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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
Where the moments come from
- Write with independent Exponential() waits (integer ).
- and, by independence, .
- The MGF is the exponential MGF to the th power, , and that form holds for ALL .
- Differentiating the MGF twice at 0 gives and , confirming the variance.
- 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.
- The wait for the 4th event in a Poisson() process is Gamma(, ).
- .
- 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.