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 \(n\)th 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 \(n\)th 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 \(n\)th 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 \(n\)th 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 \(n\)th 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 \(n\)th 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 \(n\)th arrivals, to get \(\E N_t = \sum_{n\in\N}F_n(t) = e^{-\lambda t}\sum_{n\in\N}\sum_{k \ge n}\frac{(\lambda t)^k}{k!} = \lambda t e^{-\lambda t}\sum_{k\in\N}\frac{(\lambda t)^{k-1}}{(k-1)!} = \lambda t\).
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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.