Exam PDiscreteFree to read

Poisson distribution

The number of events in a fixed interval when events occur independently at a constant average rate.

Parameters and support

λ\lambda
expected number of events in the interval - λ>0\lambda > 0
Support
x∈{0,1,2,… }x \in \{0, 1, 2, \dots\}

The formulas

p(x)
p(x)=e−λλxx!p(x) = \dfrac{e^{-\lambda}\lambda^{x}}{x!}
F(x)
F(x)=e−λ∑k=0⌊x⌋λkk!F(x) = e^{-\lambda}\displaystyle\sum_{k=0}^{\lfloor x \rfloor} \dfrac{\lambda^{k}}{k!}
Mean
E[X]=λE[X] = \lambda
Variance
Var⁡(X)=λ\operatorname{Var}(X) = \lambda
MGF
M(t)=eλ(et−1)M(t) = e^{\lambda\left(e^{t}-1\right)}
Memory hook. Mean = variance = λ. Rescaling the interval rescales λ proportionally: 3 per hour over 20 minutes is λ = 1, not 3.

Where the moments come from

  1. Take Binomial(nn, λ/n\lambda/n) and let n→∞n \to \infty: (nx)(λ/n)x(1−λ/n)n−x→e−λλxx!\binom{n}{x}(\lambda/n)^x(1-\lambda/n)^{n-x} \to \frac{e^{-\lambda}\lambda^x}{x!}.
  2. Mean: E[X]=lim⁡n⋅λn=λE[X] = \lim n \cdot \frac{\lambda}{n} = \lambda.
  3. Variance: lim⁡nλn(1−λn)=λ\lim n\frac{\lambda}{n}(1-\frac{\lambda}{n}) = \lambda, since the correction vanishes.
  4. Directly from the MGF: M(t)=eλ(et−1)M(t) = e^{\lambda(e^t-1)}, so M′(0)=λM'(0) = \lambda and M′′(0)=λ+λ2M''(0) = \lambda + \lambda^2, giving Var⁡=λ\operatorname{Var} = \lambda.
  5. Independent Poissons ADD: if X∼Poisson(λ1)X \sim \text{Poisson}(\lambda_1) and Y∼Poisson(λ2)Y \sim \text{Poisson}(\lambda_2) are independent, X+Y∼Poisson(λ1+λ2)X + Y \sim \text{Poisson}(\lambda_1+\lambda_2).

Worked example

Claims arrive at a call centre as a Poisson process at 3 per hour. Find the probability of exactly 2 claims in a 20-minute window.

  1. Rescale the rate to the window: 20 minutes is 13\tfrac{1}{3} of an hour, so λ=3×13=1\lambda = 3 \times \tfrac{1}{3} = 1.
  2. P(X=2)=e−1⋅122!=e−12P(X = 2) = \dfrac{e^{-1} \cdot 1^{2}}{2!} = \dfrac{e^{-1}}{2}.
  3. e−1=0.367879e^{-1} = 0.367879, so P(X=2)=0.18394P(X = 2) = 0.18394.

Answer: 0.1839

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run - summed over the support for a discrete distribution, integrated by quadrature for a continuous one - and compared with the closed forms printed here. A typo on this page fails that test run.

Traps

  • Leaving λ at the stated per-unit rate when the question asks about a different-length interval.
  • Adding Poisson parameters for DEPENDENT streams - the addition rule needs independence.
  • Confusing the count (Poisson) with the waiting time between events (exponential).

Related

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.