Math 311

Class 20

Poisson Processes

A stochastic process \(\{N(t), t \ge 0\}\) is called a counting process if \(N(t)\) represents a number of occurrences of certain event by the time \(t\).

  • \(N(t) \ge 0\)
  • \(N(t)\) is an integer for each \(t\)
  • If \(s < t\) then \(N(s) \le N(t)\)
  • If \(s < t\) then \(N(t) - N(s)\) is the number of occurrences during the time interval \((s, t]\)

A counting process \(\{N(t), t \ge 0\}\) is called a Poisson process with rate \(\lambda\) if all of the following are true:

  1. \(N(0) = 0\)
  2. \(\{N(t), t \ge 0\}\) has independent increments
  3. \(\operatorname{P}(N(t + h) - N(t) = 1) = \lambda h + \mathcal{o}(h)\) as \(h \to 0\)
  4. \(\operatorname{P}(N(t + h) - N(t) \ge 2) = \mathcal{o}(h)\) as \(h \to 0\)

Distribution of Increments

If \(\{N(t), t \ge 0\}\) is a Poisson process with rate \(\lambda\) then for each \(s \ge 0\), \(t > 0\), \[N(s+t) - N(s) \sim \operatorname{Pois}(\lambda t).\]

\[\operatorname{P}\left(N(s+t) - N(s) = n\right) = \frac{(\lambda t)^n e^{-\lambda t}}{n!}\]

Each Poisson process has stationary increments

Interarrival Times

Let \(\{N(t), t \ge 0\}\) be a Poisson process with rate \(\lambda\).

\(T_1 = {}\) the time of the first event.

\(T_2 = {}\) the time between the first and second event.

\(T_3 = {}\) the time between the second and third event.

\(\vdots\)

\(T_n = {}\) the time between the \((n-1)\)-st and \(n\)-th event.

\(\vdots\)

\(\displaystyle S_n = T_1 + T_2 + T_3 + \cdots + T_n = \sum_{i=1}^n T_i = {}\) the time of the \(n\)-th event

How are all these variables distributed?

Interarrival Times cont.

Suppose \(T_i \stackrel{\text{iid}}{\sim} \operatorname{Exp}(\lambda)\) for \(i = 1, 2, \ldots\).

Define \(\displaystyle S_n = T_1 + T_2 + T_3 + \cdots + T_n = \sum_{i=1}^n T_i\).

Let \(N(0) = 0\) and \(N(t) = \max\{n; S_n \le t\}\) for \(t > 0\).

Then \(\{N(t), t\ge 0\}\) is a Poisson process with rate \(\lambda\).

  1. \(N(0) = 0\)
  2. \(\{N(t), t \ge 0\}\) has independent increments
  3. \(\operatorname{P}(N(t + h) - N(t) = 1) = \lambda h + \mathcal{o}(h)\) as \(h \to 0\)
  4. \(\operatorname{P}(N(t + h) - N(t) \ge 2) = \mathcal{o}(h)\) as \(h \to 0\)

Splitting a Poisson Process

Let \(\{N(t), t \ge 0\}\) be a Poisson process with rate \(\lambda\).

Suppose each of the arriving events satisfy a condition \(C\) with a probability \(p\), independently from all the other events.

Define \(N_C(t) = {}\) the number of events satisfying the condition \(C\) that arrived by time \(t\).

Then \(\{N_C(t), t \ge 0\}\) is a Poisson process with rate \(p\lambda\).

Merging Poisson Processes

Let \(\{N_i(t), t\ge 0\}\) for \(i = 1, 2, \ldots, n\) be \(n\) independent Poisson processes with rates \(\lambda_1, \lambda_2, \ldots, \lambda_n\).

Define \(\displaystyle N(t) = \sum_{k = 1}^n N_k(t)\).

Then \(\{N(t), t \ge 0\}\) is a Poisson process with rate \(\lambda_1 + \lambda_2 + \cdots + \lambda_n\).