Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Beta distribution ▷ Mode

Theorem: Let $X$ be a random variable following a beta distribution:

\[\label{eq:beta} X \sim \mathrm{Bet}(\alpha, \beta) \; .\]

Then, the mode of $X$ is

\[\label{eq:beta-mode-p1} \mathrm{mode}(X) \in \left\{ \begin{array}{rl} \left\lbrace 0, 1 \right\rbrace \; , & \text{if} \quad \alpha < 1 \quad \text{and} \quad \beta < 1 \\ \left[ 0, 1 \right] \; , & \text{if} \quad \alpha = 1 \quad \text{and} \quad \beta = 1 \end{array} \right.\]

and

\[\label{eq:beta-mode-p2} \mathrm{mode}(X) = \left\{ \begin{array}{rl} 0 \; \text{or} \; 1 \; , & \text{if} \quad \alpha < 1 \quad \text{or} \quad \beta < 1 \quad (\text{but not} \; \alpha < 1 \; \text{and} \; \beta < 1) \\ \frac{\alpha-1}{\alpha+\beta-2} \; , & \text{if} \quad \alpha \geq 1 \quad \text{and} \quad \beta \geq 1 \quad (\text{but not} \; \alpha = 1 \; \text{and} \; \beta = 1) \; . \end{array} \right.\]

Proof: The mode is the value which maximizes the probability density function:

\[\label{eq:mode} \mathrm{mode}(X) = \operatorname*{arg\,max}_x f_X(x) \; .\]

The probability density function of the beta distribution is:

\[\label{eq:beta-pdf} f_X(x) = \frac{1}{\mathrm{B}(\alpha, \beta)} \, x^{\alpha-1} \, (1-x)^{\beta-1} \; .\]

1) If $\alpha < 1$, then 0 is the mode, since

\[\label{eq:beta-mode-p2a-s1} \lim_{\substack{x \rightarrow 0 \\ \alpha < 1}} f_X(x) = \infty \; , \quad \text{because} \quad \lim_{x \rightarrow 0} x^{\alpha-1} = \infty \quad \text{for} \quad \alpha < 1 \; ,\]

and if $\beta < 1$, then 1 is the mode, since

\[\label{eq:beta-mode-p2a-s2} \lim_{\substack{x \rightarrow 1 \\ \beta < 1}} f_X(x) = \infty \; , \quad \text{because} \quad \lim_{x \rightarrow 1} (1-x)^{\beta-1} = \infty \quad \text{for} \quad \beta < 1 \; .\]

2) If both $\alpha < 1$ and $\beta < 1$, then

\[\label{eq:beta-mode-p1a} \lim_{x \rightarrow 0} f_X(x) = \infty \quad \text{and} \quad \lim_{x \rightarrow 1} f_X(x) = \infty \; ,\]

so any value from the set $\left\lbrace 0, 1 \right\rbrace$ may be considered the mode.

3) If both $\alpha = 1$ and $\beta = 1$, then

\[\label{eq:beta-mode-p1b} \begin{split} f_X(x) &= \frac{1}{\mathrm{B}(1,1)} \, x^{1-1} \, (1-x)^{1-1} \\ &= \frac{\Gamma(2)}{\Gamma(1) \Gamma(1)} x^0 \, (1-x)^0 \\ &= 1 = \mathrm{const.} \; , \end{split}\]

i.e. the distribution becomes equivalent to the (standard) continuous uniform distribution with parameters $a = 0$ and $b = 1$ which has a constant probability density function. Thus, any value from the interval $\left[ 0,1 \right]$ may be considered the mode.

4) For the remaining cases, we must analyze the probability density function. The first two deriatives of this function are:

\[\label{eq:beta-pdf-der1} \begin{split} f'_X(x) = \frac{\mathrm{d}f_X(x)}{\mathrm{d}x} &= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ - (\beta-1) x^{\alpha-1} (1-x)^{\beta-2} + (\alpha-1) x^{\alpha-2} (1-x)^{\beta-1} \right] \\ &= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ (\alpha-1) x^{\alpha-2} (1-x)^{\beta-1} - (\beta-1) x^{\alpha-1} (1-x)^{\beta-2} \right] \end{split}\] \[\label{eq:beta-pdf-der2} \begin{split} f''_X(x) = \frac{\mathrm{d}^2f_X(x)}{\mathrm{d}x^2} &= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ (\alpha-1) \left( (\alpha-2) x^{\alpha-3} (1-x)^{\beta-1} - (\beta-1) x^{\alpha-2} (1-x)^{\beta-2} \right) - \right. \\ &\hphantom{=} \quad\quad\quad\quad\; \left. (\beta-1) \left( (\alpha-1) x^{\alpha-2} (1-x)^{\beta-2} - (\beta-2) x^{\alpha-1} (1-x)^{\beta-3} \right) \right] \\ &= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ (\alpha-1) (\alpha-2) x^{\alpha-3} (1-x)^{\beta-1} - 2 (\alpha-1) (\beta-1) x^{\alpha-2} (1-x)^{\beta-2} + \right. \\ &\hphantom{=} \quad\quad\quad\quad\; \left. (\beta-1) (\beta-2) x^{\alpha-1} (1-x)^{\beta-3} \right] \; . \end{split}\]

We now calculate the root of the first derivative \eqref{eq:beta-pdf-der1}:

\[\label{eq:beta-mode-p2b-s1} \begin{split} f'_X(x) = 0 &= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ (\alpha-1) x^{\alpha-2} (1-x)^{\beta-1} - (\beta-1) x^{\alpha-1} (1-x)^{\beta-2} \right] \\ &\Leftrightarrow \\ (\beta-1) x^{\alpha-1} (1-x)^{\beta-2} &= (\alpha-1) x^{\alpha-2} (1-x)^{\beta-1} \\ (\beta-1) x &= (\alpha-1) (1-x) \\ x [(\beta-1) + (\alpha-1)] &= \alpha-1 \\ x &= \frac{\alpha-1}{\alpha+\beta-2} \; . \end{split}\]

Note that for this quantity, we have

\[\label{eq:beta-mode-p2b-s2} \begin{split} \frac{\alpha-1}{\alpha+\beta-2} &< 0 \; , \quad \text{if} \quad \alpha < 1 \quad \text{and} \quad \beta > 2 - \alpha \\ \frac{\alpha-1}{\alpha+\beta-2} &> 1 \; , \quad \text{if} \quad \beta < 1 \quad \text{and} \quad \alpha > 2 - \beta \; . \end{split}\]

Also note that the following holds:

\[\label{eq:beta-mode-p2b-s3} 1 - x = 1 - \frac{\alpha-1}{\alpha+\beta-2} = \frac{\alpha+\beta-2}{\alpha+\beta-2} - \frac{\alpha-1}{\alpha+\beta-2} = \frac{\beta-1}{\alpha+\beta-2} \; .\]

By plugging $x$ and $1-x$ into the second deriative \eqref{eq:beta-pdf-der2}, we find:

\[\label{eq:beta-mode-p2b-s4} \begin{split} f''_X\left( \frac{\alpha-1}{\alpha+\beta-2} \right) &= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ (\alpha-1) (\alpha-2) \left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-3} \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-1} - \right. \\ &\hphantom{=} \quad\quad\quad\quad \left. 2 (\alpha-1) (\beta-1) \left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-2} \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-2} + \right. \\ &\hphantom{=} \quad\quad\quad\quad\; \left. (\beta-1) (\beta-2) \left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-1} \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-3} \right] \; . \end{split}\]

Multiplying with factors which are certainly positive, we can focus on those parts of the second derivative which determine its sign:

\[\label{eq:beta-mode-p2b-s5} \begin{split} \frac{f''_X(x) \cdot \mathrm{B}(\alpha, \beta)}{\left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-3} \cdot \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-3}} &= (\alpha-1) (\alpha-2) \left( \frac{\beta-1}{\alpha+\beta-2} \right)^2 - \\ &\hphantom{=} 2 (\alpha-1) (\beta-1) \left( \frac{\alpha-1}{\alpha+\beta-2} \right) \left( \frac{\beta-1}{\alpha+\beta-2} \right) + \\ &\hphantom{=} (\beta-1) (\beta-2) \left( \frac{\alpha-1}{\alpha+\beta-2} \right)^2 \; . \end{split}\]

Further multiplying with or dividing by terms which are necessarily positive and thus do not change the sign of the function value, we get:

\[\label{eq:beta-mode-p2b-s6} \begin{split} \frac{f''_X(x) \cdot \mathrm{B}(\alpha, \beta)}{\left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-3} \cdot \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-3}} \cdot \frac{(\alpha+\beta-2)^2}{(\alpha-1) (\beta-1)} &= (\alpha-2) (\beta-1) - 2 (\alpha-1) (\beta-1) + (\alpha-1) (\beta-2) \\ &= (\alpha-1) [(\beta-2)-(\beta-1)] + (\beta-1) [(\alpha-1)-(\alpha-2)] \\ &= - (\alpha-1) - (\beta-1) \\ &< 0, \quad \text{if} \quad \alpha > 1 \quad \text{and} \quad \beta > 1 \; . \end{split}\]

Thus, $f’'_X(x)$ is negative for $x = \frac{\alpha-1}{\alpha+\beta-2}$, demonstrating that this is a maximum. To summarize:

  • If $\alpha < 1$ and $\beta < 1$, then $f_X(x)$ diverges at both ends and both values from the set $\left\lbrace 0, 1 \right\rbrace$ can be seen as the mode of $X$.

  • If $\alpha < 1$ or $\beta < 1$ (but not $\alpha < 1$ and $\beta < 1$), then the mode of $X$ is 0 or 1, because $f_X(x)$ tends towards infinity at $x = 0$ or $x = 1$.

  • If $\alpha = 1$ and $\beta = 1$, then $f_X(x)$ is constant and any value in the interval $\left[ 0,1 \right]$ can be seen as the mode of $X$.

  • If $\alpha \geq 1$ and $\beta \geq 1$ (but not $\alpha = 1$ and $\beta = 1$), then $0 < x = \frac{\alpha-1}{\alpha+\beta-2} < 1$ and $f’_X(x) = 0$ and $f’'_X(x) < 0$, such that $f_X(x)$ reaches its machimum at $\frac{\alpha-1}{\alpha+\beta-2} = \mathrm{mode}(X)$.

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Metadata: ID: P520 | shortcut: beta-mode | author: JoramSoch | date: 2025-10-24, 13:48.