Poisson processes
A Poisson process is a basic model for counting events that occur randomly through continuous time. It is useful when events occur independently and at a constant average rate.
Start with an intuitive example
Suppose cells of a particular type arrive at a detector randomly at an average rate of
\[\lambda=2\text{ events per hour}.\]We do not know the exact arrival times. In one hour there might be 0 arrivals, 2 arrivals, 5 arrivals, or another number.
Let
\[N(t)=\text{number of arrivals observed from time 0 up to time }t.\]Then \(N(t)\) is a stochastic process. It starts at
\[N(0)=0\]and increases by one whenever an event occurs.
What does one Poisson-process path look like?
Between events, the count remains unchanged. At each event time, the count increases by one.
The horizontal axis is continuous time, but the count is integer-valued. Therefore a Poisson process is a continuous-time, discrete-state stochastic process.
The rate \(\lambda\)
The parameter \(\lambda>0\) is the event rate. Its units are
\[\frac{\text{events}}{\text{time}}.\]If \(\lambda=2\) per hour, the process produces an average of 2 events per hour over many repetitions or long periods under the model assumptions.
From rate to expected number of events
Over a time interval of length \(t\), the expected number of events is
\[\boxed{E[N(t)]=\lambda t}.\]For \(\lambda=2\) per hour:
| Time \(t\) | Expected count \(\lambda t\) |
|---|---|
| 0.5 hour | 1 |
| 1 hour | 2 |
| 3 hours | 6 |
These are expected values, not guaranteed observed counts.
The Poisson distribution gives the event-count probabilities
For a homogeneous Poisson process, the number of events during an interval of length \(t\) follows a Poisson distribution with parameter \(\lambda t\):
\[\boxed{N(t)\sim\operatorname{Poisson}(\lambda t)}.\]Therefore
\[\boxed{P(N(t)=k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}},\qquad k=0,1,2,\ldots\]| Symbol | Meaning |
|---|---|
| \(k\) | particular number of events |
| \(\lambda\) | event rate per unit time |
| \(t\) | length of time observed |
| \(\lambda t\) | expected number of events in that interval |
A complete numerical probability example
Suppose \(\lambda=2\) events per hour. What is the probability of exactly 3 events in one hour?
Here
\[\lambda t=2(1)=2.\]Therefore
\[P(N(1)=3)=e^{-2}\frac{2^3}{3!}.\]Since \(3!=6\),
\[P(N(1)=3)\approx0.180.\]So there is about an 18.0% probability of observing exactly three events during that hour under this model.
The whole count distribution
The most likely counts are around the expected value, but smaller and larger counts remain possible.
Mean and variance
For a Poisson process,
\[\boxed{E[N(t)]=\lambda t},\qquad\boxed{\operatorname{Var}(N(t))=\lambda t}.\]Thus the mean and variance of the event count are equal.
The standard deviation is
\[\operatorname{SD}(N(t))=\sqrt{\lambda t}.\]What happens in a very short interval?
Consider a short interval of length \(\Delta t\). For a Poisson process,
\[P(\text{one event in }\Delta t)=\lambda\Delta t+o(\Delta t),\]so for sufficiently small \(\Delta t\),
\[\boxed{P(\text{one event})\approx\lambda\Delta t}.\]Also,
\[P(\text{no event})\approx1-\lambda\Delta t,\]while the probability of two or more events is much smaller:
\[P(\text{two or more events})=o(\Delta t).\]Why two events become negligible in a tiny interval
For a Poisson count over \(\Delta t\),
\[P(N(\Delta t)=2)=e^{-\lambda\Delta t}\frac{(\lambda\Delta t)^2}{2}.\]The leading dependence is proportional to \((\Delta t)^2\), whereas the probability of one event is proportional to \(\Delta t\). As \(\Delta t\) becomes very small, \((\Delta t)^2\) shrinks much faster.
This is why CTMC models can treat one event as the dominant possible change during an infinitesimal interval.
Events in non-overlapping intervals are independent
A homogeneous Poisson process assumes independent increments. Counts in non-overlapping time intervals are independent.
This is a modelling assumption. It may or may not be biologically reasonable for a particular application.
The distribution depends only on interval length
A homogeneous Poisson process also has stationary increments. If two intervals have the same length, their event counts have the same distribution, regardless of where the intervals occur in time.
For example, with constant \(\lambda\), the number of events between hours 1 and 2 has the same distribution as the number between hours 8 and 9.
The main assumptions
A homogeneous Poisson process is appropriate when the following simplified assumptions are reasonable:
| Assumption | Meaning |
|---|---|
| constant rate | events occur with the same rate \(\lambda\) through time |
| independent increments | counts in non-overlapping intervals are independent |
| single events | simultaneous multiple events have negligible probability in a sufficiently short interval |
| counting starts at zero | \(N(0)=0\) |
Waiting time to the first event
Let \(T_1\) be the time until the first event. The event \(T_1>t\) means that no events have occurred by time \(t\). Therefore
\[P(T_1>t)=P(N(t)=0).\]Using the Poisson formula with \(k=0\),
\[P(N(t)=0)=e^{-\lambda t}.\]Hence
\[\boxed{P(T_1>t)=e^{-\lambda t}},\]which means
\[\boxed{T_1\sim\operatorname{Exp}(\lambda)}.\]Time between consecutive events
Not only the first waiting time, but every inter-event time is independently exponentially distributed:
\[T_1,T_2,T_3,\ldots\sim\operatorname{Exp}(\lambda).\]The mean waiting time between events is
\[E[T]=\frac1\lambda.\]If \(\lambda=2\) events per hour, the mean waiting time is
\[\frac12\text{ hour}=30\text{ minutes}.\]This does not mean events occur regularly every 30 minutes. Individual waiting times remain random.
Count view and waiting-time view
| Count view | Waiting-time view |
|---|---|
| How many events occur by time \(t\)? | How long until the next event? |
| \(N(t)\sim\operatorname{Poisson}(\lambda t)\) | \(T\sim\operatorname{Exp}(\lambda)\) |
| mean count \(=\lambda t\) | mean waiting time \(=1/\lambda\) |
Connection to CTMCs
Suppose a CTMC is currently in state \(i\) and its total event rate is
\[a(i)=b(i)+d(i).\]While the state remains fixed, events are generated at this total rate. Therefore the waiting time to the next event is exponential with rate \(a(i)\).
After the event, however, the state may change, so \(a(i)\) may change too. A general birth–death CTMC is therefore not one homogeneous Poisson process with a single constant rate.
Connection to epidemic modelling
A Poisson process can be useful for random event arrivals when a constant-rate approximation is reasonable. But epidemic infection events usually depend on the changing numbers of susceptible and infectious individuals.
For example, in an SIS CTMC,
\[b(i)=\beta\frac{(N-i)i}{N}\]changes when \(i\) changes. Therefore the full epidemic is not generally a homogeneous Poisson process.
Nevertheless, the Poisson-process idea is fundamental because it explains how continuous-time event rates generate random event counts and exponential waiting times.
When a homogeneous Poisson model may fail
The simple model can be inappropriate when events cluster, inhibit one another, occur in bursts, have a changing rate, or depend strongly on the current biological state.
| Feature | Possible issue |
|---|---|
| seasonality | rate changes with time |
| epidemic growth | infection rate changes with population state |
| superspreading / clustering | event counts may be more variable than a simple Poisson model predicts |
| refractory period | one event changes the chance of another soon afterwards |
More general stochastic processes can be used when these effects matter.
The full picture
What comes next?
The next section focuses directly on the other side of this relationship: exponential waiting times, their memoryless property, and why they arise naturally in continuous-time Markov chains.