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Gamma distribution

The waiting time until the αth event in a Poisson process; a flexible positive, right-skewed severity model.

Parameters and support

shape — number of events waited for
rate (inverse scale)
Support

The formulas

f(x)
F(x)

Elementary only when α is a positive integer, where it collapses to a finite Poisson sum: F(x) = 1 − Σ e^(−λx)(λx)^k/k! for k = 0…α−1.

Mean
Variance
MGF
Memory hook. It is α independent Exponential(λ) waits added up, so mean and variance are α times the exponential's. Check the parameterisation: this site uses the RATE λ, so the mean is α/λ, not αλ.

Where the moments come from

  1. Write with independent Exponential() waits (integer ).
  2. and, by independence, .
  3. The MGF is the exponential MGF to the th power, , and that form holds for ALL .
  4. Differentiating the MGF twice at 0 gives and , confirming the variance.
  5. Gamma() is exponential; Gamma(, ) is chi-square with degrees of freedom.

Worked example

Claims arrive as a Poisson process at 2 per day. Find the variance of the waiting time until the 4th claim.

  1. The wait for the 4th event in a Poisson() process is Gamma(, ).
  2. .
  3. The variance is 1 day².

Answer: 1 day²

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 the build.

Traps

  • Mixing the rate parameterisation (mean α/λ) with the scale parameterisation (mean αθ). They are reciprocals.
  • Trying to integrate the CDF for non-integer α — it has no elementary form.
  • Forgetting that summing independent gammas requires the SAME rate λ.

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.