An arrival process is called a Poisson process with rate $\lambda$ if it has the following properties:
-
(Time-homogeneity) The probability $P(k, \tau) $ of $k $ arrivals is the same for all intervals of the same length $\tau $.
-
(Independence) The number of arrivals during a particular interval is independent of the history of arrivals outside this interval.
-
(Small interval probabilities) The probabilities $P(k, \tau) $ satisfy
$$\begin{align*} P(0, \tau) &= 1 - \lambda \tau + o(\tau), \br P(1, \tau) &= \lambda \tau + o_1(\tau), \br P(k, \tau) &= o_k(\tau),\text{ for } k = 2, 3, … \end{align*}$$
Here, $o(\tau) $ and $o_k(\tau) $ are functions of $\tau $ that satisfy
$$\limu{ \tau }{ o } \frac{ o(\tau)}{ \tau } = 0, \limu{ \tau }{ o } \frac{ o_k(\tau)}{ \tau } = 0 $$
The first property states that arrivals are “equally likely” at all times.
The arrivals during any time interval of length $\tau$ have the same distribution.
This a counterpart to the assumption that the success probability $p$ in a Bernoulli process is the same for all trials.
The second property, consider a particular interval $[t, t']$, of length $t' - t$.
The unconditional probability of $k$ arrivals during that interval is $P(k, t'-t)$.
Suppose now that we are given complete or partial information on the arrivals outside this interval, property $2$ states that this information is irrelevant: the conditional probability of $k$ arrivals during $[t, t']$ remains equal to the unconditional probability $P(k,t'-t)$. This property is analogous to the independence of trials in a Bernoulli process.
The third property is critical. The $o(\tau) $ and $o_k(\tau) $ terms are meant to be negligible in comparison to $\tau $, when the interval length $\tau $ is very small. They can be thought of as the $O(\tau^2) $ terms in a Taylor series expansion of $P(k, \tau) $.
Thus for small $\tau $, the probability of a single arrival is roughly $\lambda \tau $, plus a negligible term. Similarly, for small $\tau $, the probability of zero arrivals is roughly $1 - \lambda \tau $.
Finally, the probability of two or more arrivals is negligible in comparison to $P(1, \tau) $, as $\tau $ becomes smaller.
Consider a fixed time interval of length $\tau $, and partition it into $\tau / \delta $ periods of length $\delta $, where $\delta $ is a very small number. The probability of more than two arrivals during any period can be neglected, because of property $(3)$ and the preceding discussion.
Different periods are independent, by property $(2)$.
Furthermore, each period has one arrival with probability approximately equal to $\lambda \delta$, or zero arrivals with probability approximately equal to $1 - \lambda \delta $.
Therefore, the process being studied can be approximated by a Bernoulli process, with the approximation becoming more and more accurate as $\delta $ becomes smaller.
The probability $P(k, \tau) $ of $k $ arrivals in time $\tau $ is approximately the same as the (binomial) probability of $k$ successes in $n = r / \delta $ independent Bernoulli trials with success probability $p = \lambda \delta$ at each trial. While keeping the length $\tau $ of the number $n$ of periods goes to infinity, while the product $np$ remains constant and equal to $\lambda \tau$.
Under these, circustances, we saw in the previous section that the binomial PMF converges to a Poisson PMF with parameter $\lambda \tau $. We are then led to the important conclusion that
$$P(k, \tau) = e^{- \lambda \tau} \frac{(\lambda \tau)^k }{ k! }, k = 0, 1, 2, … $$
Note that a Taylor series expansion of $e^{- \lambda \tau} $ yields
$$\begin{align*} P(0, \tau) &= e^{- \lambda \tau} = 1 - \lambda \tau + o(\tau) \br P(1, \tau) &= \lambda \tau e^{- \lambda \tau} = \lambda \tau - \lambda^2 \tau^2 + O(\tau^3) = \lambda \tau + o_1(\tau) \end{align*}$$
consistent with property $(3)$.
We already know the mean and variance of the Poisson PMF.
$$E[N_ \tau] = \lambda \tau, \var{ N_ \tau } = \lambda \tau$$
where $N_ \tau $ denotes the number of arrivals during a time interval of length $\tau $.
- The Poisson with parameter $\lambda \tau$, the number $N_ \tau $ of arrivals in a Poisson process with rate $\lambda $, over an interval of length $\tau $.
$$p_{N_ \tau}(k) = P(k, \tau) = e^{ - \lambda \tau } \frac{(\lambda \tau)^k }{ k! }, k = 0, 1, 2, … $$
$$E[N_ \tau] = \lambda \tau, \var{ N_ \tau } = \lambda \tau $$
- The exponential with parameter $\lambda$, the time $T$ until the first arrival.
$$f_T(t) = \lambda e^{ - \lambda t }, t \geq 0$$
$$E [ T ] = \frac{ 1 }{ \lambda }, \var{ T } = \frac{ 1 }{ \lambda^2 } $$
You get email according to a Poisson process at a rate of $\lambda = 0.2 $ messages per hour. You check your email every hour. What is the probability of finding $0$ and $1$ new messages?
The probabilities can be found using the Poisson PMF $\frac{ e^{ - \lambda \tau } (\lambda \tau)^k }{ k! } $. We have $\tau=1$.
$$P(0, 1) = e^{-0.2} = 0.819, P(1,1) = 0.2 e^{ -0.2 } = 0.164 $$
Arrivals of customers at the local supermarket are modeled by a Poisson process with a rate of $\lambda = 10$ customers per minute. Let $M$ be the number of customers arriving between $9:00 $ and $9:10$. Also, let $N$ be the number of customers arriving between $9:30$ and $9:35$. What is the distribution of $M+N$?