This is a follow up of this previous question. I'm trying to understand the following proposition from An Introduction to the Theory of Point Processes Volume I: Elementary Theory and Methods by Daley and Vere-Jones (page 138)
Proposition 5.4.V. (a) A necessary and sufficient condition for a point process to be simple is that, for all $n = 1, 2,...,$ the associated Janossy measure $J_n(·)$ allots zero mass to the ‘diagonals’ $\{x_i = x_j\}$. (b) When $\mathcal{X} = \mathbb{R}^d$, the process is simple if for all such $n$ the Janossy measures have densities $j_n(·)$ with respect to $(nd)$-dimensional Lebesgue measure.
In particular, the (b) part confuses me. Since I'm just a Phd student in engineering with not a very strong background in measure theory, I would like to have a helping hand to understand such proposition from the point of view of the Radon-Nikodym theorem.
Background
For my purpose, we can view a point process as a collection of $n$ random points $x_1, \dots, x_n$ drawn from $\mathbb{R}^d$, where $n$ is also random. If the points $x_1, \dots, x_n$ are distinct, i.e., $i \neq j \Rightarrow x_i \neq x_j$, then the point process is considered simple. Consequently, a simple point process can be seen as a random set where the points $x_1, \dots, x_n$ represent $n$ distinct locations in $\mathbb{R}^d$. Now, if the point process is not simple, the points $x_1, \dots, x_n$ do not identify $n$ different locations, but rather $\eta < n$ different locations $\xi_1, \dots, \xi_\eta$, depending on how many repeated points we have in $x_1, \dots, x_n$.
In that case, in order to not lose the information carried out by the point process, we can see the point process as a random multiset, or as a labeled random set where the element $x_i\in\mathbb{R}^d$ is replaced by its labelled counterpart $\mathbf{x}_i\triangleq[x_i', i]' \in \mathbb{R}^d\times \mathbb{N}$.
In my humble opinion, the main difference between a simple point process and a non-simple point process is as follows:
- if a point process is simple, then it is completely identified by the locations in which the points $x_1,\dots,x_n$ are generated;
- if a point process is not simple, then it is not completely identified by the locations in which the points $x_1,\dots,x_n$ are generated. Rather, we have two possible solutions to resolve the ambiguity: (a) declare the multiplicity of each location; (b) keep the locations distinguished by appending an identifier (the label).
I don't see any problem with the existance of the probability density here.
Example
Consider a point process simulator that generates random points; however, it only provides the locations where the points are generated. For instance, suppose we are in $\mathbb{R}$, and the generator returns the values $0.123, 1.432, -2.001$.
- If we know that the point process is simple, we can say that the realization of the point process is the set ${0.123, 1.432, -2.001}$.
- If we don't know that the point process is simple, then we face difficulties because there is no way to understand how many points are generated by the simulator (in our case, we only know that there are at least 3). Hence, we need a more refined simulator that provides us with additional information, such as: (a) the number of points generated at each location; or (b) the complete list of points generated.
This example summarizes my understanding of the difference between a simple and non-simple point process. I comprehend that simple point processes are somewhat 'easier' to handle because they encode less information compared to non-simple point processes. However, I don't see any connection with characterization in terms of probability densities.
Point Process Densities
A point process can be simulated according to the following intuitive algorithm
- generate a random integer $N$ according to a probability distribution $p_n$;
- given the realization $N=n$, generated $n$ points $x_1,\dots,x_n$ according to a joint probability measure $\Pi_n(A_1,\dots,A_n)$ defined over a partition $A_1,\dots,A_n$ of the hyperspace $(\mathbb{R}^d)^n$.
Thus, a point process is completely characterized by $p_n$ and the family of distributions ${\Pi_i}{i\in N{+}}$, where $N_{+}$ is the support of $p_n$. Since we are modeling random sets or random multisets, we don't observe any difference in points generated in different orders but in the same locations. This observation translates into the requirement that $\Pi_n(A_1,\dots,A_n)$ must be symmetric, meaning that we only consider probability measures that do not change under permutations of the partition $A_1,\dots,A_n$.
Without loss of information, we can merge $p_n$ with the joint measures obtaining the Janossy measures \begin{equation*} J_n(A_1,\dots,A_n)\triangleq n!\,p_n\,\Pi_n(A_1,\dots,A_n) \end{equation*} which provides a more compact probabilistic description of a generic point process. These are non-normalized measures but are still symmetric. Note that so far, we have not addressed the 'diagonal' subspaces ${x_i=x_j}$ because we are considering the general case where the point process can be simple or not.
Now, under certain circumstances, we can simplify the current 'global' representation in terms of measures with a new 'local' one in terms of densities, i.e., the Janossy densities:
\begin{equation*}j_n(x_1,\dots,x_n)\triangleq n!,p_n,\pi_n(x_1,\dots,x_n)\end{equation*}
where $\pi_n(\cdot)$ is the probability density of the probability measure $\Pi_n(\cdot)$. The advantage, of course, is that now we work with conventional vectorial functions rather than set functions.
However, at this point, I still don't see any difference between simple point processes and non-simple point processes. In other words, if we assume that the point process is simple, we don't gain any advantage in the current probabilistic description.
On proposition 5.4.Va
Part (a) states that a point process is simple if $j_n(\cdot)$ vanishes on the diagonals ${x_i=x_j}$. I almost agree with this statement because, loosely speaking, we are saying that the probability of the event $x_i=x_j$ is zero. However, now a quite recurrent question arises:
question 1) If the probability of the event $x_i=x_j$ is zero, can we conclude that the event $x_i=x_j$ never happens?
Let's assume that we have a simulator based on $J_n(\cdot)$ that vanishes on the diagonals ${x_i=x_j}$. If the answer to the previous question is yes, then the simulator always returns a simple point process; if the answer is no, then the simulator sometimes (or perhaps it's better to say 'rarely,' since we are talking about a zero-probability event) can return a non-simple point process.
I mentioned that I 'almost' agree with Daley and Vere-Jones because, in my humble opinion, the answer to the previous question is no. In other words, I believe that a zero-probability event is not impossible, and thus it can happen (albeit rarely). Therefore, I agree with Daley and Vere-Jones, not entirely, but in proportion to the number of times the ideal simulator returns a simple point process.
On proposition 5.4.Vb
I understand part (b) as a sufficient (and not necessary) condition for a process to be simple: if $j_n(\cdot)$ exists, then the point process is simple. Am I correct so far? If yes, we are not excluding the possibility that we can characterize non-simple point processes with densities. Moreover, if I'm correct, now I have to understand when we can express measures in terms of densities. Hence, I see a connection with the Radon-Nikodym theorem.
Reviewing the Radon-Nykodim theorem
Let's consider the simple case where the underlying space is $\mathbb{R}$ and take the Lebesgue measure as the reference measure (in this simple case, it is the conventional notion of length). Then, we have the following:
- Lebesgue decomposition theorem: a $\sigma$-finite measure $\mu$ can be always uniquely decomposed as follows \begin{equation*}\mu= \mu_{\text{ac}}+\mu_{\text{s}}\end{equation*} where $\mu_{\text{ac}}$ is absolutely continuous and $\mu_{\text{s}}$ is singular. Since we are in $\mathbb{R}$ and the reference measure is Lebesgue, we have: (1) $\mu_{\text{ac}}$ vanishes on single points; (2) $\mu_{\text{s}}$ can be a mixture of Dirac measures concentrated on single points.
- Radon-Nykodim theorem: the measure $\mu_{\text{ac}}$ can be written, for any measurable set $A$, as \begin{equation*}\mu_{\text{ac}}(A)=\int_A h\,\text{d}\lambda\end{equation*} where the density $h(\cdot)$ is a measurable map and it is uniquely determined up to variations on single points.
Now, I would like to combine the Radon-Nikodym theorem with part (b) of proposition 5.4.V. Firstly, Radon-Nikodym applies only to the absolutely continuous part $\mu_{ac}$, meaning that densities, at least in the Radon-Nikodym sense, exist only for absolutely continuous measures (i.e., for measures in the form $\mu=\mu_{\text{ac}}$). Thus, $j_n(\cdot)$ exists only if $J_n(\cdot)$ is $\sigma$-finite and absolutely continuous. Consequently,
corollary 1) if $J_n(\cdot)$ is absolutely continuous and $\sigma$-finite, then the process is simple
Now, I want to give practical meaning to this statement. If I'm not mistaken, '$J_n(\cdot)$ absolutely continuous and $\sigma$-finite' means that $J_n(\cdot)$ is a $\sigma$-finite measure composed only of its absolutely continuous part. Consequently, $J_n(\cdot)$ vanishes whenever we plug in a set $A_i$ that is a single point. Hence, each individual point in $\mathbb{R}^d$ has zero probability of being selected by the ideal simulator.
From another perspective, let's consider the case where the underlying space is $\mathbb{R}$. Let's define $J_n(\cdot)$ as the integral of a density $j_n(\cdot)$ that does not include any Dirac impulses in its expression (in short, I'm saying that it is not impulsive) but is instead a conventional non-negative (and measurable) function. We should be okay because this density $j_n(\cdot)$ yields null integrals over single points $x\in\mathbb{R}$. Thus, my conclusion is the following:
corollary 2) if $j_n(\cdot)$ is not impulsive, then the process is simple.
In my view, this is merely a sufficient condition. A more precise statement, which allows understanding when a point process is not simple, involves my reinterpretation of proposition 5.4.Va in terms of densities rather than measures.
Proposition 5.4.Va revisited) a process is simple if and only if $j_n(\cdot)$ is not impulsive on the diagonals $\{x_i=x_j\}$.
(Counter) Example
Let's try to figure out how to intentionally generate a non-simple point process. In this case, the simulator should sometimes return a set with repeated elements. If we assert that a zero-probability event cannot occur, my intuition tells me that this objective can be achieved only by considering particular densities that are not absolutely continuous. To make the ideas more concrete, let's consider a point process in $\mathbb{R}$ that always has 2 elements (thus, $p_2=1$, $p_{i\neq 2}=0$) and, with probability $p\in[0,1]$, returns the multi-set ${0,0}$. I would say that an algorithm that accomplishes this task is the following:
- Generate a Bernoulli integer with success rate $p$
- If the outcome of Bernoulli is a success, then return $\{0,0\}$
- Otherwise return $\{x_1,x_2\}$, where $x_1,x_2$ are obtained by sampling a joint density $f(x_1,x_2)$ not having impulses on the line $x_1=x_2$ of $\mathbb{R}^2$.
If I'm not wrong, we can summarize this algorithm with the following probability density \begin{equation*} \pi_2(x_1,x_2)=p\,\delta_{0,0}(x_1,x_2)+(1-p)f(x_1,x_2) \end{equation*} where I'm using Dirac's delta (what I was previously referring to as an 'impulse') to represent the 'density' of the singular measure. \begin{equation*} \mu_{\text{s}}(A_1,A_2)\triangleq \begin{cases} \quad\hfil 1 & \text{ if } A_1\ni 0, A_2\ni 0\\ \quad\hfil 0 & \text{ otherwise} \end{cases} \end{equation*} Then, the corresponding Janossy density should be \begin{equation*} j_2(x_1,x_2)=2!\,p_2\,\pi_2(x_1,x_2)=2p\,\delta_{0,0}(x_1,x_2)+2(1-p)f(x_1,x_2) \end{equation*}
General conclusions
Following the line suggested by the previous example and accepting that a zero-probability event cannot occur, I conclude the following:
- A point process is simple if and only if $j_n(\cdot)$ is not impulsive on the diagonals $\{x_i=x_j\}$.
- Proposition 5.4.Vb should be re-stated as follows: When $\mathcal{X}=\mathbb{R}^d$, the process is simple if for all such $n$ the Janossy measures have densities $j_n(\cdot)$ absolutely continuous with respect to $(nd)$-dimensional Lebesgue measure.
- Allowing the use of Dirac's delta, Janossy densities always exist whenever we employ $\sigma$-finite Janossy measures..
Questions
Question 1) According to Daley and Vere-Jones, a zero-probability event seems impossible. Is it true?
Question 2) Can we correctly assert that even $\mu_{\text{s}}$ can be expressed as the integral of a density if we utilize Dirac's delta?
Question 3) If we can always represent $\mu_{\text{s}}$ as the integral of some Dirac deltas, is it true that we can always express a $\sigma$-finite measure as the integral of a density?
Question 4) Are my general conclusions correct?
