}(\omega)$ defined from $(0,T)\times\Omega$ into $\mathbb{R}$ (the set of
real numbers) is measurable with respect to the $\sigma$-field over
$(0,T)\times\Omega$ generated by rectangles of the form

$\displaystyle(s,t]\times A;\quad 0<s\leq t\leq T,\quad A\in\mathcal{F}_{s}.$
(2)

For two measurable spaces $(\Omega_{1},\mathcal{F}_{1})$ and
$(\Omega_{2},\mathcal{F}_{2})$, the product space is denoted by
$(\Omega_{1}\times\Omega_{2},\mathcal{F}_{1}\otimes\mathcal{F}_{2})$. We say
that $A\rightleftarrows B\rightleftarrows C$ forms a Markov chain under
measure $P$ if $A$ and $C$ are conditionally independent given $B$ under $P$.
$P\ll Q$ denotes that the probability measure $P$ is absolutely continuous
with respect to the measure $Q$. $\textbf{1}\\{\mathsf{E}\\}$ denotes the
indicator function for an event $\mathsf{E}$. $\log(x)$ is the natural
logarithm of $x$. $(x)^{+}$ and $(x)^{-}$ denote the positive ($\max(x,0)$)
and the negative part ($-\min(x,0)$) of $x$ respectively. $\lceil x\rceil$
denotes the ceiling of $x$. Throughout this paper we will adopt the convention
that $0\log(0)=0$, $\exp(\log(0))=0$, and $0^{0}=1$.

###### Definition 1

$\phi(x)=x\log(x)$ with convention that $0\log(0)=0$.

We note that $\phi(x)$ is convex.

We will use the following form of Jensen’s inequality [45, Theorem 7.9, p.
149] and [45, Theorem 8.20, p. 177].

###### Lemma 1

If $f(x)$ is a convex function and $\mathbb{E}|X|<\infty$ then
$\mathbb{E}[f(X)]$ exists and for any two $\sigma$-fields $A$ and $B$,

$\displaystyle\mathbb{E}[f(X)]\geq\mathbb{E}[f(\mathbb{E}[X|A,B])]\geq\mathbb{E}[f(\mathbb{E}[X|A])]\geq
f(\mathbb{E}[X]).$

We now recall the definition of mutual information for general ensembles and
its properties. Let $A$, $B$, and $C$ be measurable mappings defined on a
given probability space $(\Omega,\mathcal{F},P)$, taking values in
$(\mathcal{A},\mathfrak{F}^{A})$, $(\mathcal{B},\mathfrak{F}^{B})$, and
$(\mathcal{C},\mathfrak{F}^{C})$ respectively. Consider partitions of
$\Omega$, $\mathfrak{Q}_{A}=\left\\{\mathtt{A}_{i},1\leq i\leq
N_{A}\right\\}\subseteq\sigma(A)$ and
$\mathfrak{Q}_{B}=\left\\{\mathtt{B}_{j},1\leq j\leq
N_{B}\right\\}\subseteq\sigma(B)$. Wyner defined the conditional mutual
information $I(A;B|C)$ as [46]

$\displaystyle
I(A;B|C)=\sup_{\mathfrak{Q}_{A},\mathfrak{Q}_{B}}\mathbb{E}\left[\sum_{i,j=1,1}^{N_{A},N_{B}}P(\mathtt{A}_{i},\mathtt{B}_{j}|C)\log\left(\frac{P(\mathtt{A}_{i},\mathtt{B}_{j}|C)}{P(\mathtt{A}_{i}|C)P(\mathtt{B}_{j}|C)}\right)\right],$
(3)

where the supremum is over all such partitions of $\Omega$. Wyner showed that
$I(A;B|C)\geq 0$ with equality if and only if $A\rightleftarrows
C\rightleftarrows B$ forms a Markov chain [46, Lemma 3.1], and that (what is
generally referred to as) Kolmogrov’s formula holds [46, Lemma 3.2]

$\displaystyle I(A,C;B)=I(A;B)+I(C;B|A).$ (4)

Hence if $I(A;B)<\infty$, then $I(C;B|A)=I(A,C;B)-I(A;B)$. The data processing
inequality can be obtained from (4) as well: if $A\rightleftarrows
C\rightleftarrows B$ forms a Markov chain, then $I(A;B)\leq I(C;B)$.

Denote by $P^{A,B}$, the joint distribution of $A$ and $B$ on the space
($\mathcal{A}\times\mathcal{B},\mathfrak{F}^{A}\otimes\mathfrak{F}^{B}$ ),
i.e.,

$\displaystyle P^{A,B}(dA\times dB)=P((A^{-1}(dA),B^{-1}(dB)),\quad
dA\in\mathfrak{F}^{A},dB\in\mathfrak{F}^{B}.$

Similarly, $P^{A}$ and $P^{B}$ denote the marginal distributions. Gelfand and
Yaglom [47] proved that if $P^{A,B}\ll P^{A}\times P^{B}$, then the mutual
information $I(A;B)$ (defined via (3) by taking $\sigma(C)$ to be the trivial
$\sigma$-field) can be computed as:

$\displaystyle I(A;B)=\mathbb{E}\left[\log\left(\frac{dP^{A,B}}{d(P^{A}\times
P^{B})}\right)\right].$ (5)

A sufficient condition for $P^{A,B}\ll P^{A}\times P^{B}$ is that
$I(A;B)<\infty$ [48, Lemma 5.2.3, p. 92]. We will also require the following
result [46, Lemma 2.1]:

######  Lemma 2 (Wyner’s Lemma)

If $M$ is a finite alphabet random variable, then

$\displaystyle I(M;U_{0}^{T})=H(M)-\mathbb{E}\left[H(M|U_{0}^{T})\right],$

where

$\displaystyle
H(M|U_{0}^{T})=-\sum_{m}P(M=m|U_{0}^{T})\log\left(P(M=m|U_{0}^{T})\right),$

and $H(M)$ is the entropy of $M$.

##  III Point Processes, Intensities, and Mutual Information

Let $\mathcal{N}_{0}^{T}$ denote the set of counting realizations (or point-
process realizations) on $[0,T]$, i.e., if
${N}^{T}_{0}\in\mathcal{N}_{0}^{T}$, then for $t\in[0,T]$,
${N}_{t}\in\mathbf{N}$ (the set of non-negative integers), is right
continuous, and has unit increasing jumps with ${N}_{0}=0$. Let
$\mathfrak{F}^{N}$ be the restriction of the $\sigma$-field generated by the
Skorohod topology on $D[0,1]$ to $\mathcal{N}_{0}^{T}$.

###### Definition 2

If $N_{0}^{T}$ is a counting process adapted to the history
$(\mathcal{F}_{t}:t\in[0,T])$, then $N_{0}^{T}$ is said to have
$(P,\mathcal{F}_{t}:t\in[0,T])$-_intensity_
${\Gamma}_{0}^{T}=(\Gamma_{t}:t\in[0,T])$, where $\Gamma_{0}^{T}$ is a non-
negative measurable process if

  * •

$\Gamma_{0}^{T}$ is $(\mathcal{F}_{t}:t\in[0,T])$-predictable,

  * •

$\int_{0}^{T}\Gamma_{t}\,dt<\infty$, $P$-a.s.,

  * •

and for all non-negative $(\mathcal{F}_{t}:t\in[0,T])$-predictable processes
$C_{0}^{T}$:111The limits of the Lebesgue-Stieltjes integral $\int_{a}^{b}$
should be interpreted as $\int_{(a,b]}$.

$\displaystyle\mathbb{E}\left[\int_{0}^{T}C_{s}\,d{N}_{s}\right]=\mathbb{E}\left[\int_{0}^{T}C_{s}\Gamma_{s}\,ds\right].$

When it is clear from the context, we will drop the probability measure $P$
from the notation and say $N_{0}^{T}$ has
$(\mathcal{F}_{t}:t\in[0,T])$-intensity $\Gamma_{0}^{T}$.

###### Definition 3

A point process $Y_{0}^{T}$ is said to be _Poisson process with rate
$\lambda$_ if its $(\mathcal{F}_{t}^{Y}:t\in[0,T])$-intensity is
$(\lambda:t\in[0,T])$.

The above definition can be shown to imply the usual definition of Poisson
process [39, Theorem T4, Chapter II, p. 25] and vice versa [39, Section 2,
Chapter II, p. 23].

###### Definition 4

$P_{0}^{Y_{0}^{T}}$ denotes the distribution of a point process $Y_{0}^{T}$
(on the space $(\mathcal{N}_{0}^{T},\mathfrak{F}^{N})$) under which
$Y_{0}^{T}$ is a Poisson process with unit rate.

A point processes with stochastic intensity and a Poisson process with unit
rate are linked via the following result.

###### Lemma 3

Let $P^{Y_{0}^{T}}$ be the distribution of a point process $Y_{0}^{T}$ such
that $P^{Y_{0}^{T}}\ll P_{0}^{Y_{0}^{T}}$. Then there exists a non-negative
predictable process $\Lambda_{0}^{T}$ such that

$\displaystyle\frac{dP^{Y_{0}^{T}}}{dP_{0}^{Y_{0}^{T}}}=\exp\left(\int_{0}^{T}\log(\Lambda_{t})\,dY_{t}-\Lambda_{t}+1\,dt\right).$

Moreover, the $(P^{Y_{0}^{T}},(\mathcal{F}_{t}^{Y}:t\in[0,T]))$-intensity of
$Y_{0}^{T}$ is $\Lambda_{0}^{T}$. Conversely, if the
$(P^{Y_{0}^{T}},\mathcal{F}_{t}^{Y}:t\in[0,T])$-intensity of $Y_{0}^{T}$ is
$\Gamma_{0}^{T}$ and
$\mathbb{E}_{P^{Y_{0}^{T}}}[\int_{0}^{T}|\phi(\Gamma_{t})|\,dt]<\infty$, then
$P^{Y_{0}^{T}}\ll P_{0}^{Y_{0}^{T}}$, and the corresponding Radon-Nikodym
derivative is given by the above expression, where

$\displaystyle\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}|\Gamma_{t}-\Lambda_{t}|\,dt\right]=0,\quad\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\mathbf{1}\\{\Gamma_{t}\neq\Lambda_{t}\\}\,dY_{t}\right]=0.$

In the latter case,

$\displaystyle\mathbb{E}_{P^{Y_{0}^{T}}}\left[\log\left(\frac{dP^{Y_{0}^{T}}}{dP_{0}^{Y_{0}^{T}}}\right)\right]=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\phi(\Gamma_{t})-\Gamma_{t}+1\,dt\right].$

###### Proof:

Please see the supplementary material. ∎

The following theorem allows us to express the mutual information involving a
point processes with intensity and other random variables in terms of the
intensity functions. The proof of the theorem is similar to the proof of
Theorem 1 in [42].

###### Theorem 1

Let $Y_{0}^{T}$ be a point process with
$(\mathcal{F}_{t}^{Y}:t\in[0,T])$-intensity $\Lambda_{0}^{T}$ such that

$\mathbb{E}[\int_{0}^{T}|\phi(\Lambda_{t})|\,dt]<\infty,$

and let $M$ be a measurable mapping on the given probability space satisfying
$I(M;Y_{0}^{T})<\infty$. Then there exists a process $\Gamma_{0}^{T}$ such
that $\Gamma_{0}^{T}$ is the $(\mathcal{G}_{t}=\sigma(M,Y_{0}^{t}):t\in[0,T])$
intensity of $Y_{0}^{T}$ and

$\displaystyle
I(M;Y_{0}^{T})=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\phi(\Lambda_{t})\,dt\right].$

###### Proof:

Let $P^{M,Y_{0}^{T}}$ denote the joint distribution of $M$ and $Y_{0}^{T}$,
and $P^{M}$ and $P^{Y_{0}^{T}}$ denote their marginals, respectively. Since
$I(M;Y_{0}^{T})<\infty$, we get that $P^{M,Y_{0}^{T}}\ll P^{M}\times
P^{Y_{0}^{T}}$ [48, Lemma 5.2.3, p. 92]. Lemma 3 says that $P^{Y_{0}^{T}}\ll
P_{0}^{Y_{0}^{T}}$, which together with [49, Chapter 1, Exercise 19, p. 22]
gives $P^{M,Y_{0}^{T}}\ll P^{M}\times P^{Y_{0}^{T}}\ll P^{M}\times
P_{0}^{Y_{0}^{T}}$.

Let $\tilde{P}^{M,Y_{0}^{T}}\triangleq{P}^{M}\times P_{0}^{Y_{0}^{T}}$ and

$\displaystyle\mathcal{L}=\frac{d{P}^{M,Y_{0}^{T}}}{d\tilde{P}^{M,Y_{0}^{T}}}$
(6)

denote the Radon-Nikodym derivative. Since under $\tilde{P}^{M,Y_{0}^{T}}$,
$M$ and $Y_{0}^{T}$ are independent, we note that the
$(\tilde{P}^{M,Y_{0}^{T}},(\mathcal{G}_{t}:t\in[0,T]))$-intensity of
$Y_{0}^{T}$ is 1 [39, E5 Exercise, Chapter II, p. 28]. Define the process
$L_{0}^{T}$ as

$\displaystyle L_{t}=\mathbb{E}_{\tilde{P}}[\mathcal{L}|\mathcal{G}_{t}],\quad
t\in[0,T],$ (7)

where $\mathbb{E}_{\tilde{P}}$ denotes that the conditional expectation is
taken with respect to the measure $\tilde{P}^{M,Y_{0}^{T}}$. Then $L_{0}^{T}$
is a $(\tilde{P}^{M,Y_{0}^{T}},\mathcal{G}_{t})$ non-negative absolutely-
integrable martingale.

By the martingale representation theorem, the process $L_{0}^{T}$ can be
written as [39, Chapter III, Theorem T17, p. 76] (where we have taken
$\sigma(M)$ to be the “germ $\sigma$-field”):

$\displaystyle L_{t}=1+\int_{0}^{t}K_{s}(d{Y}_{s}-ds),$

where $K_{0}^{T}$ is a $(\mathcal{G}_{t}:t\in[0,T])$-predictable process which
satisfies $\int_{0}^{T}|K_{t}|\,dt<\infty$ $\tilde{P}^{M,Y_{0}^{T}}$-a.s.
Applying [50, Lemma 19.5, p. 315], we can write $L_{0}^{T}$ as

$\displaystyle
L_{t}=\exp\left(\int_{0}^{t}\log(\Gamma_{s})\,dY_{s}+(1-\Gamma_{s})\,ds\right),\quad
t\in[0,T],$ (8)

where $\Gamma_{0}^{T}$ is a non-negative
$(\mathcal{G}_{t}:t\in[0,T])$-predictable process, and $\Gamma_{t}<\infty$
$\tilde{P}^{M,Y_{0}^{T}}$-a.s. for $t\in[0,T]$.

Now we can mimic the proof of [39, Chapter VI, Theorems T3, p. 166] to deduce:

###### Lemma 4

For all non-negative $(\mathcal{G}_{t}:t\in[0,T])$-predictable processes
$C_{0}^{T}$

$\displaystyle\mathbb{E}\left[\int_{0}^{T}C_{t}\Gamma_{t}\,dt\right]=\mathbb{E}\left[\int_{0}^{T}C_{t}\,dY_{t}\right],$

where the expectation is taken with respect to measure $P$.

###### Proof:

Please see the supplementary material. ∎

Taking $C_{t}=1$ in the above equality yields

$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]=\mathbb{E}\left[\int_{0}^{T}\,dY_{t}\right]=\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt\right]<\infty.$
(9)

Hence $\int_{0}^{T}\Gamma_{t}\,dt<\infty$ $P$-a.s. and we conclude that the
$({P}^{M,Y_{0}^{T}},\mathcal{G}_{t}:t\in[0,T])$-intensity of $Y_{0}^{T}$ is
$\Gamma_{0}^{T}$.

Now we will use:

###### Lemma 5

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right].$
(10)

###### Proof:

Please see the supplementary material. ∎

Since
$\mathbb{E}\left[\log\left(\frac{dP^{M,N_{0}^{T}}}{d\tilde{P}^{M,N_{0}^{T}}}\right)\right]$
is well-defined, (6), (7), and (8) yields

$\displaystyle\mathbb{E}\left[\log\left(\frac{dP^{M,N_{0}^{T}}}{d\tilde{P}^{M,N_{0}^{T}}}\right)\right]$
$\displaystyle=\mathbb{E}\left[\log(L_{T})\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}+(1-\Gamma_{t})\,dt\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]+\mathbb{E}\left[\int_{0}^{T}(1-\Lambda_{t})\,dt\right],$
(11)

where in the last line we have used Lemma 5 and
$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]=\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt\right]<\infty$
from (9). Also,

$\displaystyle\mathbb{E}\left[\log\left(\frac{d(P^{M}\times
P^{Y_{0}^{T}}}{d\tilde{P}^{M,Y_{0}^{T}}}\right)\right]$
$\displaystyle=\mathbb{E}\left[\log\left(\frac{dP^{Y_{0}^{T}}}{dP_{0}^{Y_{0}^{T}}}\right)\right]$
$\displaystyle\overset{(a)}{=}\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})+1-\Lambda_{t}\,dt\right]$
(12) $\displaystyle<\infty,$

where we have used Lemma 3 for (a). Using the above inequality and the fact
that

$\mathbb{E}\left[\log\left(\frac{dP^{M,N_{0}^{T}}}{d\tilde{P}^{M,N_{0}^{T}}}\right)\right]$

is well-defined, we can express the mutual information as

$\displaystyle I(M;Y_{0}^{T})$
$\displaystyle=\mathbb{E}\left[\log\left(\frac{d{P}^{M,Y_{0}^{T}}}{d({P}^{M}\times
P^{Y_{0}^{T}})}\right)\right]$
$\displaystyle=\mathbb{E}\left[\log\left(\frac{dP^{M,N_{0}^{T}}}{d\tilde{P}^{M,N_{0}^{T}}}\right)\right]-\mathbb{E}\left[\log\left(\frac{d(P^{M}\times
P^{N_{0}^{T}}}{d\tilde{P}^{M,N_{0}^{T}}}\right)\right],$ (13)

Now we can compute the mutual information from (11), (12), and (13),

$\displaystyle I(M;Y_{0}^{T})$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]+\mathbb{E}\left[\int_{0}^{T}(1-\Lambda_{t})\,dt\right]-\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right]-\mathbb{E}\left[\int_{0}^{T}1-\Lambda_{t}\,dt\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\phi(\Lambda_{t})\,dt\right].$

∎

We shall require several strong data processing inequalities, for which
purpose we now derive some ancillary results regarding the intensity of a
point process. Combining [39, T8 Theorem, Chapter II, p. 27] and [39, T9
Theorem, Chapter II, p. 28], we can conclude the following result.

###### Lemma 6

Let $\Gamma_{0}^{T}$ be a $(\mathcal{F}_{t}:t\in[0,T])$-predictable non-
negative process satisfying

$\int_{0}^{T}\Gamma_{t}\,dt<\infty\quad\text{a.s.}$

Let $Y_{0}^{T}$ be a point process adapted to $(\mathcal{F}_{t}:t\in[0,T])$.
Then $\Gamma_{0}^{T}$ is the $(\mathcal{F}_{t}:t\in[0,T])$-intensity of
$Y_{0}^{T}$ if and only if

$M_{t}\triangleq Y_{t}-\int_{0}^{t}\Gamma_{s}\,ds\quad t\in[0,T]$

is a $(\mathcal{F}_{t}:t\in[0,T])$-local martingale222 A process $Y_{0}^{T}$
is called a _local martingale_ with respect to a filtration
$(\mathcal{F}_{t}:t\geq 0)$ if $Y_{t}$ is $\mathcal{F}_{t}$-measurable for
each $t\in[0,T]$ and there exists an increasing sequence of stopping times
$T_{n}$, such that $T_{n}\to\infty$ and the stopped and shifted processes
$(Y_{\min\\{t,T_{n}\\}}-Y_{0}:t\in[0,T])$ are
$(\mathcal{F}_{t}:t\in[0,T])$-martingales for each $n$..

If we impose the stricter condition of finite expectation
$\mathbb{E}[\int_{0}^{T}\Gamma_{t}\,dt]<\infty$, the local martingale
condition in the above statement can be replaced by the martingale condition.

###### Lemma 7

Let $\Gamma_{0}^{T}$ be a $(\mathcal{F}_{t}:t\in[0,T])$-predictable non-
negative process satisfying

$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]<\infty.$

Let $Y_{0}^{T}$ be a point process adapted to $(\mathcal{F}_{t}:t\in[0,T])$.
Then $\Gamma_{0}^{T}$ is the $(\mathcal{F}_{t}:t\in[0,T])$-intensity of
$Y_{0}^{T}$ if and only if

$M_{t}\triangleq Y_{t}-\int_{0}^{t}\Gamma_{s}\,ds\quad t\in[0,T]$

is a $(\mathcal{F}_{t}:t\in[0,T])$-martingale.

###### Proof:

Please see the supplementary material. ∎

###### Lemma 8

If a point process $N_{0}^{T}$ has $(\mathcal{F}_{t}:t\in[0,T])$-intensity
${\Lambda}_{0}^{T}$, and $(\mathcal{G}_{t}:t\in[0,T])$ is another history for
$N_{0}^{T}$ such that $\mathcal{G}_{t}\subseteq\mathcal{F}_{t}$ for each
$t\in[0,T]$, then there exists a process $\Pi_{0}^{T}$ such that $\Pi_{0}^{T}$
is the $(\mathcal{G}_{t}:t\in[0,T])$-intensity of ${N}_{0}^{T}$, and for each
$t\in[0,T]$, $\Pi_{t}=\mathbb{E}[{\Lambda}_{t}|\mathcal{G}_{t-}]$ $P$-a.s.

###### Proof:

Please see the supplementary material. ∎

###### Lemma 9

Let $Y_{0}^{T}$ be a point process with
$(\mathcal{G}_{t}\triangleq\sigma(M,Y_{0}^{t}):t\in[0,T])$-intensity
$\Gamma_{0}^{T}$ for some $M$. Let $Z_{0}^{T}$ be obtained adding an
independent (of both $M$ and $Y_{0}^{T}$) point process $N_{0}^{T}$ with
$(\mathcal{F}_{t}^{N}:t\in[0,T])$-intensity $\Pi_{0}^{T}$ to $Y_{0}^{T}$. Then
$Z_{0}^{T}$ has a
$(\mathcal{F}_{t}\triangleq\sigma(M,Z_{0}^{t}):t\in[0,T])$-intensity
$\Theta_{0}^{T}$ which satisfies
$\Theta_{t}=\mathbb{E}[(\Gamma_{t}+\Pi_{t})|\mathcal{F}_{t-}]$ $P$-a.s. for
each $t\in[0,T]$.

###### Proof:

Please see the supplementary material. ∎

###### Theorem 2

Let $Y_{0}^{T}$ be a Poisson process with rate $\lambda$, $M$ be such that
$I(M;Y_{0}^{T})<\infty$, and $\Gamma_{0}^{T}$ be the
$(\sigma(M;Y_{0}^{t}):t\in[0,T])$-intensity of $Y_{0}^{T}$. Suppose
$Z_{0}^{T}$ is obtained by adding an independent (of $Y_{0}^{T}$ and $M$)
Poisson process with rate $\mu$ to $Y_{0}^{T}$. Then,

$\displaystyle I(M;Y_{0}^{T})$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\phi(\lambda)\,dt\right],$
$\displaystyle I(M;Z_{0}^{T})$
$\displaystyle\leq\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t}+\mu)-\phi(\lambda+\mu)\,dt\right].$

###### Proof:

Since $M\leftrightarrows Y_{0}^{T}\leftrightarrows Z_{0}^{T}$ forms a Markov
chain, the data processing inequality gives $I(M;Z_{0}^{T})\leq
I(M;Y_{0}^{T})<\infty$. Applying Theorem 1 and using the uniqueness of
intensities,

$\displaystyle I(M;Y_{0}^{T})$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\phi(\lambda)\,dt\right],\quad\text{and}$
$\displaystyle I(M;Z_{0}^{T})$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\hat{\Gamma}_{t})-\phi(\hat{\lambda}_{t})\,dt\right],$
(14)

where $\hat{\Gamma}_{0}^{T}$ and $\hat{\lambda}_{0}^{T}$ are the
$(\sigma(M;Z_{0}^{t}):t\in[0,T])$ and
$(\mathcal{F}_{t}^{Z}:t\in[0,T])$-intensities of $Z_{0}^{T}$. Due to the
uniqueness of the intensities and Lemma 9, we get for each $t\in[0,T]$,
$\hat{\Gamma}_{t}=\mathbb{E}[\Gamma_{t}|M,Z_{0}^{t-}]+\mu$, and
$\hat{\lambda}_{t}=\lambda+\mu$. Substituting this in (14) and applying
Jensen’s inequality yields

$\displaystyle I(M;Z_{0}^{T})$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\mathbb{E}[\Gamma_{t}|M,Z_{0}^{t-}]+\mu)-\phi(\lambda+\mu)\,dt\right],$
$\displaystyle\leq\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t}+\mu)-\phi(\lambda+\mu)\,dt\right].$

∎

###### Definition 5

A point process $Z_{0}^{T}$ is said to be obtained from $p$-thinning of a
point process $Y_{0}^{T}$, if each point in $Y_{0}^{T}$ is deleted with
probability $p$, independent of all other points and deletions.

###### Lemma 10

Suppose that $Y_{0}^{T}$ is a point process with
$\mathcal{G}_{t}\triangleq\sigma(M,Y_{0}^{t})$-intensity $\Gamma_{0}^{T}$ such
that $\mathbb{E}[\int_{0}^{T}\Gamma_{t}\,dt]<\infty$ and $Z_{0}^{T}$ is
obtained from $p$-thinning $Y_{0}^{T}$. Then the
$(\mathcal{F}_{t}\triangleq\sigma(M,Z_{0}^{t}):t\in[0,T])$-intensity of
$Z_{0}^{T}$ is given by $\Theta_{0}^{T}$, where $P$-a.s.
$\Theta_{t}=(1-p)\mathbb{E}[\Gamma_{t}|\mathcal{F}_{t-}],t\in[0,T]$.

###### Proof:

Please see the supplementary material. ∎

The following theorem was first proven by Wang in [44] using a property of
certain “contraction coefficient” used in strong data processing inequalities
[51]. Here, we provide a self-contained proof which uses Theorem 1 and Lemma
10.

###### Theorem 3

Let $Y_{0}^{T}$ be a Poisson process with rate $\lambda$, and $M$ be such that
$I(M;Y_{0}^{T})<\infty$. Let $Z_{0}^{T}$ obtained from $p$-thinning of
$Y_{0}^{T}$ such that the thinning operation is independent of $M$. Then

$\displaystyle I(M;Z_{0}^{T})\leq(1-p)I(M;Y_{0}^{T}).$

###### Proof:

The data processing inequality gives $I(M;Z_{0}^{T})\leq
I(M;Y_{0}^{T})<\infty$. Applying Theorem 1,

$\displaystyle
I(M;Y_{0}^{T})=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\phi(\lambda)\,dt\right],$
(15)

and

$\displaystyle
I(M;Z_{0}^{T})=\mathbb{E}\left[\int_{0}^{T}\phi(\hat{\Gamma}_{t})-\phi(\hat{\lambda}_{t})\,dt\right],$
(16)

where $\Gamma_{0}^{T}$ and $\lambda_{0}^{T}$ (respectively
$\hat{\Gamma}_{0}^{T}$ and $\hat{\lambda}_{0}^{T}$) are the
$(\sigma(M;Y_{0}^{t}):t\in[0,T])$ and
$(\sigma(Y_{0}^{t}):t\in[0,T])$-intensities (respectively
$(\sigma(M;Z_{0}^{t}):t\in[0,T])$ and
$(\sigma(Z_{0}^{t}):t\in[0,T])$-intensities) of $Y_{0}^{T}$ (respectively
$Z_{0}^{T}$). Due to the uniqueness of the intensities and Lemma 10, we can
take for each $t\in[0,T]$,

$\hat{\Gamma}_{t}=(1-p)\mathbb{E}[\Gamma_{t}|M,Z_{0}^{t-}],\quad\hat{\lambda}_{t}=(1-p)\lambda.$

Noting that $\phi((1-p)x)=(1-p)\phi(x)+x\phi(1-p)$, (16) yields

$\displaystyle I(M;Z_{0}^{T})=$
$\displaystyle(1-p)\mathbb{E}\left[\int_{0}^{T}\phi(\mathbb{E}[\Gamma_{t}|M,Z_{0}^{t-}])-\phi(\lambda)\,dt\right]$
$\displaystyle+\phi(1-p)\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}-\lambda\,dt\right]$
$\displaystyle\overset{(a)}{=}$
$\displaystyle(1-p)\mathbb{E}\left[\int_{0}^{T}\phi(\mathbb{E}[\Gamma_{t}|M,Z_{0}^{t-}])-\phi(\lambda)\,dt\right]$
$\displaystyle\overset{(b)}{\leq}$
$\displaystyle(1-p)\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\phi(\lambda)\,dt\right]$
$\displaystyle=$ $\displaystyle(1-p)I(M;Y_{0}^{T}),$

where for (a) we have used the fact that
$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]=\mathbb{E}\left[\int_{0}^{T}1\,dY_{t}\right]=\mathbb{E}\left[\int_{0}^{T}\lambda\,dt\right]$,
and
for (b) we have used Jensen’s inequality. ∎

We will require the following result [52, Theorem 2.11, p. 106].

###### Lemma 11

Suppose that $Y_{0}^{T}$ is a Poisson process with rate $\lambda$ and
$Z_{0}^{T}$ is obtained from $p$-thinning of $Y_{0}^{T}$. Let

$\displaystyle\hat{Z}_{t}=Y_{t}-Z_{t}\quad t\in[0,T].$

Then $\hat{Z}_{0}^{T}$ and $Z_{0}^{T}$ are independent Poisson processes with
rates $p\lambda$ and $(1-p)\lambda$ respectively.

The following lemma will be used repeatedly in the converse proofs of the
rate-distortion function.

###### Lemma 12

Let a point process $Y_{0}^{T}$ have an
$(\mathcal{F}_{t}:t\in[0,T])$-intensity $\Gamma_{0}^{T}$ such that

$\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]<\infty.$

Let $\hat{Y}_{0}^{T}$ be an non-negative
$(\mathcal{F}_{t}:t\in[0,T])$-predictable process satisfying
$\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt\right]<\infty$. Then

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}_{t})\,dY_{t}\right]=\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}_{t})\Gamma_{t}\,dt\right].$

###### Proof:

Please see the supplementary material. ∎

##  IV Functional Covering of Point Processes

In this section, we will consider general point processes and obtain the rate-
distortion function under the functional-covering distortion when feedforward
is present. Stronger results are obtained for Poisson processes in the next
sections.

###### Definition 6

Given a point process $y_{0}^{T}\in\mathcal{N}_{0}^{T}$, and a non-negative
function $\hat{y}_{0}^{T}$, the _functional-covering distortion_ $d$ is

$\displaystyle
d(\hat{y}_{0}^{T},y_{0}^{T})\triangleq\left(\int_{0}^{T}\hat{y}_{t}\,dt-\log(\hat{y}_{t})\,dy_{t}\right),$

whenever the expression on the right is well-defined.

We will allow the reconstruction function $\hat{Y}_{0}^{T}$ to depend on
$Y_{0}^{T}$ as well as the message, constrained via predictability. In
particular, we will call $\hat{Y}_{0}^{T}$ an _allowable reconstruction with
feedforward_ if it is non-negative and
$(\sigma(Y_{0}^{t}):t\in[0,T])$-predictable. Let
$\mathcal{\hat{Y}}_{0,\text{FF}}^{T}$ denote the set of all $\hat{y}_{0}^{T}$
processes which are allowable reconstructions with feedforward.

###### Definition 7

A _$(T,R,D)$ code with feedforward_ consists of an _encoder_ $f$

$\displaystyle
f:\mathcal{N}_{0}^{T}\rightarrow\\{1,\dots,\dots,\lceil\exp(RT)\rceil\\}$

and a _decoder_ $g$

$\displaystyle
g:\\{1,\dots,\lceil\exp(RT)\rceil\\}\times\mathcal{N}_{0}^{T}\rightarrow\mathcal{\hat{Y}}_{0,\text{FF}}^{T}$

satisfying

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt\right]<\infty$

and the distortion constraint

$\displaystyle\mathbb{E}\left[\frac{1}{T}d(\hat{Y}_{0}^{T},Y_{0}^{T})\right]\leq
D.$

We will call the encoder’s output $M=f(Y_{0}^{T})$ the _message_ and the
decoder’s output $\hat{Y}_{0}^{T}$ the _reconstruction._

###### Definition 8

The _minimum achievable distortion with feedforward at rate $R$ and
blocklength $T$_ is

$\displaystyle D_{F}^{*}(R,T)\triangleq\inf\\{D:\text{there exists a $(T,R,D)$
code with feedforward}\\}.$

###### Definition 9

The _distortion-rate function with feedforward_ is

$\displaystyle D_{F}(R)\triangleq\limsup_{T\to\infty}D_{F}^{*}(R,T).$

The minimum achievable rate at distortion $D$ and blocklength $T$ with
feedforward $R_{F}^{*}(D,T)$ and the rate-distortion function with feedforward
$R_{F}(D)$ can be defined similarly.

$D_{F}^{*}(R,T)$ can be characterized via the following theorem for certain
point processes.

###### Theorem 4

Let $Y_{0}^{T}$ be a point process with
$(\mathcal{F}_{t}^{Y}:t\in[0,T])$-intensity $\Lambda_{0}^{T}$ such that

$\mathbb{E}\left[\int_{0}^{T}|\phi(\Lambda_{t})|\,dt\right]<\infty.$

Let

$\displaystyle\Xi(Y_{0}^{T})\triangleq\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}-\phi(\Lambda_{t})\,dt\right],$

and

$\delta_{T}\triangleq P(Y_{T}=0)<1.$

Then $D_{F}^{*}(R,T)$ satisfies

$\displaystyle\Xi(Y_{0}^{T})-R-\frac{1}{T}\leq
D_{F}^{*}(R,T)\leq\Xi(Y_{0}^{T})-(1-\delta_{T})R+\frac{1}{T}.$

###### Proof:

_Achievability:_
Recall that since $\Lambda_{0}^{T}$ is the
$(\mathcal{F}_{t}^{Y}:t\in[0,T])$-intensity of $Y_{0}^{T}$, it is
$(\mathcal{F}_{t}^{Y}:t\in[0,T])$-predictable, and
$\mathbb{E}[\int_{0}^{T}|\phi(\Lambda_{t})|\,dt]<\infty$ implies
$\mathbb{E}[\int_{0}^{T}\Lambda_{t}\,dt]<\infty$. If the decoder outputs
$\Lambda_{0}^{T}$, this leads to distortion

$\displaystyle\frac{1}{T}\mathbb{E}[d(\Lambda_{0}^{T},Y_{0}^{T})]$
$\displaystyle=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt-\log(\Lambda_{t})\,dY_{t}\right]$
$\displaystyle=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}-\phi(\Lambda_{t})\,dt\right]$
$\displaystyle=\Xi(Y_{0}^{T}).$

Thus $D_{F}^{*}(0,T)\leq\Xi(Y_{0}^{T})$, and the upper bound in the statement
of the theorem holds at $R=0$.

Now consider the case $R>0$. Fix $T>0$ and let $J=\lceil\exp(RT)\rceil$. If
$Y_{T}=0$, then the encoder sends index $M=1$. Otherwise, let $\Theta$ denote
the first arrival instant of the observed point process $Y_{0}^{T}$. From
Lemma 3, we have that $P^{Y^{T}_{0}}\ll P_{0}^{Y^{T}_{0}}$. Since under
$P_{0}^{Y^{T}_{0}}$, $Y_{0}^{T}$ is a Poisson process with unit rate, it holds
that $P_{0}^{Y^{T}_{0}}(\Theta=t,Y_{T}>0)=0$ for any fixed $t\in[0,T]$. This
gives us $P(\Theta=t,Y_{T}>0)=0$ for $t\in[0,T]$. Thus conditioned on the
event $Y_{T}>0$, $\Theta$ has a continuous distribution function $F_{\Theta}$.
The encoder computes $F_{\Theta}(\Theta)$ which is uniformly distributed over
$[0,1]$, which the encoder suitably quantizes to obtain an $M$ which is
uniform in $\\{2,\dots,J\\}$. From Theorem 1, there exists a
$(\sigma(M,Y_{0}^{t}):t\in[0,T])$-predictable process $\Gamma_{0}^{T}$ which
is the $(\sigma(M,Y_{0}^{t}):t\in[0,T])$-intensity of $Y_{0}^{T}$. We note
that
$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]=\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt\right]<\infty$,
and from Theorem 1,
$\mathbb{E}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]<\infty$. Hence

$\frac{1}{T}\mathbb{E}[d(\Lambda_{0}^{T},Y_{0}^{T})]=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt-\log(\Gamma_{t})\,dY_{t}\right]$

is well-defined. The decoder outputs $\Gamma_{0}^{T}$ as its reconstruction.
Then we have

$\displaystyle\frac{1}{T}H(M)$
$\displaystyle=-\frac{1}{T}\left(\delta_{T}\log(\delta_{T})+(1-\delta_{T})\log(1-\delta_{T})\right)+\frac{1-\delta_{T}}{T}(\log(J-1))$
$\displaystyle\overset{(a)}{\geq}\frac{1-\delta_{T}}{T}(\log(J-1))$
$\displaystyle\overset{(b)}{\geq}\frac{1-\delta_{T}}{T}\log(J/\exp(1))$
$\displaystyle\overset{(c)}{\geq}(1-\delta_{T})R-\frac{1}{T},$ (17)

where for (a), we have used the bound
$-\delta_{T}\log(\delta_{T})-(1-\delta_{T})\log(1-\delta_{T})\geq 0$,
for (b), we have used the inequality $J-1\geq J/\exp(1)$ when $J\geq 2$, and
for (c), we used the fact that $RT\leq\log(J)$.

$H(M)$ also satisfies

$\displaystyle\frac{1}{T}H(M)$
$\displaystyle\overset{(a)}{=}\frac{1}{T}I(M;Y_{0}^{T})$
$\displaystyle\overset{(b)}{=}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right],$
(18)

where, for (a) we have used Lemma 2,
for (b) we have used Theorem 1.
The average distortion can be bounded as follows:

$\displaystyle\frac{1}{T}\mathbb{E}[d(\Lambda_{0}^{T},Y_{0}^{T})]$
$\displaystyle=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt-\log(\Gamma_{t})\,dY_{t}\right]$
$\displaystyle\overset{(a)}{=}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]$
$\displaystyle\overset{(b)}{=}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt\right]-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]$
$\displaystyle\overset{(c)}{=}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt\right]-\frac{1}{T}H(M)-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right]$
$\displaystyle\overset{(d)}{\leq}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}-\phi(\Lambda_{t})\,dt\right]-(1-\delta_{T})R+\frac{1}{T}$
$\displaystyle=\Xi(Y_{0}^{T})-(1-\delta_{T})R+\frac{1}{T},$

where, for (a), we have used the fact that
$\mathbb{E}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]<\infty$ due to
Theorem 1,
for (b), we used the equality
$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]=\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt\right]$,
for (c), we used (18), and
for (d), we used (17).

Thus we have shown the existence of a $(T,R,D)$ code with feedforward such
that $D=\Xi(Y_{0}^{T})-(1-\delta_{T})R+\frac{1}{T}$. This gives the upper
bound on $D_{F}^{*}(R,T)$.

### Converse:

For the given $(T,R,D)$ code with feedforward, let $J=\lceil\exp(RT)\rceil$.
Then $J\leq\exp(RT)+1\leq\exp(RT+1)$. Thus we have

$\displaystyle
R+\frac{1}{T}\geq\frac{1}{T}\log(J)\geq\frac{1}{T}H(M)\overset{(a)}{=}\frac{1}{T}I(M;Y_{0}^{T}),$
(19)

where (a) follows because of Lemma 2.

Since $I(M;Y_{0}^{T})<\infty$, we conclude from Theorem 1 that there exists a
process $\Gamma_{0}^{T}$ such that $\Gamma_{0}^{T}$ is the
$(\mathcal{F}_{t}=\sigma(M,Y_{0}^{t}):t\in[0,T])$ intensity of $Y_{0}^{T}$ and

$\displaystyle
I(M;Y_{0}^{T})=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right].$

Hence from (19)

$\displaystyle
R\geq\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right]-\frac{1}{T}.$
(20)

Let $\hat{Y}_{0}^{T}$ denote the decoder’s output. The distortion constraint
$D$ satisfies

$\displaystyle
D\geq\frac{1}{T}\mathbb{E}\left[d(\hat{Y}_{0}^{T},Y_{0}^{T})\right]$
$\displaystyle=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt-\log(\hat{Y}_{t})\,dY_{t}\right]$
$\displaystyle=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}-\log(\hat{Y}_{t})\Gamma_{t}\,dt\right]$
(21)

where in the last line we have used Lemma 12.

Using the inequality $u\log(v)\leq\phi(u)-u+v$, and noting that the individual
terms have finite expectations,

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}_{t})\Gamma_{t}\,dt\right]$
$\displaystyle\leq\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\Gamma_{t}+\hat{Y}_{t}\,dt\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]+\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt\right].$
(22)

From (21) and (20), we deduce

$\displaystyle R+D$
$\displaystyle\geq\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right]+\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt\right]$
$\displaystyle\phantom{====}-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}_{t})\,dY_{t}\right]-\frac{1}{T}$
$\displaystyle\overset{(a)}{\geq}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right]-\frac{1}{T}$
$\displaystyle\overset{(b)}{\geq}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt\right]-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Lambda_{t})\,dt\right]-\frac{1}{T}$
$\displaystyle\overset{}{=}\Xi(Y_{0}^{T})-\frac{1}{T},$

where, for (a) we have used (22), and
for (b) we used the fact that
$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]=\mathbb{E}\left[\int_{0}^{T}\,dY_{t}\right]=\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}\,dt\right]$.
Hence we have shown that for any $(T,R,D)$ code with feedforward,
$D\geq\Xi(Y_{0}^{T})-R-1/T$. This gives us the lower bound on $D_{F}^{*}(R,T)$
∎

###### Corollary 1

Let $Y_{0}^{T}$ be a point process with
$(\mathcal{F}_{t}^{Y}:t\in[0,T])$-intensity $\Lambda_{0}^{T}$ such that

  * •

$\mathbb{E}[\int_{0}^{T}|\phi(\Lambda_{t})|\,dt]<\infty$,

  * •

$\bar{\Xi}(Y)\triangleq\limsup_{T\to\infty}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Lambda_{t}-\phi(\Lambda_{t})\,dt\right]$
is finite.

  * •

$\lim_{T\to\infty}P(Y_{T}=0)=0$.

Then

$\displaystyle D_{F}(R)=\bar{\Xi}(Y)-R.$

###### Proof:

The corollary follows from the definition
$D_{F}(R)=\limsup_{T\to\infty}D_{F}^{*}(R,T)$ and from the bounds on
$D_{F}^{*}(R,T)$ in the Theorem 4. ∎

###### Remark 1

The above distortion-rate function is reminiscent of the logarithmic-loss
distortion-rate function for a DMS. Specifically, for a DMS $Y$ on alphabet
$\mathcal{Y}$ let the reconstruction be a probability distribution function
$Q$ on $\mathcal{Y}$. The logarithmic loss distortion is defined as
$d_{LL}(y,Q)\triangleq-\log(Q(y))$ and the distortion-rate function is then
given by $D(R)=(H(Y)-R)^{+}$ [38].

If the reconstruction $\hat{y}_{0}^{T}$ is assumed to be bounded then it can
be used define a probability measure on the space of point-processes
$(\mathcal{N}_{0}^{T},\mathfrak{F}^{N})$ via following Radon-Nikodym
derivative.

$\frac{dP_{\hat{y}_{0}^{T}}}{dP_{0}}(y_{0}^{T})=\exp\left(\int_{0}^{T}\log(\hat{y}_{t})\,dy_{t}-(\hat{y}_{t}-1)\,dt\right),$

where $P_{0}$ is the measure under which $Y_{0}^{T}$ is a Poisson process with
unit rate. Then the intensity of $Y_{0}^{T}$ under this measure is
$\hat{y}_{0}^{T}$ [39, Chapter VI, Theorems T2-T4] and the functional-covering
distortion is related to the above Radon-Nikodym derivative as

$d(\hat{y}_{0}^{T},y_{0}^{T})=-\log\left(\frac{dP_{\hat{y}_{0}^{T}}}{dP_{0}}(y_{0}^{T})\right)+T.$

$\Diamond$

Applying the above corollary to a Poisson process with rate $\lambda>0$, we
get that $D_{F}(R)=\lambda-\lambda\log(\lambda)-R$. As we will see in the next
section, this distortion-rate function can be achieved without feedforward.

##  V Constrained Functional-Covering of Poisson Processes

In this and the next section we focus on Poisson processes. Let
$\hat{\mathcal{Y}}_{0}^{T}$ denote the set of all functions $\hat{y}_{0}^{T}$
which are non-negative and left-continuous with right-limits. We assume that
we are given a set $\mathcal{A}\in\mathbb{R}_{+}$ with at least one positive
element. We will constrain the reconstruction function $\hat{Y}_{0}^{T}$ to
take value in $\mathcal{A}$, so that for all $t\in[0,T]$,
$\hat{Y}_{t}\in\mathcal{A}$.

###### Definition 10

A $(T,R,D)$ code consists of an _encoder_ $f$

$\displaystyle
f:\mathcal{N}_{0}^{T}\rightarrow\\{1,\dots,\lceil\exp(RT)\rceil\\}$

and a _decoder_ $g$

$\displaystyle
g:\\{1,\dots,\lceil\exp(RT)\rceil\\}\rightarrow\hat{\mathcal{Y}}_{0}^{T}$

satisfying

$\displaystyle\hat{Y}_{t}\in\mathcal{A},\,\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt\right]<\infty$

and the distortion constraint

$\displaystyle\frac{1}{T}\mathbb{E}\left[d(\hat{Y}_{0}^{T},Y_{0}^{T})\right]\leq
D.$

As before, we will call the encoder’s output $M=f(Y_{0}^{T})$ the _message_
and the decoder’s output $\hat{Y}_{0}^{T}=g(M)$ the _reconstruction._

###### Definition 11

A rate-distortion vector $(R,D)$ is said to be _achievable_ if for any
$\epsilon>0$, there exists a sequence of $(T_{n},R+\epsilon,D+\epsilon)$ codes
such that $\lim_{n\to\infty}T_{n}=\infty$.

###### Definition 12

The _rate-distortion region_ $\mathfrak{RD}^{\mathcal{P}}_{\mathcal{A}}$ is
the intersection of all achievable rate-distortion vectors $(R,D)$.

The rate-distortion region
$\mathfrak{RD}^{\mathcal{P},\text{F}}_{\mathcal{A}}$ with feedforward is
defined as in Definitions 11 and 12.

###### Theorem 5

The rate-distortion region for the constrained functional-covering of a
Poisson process with rate $\lambda>0$ is given by

$\mathfrak{RD}^{\mathcal{P}}_{\mathcal{A}}=\mathfrak{RD}^{\mathcal{P},\text{F}}_{\mathcal{A}}=\mathfrak{RD},$

where $\mathfrak{RD}$ is the convex hull of the union of sets of rate-
distortion vectors $(R,D)$ such that

$\displaystyle
R\geq\lambda\sum_{k=1}^{4}\beta_{k}\log\left(\frac{\beta_{k}}{\alpha_{k}}\right)$
$\displaystyle
D\geq\sum_{k=1}^{4}\alpha_{k}\Psi_{\mathcal{A}}\left(\frac{\lambda\beta_{k}}{\alpha_{k}}\right),$

where

$\displaystyle\Psi_{\mathcal{A}}(u)\triangleq\inf_{v\in\mathcal{A}}v-u\log(v)$

with the convention that $0\Psi(0/0)=0$, and $[\alpha_{k}]_{k=1}^{4}$ and
$[\beta_{k}]_{k=1}^{4}$ are probability vectors over $\\{1,2,3,4\\}$
satisfying $\alpha_{k}=0\Rightarrow\beta_{k}=0$.

###### Proof:

#### Achievability

Let

$\displaystyle{R}\triangleq\lambda\sum_{k=1}^{4}\beta_{k}\log\left(\frac{\beta_{k}}{\alpha_{k}}\right)$
$\displaystyle{D}\triangleq\sum_{k=1}^{4}\alpha_{k}\Psi_{\mathcal{A}}\left(\frac{\lambda\beta_{k}}{\alpha_{k}}\right).$

We will show achievability using a $(T,{R}+\epsilon,{D}+\epsilon)$ code
without feedforward. We will use discretization and results from the rate-
distortion theory for discrete memoryless sources (DMS). Define a binary-
valued discrete-time process $(\bar{Y}_{j}:j\in\\{1,\dots,n\\})$ as follows.
If there are one or more arrivals in the interval $((j-1)\Delta,j\Delta]$ of
the process $Y_{0}^{T}$ , then set $\bar{Y}_{j}$ to $1$, otherwise it equals
zero. Since $Y_{0}^{T}$ is a Poisson process with rate $\lambda$, the
components of $(\bar{Y}_{j}:j\in\\{1,\dots,n\\})$ are independent and
identically distributed with $P(\bar{Y}=1)=1-\exp(-\lambda\Delta)$. Consider
the following “test”-channel for $k\in\\{1,2,3,4\\}$,

$\displaystyle P(\bar{U}=k|\bar{Y}=1)$ $\displaystyle=\beta_{k},$
$\displaystyle P(\bar{U}=k|\bar{Y}=0)$ $\displaystyle=\alpha_{k}.$

Define the discretized distortion function

$\bar{d}(\hat{\bar{y}},\bar{y})\triangleq\hat{\bar{y}}-\frac{\log(\hat{\bar{y}})}{\Delta}\mathbf{1}\\{\bar{y}=1\\}\quad\hat{\bar{y}}\in\mathcal{A},\bar{y}\in\\{0,1\\}.$

The reconstruction $\hat{\bar{Y}}(k)$ is taken as a $v\in\mathcal{A}$
satisfying

$\displaystyle\left|\Psi_{\mathcal{A}}\left(\frac{\lambda\beta_{k}}{\alpha_{k}}\right)-\left(v-\frac{\lambda\beta_{k}}{\alpha_{k}}\log(v)\right)\right|\leq\frac{\epsilon}{4},$
(23)

where such a $v$ exists due to the definition of $\Psi_{\mathcal{A}}$. We
recall that if $\alpha_{k}=0$ then $\beta_{k}=0$, and hence $P(\bar{U}=k)=0$
for such a $k$. The scaling of the mutual information $I(\bar{U};\bar{Y})$ and
the distortion function $\bar{d}(\hat{\bar{Y}},\bar{Y})$ with respect to
$\Delta$ is given by the following lemma.

###### Lemma 13

$\displaystyle\lim_{\Delta\to 0}\frac{I(\bar{U};\bar{Y})}{\Delta}$
$\displaystyle={R}$ $\displaystyle\lim_{\Delta\to
0}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]$
$\displaystyle\leq{D}+\frac{\epsilon}{4}$

###### Proof:

Please see the supplementary material. ∎

Let

$\displaystyle\kappa\triangleq\max_{\begin{subarray}{c}k\in\\{1,2,3,4\\}\\\
\hat{\bar{Y}}(k)>0\end{subarray}}\left|\log\left(\hat{\bar{Y}}(k)\right)\right|.$
(24)

Due to [53, Theorem 9.3.2, p. 455], for a given $\Delta>0$,
$\bar{\epsilon}>0$, and all sufficiently large $n$, there exists an encoder
$\bar{f}$ and a decoder $\bar{g}$ such that

$\displaystyle\bar{f}:(\bar{Y}_{j}:j\in\\{1,\dots,n\\})\to\\{1,\dots,L\\}$
$\displaystyle\bar{g}:\\{1,\dots,L\\}\to(\hat{\bar{Y}}_{j}:j\in\\{1,\dots,n\\})$

satisfying

$\displaystyle\frac{1}{n}\log(L)\leq I(\bar{U};\bar{Y})+\bar{\epsilon},$
$\displaystyle\mathbb{E}\left[\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y_{j}})\right]\leq\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]+\bar{\epsilon}.$
(25)

Given the above setup, the encoder $f$ upon observing $Y_{0}^{T}$ obtains the
binary valued discrete time process $(\bar{Y}_{j}:j\in\\{1,\dots,n\\})$, and
sends $M=\bar{f}(\bar{Y}_{j}:j\in\\{1,\dots,n\\})$ to the decoder. The decoder
outputs the reconstruction $\hat{Y}_{0}^{T}$ as

$\displaystyle\hat{Y}_{t}\triangleq\sum_{j=1}^{n}\hat{\bar{Y}}_{j}\mathbf{1}\left\\{t\in((j-1)\Delta,j\Delta]\right\\}\quad
t\in[0,T].$

Let $\bar{\bar{Y}}_{j}$ denote the actual number of arrivals of $Y_{0}^{T}$ in
an interval $((j-1)\Delta,j\Delta]$. Then $\bar{d}$ is related to the original
distortion function via the above reconstruction as follows:

$\displaystyle\frac{1}{T}d(\hat{Y}_{0}^{T};Y_{0}^{T})$
$\displaystyle=\frac{1}{T}\int_{0}^{T}\hat{Y}_{t}\,dt-\frac{1}{T}\int_{0}^{T}\log(\hat{Y}_{t})\,dY_{t}$
$\displaystyle=\frac{1}{n}\sum_{j=1}^{n}\hat{\bar{Y}}_{j}-\frac{1}{T}\sum_{j=1}^{n}\log(\hat{\bar{Y}}_{j})\bar{\bar{Y}}_{j}$
$\displaystyle=\frac{1}{n}\sum_{j=1}^{n}\hat{\bar{Y}}_{j}-\frac{1}{n\Delta}\sum_{j=1}^{n}\log(\hat{\bar{Y}}_{j})\bar{Y}_{j}-\frac{1}{T}\sum_{j=1}^{n}\log(\hat{\bar{Y}}_{j})(\bar{\bar{Y}}_{j}-1)\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}.$
$\displaystyle=\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y}_{j})-\frac{1}{T}\sum_{j=1}^{n}\log(\hat{\bar{Y}}_{j})(\bar{\bar{Y}}_{j}-1)\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}$
$\displaystyle\overset{(a)}{\leq}\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y}_{j})+\frac{\kappa}{T}\sum_{j=1}^{n}(\bar{\bar{Y}}_{j}-1)\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}$
$\displaystyle\leq\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y}_{j})+\frac{\kappa}{T}\sum_{j=1}^{n}\bar{\bar{Y}}_{j}\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\},$

where for (a), we have used the definition of $\kappa$ in (24), since
$\bar{\bar{Y}}_{j}>1$ implies $\bar{Y}_{j}=1$ which implies
$\hat{\bar{Y}}_{j}>0$ in order for $\bar{d}(\hat{\bar{Y}}_{j},1)<\infty$,
which occurs a.s. since $\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]<\infty$ so
long as $\Delta$ is sufficiently small.
Hence taking expectations, we get

$\displaystyle\mathbb{E}\left[\frac{1}{T}d(\hat{Y}_{0}^{T},Y_{0}^{T})\right]$
$\displaystyle\leq\mathbb{E}\left[\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y}_{j})\right]+\kappa\mathbb{E}\left[\frac{1}{T}\sum_{j=1}^{n}\bar{\bar{Y}}_{j}\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}\right]$
$\displaystyle\overset{(a)}{\leq}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]+\kappa\mathbb{E}\left[\frac{1}{T}\sum_{j=1}^{n}\bar{\bar{Y}}_{j}\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}\right]+\bar{\epsilon}$
$\displaystyle\overset{(b)}{=}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]+\kappa(\lambda-\lambda\exp(-\lambda\Delta))+\bar{\epsilon}$
$\displaystyle\overset{(c)}{\leq}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]+\kappa\lambda^{2}\Delta+\bar{\epsilon},$
(26)

where, for (a), we have used (25),
for (b) we note that
$\mathbb{E}[\bar{\bar{Y}}_{j}\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}]=\lambda\Delta-\lambda\Delta\exp(-\lambda\Delta)$,
and
for (c), we have used the inequality $1-u\leq\exp(-u)$ .
Moreover using (25),

$\displaystyle\frac{1}{T}\log(L)=\frac{1}{n\Delta}\log(L)\leq\frac{I(\bar{U};\bar{Y})}{\Delta}+\frac{\bar{\epsilon}}{\Delta}.$
(27)

Now given the rate-distortion vector $({R},{D})$ and $\epsilon>0$, first
choose $\Delta<1$ sufficiently small so that

$\displaystyle\frac{I(\bar{U};\bar{Y})}{\Delta}$
$\displaystyle\leq{R}+\frac{\epsilon}{4}$
$\displaystyle\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]$
$\displaystyle\leq{D}+\frac{\epsilon}{2},$
$\displaystyle\kappa\lambda^{2}\Delta$ $\displaystyle\leq\epsilon/2.$

Then let $\bar{\epsilon}=\Delta\epsilon/4$, and choose a sufficiently large
$n$ so that (25) is satisfied. From (26) and (27) we conclude that a sequence
of $(T_{n},{R}+\epsilon,{D}+\epsilon)$ code exists with $T_{n}=n\Delta$ and
$T_{n}\to\infty$ as $n\to\infty$.

#### Converse

We will prove the converse when feedforward is present. For the given
$(T,R+\epsilon,D+\epsilon)$ code with feedforward, let $M$ denote the
encoder’s output. Since $I(M;Y_{0}^{T})<\infty$, we conclude from Theorem 1
that there exists a process $\Gamma_{0}^{T}$ such that $\Gamma_{0}^{T}$ is the
$(\mathcal{F}_{t}=\sigma(M,Y_{0}^{t}):t\in[0,T])$ intensity of $Y_{0}^{T}$ and

$I(M;Y_{0}^{T})=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-T\phi(\lambda).$
(28)

We also have

$\displaystyle\frac{1}{T}I(M;{Y}^{T}_{0})=\frac{1}{T}H(M)\leq\frac{1}{T}\log\left(\lceil\exp((R+\epsilon)T)\rceil\right)\leq
R+\epsilon+\frac{1}{T}.$

This gives

$\displaystyle
R\geq\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\phi(\lambda)-\epsilon-\frac{1}{T}.$

Let $\hat{Y}_{0}^{T}$ denote the decoder’s output. The distortion constraint
$D$ satisfies

$\displaystyle D$
$\displaystyle\geq\frac{1}{T}\mathbb{E}\left[d(\hat{Y}_{0}^{T},Y_{0}^{T})\right]-\epsilon$
$\displaystyle=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt-\log(\hat{Y}_{t})\,dY_{t}\right]-\epsilon$
$\displaystyle\overset{(a)}{=}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}-\Gamma_{t}\log(\hat{Y}_{t})\,\,dt\right]-\epsilon$
$\displaystyle\geq\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\inf_{v\in\mathcal{A}}v-\Gamma_{t}\log(v)\,dt\right]-\epsilon$
$\displaystyle\overset{(b)}{=}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Psi_{\mathcal{A}}(\Gamma_{t})\,dt\right]-\epsilon,$
(29)

where, for (a) we have used Lemma 12, and
for (b), we have used the definition of $\Psi_{\mathcal{A}}$. Defining $S$ to
be uniformly distributed on $[0,T]$, and independent of all other random
variables we have

$\displaystyle R$
$\displaystyle\geq\mathbb{E}\left[\phi(\Gamma_{S})\right]-\phi(\lambda)-\epsilon-\frac{1}{T}$
(30) $\displaystyle D$
$\displaystyle\geq\mathbb{E}\left[\Psi_{\mathcal{A}}(\Gamma_{S})\right]-\epsilon,$
(31)

Now we use Carathéodory’s theorem [54, Theorem 17.1]. There exist non-negative
$[\eta_{k}]_{k=1}^{4}$ and $[\alpha_{k}]_{k=1}^{4}$, such that
$\sum_{k=1}^{4}\alpha_{k}=1$ and

$\displaystyle\mathbb{E}\left[\phi({\Gamma}_{S})\right]$
$\displaystyle=\sum_{k=1}^{4}\alpha_{k}\phi(\eta_{k}),$ (32)
$\displaystyle\mathbb{E}\left[\Psi_{\mathcal{A}}({\Gamma}_{S})\right]$
$\displaystyle=\sum_{k=1}^{4}\alpha_{k}\Psi_{\mathcal{A}}(\eta_{k}),$ (33)
$\displaystyle\mathbb{E}\left[{\Gamma}_{S}\right]$
$\displaystyle=\sum_{k=1}^{4}\alpha_{k}\eta_{k}=\lambda,$ (34)

where in the last line we have used the fact that since ${\Gamma}_{0}^{T}$ is
the $(\sigma(M,{Y}_{0}^{T}):t\in[0,T])$-intensity of ${Y}_{0}^{T}$,
$\mathbb{E}\left[\int_{0}^{T}{\Gamma}_{t}\,dt\right]=\mathbb{E}[{Y}_{T}]=T\lambda$.
Now define

$\beta_{k}\triangleq\frac{\alpha_{k}\eta_{k}}{\lambda}.$

We note that $\beta_{k}=0$ if $\alpha_{k}=0$, and $\sum_{k=1}^{4}\beta_{k}=1$.
Substituting the above definitions in (30)-(31), we obtain

$\displaystyle R$
$\displaystyle\geq\left(\sum_{k=1}^{4}\alpha_{k}\eta_{k}\log(\eta_{k})-\lambda\log(\lambda)\right)-\epsilon-\frac{1}{T}$
$\displaystyle=\lambda\left(\sum_{k=1}^{4}\beta_{k}\log\left(\frac{\beta_{k}\lambda}{\alpha_{k}}\right)\mathbf{1}\\{\alpha_{k}>0\\}-\log(\lambda)\right)-\epsilon-\frac{1}{T}$
$\displaystyle=\lambda\sum_{k=1}^{4}\beta_{k}\log\left(\frac{\beta_{k}}{\alpha_{k}}\right)-\epsilon-\frac{1}{T}.$
(35)

Likewise,

$\displaystyle
D\geq\sum_{k=1}^{4}\alpha_{k}\Psi_{\mathcal{A}}\left(\frac{\lambda\beta_{k}}{\alpha_{k}}\right)-\epsilon.$

Since $\epsilon$ is arbitrary and $T$ can be made arbitrarily large, we obtain
the rate-distortion region in the statement of the theorem. ∎

If we do not place any restrictions on $\mathcal{A}$, i.e., if $\mathcal{A}$
is the set of all non-negative reals, then we obtain the functional-covering
distortion.

######  Corollary 2 (Functional Covering of Poisson Processes)

The rate-distortion function for functional-covering distortion is given by
$R_{\text{FC}}(D)=(\lambda-\lambda\log(\lambda)-D)^{+}$.

###### Proof:

For the functional-covering distortion, $\mathcal{A}$ is the set of non-
negative reals. Hence

$\displaystyle\Psi_{\mathcal{A}}(u)=\inf_{v\geq 0}v-u\log(v)=u-u\log(u).$

For any achievable $(R,D)$ we have

$\displaystyle R$
$\displaystyle\geq\lambda\sum_{k=1}^{4}\beta_{k}\log\left(\frac{\beta_{k}}{\alpha_{k}}\right),$
(36)

and

$\displaystyle D$
$\displaystyle\geq\sum_{k=1}^{4}\alpha_{k}\Psi_{\mathcal{A}}\left(\frac{\lambda\beta_{k}}{\alpha_{k}}\right)$
$\displaystyle=\sum_{k=1}^{4}\alpha_{k}\left(\frac{\lambda\beta_{k}}{\alpha_{k}}-\frac{\lambda\beta_{k}}{\alpha_{k}}\log\left(\frac{\lambda\beta_{k}}{\alpha_{k}}\right)\right)$
$\displaystyle=\lambda-\lambda\log(\lambda)-\lambda\sum_{k=1}^{4}\beta_{k}\log\left(\frac{\beta_{k}}{\alpha_{k}}\right).$

Hence

$\displaystyle R+D\geq\lambda-\lambda\log(\lambda),$

and this is achieved by $[\alpha_{k}]_{k=1}^{4}$ and $[\beta_{k}]_{k=1}^{4}$
that yield equality in (36). ∎

If take $\mathcal{A}=\\{0,1\\}$, then we recover the covering distortion in
[35, Theorem 1].

######  Corollary 3 (Covering Distortion [35])

The rate-distortion function for the covering distortion is given by
$R_{\text{C}}(D)=(-\lambda\log(D))^{+}$.

###### Proof:

For the covering distortion, $\mathcal{A}=\\{0,1\\}$. Hence

$\displaystyle\Psi_{\mathcal{A}}(u)=\inf_{v\in\\{0,1\\}}v-u\log(v)=\mathbf{1}\\{u>0\\}.$

Suppose $(R,D)$ is in $\mathfrak{RD}$. Then

$\displaystyle D$
$\displaystyle\geq\sum_{k=1}^{4}\alpha_{k}\Psi_{\mathcal{A}}\left(\frac{\lambda\beta_{k}}{\alpha_{k}}\right)$
$\displaystyle=\sum_{k=1}^{4}\alpha_{k}\mathbf{1}\\{\beta_{k}>0\\}$
$\displaystyle=\sum_{k\in\mathcal{B}}\alpha_{k},$

where we have defined $\mathcal{B}=\\{k:\beta_{k}>0\\}$. Similarly,

$\displaystyle R$
$\displaystyle\geq\lambda\sum_{k=1}^{4}\beta_{k}\log\left(\frac{\beta_{k}}{\alpha_{k}}\right)$
$\displaystyle=\lambda\sum_{k\in\mathcal{B}}\beta_{k}\log\left(\frac{\beta_{k}}{\alpha_{k}}\right)$
$\displaystyle\overset{(a)}{\geq}\lambda\left(\sum_{k\in\mathcal{B}}\beta_{k}\right)\log\left(\frac{\sum_{k\in\mathcal{B}}\beta_{k}}{\sum_{k\in\mathcal{B}}\alpha_{k}}\right)$
$\displaystyle=\lambda\log\left(\frac{1}{\sum_{k\in\mathcal{B}}\alpha_{k}}\right)$
$\displaystyle\geq\left(-\lambda\log(D)\right)^{+},$

where, (a) is due to the log-sum inequality, which can be achieved by setting
$\alpha_{1}=\min(1,D)$, $\alpha_{2}=1-\alpha_{1}$, $\beta_{1}=1$,
$\beta_{2}=0$. ∎

###### Remark 2

As in the general case in Theorem 4 (see Remark 1), the reconstruction
$\hat{y}_{0}^{T}$ (assuming it is bounded) can be used to define a probability
measure on the input space $(\mathcal{N}_{0}^{T},\mathfrak{F}^{N})$ via

$\frac{dP_{\hat{y}_{0}^{T}}}{dP_{0}}(y_{0}^{T})=\exp\left(\int_{0}^{T}\log(\hat{y}_{t})\,dy_{t}-(\hat{y}_{t}-1)\,dt\right),$

where $P_{0}$ is the measure under which $Y_{0}^{T}$ is a Poisson process with
unit rate. Moreover, in absence of feedforward, $\hat{y}_{0}^{T}$ is
deterministic (it depends only on the encoder’s output). Thus the input point-
process $Y_{0}^{T}$ is a non-homogeneous Poisson processes with rate
$\hat{y}_{0}^{T}$ under $P_{\hat{y}_{0}^{T}}$. As in the general case, the
functional-covering distortion is related to the above Radon-Nikodym
derivative via

$d(\hat{y}_{0}^{T},y_{0}^{T})=-\log\left(\frac{dP_{\hat{y}_{0}^{T}}}{dP_{0}}(y_{0}^{T})\right)+T$

$\Diamond$

##  VI The Poisson CEO Problem

Figure 1: Poisson CEO Problem.

We now consider the distributed problem shown Figure 1. Our goal is to
compress $Y_{0}^{T}$, which is a Poisson process with rate $\lambda>0$. Each
of the two encoders observes a degraded version of $Y_{0}^{T}$, denoted by
$Y_{0}^{(i),T}$, $i\in\\{1,2\\}$. $Y_{0}^{T}$ is first $p^{(i)}$-thinned to
obtain $\tilde{Y}_{0}^{(i),T}$, and then an independent Poisson process
$N_{0}^{(i),T}$ with rate $\mu^{(i)}$ is added to $\tilde{Y}_{0}^{(i),T}$ to
obtain $Y_{0}^{(i),T}$.

Recall that $\hat{\mathcal{Y}}_{0}^{T}$ is the set of all non-negative
functions $\hat{y}_{0}^{T}$ which are left-continuous with right-limits, and

$d(\hat{y}_{0}^{T},y_{0}^{T})=\int_{0}^{T}\hat{y}_{t}\,dt-\log(\hat{y}_{t})\,dy_{t}.$

###### Definition 13

A $(T,R^{(1)},R^{(2)},D)$ code for the Poisson CEO problem consists of
_encoders_ $f^{(1)}$ and $f^{(2)}$,

$\displaystyle
f^{(1)}:\mathcal{N}_{0}^{T}\rightarrow\\{1,\dots,\lceil\exp(R^{(1)}T)\rceil\\}$
$\displaystyle
f^{(2)}:\mathcal{N}_{0}^{T}\rightarrow\\{1,\dots,\lceil\exp(R^{(2)}T)\rceil\\},$

and a _decoder_ $g$,

$\displaystyle
g:\\{1,\dots,\lceil\exp(R^{(1)}T)\rceil\\}\times\\{1,\dots,\lceil\exp(R^{(2)}T)\rceil\\}\rightarrow\hat{\mathcal{Y}}_{0}^{T},$

satisfying

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt\right]<\infty,$

and the distortion constraint

$\displaystyle\frac{1}{T}\mathbb{E}\left[d(\hat{Y}_{0}^{T},Y_{0}^{T})\right]\leq
D.$

###### Definition 14

A rate-distortion vector $(R^{(1)},R^{(2)},D)$ is said to be _achievable_ for
the Poisson CEO problem if for any $\epsilon>0$, there exists a sequence
$(T_{n},R^{(1)}+\epsilon,R^{(2)}+\epsilon,D+\epsilon)$ codes $T_{n}\to\infty$
.

###### Definition 15

The _rate-distortion region_ for the Poisson CEO problem
$\mathfrak{RD}^{\mathcal{P}}$ is the intersection of all achievable rate-
distortion vectors $(R^{(1)},R^{(2)},D)$.

The rate-distortion region for the Poisson CEO problem with feedforward,
denoted by $\mathfrak{RD}^{\mathcal{P}}_{F}$, is defined analogously.

###### Theorem 6

The rate-distortion region for the Poisson CEO problem is given by

$\mathfrak{RD}^{\mathcal{P}}=\mathfrak{RD}^{\mathcal{P}}_{F}=\mathfrak{RD},$

where $\mathfrak{RD}$ is the convex hull of union of sets of rate-distortion
vectors $(R^{(1)},R^{(2)},D)$ such that

$\displaystyle
R^{(1)}\geq\left((1-p^{(1)})\lambda+\mu^{(1)}\right)\sum_{k=1}^{4}\beta^{(1)}_{k}\log\left(\frac{\beta^{(1)}_{k}}{\alpha^{(1)}_{k}}\right),$
$\displaystyle
R^{(2)}\geq\left((1-p^{(2)})\lambda+\mu^{(2)}\right)\sum_{k=1}^{4}\beta^{(2)}_{k}\log\left(\frac{\beta^{(2)}_{k}}{\alpha^{(2)}_{k}}\right),$
$\displaystyle
D\geq\lambda-\phi(\lambda)-\lambda\left(\sum_{k=1}^{4}\gamma^{(1)}_{k}\log\left(\frac{\gamma^{(1)}_{k}}{\alpha^{(1)}_{k}}\right)+\sum_{k=1}^{4}\gamma^{(2)}_{k}\log\left(\frac{\gamma^{(2)}_{k}}{\alpha^{(2)}_{k}}\right)\right)$

for some probability vectors $[\alpha^{(i)}_{k}]_{k=1}^{4}$,
$[\beta^{(i)}_{k}]_{k=1}^{4}$, and $[\gamma^{(i)}_{k}]_{k=1}^{4}$, where for
$k\in\\{1,2,3,4\\}$ and $i\in\\{1,2\\}$

$\displaystyle\begin{rcases}\gamma^{(i)}_{k}=p^{(i)}\alpha^{(i)}_{k}+(1-p^{(i)})\beta^{(i)}_{k}\\\
\alpha^{(i)}_{k}=0\Rightarrow\beta^{(i)}_{k}=0\end{rcases}\quad\text{if
}p^{(i)}<1,$
$\displaystyle\alpha^{(i)}_{k}=\beta^{(i)}_{k}=\gamma^{(i)}_{k}\quad\qquad\qquad\quad\quad\,\,\,\text{if
}p^{(i)}=1.$

###### Proof:

Please see the supplementary material. ∎

###### Remark 3

Note that there is no sum-rate constraint in the rate-distortion region of the
above theorem. This occurs due to the sparsity of points in a Poisson process.
After discretizing a Poisson process with rate $\lambda$, the expected number
of ones in the resulting binary process is roughly $\lambda T$, and the
remaining $T/\Delta-\lambda T$ bits are zeroes. When such a sparse binary
process is sent via two independent parallel channels as in (46)-(47), the
resulting output processes are almost independent. This implies that the
encoders do not need to bin their messages in the achievability argument.

######  Corollary 4 (Poisson CEO Problem without Thinning)

If $p^{(1)}=p^{(2)}=0$, then the rate-distortion region in Theorem 6 takes a
simple form

$\frac{\lambda}{\lambda+\mu^{(1)}}R^{(1)}+\frac{\lambda}{\lambda+\mu^{(2)}}R^{(2)}+D\geq\lambda-\phi(\lambda).$

######  Corollary 5 (Remote Poisson Source)

Consider a scenario where an encoder wishes to compress a Poisson process with
rate $\lambda>0$, but observes a degraded version of it, where the points are
first erased with independent probability $1-p$ and then an independent
Poisson process with rate $\mu$ is added to it. Then the rate-distortion
region $(R,D)$ is the convex hull of the union of all rate-distortion vectors
satisfying

$\displaystyle
R\geq\left((1-p)\lambda+\mu\right)\sum_{k=1}^{4}\beta_{k}\log\left(\frac{\beta_{k}}{\alpha_{k}}\right),$
$\displaystyle
D\geq\lambda-\phi(\lambda)-\lambda\cdot\sum_{k=1}^{4}\gamma_{k}\log\left(\frac{\gamma_{k}}{\alpha_{k}}\right),$

for some probability vectors $[\alpha_{k}]_{k=1}^{4}$,
$[\beta_{k}]_{k=1}^{4}$, and $[\gamma_{k}]_{k=1}^{4}$, where for
$k\in\\{1,2,3,4\\}$

$\displaystyle\gamma_{k}=p\alpha_{k}+(1-p)\beta_{k},\quad\alpha_{k}=0\Rightarrow\beta_{k}=0.$

## References

  * [1] N. V. Shende and A. B. Wagner, “Functional covering of point processes,” in _IEEE Int. Symp. Info. Theory_ , 2019, pp. 2039–2043.
  * [2] T. Berger, _Rate Distortion Theory: A Mathematical Basis for Data Compression_. Englewood Cliffs, NJ: Prentice Hall, 1971.
  * [3] D. H. Johnson, “Point process models of single-neuron discharges,” _Journal of Computational Neuroscience_ , vol. 3, no. 4, pp. 275–299, 1996\.
  * [4] J. H. Goldwyn, J. T. Rubinstein, and E. Shea-Brown, “A point process framework for modeling electrical stimulation of the auditory nerve,” _Journal of Neurophysiology_ , vol. 108, no. 5, pp. 1430–1452, 2012.
  * [5] F. Farkhooi, M. F. Strube-Bloss, and M. P. Nawrot, “Serial correlation in neural spike trains: Experimental evidence, stochastic modeling, and single neuron variability,” _Physical Review E_ , vol. 79, no. 2, p. 021905, 2009\.
  * [6] S. V. Sarma, U. T. Eden, M. L. Cheng, Z. M. Williams, R. Hu, E. Eskandar, and E. N. Brown, “Using point process models to compare neural spiking activity in the subthalamic nucleus of Parkinson’s patients and a healthy primate,” _IEEE Transactions on Biomedical Engineering_ , vol. 57, no. 6, pp. 1297–1305, 2010.
  * [7] E. N. Brown, R. E. Kass, and P. P. Mitra, “Multiple neural spike train data analysis: state-of-the-art and future challenges,” _Nature Neuroscience_ , vol. 7, no. 5, pp. 456–461, 2004.
  * [8] F. Rieke, D. Warland, R. de Ruyter van Steveninck, and W. Bialek, _Spikes: Exporing the Neural Code_. MIT Press, 1997\.
  * [9] J. Giles and B. Hajek, “An information-theoretic and game-theoretic study of timing channels,” _IEEE Trans. on Inf. Theory_ , vol. 48, no. 9, pp. 2455–2477, 2002.
  * [10] M. Shahzad and A. X. Liu, “Accurate and efficient per-flow latency measurement without probing and time stamping,” _IEEE/ACM Trans. Networking_ , vol. 24, no. 6, pp. 3477–3492, 2016.
  * [11] Y. Zhu, X. Fu, B. Graham, R. Bettati, and W. Zhao, “On flow correlation attacks and countermeasures in mix networks,” in _Proc. 4th Privacy Enhancement Technology Workshop (PET)_ , 2004.
  * [12] C. B. Attila Börcs, “A marked point process model for vehicle detection in aerial LIDAR point clouds,” in _ISPRS Ann. Photogrammetry, Remote Sens. and Spatial Inf. Sci_ , vol. 1-3, 2012, pp. 93–98.
  * [13] Y. Yu, J. Li, H. Guan, C. Wang, and M. Cheng, “A marked point process for automated tree detection from mobile laser scanning point cloud data,” in _2012 Intl. Conf. Comp. Vision in Remote Sensing_ , 2012, pp. 140–145.
  * [14] S. Nakamoto, “A peer-to-peer electronic cash system,” 2008. [Online]. Available: bitcoin.org/bitcoin
  * [15] Y. Lewenberg, Y. Bachrach, Y. Sompolinsky, A. Zohar, and J. S. Rosenschein, “Bitcoin mining pools: A cooperative game theoretic analysis,” in _Proc. 2015 Int. Conf. Autonomous Agents and Multiagent Sys._ , 2015, p. 919–927.
  * [16] Y. Kawase and S. Kasahara, “Transaction-confirmation time for bitcoin: A queueing analytical approach to blockchain mechanism,” in _Intl. Conf. on Queueing Theory and Network App._ , 2017, p. 75–88.
  * [17] C. Decker and R. Wattenhofer, “Information propagation in the bitcoin network,” in _IEEE P2P 2013 Proc._ , 2013, p. 1–10.
  * [18] A. Laourine and A. B. Wagner, “Secrecy capacity of the degraded Poisson wiretap channel,” in _Proc. IEEE Intl. Symp. Inf. Theory_ , Jun. 2010, pp. 2553–2557.
  * [19] A. D. Wyner, “Capacity and error exponent for the direct detection photon channel—Part I,” _IEEE Trans. Inf. Theory_ , vol. 34, no. 6, pp. 1449–1461, Nov. 1988.
  * [20] ——, “Capacity and error exponent for the direct detection photon channel—Part II,” _IEEE Trans. Inf. Theory_ , vol. 34, no. 6, pp. 1462–1471, Nov. 1988.
  * [21] A. Lapidoth, “On the reliability function of the ideal Poisson channel with noiseless feedback,” _IEEE Trans. Inf. Theory_ , vol. 39, no. 2, pp. 491–503, Mar. 1993.
  * [22] N. Shende and A. B. Wagner, “The stochastic-calculus approach to multiple-decoder Poisson channels,” _IEEE Trans. Inf. Theory_ , vol. 65, no. 8, pp. 5007–5027, Aug. 2019.
  * [23] F. Baccelli and P. Brémaud, _Palm Probabilities and Stationary Queues_. Springer-Verlag, 1987.
  * [24] C. Sutardja and J. M. Rabaey, “Isolator-less near-field RFID reader for sub-cranial powering/data link of millimeter-sized implants,” _IEEE Journal of Solid-State Circuits_ , vol. 53, no. 7, pp. 2032–2042, 2018.
  * [25] A. K. Skrivervik, A. J. M. Montes, I. V. Trivino, M. Bosiljevac, M. Veljovic, and Z. Sipus, “Antenna design for a cranial implant,” in _2020 Intl. Work. Antenna Tech. (iWAT)_ , 2020, pp. 1–4.
  * [26] R. L. de Queiroz and P. A. Chou, “Compression of 3D point clouds using a region-adaptive hierarchical transform,” _IEEE Trans. on Image Proc._ , vol. 25, no. 8, pp. 3947–3956, 2016.
  * [27] T. Golla and R. Klein, “Real-time point cloud compression,” in _2015 IEEE/RSJ International Conference on Intelligent Robots and Systems (IROS)_ , 2015, pp. 5087–5092.
  * [28] C. Tu, E. Takeuchi, C. Miyajima, and K. Takeda, “Compressing continuous point cloud data using image compression methods,” in _2016 IEEE 19th Intl. Conf. Intell. Transportation Sys. (ITSC)_ , 2016, pp. 1712–1719.
  * [29] R. Gallager, “Basic limits on protocol information in data communication networks,” _IEEE Trans. Info. Theory_ , vol. 22, no. 4, pp. 385–398, 1976\.
  * [30] A. S. Bedekar, “On the information about message arrival times required for in-order decoding,” in _IEEE Int. Symp. Inf. Theory_ , June 2001, p. 227\.
  * [31] S. Verdú, “The exponential distribution in information theory.” _Probl. Inf. Transm._ , vol. 32, no. 1, pp. 86–95, 1996.
  * [32] T. P. Coleman, N. Kiyavash, and V. G. Subramanian, “The rate-distortion function of a Poisson process with a queueing distortion measure,” in _Data Compress. Conf_ , Mar 2008, pp. 63–72.
  * [33] I. Rubin, “Information rates and data-compression schemes for Poisson processes,” _IEEE Trans. Info. Theory_ , vol. 20, no. 2, pp. 200–210, 1974\.
  * [34] G. Koliander, D. Schuhmacher, and F. Hlawatsch, “Rate-distortion theory of finite point processes,” _IEEE Trans. Info. Theory_ , vol. 64, no. 8, pp. 5832–5861, 2018.
  * [35] A. Lapidoth, A. Malar, and L. Wang, “Covering point patterns,” _IEEE Trans. Info. Theory_ , vol. 61, no. 9, pp. 4521–4533, 2015.
  * [36] H.-A. Shen, S. M. Moser, and J.-P. Pfister, “Rate-distortion problems of the Poisson process based on a group-theoretic approach,” 2022. [Online]. Available: https://arxiv.org/abs/2202.13684
  * [37] T. A. Courtade and R. D. Wesel, “Multiterminal source coding with an entropy-based distortion measure,” in _IEEE Int. Symp. Info. Theory_ , Jul 2011, pp. 2040–2044.
  * [38] T. A. Courtade and T. Weissman, “Multiterminal source coding under logarithmic loss,” _IEEE Trans. Info. Theory_ , vol. 60, no. 1, pp. 740–761, 2014.
  * [39] P. Brémaud, _Point Procceses and Queues: Martingale Dynamics_. Springer-Verlag, 1981.
  * [40] Y. Kabanov, “The capacity of a channel of the Poisson type,” _Theory of Probabilty and Applications_ , vol. 23, pp. 143–147, 1978.
  * [41] M. Davis, “Capacity and cutoff rate for Poisson-type channels,” _IEEE Trans. Info. Theory_ , vol. 26, no. 6, pp. 710–715, Nov 1980.
  * [42] N. V. Shende and A. B. Wagner, “The stochastic-calculus approach to multi-receiver Poisson channels,” _IEEE Trans. Info. Theory_ , vol. 65, no. 8, pp. 5007–5027, Aug 2019.
  * [43] R. Sundaresan and S. Verdú, “Capacity of queues via point-process channels,” _IEEE Trans. Info. Theory_ , vol. 52, no. 6, pp. 2697–2709, June 2006.
  * [44] L. Wang, “A strong data processing inequality for thinning Poisson processes and some applications,” in _IEEE Int. Symp. Info. Theory_ , June 2017, pp. 3180–3184.
  * [45] A. Klenke, _Probability Theory: A Comprehensive Course_ , 2nd ed. Springer London, 2013.
  * [46] A. Wyner, “A definition of conditional mutual information for arbitrary ensembles,” _Information and Control_ , vol. 38, no. 1, pp. 51 – 59, 1978\.
  * [47] I. M. Gel’fand and A. M. Yaglom, “Computation of the amount of information about a stochastic function contained in another such function,” _Uspekhi Mat. Nauk_ , vol. 12, no. 1, pp. 3–52, 1957.
  * [48] R. M. Gray, _Entropy and Information Theory_. Springer-Verlag, 1990.
  * [49] O. Kallenberg, _Foundations of Modern Probability_ , 2nd ed. Springer-Verlag, New York, 2002.
  * [50] R. S. Liptser and A. N. Shiryaev, _Statistics of Random Processes II_ , 2nd ed. Springer-Verlag Berlin Heidelberg, 2001.
  * [51] Y. Polyanskiy and Y. Wu, “Strong data-processing inequalities for channels and Bayesian networks,” in _Convexity and Concentration_. New York, NY: Springer New York, 2017, pp. 211–249.
  * [52] R. Durrett, _Essentials of Stochastic Processes_ , ser. Springer Texts in Statistics. Springer International Publishing, 2016.
  * [53] R. G. Gallager, _Information Theory and Reliable Communication_. New York, NY, USA: John Wiley & Sons, Inc., 1968\.
  * [54] R. Rockafellar, _Convex Analysis_. Princeton University Press, 1997.
  * [55] C. Dellacherie and P. A. Meyer, _Probabilities and Potential B: Theory of Martingales_ , ser. North-Holland Mathematics Studies. North-Holland, 1982, vol. 72.
  * [56] A. E. Gamal and Y.-H. Kim, _Network Information Theory_. Cambridge University Press, 2011.

###### Proof:

The first part of the lemma is due to [39, T12 Theorem, Chapter VI, p. 187].
To prove the second part we note that
$\mathbb{E}_{P^{Y_{0}^{T}}}[\int_{0}^{T}|\phi(\Gamma_{t})|\,dt]<\infty$
implies $\mathbb{E}_{P^{Y_{0}^{T}}}[\int_{0}^{T}\Gamma_{t}\,dt]<\infty$, which
in turn gives

$\int_{0}^{T}(1-\sqrt{\Gamma_{t}}))^{2}\leq\int_{0}^{T}(\Gamma_{t}+1)<\infty,$

${P^{Y_{0}^{T}}}$-a.s. Thus applying [50, Theorem 19.7, p. 343], we conclude
that $P^{Y_{0}^{T}}\ll P_{0}^{Y_{0}^{T}}$. Hence, from the first part of the
lemma

$\displaystyle\frac{dP^{Y_{0}^{T}}}{dP_{0}^{Y_{0}^{T}}}=\exp\left(\int_{0}^{T}\log(\Lambda_{t})\,dY_{t}-\Lambda_{t}+1\,dt\right),$

where the uniqueness of intensity [39, T12 Theorem, Chapter II, p. 31] gives
us

$\displaystyle\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}|\Gamma_{t}-\Lambda_{t}|\,dt\right]=0,\quad\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\mathbf{1}\\{\Gamma_{t}\neq\Lambda_{t}\\}\,dY_{t}\right]=0.$

Since

$\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}|\phi(\Gamma_{t})|\,dt\right]<\infty,$

we have

$\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}(\log(\Gamma_{t}))^{+}\,dY_{t}\right]=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}(\log(\Gamma_{t}))^{+}\Gamma_{t}\,dt\right]=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}(\phi(\Gamma_{t}))^{+}\,dt\right]<\infty,$

and

$\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}(\log(\Gamma_{t}))^{-}\,dY_{t}\right]=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}(\log(\Gamma_{t}))^{-}\Gamma_{t}\,dt\right]=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}(\phi(\Gamma_{t}))^{-}\,dt\right]<\infty.$

Hence

$\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]<\infty.$

Finally,

$\displaystyle\mathbb{E}_{P^{Y_{0}^{T}}}\left[\log\left(\frac{dP^{Y_{0}^{T}}}{dP_{0}^{Y_{0}^{T}}}\right)\right]$
$\displaystyle=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\log(\Lambda_{t})\,dY_{t}+\Lambda_{t}-1\,dt\right]$
$\displaystyle\mathbf{\overset{(a)}{=}}\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}+\Gamma_{t}-1\,dt\right]$
$\displaystyle=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]-\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}-1\,dt\right]$
$\displaystyle=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}-1\,dt\right]$
$\displaystyle=\mathbb{E}_{P^{Y_{0}^{T}}}\left[\int_{0}^{T}\phi(\Gamma_{t})-\Gamma_{t}+1\,dt\right].$

Here, for (a) we have used the uniqueness of the intensity and in the
remaining equalities, we have used the finiteness of the expectations
$\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]$,
$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]$. ∎

###### Proof:

Recall that $L_{0}^{T}$ can be written as

$\displaystyle
L_{t}=\exp\left(\int_{0}^{t}\log(\Gamma_{s})\,dY_{s}+(1-\Gamma_{s})\,ds\right),\quad
t\in[0,T].$

We note that for $t\in[0,T]$ $L_{t}$ satisfies

$\displaystyle L_{t}=\begin{cases}L_{t-}&\text{if }Y_{t}-Y_{t-}=0,\\\
\Gamma_{t}L_{t-}&\text{if }Y_{t}-Y_{t-}=1.\end{cases}$ (37)

Let $C_{0}^{T}$ be a non-negative $(\mathcal{G}_{t}:t\in[0,T])$-predictable
process. Then

$\displaystyle\mathbb{E}\left[\int_{0}^{T}C_{t}\,dY_{t}\right]$
$\displaystyle\overset{(a)}{=}\mathbb{E}_{\tilde{P}^{M,Y_{0}^{T}}}\left[L_{T}\int_{0}^{T}C_{t}\,dY_{t}\right]$
$\displaystyle\overset{(b)}{=}\mathbb{E}_{\tilde{P}^{M,Y_{0}^{T}}}\left[\int_{0}^{T}L_{t}C_{t}\,dY_{t}\right]$
$\displaystyle\overset{(c)}{=}\mathbb{E}_{\tilde{P}^{M,Y_{0}^{T}}}\left[\int_{0}^{T}\Gamma_{t}L_{t-}C_{t}\,dY_{t}\right]$
$\displaystyle\overset{(d)}{=}\mathbb{E}_{\tilde{P}^{M,Y_{0}^{T}}}\left[\int_{0}^{T}\Gamma_{t}L_{t-}C_{t}\,dt\right]$
$\displaystyle\overset{(e)}{=}\mathbb{E}_{\tilde{P}^{M,Y_{0}^{T}}}\left[\int_{0}^{T}\Gamma_{t}L_{t}C_{t}\,dt\right]$
$\displaystyle\overset{(f)}{=}\mathbb{E}_{\tilde{P}^{M,Y_{0}^{T}}}\left[L_{T}\int_{0}^{T}\Gamma_{t}C_{t}\,dt\right]$
$\displaystyle\overset{(g)}{=}\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}C_{t}\,dt\right],$

where, (a) follows since $L_{T}$ is the Radon-Nikodym derivative
$\frac{d{P}^{M,Y_{0}^{T}}}{d\tilde{P}^{M,Y_{0}^{T}}}$,
(b) follows due to [39, T19 Theorem, Appendix A2, p. 302],
(c) follows due to (37),
(d) follows since the
$(\tilde{P}^{M,Y_{0}^{T}},\mathcal{G}_{t}:t\in[0,T])$-intensity of $Y_{0}^{T}$
is 1, and $L_{t-}$ being a left-continuous adapted process is
$(\mathcal{G}_{t}:t\in[0,T])$-predictable,
(e) follows since the Lebesgue measure of the set $\\{t:t\in[0,T],L_{t-}\neq
L_{t}\\}$ is zero due to (37),
(f) again follows again due to [39, T19 Theorem, Appendix A2, p. 302], and
(g) again follows since $L_{T}$ is the Radon-Nikodym derivative
$\frac{d{P}^{M,Y_{0}^{T}}}{d\tilde{P}^{M,Y_{0}^{T}}}$. ∎

###### Proof:

We will first show that

$\mathbb{E}\left[\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\,dY_{t}\right]=\mathbb{E}\left[\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\Gamma_{t}\,dt\right]<\infty.$

Define $\Gamma^{1+}_{t}\triangleq\max(\Gamma_{t},1)$ and
$\Gamma^{1-}_{t}\triangleq\min(\Gamma_{t},1)$. We note that
$\Gamma_{t}\leq\Gamma^{1+}_{t}\leq\Gamma_{t}+1$ and
$\Gamma_{t}=\Gamma^{1+}_{t}\Gamma^{1-}_{t}$. Define the process $\mu_{0}^{T}$
as

$\displaystyle\mu_{t}\triangleq\frac{\Gamma^{1+}_{t}}{\Gamma_{t}}\mathbf{1}\\{\Gamma_{t}>0\\},\quad
t\in[0,T].$

Then $\mu_{0}^{T}$ is a non-negative $(\mathcal{G}_{t}:t\in[0,T])$-predictable
process and

$\int_{0}^{T}\mu_{t}\Gamma_{t}\,dt=\int_{0}^{T}\Gamma^{1+}_{t}\mathbf{1}\\{\Gamma_{t}>0\\}\,dt\leq\int_{0}^{T}(\Gamma_{t}+1)\,dt<\infty$

$P$-a.s. since $\mathbb{E}[\int_{0}^{T}\Gamma_{t}\,dt]<\infty$. Hence the
process $\hat{L}_{0}^{T}$ defined as

$\displaystyle\hat{L}_{t}\triangleq\exp\left(\int_{0}^{T}\log(\mu_{t})\,dY_{t}+(1-\mu_{t})\Gamma_{t}\,dt\right),\quad
t\in[0,T]$

is a $(P,G_{t}:t\in[0,T])$ non-negative super-martingale [39, T2 Theorem,
Chapter VI, p. 165]. Hence the following chain of inequalities hold

$\displaystyle\mathbb{E}\left[\log(\hat{L}_{T})\right]$
$\displaystyle\overset{(a)}{\leq}\log\left(\mathbb{E}[\hat{L}_{T}]\right)$
$\displaystyle\overset{(b)}{\leq}\log\left(\mathbb{E}[\hat{L}_{0}]\right)$
$\displaystyle=0.$ (38)

Here, for (a) we have used the fact that since $L_{0}^{T}$ is a super-
martingale, $L_{T}$ is integrable, and then Jensen’s inequality and
for (b), we have used the fact that $\hat{L}_{0}^{T}$ is a super-martingale,
hence $\mathbb{E}[\hat{L}_{T}]\leq\mathbb{E}[\hat{L}_{0}]$.
Let $\tau_{k}$ denote the $k$th arrival instant of the process $Y_{0}^{T}$,
i.e.,

$\displaystyle\tau_{k}=\inf\\{t\in[0,T]:Y_{t}=k\\},$

where the infimum of the null set is taken as $\infty$. Then if $\tau_{k}\leq
T$, $\Gamma_{\tau_{k}}>0$ $P$-a.s. [39, T12 Theorem, Chapter II, p. 31]. Hence
for $\tau_{k}\leq T$,

$\displaystyle\log(\mu_{\tau_{k}})=\log(\Gamma^{1+}_{\tau_{k}})-\log(\Gamma_{\tau_{k}})=-\log(\Gamma^{1-}_{\tau_{k}})=\left(\log(\Gamma_{\tau_{k}})\right)^{-}\quad
P-\text{a.s.},$

Thus we can write

$\displaystyle\log(\hat{L}_{T})=\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\,dY_{t}+\int_{0}^{T}(\Gamma_{t}-\Gamma^{1+}_{t})\mathbf{1}\\{\Gamma_{t}>0\\}\,dt.$

Using (38) we obtain

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\,dY_{t}+\int_{0}^{T}(\Gamma_{t}-\Gamma^{1+}_{t})\mathbf{1}\\{\Gamma_{t}>0\\}\,dt\right]=\mathbb{E}[\log(\hat{L_{T}})]\leq
0.$

We note that $\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\,dY_{t}$ is a non-
negative random variable, and

$\left|\mathbb{E}\left[\int_{0}^{T}(\Gamma_{t}-\Gamma^{1+}_{t})\mathbf{1}\\{\Gamma_{t}>0\\}\,dt\right]\right|\leq\mathbb{E}\left[\int_{0}^{T}(\Gamma_{t}+\Gamma^{1+}_{t})\,dt\right]\leq\mathbb{E}\left[\int_{0}^{T}(2\Gamma_{t}+1)\,dt\right]<\infty.$

Hence we can split the expectation to get

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\,dY_{t}\right]+\mathbb{E}\left[\int_{0}^{T}(\Gamma_{t}-\Gamma^{1+}_{t})\mathbf{1}\\{\Gamma_{t}>0\\}\,dt\right]\leq
0,$

which gives

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\,dY_{t}\right]\leq-\mathbb{E}\left[\int_{0}^{T}(\Gamma_{t}-\Gamma^{1+}_{t})\mathbf{1}\\{\Gamma_{t}>0\\}\,dt\right]<\infty.$
(39)

Hence

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\log(\Gamma_{t})\,dY_{t}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{+}\,dY_{t}\right]-\mathbb{E}\left[\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\,dY_{t}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}(\log(\Gamma_{t}))^{+}\Gamma_{t}\,dt\right]-\mathbb{E}\left[\int_{0}^{T}\left(\log(\Gamma_{t})\right)^{-}\Gamma_{t}\,dt\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right].$ (40)

∎

###### Proof:

Suppose that $\Gamma_{0}^{T}$ is the $(\mathcal{F}_{t}:t\in[0,T])$-intensity
of $Y_{0}^{T}$. Then applying [39, T8 Theorem, Chapter II, p. 27] with
$X_{s}=1$ proves $M_{0}^{T}$ is a $(\mathcal{F}_{t}:t\in[0,T])$-martingale.
Now suppose that $M_{0}^{T}$ is a $(\mathcal{F}_{t}:t\in[0,T])$-martingale.
Consider a simple $(\mathcal{F}_{t}:t\in[0,T])$-predictable process
$C_{0}^{T}$ of the form

$\displaystyle C_{t}=\mathbf{1}\\{\mathcal{E}\\}\mathbf{1}\\{u<t\leq v\leq
T\\}\quad\mathcal{E}\in\mathcal{F}_{u}.$

Then

$\displaystyle\mathbb{E}\left[\int_{0}^{T}C_{s}\,d{Y}_{s}\right]$
$\displaystyle=\mathbb{E}\left[\mathbf{1}\\{\mathcal{E}\\}(Y_{v}-Y_{u})\right]$
$\displaystyle=\mathbb{E}\left[\mathbf{1}\\{\mathcal{E}\\}\mathbb{E}[(Y_{v}-Y_{u})|\mathcal{F}_{u}]\right]$
$\displaystyle\overset{(a)}{=}\mathbb{E}\left[\mathbf{1}\\{\mathcal{E}\\}\mathbb{E}\left[\int_{u}^{v}\Gamma_{s}\,ds\middle|\mathcal{F}_{u}\right]\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}C_{s}\Gamma_{s}\,d{s}\right],$ (41)

where for (a) we have used the martingale property of $M_{0}^{T}$. Thus by the
monotone class theorem, for all bounded
$(\mathcal{F}_{t}:t\in[0,T])$-predictable processes $C_{0}^{T}$, (41) holds
(see [39, App. A1, Theorem T5, p. 264]). Then by applying the monotone
convergence theorem, we can show that (41) holds for all non-negative
$(\mathcal{F}_{t}:t\in[0,T])$-predictable processes as well, so that
$\Gamma_{0}^{T}$ is the $(\mathcal{F}_{t}:t\in[0,T])$-intensity of
$Y_{0}^{T}$. ∎

###### Proof:

There exists a $(\mathcal{G}_{t}:t\in[0,T])$-predictable process $\Pi_{0}^{T}$
such that $P$-a.s. $\Pi_{t}=\mathbb{E}[{\Lambda}_{t}|\mathcal{G}_{t-}]$,
$t\in[0,T]$ [55, Chapter 6, Theorem 43, p. 103]. We will show that
$\Pi_{0}^{T}$ is the $(\mathcal{G}_{t}:t\in[0,T])$-intensity of ${N}_{0}^{T}$.
Let ${D}_{0}^{T}$ be a non-negative $(\mathcal{G}_{t}:t\in[0,T])$-predictable
process. As $\mathcal{G}_{t}\subseteq\mathcal{F}_{t}$, it is also
$(\mathcal{F}_{t}:t\in[0,T])$-predictable. Thus

$\displaystyle\mathbb{E}\left[\int_{0}^{T}D_{s}\,d{{N}}_{s}\right]=\mathbb{E}\left[\int_{0}^{T}D_{s}{\Lambda}_{s}\,ds\right].$
(42)

Hence

$\displaystyle\mathbb{E}\left[\int_{0}^{T}D_{s}\Pi_{s}\,ds\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}D_{s}\mathbb{E}[{\Lambda}_{s}|\mathcal{G}_{s-}]\,ds\right]$
$\displaystyle\stackrel{{\scriptstyle(a)}}{{=}}\mathbb{E}\left[\int_{0}^{T}\mathbb{E}[D_{s}{\Lambda}_{s}|\mathcal{G}_{s-}]\,ds\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}D_{s}{\Lambda}_{s}\,ds\right]$
$\displaystyle\stackrel{{\scriptstyle(b)}}{{=}}\mathbb{E}\left[\int_{0}^{T}D_{s}\,d{{N}}_{s}\right].$

Here, (a) is due to the fact that $D_{s}$ is $\mathcal{G}_{s-}$ measurable
[39, Exercise E10, Chapter I, p. 9], and
(b) is due to (42).

Hence the $\left(\mathcal{G}_{t}:t\in[0,T]\right)$-intensity of
${N}_{0}^{{T}}$ is $\Pi_{0}^{{T}}$. ∎

###### Proof:

We first note that since $Y_{0}^{T}$ and $N_{0}^{T}$ are independent,
trajectories of $Z_{0}^{T}$ are a.s. in $\mathcal{N}_{0}^{T}$. The
$(\tilde{\mathcal{F}}_{t}\triangleq\sigma(M,Y_{0}^{t},N_{0}^{t}):t\in[0,T])$-intensities
of $Y_{0}^{T}$ and $N_{0}^{T}$ are $\Gamma_{0}^{T}$ and $\Pi_{0}^{T}$
respectively [39, E5 Exercise, Chapter II, p. 28]. Then for a non-negative
$(\tilde{\mathcal{F}}_{t}:t\in[0,T])$-predictable process $C_{0}^{T}$:

$\displaystyle\mathbb{E}\left[\int_{0}^{T}C_{s}\,d{Z}_{s}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}C_{s}\,d{Y}_{s}\right]+\mathbb{E}\left[\int_{0}^{T}C_{s}\,d{N}_{s}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}C_{s}\Gamma_{s}\,d{s}\right]+\mathbb{E}\left[\int_{0}^{T}C_{s}\Pi_{s}\,ds\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}C_{s}(\Gamma_{s}+\Pi_{s})\,d{s}\right].$

Hence the $(\tilde{\mathcal{F}}_{t}:t\in[0,T])$-intensity of $Z_{0}^{T}$ is
$(\Gamma_{t}+\Pi_{t}:t\in[0,T])$. Since
$\mathcal{F}_{t}\subseteq\tilde{\mathcal{F}}_{t}$ the statement of the lemma
follows from an application of Lemma 8. ∎

###### Proof:

Let $\mathcal{H}_{t}\triangleq\sigma(M,Y_{0}^{t},Z_{0}^{t})$. We note that the
$(\mathcal{H}_{t}:t\in[0,T])$-intensity of $Y_{0}^{T}$ is $\Gamma_{0}^{T}$.
Now we will compute the $(\mathcal{H}_{t}:t\in[0,T])$-intensity of
$Z_{0}^{T}$. Let $(\chi_{i}:i\in\\{1,\dots\\})$ denote the sequence of
independent and identically distributed Bernoulli random variables which
indicate if a particular point in point process ${Y}_{0}^{T}$ is erased or
not. In particular, if $\chi_{j}=1$, then the $j$th point in $Y_{0}^{T}$ is
retained, so that $\mathbb{E}[\chi_{j}]=1-p$. Then for $0\leq u<v\leq T$

$\displaystyle
Z_{v}-Z_{u}=\sum_{k=Y_{u}+1}^{Y_{v}}\chi_{k}=\sum_{k=1}^{\infty}\chi_{k}\mathbf{1}\\{Y_{u}<k\leq
Y_{v}\\}.$

Using the monotone convergence theorem for the conditional expectation,

$\displaystyle\mathbb{E}\left[(Z_{v}-Z_{u})|\mathcal{H}_{u}\right]$
$\displaystyle=\sum_{k=1}^{\infty}\mathbb{E}[\chi_{k}\mathbf{1}\\{Y_{u}<k\leq
Y_{v}\\}|\mathcal{H}_{u}]$
$\displaystyle\overset{(a)}{=}\sum_{k=1}^{\infty}\mathbb{E}[\chi_{k}|\mathcal{H}_{u}]\mathbb{E}[\mathbf{1}\\{Y_{u}<k\leq
Y_{v}\\}|\mathcal{H}_{u}]$
$\displaystyle\overset{(b)}{=}(1-p)\mathbb{E}[(Y_{v}-Y_{u})|\mathcal{H}_{u}]$
$\displaystyle\overset{(c)}{=}(1-p)\mathbb{E}\left[\int_{u}^{v}\Gamma_{s}\,ds\middle|\mathcal{H}_{u}\right],$

where, for (a) we have used the fact that given $\mathcal{H}_{u}$, $\chi_{k}$
is independent of $Y_{0}^{T}$,
for (b), we use note that
$\mathbb{E}[\chi_{k}|\mathcal{H}_{u}]=\chi_{k}\mathbf{1}\\{k\leq
Y_{u}\\}+(1-p)\mathbf{1}\\{k>Y_{u}\\}$, and
for (c), we have used the martingale property of $M_{0}^{T}$.
Then

$\tilde{M}_{t}\triangleq Z_{t}-\int_{0}^{t}(1-p)\Gamma_{s}\,ds\quad t\in[0,T]$

is a $(\mathcal{H}_{t}:t\in[0,T])$-martingale. Hence from Lemma 7, the
$(\mathcal{H}_{t}:t\in[0,T])$-intensity of $Z_{0}^{T}$ is
$((1-p)\Gamma_{t}:t\in[0,T])$. An application of Lemma 8 the proves the
statement of the lemma. ∎

###### Proof:

We will require the following inequality

$\displaystyle u\log(v)\leq\phi(u)-u+v,\quad 0\leq u,v<\infty.$ (43)

The inequality can be verified to be true if either or both $u$, $v$ are zero.
If $u,v>0$, the inequality follows from $\log(u/v)\geq(1-v/u)$.

Defining $\hat{Y}^{1+}_{t}\triangleq\max(1,\hat{Y}_{t})$, we note that
$\hat{Y}^{1+}_{t}\leq\hat{Y}_{t}+1$. Consider

$\displaystyle\mathbb{E}\left[\int_{0}^{T}(\log(\hat{Y}_{t}))^{+}\,dY_{t}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}^{1+}_{t})\,dY_{t}\right]$
$\displaystyle\overset{(a)}{=}\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}^{1+}_{t})\Gamma_{t}\,dt\right]$
$\displaystyle\overset{(b)}{\leq}\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\Gamma_{t}+\hat{Y}^{1+}_{t}\,dt\right]$
$\displaystyle\overset{(c)}{=}\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]+\mathbb{E}\left[\int_{0}^{T}\hat{Y}^{1+}_{t}\,dt\right]$
$\displaystyle<\infty,$ (44)

where, for (a), we have used the facts that $(\hat{Y}^{1+}_{t}:t\in[0,T])$ is
$(\mathcal{F}_{t}:t\in[0,T])$-predictable, $\log(\hat{Y}^{1+}_{t})$ is non-
negative, and $\Gamma_{0}^{T}$ is the $(\mathcal{F}_{t}:t\in[0,T])$-intensity
of $Y_{0}^{T}$,
for (b), we note that $\hat{Y}_{t}^{1+}$ and $\Gamma_{t}$ are $P$-a.s finite,
and then use the inequality in (43), and
for (c), we have used the facts that
$\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]<\infty$ (via Theorem
1), $\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]<\infty$, and
$\mathbb{E}\left[\int_{0}^{T}\hat{Y}^{1+}_{t}\,dt\right]\leq\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}+1\,dt\right]<\infty$.
Hence we can write

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}_{t})\,dY_{t}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}(\log(\hat{Y}_{t}))^{+}\,dY_{t}\right]-\mathbb{E}\left[\int_{0}^{T}(\log(\hat{Y}_{t}))^{-}\,dY_{t}\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}(\log(\hat{Y}_{t}))^{+}\Gamma_{t}\,dt\right]-\mathbb{E}\left[\int_{0}^{T}(\log(\hat{Y}_{t}))^{-}\Gamma_{t}\,dt\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}_{t})\Gamma_{t}\,dt\right].$
(45)

∎

###### Proof:

The first part of the lemma follows directly from L’Hôpital’s rule. For the
second part

$\displaystyle\lim_{\Delta\to 0}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]$
$\displaystyle=\lim_{\Delta\to
0}\sum_{k=1}^{4}\hat{\bar{Y}}(k)\exp(-\lambda\Delta)\alpha_{k}+\left(\hat{\bar{Y}}(k)-\frac{\log(\hat{\bar{Y}}(k))}{\Delta}\right)(1-\exp(-\lambda\Delta)\beta_{k}$
$\displaystyle=\sum_{k=1}^{4}\hat{\bar{Y}}(k)\alpha_{k}-\lambda\log(\hat{\bar{Y}}(k))\beta_{k}$
$\displaystyle=\sum_{k=1}^{4}\alpha_{k}\left(\hat{\bar{Y}}(k)-\frac{\lambda\beta_{k}}{\alpha_{k}}\log(\hat{\bar{Y}}(k))\right)\mathbf{1}\\{\alpha_{k}>0\\}$
$\displaystyle\overset{(a)}{\leq}\sum_{k=1}^{4}\alpha_{k}\left(\Psi_{\mathcal{A}}\left(\frac{\lambda\beta_{k}}{\alpha_{k}}\right)+\frac{\epsilon}{4}\right)$
$\displaystyle={D}+\frac{\epsilon}{4},$

where for (a), we have used the definition in (23).
∎

###### Proof:

The first limit can be evaluated using L’Hôpital’s rule. To compute the second
limit, consider

$\displaystyle\lim_{\Delta\to 0}{P(\bar{U}^{(i)}=k|\bar{Y}=1)}$
$\displaystyle=\lim_{\Delta\to
0}\sum_{l=0}^{1}{P(\bar{U}^{(i)}=k,\bar{Y}^{(i)}=l|\bar{Y}=1)}$
$\displaystyle=\lim_{\Delta\to
0}\sum_{l=0}^{1}{P(\bar{Y}^{(i)}=l|\bar{Y}=1)}(\bar{U}^{(i)}=k|\bar{Y}^{(i)}=l)$
$\displaystyle=p^{(i)}\alpha^{(i)}_{k}+(1-p^{(i)})\beta^{(i)}_{k}$
$\displaystyle={\gamma^{(i)}_{k}}.$

Then we have

$\displaystyle\lim_{\Delta\to
0}P(\bar{U}_{1}^{(1)}=k_{1},\bar{U}_{1}^{(2)}=k_{2}|\bar{Y}=1))$
$\displaystyle=\lim_{\Delta\to
0}P(\bar{U}_{1}^{(1)}=k_{1}|\bar{Y}=1)P(\bar{U}_{1}^{(2)}=k_{2}|\bar{Y}=1)$
$\displaystyle=\gamma^{(1)}_{k_{1}}\gamma^{(2)}_{k_{2}}.$

Recalling that $\alpha^{(i)}_{k}=0$ implies
$\beta^{(i)}_{k}=\gamma^{(i)}_{k}=0$, we have

$\displaystyle\lim_{\Delta\to
0}\frac{\mathbb{E}[\log(\hat{\bar{Y}})\mathbf{1}\\{\bar{Y}=1\\}]}{\Delta}$
$\displaystyle=\lim_{\Delta\to 0}\frac{P(\bar{Y}=1)}{\Delta}\lim_{\Delta\to
0}\mathbb{E}[\log(\hat{\bar{Y}})|\bar{Y}=1]$
$\displaystyle=\lambda\sum_{k_{1},k_{2}}\lim_{\Delta\to
0}P(\bar{U}=k_{1},\bar{U}_{2}=k_{2}|\bar{Y}=1))\log\left(\hat{\bar{Y}}(k_{1},k_{2})\right)$
$\displaystyle=\lambda\sum_{k_{1},k_{2}}\gamma^{(1)}_{k_{1}}\gamma^{(2)}_{k_{2}}\log\left(\lambda\frac{\gamma^{(1)}_{k_{1}}\gamma^{(2)}_{k_{2}}}{\alpha^{(1)}_{k_{1}}\alpha^{(2)}_{k_{2}}}\right)\mathbf{1}\\{\gamma^{(1)}_{k_{1}}\gamma^{(2)}_{k_{2}}>0\\}$
$\displaystyle=\lambda\sum_{k_{1}=1}^{4}\gamma^{(1)}_{k_{1}}\log\left(\frac{\gamma^{(1)}_{k_{1}}}{\alpha^{(1)}_{k_{1}}}\right)+\lambda\sum_{k_{2}=1}^{4}\gamma^{(1)}_{k_{2}}\log\left(\frac{\gamma^{(2)}_{k_{2}}}{\alpha^{(2)}_{k_{2}}}\right)+\phi(\lambda).$

Now to compute $\lim_{\Delta\to 0}\mathbb{E}[\hat{\bar{Y}}]$, we first
calculate

$\displaystyle\lim_{\Delta\to
0}P(\bar{U}_{1}^{(1)}=k_{1},\bar{U}_{1}^{(2)}=k_{2})$
$\displaystyle=\lim_{\Delta\to
0}P(\bar{U}_{1}^{(1)}=k_{1},\bar{U}_{1}^{(2)}=k_{2}|\bar{Y}=0)P(\bar{Y}=0)$
$\displaystyle\quad+\lim_{\Delta\to
0}P(\bar{U}_{1}^{(1)}=k_{1},\bar{U}_{1}^{(2)}=k_{2}|\bar{Y}=1)P(\bar{Y}=1)$
$\displaystyle=\lim_{\Delta\to
0}P(\bar{U}_{1}^{(1)}=k_{1},\bar{U}_{1}^{(2)}=k_{2}|\bar{Y}=0)$
$\displaystyle=\lim_{\Delta\to
0}P(\bar{U}_{1}^{(1)}=k_{1}|\bar{Y}=0)P(\bar{U}_{1}^{(2)}=k_{2}|\bar{Y}=0)$
$\displaystyle=\alpha^{(1)}_{k_{1}}\alpha^{(2)}_{k_{2}}.$

This gives

$\displaystyle\lim_{\Delta\to 0}\mathbb{E}[\hat{\bar{Y}}]$
$\displaystyle=\sum_{k_{1},k_{2}}\lim_{\Delta\to
0}P(\bar{U}=k_{1},\bar{U}_{2}=k_{2})\hat{\bar{Y}}(k_{1},k_{2})$
$\displaystyle=\lambda\sum_{k_{1},k_{2}}\alpha^{(1)}_{k_{1}}\alpha^{(2)}_{k_{2}}\frac{\gamma^{(1)}_{k_{1}}\gamma^{(2)}_{k_{2}}}{\alpha^{(1)}_{k_{1}}\alpha^{(2)}_{k_{2}}}\mathbf{1}\\{\alpha^{(1)}_{k_{1}}\alpha^{(2)}_{k_{2}}>0\\}$
$\displaystyle=\lambda.$

Thus

$\displaystyle\lim_{\Delta\to 0}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]$
$\displaystyle=\lambda-\phi(\lambda)-\lambda\left(\sum_{k_{1}=1}^{4}\gamma^{(1)}_{k_{1}}\log\left(\frac{\gamma^{(1)}_{k_{1}}}{\alpha^{(1)}_{k_{1}}}\right)+\sum_{k_{2}=1}^{4}\gamma^{(2)}_{k_{2}}\log\left(\frac{\gamma^{(2)}_{k_{2}}}{\alpha^{(2)}_{k_{2}}}\right)\right)$
$\displaystyle={D}.$

∎

###### Proof:

### Achievability:

Let

$\displaystyle{R}^{(1)}\triangleq\left((1-p^{(1)})\lambda+\mu^{(1)}\right)\sum_{k=1}^{4}\beta^{(1)}_{k}\log\left(\frac{\beta^{(1)}_{k}}{\alpha^{(1)}_{k}}\right)$
$\displaystyle{R}^{(2)}\triangleq\left((1-p^{(2)})\lambda+\mu^{(2)}\right)\sum_{k=1}^{4}\beta^{(2)}_{k}\log\left(\frac{\beta^{(2)}_{k}}{\alpha^{(2)}_{k}}\right)$
$\displaystyle{D}\triangleq\lambda-\phi(\lambda)-\lambda\cdot\left(\sum_{k=1}^{4}\gamma^{(1)}_{k}\log\left(\frac{\gamma^{(1)}_{k}}{\alpha^{(1)}_{k}}\right)+\gamma^{(2)}_{k}\log\left(\frac{\gamma^{(2)}_{k}}{\alpha^{(2)}_{k}}\right)\right).$

We will show achievability using a
$(T,{R}^{(1)}+\epsilon,{R}^{(2)}+\epsilon,{D}+\epsilon)$ code without
feedforward. We will use discretization and results from the rate-distortion
theory for discrete memoryless sources (DMS).

First consider the case when for each $i\in\\{1,2\\}$, at least one of the
following conditions is satisfied

  1. C.1

$\beta^{(i)}_{k}>0$ for all $k$,

  2. C.2

$p^{(i)}>0$.

Fix $\Delta>0$, and let $T\triangleq n\Delta$ for an integer $n$. For each
$i\in\\{1,2\\}$, define a binary valued discrete time process
$(\bar{Y}^{(i)}_{j}:j\in\\{1,\dots,n\\})$ as follows. If there are one or more
arrivals in the interval $((j-1)\Delta,j\Delta]$ of the process
$Y_{0}^{(i),T}$, then set $\bar{Y}^{(i)}_{j}$ to $1$, otherwise set it equal
to zero. Since $Y_{0}^{(i),T}$ is a Poisson process with rate
$\lambda^{(i)}\triangleq(1-p^{(i)})\lambda+\mu^{(i)}$, the components of
$(\bar{Y}^{(i)}_{j}:j\in\\{1,\dots,n\\})$ are independent and identically
distributed with $P(\bar{Y}^{(i)}=1)=1-\exp(-\lambda^{(i)}\Delta)$. Similarly,
if $(\bar{Y}_{j}:j\in\\{1,\dots,n\\})$ denotes the discretized process
$Y_{0}^{T}$, then we have

$\displaystyle
P\left(\bar{Y}^{(i)}_{j}:j\in\\{1,\dots,n\\}\middle|\bar{Y}_{j}:j\in\\{1,\dots,n\\}\right)=\prod_{j=1}^{n}P(\bar{Y}^{(i)}_{j}|\bar{Y}_{j})$

due to the memoryless property of Poisson processes and independent thinning.
Consider the following “test”-channel for $k\in\\{1,2,3,4\\}$,

$\displaystyle P(\bar{U}^{(i)}=k|\bar{Y}^{(i)}=1)=\beta^{(i)}_{k},$ (46)
$\displaystyle P(\bar{U}^{(i)}=k|\bar{Y}^{(i)}=0)=\alpha^{(i)}_{k}.$ (47)

Define the discretized distortion function

$\displaystyle\bar{d}(\hat{\bar{y}},\bar{y})\triangleq\hat{\bar{y}}-\frac{\log(\hat{\bar{y}})}{\Delta}\mathbf{1}\\{\bar{y}=1\\}\quad\hat{\bar{y}}\geq
0,\bar{y}\in\\{0,1\\}.$ (48)

The reconstruction $\hat{\bar{Y}}$ is taken as

$\displaystyle\hat{\bar{Y}}(\bar{U}^{(1)},\bar{U}^{(2)})=\lambda\hat{\bar{Y}}^{(1)}(\bar{U}^{(1)})\hat{\bar{Y}}^{(2)}(\bar{U}^{(2)}),$

where

$\displaystyle\hat{\bar{Y}}^{(i)}(k)=\begin{cases}\frac{\gamma^{(i)}_{k}}{\alpha^{(i)}_{k}}&\text{if}\,\alpha^{(i)}_{k}>0,\\\
1&\text{otherwise}.\end{cases}$

We note that since
$\gamma^{(i)}_{k}=p^{(i)}\alpha^{(i)}_{k}+(1-p^{(i)})\beta^{(i)}_{k}$, and at
least one of C.1-C.2 is satisfied, $\hat{\bar{Y}}^{(i)}(k)>0$, and hence
$\hat{\bar{Y}}>0$. Thus the distortion function
$\bar{d}(\hat{\bar{Y}},\bar{Y})$ in (48) is bounded. Let

$\displaystyle\kappa\triangleq\max_{k_{1},k_{2}}\left|\log\left(\hat{\bar{Y}}(k_{1},k_{2})\right)\right|.$
(49)

Due to the Berger-Tung inner bound [56, Theorem 12.1, p. 295], for a given
$\Delta>0$, $\bar{\epsilon}>0$, and all sufficiently large $n$, there exists
encoders $\bar{f}^{(1)}$ and $\bar{f}^{(2)}$, and a decoder $\bar{g}$ such
that for $i\in\\{1,2\\}$

$\displaystyle\bar{f}^{(i)}:(\bar{Y}^{(i)}_{j}:j\in\\{1,\dots,n\\})\to\\{1,\dots,L^{(i)}\\}$
$\displaystyle\bar{g}:\\{1,\dots,L^{(1)}\\}\times\\{1,\dots,L^{(2)}\\}\to(\hat{\bar{Y}}_{j}:j\in\\{1,\dots,n\\}),$

satisfying

$\displaystyle\frac{1}{n}\log(L^{(i)})\leq
I(\bar{U}^{(i)};\bar{Y}^{(i)})+\bar{\epsilon},$ (50)
$\displaystyle\mathbb{E}\left[\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y_{j}})\right]\leq\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]+\bar{\epsilon}.$
(51)

It is noteworthy that the Berger-Tung inner bound has a conditioning term in
the mutual-information expression, which in general is a stronger bound than
that presented here. However, in our setting we can drop this conditioning as
explained in Remark 3 in the main paper.

Given the above setup, each encoder $f^{(i)}$ upon observing $Y_{0}^{(i),T}$
obtains the binary valued discrete-time process
$(\bar{Y}^{(i)}_{j}:j\in\\{1,\dots,n\\})$, and sends
$M^{(i)}=\bar{f}^{(i)}(\bar{Y}^{(i)}_{j}:j\in\\{1,\dots,n\\})$ to the decoder.
The decoder outputs the reconstruction $\hat{Y}_{0}^{T}$ as

$\displaystyle\hat{Y}_{t}\triangleq\sum_{j=1}^{n}\hat{\bar{Y}}_{j}\mathbf{1}\left\\{t\in((j-1)\Delta,j\Delta]\right\\}\quad
t\in[0,T].$

Let $\bar{\bar{Y}}_{j}$ denote the actual number of arrivals of $Y_{0}^{T}$ in
an interval $((j-1)\Delta,j\Delta]$. Then $\bar{d}$ is related to the original
distortion function via the above reconstruction as follows:

$\displaystyle\frac{1}{T}d(\hat{Y}_{0}^{T};Y_{0}^{T})$
$\displaystyle=\frac{1}{T}\int_{0}^{T}\hat{Y}_{t}\,dt-\frac{1}{T}\int_{0}^{T}\log(\hat{Y}_{t})\,dY_{t}$
$\displaystyle=\frac{1}{n}\sum_{j=1}^{n}\hat{\bar{Y}}_{j}-\frac{1}{T}\sum_{j=1}^{n}\log(\hat{\bar{Y}}_{j})\bar{\bar{Y}}_{j}$
$\displaystyle=\frac{1}{n}\sum_{j=1}^{n}\hat{\bar{Y}}_{j}-\frac{1}{n\Delta}\sum_{j=1}^{n}\log(\hat{\bar{Y}}_{j})\bar{Y}_{j}-\frac{1}{T}\sum_{j=1}^{n}\log(\hat{\bar{Y}}_{j})(\bar{\bar{Y}}_{j}-1)\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}.$
$\displaystyle=\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y}_{j})-\frac{1}{T}\sum_{j=1}^{n}\log(\hat{\bar{Y}}_{j})(\bar{\bar{Y}}_{j}-1)\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}$
$\displaystyle\overset{(a)}{\leq}\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y}_{j})+\frac{\kappa}{T}\sum_{j=1}^{n}(\bar{\bar{Y}}_{j}-1)\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}$
$\displaystyle\leq\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y}_{j})+\frac{\kappa}{T}\sum_{j=1}^{n}\bar{\bar{Y}}_{j}\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\},$

where for (a), we have used the definition of $\kappa$ in (49).
Hence taking the expectation, we get

$\displaystyle\mathbb{E}\left[\frac{1}{T}d(\hat{Y}_{0}^{T};Y_{0}^{T})\right]$
$\displaystyle\leq\mathbb{E}\left[\frac{1}{n}\sum_{j=1}^{n}\bar{d}(\hat{\bar{Y}}_{j},\bar{Y}_{j})\right]+\kappa\mathbb{E}\left[\frac{1}{T}\sum_{j=1}^{n}\bar{\bar{Y}}_{j}\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}\right]$
$\displaystyle\overset{(a)}{\leq}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]+\kappa\mathbb{E}\left[\frac{1}{T}\sum_{j=1}^{n}\bar{\bar{Y}}_{j}\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}\right]+\bar{\epsilon}$
$\displaystyle\overset{(b)}{=}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]+\kappa(\lambda-\lambda\exp(-\lambda\Delta))+\bar{\epsilon}$
$\displaystyle\overset{(c)}{\leq}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]+\kappa\lambda^{2}\Delta+\bar{\epsilon},$
(52)

where, for (a), we have used (51),
for (b) we note that
$\mathbb{E}[\bar{\bar{Y}}_{j}\mathbf{1}\\{\bar{\bar{Y}}_{j}>1\\}]=\lambda\Delta-\lambda\Delta\exp(-\lambda\Delta)$,
and
for (c), we have used the inequality $1-u\leq\exp(-u)$.
Moreover using (50), for $i\in\\{1,2\\}$

$\displaystyle\frac{1}{T}\log(L^{(i)})=\frac{1}{n\Delta}\log(L^{(i)})\leq\frac{I(\bar{U}^{(i)};\bar{Y}^{(i)})}{\Delta}+\frac{\bar{\epsilon}}{\Delta}.$
(53)

The scaling of the mutual information $I(\bar{U}^{(i)};\bar{Y}^{(i)})$ and the
distortion function $\bar{d}(\hat{\bar{Y}},\bar{Y})$ with respect to $\Delta$
is given by the following lemma.

###### Lemma 14

For $i\in\\{1,2\\}$

$\displaystyle\lim_{\Delta\to 0}\frac{I(\bar{U}^{(i)};\bar{Y}^{(i)})}{\Delta}$
$\displaystyle={R}^{(i)},$ $\displaystyle\lim_{\Delta\to
0}\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]$ $\displaystyle={D}.$

###### Proof:

Please see the supplementary material. ∎

Now given the rate-distortion vector $({R}^{(1)},{R}^{(2)},{D})$ and
$\epsilon>0$, first choose $\Delta$ sufficiently small so that

$\displaystyle\frac{I(\bar{U}^{(i)};\bar{Y}^{(i)})}{\Delta}$
$\displaystyle\leq{R}^{(i)}+\frac{\epsilon}{4},$
$\displaystyle\mathbb{E}[\bar{d}(\hat{\bar{Y}},\bar{Y})]$
$\displaystyle\leq{D}+\frac{\epsilon}{4},$
$\displaystyle\kappa\lambda^{2}\Delta$ $\displaystyle\leq\epsilon/4.$

Then let $\bar{\epsilon}=\Delta\epsilon/4$, and choose a sufficiently large
$n$ so that (50) and (51) are satisfied. From (52) and (53) we conclude that a
sequence of $(T_{n},{R}^{(1)}+\epsilon,{R}^{(2)}+\epsilon,{D}+\epsilon)$ code
exists with $T_{n}=n\Delta$ when at least one of the conditions C.1 or C.2 is
satisfied.

Now consider the case when $p^{(i)}=0$ some $i\in\\{1,2\\}$, and for that $i$,
$\beta^{(i)}_{k}=0$ for some $k$’s. Say $p^{(1)}=0$ and $p^{(2)}>0$. This
gives us $\gamma^{(1)}_{k}=\beta^{(1)}_{k}$ for $k\in\\{1,2,3,4\\}$. Then we
need to show that the rate-distortion vector

$\displaystyle{R}^{(1)}=\left(\lambda+\mu^{(1)}\right)\sum_{k=1}^{4}\beta^{(1)}_{k}\log\left(\frac{\beta^{(1)}_{k}}{\alpha^{(1)}_{k}}\right)$
$\displaystyle{R}^{(2)}=\left((1-p^{(2)})\lambda+\mu^{(2)}\right)\sum_{k=1}^{4}\beta^{(2)}_{k}\log\left(\frac{\beta^{(2)}_{k}}{\alpha^{(2)}_{k}}\right)$
$\displaystyle{D}=\lambda-\phi(\lambda)-\lambda\left(\sum_{k=1}^{4}\beta^{(1)}_{k}\log\left(\frac{\beta^{(1)}_{k}}{\alpha^{(1)}_{k}}\right)+\sum_{k=1}^{4}\gamma^{(2)}_{k}\log\left(\frac{\gamma^{(2)}_{k}}{\alpha^{(2)}_{k}}\right)\right)$
(54)

is achievable. Let $[\hat{\beta}^{(1)}_{k}]_{k=1}^{4}=[1/4,1/4,1/4]$ and
$[\hat{\alpha}^{(1)}_{k}]_{k=1}^{4}=[1/4,1/4,1/4-\nu,1/4+\nu]$ for some
$\nu\in[0,1/3)$. Then the term

$\sum_{k=1}^{4}\hat{\beta}^{(1)}_{k}\log\left(\frac{\hat{\beta}^{(1)}_{k}}{\hat{\alpha}^{(1)}_{k}}\right)$

is continuous in $\nu$ and goes from zero to infinity as $\nu$ is increased
from zero to $1/4$, hence there exists some $\hat{\nu}\in[0,1/4)$ such that
with
$[\hat{\alpha}^{(1)}_{k}]_{k=1}^{4}=[1/4,1/4,1/4-\hat{\nu},1/4+\hat{\nu}]$,

$\displaystyle\sum_{k=1}^{4}\hat{\beta}^{(1)}_{k}\log\left(\frac{\hat{\beta}^{(1)}_{k}}{\hat{\alpha}^{(1)}_{k}}\right)=\sum_{k=1}^{4}\beta^{(1)}_{k}\log\left(\frac{\beta^{(1)}_{k}}{\alpha^{(1)}_{k}}\right).$
(55)

We note that this $[\hat{\beta}^{(1)}_{k}]_{k=1}^{4}$ satisfies condition C.1.
Hence the rate-distortion vector in (54) is achievable by using
$[\hat{\alpha}^{(1)}_{k}]_{k=1}^{4}$ that satisfies (55). The case when
$p^{(2)}=0$ or both $p^{(1)}=p^{(2)}=0$ can be handled similarly.

### Converse:

We will prove the converse when feedforward is present. For the given
$(T,R^{(1)}+\epsilon,R^{(2)}+\epsilon,D+\epsilon)$ code with feedforward, let
$M^{(1)}$ and $M^{(2)}$ denote the first and second encoder’s output
respectively. We essentially repeat the steps in the converse proof of Theorem
4 to show that

$\displaystyle\frac{1}{T}I(M^{(1)},M^{(2)};Y_{0}^{T})+D\geq\lambda-\phi(\lambda)-\epsilon.$

Since $I(M^{(1)},M^{(2)};Y_{0}^{T})<\infty$, we conclude from Theorem 1 that
there exists a process $\Gamma_{0}^{T}$ such that $\Gamma_{0}^{T}$ is the
$(\mathcal{F}_{t}=\sigma(M^{(1)},M^{(2)},Y_{0}^{t}):t\in[0,T])$ intensity of
$Y_{0}^{T}$ and

$\displaystyle
I(M^{(1)},M^{(2)};Y_{0}^{T})=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-T\phi(\lambda),$
(56)

Let $\hat{Y}_{0}^{T}$ denote the decoder’s output. The distortion constraint
$D$ satisfies

$\displaystyle
D\geq\frac{1}{T}\mathbb{E}\left[d(\hat{Y}_{0}^{T},Y_{0}^{T})\right]-\epsilon$
$\displaystyle=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt-\log(\hat{Y}_{t})\,dY_{t}\right]-\epsilon$
$\displaystyle=\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}-\log(\hat{Y}_{t})\Gamma_{t}\,dt\right]-\epsilon,$
(57)

where for the last equality we have used Lemma 12. Once again using the
inequality $u\log(v)\leq\phi(u)-u+v,\quad 0\leq u,v<\infty$, and noting that
the individual terms have finite expectations,

$\displaystyle\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}_{t})\Gamma_{t}\,dt\right]$
$\displaystyle\leq\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})-\Gamma_{t}+\hat{Y}_{t}\,dt\right]$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]+\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt\right].$
(58)

Combining these inequalities, we obtain

$\displaystyle\frac{1}{T}I(M^{(1)},M^{(2)};Y_{0}^{T})+D$
$\displaystyle\geq\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\Gamma_{t})\,dt\right]-\phi(\lambda)$
$\displaystyle\qquad+\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\hat{Y}_{t}\,dt\right]-\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\log(\hat{Y}_{t})\,dY_{t}\right]-\epsilon$
$\displaystyle\overset{(a)}{\geq}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]-\phi(\lambda)-\epsilon$
$\displaystyle\overset{(b)}{=}\lambda-\phi(\lambda)-\epsilon,$ (59)

where, for (a) we have used (57) and (58) and
for (b) we use the fact that
$\mathbb{E}\left[\int_{0}^{T}\Gamma_{t}\,dt\right]=\mathbb{E}\left[\int_{0}^{T}\,dY_{t}\right]=\lambda
T$.

We can upper bound the term $I(M^{(1)},M^{(2)};Y_{0}^{T})$ as

$\displaystyle I(M^{(1)},M^{(2)};Y_{0}^{T})$
$\displaystyle\overset{(a)}{=}H(M^{(1)},M^{(2)})-\mathbb{E}\left[H(M^{(1)},M^{(2)}|Y_{0}^{T})\right]$
$\displaystyle\overset{(b)}{=}H(M^{(1)},M^{(2)})-\mathbb{E}\left[H(M^{(1)}|Y_{0}^{T})\right]-\mathbb{E}\left[H(M^{(2)}|Y_{0}^{T})\right]$
$\displaystyle\overset{}{\leq}H(M^{(1)})+H(M^{(2)})-\mathbb{E}\left[H(M^{(1)}|Y_{0}^{T})\right]-\mathbb{E}\left[H(M^{(2)}|Y_{0}^{T})\right]$
$\displaystyle=I(M^{(1)};Y_{0}^{T})+I(M^{(2)};Y_{0}^{T}),$ (60)

where, for (a) we have used Lemma 2 and
for (b), we used the Markov chain
$M^{(1)}\leftrightarrows{Y}_{0}^{T}\leftrightarrows M^{(2)}$.
Combining (59) and (60) we get

$\displaystyle
D\geq\lambda-\phi(\lambda)-\frac{1}{T}I(M^{(1)};Y_{0}^{T})-\frac{1}{T}I(M^{(2)};Y_{0}^{T})-\epsilon.$
(61)

For $i\in\\{1,2\\}$, using Lemma 2

$\displaystyle\frac{1}{T}I(M^{(i)};{Y}^{(i),T}_{0})=\frac{1}{T}H(M^{(i)})\leq\frac{1}{T}\log\left(\lceil\exp((R^{(i)}+\epsilon)T)\rceil\right)\leq
R^{(i)}+\epsilon+\frac{1}{T}.$ (62)

Figure 2: Thinning and superposition operations defined in the proof for the
first encoder. Note that the joint distribution of
$(Y_{0}^{(1),T},\tilde{Y}_{0}^{(1),T},Y_{0}^{T})$ is same as that of
$(Y_{0}^{(1),T},\tilde{Z}_{0}^{(1),T},Z_{0}^{T})$.

We will first consider the case when $p^{(i)}<1$ for $i\in\\{1,2\\}$. We shall
proceed by defining certain auxiliary processes (see Figure 2). Let
$\tilde{Z}_{0}^{(i),T}$ be obtained from $\tilde{p}^{(i)}$-thinning of
$Y_{0}^{(i),T}$, where

$\tilde{p}^{(i)}=\frac{\mu^{(i)}}{((1-p^{(i)})\lambda+\mu^{(i)})}.$

Then using Lemma 11 we can write

$Y_{t}^{(i)}=\tilde{Z}_{t}^{(i)}+\hat{Z}_{t}^{(i)}\quad t\in[0,T],$

where $\tilde{Z}_{0}^{(i),T}$ and $\hat{Z}_{0}^{(i),T}$ are independent
Poisson processes with rates $(1-p^{(i)})\lambda$ and $\mu^{(i)}$
respectively. Whereas, by definition

$Y_{t}^{(i)}=\tilde{Y}_{t}^{(i)}+{N}_{t}^{(i)}\quad t\in[0,T],$

where $\tilde{Y}_{t}^{(i)}$ and $N_{t}^{(i)}$ are independent Poisson
processes with rates $(1-p^{(i)})\lambda$ and $\mu^{(i)}$ respectively. Hence
we conclude that the joint distribution of
$(Y_{0}^{(i),T},\tilde{Z}_{0}^{(i),T})$ is identical to the joint distribution
of $(Y_{0}^{(i),T},\tilde{Y}_{0}^{(i),T})$. Let $Z_{0}^{(i),T}$ be obtained by
adding an independent Poisson process $\hat{N}_{0}^{(i),T}$ with rate
$p^{(i)}\lambda$ to $\tilde{Z}^{(i),T}_{0}$,

$Z_{t}^{(i)}=\tilde{Z}_{t}^{(i)}+\hat{N}_{t}^{(i)}\quad t\in[0,T].$

Also using Lemma 11 we have

$Y_{t}=\tilde{Y}^{(i)}_{t}+\tilde{\tilde{Y}}^{(i)}_{t}\quad t\in[0,T],$

where $\tilde{Y}_{0}^{(i),T}$ and $\tilde{\tilde{Y}}^{(i),T}_{0}$ are
independent Poisson processes with rates $(1-p^{(i)})\lambda$ and
$p^{(i)}\lambda$. Hence the joint distribution of
$(Z_{0}^{(i),T},\tilde{Z}^{(i),T}_{0})$ and
$(Y_{0}^{T},\tilde{Y}^{(i),T}_{0})$ are identical. Moreover,
$M^{(i)}\rightleftarrows
Y_{0}^{(i),T}\rightleftarrows\tilde{Y}_{0}^{(i),T}\rightleftarrows Y_{0}^{T}$
forms a Markov chain and $M^{(i)}\rightleftarrows
Y_{0}^{(i),T}\rightleftarrows\tilde{Z}_{0}^{(i),T}\rightleftarrows
Z_{0}^{(i),T}$ forms a Markov chain. This allows us to write

$\displaystyle I\left(M^{(i)};\tilde{Z}_{0}^{(i),T}\right)$
$\displaystyle=I\left(M^{(i)};\tilde{Y}_{0}^{(i),T}\right),$ $\displaystyle
I\left(M^{(i)};{Z}_{0}^{(i),T}\right)$ $\displaystyle=I(M^{(i)};Y_{0}^{T}).$
(63)

Since $\tilde{Z}_{0}^{(i),T}$ is a
$\frac{\mu^{(i)}}{((1-p^{(i)})\lambda+\mu^{(i)})}$-thinning of
${Y}^{(i),T}_{0}$, Theorem 3 gives

$\displaystyle
I(M^{(i)};\tilde{Z}_{0}^{(i),T})\leq\left(1-\frac{\mu^{(i)}}{(1-p^{(i)})\lambda+\mu^{(i)}}\right)I(M^{(i)};{Y}^{(i),T}_{0}).$
(64)

Also $Z_{0}^{(i),T}$ is obtained by adding an independent Poisson process with
rate $p^{(i)}\lambda$ to $\tilde{Z}^{(i),T}_{0}$, Theorem 2 yields

$\displaystyle I(M^{(i)};\tilde{Z}_{0}^{(i),T})$
$\displaystyle=\mathbb{E}\left[\int_{0}^{T}\phi(\tilde{\Gamma}_{t}^{(i)})-\phi((1-p^{(i)})\lambda)\,dt\right],$
$\displaystyle I(M^{(i)};{Z}_{0}^{(i),T})$
$\displaystyle\leq\mathbb{E}\left[\int_{0}^{T}\phi(\tilde{\Gamma}_{t}^{(i)}+p^{(i)}\lambda)-\phi(\lambda)\,dt\right],$
(65)

where, $\tilde{\Gamma}_{0}^{(i),T}$ is the
$(\sigma(M^{(i)},\tilde{Z}_{0}^{(i),T}):t\in[0,T])$-intensity of
$\tilde{Z}_{0}^{(i),T}$. Then we can further lower bound $D$ in (61) as

$\displaystyle D$
$\displaystyle\geq\lambda-\phi(\lambda)-\frac{1}{T}I(M^{(1)};Y_{0}^{T})-\frac{1}{T}I(M^{(2)};Y_{0}^{T})-\epsilon$
$\displaystyle\overset{(a)}{=}\lambda-\phi(\lambda)-\frac{1}{T}I(M^{(1)};{Z}_{0}^{(1),T})-\frac{1}{T}I(M^{(2)};{Z}_{0}^{(2),T})-\epsilon$
$\displaystyle\overset{(b)}{\geq}\lambda-\phi(\lambda)-\frac{1}{T}\left(\mathbb{E}\left[\int_{0}^{T}\phi(\tilde{\Gamma}_{t}^{(1)}+p^{(1)}\lambda)-\phi(\lambda)\,dt\right]\right)$
$\displaystyle\phantom{===}-\frac{1}{T}\left(\mathbb{E}\left[\int_{0}^{T}\phi(\tilde{\Gamma}_{t}^{(2)}+p^{(2)}\lambda)-\phi(\lambda)\,dt\right]\right)-\epsilon$
$\displaystyle\overset{(c)}{=}\lambda+\phi(\lambda)-\mathbb{E}\left[\phi(\tilde{\Gamma}_{S_{1}}^{(1)}+p^{(1)}\lambda)\right]-\mathbb{E}\left[\phi(\tilde{\Gamma}^{(2)}_{S_{2}}+p^{(2)}\lambda)\right]-\epsilon,$

where for (a), we have used (63),
for (b), we have used (65), and
for (c), we define $S_{1}$ and $S_{2}$ to be uniformly distributed on $[0,T]$,
independent of all other random variables and independent of each other as
well.
For each $i\in\\{1,2\\}$, $R^{(i)}$ in (62) can be lower bounded as

$\displaystyle R^{(i)}$
$\displaystyle\geq\frac{1}{T}I(M^{(i)};{Y}^{(i),T}_{0})-\epsilon-\frac{1}{T}$
$\displaystyle\overset{(a)}{\geq}\frac{(1-p^{(i)})\lambda+\mu^{(i)}}{(1-p^{(i)})\lambda}\frac{1}{T}I(M^{(i)};\tilde{Z}_{0}^{(i),T})-\epsilon-\frac{1}{T}$
$\displaystyle\overset{(b)}{=}\frac{(1-p^{(i)})\lambda+\mu^{(i)}}{(1-p^{(i)})\lambda}\frac{1}{T}\mathbb{E}\left[\int_{0}^{T}\phi(\tilde{\Gamma}_{t}^{(i)})-\phi((1-p^{(i)})\lambda)\,dt\right]-\epsilon-\frac{1}{T}$
$\displaystyle\overset{(c)}{=}\frac{(1-p^{(i)})\lambda+\mu^{(i)}}{(1-p^{(i)})\lambda}\mathbb{E}\left[\phi(\tilde{\Gamma}_{S_{i}}^{(i)})-\phi((1-p^{(i)})\lambda)\right]-\epsilon-\frac{1}{T},$

where for (a), we have used (64),
for (b), we have used (65), and
for (c), recall that $S_{1}$ and $S_{2}$ are uniformly distributed on $[0,T]$,
independent of all other random variables and independent of each other.
Now we use Carathéodory’s theorem [54, Theorem 17.1]. For each
$i\in\\{1,2\\}$, there exist non-negative $[\eta^{(i)}_{k}]_{k=1}^{4}$ and
$[\alpha^{(i)}_{k}]_{k=1}^{4}$, such that $\sum_{k=1}^{4}\alpha^{(i)}_{k}=1$
and

$\displaystyle\mathbb{E}\left[\phi(\tilde{\Gamma}_{S_{i}}^{(i)})\right]$
$\displaystyle=\sum_{k=1}^{4}\alpha^{(i)}_{k}\phi(\eta^{(i)}_{k}),$
$\displaystyle\mathbb{E}\left[\phi(\tilde{\Gamma}_{S_{i}}^{(i)}+p^{(i)}\lambda)\right]$
$\displaystyle=\sum_{k=1}^{4}\alpha^{(i)}_{k}\phi(\eta^{(i)}_{k}+p^{(i)}\lambda),$
$\displaystyle\mathbb{E}\left[\tilde{\Gamma}_{S_{i}}^{(i)}\right]$
$\displaystyle=\sum_{k=1}^{4}\alpha^{(i)}_{k}\eta^{(i)}_{k}=(1-p^{(i)})\lambda,$

where in the last line we have used the fact that since
$\tilde{\Gamma}_{0}^{(i),T}$ is the
$(\sigma(M^{(i)},\tilde{Z}_{0}^{(i),T}):t\in[0,T])$-intensity of
$\tilde{Z}_{0}^{(i),T}$,
$\mathbb{E}\left[\int_{0}^{T}\tilde{\Gamma}_{t}^{(i)}\,dt\right]=\mathbb{E}[\tilde{Z}^{(i)}_{T}]=T(1-p^{(i)})\lambda$.
Hence we have

$\displaystyle R^{(i)}$
$\displaystyle\geq\frac{(1-p^{(i)})\lambda+\mu^{(i)}}{(1-p^{(i)})\lambda}\left(\sum_{k=1}^{4}\alpha^{(i)}_{k}\phi(\eta^{(i)}_{k})-\phi((1-p^{(i)})\lambda)\right)-\epsilon-\frac{1}{T},$
(66) $\displaystyle D$
$\displaystyle\geq\lambda+\phi(\lambda)-\sum_{k=1}^{4}\alpha^{(1)}_{k}\phi(\eta^{(1)}_{k}+p^{(1)}\lambda)-\sum_{k=1}^{4}\alpha^{(2)}_{k}\phi(\eta^{(2)}_{k}+p^{(2)}\lambda)-\epsilon.$
(67)

Now define

$\beta^{(i)}_{k}\triangleq\frac{\alpha^{(i)}_{k}\eta^{(i)}_{k}}{(1-p^{(i)})\lambda},\quad\gamma^{(i)}_{k}\triangleq
p^{(i)}\alpha^{(i)}_{k}+(1-p^{(i)})\beta^{(i)}_{k}.$

We note that $\beta^{(i)}_{k}=0$ if $\alpha^{(i)}_{k}=0$, and
$\sum_{k=1}^{4}\beta^{(i)}_{k}=1$. Substituting the above definitions in (66)

$\displaystyle R^{(i)}$
$\displaystyle\geq\frac{(1-p^{(i)})\lambda+\mu^{(i)}}{(1-p^{(i)})\lambda}\left(\sum_{k=1}^{4}\alpha^{(i)}_{k}\eta^{(i)}_{k}\log(\eta^{(i)}_{k})-\phi((1-p^{(i)})\lambda)\right)-\epsilon-\frac{1}{T}$
$\displaystyle=((1-p^{(i)})\lambda+\mu^{(i)})\left(\sum_{k=1}^{4}\beta^{(i)}_{k}\log\left(\frac{\beta^{(i)}_{k}(1-p^{(i)})\lambda}{\alpha^{(i)}_{k}}\right)\mathbf{1}\\{\alpha^{(i)}_{k}>0\\}-\log((1-p^{(i)})\lambda)\right)$
(68) $\displaystyle\phantom{=========}-\epsilon-\frac{1}{T}$
$\displaystyle=((1-p^{(i)})\lambda+\mu^{(i)})\sum_{k=1}^{4}\beta^{(i)}_{k}\log\left(\frac{\beta^{(i)}_{k}}{\alpha^{(i)}_{k}}\right)-\epsilon-\frac{1}{T}.$
(69)

Likewise,

$\displaystyle\sum_{k=1}^{4}\alpha^{(i)}_{k}\phi(\eta^{(i)}_{k}+p^{(i)}\lambda)$
$\displaystyle=\sum_{k=1}^{4}\alpha^{(i)}_{k}\phi\left(\frac{\beta^{(i)}_{k}(1-p^{(i)})\lambda}{\alpha^{(i)}_{k}}+p^{(i)}\lambda\right)\mathbf{1}\\{\alpha^{(i)}_{k}>0\\}$
$\displaystyle=\sum_{k=1}^{4}\alpha^{(i)}_{k}\phi\left(\frac{\gamma^{(i)}_{k}}{\alpha^{(i)}_{k}}\lambda\right)\mathbf{1}\\{\alpha^{(i)}_{k}>0\\}$
$\displaystyle=\lambda\sum_{k=1}^{4}\gamma^{(i)}_{k}\log\left(\frac{\gamma^{(i)}_{k}}{\alpha^{(i)}_{k}}\right)+\phi(\lambda).$

Substituting the above in (67), we get

$\displaystyle
D\geq\lambda-\phi(\lambda)-\lambda\sum_{k=1}^{4}\gamma^{(1)}_{k}\log\left(\frac{\gamma^{(1)}_{k}}{\alpha^{(1)}_{k}}\right)-\lambda\sum_{k=1}^{4}\gamma^{(2)}_{k}\log\left(\frac{\gamma^{(2)}_{k}}{\alpha^{(2)}_{k}}\right)-\epsilon.$
(70)

If either $p^{(i)}$, say $p^{(1)}$, equals 1, then $M^{(1)}$ and $Y_{0}^{T}$
are independent so that $I(M^{(1)};Y_{0}^{T})=0$, and we can repeat the above
steps to show that

$\displaystyle R^{(2)}$
$\displaystyle\geq((1-p^{(2)})\lambda+\mu^{(2)})\sum_{k=1}^{4}\beta^{(2)}_{k}\log\left(\frac{\beta^{(2)}}{\alpha^{(2)}_{k}}\right)-\epsilon-\frac{1}{T},$
$\displaystyle D$
$\displaystyle\geq\lambda-\phi(\lambda)-\lambda\sum_{k=1}^{4}\gamma^{(2)}_{k}\log\left(\frac{\gamma^{(2)}_{k}}{\alpha^{(2)}_{k}}\right)-\epsilon,$

which is the region in (69)-(70) with
$\alpha^{(1)}_{k}=\beta^{(1)}_{k}=\gamma^{(1)}_{k}$ for $k\in\\{1,2,3,4\\}$.

Since $\epsilon$ is arbitrary, taking $\epsilon\to 0$ and $T\to\infty$ gives
us the rate region in the statement of the theorem.

∎