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Beta distribution

A proportion or probability constrained to [0, 1], with a shape set by two counts.

Parameters and support

α\alpha
first shape parameter - α>0\alpha > 0
β\beta
second shape parameter - β>0\beta > 0
Support
0<x<10 < x < 1

The formulas

f(x)
f(x)=Γ(α+β)Γ(α)Γ(β)xα−1(1−x)β−1f(x) = \dfrac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}
F(x)
F(x)=Ix(α,β)F(x) = I_{x}(\alpha, \beta)

The regularized incomplete beta function. Elementary only when α or β is a small positive integer.

Mean
E[X]=αα+βE[X] = \dfrac{\alpha}{\alpha+\beta}
Variance
Var⁡(X)=αβ(α+β)2(α+β+1)\operatorname{Var}(X) = \dfrac{\alpha\beta}{(\alpha+\beta)^{2}(\alpha+\beta+1)}
MGF

Not examinable for this distribution.

The MGF is a confluent hypergeometric series and is not examinable. Use E[Xᵏ] = ∏(α+j)/(α+β+j) for j = 0…k−1.

Memory hook. Read α and β as pseudo-counts of successes and failures: the mean α/(α+β) is exactly the sample proportion they imply. Beta(1,1) is the standard uniform.

Where the moments come from

  1. The normalising constant is 1/B(α,β)1/B(\alpha,\beta) with B(α,β)=Γ(α)Γ(β)Γ(α+β)B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.
  2. E[Xk]=B(α+k,β)B(α,β)=Γ(α+k)Γ(α+β)Γ(α)Γ(α+β+k)E[X^k] = \frac{B(\alpha+k,\beta)}{B(\alpha,\beta)} = \frac{\Gamma(\alpha+k)\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\alpha+\beta+k)}.
  3. k=1k = 1 gives E[X]=αα+βE[X] = \frac{\alpha}{\alpha+\beta}.
  4. k=2k = 2 gives E[X2]=α(α+1)(α+β)(α+β+1)E[X^2] = \frac{\alpha(\alpha+1)}{(\alpha+\beta)(\alpha+\beta+1)}.
  5. Subtracting the squared mean collapses to αβ(α+β)2(α+β+1)\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}.

Worked example

The proportion of a portfolio that lapses in a year follows a Beta distribution with α = 2 and β = 8. Find the mean and variance of the lapse proportion.

  1. E[X]=αα+β=210=0.2E[X] = \dfrac{\alpha}{\alpha+\beta} = \dfrac{2}{10} = 0.2.
  2. Var⁡(X)=αβ(α+β)2(α+β+1)=2×8102×11\operatorname{Var}(X) = \dfrac{\alpha\beta}{(\alpha+\beta)^{2}(\alpha+\beta+1)} = \dfrac{2 \times 8}{10^{2} \times 11}.
  3. =161100=0.014545= \dfrac{16}{1100} = 0.014545.
  4. The standard deviation is 0.014545=0.1206\sqrt{0.014545} = 0.1206.

Answer: 0.014545

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

  • Writing (α+β+1) as (α+β)+1 in the wrong place - the cube-like denominator is (α+β)²(α+β+1).
  • Forgetting that Beta(1,1) is uniform, which is often the fastest sanity check on an answer.

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.