# Information divergences and likelihood ratios of Poisson processes and point
patterns

Lasse Leskelä

###### Abstract

This article develops an analytical framework for studying information
divergences and likelihood ratios associated with Poisson processes and point
patterns on general measurable spaces. The main results include explicit
analytical formulas for Kullback–Leibler divergences, Rényi divergences,
Hellinger distances, and likelihood ratios of the laws of Poisson point
patterns in terms of their intensity measures. The general results yield
similar formulas for inhomogeneous Poisson processes, compound Poisson
processes, as well as spatial and marked Poisson point patterns. Additional
results include simple characterisations of absolute continuity, mutual
singularity, and the existence of common dominating measures. The analytical
toolbox is based on Tsallis divergences of sigma-finite measures on abstract
measurable spaces. The treatment is purely information-theoretic and free of
any topological assumptions.

Keywords: Poisson random measure, inhomogeneous Poisson process, point
process, spatial point pattern, Rényi divergence, Tsallis divergence,
Hellinger distance, mutual information, Chernoff information, Bhattacharyya
distance

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Measures
    2. 2.2 Point patterns
    3. 2.3 Poisson PPs
    4. 2.4 Rényi divergences
  3. 3 Tsallis divergences of sigma-finite measures
    1. 3.1 Definition
    2. 3.2 Properties
    3. 3.3 Absolute continuity and mutual singularity
    4. 3.4 Hellinger distances of sigma-finite measures
    5. 3.5 Disintegration of Tsallis divergences
  4. 4 Likelihood ratios of Poisson PPs
    1. 4.1 Finite Poisson PPs
    2. 4.2 Sigma-finite PPs
  5. 5 Information divergences of Poisson PPs
    1. 5.1 Divergences and distances
    2. 5.2 Absolute continuity
    3. 5.3 Existence of a dominating Poisson PP
  6. 6 Applications
    1. 6.1 Poisson processes
    2. 6.2 Chernoff information for Poisson vectors
    3. 6.3 Marked Poisson PPs
    4. 6.4 Compound Poisson processes
  7. 7 Proofs
    1. 7.1 Proof of Theorem 3.1
    2. 7.2 Proof of Theorem 3.11
    3. 7.3 Proof of Theorem 4.1
    4. 7.4 Proof of Theorem 4.4
    5. 7.5 Proof of Theorem 5.1
  8. 8 Conclusions
  9. A Measure theory
    1. A.1 Densities
    2. A.2 Basic measure theory
    3. A.3 Measurable bijections
  10. B Point patterns
    1. B.1 Measurability of sigma-finite decompositions
    2. B.2 Compensated Poisson integrals
    3. B.3 Truncated Poisson PPs
  11. C Elementary analysis

##  1 Introduction

A point pattern or a point process (PP) represents a countable set of points
in space or time. Poisson PPs are fundamental statistical models for
generating randomly scattered points in an abstract measurable space $S$. They
are encountered in a wide range of applications such as archeology [Bev20],
[PRBR18], astronomy [SMSS02], forestry statistics [SP00, KL13], machine
learning [BDK+21, AS15, ABS21, AKL24], neuroscience and genomics [RBS10,
PRBR18], and queueing systems [ABDW21, Les22]. The law of a Poisson PP is a
probability measure $P_{\lambda}$ characterised by an intensity measure
$\lambda$ so that $\lambda(B)$ indicates the expected number of points in
$B\subset S$. In statistical inference, it is important to understand how the
intensity measure can be identified from data. Statistical research devoted to
this question has a long history — well summarised in standard textbooks
[Kar91, DVJ03, IPSS08, Dig13]. The main analytical approaches include
computing and estimating likelihood ratios $\frac{dP_{\lambda}}{dP_{\mu}}$ and
information divergences of laws of Poisson PPs with intensity measures
$\lambda$ and $\mu$.

Likelihood ratios are easy to compute for standard families of probability
distributions on finite-dimensional spaces, but not so for probability
measures of infinite-dimensional objects such as paths of stochastic processes
or spaces of point patterns. In fact, even verifying the absolute continuity
of a pair of probability measures, a necessary condition for the existence of
a likelihood ratio, can be nontrivial. For Poisson PPs with finite intensity
measures, a classical result [Kar91, Rei93] tells that $P_{\lambda}\ll
P_{\mu}$ if and only if $\lambda\ll\mu$, in which case a likelihood ratio is
given by

$\frac{dP_{\lambda}}{dP_{\mu}}(\eta)\ =\
\exp\left(\int_{S}\log\phi\,d\eta+\int_{S}(1-\phi)\,d\mu\right)$

with $\phi=\frac{d\lambda}{d\mu}$ being a density of the intensity measures.
Using this formula, it is easy to compute various types of information
divergences and distances for $P_{\lambda}$ and $P_{\mu}$.

For Poisson PPs with general sigma-finite intensity measures, not so much is
known, and most of the known results [Sko57, Lie75, Kar83, Tak90] are
restricted to locally compact Polish spaces, thereby ruling out e.g. infinite-
dimensional Hilbert spaces. This article aims to fill this gap by developing
general formulas for likelihood ratios and information divergences in the most
general natural setting, for Poisson PPs sigma-finite intensity measures on a
general measurable space. The approach is purely information-theoretic, with
no topological assumptions, and follows the abstract measure-theoretic
framework in [LP18]. One of the main contributions is an explicit formula
(Theorem 4.4) for the likelihood ratio $\frac{dP_{\lambda}}{dP_{\mu}}$ that is
valid in general for all Poisson PP distributions such that $P_{\lambda}\ll
P_{\mu}$. This result helps to derive a characterisation for pairs of Poisson
PPs whose laws are dominated by a Poisson PP distribution (Theorem 5.7).

Most information divergences and distances of interest can in principle be
computed using the likelihood ratio $\frac{dP_{\lambda}}{dP_{\mu}}$.
Unfortunately, the general likelihood ratio formula involves a rather
complicated stochastic integral that renders it difficult to obtain explicit
analytical expressions. However, the fact that the laws of Poisson PPs are
infinitely divisible suggests that simple formulas should be available for
information divergences that are additive with respect to product measures. It
is well known that Rényi divergences, including the Kullback–Leibler
divergence, enjoy this tensorisation property [vH14, PW24]. Furthermore, it
was recently confirmed that linear combinations of Rényi divergences are the
only divergences satisfying the tensorisation and the data processing
inequality [MPST21, Theorem 2]. This article presents a general
characterisation of Rényi divergences of Poisson PPs (Theorem 5.1), and shows
that all such Rényi divergences can be written as generalised Tsallis
divergences of associated intensity measures. Tsallis divergences of
probability measures were introduced in [Tsa98]. Theorem 3.1 shows that
Tsallis divergences admit a natural extension from probability measures to
sigma-finite measures, and gives a representation of Tsallis divergences as
linear combinations of Rényi divergences of Poisson distributions. These
Poisson–Rényi–Tsallis relationships yield simple characterisation for the
absolute continuity and mutual singularity of general Poisson PP
distributions.

The rest of the article is organised as follows. Section 2 introduces
notations and definitions. Section 3 develops the basic theoretical
foundations for Tsallis divergences of sigma-finite measures. Section 4
presents the main results concerning likelihood ratios of Poisson PP
distributions. Section 5 presents the main results about information
divergences of Poisson PP distributions. Section 6 illustrates how the main
results can applied to analyse Poisson processes, compound Poisson processes,
marked Poisson point patterns, and Chernoff information of Poisson vectors.
Section 7 contains the technical proofs of the main results, and Section 8
concludes.

##  2 Preliminaries

###  2.1 Measures

The set of nonnegative integers is denoted by
$\mathbb{Z}_{+}=\\{0,1,2,\dots\\}$. Standard conventions of measure theory and
Lebesgue integration [Kal02] are used. For measures $\lambda,\mu$ on a
measurable space $(S,\mathcal{S})$, the notation $\lambda\ll\mu$ means that
$\lambda$ is absolutely continuous with respect to $\mu$, that is, $\mu(A)=0$
for all $A\in\mathcal{S}$ such that $\lambda(A)=0$. We denote
$\lambda\perp\mu$ and say that $\lambda,\mu$ are mutually singular if there
exist a measurable set $B$ such that $\lambda(B^{c})=0$ and $\mu(B)=0$. A
measurable function $f\colon S\to\mathbb{R}_{+}$ is called a density (or
Radon–Nikodym derivative) of $\lambda$ with respect to $\mu$ when
$\mu(A)=\int_{A}f\,d\mu$ for all $A\in\mathcal{S}$; in this case we denote
$f=\frac{d\lambda}{d\mu}$. A density of probability measure is called a
likelihood ratio.

The symbol $\delta_{x}$ refers to the Dirac measure at $x$. For a number
$c\geq 0$, the symbol $\operatorname{Poi}(c)$ denotes the Poisson probability
distribution with mean $c$ and density $k\mapsto e^{-c}\frac{c^{k}}{k!}$ with
respect to the counting measure on $\mathbb{Z}_{+}$, with the standard
conventions that $\operatorname{Poi}(0)=\delta_{0}$ and
$\operatorname{Poi}(\infty)=\delta_{\infty}$.

###  2.2 Point patterns

A point pattern is a countable collection of points, possibly with
multiplicities, in a measurable space $(S,\mathcal{S})$. Such a collection is
naturally represented as a measure $\eta$ on $(S,\mathcal{S})$, so that
$\eta(B)$ equals the number of points in $B\in\mathcal{S}$. The requirement on
countability is guaranteed when $\eta=\sum_{n=1}^{\infty}\eta_{n}$ for some
finite measures $\eta_{n}$. Following [LP18], we define $N(S)$ as the set of
all measures that can be written as a countable sum of integer-valued finite
measures on $(S,\mathcal{S})$, and equip it with the sigma-algebra
$\mathcal{N}(S)$ generated by the evaluation maps
$\operatorname{ev}_{B}\colon\eta\mapsto\eta(B)$, $B\in\mathcal{S}$. An element
of $N(S)$ is called a point pattern or a point process (PP).

###  2.3 Poisson PPs

Poisson PPs are defined in the standard manners, see [Kin67, Kal02, LP18] for
general background. A PP distribution is a probability measure on
$(N(S),\mathcal{N}(S))$. The Laplace functional of a PP distribution $P$ is
the map that assigns to every measurable function $u\colon S\to[0,\infty]$ a
number $L_{P}(u)=\int_{N(S)}\exp(-\int_{X}u\,d\eta)\,P(d\eta)\in[0,1]$. Given
a sigma-finite measure $\lambda$ on $(S,\mathcal{S})$, the Poisson PP
distribution with intensity measure $\lambda$ is the unique probability
measure $P_{\lambda}$ on $(N(S),\mathcal{N}(S))$ such that

  1. (i)

$P_{\lambda}\circ(\operatorname{ev}_{B_{1}},\dots,\operatorname{ev}_{B_{n}})^{-1}=\bigotimes_{i=1}^{n}(P_{\lambda}\circ\operatorname{ev}_{B_{i}}^{-1})$
for all integers $n\geq 1$ and mutually disjoint
$B_{1},\dots,B_{n}\in\mathcal{S}$.

  2. (ii)

$P_{\lambda}\circ\operatorname{ev}_{B}^{-1}=\operatorname{Poi}(\lambda(B))$
for all $B\in\mathcal{S}$

For the existence and uniqueness, see e.g. [LP18, Proposition 2.10, Theorem
3.6] or [Kal02, Lemma 12.1–12.2, Theorem 12.7]. Samples $\eta$ from
$P_{\lambda}$ are called Poisson PPs.

###  2.4 Rényi divergences

Rényi divergences were introduced in [Rén61]. The Rényi divergence of order
$\alpha\in[0,\infty)$ for probability measures $P,Q$ on a measurable space
$(S,\mathcal{S})$ is defined [vH14, PW24] by

$R_{\alpha}(P\|Q)\ =\ \begin{cases}-\log Q(p>0),&\quad\alpha=0,\\\
\frac{1}{\alpha-1}\log\int_{S}p^{\alpha}q^{1-\alpha}\,d\nu,&\quad\alpha\notin\\{0,1\\},\\\
\int_{S}p\log\frac{p}{q}\,d\nu,&\quad\alpha=1,\end{cases}$ (2.1)

where $p=\frac{dP}{d\nu}$ and $q=\frac{dQ}{d\nu}$ are densities of $P,Q$ with
respect to a sigma-finite measure111The definition does not depend on the
choice of reference measure or densities, we may take $\nu=\frac{1}{2}(P+Q)$
for example [vH14]. $\nu$ on $S$; and for $\alpha>1$ we read
$p^{\alpha}q^{1-\alpha}=\frac{p^{\alpha}}{q^{\alpha-1}}$ and adopt the
conventions [LV06, vH14] that $\frac{0}{0}=0$ and $\frac{t}{0}=\infty$ for
$t>0$, together with $0\log\frac{0}{t}=0$ for $t\geq 0$ and
$t\log\frac{t}{0}=\infty$ for $t>0$. For $\alpha\in(1,\infty)$, note that
$\int_{S}p^{\alpha}q^{1-\alpha}\,d\nu=\int_{p>0,\,q>0}p^{\alpha}q^{1-\alpha}\,d\nu+\infty
P\\{q=0\\}$, so that

$R_{\alpha}(P\|Q)\ =\
\begin{cases}\frac{1}{\alpha-1}\log\int_{p>0,\,q>0}p^{\alpha}q^{1-\alpha}\,dm,&\qquad
P\ll Q,\\\ \infty,&\qquad P\not\ll Q.\end{cases}$

Rényi divergences also admit several variational characterisations, see for
example [Ana18, BDK+21, vH14].

Rényi divergences of orders $\frac{1}{2},1,2$ are connected to other important
information quantities as follows. $R_{1}(P,Q)$ equals the Kullback–Leibler
divergence or relative entropy.
$R_{1/2}(P\|Q)=-2\log(1-\frac{\operatorname{Hel}^{2}(P,Q)}{2})$ where
$\operatorname{Hel}(P,Q)$ is the Hellinger distance, and
$R_{2}(P\|Q)=\log(1+\chi^{2}(P,Q))$ where $\chi^{2}(P,Q)$ refers to the
$\chi^{2}$-divergence [GS02].

##  3 Tsallis divergences of sigma-finite measures

This section introduces a theoretical framework of Tsallis divergences of
sigma-finite measures. Section 3.1 provides a definition and a representation
formula as a Poisson–Rényi integral, and Section 3.2 summarises some basic
properties. Section 3.3 shows how Tsallis divergences characterise absolute
continuity and mutual singularity. Section 3.4 establishes a connection with
Hellinger distances, and Section 3.5 presents a disintegration formula.

###  3.1 Definition

The following definition generalises the notion of Tsallis divergence from
probability measures [Tsa98, NN11] to arbitrary sigma-finite measures. Let
$\lambda,\mu$ be sigma-finite measures on a measurable space $S$ admitting
densities $f=\frac{d\lambda}{d\nu}$ and $g=\frac{d\mu}{d\nu}$ with respect to
a sigma-finite measure $\nu$. We define the Tsallis divergence of order
$\alpha\in\mathbb{R}_{+}$ by

$T_{\alpha}(\lambda\|\mu)=\begin{cases}\mu\\{f=0\\},&\quad\alpha=0,\\\
\int_{S}\big{(}\frac{\alpha
f+(1-\alpha)g-f^{\alpha}g^{1-\alpha}}{1-\alpha}\big{)}\,d\nu,&\quad\alpha\notin\\{0,1\\},\\\
\int_{S}\big{(}f\log\frac{f}{g}+g-f\big{)}\,d\nu,&\quad\alpha=1,\end{cases}$
(3.1)

where for $\alpha>1$ we read $f^{\alpha}g^{1-\alpha}$ as
$\frac{f^{\alpha}}{g^{\alpha-1}}$ and adopt the conventions that
$\frac{0}{0}=0$ and $\frac{t}{0}=\infty$ for $t>0$, as well as
$t\log\frac{t}{0}=\infty$ for $t>0$ and $0\log\frac{0}{t}=0$ for $t\geq 0$.

######  Theorem 3.1.

$\alpha\mapsto T_{\alpha}(\lambda\|\mu)$ is a well-defined nondecreasing
function from $\mathbb{R}_{+}$ into $[0,\infty]$ that is continuous on the
interval $\\{\alpha\colon T_{\alpha}(\lambda\|\mu)<\infty\\}$, and admits a
representation

$T_{\alpha}(\lambda\|\mu)\ =\
\int_{S}R_{\alpha}(p_{f(x)}\|p_{g(x)})\,\nu(dx),$ (3.2)

where $p_{s}$ refers to the Poisson distribution $k\mapsto
e^{-sk}\frac{s^{k}}{k!}$ on the nonnegative integers with mean $s$.
Furthermore, the value of $T_{\alpha}(\lambda\|\mu)$ does not depend on the
choice of the densities $f,g$ nor the measure $\nu$.

###### Proof.

Section 7.1. ∎

######  Remark 3.2 (Probability measures).

In the special case with $\lambda(S)=\mu(S)=1$, we find that for
$\alpha\in(0,1)$,
$T_{\alpha}(\lambda\|\mu)=\frac{1-\int_{S}f^{\alpha}g^{1-\alpha}\,d\nu}{1-\alpha}$
agrees with the classical definition of the Tsallis divergence for probability
measures [Tsa98]. In this case
$T_{1}(\lambda\|\mu)=\operatorname{KL}(\lambda\|\mu)$ equals the
Kullback–Leibler divergence, and the Tsallis divergence of order
$\alpha\notin\\{0,1\\}$ is related to the Rényi divergence by
$T_{\alpha}(\lambda\|\mu)=\frac{1-e^{-(1-\alpha)R_{\alpha}(\lambda\|\mu)}}{1-\alpha}$.

######  Remark 3.3.

$T_{1/2}(\lambda\|\mu)=2H^{2}(\lambda,\mu)$ where $H(\lambda,\mu)$ refers to
the Hellinger distance (see Section 3.4), and $T_{1}(\lambda\|\mu)$
corresponds to the generalized KL divergence discussed in [MFWF23].

###  3.2 Properties

Tsallis divergences share several properties in common with Rényi divergences.
This section summarises some of the most important.

######  Proposition 3.4.

The Tsallis divergence for sigma-finite measures $\lambda\ll\mu$ with density
$\phi=\frac{d\lambda}{d\mu}$ can be written as

$T_{\alpha}(\lambda\|\mu)=\begin{cases}\mu\\{\phi=0\\},&\qquad\alpha=0,\\\
\int_{S}\big{(}\frac{\alpha\phi+1-\alpha-\phi^{\alpha}}{1-\alpha}\big{)}d\mu,&\qquad\alpha\notin\\{0,1\\},\\\
\int_{S}\left(\phi\log\phi+1-\phi\right)d\mu,&\qquad\alpha=1.\end{cases}$
(3.3)

###### Proof.

Theorem 3.1 tells that we are free to choose densities and reference measures
in (3.1). The claim follows by choosing densities $f=\phi$, $g=1$ of
$\lambda,\mu$ with respect to reference measure $\nu=\mu$. ∎

######  Proposition 3.5.

$(1-\alpha)T_{\alpha}(\mu\|\lambda)=\alpha T_{1-\alpha}(\lambda\|\mu)$ for all
$\alpha\in(0,1)$.

###### Proof.

Immediate from formula (3.1). ∎

######  Proposition 3.6.

$\frac{\alpha}{\beta}\frac{1-\beta}{1-\alpha}T_{\beta}(\lambda\|\mu)\leq
T_{\alpha}(\lambda\|\mu)\leq T_{\beta}(\lambda\|\mu)$ for all
$\alpha\leq\beta$ in $(0,1)$.

###### Proof.

The second inequality follows by the monotonicity of Tsallis divergences
(Theorem 3.1). The first inequality follows by applying Proposition 3.5 and
monotonicity to conclude that

$\displaystyle\frac{\alpha}{\beta}\frac{1-\beta}{1-\alpha}T_{\beta}(\lambda\|\mu)$
$\displaystyle\ =\ \frac{\alpha}{1-\alpha}T_{1-\beta}(\mu\|\lambda)$
$\displaystyle\ \leq\ \frac{\alpha}{1-\alpha}T_{1-\alpha}(\mu\|\lambda)\ \leq\
T_{\alpha}(\mu\|\lambda).$

∎

###  3.3 Absolute continuity and mutual singularity

The following is a characterisation of absolute continuity for sigma-finite
measures in terms of Tsallis divergences. It is similar in spirit for an
analogous characterisation of probability measures using Rényi divergences:
$P\ll Q$ iff $R_{0}(Q\|P)=0$ (see [Shi96, Theorem III.9.2], [vH14, Theorem
23]).

######  Proposition 3.7.

The following are equivalent for all sigma-finite measures:

  1. (i)

$\lambda\ll\mu$.

  2. (ii)

$T_{0}(\mu\|\lambda)=0$.

###### Proof.

Let $\lambda,\mu$ be measures on a measurable space $(S,\mathcal{S})$
admitting densities $f=\frac{d\lambda}{d\nu}$ and $g=\frac{d\mu}{d\nu}$ with
respect to a measure $\nu$.

(i)$\iff$(ii). This follows by applying Lemma A.1 and noting that
$\lambda\\{g=0\\}=T_{0}(\mu\|\lambda)$ by definition (3.1).

∎

The following is a characterisation of mutual singularity for finite measures
in terms of Tsallis divergences. It is similar in spirit for an analogous
characterisation of probability measures using Rényi divergences: $P\perp Q$
iff $R_{0}(P\|Q)=\infty$ iff $R_{0}(Q\|P)=\infty$ (see [Shi96, Theorem
III.9.3], [vH14, Theorem 24]).

######  Proposition 3.8.

The following are equivalent for all finite measures:

  1. (i)

$\lambda\perp\mu$.

  2. (ii)

$T_{0}(\lambda\|\mu)=\mu(S)$.

  3. (iii)

$T_{0}(\mu\|\lambda)=\lambda(S)$.

###### Proof.

Let $\lambda,\mu$ be measures on a measurable space $(S,\mathcal{S})$
admitting densities $f=\frac{d\lambda}{d\nu}$ and $g=\frac{d\mu}{d\nu}$ with
respect to a measure $\nu$.

(i)$\iff$(ii). By (3.1), $T_{0}(\lambda\|\mu)=\mu\\{f=0\\}$. Hence
$\mu(S)-T_{0}(\lambda\|\mu)=\mu\\{f>0\\}$. The claim follows because
$\lambda\perp\mu$ is equivalent to $\mu\\{f>0\\}$ (Lemma A.1).

(i)$\iff$(iii). Analogously,
$\lambda(S)-T_{0}(\mu\|\lambda)=\lambda(S)-\lambda\\{g=0\\}=\lambda\\{g>0\\}$.
The claim follows because $\lambda\perp\mu$ is equivalent to
$\lambda\\{g>0\\}$ (Lemma A.1).

∎

###  3.4 Hellinger distances of sigma-finite measures

The Hellinger distance between sigma-finite measures $\lambda$ and $\mu$ is
defined by

$H(\lambda,\mu)\ =\
\left(\frac{1}{2}\int_{S}\left(\sqrt{f}-\sqrt{g}\right)^{2}d\nu\right)^{1/2},$
(3.4)

where $f=\frac{d\lambda}{d\nu}$ and $g=\frac{d\mu}{d\nu}$ are densities with
respect to a sigma-finite measure $\nu$. Hellinger distances take values in
$[0,1]$ for probability measures, and in $[0,\infty]$ for general sigma-finite
measures. By writing $(\sqrt{f}-\sqrt{g})^{2}=f+g-2f^{1/2}g^{1/2}$, we see by
comparing (3.1) and (3.4) that

$T_{1/2}(\lambda\|\mu)\ =\ 2H^{2}(\lambda,\mu).$ (3.5)

In particular, we see see that Tsallis divergences of order $\frac{1}{2}$ are
symmetric according to $T_{1/2}(\lambda\|\mu)=T_{1/2}(\mu\|\lambda)$. When
$\lambda,\mu$ are probability measures, we note that
$T_{1/2}(\lambda\|\mu)=2(1-\int_{S}f^{1/2}g^{1/2}\,d\nu)=2(1-\exp(-\frac{1}{2}R_{1/2}(\lambda\|\mu))$.
Hence for probability measures $\lambda,\mu$,

$H^{2}(\lambda,\mu)\ =\ 1-\exp\left(-\frac{1}{2}R_{1/2}(\lambda\|\mu)\right).$
(3.6)

######  Proposition 3.9.

The right side of (3.4) does not depend on the choice of the densities $f,g$
nor the reference measure $\nu$. Furthermore, if $\lambda\ll\mu$, then the
Hellinger distance can also be written as

$H(\lambda,\mu)\ =\
\left(\frac{1}{2}\int_{S}\left(\sqrt{\phi}-1\right)^{2}\,d\mu\right)^{1/2},$

where $\phi\colon S\to\mathbb{R}_{+}$ is a density of $\lambda$ with respect
to $\mu$.

###### Proof.

The first claim follows by applying Theorem 3.1 with (3.5). The second claim
follows by applying (3.4) with $f=\phi$, $g=1$, and $\nu=\mu$. ∎

######  Proposition 3.10.

$H(\lambda,\xi)\leq H(\lambda,\mu)+H(\mu,\xi)$ for all sigma-finite measures.

###### Proof.

Let $f,g,h$ be densities of $\lambda,\mu,\xi$ with respect to the sigma-finite
measure $\nu=\lambda+\mu+\xi$. We note that

$\displaystyle H(\lambda,\xi)$ $\displaystyle\ =\
\frac{1}{\sqrt{2}}{\lVert\sqrt{f}-\sqrt{h}\rVert}_{L^{2}(\nu)}$
$\displaystyle\ =\
\frac{1}{\sqrt{2}}{\lVert(\sqrt{f}-\sqrt{g})+(\sqrt{g}-\sqrt{h})\rVert}_{L^{2}(\nu)}.$

Therefore, the claim follows by applying Minkowski’s inequality ${\lVert
u+v\rVert}_{L^{2}(\nu)}\leq{\lVert u\rVert}_{L^{2}(\nu)}+{\lVert
v\rVert}_{L^{2}(\nu)}$ which is true for all measurable functions $u,v$,
regardless of whether ${\lVert u\rVert}_{L^{2}(\nu)},{\lVert
v\rVert}_{L^{2}(\nu)}$ are finite or not. ∎

###  3.5 Disintegration of Tsallis divergences

The disintegration of a measure $\Lambda$ on a product space $(S_{1}\times
S_{2},\mathcal{S}_{1}\otimes\mathcal{S}_{2})$ refers to representing as
$\Lambda=\lambda\otimes K$ where $\lambda$ is a measure on
$(S_{1},\mathcal{S}_{1})$ corresponding to the first marginal of $\Lambda$ and
$K$ is a kernel from $(S_{1},\mathcal{S}_{1})$ into $(S_{2},\mathcal{S}_{2})$.
Equivalently,

$\Lambda(C)\ =\ \int_{S_{1}}\int_{S_{2}}1_{C}(x,y)\,K(x,dy)\,\lambda(dx),\quad
C\subset\mathcal{S}_{1}\otimes\mathcal{S}_{2}.$

Informally, we write

$\Lambda(dx,dy)\ =\ \lambda(dx)K(x,dy).$

If $\Lambda$ disintegrates according to $\Lambda=\lambda\otimes K$, then it
also disintegrates according to $\Lambda=\tilde{\lambda}\otimes\tilde{K}$
where $\tilde{\lambda}=2\lambda$ and $\tilde{K}=\frac{1}{2}K$. To rule out
such unidentifiability issues, in applications it is natural to require $K$ to
be probability kernel, that is, a kernel such that $K(x,S_{2})=1$ for all
$x\in S_{1}$. The following result characterises Tsallis divergences of
disintegrated measures that helps to compute information divergences of
compound Poisson processes and marked Poisson PPs (see Sections 6.3–6.4). See
[PW24, Theorem 2.13, Equation 7.71] for similar results concerning Rényi
divergences.

######  Theorem 3.11.

Let $\lambda,\mu$ be sigma-finite measures on a measurable space
$(S_{1},\mathcal{S}_{1})$, and let $K,L$ be probability kernels from
$(S_{1},\mathcal{S}_{1})$ into measurable space $(S_{2},\mathcal{S}_{2})$.
Assume that there exists measurable functions $k,\ell\colon S_{1}\times
S_{2}\to\mathbb{R}_{+}$ and a kernel $M$ from $(S_{1},\mathcal{S}_{1})$ into
$(S_{2},\mathcal{S}_{2})$ such that

$\displaystyle K_{t}(dx)$ $\displaystyle=k_{t}(x)M_{t}(dx),$ (3.7)
$\displaystyle L_{t}(dx)$ $\displaystyle=\ell_{t}(x)M_{t}(dx),$

for all $t\in S_{1}$. Then the Tsallis divergence of order
$\alpha\in\mathbb{R}_{+}$ equals

$\displaystyle T_{\alpha}(\lambda\otimes K\|\mu\otimes L)$ (3.8)
$\displaystyle\ =\ \begin{cases}T_{0}(\lambda\|\mu)+\int_{f\neq
0}T_{0}(K_{t}\|L_{t})\,\mu(dt),&\quad\alpha=0,\\\
T_{\alpha}(\lambda\|\mu)+\int_{S_{1}}T_{\alpha}(K_{t}\|L_{t})\,f_{t}^{\alpha}g_{t}^{1-\alpha}\,\nu(dt),&\quad\alpha\notin\\{0,1\\},\\\
T_{1}(\lambda\|\mu)+\int_{S_{1}}T_{1}(K_{t}\|L_{t})\,\lambda(dt),&\quad\alpha=1,\end{cases}$

where $f_{t}=\frac{d\lambda}{d\nu}(t)$ and $g_{t}=\frac{d\mu}{d\nu}(t)$ are
densities of $\lambda$ and $\mu$ with respect to a sigma-finite measure $\nu$.

A sufficient condition for (3.7) is to assume that $(S_{2},\mathcal{S}_{2})$
is separable in the sense that $\mathcal{S}_{2}$ is generated by a countable
set family. In this case there exists [DM82, Theorem 58] measurable functions
$k,\ell\colon S_{1}\times S_{2}\to\mathbb{R}_{+}$ such that (3.7) holds for
the probability kernel $M=\frac{1}{2}(K+L)$. It might be that (3.7) is not
needed for Proposition 3.11. This might require a measurable selection theorem
that is different from the usual ones for which it is assumed that at least
one of the sigma-algebras is generated by a regular topology [Les10, LV17].

##  4 Likelihood ratios of Poisson PPs

This section presents a general likelihood ratio formula for Poisson PPs.
Section 4.1 first focuses on the case with finite intensity measures, and
Section 4.2 then provides a general formula.

###  4.1 Finite Poisson PPs

Poisson PPs with finite intensity measures generated finite PPs. Likelihood
ratio formulas for Poisson PPs with finite intensity measures are classical
(e.g. [LS78, Kar91, Rei93, Bir07]), although they are usually restricted to
Polish spaces. The proof of the following general is included in Section 7.3
for completeness.

######  Theorem 4.1.

Any Poisson PPs with finite intensity measures $\lambda\ll\mu$ satisfy
$P_{\lambda}\ll P_{\mu}$, and a likelihood ratio is given by

$\frac{dP_{\lambda}}{dP_{\mu}}(\eta)\ =\
1_{M_{\lambda,\mu}}(\eta)\,\exp\bigg{(}\int_{S}(1-\phi)\,d\mu+\int_{S}\log\phi\,d\eta\bigg{)},$
(4.1)

where $\phi=\frac{d\lambda}{d\mu}$ is a density of $\lambda$ with respect to
$\mu$, and

$M_{\lambda,\mu}\ =\ \Big{\\{}\eta\in N(S)\colon\eta\\{\phi=0\\}=0\Big{\\}}.$
(4.2)

######  Remark 4.2.

For every finite point pattern $\eta\in M_{\lambda,\mu}$, the integral
$\int_{S}\log\phi\,d\eta$ in (4.1) is a well-defined real number because
$\log\phi\in\mathbb{R}$ outside the set $\\{\phi=0\\}$ of $\eta$-measure zero.
Also recall that every point pattern generated by a Poisson PP with a finite
intensity measure is finite almost surely. Therefore, the right side in (4.1)
is well defined for $P_{\mu}$-almost every $\nu$.

######  Remark 4.3.

The set $M_{\lambda,\mu}$ in (4.2) indicates the set of point patterns that
contain no points in the region where $\phi=\frac{d\lambda}{d\mu}$ vanishes.
If we assume that $\lambda$ and $\mu$ are mutually absolutely continuous, then
we may omit $1_{M_{\lambda,\mu}}(\eta)$ from (4.2). To see why, observe that
$\lambda\\{\phi=0\\}=\int_{\\{\phi=0\\}}\phi\,d\mu=0$ together with
$\mu\ll\lambda$ implies that $\mu\\{\phi=0\\}=0$. Therefore, by Markov’s
inequality,

$\displaystyle P_{\mu}(M^{c})$ $\displaystyle\ =\
P_{\mu}\\{\eta\\{\phi=0\\}\geq 1\\}$ $\displaystyle\ \leq\
E_{\mu}\eta\\{\phi=0\\}\ =\ \mu\\{\phi=0\\}\ =\ 0.$

###  4.2 Sigma-finite PPs

The simple density formula of Theorem 4.1 is not in general valid for Poisson
PPs with infinite intensity measures, because the integral
$\int_{S}\log\phi\,d\eta$ might not converge for infinite point patterns
$\eta$. In the general setting, we need to work with carefully compensated
integrals. A key observation is that a compensated Poisson integral
$\int_{\\{{\lvert\log\phi\rvert}\leq 1\\}}\log\phi\,d(\eta-\mu)$ and the
ordinary Poisson integral
$\int_{\\{{\lvert\log\phi\rvert}>1\\}}\log\phi\,d\eta$ of the logarithm of
$\phi=\frac{d\lambda}{d\mu}$ converge for $P_{\mu}$-almost every $\eta$
whenever $P_{\lambda}\ll P_{\mu}$. See Appendix B.2 for the definition of the
compensated integral and details. The following theorem confirms that a
density of $P_{\lambda}$ with respect to $P_{\mu}$ can be written using these
integrals.

######  Theorem 4.4.

Any Poisson PPs with sigma-finite intensity measures such that $\lambda\ll\mu$
and $H(\lambda,\mu)<\infty$ satisfy $P_{\lambda}\ll P_{\mu}$, and a likelihood
ratio is given by

$\frac{dP_{\lambda}}{dP_{\mu}}(\eta)\ =\
1_{M_{\lambda,\mu}}(\eta)\exp(\ell_{\lambda,\mu}(\eta)),$ (4.3)

where

$\displaystyle\ell_{\lambda,\mu}(\eta)$ $\displaystyle\
=\int\displaylimits_{{\lvert\log\phi\rvert}\leq
1}\log\phi\,d(\eta-\mu)\,+\int\displaylimits_{{\lvert\log\phi\rvert}>1}\log\phi\,d\eta$
(4.4) $\displaystyle\quad+\int\displaylimits_{{\lvert\log\phi\rvert}\leq
1}(\log\phi+1-\phi)\,d\mu+\int\displaylimits_{{\lvert\log\phi\rvert}>1}(1-\phi)\,d\mu$

and

$M_{\lambda,\mu}\ =\ \\{\eta\in N(S)\colon\eta\\{\phi=0\\}=0\\}$ (4.5)

are defined in terms of a density $\phi=\frac{d\lambda}{d\mu}$.

###### Proof.

Section 7.4. ∎

######  Remark 4.5.

When $\lambda$ and $\mu$ are mutually absolutely continuous, the factor
$1_{M_{\lambda,\mu}}(\eta)$ may be omitted from (4.3), as explained in Remark
4.3.

##  5 Information divergences of Poisson PPs

This section provides formulas for Rényi divergences and Hellinger distances
of Poisson PPs. Section 5.1 summaries formulas Rényi divergences,
Kullback–Leibler divergences, and Hellinger distances. Section 5.2
characterises the absolute continuity of Poisson PPs using Tsallis divergences
of their intensity measures. Section 5.3 characterises pairs of Poisson PPs
whose laws admit a common dominating measure corresponding to a Poisson PP.

###  5.1 Divergences and distances

In what follows, $P_{\lambda}$ and $P_{\mu}$ are Poisson PP distributions with
sigma-finite intensity measures $\lambda$ and $\mu$ on a measurable space $S$.

######  Theorem 5.1.

The Rényi divergence of order $\alpha\in(0,\infty)$ for Poisson PPs
$P_{\lambda}$ and $P_{\mu}$ is given by the Tsallis divergence of their
intensity measures according to

$R_{\alpha}(P_{\lambda}\|P_{\mu})\ =\ T_{\alpha}(\lambda\|\mu).$ (5.1)

If $T_{\alpha}(\lambda\|\mu)<\infty$ for some $\alpha>0$, then (5.1) also
holds for $\alpha=0$.

###### Proof.

Section 7.5. ∎

######  Corollary 5.2.

The Kullback–Leibler divergence for Poisson PPs with sigma-finite intensity
measures $\lambda$ and $\mu$ is given by

$\operatorname{KL}(P_{\lambda}\|P_{\mu})\ =\
\int_{S}\left(f\log\frac{f}{g}+g-f\right)d\nu,$ (5.2)

where $f=\frac{d\lambda}{d\nu}$ and $g=\frac{d\mu}{d\nu}$ are densities with
respect to a measure $\nu$.

###### Proof.

Immediate corollary of Theorem 5.1. ∎

As another corollary of Theorem 5.1, we obtain a simple formula for the
Hellinger distance between Poisson PP distributions. This result was proved in
[Tak90] for locally finite intensity measures on locally compact Polish
spaces.

######  Corollary 5.3.

The Hellinger distance between Poisson PPs $P_{\lambda}$ and $P_{\mu}$ is
given by $H^{2}(P_{\lambda},P_{\mu})=1-e^{-H^{2}(\lambda,\mu)}$.

###### Proof.

Theorem 5.1 combined with formula (3.5) implies that
$R_{1/2}(P_{\lambda}\|P_{\mu})=T_{1/2}(\lambda\|\mu)=2H^{2}(\lambda,\mu)$. For
probability measures $P_{\lambda}$ and $P_{\mu}$, we find that by (2.1) that
$H^{2}(P_{\lambda},P_{\mu})=1-e^{-\frac{1}{2}R_{1/2}(P_{\lambda},P_{\mu})}$.
By combining these findings, we conclude that
$H^{2}(P_{\lambda},P_{\mu})=1-e^{-H^{2}(\lambda,\mu)}$. ∎

###  5.2 Absolute continuity

######  Theorem 5.4.

The following are equivalent for any Poisson PPs with sigma-finite intensity
measures:

  1. (i)

$P_{\lambda}\ll P_{\mu}$.

  2. (ii)

$T_{0}(\mu\|\lambda)=0$ and $T_{\alpha}(\mu\|\lambda)<\infty$ for all
$\alpha\in\mathbb{R}_{+}$.

  3. (iii)

$T_{0}(\mu\|\lambda)=0$ and $T_{\alpha}(\mu\|\lambda)<\infty$ for some
$0<\alpha<\infty$.

###### Proof.

(i)$\implies$(ii). Assume that $P_{\lambda}\ll P_{\mu}$. Fix a measurable set
$A\subset S$ such that $\mu(A)=0$. Let $C=\\{\eta\in N(S)\colon\eta(A)>0\\}$.
Note that $\eta(A)$ is Poisson-distributed with mean $\lambda(A)$ (resp.
$\mu(A)$) when $\eta$ is sampled from $P_{\lambda}$ (resp. $P_{\mu}$).
Therefore $P_{\mu}(C)=1-e^{-\mu(A)}=0$. Because $P_{\lambda}\ll P_{\mu}$, it
follows that $0=P_{\lambda}(C)=1-e^{-\lambda(A)}$, and we conclude that
$\lambda(A)=0$. Hence $\lambda\ll\mu$, and Proposition 3.7 implies that
$T_{0}(\mu\|\lambda)=0$. Furthermore, because $P_{\lambda}$ and $P_{\mu}$ are
not mutually singular, we know [vH14, Theorem 24] that
$R_{\alpha}(P_{\mu}\|P_{\lambda})<\infty$ for all $\alpha\in\mathbb{R}_{+}$.
By Theorem 5.1, it follows that $T_{\alpha}(\mu\|\lambda)<\infty$ for all
$\alpha\in\mathbb{R}_{+}$.

(ii)$\implies$(iii). Immediate.

(iii)$\implies$(i). Theorem 5.1 implies that
$R_{0}(P_{\mu}\|P_{\lambda})=T_{0}(\mu\|\lambda)=0$. By formula 2.1, we see
that $-\log P_{\lambda}\\{G>0\\}=0$ where $G=\frac{dP_{\mu}}{dm}$ is a density
of $P_{\mu}$ with respect to an arbitrary measure $m$ such that
$P_{\lambda},P_{\mu}\ll m$. Hence $P_{\lambda}\\{G=0\\}=0$, and Lemma A.1
confirms that $P_{\lambda}\ll P_{\mu}$. ∎

As a corollary of Theorem 5.4 we obtain a simple proof of the following result
extending [Tak90] to general nontopological spaces.

######  Corollary 5.5.

The following are equivalent for Poisson PPs with sigma-finite intensity
measures:

  1. (i)

$P_{\lambda}\ll P_{\mu}$

  2. (ii)

$\lambda\ll\mu$ and $H(\lambda,\mu)<\infty$.

###### Proof.

Assume that $\lambda\ll\mu$ and $H(\lambda,\mu)<\infty$. Then
$T_{0}(\mu\|\lambda)=0$ by Proposition 3.7. Formula (3.5) implies that
$T_{1/2}(\mu\|\lambda)=2H^{2}(\mu,\lambda)=2H^{2}(\lambda,\mu)$ is finite.
Theorem 5.4 now implies that $P_{\lambda}\ll P_{\mu}$.

Assume that $P_{\lambda}\ll P_{\mu}$. Theorem 5.4 then implies that
$T_{0}(\mu\|\lambda)=0$ and $T_{1/2}(\mu\|\lambda)<\infty$. Then Proposition
3.7 implies that $\lambda\ll\mu$, and formula (3.5) implies that
$H(\lambda,\mu)<\infty$. ∎

Kakutani’s famous dichotomy [Kak48] tells that infinite products of
probability measures $\prod_{i}P_{i}$ and $\prod_{i}Q_{i}$, such that $P_{i}$
and $Q_{i}$ are mutually absolutely continuous for all $i$, are either
mutually absolutely continuous or mutually singular—there is no middle ground.
Earlier results in this and the previous section yield a simple proof of an
analogue of Kakutani’s dichotomy for Poisson PPs. This result is well known
for Polish spaces [Lie75, Kar91], and has also been presented in [Bro71] in
terms of a more complicated criterion for intensity measures that is
equivalent to $H(\lambda,\mu)<\infty$.

######  Corollary 5.6.

Let $\lambda$ and $\mu$ be mutually absolutely continuous. Then $P_{\lambda}$
and $P_{\mu}$ are either mutually absolutely continuous, or mutually singular,
according to $H(\lambda,\mu)$ being finite or infinite.

###### Proof.

Assume that $H(\lambda,\mu)<\infty$. Corollary 5.5 then shows that
$P_{\lambda}\ll P_{\mu}$ and $P_{\mu}\ll P_{\lambda}$, so that $P_{\lambda}$
and $P_{\mu}$ are mutually absolutely continuous.

Assume next that $H(\lambda,\mu)=\infty$. Corollary 5.3 then implies that
$H(P_{\lambda},P_{\mu})=1$. Therefore
$Z_{1/2}(P_{\lambda}\|P_{\mu})=1-H^{2}(P_{\lambda},P_{\mu}) =0$, or
equivalently
$R_{1/2}(P_{\lambda}\|P_{\mu})=-2\log(1-H^{2}(P_{\lambda},P_{\mu}))=\infty$.
It follows [vH14, Theorem 24] that $P_{\lambda}$ and $P_{\mu}$ are mutually
singular. ∎

###  5.3 Existence of a dominating Poisson PP

By the Radon–Nikodym theorem, for any pair of Poisson PP distributions
$P_{\lambda},P_{\mu}$ there exists a probability measure $Q$ on
$(N(S),\mathcal{N}(S))$ such that $P_{\lambda}\ll Q$ and $P_{\mu}\ll Q$, for
example $Q=\frac{1}{2}(P_{\lambda}+P_{\mu})$. For practical purposes (e.g.
Monte Carlo simulation), it would be helpful to find a Poisson PP
distribution, such as $Q=P_{\lambda+\mu}$, that would also serve as a common
dominating probability measure. This is always possible finite intensity
measures (Remark 5.8) but fails in general. Remarkably, even when a dominating
Poisson PP exists, $\lambda+\mu$ might not be a feasible choice for its
intensity.

######  Theorem 5.7.

Given Poisson PP distribution $P_{\lambda},P_{\mu}$ with sigma-finite
intensity measures $\lambda,\mu$, there exists a Poisson PP distribution
$P_{\xi}$ such that $P_{\lambda},P_{\mu}\ll P_{\xi}$ if and only if
$H(\lambda,\mu)<\infty$.

###### Proof.

Assume that there exists a Poisson PP distribution $P_{\xi}$ with a sigma-
finite intensity measure $\xi$ such that $P_{\lambda},P_{\mu}\ll P_{\xi}$.
Proposition 5.5 implies that $H(\lambda,\xi)<\infty$ and $H(\xi,\mu)<\infty$.
The triangle inequality (Proposition 3.10) implies that $H(\lambda,\mu)\leq
H(\lambda,\xi)+H(\xi,\mu)$. Hence $H(\lambda,\mu)<\infty$.

Assume that $H(\lambda,\mu)<\infty$. Let $f,g$ be densities of $\lambda,\mu$
with respect to the sigma-finite measure $\nu=\lambda+\mu$. Define a measure
$\xi(dx)=h(x)\nu(dx)$ where $h=\frac{1}{4}(\sqrt{f}+\sqrt{g})^{2}$. Observe
that $f\leq 4h$ and $g\leq 4h$ pointwise, and therefore we see that
$\lambda\ll\xi$ and $\mu\ll\xi$. Furthermore, because
$\sqrt{h}=\frac{1}{2}(\sqrt{f}+\sqrt{g})$, we see that
$\sqrt{h}-\sqrt{f}=\frac{1}{2}(\sqrt{g}-\sqrt{f})$ and
$\sqrt{h}-\sqrt{g}=\frac{1}{2}(\sqrt{f}-\sqrt{g})$. Then by formula (3.4),

$\displaystyle H^{2}(\lambda,\xi)\ =\
\frac{1}{2}\int(\sqrt{h}-\sqrt{f})^{2}\,d\nu$ $\displaystyle\ =\
\frac{1}{8}\int(\sqrt{f}-\sqrt{g})^{2}\,d\nu$ $\displaystyle\ =\
\frac{1}{4}H^{2}(\lambda,\mu).$

Hence $H(\lambda,\xi)=\frac{1}{\sqrt{2}}H(\lambda,\mu)$. By symmetry,
$H(\mu,\xi)=\frac{1}{\sqrt{2}}H(\lambda,\mu)$. We conclude that
$H(\lambda,\xi)$ and $H(\mu,\xi)$ are finite. Corollary 5.5 now confirms that
$P_{\lambda}\ll P_{\xi}$ and $P_{\mu}\ll P_{\xi}$. ∎

######  Remark 5.8.

$P_{\lambda},P_{\mu}\ll P_{\lambda+\mu}$ if and only if
$H(\lambda,\lambda+\mu)+H(\mu,\lambda+\mu)<\infty$. The latter requirement is
stronger than $H(\lambda,\mu)<\infty$. To see why, observe that (see
Proposition 3.9)
$H^{2}(\lambda,2\lambda)=\frac{1}{2}\int_{S}(\sqrt{2}-\sqrt{1})^{2}\,d\lambda=\frac{1}{2}(\sqrt{2}-1)^{2}\lambda(S)$.
Therefore $H(\lambda,\mu)=0$ and $H(\lambda,\lambda+\mu)=\infty$ whenever
$\lambda$ is an infinite measure and $\mu=\lambda$. Let us also note that
$P_{\lambda},P_{\mu}\ll P_{\lambda+\mu}$ is always true for finite intensity
measures $\lambda,\mu$ because Hellinger distances between finite measures are
finite; as is seen from (3.4) combined with the inequality
$(\sqrt{f}-\sqrt{g})^{2}\leq f+g$.

##  6 Applications

This section lists various applications of the general formulas derived in the
previous sections. Section 6.1 discusses Poisson processes, Section 6.2
Chernoff information of Poisson vectors, Section 6.3 marked Poisson PPs, and
Section 6.4 concludes with compound Poisson processes.

###  6.1 Poisson processes

The counting process of a point pattern $\eta\in N(\mathbb{R}_{+})$ is a
function $X\colon\mathbb{R}_{+}\to\mathbb{Z}_{+}$ defined by

$X_{t}\ =\ \int_{\mathbb{R}_{+}}1(s\leq t)\,\eta(ds)\ =\ \eta([0,t]).$ (6.1)

Denote by $\operatorname{cou}\colon\eta\mapsto X$ the map induced by (6.1). A
Poisson process with intensity measure $\lambda$ is the counting process
$X=\operatorname{cou}(\eta)$ of a point pattern $\eta$ sampled from a Poisson
PP distribution $P_{\lambda}$ with a sigma-finite intensity measure $\lambda$
on $\mathbb{R}_{+}$. The law of the Poisson process is the pushforward measure
$\mathcal{L}(X)=P_{\lambda}\circ\operatorname{cou}^{-1}$. When $\lambda$
admits a density $f$ with respect to the Lebesgue measure on $\mathbb{R}_{+}$,
the process $X=(X_{t})_{t\in\mathbb{R}_{+}}$ is called an inhomogeneous
Poisson process with intensity function $f$.

######  Corollary 6.1.

For Poisson processes $X=(X_{t})_{t\in\mathbb{R}_{+}}$ and
$Y=(Y_{t})_{t\in\mathbb{R}_{+}}$ with intensity functions $f$ and $g$, the
Kullback–Leibler divergence is given by

$\displaystyle R_{1}(\mathcal{L}(X)\|\mathcal{L}(Y))\ =\
\int_{0}^{\infty}\left(f_{t}\log\frac{f_{t}}{g_{t}}+g_{t}-f_{t}\right)dt,$

the Rényi divergence or order $\alpha\neq 1$ is given by

$\displaystyle R_{\alpha}(\mathcal{L}(X)\|\mathcal{L}(Y))\ =\
\int_{0}^{\infty}\Big{(}\frac{\alpha
f_{t}+(1-\alpha)g_{t}-f_{t}^{\alpha}g_{t}^{1-\alpha}}{1-\alpha}\Big{)}dt,$

and the Hellinger distance is given by

$\operatorname{Hel}(\mathcal{L}(X),\mathcal{L}(Y))\ =\
\sqrt{1-e^{-\frac{1}{2}T_{1/2}(\lambda\|\mu)}},$

where
$T_{1/2}(\lambda\|\mu)=\int_{0}^{\infty}\left(\sqrt{f_{t}}-\sqrt{g_{t}}\right)^{2}dt.$

###### Proof.

Denote by $F(\mathbb{R}_{+},\mathbb{Z}_{+})$ set of nondecreasing functions
$X\colon\mathbb{R}_{+}\to\mathbb{Z}_{+}$ that are right-continuous with left
limits (càdlàg), equipped with the sigma-algebra generated by the evaluation
maps $X\mapsto X_{t}$. It follows from (6.1) that the map
$\operatorname{cou}\colon N(\mathbb{R}_{+})\to F(\mathbb{R}_{+})$ is a
measurable bijection with a measurable inverse. Therefore (Lemma A.4),
$R_{\alpha}(\mathcal{L}(X)\|\mathcal{L}(Y))=R_{\alpha}(P_{\lambda}\|P_{\mu})$
with $\lambda(dt)=f(t)dt$ and $\mu(dt)=g(t)dt$. Theorem 5.1 implies that
$R_{\alpha}(P_{\lambda}\|P_{\mu})=T_{\alpha}(\lambda\|\mu)$, and the first two
claims follow by (3.1).

Next, we note by (3.6) that
$H^{2}(\mathcal{L}(X),\mathcal{L}(Y))=1-\exp\left(-\frac{1}{2}R_{1/2}(\mathcal{L}(X)\|\mathcal{L}(Y))\right)$,
so that the last claim follows from
$R_{1/2}(\mathcal{L}(X)\|\mathcal{L}(Y))=R_{1/2}(P_{\lambda}\|P_{\mu})=T_{1/2}(\lambda\|\mu)$.

∎

###  6.2 Chernoff information for Poisson vectors

In classical binary hypothesis testing, the task is to estimate a parameter
$\theta\in\\{0,1\\}$ from $n$ independent samples from a probability
distribution $F_{\theta}$ on a measurable space $S$. The error rate (i.e.
Bayes risk) of a decision rule $\hat{\theta}_{n}\colon S^{n}\to\\{0,1\\}$,
averaged with respect to prior probabilities $\pi_{0},\pi_{1}>0$ is given by
$R_{\pi}(\hat{\theta}_{n})=\pi_{0}F_{0}^{\otimes
n}(\hat{\theta}_{n}=1)+\pi_{1}F_{1}^{\otimes n}(\hat{\theta}_{n}=0)$.
Chernoff’s famous theorem [Che52] tells for $n\gg 1$, the minimum error rate
scales as

$\inf_{\hat{\theta}_{n}}R_{\pi}(\hat{\theta}_{n})\ =\
e^{-(1+o(1))C(F_{0}\|F_{1})n},$

where

$C(F_{0}\|F_{1})\ =\ \sup_{\alpha\in(0,1)}(1-\alpha)R_{\alpha}(F_{0}\|F_{1}).$

is called the Chernoff information of the test [Nie13].

In a network community detection problem related to a stochastic block model,
Abbe and Sandon [AS15] reduced a key estimation task into a binary hypothesis
test for the law of a random vector with independent Poisson-distributed
components, either having mean vector $H_{0}:(\lambda_{1},\dots,\lambda_{K})$
or $H_{1}:(\mu_{1},\dots,\mu_{K})$. Equivalently, the law of this random
vector can be seen as a Poisson PP distribution on the finite set
$\\{1,\dots,K\\}$ with intensity measure admitting density
$k\mapsto\lambda_{k}$ or $k\mapsto\mu_{k}$ with respect to the counting
measure. Hence we are looking at a hypothesis test between Poisson PP
distributions $F_{0}=P_{\lambda}$ and $F_{1}=P_{\mu}$. With the help of
Theorem 5.1 and formula (3.1) we see that the Chernoff information of the test
is given by

$C(P_{\lambda}\|P_{\mu})\ =\
\sup_{\alpha\in(0,1)}\sum_{k=1}^{K}\big{(}\alpha\lambda_{k}+(1-\alpha)\mu_{k}-\lambda_{k}^{\alpha}\mu_{k}^{1-\alpha}\big{)}.$

This is what is called Chernoff–Hellinger divergence in [AS15, RB17, ZT23].

###  6.3 Marked Poisson PPs

A marking of a (random or nonrandom) set of points with locations
$x_{1},x_{2},\dots$ in $S_{1}$ associates each point $x_{i}$ a random variable
$y_{i}$ in $S_{2}$, called a mark, so that the conditional distribution of
$y_{i}$ given $x_{i}=x$ is determined by $x$, and the marks are conditionally
independent given the locations. The statistics of the marking mechanism is
parameterised by the collection of conditional distributions $K\colon x\mapsto
K_{x}=\mathcal{L}(y_{i}\,|\,x_{i}=x)$ that constitutes a probability kernel
from $S_{1}$ into $S_{2}$. Such a marking mechanism can also be defined for
random point patterns that admit a proper enumeration. This is the case for
point patterns sampled from a Poisson PP distribution [LP18, Corollary 3.7].

Let $P_{\lambda}$ be a Poisson PP distribution with a sigma-finite intensity
measure $\lambda$ on a measurable space $(S_{1},\mathcal{S}_{1})$, and let $K$
be a probability kernel from $(S_{1},\mathcal{S}_{1})$ into
$(S_{2},\mathcal{S}_{2})$. A marked Poisson PP with intensity measure
$\lambda$ and mark kernel $K$ is then defined by sampling a point pattern
$\xi$ on $S_{1}\times S_{2}$ from a Poisson PP distribution $P_{\lambda\otimes
K}$ with intensity measure $\lambda\otimes K$ defined by

$(\lambda\otimes K)(C)\ =\
\int_{S_{1}}\int_{S_{2}}1_{C}(x,y)\,K_{x}(dy)\,\lambda(dx),\qquad
C\in\mathcal{S}_{1}\otimes\mathcal{S}_{2}.$

Then the Poisson PP distribution $P_{\lambda}$ equals the law of the point
pattern $A\mapsto\xi(A\times S_{2})$ corresponding to the locations of the
points in $S_{1}$, and the probability kernel $K$ yields the conditional
distributions of the marks [LP18, Section 5.2]. Hence $\xi$ is a marked
Poisson PP with intensity measure $\lambda$ and marking kernel $K$.

######  Corollary 6.2.

If $\xi$ and $\zeta$ and marked Poisson PPs with sigma-finite intensity
measures $\lambda$ and $\mu$ and mark kernels $K$ and $L$, respectively, then
the Rényi divergence of order $\alpha>0$ is given by
$R_{\alpha}(\mathcal{L}(\xi)\|\mathcal{L}(\zeta))=T_{\alpha}(\lambda\otimes
K\|\mu\otimes L)$.

###### Proof.

Because $\mathcal{L}(\xi)=P_{\lambda\otimes K}$ and
$\mathcal{L}(\zeta)=P_{\mu\otimes L}$, the claim follows by Theorem 5.1. ∎

Corollary 6.2 combined with Proposition 3.11 now allows to compute
Kullback–Leibler and Rényi divergences of marked Poisson PPs in terms of the
intensity measures $\lambda,\mu$ and the kernels $K,L$.

###  6.4 Compound Poisson processes

A set of points $(t_{i},x_{i})$ associated with time stamp
$t_{i}\in\mathbb{R}_{+}$ and label $x_{i}\in\mathbb{R}^{d}$ can be modelled as
a point pattern on $\mathbb{R}_{+}\times\mathbb{R}^{d}$, or as a marked point
pattern on $\mathbb{R}_{+}$ with mark space $\mathbb{R}^{d}$. Denote by
$N(\mathbb{R}_{+}\times\mathbb{R}^{d})$ the set of point patterns $\eta$ on
$\mathbb{R}_{+}\times\mathbb{R}^{d}$ such that
$\eta([0,t]\times\mathbb{R}^{d})<\infty$ for all $t\in\mathbb{R}_{+}$. The
cumulative process of such a point pattern is a function
$X\colon\mathbb{R}_{+}\to\mathbb{R}^{d}$ defined by

$X_{t}\ =\ \int_{[0,t]\times\mathbb{R}^{d}}x\,\eta(ds,dx).$ (6.2)

Denote by $\operatorname{agg}\colon\eta\mapsto X$ the map induced by (6.2). A
compound Poisson process with event intensity measure $\lambda$ and increment
probability kernel $K$ is the cumulative process $X=\operatorname{agg}(\eta)$
of a point pattern $\eta$ sampled from a Poisson PP distribution
$P_{\lambda\otimes K}$ on $N(\mathbb{R}_{+}\times\mathbb{R}^{d})$ with
intensity measure $(\lambda\otimes K)(dt,dx)=\lambda(dt)K_{t}(dx)$ where
$\lambda$ is a locally finite measure $\mathbb{R}_{+}$ and $K$ is a
probability kernel from $\mathbb{R}_{+}$ into $\mathbb{R}^{d}$. The law of the
compound Poisson process is the pushforward measure
$\mathcal{L}(X)=P_{\lambda\otimes K}\circ\operatorname{agg}^{-1}$. The
assumption that $\lambda$ is locally finite implies that $P_{\lambda\otimes
K}(N(\mathbb{R}_{+}\times\mathbb{R}^{d}))=1$, so that the right side of (6.2)
is well defined for $P_{\lambda\otimes K}$-almost every $\eta$.

The map $\operatorname{agg}\colon\eta\mapsto X$ induced by (6.2) maps
$N(\mathbb{R}_{+}\times\mathbb{R}^{d})$ into the set
$F(\mathbb{R}_{+},\mathbb{R}^{d})$ of right-continuous and piecewise constant
functions from $\mathbb{R}_{+}$ into $\mathbb{R}^{d}$. However, this map is
not bijective because: (i) two simultaneous events $(t,x)$ and $(t,y)$ have
the same contribution as a single event $(t,x+y)$ to the cumulative process;
and (ii) points $(t,0)$ with zero mark do not contribute anything. To rule out
such identifiability issues, we will restrict to a set
$N_{s}(\mathbb{R}_{+}\times\mathbb{R}^{d})$ of point patterns $\eta\in
N(\mathbb{R}_{+}\times\mathbb{R}^{d})$ such that
$\eta(\\{t\\}\times\mathbb{R}^{d})\in\\{0,1\\}$ for all $t$, and
$\eta(\mathbb{R}_{+}\times\\{0\\})=0$. This is why we will assume that
$\lambda$ is locally finite, diffuse in the sense that $\lambda(\\{t\\})=0$
for all $t$, and that the probability kernel $K$ satisfies $K_{t}(\\{0\\})=0$
for all $t$. Under these assumptions, it follows [LP18, Proposition 6.9] that
$P_{\lambda\otimes K}(N_{s}(\mathbb{R}_{+}\times\mathbb{R}^{d}))=1$.

######  Corollary 6.3.

For compound Poisson processes $X=(X_{t})_{t\in\mathbb{R}_{+}}$ and
$Y=(Y_{t})_{t\in\mathbb{R}_{+}}$ with diffuse locally finite event intensity
measures $\lambda$ and $\mu$, and increment probability kernels $K$ and $L$
such that $K_{t}(\\{0\\})=0$ and $L_{t}(\\{0\\})=0$, the Rényi divergence of
order $\alpha>0$ is given by
$R_{\alpha}(\mathcal{L}(X)\|\mathcal{L}(Y))=T_{\alpha}(\lambda\otimes
K\|\mu\otimes L)$.

###### Proof.

Let us equip $F(\mathbb{R}_{+},\mathbb{R}^{d})$ with the sigma-algebra
generated by the evaluation maps $X\mapsto X_{t}$. The aggregation map
$\operatorname{agg}\colon N(\mathbb{R}_{+}\times\mathbb{R}^{d})\to
F(\mathbb{R}_{+},\mathbb{R}^{d})$ restricted to
$N_{s}(\mathbb{R}_{+}\times\mathbb{R}^{d})$ is bijective, with inverse map
given by

$(\operatorname{agg}^{-1}(X))(C)\ =\ \\#\\{(t,x)\in C\colon
X_{t}-X_{t-}=x,\,x\neq 0\\},$

where $X_{t-}=\lim_{s\uparrow t}X_{s}$ for $t>0$ and $X_{t-}=X_{0}$ for $t=0$.
Standard techniques (as in the proof of Theorem 5.1) imply that
$\operatorname{agg}\colon N_{s}(\mathbb{R}_{+}\times\mathbb{R}^{d})\to
F(\mathbb{R}_{+},\mathbb{R}^{d})$ is measurable with a measurable inverse.
Because $P_{\lambda\otimes K}$ and $P_{\mu\otimes L}$ have all their mass
supported on $N_{s}(\mathbb{R}_{+}\times\mathbb{R}^{d})$, it follows (Lemma
A.4) that
$R_{\alpha}(\mathcal{L}(X)\|\mathcal{L}(Y))=R_{\alpha}(P_{\lambda\otimes
K}\|P_{\mu\otimes L})$. Theorem 5.1 then implies that
$R_{\alpha}(\mathcal{L}(X)\|\mathcal{L}(Y))=T_{\alpha}(\lambda\|\mu)$.

∎

Corollary 6.3 combined with Proposition 3.11, we find that for example that
for $\alpha\notin\\{0,1\\}$, the Rényi divergence of compound Poisson process
$X=(X_{t})_{t\in\mathbb{R}_{+}}$ and $Y=(Y_{t})_{t\in\mathbb{R}_{+}}$ is given
by

$\displaystyle R_{\alpha}(\mathcal{L}(X)\|\mathcal{L}(Y))$ $\displaystyle\ =\
R_{\alpha}(P_{\lambda}\|P_{\mu})+\int_{S_{1}}T_{\alpha}(K_{t}\|L_{t})\,f_{t}^{\alpha}g_{t}^{1-\alpha}\,\nu(dt),$

This shows that how the information content decomposes into two parts:
$R_{\alpha}(P_{\lambda}\|P_{\mu})$ associated with only observing the jump
instants of the compound Poisson processes, and the additional term
$\int_{S_{1}}T_{\alpha}(K_{t}\|L_{t})\,f_{t}^{\alpha}g_{t}^{1-\alpha}\,\nu(dt)$
characterising the information gain when we also observe the jump sizes.

##  7 Proofs

This section contains proofs of the main results, with some of the technical
parts postponed in the appendix.

###  7.1 Proof of Theorem 3.1

######  Lemma 7.1.

The Rényi divergence of Poisson distributions $p_{s}$ and $p_{t}$ with means
$s,t\in\mathbb{R}_{+}$ is given by222$R_{\alpha}(p_{s}\|p_{t})=\infty$ when
$\alpha\geq 1$, $s>0$, and $t=0$ by our 0-division conventions.

$R_{\alpha}(p_{s}\|p_{t})\ =\ \begin{cases}1(s=0)t,&\quad\alpha=0,\\\
\frac{\alpha
s+(1-\alpha)t-s^{\alpha}t^{1-\alpha}}{1-\alpha},&\quad\alpha\notin\\{0,1\\},\\\
s\log\frac{s}{t}+t-s,&\quad\alpha=1.\end{cases}$ (7.1)

###### Proof.

Assume first that $s,t>0$. By definition, the Rényi divergence or order
$\alpha\notin\\{0,1\\}$ equals
$R_{\alpha}(p_{s}\|p_{t})=\frac{1}{\alpha-1}\log Z_{\alpha}(p_{s}\|p_{t})$,
where

$\displaystyle Z_{\alpha}(p_{s}\|p_{t})$ $\displaystyle\ =\
\sum_{x=0}^{\infty}\Big{(}e^{-s}\frac{s^{x}}{x!}\Big{)}^{\alpha}\Big{(}e^{-t}\frac{t^{x}}{x!}\Big{)}^{1-\alpha}$
$\displaystyle\ =\ \exp\Big{(}-\Big{(}\alpha
s+(1-\alpha)t-s^{\alpha}t^{1-\alpha}\Big{)}\Big{)}.$

By taking logarithms, (7.1) follows for $\alpha\notin\\{0,1\\}$. We also note
that
$\log\frac{e^{-s}\frac{s^{x}}{x!}}{e^{-t}\frac{t^{x}}{x!}}=x\log\frac{s}{t}+t-s$,
and taking expectations with respect to $p_{s}$ yields (7.1) for $\alpha=1$.
Furthermore, when $s,t>0$, both $p_{s}$ and $p_{t}$ assign strictly positive
probabilities to all nonnegative integers. Therefore,
$R_{0}(p_{s}\|p_{t})=-\log p_{t}(\mathbb{Z}_{+})=0$, confirming (7.1) for
$\alpha=0$.

We also note that the Poisson distribution with mean 0 equals the Dirac
measure at 0. Therefore, a simple computation shows that for all $s,t>0$,

$R_{\alpha}(p_{s}\|\delta_{0})\ =\ \begin{cases}0,&\quad\alpha=0,\\\
\frac{\alpha}{1-\alpha}s,&\quad\alpha\in(0,1),\\\ \infty,&\quad\alpha\geq
1,\end{cases}$

and $R_{\alpha}(\delta_{0}\|p_{t})=t$ for all $\alpha\in[0,\infty)$. Finally,
$R_{\alpha}(\delta_{0}\|\delta_{0})=0$ for all $\alpha\in[0,\infty)$. By
recalling our conventions with dividing by 0, we may conclude that (7.1) holds
for all $s,t,\alpha\in[0,\infty)$. ∎

###### Proof of Theorem 3.1.

Let $\lambda,\mu$ be sigma-finite measures admitting densities $f,g\colon
S\to\mathbb{R}_{+}$ with respect to a sigma-finite measure $\nu$ on a
measurable space $(S,\mathcal{S})$. Let $p_{f(x)}$ and $p_{g(x)}$ be Poisson
distributions with means $f(x)$ and $g(x)$. By applying the Rényi divergence
formula for Poisson distributions in Lemma 7.1, we find that

$\displaystyle\int_{S}R_{\alpha}(p_{f(x)}\|p_{g(x)})\,\nu(dx)$
$\displaystyle\quad\ =\ \begin{cases}\int_{S}1(f=0)g\,d\nu,&\quad\alpha=0,\\\
\int_{S}\frac{\alpha
f+(1-\alpha)g-f^{\alpha}g^{1-\alpha}}{1-\alpha}\,d\nu,&\quad\alpha\notin\\{0,1\\},\\\
\int_{S}\left(f\log\frac{f}{g}+g-f\right)\,d\nu,&\quad\alpha=1.\end{cases}$

By comparing this with the definition of the Tsallis divergence (3.1), we find
that

$T_{\alpha}(\lambda\|\mu)\ =\
\int_{S}R_{\alpha}(p_{f(x)}\|p_{g(x)})\,\nu(dx).$ (7.2)

Because Rényi divergences of probability measures are nonnegative, (7.2) shows
that $T_{\alpha}(\lambda\|\mu)$ is a well-defined element in $[0,\infty]$ for
all $\alpha\in\mathbb{R}_{+}$. We also know [vH14, Theorem 3] that Rényi
divergences of probability measures are nondecreasing in $\alpha$, so it also
follows from (7.2) that $T_{\alpha}(\lambda\|\mu)$ is nondecreasing in
$\alpha$.

Let us next verify that $\alpha\mapsto T_{\alpha}(\lambda\|\mu)$ is continuous
on $A=\\{\alpha\in\mathbb{R}_{+}\colon T_{\alpha}(\lambda\|\mu)<\infty\\}$. To
this end, assume that $A$ is nonempty and $\alpha_{n}\to\alpha$ for some
$\alpha_{n},\alpha\in A$. Because $T_{\alpha}(\lambda\|\mu)$ is nondecreasing
in $\alpha$, we see that $A$ is an interval, and that there exists a number
$\beta\in A$ such that $\alpha_{n},\alpha\leq\beta$ for all $n$, and let
$U=\\{x\in S\colon r_{\beta}(x)<\infty\\}$. Because
$T_{\beta}(\lambda\|\mu)=\int_{S}r_{\beta}\,d\nu$ is finite, it follows that
$\nu(U^{c})=0$. Therefore,

$T_{\alpha_{n}}(\lambda\|\mu)\ =\ \int_{S}r_{\alpha_{n}}\,d\nu\ =\
\int_{U}r_{\alpha_{n}}\,d\nu.$

For any $x\in U$, the monotonicity of Rényi divergences [vH14, Theorem 3]
implies that $r_{\alpha}(x),r_{\alpha_{n}}(x)\leq r_{\beta}(x)$. In
particular, $r_{\alpha}(x),r_{\alpha_{n}}(x)$ are finite for $x\in U$. The
continuity of Rényi divergences [vH14, Theorem 7] then implies that
$r_{\alpha_{n}}\to r_{\alpha}$ pointwise on $U$. Lebesgue’s dominated
convergence theorem then implies that

$T_{\alpha_{n}}(\lambda\|\mu)\ =\ \int_{U}r_{\alpha_{n}}\,d\nu\ \to\
\int_{U}r_{\alpha}\,d\nu\ =\ T_{\alpha}(\lambda\|\mu).$

Finally, let us verify that the right side of (3.1) does not depend on the
choice of the densities nor the reference measure. Assume that $\lambda,\mu$
admit densities $f_{1},g_{1}\colon S\to\mathbb{R}_{+}$ with respect to a
sigma-finite measure $\nu_{1}$, and densities $f_{2},g_{2}\colon
S\to\mathbb{R}_{+}$ with respect to a sigma-finite measure $\nu_{2}$. Define
$\nu=\nu_{1}+\nu_{2}$. Then $\nu_{1},\nu_{2}\ll\nu$ and $\nu$ is sigma-finite.
The Radon–Nikodym theorem [Kal02, Theorem 2.10] implies that there exist
densities $h_{1},h_{2}\colon S\to\mathbb{R}_{+}$ of $\nu_{1},\nu_{2}$ with
respect to $\nu$. Then

$\lambda(A)\ =\ \int_{A}f_{i}\,d\nu_{i}\ =\
\int_{A}\tilde{f}_{i}\,d\nu\quad\text{for $i=1,2$},$

where $\tilde{f}_{i}=f_{i}h_{i}$. We see that both $\tilde{f}_{1}$ and
$\tilde{f}_{2}$ are densities of $\lambda$ with respect to $\nu$. The
Radon–Nikodym theorem [Kal02, Theorem 2.10] implies that
$\tilde{f}_{1}=\tilde{f}_{2}$ $\nu$-almost everywhere. Similarly, we see that
both $\tilde{g}_{1}$ and $\tilde{g}_{2}$ are densities of $\mu$ with respect
to $\nu$, so that $\tilde{g}_{1}=\tilde{g}_{2}$ $\nu$-almost everywhere.
Formula (7.1) shows that Rényi divergences of Poisson distributions are
homogeneous in the sense that
$R_{\alpha}(p_{cs}\|p_{ct})=cR_{\alpha}(p_{s}\|p_{t})$ for all
$c\in\mathbb{R}_{+}$. As a consequence, we see that We also see that

$\displaystyle\int_{S}R_{\alpha}(p_{\tilde{f}_{1}(x)}\|p_{\tilde{g}_{1}(x)})\,\nu(dx)$
$\displaystyle\ =\
\int_{S}R_{\alpha}(p_{f_{1}(x)}\|p_{g_{1}(x)})\,\nu_{1}(dx),$
$\displaystyle\int_{S}R_{\alpha}(p_{\tilde{f}_{2}(x)}\|p_{\tilde{g}_{2}(x)})\,\nu(dx)$
$\displaystyle\ =\
\int_{S}R_{\alpha}(p_{f_{2}(x)}\|p_{g_{2}(x)})\,\nu_{2}(dx).$

Because $\tilde{f}_{1}=\tilde{f}_{2}$ and $\tilde{g}_{1}=\tilde{g}_{2}$
$\nu$-almost everywhere, it follows that all of the above integrals are equal
to each other. In particular,

$\int_{S}R_{\alpha}(p_{f_{1}(x)}\|p_{g_{1}(x)})\,\nu_{1}(dx)\ =\
\int_{S}R_{\alpha}(p_{f_{2}(x)}\|p_{g_{2}(x)})\,\nu_{2}(dx).$

In light of (7.2), we conclude that the value of $T_{\alpha}(\lambda\|\mu)$ as
defined by formula (7.2) is the same for both triples $(f_{1},g_{1},\nu_{1})$
and $(f_{2},g_{2},\nu_{2})$.

∎

###  7.2 Proof of Theorem 3.11

###### Proof of Theorem 3.11.

Define $\nu=\lambda+\mu$. Then $\nu$ is a sigma-finite measure such that
$\lambda,\mu\ll\nu$. Then there exist densities $f,g$ of $\lambda,\mu$ with
respect to $\nu$. Then

$\displaystyle(\lambda\otimes K)(C)$ $\displaystyle\ =\ \int_{S_{1}\times
S_{2}}1_{C}(t,x)\,\lambda(dt)K_{t}(dx)$ $\displaystyle\ =\ \int_{S_{1}\times
S_{2}}1_{C}(t,x)\,f_{t}k_{t}(x)\,\nu(dt)M_{t}(dx)$

shows that $(t,x)\mapsto f_{t}k_{t}(x)$ is a density of $\lambda\otimes K$
with respect to $\nu\otimes M$. Similarly, $t\mapsto g_{t}\ell_{t}(x)$ is a
density of $\mu\otimes L$ with respect to $\nu\otimes M$.

By (3.1), the Tsallis divergence of order $\alpha\neq 0,1$ is given by

$T_{\alpha}(\lambda\otimes K\|\mu\otimes L)\ =\
\int_{S_{1}}\left(\int_{S_{2}}\tau_{\alpha}(t,x)\,M_{t}(dx)\right)\,\nu(dt),$
(7.3)

where the integrand equals

$\tau_{\alpha}(t,x)\ =\ \frac{\alpha
f_{t}k_{t}(x)+(1-\alpha)g_{t}\ell_{t}(x)-f_{t}^{\alpha}g_{t}^{1-\alpha}k_{t}(x)^{\alpha}\ell_{t}(x)^{1-\alpha}}{1-\alpha}.$

Using the representation in Theorem 3.1 we see that $\tau_{\alpha}(t,x)\geq
0$. This is why the order of integration in (7.3) does not matter. We also
note that

$\displaystyle\tau_{\alpha}(t,x)$ $\displaystyle\ =\ \frac{\alpha
f_{t}k_{t}(x)+(1-\alpha)g_{t}\ell_{t}(x)}{1-\alpha}$
$\displaystyle\qquad-\left(\frac{\alpha
k_{t}(x)+(1-\alpha)\ell_{t}(x)}{1-\alpha}\right)f_{t}^{\alpha}g_{t}^{1-\alpha}$
$\displaystyle\qquad+\left(\frac{\alpha
k_{t}(x)+(1-\alpha)\ell_{t}(x)-k_{t}(x)^{\alpha}\ell_{t}(x)^{1-\alpha}}{1-\alpha}\right)f_{t}^{\alpha}g_{t}^{1-\alpha}.$

We note that $\int_{S_{2}}k_{t}(x)\,M_{t}(dx)=1$ and
$\int_{S_{2}}\ell_{t}(x)\,M_{t}(dx)=1$, and that

$\int_{S_{2}}\left(\frac{\alpha
k_{t}(x)+(1-\alpha)\ell_{t}(x)-k_{t}(x)^{\alpha}\ell_{t}(x)^{1-\alpha}}{1-\alpha}\right)M_{t}(dx)\
=\ T_{\alpha}(K_{t}\|L_{t}).$

It follows that the inner integral in (7.3) equals

$\displaystyle\int_{S_{2}}\tau_{\alpha}(t,x)\,M_{t}(dx)$ $\displaystyle\ =\
\frac{\alpha
f_{t}+(1-\alpha)g_{t}-f_{t}^{\alpha}g_{t}^{1-\alpha}}{1-\alpha}+T_{\alpha}(K_{t}\|L_{t})\,f_{t}^{\alpha}g_{t}^{1-\alpha}.$

By integrating both sides against $\nu(dt)$, we obtain (3.8) for
$\alpha\notin\\{0,1\\}$.

Let us now consider the case with $\alpha=1$. In this case
$T_{\alpha}(\lambda\otimes K\|\mu\otimes L)$ is again given by (7.3), but now
we replace $\tau_{\alpha}$ by

$\displaystyle\tau_{1}(t,x)$ $\displaystyle\ =\
f_{t}k_{t}(x)\log\frac{f_{t}k_{t}(x)}{g_{t}\ell_{t}(x)}+g_{t}\ell_{t}(x)-f_{t}k_{t}(x)$
$\displaystyle\ =\
f_{t}k_{t}(x)\log\frac{f_{t}}{g_{t}}+f_{t}k_{t}(x)\log\frac{k_{t}(x)}{\ell_{t}(x)}+g_{t}\ell_{t}(x)-f_{t}k_{t}(x).$

By integrating the above equation against $M_{t}(dx)$, we find that

$\displaystyle\int_{S_{2}}\tau_{1}(t,x)\,M_{t}(dx)$ $\displaystyle\ =\
f_{t}\log\frac{f_{t}}{g_{t}}+g_{t}-f_{t}+f_{t}T_{1}(K_{t}\|L_{t}).$

By further integrating this $\nu(dt)$, it follows that

$\displaystyle\int_{S_{1}}\int_{S_{2}}\tau_{1}(t,x)\,M_{t}(dx)\,\nu(dt)\ =\
T_{1}(\lambda\|\mu)+\int_{S_{1}}T_{1}(K_{t}\|L_{t})\,\lambda(dt),$

from which conclude the validity of (3.8) for $\alpha=1$.

Finally, for $\alpha=0$ we note that

$\displaystyle\\{(t,x)\colon f_{t}k_{t}(x)=0\\}$ $\displaystyle\ =\
(\\{f=0\\}\times S_{2})\cup\\{(t,x)\colon f_{t}\neq 0,\,k_{t}(x)=0\\}.$

Hence by (3.1),

$\displaystyle T_{0}(\lambda\otimes K\|\mu\otimes L)$ $\displaystyle\ =\
(\mu\otimes L)\\{(t,x)\colon f_{t}k_{t}(x)=0\\}$ $\displaystyle\ =\
\mu\\{f=0\\}+\int_{f\neq 0}L_{t}\\{k_{t}=0\\}\,\mu(dt)$ $\displaystyle\ =\
T_{0}(\lambda\|\mu)+\int_{f\neq 0}T_{0}(K_{t}\|L_{t})\,\mu(dt).$

∎

###  7.3 Proof of Theorem 4.1

The following result shows that embeddings
$(x_{1},\dots,x_{n})\mapsto\sum_{i=1}^{n}\delta_{x_{i}}$ are measurable even
when singleton sets in $(S,\mathcal{S})$ might be nonmeasurable. This is the
reason why there is no need to deal with ‘chunks’ as in [Bro71], and we get a
conceptually simplified proof of Theorem 4.1.

######  Lemma 7.2.

For any $n\geq 1$, the function $\iota_{n}\colon S^{n}\to N(S)$ by
$\iota_{k}(x_{1},\dots,x_{n})=\sum_{i=1}^{n}\delta_{x_{i}}$ is measurable.

###### Proof.

Fix a set $A\in\mathcal{S}$ and an integer $k\geq 0$. Let
$C=\\{\eta\colon\eta(A)=k\\}$. Then the set

$\displaystyle\iota_{n}^{-1}(C)$ $\displaystyle\ =\ \\{(x_{1},\dots,x_{n})\in
S^{n}\colon\sum_{i=1}^{n}\delta_{x_{i}}(A)=k\\}.$

consists of the $n$-tuples in $S$ for which exactly $k$ members belong to $A$,
and the remaining members belong to $A^{c}$. The set of all such $n$-tuples
can be written as

$\iota_{n}^{-1}(C)\ =\ \bigcup_{b\in\\{0,1\\}^{n}:\sum
b_{i}=k}1_{A}^{-1}(\\{b_{1}\\})\times\cdots\times 1_{A}^{-1}(\\{b_{n}\\}),$

where $1_{A}^{-1}(\\{b_{k}\\})$ denotes the preimage of the indicator function
$1_{A}\colon S\to\\{0,1\\}$ for $\\{b_{k}\\}$, and equals $A$ for $b_{k}=1$
and $A^{c}$ for $b_{k}=0$. Because all sets appearing in the finite union on
the right are products of $A$ and $A^{c}$, we conclude that
$\iota_{n}^{-1}(C)\in\mathcal{S}^{\otimes n}$. Because $\mathcal{N}(S)$ is the
sigma-algebra generated by sets of form $C$, we conclude that
$\iota_{n}^{-1}(C)\in\mathcal{S}^{\otimes n}$ for all $C\in\mathcal{N}(S)$. ∎

###### Proof of Theorem 4.1.

Assume that $P_{\lambda}\ll P_{\mu}$ are Poisson PP distributions with finite
intensity measures $\lambda\ll\mu$. To rule out trivialities, we assume that
$\lambda,\mu$ are nonzero. The Poisson PP distributions may [LP18, Proposition
3.5] then be represented as

$P_{\lambda}\ =\ \sum_{n\geq
0}e^{-\lambda(S)}\frac{\lambda(S)^{n}}{n!}\,\lambda_{1}^{\otimes
n}\\!\circ\\!\iota_{n}^{-1},\qquad P_{\mu}\ =\ \sum_{n\geq
0}e^{-\mu(S)}\frac{\mu(S)^{n}}{n!}\,\mu_{1}^{\otimes
n}\\!\circ\\!\iota_{n}^{-1},$ (7.4)

where $\lambda_{1}^{\otimes n},\mu_{1}^{\otimes n}$ are $n$-fold products of
probability measures $\lambda_{1}=\lambda/\lambda(S)$ and $\mu_{1}=\mu/\mu(S)$
on $S$, and $\iota_{n}(x_{1},\dots,x_{n})=\sum_{i=1}^{n}\delta_{x_{i}}$.
(Lemma 7.2 guarantees that the maps $\iota_{n}\colon S^{n}\to N(S)$ are
measurable, and therefore $P_{\lambda},P_{\mu}$ are well-defined probability
measures on $N(S)$.)

Let $\phi\colon S\to\mathbb{R}_{+}$ be a density of $\lambda$ with respect to
$\mu$, and consider the function $\Phi\colon N(S)\to\mathbb{R}_{+}$ such that

$\Phi(\eta)=e^{\mu(S)-\lambda(S)}\prod_{i=1}^{n}\phi(x_{i})\qquad\text{for}\quad\eta=\sum_{i=1}^{n}\delta_{x_{i}},$
(7.5)

and $\Phi(\eta)=0$ for $\eta(S)=\infty$. We will show that $\Phi$ is a density
of $P_{\lambda}$ with respect to $P_{\mu}$. To do this, fix a measurable set
$C\subset N(S)$, and note that

$\displaystyle\int_{C}\Phi(\eta)\,P_{\mu}(d\eta)$ $\displaystyle\ =\
\sum_{n\geq
0}e^{-\mu(S)}\frac{\mu(S)^{n}}{n!}\int_{C}\Phi(\eta)\,\mu_{1}^{\otimes
n}\\!\circ\\!\iota_{n}^{-1}(d\eta)$ $\displaystyle\ =\ \sum_{n\geq
0}e^{-\mu(S)}\frac{\mu(S)^{n}}{n!}\int_{\iota_{n}^{-1}(C)}\Phi(\iota_{n}(x))\,\mu_{1}^{\otimes
n}(dx)$ $\displaystyle\ =\ \sum_{n\geq
0}e^{-\lambda(S)}\frac{\mu(S)^{n}}{n!}\int_{\iota_{n}^{-1}(C)}\phi^{\otimes
n}(x)\,\mu_{1}^{\otimes n}(dx),$

where $\phi^{\otimes n}(x_{1},\dots,x_{n})=\prod_{i=1}^{n}\phi(x_{i})$. We
also note that $\frac{\mu(S)}{\lambda(S)}\phi$ is a density of $\lambda_{1}$
with respect to $\mu_{1}$, and therefore,
$(\frac{\mu(S)}{\lambda(S)})^{n}\phi^{\otimes n}$ is a density of
$\lambda_{1}^{\otimes n}$ with respect to $\mu_{1}^{\otimes n}$. Then

$\lambda_{1}^{\otimes n}(\iota_{n}^{-1}(C))\ =\
\int_{\iota_{n}^{-1}(C)}\left(\frac{\mu(S)}{\lambda(S)}\right)^{n}\phi^{\otimes
n}(x)\,\mu_{1}^{\otimes n}(dx),$

and it follows that

$\displaystyle\int_{C}\Phi(\eta)\,P_{\mu}(d\eta)$ $\displaystyle\ =\
\sum_{n\geq
0}e^{-\lambda(S)}\frac{\mu(S)^{n}}{n!}\left(\frac{\lambda(S)}{\mu(S)}\right)^{n}\lambda_{1}^{\otimes
n}(\iota_{n}^{-1}(C))$ $\displaystyle\ =\ \sum_{n\geq
0}e^{-\lambda(S)}\frac{\lambda(S)^{n}}{n!}\lambda_{1}^{\otimes
n}(\iota_{n}^{-1}(C)).$

In light of (7.4), we conclude that
$\int_{C}\Phi(\eta)\,P_{\mu}(d\eta)=P_{\lambda}(C)$. Hence $\Phi$ is a density
of $P_{\lambda}$ with respect to $P_{\mu}$.

Finally, let us verify that the function $\Phi$ can be written in form (4.1).
By definition (7.5), we see that $\Phi(\eta)=0$ whenever $\eta(S)=\infty$ or
$\eta\\{\phi=0\\}>0$. Hence $\Phi$ vanishes outside the set $M\cap\Omega$
where $M=\\{\eta\in N(S)\colon\eta\\{\phi=0\\}=0\\}$ and $\Omega=\\{\eta\in
N(S)\colon\eta(S)<\infty\\}$. On the other hand,
$\prod_{i=1}^{n}\phi(x_{i})=\exp(\int_{S}\log\phi\,d\eta)$ for every point
pattern in $M\cap\Omega$ of form $\eta=\sum_{i=1}^{n}\delta_{x_{i}}$. By
noting that $\mu(S)-\lambda(S)=\int_{S}(1-\phi)\,d\mu$, we see that $\Phi$ may
be written as in (4.1), but with $1_{M}$ replaced by $1_{M\cap\Omega}$.
Because Campbell’s theorem implies that $P_{\mu}(\Omega)=1$, we see that
$\Phi$ is equal to the function in (4.1) as an element of
$L_{1}(N(S),\mathcal{N}(S),P_{\mu})$. ∎

###  7.4 Proof of Theorem 4.4

We start by proving the following simple upper bound that confirms that
$\lambda$ is finite whenever $\lambda\ll\mu$ and $H(\lambda,\mu)<\infty$ for
some finite measure $\mu$.

######  Lemma 7.3.

Assume that $\lambda\ll\mu$ and that $\phi\colon S\to\mathbb{R}_{+}$ is a
density of $\lambda$ with respect to $\mu$. Then $\lambda(B)\leq
4\mu(B)+3\int_{B}(\sqrt{\phi}-1)^{2}\,d\mu$ for any measurable $B\subset S$.
Especially, $\lambda(S)\leq 4\mu(S)+6H(\lambda,\mu)^{2}$.

###### Proof.

By writing $\phi=1+(\phi-1)\leq 1+(\phi-1)_{+}$, we see that
$\lambda(B)=\int_{B}\phi\,d\mu$ is bounded by

$\lambda(B)\ \leq\ \mu(B)+\int_{B}(\phi-1)_{+}\,d\mu.$ (7.6)

We also note that $t\mapsto\frac{t-1}{t+1}$ is increasing on $\mathbb{R}_{+}$,
so that

$(\sqrt{\phi}-1)^{2}\ =\ \frac{\sqrt{\phi}-1}{\sqrt{\phi}+1}(\phi-1)\ \geq\
\frac{1}{3}(\phi-1)\qquad\text{for $\phi>4$}.$

Because $(\phi-1)_{+}\leq 3$ for $\phi\leq 4$, it follows that

$\displaystyle\int\displaylimits_{B}(\phi-1)_{+}\,d\mu$ $\displaystyle\ =\
\int\displaylimits_{B\cap\\{\phi\leq 4\\}}(\phi-1)_{+}\,d\mu\
+\int\displaylimits_{B\cap\\{\phi>4\\}}(\phi-1)_{+}\,d\mu$ $\displaystyle\
\leq\ 3\mu(B)+3\int\displaylimits_{B}(\sqrt{\phi}-1)^{2}\,d\mu.$

The first claim follows by combining this with (7.6). The second claim follows
by noting that $\int_{S}(\sqrt{\phi}-1)^{2}\,d\mu=2H(\lambda,\mu)^{2}$ due to
Proposition 3.9. ∎

###### Proof of Theorem 4.4.

Consider sigma-finite intensity measures such that $\lambda\ll\mu$ and
$H(\lambda,\mu)<\infty$. Fix a density $\phi=\frac{d\lambda}{d\mu}$, and a
sequence $S_{n}\uparrow S$ such that $\mu(S_{n})<\infty$ for all $n$.

(i) _Computing a truncated density_. Let $P_{\lambda_{n}}$ and $P_{\mu_{n}}$
be Poisson PP distributions with truncated intensity measures
$\lambda_{n}(B)=\lambda(B\cap S_{n})$ and $\mu_{n}(B)=\mu(B\cap S_{n})$ on
$(S,\mathcal{S})$. We also employ the same notation $\eta_{n}(B)=\eta(B\cap
S_{n})$ for truncations of point patterns $\eta\in N(S)$. We note that
$\lambda_{n}\ll\mu_{n}$, and that $\phi$ serves also as a density of
$\lambda_{n}$ with respect to $\mu_{n}$. Our choice of $S_{n}$ implies that
$\mu_{n}$ is a finite measure. We also note that $\lambda_{n}$ is finite
because $\lambda_{n}(S)\leq
4\mu_{n}(S)+3\int_{S}(\sqrt{\phi}-1)^{2}\,d\mu_{n}=4\mu(S_{n})+6H^{2}(\lambda,\mu)$
due to Lemma 7.3. Theorem 4.1 now implies that $P_{\lambda_{n}}\ll
P_{\mu_{n}}$, with a likelihood ratio given by

$\Phi_{n}(\eta)\ =\
1_{M}(\eta)\,\exp\bigg{(}\int_{S}(1-\phi)\,d\mu_{n}+\int_{S}\log\phi\,d\eta\bigg{)},$
(7.7)

where

$M\ =\ \\{\eta\in N(S)\colon\eta\\{\phi=0\\}=0\\}.$ (7.8)

(ii) _Approximate Laplace functional._ Fix a measurable function $u\colon
S\to\mathbb{R}_{+}$. Monotone convergence of integrals then implies that
$\eta_{n}(u)=\eta(u1_{S_{n}})\uparrow\eta(u)$ for all $\eta\in N(S)$.
Lebesgue’s dominated convergence theorem then implies that

$\int_{N(S)}e^{-\eta(u)}P_{\lambda}(d\eta)\ =\
\lim_{n\to\infty}\int_{N(S)}e^{-\eta_{n}(u)}P_{\lambda}(d\eta).$ (7.9)

By [LP18, Theorem 5.2],

$\displaystyle\int_{N(S)}e^{-\eta_{n}(u)}P_{\lambda}(d\eta)$ $\displaystyle\
=\ \int_{N(S)}e^{-\eta(u)}P_{\lambda_{n}}(d\eta)$ $\displaystyle\ =\
\int_{N(S)}e^{-\eta(u)}\Phi_{n}(\eta)P_{\mu_{n}}(d\eta)$ $\displaystyle\ =\
\int_{N(S)}e^{-\eta_{n}(u)}\Phi_{n}(\eta_{n})P_{\mu}(d\eta).$

Together with (7.9), we conclude that

$\int_{N(S)}e^{-\eta(u)}\,P_{\lambda}(d\eta)\ =\
\lim_{n\to\infty}\int_{N(S)}e^{-\eta_{n}(u)}\Phi_{n}(\eta_{n})\,P_{\mu}(d\eta).$
(7.10)

(iii) _Identifying the limiting density._ Let us next identify the limit of
$\Phi_{n}(\eta_{n})$ as $n\to\infty$. In light of (7.7), we see that

$\Phi_{n}(\eta_{n})\ =\
1_{M}(\eta_{n})\exp\bigg{(}\int_{S_{n}}(1-\phi)\,d\mu+\int_{S_{n}}\log\phi\,d\eta\bigg{)}.$
(7.11)

Even though $S_{n}\uparrow S$, the integrals on the right side above may not
converge as expected because $\int_{S}(1-\phi)\,d\mu$ and
$\int_{S}\log\phi\,d\eta$ are not necessarily well defined. Also, the
compensated integral $\int_{S}\log\phi\,d(\eta-\mu)$ might diverge. A key
observation (proven soon) is that the compensated integral
$\int_{A}\log\phi\,d(\eta-\mu)$ will converge for $P_{\mu}$-almost every
$\eta$, where

$A\ =\ \\{x\in S\colon{\lvert\log\phi(x)\rvert}\leq 1\\}.$

With this target in mind, we will reorganise the integral terms of (7.11)
according to

$\int_{S_{n}}(1-\phi)\,d\mu+\int_{S_{n}}\log\phi\,d\eta\ =\
W_{n}(\eta)+Z_{n}(\eta)+w_{n}+z_{n},$ (7.12)

where

$\begin{aligned} W_{n}(\eta)&=\int_{A\cap S_{n}}\log\phi\,d\eta-\int_{A\cap
S_{n}}\log\phi\,d\mu,\\\ Z_{n}(\eta)&=\int_{A^{c}\cap
S_{n}}\log\phi\,d\eta,\end{aligned}\qquad\begin{aligned} w_{n}&=\int_{A\cap
S_{n}}(\log\phi+1-\phi)\,d\mu,\\\ z_{n}&=\int_{A^{c}\cap
S_{n}}(1-\phi)\,d\eta.\end{aligned}$

We will show that all terms on the right side of (7.12) converge for all
$\eta\in M\cap\Omega$, where $M$ is defined by (7.8) and
$\Omega=\Omega_{1}\cap\Omega_{2}$, where

$\Omega_{1}\ =\ \left\\{\eta\in
N(S)\colon\int_{A^{c}\cap\\{\phi>0\\}}{\lvert\log\phi\rvert}\,d\eta<\infty,\
\eta(A^{c})<\infty\right\\}$

and

$\Omega_{2}\ =\ \left\\{\eta\in
N(S)\colon\text{$\int_{A}\log\phi\,d(\eta-\mu)$ converges},\
\eta(S_{n})<\infty\ \text{for all $n$}\right\\}.$

First, the functions $1_{A^{c}\cap S_{n}}\log\phi$ are dominated in absolute
value by $1_{A^{c}}{\lvert\log\phi\rvert}$ for all $n$. The dominating
function is integrable with respect to any $\eta\in M\cap\Omega$ by the
definition of $\Omega_{1}$. Lebesgue’s dominated convergence theorem then
implies that

$Z_{n}(\eta)\to\int_{A^{c}}\log\phi\,d\eta$ (7.13)

for every $\eta\in M\cap\Omega_{1}$. The definition of $\Omega_{2}$ in turn
implies that

$W_{n}(\eta)\to\int_{A}\log\phi\,d(\eta-\mu)$ (7.14)

for all $\eta\in\Omega_{2}$. We also note that the functions associated with
the definitions of $z_{n}$ and $w_{n}$ converge pointwise according to

$\displaystyle(1-\phi)1_{A^{c}\cap S_{n}}$
$\displaystyle\to(1-\phi)1_{A^{c}},$ $\displaystyle(\log\phi+1-\phi)1_{A\cap
S_{n}}$ $\displaystyle\to(\log\phi+1-\phi)1_{A}.$

By (C.6), $\phi+1\ \leq\ \frac{e+1}{(e^{1/2}-1)^{2}}(\sqrt{\phi}-1)^{2}$ on
$A^{c}$, so that the functions $(1-\phi)1_{A^{c}\cap S_{n}}$ are dominated in
absolute value by $\frac{e+1}{(e^{1/2}-1)^{2}}(\sqrt{\phi}-1)^{2}$. Similarly,
by (C.7), the functions $(\log\phi+1-\phi)1_{A\cap S_{n}}$ are dominated in
absolute value by $2e^{3}(\sqrt{\phi}-1)^{2}$. Both dominating functions are
integrable due to
$\int_{S}(\sqrt{\phi}-1)^{2}\,d\mu=2H^{2}(\lambda,\mu)<\infty$ (recall
Proposition 3.9). Therefore, by dominated convergence, we see that

$\displaystyle z_{n}$ $\displaystyle\ \to\ \int_{A^{c}}(1-\phi)\,d\mu,$ (7.15)
$\displaystyle w_{n}$ $\displaystyle\ \to\ \int_{A}(\log\phi+1-\phi)\,d\mu.$
(7.16)

By plugging (7.13)–(7.16) into (7.12), we find that for all $\eta\in
M\cap\Omega$,

$\lim_{n\to\infty}\left(\int_{S_{n}}(1-\phi)\,d\mu+\int_{S_{n}}\log\phi\,d\eta\right)\
=\ \ell(\eta)$

where

$\displaystyle\ell(\eta)$ $\displaystyle\ =\
\int_{A^{c}}\log\phi\,d\eta+\int_{A}\log\phi\,d(\eta-\mu)$ (7.17)
$\displaystyle\qquad+\int_{A^{c}}(1-\phi)\,d\mu+\int_{A}(\log\phi+1-\phi)\,d\mu.$

By noting that $\eta_{n}\in M$ whenever $\eta\in M\cap\Omega$, we see in light
of (7.11) that

$\lim_{n\to\infty}\Phi_{n}(\eta_{n})\ =\ e^{\ell(\eta)}$

for all $\eta\in M\cap\Omega$. Furthermore, for any $\eta\in M^{c}\cap\Omega$,
we note that $\eta\\{\phi=0\\}\geq 1$, and the fact that $\\{\phi=0\\}\cap
S_{n}\uparrow\\{\phi=0\\}$ then implies that $\eta(\\{\phi=0\\}\cap S_{n})\geq
1$ eventually for all large $n$. Hence for any $\eta\in M^{c}\cap\Omega$,
$1_{M}(\eta_{n})=0$ eventually for all sufficiently large values of $n$. We
also note that $\eta_{n}(u)=\eta(u1_{S_{n}})\uparrow\eta(u)$ by monotone
convergence of integrals. By denoting
$\Phi(\eta)=1_{M\cap\Omega}(\eta)e^{-\eta(u)}e^{\ell(\eta)}$, we see that

$\lim_{n\to\infty}1_{\Omega}(\eta)e^{-\eta_{n}(u)}\Phi_{n}(\eta_{n})\ =\
e^{-\eta(u)}\Phi(\eta)\qquad\text{for all $\eta$}.$ (7.18)

(iv) _Exchanging the limit and integral._ Let just justify that we may
interchange the limit and the integral in (7.10). We know by Lemma B.3 that
$P_{\mu}(\Omega)=1$. Therefore, (7.10) can be written as

$\int_{N(S)}e^{-\eta(u)}\,P_{\lambda}(d\eta)\ =\
\lim_{n\to\infty}\int_{N(S)}1_{\Omega}(\eta)e^{-\eta_{n}(u)}\Phi_{n}(\eta_{n})\,P_{\mu}(d\eta).$
(7.19)

We wish to take the limit inside the integral in (7.19). To justify this, we
note that the functions
$f_{n}=1_{\Omega}(\eta)e^{-\eta_{n}(u)}\Phi_{n}(\eta_{n})$ are bounded by
$0\leq f_{n}\leq g_{n}$, where $g_{n}=\Phi_{n}(\eta_{n})$. We also note [LP18,
Theorem 5.2], that

$\int_{N(S)}g_{n}\,dP_{\mu}\ =\ \int_{N(S)}\Phi_{n}(\eta)\,P_{\mu_{n}}(d\eta)\
=\ \int_{N(S)}P_{\lambda_{n}}(d\eta)\ =\ 1,$

because $\Phi_{n}=\frac{dP_{\lambda_{n}}}{dP_{\mu_{n}}}$. Especially, ${\lvert
f_{n}\rvert}\leq g_{n}$ for all $n$, and
$\sup_{n}\int_{N(S)}g_{n}\,dP_{\mu}<\infty$. A modified version of Lebesgue’s
dominated convergence theorem (Lemma A.3) then justifies exchanging the limit
and integral on the right side of (7.19), and plugging in the limit of (7.18)
shows that for all measurable $u\colon S\to\mathbb{R}_{+}$,

$\int_{N(S)}e^{-\eta(u)}\,P_{\lambda}(d\eta)\ =\
\int_{N(S)}e^{-\eta(u)}\,\Phi(\eta)P_{\mu}(d\eta).$ (7.20)

(v) _Conclusion._ Finally, we note that the formula
$Q(d\eta)=\Phi(\eta)P_{\mu}(d\eta)$ defines a measure on
$(N(S),\mathcal{N}(S))$. By applying (7.20) with $u=0$, we see that
$Q(N(S))=\int_{N(S)}\Phi(\eta)P_{\mu}(d\eta)=P_{\lambda}(N(S))=1$, so that $Q$
is a probability measure. Because the Laplace functional uniquely
characterises [LP18, Proposition 2.10] a probability measure on
$(N(S),\mathcal{N}(S))$, we conclude from (7.20) that $Q=P_{\lambda}$. In
other words, $\Phi$ is a density of $P_{\lambda}$ with respect to $P_{\mu}$.
As an element of $L_{1}(N(S),\mathcal{N}(S),P_{\mu})$, we see that
$\Phi(\eta)=1_{M}(\eta)e^{\ell(\eta)}$, because $P_{\mu}(\Omega)=1$. ∎

###  7.5 Proof of Theorem 5.1

First, Lemma 7.5 proves the claim under an additional condition that $\lambda$
and $\mu$ are finite measures on $S$. This proof is different from the usual
topological approach that is based approximating the measurable sets of $S$ by
a finite sigma-algebra [Lie75, Kar83, Kar91], which usually requires $S$ to be
a separable metric space. Instead, the following proof is based on (i)
representing a Poisson PP distribution with a finite intensity measure using a
Poisson-distributed number of IID random variables (see [Kin67, Rei93, LP18]);
and (ii) representing a Poisson PP distribution with a sigma-finite intensity
measure using a decomposition with respect to a countable partition.

######  Lemma 7.4.

$P_{\lambda}\ll P_{\mu}$ $\implies$ $\lambda\ll\mu$ for any Poisson PP
distributions with sigma-finite intensity measures. Furthermore, the converse
implication $\lambda\ll\mu$ $\implies$ $P_{\lambda}\ll P_{\mu}$ when the
intensity measures are finite.

###### Proof.

Assume that $P_{\lambda}\ll P_{\mu}$. Consider a set $B\subset S$ such that
$\mu(B)=0$. Define $C=\\{\eta\colon\eta(B)>0\\}$. Recall that $\eta(B)$ is
Poisson distributed with mean $\mu(B)$ when $\eta$ is sampled from $P_{\mu}$.
Therefore, $P_{\mu}(C)=1-e^{-\mu(B)}=0$. Now $P_{\lambda}\ll P_{\mu}$ implies
that $0=P_{\lambda}(C)=1-e^{-\lambda(B)}$, from which we conclude that
$\lambda(B)=0$. Hence $\lambda\ll\mu$.

For finite intensity measures $\lambda,\mu$, Theorem 4.1 confirms the reverse
implication $\lambda\ll\mu$ $\implies$ $P_{\lambda}\ll P_{\mu}$. ∎

######  Lemma 7.5.

$R_{\alpha}(P_{\lambda}\|P_{\mu})=T_{\alpha}(\lambda\|\mu)$ for all
$\alpha\in\mathbb{R}_{+}$ and all Poisson PP distributions
$P_{\lambda},P_{\mu}$ with finite intensity measures $\lambda,\mu$.

###### Proof.

Let $P_{\nu}$ be a Poisson PP distribution with intensity measure
$\nu=\lambda+\mu$. Let $f=\frac{d\lambda}{d\nu},g=\frac{d\mu}{d\nu}$ be
densities of $\lambda,\mu$ with respect to $\nu$. Such functions exist by the
Radon–Nikodym theorem [Kal02, Theorem 2.10]. Theorem 4.1 tells that
$P_{\lambda},P_{\mu}$ are absolutely continuous with respect to $P_{\nu}$,
admitting likelihood ratios $F=\frac{dP_{\lambda}}{dP_{\nu}}$ and
$G=\frac{dP_{\mu}}{dP_{\nu}}$ given by

$\displaystyle F(\eta)$ $\displaystyle\ =\
1_{M_{f}}(\eta)\,e^{\nu(1-f)+\eta(\log f)},$ (7.21) $\displaystyle G(\eta)$
$\displaystyle\ =\ 1_{M_{g}}(\eta)\,e^{\nu(1-g)+\eta(\log g)},$

where $M_{f}=\\{\eta\in N(S)\colon\eta\\{f=0\\}=0,\,\eta(S)<\infty\\}$ and
$M_{g}$ is defined similarly, and we abbreviate $\nu(f)=\int f\,d\nu$.

Let us first consider the case with $\alpha\notin\\{0,1\\}$. Then
$R_{\alpha}(P_{\lambda}\|P_{\mu})=\frac{1}{\alpha-1}\log Z_{\alpha}$ where
$Z_{\alpha}=\int_{N(S)}F^{\alpha}G^{1-\alpha}\,dP_{\nu}$. By the standard
conventions $0\cdot\infty=\frac{0}{0}=0$ and $\frac{1}{0}=\infty$, we see that

$Z_{\alpha}\ =\ \begin{cases}\tilde{Z}_{\alpha},&\quad\alpha\in(0,1),\\\
\tilde{Z}_{\alpha}+\infty\cdot
P_{\nu}\\{F>0,G=0\\},&\quad\alpha\in(1,\infty),\end{cases}$ (7.22)

where

$\tilde{Z}_{\alpha}\ =\ \int_{F>0,\,G>0}F^{\alpha}G^{1-\alpha}\,dP_{\nu}.$

To derive a simplified expression for $\tilde{Z}_{\alpha}$, define
$U=\\{f>0,\,g>0\\}$ and consider a set of point patterns $N(U)=\\{\eta\in
N(S)\colon\eta(S)<\infty,\,\eta(U^{c})=0\\}$. In light of (7.21), we see that
$\\{F>0,\,G>0\\}\cap\\{\eta\colon\eta(S)<\infty\\}=M_{f}\cap M_{g}=N(U)$.
Because $P_{\nu}\\{\eta\colon\eta(S)<\infty\\}=1$, it follows that

$\tilde{Z}_{\alpha}\ =\ \int_{N(U)}F^{\alpha}G^{1-\alpha}\,dP_{\nu}.$

We also note that for $\eta\in N(U)$,

$\displaystyle F(\eta)^{\alpha}G(\eta)^{1-\alpha}$ $\displaystyle\ =\
e^{\alpha\nu(1-f)+(1-\alpha)\nu(1-g)}e^{\eta(\log h_{\alpha})},$ (7.23)

where $h_{\alpha}=f^{\alpha}g^{1-\alpha}$. The conditional distribution of a
point pattern $\eta$ sampled from $P_{\nu}$ given $\eta\in N(U)$ equals
(Proposition B.4) $P_{\nu_{U}}$. By also noting that
$P_{\nu}(N(U))=e^{-\nu(U^{c})}$, it follows that

$\int_{N(U)}e^{\eta(\log h_{\alpha})}\,P_{\nu}(d\eta)\ =\
e^{-\nu(U^{c})}\int_{N(S)}e^{\eta(\log h_{\alpha})}\,P_{\nu_{U}}(d\eta).$

Because $\int_{S}({\lvert\log h_{\alpha}\rvert}\wedge
1)\,d\nu_{U}\leq\nu(U)<\infty$ and $x\mapsto\log h(x)$ restricted to $U$ is
$\mathbb{R}$-valued, the Laplace functional formula of Poisson point patterns
[Kal02, Lemma 12.2] implies that $\int_{N(S)}e^{\eta(\log
h_{\alpha})}\,P_{\nu_{U}}(d\eta)=e^{\nu_{U}(h_{\alpha}-1)}$. By integrating
(7.23) with respect to $P_{\nu}$, it follows that

$\displaystyle\tilde{Z}_{\alpha}$ $\displaystyle\ =\
e^{\alpha\nu(1-f)+(1-\alpha)\nu(1-g)}e^{-\nu(U^{c})}e^{\nu_{U}(h_{\alpha}-1)}$
(7.24) $\displaystyle\ =\
e^{-\alpha\nu(f)-(1-\alpha)\nu(g)+\nu_{U}(h_{\alpha})}.$

We are now ready to verify the claim by considering the following five cases
one by one:

  1. (i)

Assume now that $\alpha\in(0,1)$. Then by (7.22), we see that
$R_{\alpha}(P_{\lambda}\|P_{\mu})=\frac{1}{\alpha-1}\log\tilde{Z}_{\alpha}$.
We also note that $\nu_{U}(h_{\alpha})=\nu(h_{\alpha})$, so that by (7.24), we
conclude that

$R_{\alpha}(P_{\lambda}\|P_{\mu})\ =\ \frac{\nu(\alpha
f+(1-\alpha)g-h_{\alpha})}{1-\alpha}.$ (7.25)

The claim follows because the right side equals $T_{\alpha}(\lambda\|\mu)$ by
(3.1).

  2. (ii)

Assume now that $\alpha\in(1,\infty)$ and $\lambda\ll\mu$. Then
$P_{\lambda}\ll P_{\mu}$ by Lemma 7.4. Then $\nu\\{f>0,g=0\\}=0$ and
$P_{\nu}\\{F>0,G=0\\}=0$ (Lemma A.1). In this case we again find that
$\nu_{U}(h_{\alpha})=\nu(h_{\alpha})$. Therefore, in light of (7.22) and
(7.24), we see that (7.25) holds also in this case, and the claim follows.

  3. (iii)

Assume now that $\alpha\in(1,\infty)$ and $\lambda\not\ll\mu$. Then
$P_{\lambda}\not\ll P_{\mu}$ by Lemma 7.4. Then $\nu\\{f>0,g=0\\}>0$ and
$P_{\nu}\\{F>0,G=0\\}>0$ (Lemma A.1). Hence by (7.22),
$R_{\alpha}(P_{\lambda}\|P_{\mu})=\infty$. The assumption $\alpha>1$ implies
that $f^{\alpha}g^{1-\alpha}=\infty$ on the set $\\{f>0,\,g=0\\}$. Therefore
$\int_{S}f^{\alpha}g^{1-\alpha}\,d\nu=\infty$, and we find that
$T_{\alpha}(\lambda\|\mu)=\infty$ by (3.1), and the claim follows.

  4. (iv)

Assume that $\alpha=1$. Now $R_{1}(P_{\lambda}\|P_{\mu})=\lim_{\alpha\uparrow
1}R_{\alpha}(P_{\lambda}\|P_{\mu})$ [vH14]. By (i), we know that
$R_{\alpha}(\lambda\|\mu)=T_{\alpha}(\lambda\|\mu)$ for all $\alpha\in(0,1)$.
The claim follows by letting $\alpha\uparrow 1$ and noting that
$T_{1}(P_{\lambda}\|P_{\mu})=\lim_{\alpha\uparrow
1}T_{\alpha}(P_{\lambda}\|P_{\mu})$ by Theorem 3.1.

  5. (v)

Assume that $\alpha=0$. Formula (2.1) shows that
$R_{0}(P_{\lambda}\|P_{\mu})=-\log P_{\mu}(F>0)=-\log P_{\mu}(M_{f})$. Because
$P_{\mu}(M_{f})=e^{-\mu\\{f=0\\}}$, it follows that
$R_{0}(P_{\lambda}\|P_{\mu})=\mu\\{f=0\\}$. By formula (3.1), we see that
$R_{0}(P_{\lambda}\|P_{\mu})=T_{0}(\lambda\|\mu)$.

∎

With the help of Lemma 7.5 we will prove Theorem 5.1 in the general case where
$\lambda,\mu$ are sigma-finite measures on $S$.

###### Proof of Theorem 5.1.

(i) Fix $\alpha\in(0,\infty)$. Define $\nu=\lambda+\mu$. Then $\nu$ is sigma-
finite. Select a partition (see Lemma A.2) $S=\cup_{n\geq 1}S_{n}$ such that
$\nu(S_{n})<\infty$ for all $n$. Denote $N(S_{n})=\\{\eta\in
N(S)\colon\eta(S_{n}^{c})=0\\}$. Define $\tau\colon
N(S)\to\prod_{n=1}^{\infty}N(S_{n})$ by

$\tau(\eta)\ =\ (\eta_{S_{1}},\eta_{S_{2}},\dots),$ (7.26)

where $\eta_{S_{n}}\in N(S_{n})$ is defined by $\eta_{S_{n}}(B)=\eta(B\cap
S_{n})$. A restriction theorem [LP18, Theorem 5.2] implies that when $\eta$ is
sampled from $P_{\lambda}$, then the restrictions
$\eta_{S_{1}},\eta_{S_{2}},\dots$ are mutually independent Poisson PPs with
intensity measures $\lambda_{S_{1}},\lambda_{S_{2}},\dots$ defined by
$\lambda_{S_{n}}(B)=\lambda(B\cap S_{n})$. Therefore, the pushforward
probability measure $P_{\lambda}\circ\tau^{-1}$ can be written as a product of
Poisson PP distributions $P_{\lambda_{S_{n}}}$, $n\geq 1$. The same reasoning
is valid also for $P_{\mu}$. Hence

$P_{\lambda}\circ\tau^{-1}\ =\
\bigotimes_{n=1}^{\infty}P_{\lambda_{S_{n}}}\qquad\text{and}\qquad
P_{\mu}\circ\tau^{-1}\ =\ \bigotimes_{n=1}^{\infty}P_{\mu_{S_{n}}}.$

Because the sets $S_{1},S_{2},\dots$ form a partition of $S$, we see that the
map $\tau$ defined by (7.26) is a bijection with inverse
$\tau^{-1}(\eta_{1},\eta_{2},\dots)=\sum_{n\geq 1}\eta_{n}$. Standard
arguments show that $\tau$ and $\tau^{-1}$ are measurable mappings (see
Section B.1). Lemma A.4 then implies that
$R_{\alpha}(P_{\lambda}\|P_{\mu})=R_{\alpha}(P_{\lambda}\circ\tau^{-1}\|P_{\mu}\circ\tau^{-1})$.
Because Rényi divergences of order $\alpha>0$ factorise over tensor products
[vH14, Theorem 28] it follows that

$R_{\alpha}(P_{\lambda}\|P_{\mu})\ =\
\sum_{n=1}^{\infty}R_{\alpha}(P_{\lambda_{S_{n}}}\|P_{\mu_{S_{n}}}).$ (7.27)

Because $\lambda,\mu\ll\nu$ and $\nu$ is sigma-finite, the Radon–Nikodym
theorem [Kal02, Theorem 2.10] implies that there exist densities
$f=\frac{d\lambda}{d\nu}$ and $g=\frac{d\mu}{d\nu}$ of $\lambda$ and $\mu$
with respect to $\nu$. Observe now that $\lambda_{S_{n}}(A)=\int_{A\cap
S_{n}}f\,d\nu=\int_{A}f\,d\nu_{S_{n}}$ for all measurable $A\subset S$.
Similarly, $\mu_{S_{n}}(A)=\int_{A}g\,d\nu_{S_{n}}$. We conclude that the
functions $f$ and $g$ also act as densities
$f=\frac{d\lambda_{S_{n}}}{d\nu_{S_{n}}}$ and
$g=\frac{d\mu_{S_{n}}}{d\nu_{S_{n}}}$ of the finite measures
$\lambda_{S_{n}},\mu_{S_{n}}$ with respect to $\nu_{S_{n}}$. Lemma 7.5 now
implies that

$R_{\alpha}(P_{\lambda_{S_{n}}}\|P_{\mu_{S_{n}}})=T_{\alpha}(\lambda_{S_{n}}\|\mu_{S_{n}}).$
(7.28)

Furthermore, by (3.2) in Theorem 3.1, we see that

$T_{\alpha}(\lambda_{S_{n}}\|\mu_{S_{n}})\ =\
\int_{S}R_{\alpha}(p_{f(x)}\|p_{g(x)})\,\nu_{S_{n}}(dx),$

where $p_{s}$ refers to the Poisson distribution $k\mapsto
e^{-sk}\frac{s^{k}}{k!}$ with mean $s$. Observe that the integrand on the
right side above is nonnegative, $d\nu_{S_{n}}=1_{S_{n}}d\nu$, and
$\sum_{n}1_{S_{n}}=1$. Fubini’s theorem combined with (3.2) then shows that

$\displaystyle\sum_{n=1}^{\infty}T_{\alpha}(\lambda_{S_{n}}\|\mu_{S_{n}})\ =\
\int_{S}R_{\alpha}(p_{f(x)}\|p_{g(x)})\,\nu(dx)\ =\ T_{\alpha}(\lambda\|\mu),$

By combining this with (7.27) and (7.28), it follows that
$R_{\alpha}(P_{\lambda}\|P_{\mu})=T_{\alpha}(\lambda\|\mu)$.

(ii) Finally, let use verify that
$R_{0}(P_{\lambda}\|P_{\mu})=T_{0}(\lambda\|\mu)$ under the additional
assumption that $T_{\beta}(\lambda\|\mu)<\infty$ for some $\beta>0$. We saw in
Part (i) of the proof that

$R_{\alpha}(\lambda\|\mu)=T_{\alpha}(\lambda\|\mu)\quad\text{for all
$\alpha\in(0,\infty)$}.$ (7.29)

Now [vH14, Theorem 7] implies that $\alpha\mapsto
R_{\alpha}(P_{\lambda}\|P_{\mu})$ is continuous on $[0,1]$, and Theorem 3.1
implies that $\alpha\mapsto T_{\alpha}(\lambda\|\mu)$ is continuous on
$[0,\beta]$. Hence the claim follows by taking limits $\alpha\to 0$ in (7.29).
∎

##  8 Conclusions

By developing an analytical toolbox of generalised Tsallis divergences for
sigma-finite measures, a framework was derived for analysing likelihood ratios
and Rényi divergences of Poisson PPs on general measurable spaces. The main
advantage of this approach is that it is purely information-theoretic and
completely free of any topological assumptions. This allowed to derive
explicit descriptions of Kullback–Leibler divergences, Rényi divergences,
Hellinger distances, and likelihood ratios for statistical models that admit a
measurable one-to-one map into space of point patterns governed by a Poisson
PP distribution. Marked Poisson PPs corresponding to Poisson PPs on abstract
product spaces provide a rich context for various applications. The
disintegrated Tsallis divergence formula in Section 3.5 is key to
understanding their information-theoretic features. For completing the general
theory, understanding whether the technical condition (3.7) is necessary in
Theorem 3.11 remains an important open problem.

Future directions of extending this work include deriving similar results for
a wider class of statistical models derived from Poisson processes and marked
point patterns, for example Cox processes, Hawkes processes, Poisson shot
noise random measures, and Matérn point patterns. A challenge here is that
mechanisms used to derive the model from a marked point pattern tend to lose
information. For example, Poisson shot noise models in certain limiting
regimes reduce to Gaussian white noises [KLNS07]. The thorough
characterisation of Poisson PP distributions derived in this article is
expected to serve as a cornerstone for this type of further studies.

##  Appendix A Measure theory

###  A.1 Densities

Recall notations from Section 2.1.

######  Lemma A.1.

Let $\lambda,\mu$ be measures on a measurable space $(S,\mathcal{S})$
admitting densities $f=\frac{d\lambda}{d\nu}$ and $g=\frac{d\mu}{d\nu}$ with
respect to a measure $\nu$.

  1. (i)

$\lambda\\{f=0\\}=0$ and $\mu\\{g=0\\}=0$.

  2. (ii)

For any measurable set $B$, $\lambda(B)=0$ if and only if $\nu(\\{f>0\\}\cap
B)=0$.

  3. (iii)

$\lambda\ll\mu$ if and only if $\lambda\\{g=0\\}=0$.

  4. (iv)

$\lambda\ll\mu$ if and only if $\nu\\{f>0,\,g=0\\}=0$.

  5. (v)

$\lambda\perp\mu$ if and only if $\nu\\{f>0,\,g>0\\}=0$.

  6. (vi)

$\lambda\perp\mu$ if and only if $\lambda\\{g>0\\}=0$.

###### Proof.

(i) $\lambda\\{f=0\\}=\int_{\\{f=0\\}}f\,d\nu=0.$ Analogously we see that
$\mu\\{g=0\\}=0$.

(ii) Fix a measurable set $B$, and denote $A=\\{f>0\\}\cap B$. Note that (i)
implies that $\lambda(B)=\lambda(A)=\int_{A}f\,d\nu$. Hence $\nu(A)=0$ implies
that $\lambda(B)=0$. Assume next that $\nu(A)>0$. Then $\nu(A_{n})>0$ for some
integer $n\geq 1$, where $A_{n}=\\{f\geq n^{-1}\\}\cap B$. Hence
$\lambda(B)\geq\lambda(A_{n})=\int_{A_{n}}f\,d\nu\geq n^{-1}\nu(A_{n})>0$.

(iii) Assume that $\lambda\\{g=0\\}=0$, and consider a set $B$ such that
$\mu(B)=0$. By applying (ii) for $\mu$, we find that $\nu(\\{g>0\\}\cap B)=0$.
Then $\lambda\ll\nu$ implies that $\lambda(\\{g>0\\}\cap B)=0$. Then
$\lambda(B)=\lambda(\\{g>0\\}\cap B)+\lambda(\\{g=0\\}\cap B)=0$ due to
assumption $\lambda\\{g=0\\}=0$. Hence $\lambda(B)=0$, and we conclude that
$\lambda\ll\mu$. The converse implication
$\lambda\ll\mu\implies\lambda\\{g=0\\}=0$ is immediate from (i).

(iv) By applying (ii) with $B=\\{g=0\\}$ we find that $\lambda\\{g=0\\}$ is
equivalent to $\nu\\{f>0,\,g=0\\}=0$. The claim now follows by (iii).

(v) Assume that $\nu\\{f>0,\,g>0\\}=0$. Let $B=\\{f>0\\}$. Then
$\lambda(B^{c})=0$ due to (i). Furthermore by (i),
$\mu(B)=\mu(B\cap\\{g>0\\})=\mu\\{f>0,\,g>0\\}$. Hence $\mu(B=0)$ due to
$\mu\ll\nu$. Hence $\lambda\perp\mu$. Assume now that $\lambda\perp\mu$. Then
there exists a set $B$ such that $\lambda(B^{c})=0$ and $\mu(B)=0$. Then (ii)
implies that $\nu(\\{f>0\\}\cap B^{c})=0$ and $\nu(\\{g>0\\}\cap B)=0$. Then

$\displaystyle\nu(\\{f>0,g>0\\})\ \leq\ \nu(\\{f>0\\}\cap
B^{c})+\nu(\\{g>0\\}\cap B)\ =\ 0.$

(vi) Assume that $\lambda\\{g>0\\}=0$. Let $B=\\{g=0\\}$. Then
$\lambda(B^{c})=0$, and $\mu(B)=0$ due to (i). Hence $\lambda\perp\mu$. Assume
now that $\lambda\perp\mu$. Then there exists a set $B$ such that
$\lambda(B^{c})=0$ and $\mu(B)=0$. Then by (ii), we see that
$\nu(\\{g>0\\}\cap B)=0$. Then $\lambda\ll\nu$ implies that
$\lambda(\\{g>0\\}\cap B)=0$. Then

$\displaystyle\lambda\\{g>0\\}\ =\ \lambda(\\{g>0\\}\cap
B)+\lambda(\\{g>0\\}\cap B^{c})\ =\ 0.$

∎

###  A.2 Basic measure theory

######  Lemma A.2.

If $\nu$ is a sigma-finite measure on a measurable space $(S,\mathcal{S})$,
then there exists a partition $S=\cup_{n\geq 1}S_{n}$ such that
$\nu(S_{n})<\infty$ for all $n$.

###### Proof.

Because $\nu$ is sigma-finite, there exist measurable sets such that
$\cup_{n\geq 1}C_{n}=S$ and $\nu(C_{n})<\infty$ for all $n$. Define
$S_{0}=\emptyset$ and $S_{n}=C_{n}\setminus S_{n-1}$ for $n\geq 1$. Then the
sets $S_{1},S_{2},\dots$ are mutually disjoint, and $S=\cup_{n\geq 1}S_{n}$,
together with $\nu(S_{n})\leq\nu(C_{n})<\infty$ for all $n$. ∎

The following result is a convenient alternative form of Lebesgue’s dominated
convergence theorem that quantifies uniform integrability (boundedness in the
increasing convex stochastic order [LV13]) in a flexible manner.

######  Lemma A.3.

Let $f,f_{1},f_{2},\dots$ and $g_{1},g_{2},\dots$ be measurable real-valued
functions on $(S,\mathcal{S})$ such that $f_{n}\to f$, $g_{n}\to g$, ${\lvert
f_{n}\rvert}\leq g_{n}$ for all $n$, and $\sup_{n}\int_{S}g_{n}\,d\mu<\infty$.
Then $\int_{S}f_{n}\,d\mu\to\int_{S}f\,d\mu$, and $\int_{S}{\lvert
f\rvert}\,d\mu\leq\sup_{n}\int_{S}g_{n}\,d\mu$.

###### Proof.

We note that $g=\liminf_{n\to\infty}g_{n}$, and that Fatou’s lemma [Kal02,
Lemma 1.20] implies $\int
g\,d\mu=\int(\liminf_{n\to\infty}g_{n})\,d\mu\leq\liminf_{n\to\infty}\int
g_{n}\,d\mu\leq\sup_{n}\int g_{n}\,d\mu$. Then $\int g\,d\mu$ is finite, and
Kallenberg’s version of Lebesgue’s dominated convergence theorem [Kal02,
Theorem 1.21] yields the first claim. For the second claim, we note by Fatou’s
lemma that $\int_{S}{\lvert f\rvert}\,d\mu=\int_{S}\liminf{\lvert
f_{n}\rvert}\,d\mu\leq\liminf\int_{S}{\lvert
f_{n}\rvert}\,d\mu\leq\sup_{n}\int_{S}g_{n}\,d\mu$. ∎

###  A.3 Measurable bijections

######  Lemma A.4.

Let $S,T$ be measurable spaces, and let $\phi\colon S\to T$ be a measurable
bijection with a measurable inverse. Then
$R_{\alpha}(P\circ\phi^{-1}\|\,Q\circ\phi^{-1})=R_{\alpha}(P\|Q)$ for all
$\alpha>0$.

###### Proof.

Let $P,Q$ be probability measures on $S$. Fix densities $p=\frac{dP}{dm}$ and
$q=\frac{dQ}{dm}$ with respect to $m=P+Q$. Define a function $\tilde{p}\colon
T\to\mathbb{R}_{+}$ by $\tilde{p}=p\circ\phi^{-1}$ and a measure $\tilde{m}$
on $T$ by $\tilde{m}=m\circ\phi^{-1}$. Note that for any measurable $A\subset
T$,

$\displaystyle\int_{A}\tilde{p}\,d\tilde{m}\ =\
\int_{T}1_{A}(y)\tilde{p}(y)\,\tilde{m}(dy)\ =\
\int_{S}1_{A}(\phi(x))\tilde{p}(\phi(x))\,m(dx).$

Because $\tilde{p}(\phi(x))=p(x)$ for all $x$, we see that

$\displaystyle\int_{A}\tilde{p}\,d\tilde{m}\ =\
\int_{\phi^{-1}(A)}p(x)\,m(dx)\ =\ P(\phi^{-1}(A)).$

We conclude that $\tilde{p}=\frac{dP\circ\phi^{-1}}{dm\circ\phi^{-1}}$ is a
density of $P\circ\phi^{-1}$ with respect to $\tilde{m}$. Similarly, we see
that $\tilde{q}=q\circ\phi^{-1}$ is a density of $Q\circ\phi^{-1}$ with
respect to $\tilde{m}$. Hence,

$\displaystyle\int_{T}(p\circ\phi^{-1})^{\alpha}(q\circ\phi^{-1})^{1-\alpha}d\tilde{m}$
$\displaystyle\ =\
\int_{S}(p\circ\phi^{-1}(\phi(x)))^{\alpha}(q\circ\phi^{-1}(\phi(x))^{1-\alpha}m(dx)$
$\displaystyle\ =\ \int_{S}(p(x))^{\alpha}(q(x))^{1-\alpha}m(dx)$
$\displaystyle\ =\ \int_{S}p^{\alpha}q^{1-\alpha}\,dm.$

From this the claim follows for $\alpha\neq 1$. The case with $\alpha=1$ is
similar. ∎

##  Appendix B Point patterns

###  B.1 Measurability of sigma-finite decompositions

This section discusses a decomposition of a point pattern with respect to a
countable partition of the ground space $S$. Let $(S,\mathcal{S})$ be a
measurable space. Let $N(S)$ be the set of point patterns (measures with
values in $\mathbb{Z}_{+}\cup\\{\infty\\}$) on $(S,\mathcal{S})$ equipped with
the sigma-algebra $\mathcal{N}(S)=\sigma(\operatorname{ev}_{B}\colon
B\in\mathcal{S})$ generated by the evaluation maps
$\operatorname{ev}_{B}\colon\eta\mapsto\eta(B)$.

Assume that $S_{1},S_{2},\dots\in\mathcal{S}$ are disjoint and such that
$S=\cup_{n=1}^{\infty}S_{n}$. We define $N(S_{n})=\\{\eta\in
N(S)\colon\eta(S_{n}^{c})=0\\}$ and equip this set with the trace sigma-
algebra $\mathcal{N}(S_{n})=\mathcal{N}(S)\cap N(S_{n})$. We define the
truncation map $\tau_{S_{n}}\colon N(S)\to N(S_{n})$ by

$(\tau_{S_{n}}\eta)(B)\ =\ \eta(B\cap S_{n}),\quad B\in\mathcal{S}.$

Then we define $\tau\colon N(S)\to\prod_{n=1}^{\infty}N(S_{n})$ by

$\tau(\eta)\ =\ (\tau_{S_{1}}(\eta),\tau_{S_{2}}(\eta),\dots).$ (B.1)

We find that $\tau$ is a bijection with inverse

$\tau^{-1}(\eta_{1},\eta_{2},\dots)\ =\ \sum_{n=1}^{\infty}\eta_{n}.$

We equip $\prod_{n=1}^{\infty}N(S_{n})$ with the product sigma-algebra
$\bigotimes_{n=1}^{\infty}\mathcal{N}(S_{n})$.

######  Lemma B.1.

$\tau\colon(N(S),\mathcal{N}(S))\to(\prod_{n=1}^{\infty}N(S_{n}),\bigotimes_{n=1}^{\infty}\mathcal{N}(S_{n}))$
is a measurable bijection with a measurable inverse.

###### Proof.

Denote $\mathcal{C}(S)=\\{\operatorname{ev}_{B}^{-1}(\\{k\\})\colon
B\in\mathcal{S},\,k\in\mathbb{Z}_{+}\\}$ and note that this set family
generates the sigma-algebra $\mathcal{N}(S)$. By [Shi96, Lemma II.3.3], we
know that the set family $\mathcal{C}(S)\cap N(S_{n})$ generates the trace
sigma-algebra $\mathcal{N}(S_{n})=\mathcal{N}(S)\cap N(S_{n})$.

We start by by verifying that $\tau_{S_{n}}\colon N(S)\to N(S_{n})$ is
measurable. Fix a set $B\in\mathcal{S}$ and an integer $k\geq 0$, and consider
a set in $\mathcal{C}(S)\cap N(S_{n})$ of form

$C\ =\ \\{\eta\in N(S)\colon\eta(S_{n}^{c})=0,\,\eta(B)=k\\}.$

Then

$\displaystyle\tau_{S_{n}}^{-1}(C)$ $\displaystyle\ =\ \\{\eta\in
N(S)\colon\eta(S_{n}^{c}\cap S_{n})=0,\,\eta(B\cap S_{n})=k\\}$
$\displaystyle\ =\ \\{\eta\in N(S)\colon\eta(B\cap S_{n})=k\\}$

shows that $\tau_{S_{n}}^{-1}(C)\in\mathcal{N}(S)$. Because such sets $C$
generate $\mathcal{N}(S_{n})$, it follows [Kal02, Lemma 1.4] that
$\tau_{S_{n}}$ is measurable. Because each coordinate map $\tau_{S_{n}}$, it
follows [Kal02, Lemma 1.8] that $\tau$ is measurable.

Let us now verify that the inverse map
$\tau^{-1}\colon\prod_{n=1}^{\infty}N(S_{n})\to N(S)$ is measurable. Fix a set
$B\in\mathcal{S}$ and an integer $k\geq 0$, and consider a set in
$\mathcal{C}(S)$ of form

$C\ =\ \\{\eta\in N(S)\colon\eta(B)=k\\}.$

Then

$\displaystyle(\tau^{-1})^{-1}(C)$ $\displaystyle\ =\
\\{(\eta_{1},\eta_{2},\dots)\colon\sum_{n=1}^{\infty}\eta_{n}(B)=k\\}.$

Let $Z_{k}$ be the collection of integer-valued measures
$z=\sum_{n=1}^{\infty}z_{n}\delta_{n}$ on $\mathbb{N}=\\{1,2,\dots\\}$ with
total mass $\sum_{n=1}^{\infty}z_{n}=k$. Then

$\displaystyle(\tau^{-1})^{-1}(C)$ $\displaystyle\ =\ \bigcup_{z\in
Z_{k}}\\{(\eta_{1},\eta_{2},\dots)\colon\eta_{n}(B)=z_{n}\ \text{for all
$n$}\\}$ $\displaystyle\ =\ \bigcup_{z\in
Z_{k}}\bigcap_{n=1}^{\infty}\\{(\eta_{1},\eta_{2},\dots)\colon\eta_{n}(B)=z_{n}\\}$

shows that
$(\tau^{-1})^{-1}(C)\in\bigotimes_{n=1}^{\infty}\mathcal{N}(S_{n})$. Because
such sets $C$ generate $\mathcal{N}(S)$, it follows [Kal02, Lemma 1.4] that
$\tau^{-1}$ is measurable. ∎

###  B.2 Compensated Poisson integrals

Given measurable sets $S_{n}\uparrow S$ and measures $\eta,\mu$ on a
measurable space $(S,\mathcal{S})$, we say that the compensated integral

$\int_{A}f\,d(\eta-\mu)\ =\ \lim_{n\to\infty}\left(\int_{A\cap
S_{n}}f\,d\eta-\int_{A\cap S_{n}}f\,d\mu\right)$ (B.2)

of a measurable function $f\colon S\to\mathbb{R}$ over a measurable set
$A\subset S$ converges if $\int_{A\cap S_{n}}{\lvert
f\rvert}\,d\eta+\int_{A\cap S_{n}}{\lvert f\rvert}\,d\mu<\infty$ for all $n$,
and the limit in (B.2) exists in $\mathbb{R}$. The following two results
characterise the convergence of compensated integrals when $\eta$ is sampled
from a Poisson PP distribution with intensity measure $\mu$. These are needed
for proving Theorem 4.4.

######  Lemma B.2.

Let $P_{\mu}$ be a Poisson PP distribution with a sigma-finite intensity
measure $\mu$ such that $\mu(S_{n})<\infty$ for all $n$. For any bounded
function $f\colon S\to\mathbb{R}$ such that $\int_{A}f^{2}\,d\mu<\infty$, the
compensated integral $\int_{A}f\,d(\eta-\mu)$ converges for $P_{\mu}$-almost
every $\eta\in N(S)$.

###### Proof.

Define $U_{n}=S_{n}\setminus S_{n-1}$ for $n\geq 1$, where $S_{0}=\emptyset$.
Because $f$ is bounded, we see that $\int_{U_{n}}{\lvert
f\rvert}\,d\mu\leq\|f\|_{\infty}\mu(S_{n})<\infty$. Campbell’s theorem [Kin93,
Section 3.2] then implies that

$W_{n}(\eta)\ =\ \int_{U_{n}}f\,d\eta-\int_{U_{n}}f\,d\mu$

defines a real-valued random variable on probability space
$(N(S),\mathcal{N}(S),P_{\mu})$ with mean $E_{\mu}W_{n}=0$ and variance
$E_{\mu}W_{n}^{2}=\int_{U_{n}}f^{2}\,\mu$. Because the sets $U_{n}$ are
disjoint, the random variables $W_{n}$ are independent. Furthermore,
$E_{\mu}\sum_{n=1}^{\infty}W_{n}^{2}=\int_{S}f^{2}\,d\mu<\infty$. The
Khinchin–Kolmogorov variance criterion [Kal02, Lemma 4.16] then implies that
the sum $W=\sum_{n=1}^{\infty}W_{n}$ converges almost surely. Hence $W$, or
equivalently the right side of formula (B.2), is a well-defined real-valued
random variable on the probability space $(N(S),\mathcal{N}(S),P_{\mu})$. In
particular $W(\eta)\in\mathbb{R}$ for $P_{\mu}$-almost every $\eta$. ∎

######  Lemma B.3.

Let $P_{\mu}$ be a Poisson PP distribution with a sigma-finite intensity
measure $\mu$. Let $\phi\colon S\to\mathbb{R}_{+}$ be such that
$\int_{S}(\sqrt{\phi}-1)^{2}\,d\mu<\infty$, and denote $A=\\{x\in
S\colon{\lvert\log\phi(x)\rvert}\leq 1\\}$. Assume that $S_{n}\uparrow S$ and
$\mu(S_{n})<\infty$ for all $n$. Then the sets

$\Omega_{1}\ =\ \left\\{\eta\in
N(S)\colon\int_{A^{c}\cap\\{\phi>0\\}}{\lvert\log\phi\rvert}\,d\eta<\infty,\
\eta(A^{c})<\infty\right\\}$

and

$\Omega_{2}\ =\ \left\\{\eta\in
N(S)\colon\text{$\int_{A}\log\phi\,d(\eta-\mu)$ converges},\
\eta(S_{n})<\infty\ \text{for all $n$}\right\\}$

satisfy $P_{\mu}(\Omega_{1})=1$ and $P_{\mu}(\Omega_{2})=1$.

###### Proof.

We note that $\mu(A^{c})=\int_{A^{c}}d\mu\leq\int_{A^{c}}(\phi+1)\,d\mu$. By
(C.6), $\phi+1\ \leq\ \frac{e+1}{(e^{1/2}-1)^{2}}(\sqrt{\phi}-1)^{2}$ on
$A^{c}$. It follows that $\mu(A^{c})$ is finite. Campbell’s formula then
implies $\int_{N(S)}\eta(A^{c})P_{\mu}(d\eta)=\mu(A^{c})<\infty$. Hence the
set $\Omega_{1}^{\prime}=\\{\eta\colon\eta(A^{c})<\infty\\}$ satisfies
$P_{\mu}(\Omega_{1}^{\prime})=1$. We also note that for any
$\eta\in\Omega_{1}^{\prime}$, the restriction of $\eta$ into $A^{c}$ can be
written as a finite sum $\eta_{A^{c}}=\sum_{i}\delta_{x_{i}}$ with $x_{i}\in
S$ such that $\phi(x_{i})>0$, so that
$\int_{A^{c}\cap\\{\phi>0\\}}{\lvert\log\phi\rvert}\,d\eta=\sum_{i}{\lvert\log\phi(x_{i})\rvert}$
is finite. Therefore, $\Omega_{1}^{\prime}=\Omega_{1}$, and we conclude that
$P_{\mu}(\Omega_{1})=1$.

Let

$\Omega_{2}^{\prime}\ =\ \Big{\\{}\eta\in N(S)\colon\eta(S_{n})<\infty\
\text{for all $n$}\Big{\\}}.$

Because $\mu(S_{n})<\infty$ for all $n$, we see that
$P_{\mu}(\Omega_{2}^{\prime})=1$. Observe that $f=1_{A}\log\phi$ is bounded,
and by (C.8),

$\int_{S}f^{2}\,d\mu\ =\ \int_{A}\log^{2}\phi\,d\mu\ \leq\
4e^{3}\int_{S}(\sqrt{\phi}-1)^{2}\ <\ \infty.$

Lemma B.2 tells that $P_{\mu}(\Omega_{2})=1$. ∎

###  B.3 Truncated Poisson PPs

Let $(S,\mathcal{S})$ be a measurable space. Given a set $U\in\mathcal{S}$ and
a measure $\lambda$, define $\lambda_{U}(A)=\lambda(A\cap U)$ for
$A\in\mathcal{S}$. Then $\lambda_{U}$ is a measure on $(S,\mathcal{S})$. We
denote the truncation map by $\pi_{U}\colon\lambda\mapsto\lambda_{U}$.

When $\eta$ is sampled from a Poisson PP distribution $P_{\lambda}$ with a
sigma-finite intensity measure $\lambda$ on $S$, then the probability
distribution of the random point pattern $\eta_{U}$ is given by
$P_{\lambda}\circ\pi_{U}^{-1}$. The following proposition confirms that the
law of $\eta_{U}$ is a Poisson PP distribution with truncated intensity
measure $\lambda_{U}$, and that that $P_{\lambda_{U}}$ is also the conditional
distribution of $\eta$ sampled from $P_{\lambda}$ given that $\eta(U^{c})=0$.

######  Proposition B.4.

For any measurable set $U\subset S$:

  1. (i)

$P_{\lambda}\circ\pi_{U}^{-1}=P_{\lambda_{U}}$.

  2. (ii)

$P_{\lambda}\\{\eta\colon\eta\in
A,\,\eta(U^{c})=0\\}=e^{-\lambda(U^{c})}P_{\lambda_{U}}(A)$ for all measurable
$A\subset S$.

###### Proof.

(i) This follows by [LP18, Theorem 5.2].

(ii) We note that $\eta=\eta_{U}$ for any $\eta$ such that $\eta(U^{c})=0$,
and that $\eta(U^{c})=0$ if and only if $\eta_{U^{c}}=0$. We also note ([LP18,
Theorem 5.2]) that the random point patterns $\eta_{U}$ and $\eta_{U^{c}}$ are
independent when $\eta$ is sampled from $P_{\lambda}$. Therefore, by (i),

$\displaystyle P_{\lambda}\\{\eta\colon\eta\in A,\,\eta(U^{c})=0\\}$
$\displaystyle\ =\ P_{\lambda}\\{\eta\colon\eta_{U}\in A,\,\eta_{U^{c}}=0\\}$
$\displaystyle\ =\ P_{\lambda_{U}}(A)\,P_{\lambda_{U^{c}}}(\\{0\\}).$

By noting that $P_{\lambda_{U^{c}}}(\\{0\\})=P_{\lambda_{U^{c}}}(\eta(S)=0)$
and noting that $\eta(S)$ is Poisson-distributed with parameter
$\lambda(U^{c})$ when $\eta$ is sampled from $P_{\lambda_{U^{c}}}$, we see
that $P_{\lambda_{U^{c}}}(\\{0\\})=e^{-\lambda(U^{c})}$. Hence the claim
follows. ∎

##  Appendix C Elementary analysis

######  Lemma C.1.

For all $x>-1$,

$\displaystyle\frac{x}{1+x}$ $\displaystyle\ \leq\ \log(1+x)\ \leq\ x,$ (C.1)
$\displaystyle x-\frac{x^{2}}{2(1-x_{-})^{2}}$ $\displaystyle\ \leq\
\log(1+x)\ \leq\ x-\frac{x^{2}}{2(1+x_{+})^{2}},$ (C.2)

and for all $y>0$,

${\lvert\log y\rvert}\ \leq\ \frac{{\lvert y-1\rvert}}{y\wedge 1}$ (C.3)

and

$0\ \leq\ y-1-\log y\ \leq\ \frac{(y-1)^{2}}{2(y\wedge 1)^{2}}.$ (C.4)

###### Proof.

Fix a number $x>-1$. Define $f(t)=\log(1+tx)$ for $t\in[0,1]$. Note that
$f^{\prime}(t)=x(1+tx)^{-1}$ and $f^{\prime\prime}(t)=-x^{2}(1+tx)^{-2}$. Then
the formula $f(1)=f(0)+\int_{0}^{1}f^{\prime}(r)\,dr$ implies that

$\log(1+x)\ =\ \int_{0}^{1}\frac{x}{1+rx}\,dr.$

Then (C.1) follows by noting that that $\frac{x}{1+x}\leq\frac{x}{1+rx}\leq x$
for all $0\leq r\leq 1$. Similarly, the formula
$f(1)=f(0)+f^{\prime}(0)+\int_{0}^{1}\int_{0}^{s}f^{\prime\prime}(r)\,dr\,ds$
implies that

$\log(1+x)\ =\ x-\int_{0}^{1}\int_{0}^{s}\frac{x^{2}}{(1+rx)^{2}}\,dr\,ds.$

Then (C.2) follows by noting that $1-x_{-}\leq 1+rx\leq 1+x_{+}$ for all
$0\leq r\leq 1$.

Fix a number $y>0$. By substituting $x=y-1$ into (C.1), we see that
$\frac{y-1}{y}\leq\log y\leq y-1$. Hence $-\frac{{\lvert
y-1\rvert}}{y}\leq\log y\leq{\lvert y-1\rvert}$, and (C.3) follows. By
substituting $x=y-1$ into (C.2), we see that

$\frac{(y-1)^{2}}{2(1+(y-1)_{+})^{2}}\ \leq\ y-1-\log y\ \leq\
\frac{(y-1)^{2}}{2(1-(y-1)_{-})^{2}}.$

Now (C.4) follows by noting that $1-(y-1)_{-}=y\wedge 1$. ∎

######  Lemma C.2.

If $t\geq 0$ satisfies ${\lvert t-1\rvert}\geq c$ for some $c>0$, then
$t+1\leq C(\sqrt{t}-1)^{2}$ where

$C\ =\ \begin{cases}\frac{2-c}{(\sqrt{1-c}-1)^{2}},&\quad 0<c\leq 1,\\\
\frac{2+c}{(\sqrt{1+c}-1)^{2}},&\quad c>1.\end{cases}$

###### Proof.

Differentiation shows that
$r(t)=\frac{(\sqrt{t}-1)^{2}}{t+1}=1-\frac{2\sqrt{t}}{t+1}$ is strictly
decreasing on $[0,1]$ and strictly increasing on $[1,\infty)$.

(i) Assume that $0<c\leq 1$. Then ${\lvert t-1\rvert}\geq c$ implies that
either $t\leq 1-c$ or $t\geq 1+c$. In the former case $r(t)\geq r(1-c)$, and
in the latter case $r(t)\geq r(1+c)$. Hence $t+1\leq(r(1-c)^{-1}\wedge
r(1+c)^{-1})(\sqrt{t}-1)^{2}=r(1-c)^{-1}(\sqrt{t}-1)^{2}$.

(ii) Assume that $c>1$. Then ${\lvert t-1\rvert}\geq c$ implies that $t\geq
1+c$, so that $r(t)\geq r(1+c)$. Hence $t+1\leq r(1+c)^{-1}(\sqrt{t}-1)^{2}$.
∎

######  Lemma C.3.

For all $0\leq t\leq c$,

${\lvert t-1\rvert}\ \leq\ (1+c^{1/2}){\lvert\sqrt{t}-1\rvert}.$ (C.5)

For all $0\leq t\leq c^{-1}$ and all $t\geq c$ with $c>1$,

$t+1\ \leq\ \frac{c+1}{(c^{1/2}-1)^{2}}(\sqrt{t}-1)^{2}.$ (C.6)

###### Proof.

(i) Because ${\lvert
t-1\rvert}={\lvert\sqrt{t}+1\rvert}{\lvert\sqrt{t}-1\rvert}$, we see that
(C.5) follows by noting that ${\lvert\sqrt{t}+1\rvert}\leq 1+c^{1/2}$ for all
$0\leq t\leq c$.

(ii) Differentiation shows that
$r(t)=\frac{(\sqrt{t}-1)^{2}}{t+1}=1-\frac{2\sqrt{t}}{t+1}$ is decreasing on
$[0,1]$ and increasing on $[1,\infty)$. Hence $r(t)\geq r(c)$ for $t\geq c$
and $r(t)\geq r(c^{-1})$ for $0\leq t\leq c^{-1}$. Because
$r(c)=r(c^{-1})=\frac{(c^{1/2}-1)^{2}}{c+1}$, we conclude (C.6). ∎

######  Lemma C.4.

For all $t$ such that ${\lvert\log t\rvert}\leq 1$,

$\displaystyle{\lvert\log t+1-t\rvert}$ $\displaystyle\ \leq\
2e^{3}(\sqrt{t}-1)^{2},$ (C.7) $\displaystyle\log^{2}t$ $\displaystyle\ \leq\
4e^{3}(\sqrt{t}-1)^{2},$ (C.8)

###### Proof.

Fix a number $t$ such that ${\lvert\log t\rvert}\leq 1$. Then $e^{-1}\leq
t\leq e$. Then (C.4) implies that ${\lvert\log
t+1-t\rvert}\leq\frac{(t-1)^{2}}{2(t\wedge
1)^{2}}\leq\frac{1}{2}e^{2}(t-1)^{2}.$ By combining this with (C.5),
inequality (C.7) follows. Furthermore, (C.3) implies that ${\lvert\log
t\rvert}\leq\frac{{\lvert t-1\rvert}}{t\wedge 1}\leq e{\lvert t-1\rvert}$. By
combining this with (C.5), we that (C.8) is valid. ∎

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