Poisson process on the half-line
Simple point processes on the half-line
A stochastic process defined on the half-line N: \Omega \to \Z_+^{\R_+} is a counting process if
N_0 = 0, and
for each \omega \in \Omega, the sample path N(\omega): \R_+ \to \Z_+ is non-decreasing, integer valued, and right continuous function of time t \in \R_+.
Each discontinuity of the sample path of the counting process can be thought of as a jump of the process, as shown in Figure 1. A simple counting process has the unit jump size almost surely. General point processes in higher dimension don’t have any inter-arrival time interpretation.
Definition 1. The points of discontinuity are also called the arrival instants of the counting process N. The nth arrival instant is a random variable denoted \tS_n: \Omega \to \R_+, defined inductively as
Definition 2. The inter arrival time between (n-1)th and nth arrival is denoted by X_n and written as X_n \triangleq \tS_n - \tS_{n-1}.
Remark 1. For a simple point process, we have P\set{X_{n} = 0} = P\set{X_n\le 0} = 0.
Lemma 3. Simple counting process N:\Omega\to\Z_+^{\R_+} and arrival process \tS:\Omega\to\R_+^\N are inverse processes, i.e.
Proof. Proof. Let \omega \in \set{\tS_n \le t}, then N_{\tS_n} = n by definition. Since N is a non-decreasing process, we have N_t \ge N_{\tS_n} = n. Conversely, let \omega \in \set{N_t \ge n}, then it follows from definition that \tS_n \le t. ◻
Corollary 4. For arrival instants \tS:\Omega\to\R_+^\N associated with a counting process N:\Omega\to\Z_+^{\R_+} we have \set{\tS_n \le t, \tS_{n+1} > t} = \set{N_t = n} for all n \in \Z_+ and t \in \R_+.
Proof. Proof. It is easy to see that \set{\tS_{n+1} > t } = \set{\tS_{n+1} \le t}^c = \set{N_t \ge n+1}^c = \set{N_t < n+1}. Hence, ◻
Lemma 5. Let F_n(x) be the distribution function for S_n, then P_n(t) \triangleq P\set{N_t = n} = F_{n}(t)-F_{n+1}(t).
Proof. Proof. It suffices to observe that following is a union of disjoint events, ◻
IID exponential inter-arrival times characterization
Proposition 6. The counting process N:\Omega\to\Z_+^{\R_+} associated with a simple Poisson point process S:\Omega\to\R_+^\N is Markov.
Proof. Proof. We define the event space \sF_t \triangleq \sigma(N_s: s \le t) as the history of the process until time t \in \R_+. Then, from the independent increment property of Poisson processes, we have for any historical event H_s \in \sF_s The transition probability matrix is P(s,t) with its (k,n)th entry given by e^{-\Lambda(s,t]}\frac{(\Lambda(s,t])^{n-k}}{(n-k)!}. ◻
Remark 2. A Markov process X:\Omega\to\sX^\R is time homogeneous if the transition matrix P(s,t) = P(t-s) for all t \ge s. Thus the counting process for a homogeneous Poisson point process is time homogeneous Markov process, as the transition probability matrix P(s,t) = P(t-s) with its (k,n)th entry given by e^{-\lambda(t-s)}\frac{(\lambda(t-s))^{n-k}}{(n-k)!}.
Theorem 7. The counting process N:\Omega\to\Z_+^{\R_+} associated with a simple Poisson point process S:\Omega\to\R_+^\N is strongly Markov.
Proposition 8. A simple counting process N:\Omega\to\Z_+^{\R_+} is associated with a homogeneous Poisson process with a constant intensity density \lambda, iff the inter-arrival time sequence X:\Omega\to\R_+^\N are random variables with an exponential distribution of rate \lambda.
Proof. Proof. Let N_t be a counting process associated with a homogeneous Poisson point process on half-line with constant intensity density \lambda. From equivalence iii\_ in Theorem [thm:equiv], we obtain for any positive integer t, It suffices to show that inter-arrivals time sequence X:\Omega\to\R_+^\N is . We can show that N is Markov process with strong Markov property. Since the sequence of ordered points \tS:\Omega\to\R_+^\N is a sequence of stopping times for the counting process, it follows from the strong Markov property of this process that (N_{\tS_n+t} - N_{\tS_n}: t \ge 0) is independent of \sigma(N_s: s \le \tS_n) and hence of \tS_n and N_{\tS_n}. Further, we see that It follows that X:\Omega\to\R_+^\N is an independent sequence. For homogeneous Poisson point process, we have N_{\tS_n+t}-N_{\tS_n} = N_t in distribution, and hence X_{n+1} has same distribution as X_1 for each n \in \N.
For the given inter-arrival time sequence X:\Omega\to\R_+^\N distributed exponentially with rate \lambda, we define the nth arrival instant \tS_n \triangleq \sum_{i=1}^nX_i for each n \in \N, and the number of arrivals in time duration (0,t] as N_t\triangleq \sum_{n\in\N}\SetIn{\tS_n\le t} for all t\in \R_+. It follows that N_t is path wise non-decreasing, integer-valued, right continuous, and simple since P\set{X_1 \le 0} = 0. Therefore, N is a simple counting process such that It follows that the void probabilities are exponential and hence the random variable N_t is Poisson with parameter \lambda t for all t \in \R_+. Hence, N is a counting process associated with a homogeneous Poisson process with the constant intensity density \lambda from the equivalence ii\_ in Theorem [thm:equiv]. ◻
For many proofs regarding Poisson processes, we partition the sample space with the disjoint events \set{N_t = n} for n \in \Z_+. We need the following lemma that enables us to do that.
Lemma 9. For any finite time t > 0, the number of points on the interval (0,t] from a Poisson process is finite almost surely.
Proof. Proof. By strong law of large numbers, we have \lim_{n \to \infty} \frac{S_{n}}{n} = \E[X_{1}] = \frac{1}{\lambda} almost surely. Fix t > 0 and we define a sample space subset M = \set{\omega \in \Omega: N(\omega, t) = \infty }. For any \omega \in M, we have S_{n}(\omega)\le t for all n \in \N. This implies \lim\sup_n\frac{S_{n}}{n} = 0 and \omega \not\in \set{\lim_n \frac{S_{n}}{n} = \frac{1}{\lambda} }. Hence, the probability measure for set M is zero. ◻
Distribution functions
Lemma 10. The following are true for the nth arrival instant \tS_n of the Poisson arrival process \tS:\Omega\to\R_+^\N with constant intensity density \lambda.
The moment generating function is M_{\tS_n}(\theta) = \E[e^{\theta \tS_n} ] = \frac{\lambda^n}{(\lambda-\theta)^n}\SetIn{\theta < \lambda} + \infty\SetIn{\theta \ge \lambda}. %\infty, & \theta \ge \lambda. %\begin{cases} %\frac{\lambda^n}{(\lambda-\theta)^n}, & \theta < \lambda \\ %\infty, & \theta \ge \lambda. %\end{cases}
The distribution function is F_n(t) \triangleq P\set{\tS_n \le t} = 1 - e^{-\lambda t}\sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!}.
The density function is Gamma distributed with parameters n and \lambda. That is, f_{n}(s) =\frac{\lambda (\lambda s)^{n-1}} {(n-1)!} e^{-\lambda s}.
Corollary 11. Consider the counting process N:\Omega\to\Z_+^{\R_+} associated with the Poisson arrival process \tS:\Omega\to\R_+^\N having constant intensity density \lambda. The following are true.
The relation between distribution of nth arrival instant and probability mass function for the counting process is given by F_n(t)= \sum_{j \ge n}P_j(t).
For each t\in\R_+, the probability mass function P_{N_t}\in\cM(\Z_+) for discrete random variable N_t:\Omega\to\Z_+ is given by P_n(t) \triangleq P_{N_t}(n) = P\set{N_t=n)}= e^{-\lambda t}\frac{(\lambda t)^{n}}{n!}.
The relation between distribution of nth arrival instant and the mean of the counting process is given by \sum_{n \in \N}F_n(t) = \E N_t.
For each t\in\R_+, the mean \E[N_t] = \lambda t, explaining the rate parameter \lambda for the Poisson process.
Proof. Proof. We observe the inverse relationship \set{\tS_n \le t} = \set{N_t \ge n} for all n \in \Z_+ and t\in\R_+.
The result follows by taking the probability on both sides of the inverse relationship, to get F_n(t) = P\set{\tS_n \le t} = P\set{N_t \ge n} = \sum_{j \ge n}P\set{N_t = j} = \sum_{j \ge n}P_j(t).
The result follows from the explicit from for the distribution of \tS_n and recognizing that P_n(t) = F_n(t) - F_{n+1}(t).
The result from the following observation \sum_{n \in \N}F_n(t) = \E\sum_{n \in \N}\SetIn{N_t \ge n} = \sum_{n \in \N}P\set{N_t \ge n} = \E N_t.
The result follows by summing the distribution function of the nth arrivals, to get \E N_t = \sum_{n\in\N}F_n(t) = e^{-\lambda t}\sum_{n\in\N}\sum_{k \ge n}\frac{(\lambda t)^k}{k!} = \lambda t e^{-\lambda t}\sum_{k\in\N}\frac{(\lambda t)^{k-1}}{(k-1)!} = \lambda t.
◻
Remark 3. A Poisson process is not a stationary process. That is, the finite dimensional distributions are not shift invariant. This is clear from looking at the first moment \E N_t = \lambda t, which is linearly increasing in time.