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‎P/beta-med.md‎

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---
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layout: proof
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mathjax: true
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author: "Joram Soch"
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affiliation: "BCCN Berlin"
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e_mail: "joram.soch@bccn-berlin.de"
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date: 2025-10-24 13:45:52
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title: "Median of the beta distribution"
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chapter: "Probability Distributions"
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section: "Univariate continuous distributions"
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topic: "Beta distribution"
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theorem: "Median"
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sources:
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- authors: "Wikipedia"
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year: 2025
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title: "Beta distribution"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2025-10-24"
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url: "https://en.wikipedia.org/wiki/Beta_distribution#Median"
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proof_id: "P519"
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shortcut: "beta-med"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [random variable](/D/rvar) following a [beta distribution](/D/beta):
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$$ \label{eq:beta}
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X \sim \mathrm{Bet}(\alpha, \beta) \; .
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$$
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Then, the [median](/D/med) of $X$ is
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$$ \label{eq:beta-med}
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\mathrm{median}(X) = \mathrm{B}_{\alpha,\beta}^{-1}\left( \frac{\mathrm{B}(\alpha,\beta)}{2} \right) = \mathrm{I}_{1/2}^{-1}(\alpha,\beta)
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$$
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where $\mathrm{B}(a,b)$ is the beta function, $\mathrm{B}_{a,b}^{-1}(y)$ is the inverse function of the incomplete beta function $\mathrm{B}(x; a, b)$ and $\mathrm{I}_y^{-1}(a,b)$ is the inverse function of the regularized incomplete beta function.
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**Proof:** The [median](/D/med) is the value at which the [cumulative distribution function](/D/cdf) is $1/2$:
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$$ \label{eq:median}
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F_X(\mathrm{median}(X)) = \frac{1}{2} \; .
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$$
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The [cumulative distribution function of the beta distribution](/P/beta-cdf) is
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$$ \label{eq:beta-cdf}
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F_X(x) = \frac{\mathrm{B}(x; \alpha, \beta)}{\mathrm{B}(\alpha, \beta)}
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$$
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where the incomplete beta function $\mathrm{B}(x; a, b)$ is given by
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$$ \label{eq:inc-beta-fct}
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\mathrm{B}(x; a, b) = \int_0^x t^{a-1} (1-t)^{b-1} \, \mathrm{d}t \; .
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$$
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Thus, the inverse CDF, i.e. the [quantile function](/D/qf), is
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$$ \label{eq:beta-cdf-inv}
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x = \mathrm{B}_{\alpha,\beta}^{-1}(p \, \mathrm{B}(\alpha,\beta))
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$$
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where $\mathrm{B}_{a,b}^{-1}(y)$ is the inverse function of $\mathrm{B}(x; a, b)$. Setting $p = 1/2$, we obtain:
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$$ \label{eq:beta-med-qed}
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\mathrm{median}(X) = \mathrm{B}_{\alpha,\beta}^{-1}\left( \frac{\mathrm{B}(\alpha,\beta)}{2} \right) \; .
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$$
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Alternatively, the cumulative distribution function may be written as
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$$ \label{eq:beta-cdf-alt}
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F_X(x) = \mathrm{I}_x(a,b)
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$$
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using the regularized incomplete beta function
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$$ \label{eq:reg-inc-beta-fct}
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\mathrm{I}_x(a,b) = \frac{\mathrm{B}(x; \alpha, \beta)}{\mathrm{B}(\alpha, \beta)}
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$$
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in which case the inverse CDF is
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$$ \label{eq:beta-cdf-inv-alt}
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x = \mathrm{I}_p^{-1}(a,b) \; ,
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$$
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such that the median becomes
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$$ \label{eq:beta-med-qed-alt}
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\mathrm{median}(X) = \mathrm{I}_{1/2}^{-1}(\alpha,\beta) \; .
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$$

‎P/beta-mode.md‎

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---
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layout: proof
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mathjax: true
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author: "Joram Soch"
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affiliation: "BCCN Berlin"
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e_mail: "joram.soch@bccn-berlin.de"
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date: 2025-10-24 13:48:56
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title: "Mode of the beta distribution"
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chapter: "Probability Distributions"
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section: "Univariate continuous distributions"
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topic: "Beta distribution"
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theorem: "Mode"
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sources:
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- authors: "Wikipedia"
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year: 2025
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title: "Beta distribution"
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in: "Wikipedia, the free encyclopedia"
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pages: "retrieved on 2025-10-24"
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url: "https://en.wikipedia.org/wiki/Beta_distribution#Mode"
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proof_id: "P520"
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shortcut: "beta-mode"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [random variable](/D/rvar) following a [beta distribution](/D/beta):
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$$ \label{eq:beta}
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X \sim \mathrm{Bet}(\alpha, \beta) \; .
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$$
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Then, the [mode](/D/mode) of $X$ is
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$$ \label{eq:beta-mode-p1}
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\mathrm{mode}(X) \in \left\{
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\begin{array}{rl}
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\left\lbrace 0, 1 \right\rbrace \; , & \text{if} \quad \alpha < 1 \quad \text{and} \quad \beta < 1 \\
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\left[ 0, 1 \right] \; , & \text{if} \quad \alpha = 1 \quad \text{and} \quad \beta = 1
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\end{array}
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\right.
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$$
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and
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$$ \label{eq:beta-mode-p2}
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\mathrm{mode}(X) = \left\{
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\begin{array}{rl}
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0 \; \text{or} \; 1 \; ,
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& \text{if} \quad \alpha < 1 \quad \text{or} \quad \beta < 1 \quad (\text{but not} \; \alpha < 1 \; \text{and} \; \beta < 1) \\
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\frac{\alpha-1}{\alpha+\beta-2} \; ,
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& \text{if} \quad \alpha \geq 1 \quad \text{and} \quad \beta \geq 1 \quad (\text{but not} \; \alpha = 1 \; \text{and} \; \beta = 1) \; .
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\end{array}
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\right.
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$$
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**Proof:** The [mode](/D/mode) is the value which maximizes the [probability density function](/D/pdf):
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$$ \label{eq:mode}
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\mathrm{mode}(X) = \operatorname*{arg\,max}_x f_X(x) \; .
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$$
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The [probability density function of the beta distribution](/P/beta-pdf) is:
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$$ \label{eq:beta-pdf}
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f_X(x) = \frac{1}{\mathrm{B}(\alpha, \beta)} \, x^{\alpha-1} \, (1-x)^{\beta-1} \; .
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$$
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1) If $\alpha < 1$, then 0 is the mode, since
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$$ \label{eq:beta-mode-p2a-s1}
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\lim_{\substack{x \rightarrow 0 \\ \alpha < 1}} f_X(x) = \infty \; ,
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\quad \text{because} \quad
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\lim_{x \rightarrow 0} x^{\alpha-1} = \infty
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\quad \text{for} \quad
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\alpha < 1 \; ,
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$$
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and if $\beta < 1$, then 1 is the mode, since
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$$ \label{eq:beta-mode-p2a-s2}
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\lim_{\substack{x \rightarrow 1 \\ \beta < 1}} f_X(x) = \infty \; ,
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\quad \text{because} \quad
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\lim_{x \rightarrow 1} (1-x)^{\beta-1} = \infty
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\quad \text{for} \quad
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\beta < 1 \; .
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$$
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2) If both $\alpha < 1$ and $\beta < 1$, then
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$$ \label{eq:beta-mode-p1a}
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\lim_{x \rightarrow 0} f_X(x) = \infty
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\quad \text{and} \quad
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\lim_{x \rightarrow 1} f_X(x) = \infty \; ,
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$$
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so any value from the set $\left\lbrace 0, 1 \right\rbrace$ may be considered the mode.
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3) If both $\alpha = 1$ and $\beta = 1$, then
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$$ \label{eq:beta-mode-p1b}
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\begin{split}
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f_X(x)
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&= \frac{1}{\mathrm{B}(1,1)} \, x^{1-1} \, (1-x)^{1-1} \\
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&= \frac{\Gamma(2)}{\Gamma(1) \Gamma(1)} x^0 \, (1-x)^0 \\
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&= 1 = \mathrm{const.} \; ,
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\end{split}
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$$
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i.e. the distribution becomes equivalent to the (standard) [continuous uniform distribution](/D/cuni) with parameters $a = 0$ and $b = 1$ which has a [constant probability density function](/P/cuni-pdf). Thus, any value from the interval $\left[ 0,1 \right]$ [may be considered the mode](/P/cuni-mode).
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4) For the remaining cases, we must analyze the probability density function. The first two deriatives of this function are:
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$$ \label{eq:beta-pdf-der1}
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\begin{split}
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f'_X(x)
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= \frac{\mathrm{d}f_X(x)}{\mathrm{d}x}
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&= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ - (\beta-1) x^{\alpha-1} (1-x)^{\beta-2} + (\alpha-1) x^{\alpha-2} (1-x)^{\beta-1} \right] \\
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&= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ (\alpha-1) x^{\alpha-2} (1-x)^{\beta-1} - (\beta-1) x^{\alpha-1} (1-x)^{\beta-2} \right]
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\end{split}
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$$
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$$ \label{eq:beta-pdf-der2}
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\begin{split}
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f''_X(x)
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= \frac{\mathrm{d}^2f_X(x)}{\mathrm{d}x^2}
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&= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[
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(\alpha-1) \left( (\alpha-2) x^{\alpha-3} (1-x)^{\beta-1} - (\beta-1) x^{\alpha-2} (1-x)^{\beta-2} \right) - \right. \\
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&\hphantom{=} \quad\quad\quad\quad\;
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\left. (\beta-1) \left( (\alpha-1) x^{\alpha-2} (1-x)^{\beta-2} - (\beta-2) x^{\alpha-1} (1-x)^{\beta-3} \right)
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\right] \\
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&= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[
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(\alpha-1) (\alpha-2) x^{\alpha-3} (1-x)^{\beta-1} - 2 (\alpha-1) (\beta-1) x^{\alpha-2} (1-x)^{\beta-2} + \right. \\
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&\hphantom{=} \quad\quad\quad\quad\;
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\left. (\beta-1) (\beta-2) x^{\alpha-1} (1-x)^{\beta-3}
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\right] \; .
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\end{split}
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$$
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We now calculate the root of the first derivative \eqref{eq:beta-pdf-der1}:
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$$ \label{eq:beta-mode-p2b-s1}
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\begin{split}
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f'_X(x)
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= 0
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&= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[ (\alpha-1) x^{\alpha-2} (1-x)^{\beta-1} - (\beta-1) x^{\alpha-1} (1-x)^{\beta-2} \right] \\
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&\Leftrightarrow \\
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(\beta-1) x^{\alpha-1} (1-x)^{\beta-2} &= (\alpha-1) x^{\alpha-2} (1-x)^{\beta-1} \\
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(\beta-1) x &= (\alpha-1) (1-x) \\
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x [(\beta-1) + (\alpha-1)] &= \alpha-1 \\
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x &= \frac{\alpha-1}{\alpha+\beta-2} \; .
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\end{split}
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$$
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Note that for this quantity, we have
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$$ \label{eq:beta-mode-p2b-s2}
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\begin{split}
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\frac{\alpha-1}{\alpha+\beta-2} &< 0 \; ,
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\quad \text{if} \quad \alpha < 1
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\quad \text{and} \quad \beta > 2 - \alpha \\
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\frac{\alpha-1}{\alpha+\beta-2} &> 1 \; ,
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\quad \text{if} \quad \beta < 1
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\quad \text{and} \quad \alpha > 2 - \beta \; .
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\end{split}
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$$
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Also note that the following holds:
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$$ \label{eq:beta-mode-p2b-s3}
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1 - x
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= 1 - \frac{\alpha-1}{\alpha+\beta-2}
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= \frac{\alpha+\beta-2}{\alpha+\beta-2} - \frac{\alpha-1}{\alpha+\beta-2}
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= \frac{\beta-1}{\alpha+\beta-2} \; .
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$$
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By plugging $x$ and $1-x$ into the second deriative \eqref{eq:beta-pdf-der2}, we find:
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$$ \label{eq:beta-mode-p2b-s4}
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\begin{split}
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f''_X\left( \frac{\alpha-1}{\alpha+\beta-2} \right)
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&= \frac{1}{\mathrm{B}(\alpha, \beta)} \left[
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(\alpha-1) (\alpha-2) \left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-3} \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-1} - \right. \\
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&\hphantom{=} \quad\quad\quad\quad
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\left. 2 (\alpha-1) (\beta-1) \left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-2} \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-2} + \right. \\
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&\hphantom{=} \quad\quad\quad\quad\;
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\left. (\beta-1) (\beta-2) \left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-1} \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-3} \right] \; .
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\end{split}
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$$
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Multiplying with factors which are certainly positive, we can focus on those parts of the second derivative which determine its sign:
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$$ \label{eq:beta-mode-p2b-s5}
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\begin{split}
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\frac{f''_X(x) \cdot \mathrm{B}(\alpha, \beta)}{\left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-3} \cdot \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-3}}
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&= (\alpha-1) (\alpha-2) \left( \frac{\beta-1}{\alpha+\beta-2} \right)^2 - \\
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&\hphantom{=} 2 (\alpha-1) (\beta-1) \left( \frac{\alpha-1}{\alpha+\beta-2} \right) \left( \frac{\beta-1}{\alpha+\beta-2} \right) + \\
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&\hphantom{=} (\beta-1) (\beta-2) \left( \frac{\alpha-1}{\alpha+\beta-2} \right)^2 \; .
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\end{split}
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$$
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Further multiplying with or dividing by terms which are necessarily positive and thus do not change the sign of the function value, we get:
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$$ \label{eq:beta-mode-p2b-s6}
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\begin{split}
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\frac{f''_X(x) \cdot \mathrm{B}(\alpha, \beta)}{\left( \frac{\alpha-1}{\alpha+\beta-2} \right)^{\alpha-3} \cdot \left( \frac{\beta-1}{\alpha+\beta-2} \right)^{\beta-3}} \cdot \frac{(\alpha+\beta-2)^2}{(\alpha-1) (\beta-1)}
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&= (\alpha-2) (\beta-1) - 2 (\alpha-1) (\beta-1) + (\alpha-1) (\beta-2) \\
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&= (\alpha-1) [(\beta-2)-(\beta-1)] + (\beta-1) [(\alpha-1)-(\alpha-2)] \\
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&= - (\alpha-1) - (\beta-1) \\
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&< 0,
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\quad \text{if} \quad \alpha > 1
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\quad \text{and} \quad \beta > 1 \; .
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\end{split}
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$$
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Thus, $f''_X(x)$ is negative for $x = \frac{\alpha-1}{\alpha+\beta-2}$, demonstrating that this is a maximum. To summarize:
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* 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 $X$.
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* 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 $x = 1$.
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* 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 $X$.
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* If $\alpha \geq 1$ and $\beta \geq 1$ (but not $\alpha = 1$ and $\beta = 1$), then $0 < x = < 1$ and $f'_X(x) = 0$ and $f''_X(x) < 0$, such that $f_X(x)$ reaches its machimum at $\mathrm{mode}(X) = \frac{\alpha-1}{\alpha+\beta-2}$.

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