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Beta distribution

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

Parameters and support

$\alpha$

first shape parameter — $\alpha >0$

$\beta$

second shape parameter — $\beta >0$

Support

$0<x<1$

The formulas

f(x)

$$f(x)=\frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}{x}^{\alpha -1}(1-x{)}^{\beta -1}$$

F(x)

$$F(x)=I_x(\alpha ,\beta )$$

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

Mean

$$E[X]=\frac{\alpha }{\alpha +\beta }$$

Variance

$$\mathrm{Var}(X)=\frac{\alpha \beta }{(\alpha +\beta {)}^2(\alpha +\beta +1)}$$

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.

Where the moments come from

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

Worked example

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

• 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

• Continuous uniform distribution
• Gamma distribution
• Binomial distribution

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.