Poisson processes: Conditional distribution
Joint conditional distribution of points in a finite window
Let \sX =\R^d be a d-dimensional Euclidean space, and S:\Omega\to\sX^\N be a Poisson point process with intensity measure \Lambda: \sB(\sX) \to \R_+ and associated counting process N:\Omega\to\Z_+^{\sB(\sX)}.
Proposition 1. Let k \in \N be any positive integer. Consider a Poisson point process S: \Omega \to \sX^\N with intensity measure \Lambda:\sB(\sX)\to\R_+, a finite partition A\in \sB(\sX)^k that partitions a bounded set B \in \sB(\sX), and a vector n \in \Z_+^k that partitions a non negative integer m \in \Z_+. Then,
Proof. Proof. From the definition of conditional probability and the fact that \cap_{i=1}^k\set{N(A_i) = n_i} \subseteq \set{N(B) =m}, we can write the conditional probability on LHS as the ratio \begin{equation*} %P(\set{N(A_1) = n_1, \dots, N(A_k) = n_k}\given\set{N(A) = n}) = \frac{P\set{N(A_1) = n_1, \dots, N(A_k) = n_k, N(B) = m}}{P\set{N(B) = m}} = \frac{P\set{N(A_1) = n_1, \dots, N(A_k) = n_k}}{P\set{N(B) = m}}. \end{equation*} From the complete independence property and Poisson marginals for the joint distribution of (N(A_1), \dots, N(A_k)) for the partition A \in \sB(\sX)^k, and the fact that the intensity measure sums over disjoint sets, i.e. \Lambda(B) = \sum_{i=1}^k\Lambda(A_i), we can rewrite the RHS of the above equation as \begin{equation*} \frac{P\set{N(A_1) = n_1, \dots, N(A_k) = n_k}}{P\set{N(B) = m}} = \frac{\prod_{i=1}^ke^{-\Lambda(A_i)}\frac{\Lambda(A_i)^{n_i}}{n_i!}}{e^{-\Lambda(B)}\frac{\Lambda(B)^m}{m!}} = \binom{m}{n_1, \dots, n_k}\prod_{i=1}^k\left(\frac{\Lambda(A_i)}{\Lambda(B)}\right)^{n_i}. \end{equation*} ◻
Remark 1. Consider a Poisson point process S:\Omega\to\sX^\N with intensity measure \Lambda: \sB(\sX) \to \R_+ and counting process N:\Omega\to\Z_+^{\sB(\sX)}. Let A \in \sB(\sX)^k be a partition for bounded set B \in \sB(\sX).
Defining p_i \triangleq \frac{\Lambda(A_i)}{\Lambda(B)}, we see that (p_1, \dots, p_k) \in \cM([k]) is a probability distribution. We also observe that When N(B)=1, we can call the point of S in B as S_1 without any loss of generality. That is, if we call \set{S_1} = S \cap B, then we have Similarly, when N(B) = n_i, we call the points of S in B as S_1, \dots, S_{n_i} and denote S\cap B = \set{S_1, \dots, S_{n_i}}. For this case, we observe
We can rewrite the Equation [eqn:multi] as a multinomial distribution, where
For any finite set F\subseteq \N of size m \in \Z_+ and n\in\Z_+^k a partition of m, we define \cP_k(F, n) to be the collection of all k-partitions E \in \cP(\N)^k of F such that \abs{E_i} = n_i for i \in [k]. Then, the multinomial coefficient accounts for number of partitions of m points into sets with n_1, \dots, n_k points. That is,
Recall that the event \set{N(A_i) = n_i} = \set{\abs{S \cap A_i} = n_i}. Hence, we can write
When N(B) = m, we denote S\cap B by F = \set{S_1, \dots, S_m} without any loss of generality. We further observe that when N(A_i) = n_i for all i \in [k], then (S \cap A_1, \dots, S \cap A_k) \in \cP_k(S\cap B,n). Therefore, we can re-write the event That is, we can write the conditional probability conditioned on S\cap B = F, as
Equating the RHS of the above equation term-wise, we obtain that conditioned on each of these points falling inside the window B, the conditional probability of each point falling in partition A_i is independent of all other points and given by p_i. That is, we have It means that given m points in the window B, the location of these points are independently and identically distributed in B according to the distribution \frac{\Lambda(\cdot)}{\Lambda(B)}.
If the Poisson process is homogeneous, the distribution is uniform over the window B.
For a Poisson process with intensity measure \Lambda and any bounded set A \in \sB, the number of points N(A) in the set A is a Poisson random variable with parameter \Lambda(A). Given the number of points N(A), the location of all the points in S\cap A are with density \frac{\lambda(x)}{\Lambda(A)} for all x \in A.
Remark 2 (Simulating a homogeneous Poisson point process). If we are interested in simulating a two dimensional homogeneous Poisson point process with density \lambda in a uniform square A = [0,1]\times[0,1]. Then, we first generate the random variable N(A):\Omega\to\Z_+ that takes value n with probability e^{-\lambda}\frac{\lambda^n}{n!}. Next, for each of the N(A)=n points, we generate the location (X_i, Y_i) \in \R^2 uniformly at random. That is, X:\Omega\to[0,1]^n and Y:\Omega\to[0,1]^n are independent uniform sequences.
Corollary 2. For a homogeneous Poisson point process on the half-line with ordered set of points \tilde{S} = (S_{(n)} \in \R_+: n \in \N), we can write the conditional density of ordered points (S_{(1)}, \dots, S_{(k)}) given the event \set{N_t = k} as the ordered statistics of uniformly distributed random variables. Specifically, we have
Proof. Proof. Given \set{N_t = k}, we can denote the points of the Poisson process in (0,t] by S_1, \dots, S_k. From the above remark, we know that S_1, \dots, S_k are uniform in (0, t], conditioned on the number of points N_t = k. Hence, we can write Therefore, for 0 < t_1 < \dots < t_k < 1 and (h_1, \dots, h_k) sufficiently small, we have Since (S_1, \dots, S_k) are conditionally independent given S\cap A = \set{S_1, \dots, S_k}, it follows that any permutation \sigma: [k] \to [k], the conditional joint distribution of (S_{\sigma(1)}, \dots, S_{\sigma(k)}) is identical to that of (S_1, \dots, S_k) given S\cap A = \set{S_1, \dots, S_k}. Further, we observe that the order statistics of (S_{\sigma(1)}, \dots, S_{\sigma(k)}) is identical to that of (S_{1}, \dots, S_{k}). Therefore, we can write the following equality for the events The result follows since the number of permutations \sigma: [k] \to [k] is k!. ◻