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Risk Modelling and Survival Analysis · Stochastic processes

Poisson Processes: Properties, Inter-Arrival Times and Compound Poisson

Updated 11 October 2026 · Fact-checked

A Poisson process with rate λ counts events in continuous time. Counts in disjoint intervals are independent, and N(t) ~ Poisson(λt). Waiting times between events are independent Exponential(λ). To solve questions, identify the interval, compute λt, then use the Poisson or exponential formula. For compound sums, use E[S] = λt E[X] and Var[S] = λt E[X²].

Understand Poisson Processes

A Poisson process models events that occur one at a time at random moments, such as claims reaching an insurer or deaths in a group. Let N(t) be the number of events up to time t. The process is described by a rate λ, the average number of events per unit time.

The usual definition has three parts. N(0) = 0. The process has independent increments: counts in non-overlapping intervals are independent. It has stationary increments: the number of events in an interval depends only on its length. Then N(t) ~ Poisson(λt). An equivalent definition works with small intervals h: P(one event in h) = λh + o(h), and P(two or more events in h) = o(h).

The time to the first event, and the gaps between later events, are the inter-arrival times. They are independent and each is Exponential(λ), with mean 1/λ. The exponential has the memoryless property, so the time already waited does not change the remaining wait. The time to the nth event is the sum of n such gaps, which is Gamma(n, λ).

Superposition: if you add independent Poisson processes with rates λ₁ and λ₂, the combined process is Poisson with rate λ₁ + λ₂. Thinning: if each event is independently of type 1 with probability p, the type 1 events form a Poisson process with rate pλ. The type 1 and type 2 processes are independent of each other. Given the combined process has an event, it is type 1 with probability λ₁ ÷ (λ₁ + λ₂).

A compound Poisson process is S(t) = X₁ + X₂ + ... + X_N(t), where N(t) is a Poisson process and the X's are independent, identically distributed claim sizes, independent of N. This is the standard aggregate claims model. It has independent and stationary increments as well.

Key rules to remember

Poisson count
P(N(t) = n) = e^(−λt) (λt)ⁿ ÷ n!
Mean and variance of N(t) both equal λt.
Inter-arrival time
T ~ Exponential(λ); f(t) = λe^(−λt); P(T > t) = e^(−λt); E[T] = 1/λ; Var[T] = 1/λ²
Gaps are independent. Time to the nth event is Gamma(n, λ) with mean n/λ.
Memoryless property
P(T > s + t | T > s) = P(T > t)
Holds for the exponential only, among continuous distributions.
Superposition
N₁(t) + N₂(t) is Poisson process with rate λ₁ + λ₂
Requires independent processes.
Thinning
Type 1 events: rate pλ; type 2 events: rate (1 − p)λ
Each event classified independently. The two resulting processes are independent.
Compound Poisson mean
E[S(t)] = λt E[X]
X is a single claim size.
Compound Poisson variance
Var[S(t)] = λt E[X²]
Uses the second moment, not the variance, of X.
Compound Poisson MGF
M_S(r) = exp{λt (M_X(r) − 1)}
Useful for skewness and for proving closure results.
Third central moment
E[(S − E[S])³] = λt E[X³]
Gives the skewness as λt E[X³] ÷ (λt E[X²])^(3/2).

How to solve Poisson Processes questions

Use this method for any Poisson process question, whether on counts, waiting times, splitting or aggregate claims.

  1. 1Write down the rate λ and the time unit. Convert all times to the same unit before doing anything else.
  2. 2Decide what is asked: a count in an interval (Poisson), a waiting time (exponential or gamma), a split by type (thinning), or a total claim amount (compound).
  3. 3For counts, compute the mean λt for the interval given and use the Poisson probability formula. Use independence for disjoint intervals.
  4. 4For waiting times, use P(T > t) = e^(−λt). For the nth event, sum n gaps or use the gamma, or convert to a count: the nth event occurs after t if N(t) ≤ n − 1.
  5. 5For several event types, use thinning to get each rate, or superposition to add rates. Check the independence conditions are met.
  6. 6For aggregate claims, find E[X] and E[X²] from the claim distribution, then apply the compound formulas for mean and variance.
  7. 7If a conditional probability is asked, use independent increments to separate intervals, or use the memoryless property to restart the clock.
  8. 8State the result with units and check it is sensible: probabilities between 0 and 1, variance at least zero.

Quickest way: Rate-scaling shortcut

When to use it: Use this for MCQs and for the first marks of written parts, where you need a quick probability or moment.

  1. Scale the rate to the interval: μ = λ × length. Everything then depends on μ.
  2. Use the right tool: Poisson(μ) for counts, e^(−λt) for survival of a wait, adding rates for merging, multiplying rate by p for thinning.
  3. For compound sums, go straight to λt E[X] and λt E[X²]. Do not compute Var[X] first.
  4. For P(at least one) use 1 − e^(−μ). For P(none) use e^(−μ).
  5. Sense check: mean waiting time is 1/λ, and compound variance is always larger than the variance of the count times E[X]².

Common mistakes in Poisson Processes

  • Using Var[S] = λt Var[X] for a compound Poisson process.

    Students mix this up with the general compound formula E[N]Var[X] + Var[N]E[X]².

    Fix: For Poisson N, Var[N] = E[N] = λt, so the result simplifies to λt E[X²]. Always use the second moment.

  • Forgetting to change the time unit, for example using a monthly rate for a yearly interval.

    The rate and the interval are given in different parts of the question.

    Fix: Write the unit next to λ and convert the interval length before computing λt.

  • Treating the mean wait as λ instead of 1/λ.

    The exponential density is λe^(−λt), so λ is easy to confuse with the mean.

    Fix: Remember λ is a rate (events per unit time), so the mean gap is 1/λ.

  • Applying independence to overlapping intervals, such as N(2) and N(5).

    Students assume all counts in a Poisson process are independent.

    Fix: Only disjoint intervals are independent. Split N(5) = N(2) + (N(5) − N(2)) and use the independent pieces.

  • Saying the nth arrival time is exponential.

    The single gap is exponential, and students carry this over.

    Fix: The nth arrival time is a sum of n independent exponentials, so it is Gamma(n, λ).

  • Adding rates for thinned processes instead of multiplying by p.

    Superposition and thinning are learned together and get confused.

    Fix: Merging independent processes adds rates. Splitting one process by probability p multiplies its rate by p.

Worked examples

Example 1

Claims arrive at an insurer as a Poisson process at rate 6 per day. (a) Find the probability of exactly 2 claims in a 4-hour period, treating a day as 24 hours. (b) Find the probability that the wait for the next claim exceeds 6 hours.

Show the solution
  1. Rate per hour = 6 ÷ 24 = 0.25.
  2. (a) Mean over 4 hours μ = 0.25 × 4 = 1.
  3. P(N = 2) = e^(−1) × 1² ÷ 2! = e^(−1) ÷ 2 = 0.36788 ÷ 2 = 0.18394.
  4. (b) Wait T ~ Exponential(0.25). P(T > 6) = e^(−0.25 × 6) = e^(−1.5).
  5. e^(−1.5) = 0.22313.

Answer: (a) 0.1839 (b) 0.2231

Example 2

Claims arrive as a Poisson process at rate 10 per year. Claim sizes are independent and identically distributed, with mean ₹40,000 and standard deviation ₹30,000, and are independent of the claim count. Find the mean and standard deviation of total claims over 2 years.

Show the solution
  1. λt = 10 × 2 = 20.
  2. E[X] = 40,000. Var[X] = 30,000² = 900,000,000.
  3. E[X²] = Var[X] + (E[X])² = 900,000,000 + 1,600,000,000 = 2,500,000,000.
  4. E[S] = 20 × 40,000 = ₹8,00,000.
  5. Var[S] = 20 × 2,500,000,000 = 50,000,000,000.
  6. SD = √(5 × 10¹⁰) = 223,607 approximately.

Answer: Mean = ₹8,00,000; standard deviation ≈ ₹2,23,607

Exam tips

  • Write the definition clues you use: independent increments, stationary increments, Poisson(λt). Written questions give marks for stating them.
  • In compound questions, compute E[X²] explicitly and show it. Examiners look for the second moment.
  • For waiting-time questions, try converting to a count: the time to the nth event exceeds t exactly when N(t) ≤ n − 1.
  • Always check units and the rate per period before substituting. This is the most common lost mark.
  • When asked to prove a result such as thinning or superposition, work with the MGF or the small-interval definition and state your assumptions.

Practice questions from Stochastic processes

Poisson Processes: frequently asked questions

Why are Poisson process inter-arrival times exponential?

The time to the first event exceeds t exactly when no events occur in (0, t). That probability is e^(−λt), which is the exponential survival function. Independent and stationary increments make later gaps independent with the same distribution.

What is the mean and variance of a compound Poisson process?

For S(t) with Poisson rate λ and claim size X, the mean is λt E[X] and the variance is λt E[X²]. The variance uses the second moment about zero, not the variance of X.

What is the difference between superposition and thinning?

Superposition merges independent Poisson processes, and the new rate is the sum of the rates. Thinning splits one Poisson process by independent classification of events, and each type has rate equal to p times λ.

Is the Poisson process memoryless?

Yes. The future of the process after any fixed time is independent of the past and has the same law as the original process. In particular, the remaining wait for the next event is again Exponential(λ).