Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Beta distribution ▷ Special case of Dirichlet distribution

Theorem: The beta distribution with shape parameters $\alpha$ and $\beta$ is a special case of the Dirichlet distribution with concentration parameters $\alpha_1 = \alpha$ and $\alpha_2 = \beta$:

\[\label{eq:beta-dir} X \sim \mathrm{Dir}(\left[ \alpha, \beta \right]) \quad \Rightarrow \quad X \sim \mathrm{Bet}(\alpha, \beta) \; .\]

Proof: The probability density function of the Dirichlet distribution, where $x$ is a $1 \times k$ vector with $x_i \in [0,1]$ for $i = 1,\ldots,k$, is as follows:

\[\label{eq:dir-pdf} \mathrm{Dir}(x; \left[ \alpha_1, \ldots, \alpha_k \right]) = \frac{\Gamma\left( \sum_{i=1}^k \alpha_i \right)}{\prod_{i=1}^k \Gamma(\alpha_i)} \, \prod_{i=1}^k {x_i}^{\alpha_i-1} \; .\]

If we let $k = 2$, $\alpha_1 = \alpha$ and $\alpha_2 = \beta$, we obtain

\[\label{eq:dir-pdf-beta-s1} \mathrm{Dir}(x; \left[ \alpha, \beta \right]) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \, \Gamma(\beta)} \, x_1^{\alpha-1} \, x_2^{\beta-1} \; .\]

Applying the beta function $\mathrm{B}(x,y) = \Gamma(x) \Gamma(y) / \Gamma(x+y)$, we have

\[\label{eq:dir-pdf-beta-s2} \mathrm{Dir}(x; \left[ \alpha, \beta \right]) = \frac{1}{\mathrm{B}(\alpha,\beta)} \, x_1^{\alpha-1} \, x_2^{\beta-1} \; .\]

Recognizing that $\sum_{i=1}^{k} x_i = 1$, such that $x_2 = 1 - x_1$, we get

\[\label{eq:dir-pdf-beta-s3} \mathrm{Dir}(x; \left[ \alpha, \beta \right]) = \frac{1}{\mathrm{B}(\alpha,\beta)} \, x_1^{\alpha-1} \, (1-x_1)^{\beta-1}\]

which is equivalent to the probability density function of the beta distribution.

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Metadata: ID: P496 | shortcut: beta-dir | author: JoramSoch | date: 2025-04-04, 14:41.