The Central Idea
Let
\[ N(t) \]denote the number of events that have occurred by time \(t\).
A homogeneous Poisson process with rate
\[ \lambda>0 \]satisfies
\[ \boxed{ N(t)\sim\operatorname{Poisson}(\lambda t) }. \]Therefore, the number of events observed over a time interval depends on the interval length through the mean count \(\lambda t\).
The Counting-Process View
A Poisson process is a counting process.
Its sample paths begin at
\[ N(0)=0 \]and increase by one whenever an event occurs.
Between events, the process remains constant.
The State Space
At any time \(t\),
\[ N(t)\in\{0,1,2,\ldots\}. \]The process therefore has a countably infinite state space.
Unlike the earlier discrete-time Markov-chain examples, however, the index \(t\) now ranges over continuous time.
The Rate Parameter
The parameter
\[ \lambda \]represents the average event rate per unit time.
If
\[ \lambda=4 \]events per hour, then the expected number of events in \(t\) hours is
\[ \boxed{E[N(t)]=4t}. \]Units Matter
The rate and the time interval must use compatible units.
If
\[ \lambda=6 \]events per hour and the interval is \(30\) minutes, then
\[ t=0.5\text{ hour}. \]Thus,
\[ \lambda t=6(0.5)=3. \]Poisson Count Probabilities
If
\[ N(t)\sim\operatorname{Poisson}(\lambda t), \]then
\[ \boxed{ P(N(t)=k) = e^{-\lambda t} \frac{(\lambda t)^k}{k!} }, \qquad k=0,1,2,\ldots \]This gives the probability of exactly \(k\) events by time \(t\).
Example: Exactly Three Events
Suppose events occur at rate
\[ \lambda=2 \]per hour. Find the probability of exactly three events in one hour.
Since
\[ \lambda t=2, \]we have
\[ P(N(1)=3) = e^{-2}\frac{2^3}{3!}. \]Therefore,
\[ \boxed{ P(N(1)=3) = \frac{4}{3}e^{-2} }. \]Probability of No Events
Setting \(k=0\) gives
\[ P(N(t)=0) = e^{-\lambda t}. \]Thus,
\[ \boxed{ P(\text{no events in length }t) = e^{-\lambda t} }. \]Probability of At Least One Event
Using the complement rule,
\[ P(N(t)\ge1) = 1-P(N(t)=0). \]Therefore,
\[ \boxed{ P(N(t)\ge1) = 1-e^{-\lambda t} }. \]Mean and Variance
For a Poisson random variable with parameter \(\lambda t\),
\[ \boxed{ E[N(t)]=\lambda t } \]and
\[ \boxed{ \operatorname{Var}(N(t))=\lambda t }. \]The mean and variance are equal.
Independent Increments
A Poisson process has independent increments.
Counts occurring in disjoint time intervals are independent.
For example,
\[ N(2)-N(1) \]and
\[ N(5)-N(3) \]are independent because the intervals \((1,2]\) and \((3,5]\) do not overlap.
Stationary Increments
A homogeneous Poisson process also has stationary increments.
The distribution of the number of events in an interval depends only on the interval's length, not on its location in time.
Thus,
\[ N(t+h)-N(t) \sim \operatorname{Poisson}(\lambda h). \]Increment Formula
More generally, for
\[ 0\le sThis is the core count formula for arbitrary time intervals.
Example: Events Between Two Times
Suppose
\[ \lambda=3 \]events per hour. The number of events between hour \(2\) and hour \(4\) is
\[ N(4)-N(2). \]The interval has length \(2\), so
\[ N(4)-N(2) \sim \operatorname{Poisson}(6). \]Small-Interval Behavior
One way to characterize the Poisson process is through behavior over a very short interval of length \(h\).
For small \(h\),
\[ P(\text{one event in }h) \approx \lambda h, \]while the probability of two or more events is much smaller than \(h\).
This local behavior leads to the Poisson count distribution.
More Precise Small-Interval Statements
As \(h\to0\),
\[ P(N(h)=1) = \lambda h+o(h), \]and
\[ P(N(h)\ge2) = o(h). \]The notation \(o(h)\) indicates a quantity that becomes negligible relative to \(h\).
Arrival Times
Let
\[ S_n \]denote the time of the \(n\)-th event.
Then
\[ S_1Interarrival Times
Define
\[ T_1=S_1 \]and, for \(n\ge2\),
\[ T_n=S_n-S_{n-1}. \]The variables
\[ T_1,T_2,\ldots \]are the waiting times between consecutive events.
Exponential Waiting Times
In a homogeneous Poisson process,
\[ \boxed{ T_1,T_2,\ldots \overset{\text{i.i.d.}}{\sim} \operatorname{Exponential}(\lambda) }. \]Each interarrival time has density
\[ f_T(t) = \lambda e^{-\lambda t}, \qquad t\ge0. \]The Mean Waiting Time
If
\[ T\sim\operatorname{Exponential}(\lambda), \]then
\[ \boxed{ E[T]=\frac{1}{\lambda} }. \]A higher event rate therefore corresponds to a shorter average waiting time.
Variance of the Waiting Time
For an exponential waiting time,
\[ \boxed{ \operatorname{Var}(T) = \frac{1}{\lambda^2} }. \]Connecting No Events to Waiting Time
The event
\[ T_1>t \]means that no event has occurred by time \(t\).
Therefore,
\[ P(T_1>t) = P(N(t)=0) = e^{-\lambda t}. \]This is exactly the survival function of an exponential random variable.
Memoryless Waiting Times
Exponential waiting times satisfy
\[ \boxed{ P(T>s+t\mid T>s) = P(T>t) }. \]After waiting \(s\) units of time with no event, the remaining waiting-time distribution is the same as it was initially.
Count View and Waiting-Time View
A Poisson process can be described in two equivalent ways.
- Count events using \(N(t)\).
- Track independent exponential waiting times \(T_1,T_2,\ldots\).
The count formulation and the waiting-time formulation describe the same stochastic process.
The \(n\)-th Arrival Time
Since
\[ S_n=T_1+\cdots+T_n, \]the \(n\)-th arrival time is the sum of \(n\) independent exponential waiting times.
Therefore,
\[ \boxed{ S_n\sim\operatorname{Gamma}(n,\lambda) } \]when \(\lambda\) is used as the rate parameter.
Density of the \(n\)-th Arrival Time
For integer \(n\ge1\),
\[ \boxed{ f_{S_n}(t) = \frac{\lambda^n t^{n-1}e^{-\lambda t}} {(n-1)!} }, \qquad t\ge0. \]Expected \(n\)-th Arrival Time
Because \(S_n\) is the sum of \(n\) waiting times,
\[ E[S_n] = nE[T]. \]Thus,
\[ \boxed{ E[S_n]=\frac{n}{\lambda} }. \]Count and Arrival-Time Events
The statements
\[ N(t)\ge n \]and
\[ S_n\le t \]describe the same event.
Therefore,
\[ \boxed{ P(N(t)\ge n) = P(S_n\le t) }. \]Exactly \(n\) Events by Time \(t\)
The event
\[ N(t)=n \]means that the \(n\)-th event has occurred by time \(t\), but the \((n+1)\)-st has not.
Thus,
\[ \{N(t)=n\} = \{S_n\le tConditional Event Times Given the Count
Suppose exactly \(n\) events occurred during the interval \([0,t]\).
Conditional on
\[ N(t)=n, \]the \(n\) event times have the same joint distribution as the ordered values of \(n\) independent
\[ \operatorname{Uniform}(0,t) \]random variables.
One Event Given That One Occurred
In particular, if
\[ N(t)=1, \]then the event time is uniformly distributed over
\[ [0,t]. \]Given that exactly one event occurred, no location inside the interval is favored.
Example: Conditional Event Location
Suppose exactly one event occurred in a one-hour interval.
Conditional on that fact, the probability that it occurred during the first \(15\) minutes is
\[ \frac{15}{60} = \boxed{\frac14}. \]Splitting an Interval
Suppose a Poisson process has rate \(\lambda\), and an interval of length \(t\) is divided into pieces of lengths
\[ t_1,t_2,\ldots,t_r \]with
\[ t_1+\cdots+t_r=t. \]The event counts in those disjoint pieces are independent Poisson random variables with means
\[ \lambda t_1,\ldots,\lambda t_r. \]Superposition of Poisson Processes
Suppose
\[ N_1(t),\ldots,N_r(t) \]are independent Poisson processes with rates
\[ \lambda_1,\ldots,\lambda_r. \]Their combined count
\[ N(t) = N_1(t)+\cdots+N_r(t) \]is a Poisson process with rate
\[ \boxed{ \lambda_1+\cdots+\lambda_r }. \]Example: Combining Arrival Streams
Suppose one source generates events at rate
\[ 2 \]per hour and another independent source generates events at rate
\[ 5 \]per hour.
The combined process has rate
\[ \boxed{7\text{ per hour}}. \]Thinning a Poisson Process
Suppose events arrive according to a Poisson process of rate \(\lambda\), and each event is independently classified as type \(A\) with probability \(p\).
Then the type-\(A\) events form a Poisson process with rate
\[ \boxed{p\lambda}. \]The remaining events form an independent Poisson process with rate
\[ (1-p)\lambda. \]Example: Thinning
Calls arrive at rate
\[ \lambda=12 \]per hour. Suppose \(25\%\) are priority calls, independently of other arrivals.
The priority-call process has rate
\[ 0.25(12) = \boxed{3\text{ per hour}}. \]Independent Types After Thinning
A particularly useful result is that independently thinned categories are themselves independent Poisson processes.
This allows one arrival stream to be separated into several independent classes while preserving Poisson structure.
Poisson Processes Are Markov
Because future increments are independent of the past given the current count,
\[ \{N(t):t\ge0\} \]is a continuous-time Markov process.
The next behavior depends on the present count and the future waiting-time law, not on the detailed history of earlier arrival times.
Why the Count Itself Often Does Not Affect the Next Arrival
In a homogeneous Poisson process, the next waiting time is exponential with rate \(\lambda\) regardless of whether
\[ N(t)=2 \]or
\[ N(t)=200. \]The event rate is constant and does not depend on the accumulated count.
Homogeneous Versus Nonhomogeneous Poisson Processes
A homogeneous Poisson process uses a constant rate
\[ \lambda. \]A nonhomogeneous Poisson process uses a time-dependent rate
\[ \lambda(t). \]Then the expected number of events by time \(t\) is
\[ \boxed{ \Lambda(t) = \int_0^t\lambda(u)\,du }. \]Counts in a Nonhomogeneous Poisson Process
For a nonhomogeneous Poisson process,
\[ N(t)-N(s) \sim \operatorname{Poisson} \left( \int_s^t\lambda(u)\,du \right). \]Independent increments are retained, but stationary increments generally are not.
Why Stationary Increments Fail with a Changing Rate
If \(\lambda(t)\) varies over time, two intervals of equal length can have different expected event counts.
Therefore, interval length alone no longer determines the increment distribution.
Poisson Approximation to Rare Events
Poisson distributions can approximate counts of rare independent events.
If
\[ X\sim\operatorname{Binomial}(n,p) \]with large \(n\), small \(p\), and
\[ np\approx\lambda, \]then
\[ X\approx\operatorname{Poisson}(\lambda). \]From Binomial Counts to Poisson Processes
One way to motivate a Poisson process is to divide time into many short intervals.
If each interval has a small event probability and events in disjoint intervals behave independently, the total count approaches a Poisson distribution as the intervals become increasingly fine.
Applications of Poisson Processes
Poisson processes are commonly used to model:
- customer arrivals;
- telephone calls;
- network requests;
- equipment failures;
- insurance claims;
- particle detections;
- traffic arrivals;
- transactions;
- service requests;
- rare-event occurrences.
Model Assumption: Constant Rate
A homogeneous Poisson process assumes a constant event rate over time.
Daily cycles, seasonality, growth, congestion, fatigue, or scheduled operating changes can violate this assumption.
A nonhomogeneous model may then be more appropriate.
Model Assumption: Independent Increments
Events in disjoint time intervals are assumed independent.
Clustering, contagion, feedback, burstiness, or common external causes can create dependence that a standard Poisson process does not capture.
Model Assumption: No Simultaneous Events
In a standard Poisson process, the probability of two or more events occurring at exactly the same continuous time is \(0\).
Real systems that naturally generate batches or simultaneous arrivals may require compound or batch-arrival models.
Model Assumption: Event Definition Is Stable
The process counts events according to a fixed definition.
If recording rules, detection thresholds, or classification criteria change over time, the observed count process can change even if the underlying physical mechanism does not.
Common Error: Confusing Rate with Expected Count
The rate
\[ \lambda \]is measured per unit time.
The expected count over an interval of length \(t\) is
\[ \lambda t. \]These quantities are equal only when \(t=1\) in the chosen time unit.
Common Error: Forgetting to Convert Time Units
If the rate is given per hour, a \(20\)-minute interval must be written as
\[ \frac13 \]of an hour before computing
\[ \lambda t. \]Common Error: Using \(\lambda\) Instead of \(\lambda t\)
The count distribution over an interval of length \(t\) is
\[ \operatorname{Poisson}(\lambda t), \]not generally
\[ \operatorname{Poisson}(\lambda). \]Common Error: Treating Overlapping Increments as Independent
Independent increments apply to disjoint time intervals.
Counts from overlapping intervals share events and are generally dependent.
Common Error: Assuming Every Count Process Is Poisson
A process can count events without satisfying Poisson assumptions.
Event clustering, scheduled arrivals, inhibition, changing rates, and batch arrivals can all produce non-Poisson counting behavior.
Common Error: Confusing Poisson Counts with Exponential Waiting Times
The Poisson distribution describes how many events occur in a time interval.
The exponential distribution describes how long until the next event.
They are connected, but they answer different questions.
Common Error: Misreading the Memoryless Property
Memorylessness applies to exponential waiting times.
It does not mean that the number of past events is erased or that all quantities associated with the process are independent.
The precise statement concerns the conditional remaining waiting time.
A Reliable Procedure for Poisson Count Problems
- Identify the event rate \(\lambda\).
- Convert the interval length \(t\) into matching time units.
- Compute the mean count \(\lambda t\).
- Use \(N(t)\sim\operatorname{Poisson}(\lambda t)\).
- Apply the Poisson probability formula or an appropriate complement.
- Check that the numerical result lies between \(0\) and \(1\).
A Reliable Procedure for Waiting-Time Problems
- Identify the rate \(\lambda\).
- Use \(T\sim\operatorname{Exponential}(\lambda)\) for one interarrival time.
- Use \(P(T>t)=e^{-\lambda t}\) for survival beyond time \(t\).
- Use \(P(T\le t)=1-e^{-\lambda t}\) for an arrival by time \(t\).
- For the \(n\)-th arrival, use the gamma distribution of \(S_n\).
- Keep count questions and waiting-time questions conceptually separate.
A Reliable Procedure for Combined or Thinned Processes
- Verify that the component processes are independent when using superposition.
- Add rates to combine independent Poisson processes.
- For independent classification with probability \(p\), multiply the original rate by \(p\).
- Use the resulting rate in the standard Poisson-process formulas.
- Interpret the new process in the same time units as the original rate.
Example: Full Count Workflow
Calls arrive according to a Poisson process at rate
\[ \lambda=6 \]per hour. Find the probability of exactly two calls in \(30\) minutes.
Convert the interval:
\[ t=0.5\text{ hour}. \]Then
\[ \lambda t=6(0.5)=3. \]Therefore,
\[ P(N(0.5)=2) = e^{-3}\frac{3^2}{2!}. \]Thus,
\[ \boxed{ P(N(0.5)=2) = \frac92e^{-3} }. \]Example: Full Waiting-Time Workflow
Suppose events occur at rate
\[ \lambda=4 \]per hour. Find the probability that the next event takes more than \(30\) minutes to arrive.
Since
\[ t=0.5\text{ hour}, \]we use
\[ P(T>0.5) = e^{-4(0.5)} = \boxed{e^{-2}}. \]Simulation of a Poisson Process
One direct simulation method uses exponential waiting times.
- Set the current time to \(0\).
- Generate \(T_1\sim\operatorname{Exponential}(\lambda)\).
- Set the first arrival time \(S_1=T_1\).
- Generate another exponential waiting time.
- Add it to obtain the next arrival time.
- Continue until the desired time horizon is exceeded.
What Simulation Can Estimate
Repeated simulations can estimate:
- count probabilities;
- waiting-time distributions;
- arrival-time distributions;
- maximum counts over fixed horizons;
- effects of superposition and thinning;
- queueing behavior when arrivals feed a service system.
Practice 1: Expected Count
Events occur at rate
\[ \lambda=5 \]per hour. What is the expected number of events in \(3\) hours?
Practice 2: Exactly Two Events
For a Poisson process with
\[ \lambda=2 \]per hour, find
\[ P(N(1)=2). \]Practice 3: No Events
Events occur at rate
\[ \lambda=4 \]per hour. Find the probability of no events during the next \(15\) minutes.
Practice 4: At Least One Event
Using the same rate and interval as Practice 3, find the probability of at least one event.
Practice 5: Increment Distribution
If
\[ \lambda=3 \]per hour, what is the distribution of
\[ N(5)-N(2)? \]Practice 6: Mean Waiting Time
Events occur at rate
\[ \lambda=8 \]per hour. Find the mean waiting time until the next event, in hours and in minutes.
Practice 7: Waiting Beyond a Threshold
With rate
\[ \lambda=2 \]per hour, find
\[ P(T>1.5). \]Practice 8: Superposition
Independent Poisson processes have rates
\[ 3,\qquad5,\qquad2 \]events per hour. Find the rate of the combined process.
Practice 9: Thinning
Events arrive at rate
\[ 20 \]per hour. Each is independently labeled type \(A\) with probability \(0.3\).
Find the rate of type-\(A\) events.
Practice 10: Identify the Error
A student is told that events occur at rate \(6\) per hour and wants the probability of exactly two events in \(20\) minutes. The student uses
\[ e^{-6}\frac{6^2}{2!}. \]Explain the error.
Practice Answers
Practice 1
\[ E[N(3)] = 5(3) = \boxed{15}. \]Practice 2
\[ P(N(1)=2) = e^{-2}\frac{2^2}{2!} = \boxed{2e^{-2}}. \]Practice 3
Fifteen minutes is
\[ 0.25 \]hour, so
\[ \lambda t=4(0.25)=1. \]Therefore,
\[ \boxed{ P(N(0.25)=0)=e^{-1} }. \]Practice 4
\[ P(N(0.25)\ge1) = 1-e^{-1}. \]Thus,
\[ \boxed{1-e^{-1}}. \]Practice 5
The interval length is
\[ 5-2=3 \]hours, so
\[ \boxed{ N(5)-N(2) \sim \operatorname{Poisson}(9) }. \]Practice 6
\[ E[T] = \frac18 \]hour.
Since one hour is \(60\) minutes,
\[ \frac18(60) = 7.5. \]Therefore,
\[ \boxed{ E[T]=\frac18\text{ hour}=7.5\text{ minutes} }. \]Practice 7
\[ P(T>1.5) = e^{-2(1.5)} = \boxed{e^{-3}}. \]Practice 8
Add the independent rates:
\[ 3+5+2 = \boxed{10\text{ per hour}}. \]Practice 9
\[ p\lambda = 0.3(20) = \boxed{6\text{ per hour}}. \]Practice 10
The student used \(\lambda=6\) as the Poisson mean without adjusting for the shorter time interval.
Since
\[ 20\text{ minutes} = \frac13\text{ hour}, \]the correct mean count is
\[ \lambda t = 6\left(\frac13\right) = 2. \]Therefore, the correct probability is
\[ \boxed{ e^{-2}\frac{2^2}{2!} }. \]What to Remember
A homogeneous Poisson process counts random events occurring at a constant rate:
\[ \boxed{ N(t)\sim\operatorname{Poisson}(\lambda t) }. \]- \(\lambda\) is the average event rate per unit time.
- The expected count over length \(t\) is \(\lambda t\).
- The variance of the count is also \(\lambda t\).
- Counts in disjoint intervals are independent.
- For a homogeneous process, increment distributions depend only on interval length.
- \(N(t)-N(s)\sim\operatorname{Poisson}(\lambda(t-s))\).
- The probability of no events in length \(t\) is \(e^{-\lambda t}\).
- Interarrival times are independent exponential random variables with rate \(\lambda\).
- The mean interarrival time is \(1/\lambda\).
- The \(n\)-th arrival time is gamma distributed.
- Independent Poisson processes can be superposed by adding rates.
- Independent thinning with probability \(p\) produces a Poisson process of rate \(p\lambda\).
- Conditional on \(N(t)=n\), the event times behave like ordered uniform points on \([0,t]\).
- A nonhomogeneous Poisson process replaces the constant rate by \(\lambda(t)\).
- Poisson counts and exponential waiting times are two views of the same process.
The central modeling pattern is: a constant random arrival rate produces Poisson event counts and exponential waiting times.
Where We Go Next
Poisson processes model arrivals through time. Many real systems must then process those arrivals using limited service capacity, creating congestion and waiting.
The next lesson, Queues and Waiting Systems, combines stochastic arrivals with service times to study queue length, waiting time, utilization, stability, and system performance.