4.4 Further Properties of the Poisson Process
Theorem 4.4.1 (Splitting). Let \(\{N(t)\}\) be a Poisson process with rate \(\lambda \), and suppose each event is independently classified as type I with probability \(p\) and type II with probability \(1 - p\). Let \(N_1(t)\) and \(N_2(t)\) be the numbers of type I and type II events in \((0, t]\). Then \(\{N_1(t)\}\) and \(\{N_2(t)\}\) are independent Poisson processes with rates \(\lambda p\) and \(\lambda (1 - p)\).
Proof. We show that \(N_1(t)\) and \(N_2(t)\) are independent Poisson variables; the process properties follow in the same way for increments over disjoint intervals. Given \(N(t) = n_1 + n_2\), the number of type I events is binomial with \(n_1 + n_2\) trials and probability \(p\), so \begin {align*} P(N_1(t) = n_1, N_2(t) = n_2) &= \binom {n_1 + n_2}{n_1}p^{n_1}(1 - p)^{n_2}\cdot \frac {e^{-\lambda t}(\lambda t)^{n_1 + n_2}}{(n_1 + n_2)!}\\ &= \frac {e^{-\lambda pt}(\lambda pt)^{n_1}}{n_1!}\cdot \frac {e^{-\lambda (1 - p)t}(\lambda (1 - p)t)^{n_2}}{n_2!}, \end {align*}
using \(e^{-\lambda t} = e^{-\lambda pt}e^{-\lambda (1 - p)t}\). The joint probability factorises into a Poisson\((\lambda pt)\) probability for \(n_1\) and a Poisson\((\lambda (1 - p)t)\) probability for \(n_2\), which proves both the distributions and the independence. □
The converse also holds: the superposition of independent Poisson processes with rates \(\lambda _1\) and \(\lambda _2\) is a Poisson process with rate \(\lambda _1 + \lambda _2\), and each event belongs to the first process with probability \(\frac {\lambda _1}{\lambda _1 + \lambda _2}\) (example 1.4.2).
Example 4.4.2. Cars pass a checkpoint as a Poisson process with rate \(30\) per hour, and \(10\%\) of them are trucks. Find the probability that at least one truck passes in a \(15\)-minute period, and the expected number of cars that pass while waiting for the first truck.
Solution. By theorem 4.4.1, trucks form a Poisson process with rate \(3\) per hour and other cars an independent Poisson process with rate \(27\) per hour. In \(\frac 14\) hour the number of trucks is Poisson with mean \(0.75\), so \(P(\text {at least one}) = 1 - e^{-0.75} = 0.528\). The time to the first truck is exponential with mean \(\frac 13\) hour; given that time \(T\), the number of other cars is Poisson with mean \(27T\), so the expected number is \(27E(T) = 9\).
Theorem 4.4.3 (Arrival times given the count). Given \(N(t) = n\), the times of the \(n\) events are distributed as the order statistics of \(n\) independent uniform random variables on \((0, t)\).
This is why a Poisson process is described as events occurring “completely at random”: once it is known how many events happened in \((0, t]\), they are scattered over the interval as independent uniform points. For example, given one event in \((0, t]\), its time is uniform on \((0, t)\), since for \(s<t\) \[P(T_1\leq s\mid N(t) = 1) = \frac {P(N(s) = 1, N(t) - N(s) = 0)}{P(N(t) = 1)} = \frac {\lambda se^{-\lambda s}e^{-\lambda (t - s)}}{\lambda te^{-\lambda t}} = \frac st .\]
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