Poisson processes: Conditional distribution

Author

Parimal Parag

Updated

July 1, 2026

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 \(B \in \sB(\sX)\).

  1. 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

  2. We can rewrite the Equation [eqn:multi] as a multinomial distribution, where

  3. 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,

  4. Recall that the event \(\set{N(A_i) = n_i} = \set{\abs{S \cap A_i} = n_i}\). Hence, we can write

  5. 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

  6. 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 \(\frac{\Lambda(\cdot)}{\Lambda(B)}\).

  7. If the Poisson process is homogeneous, the distribution is uniform over the window \(B\).

  8. 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 \(k!\). ◻