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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

α\alpha
shape - number of events waited for - α>0\alpha > 0
λ\lambda
rate (inverse scale) - λ>0\lambda > 0
Support
x>0x > 0

The formulas

f(x)
f(x)=λαxα−1e−λxΓ(α)f(x) = \dfrac{\lambda^{\alpha} x^{\alpha - 1} e^{-\lambda x}}{\Gamma(\alpha)}
F(x)
F(x)=γ(α, λx)Γ(α)F(x) = \dfrac{\gamma(\alpha,\, \lambda x)}{\Gamma(\alpha)}

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
E[X]=αλE[X] = \dfrac{\alpha}{\lambda}
Variance
Var⁡(X)=αλ2\operatorname{Var}(X) = \dfrac{\alpha}{\lambda^{2}}
MGF
M(t)=(λλ−t)α,t<λM(t) = \left(\dfrac{\lambda}{\lambda - t}\right)^{\alpha}, \quad t < \lambda
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 X=∑k=1αEkX = \sum_{k=1}^{\alpha} E_k with independent Exponential(λ\lambda) waits (integer α\alpha).
  2. E[X]=α⋅1λE[X] = \alpha \cdot \frac{1}{\lambda} and, by independence, Var⁡(X)=α⋅1λ2\operatorname{Var}(X) = \alpha \cdot \frac{1}{\lambda^2}.
  3. The MGF is the exponential MGF to the α\alphath power, (λλ−t)α\left(\frac{\lambda}{\lambda-t}\right)^{\alpha}, and that form holds for ALL α>0\alpha > 0.
  4. Differentiating the MGF twice at 0 gives E[X]=α/λE[X] = \alpha/\lambda and E[X2]=α(α+1)λ2E[X^2] = \frac{\alpha(\alpha+1)}{\lambda^2}, confirming the variance.
  5. Gamma(α=1\alpha = 1) is exponential; Gamma(α=k/2\alpha = k/2, λ=1/2\lambda = 1/2) is chi-square with kk 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(λ=2\lambda = 2) process is Gamma(α=4\alpha = 4, λ=2\lambda = 2).
  2. Var⁡(X)=αλ2=44=1\operatorname{Var}(X) = \dfrac{\alpha}{\lambda^{2}} = \dfrac{4}{4} = 1.
  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 that test run.

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.