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

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

\[\label{eq:gam} X \sim \mathrm{Gam}(a, b) \; .\]

Then, the mode of $X$ is

\[\label{eq:gam-mode} \mathrm{mode}(X) = \left\{ \begin{array}{rl} \frac{a-1}{b} \; , & \text{if} \; a \geq 1 \\ 0 \; , & \text{if} \; a < 1 \; . \end{array} \right.\]

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 gamma distribution is:

\[\label{eq:gam-pdf} f_X(x) = \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \; .\]

The first two deriatives of this function are:

\[\label{eq:gam-pdf-der1} \begin{split} f'_X(x) = \frac{\mathrm{d}f_X(x)}{\mathrm{d}x} &= \frac{b^a}{\Gamma(a)} \left( -b x^{a-1} \exp[-b x] + (a-1) x^{a-2} \exp[-b x] \right) \\ &= \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \left( \frac{a-1}{x} - b \right) \end{split}\] \[\label{eq:gam-pdf-der2} \begin{split} f''_X(x) = \frac{\mathrm{d}^2f_X(x)}{\mathrm{d}x^2} &= \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \left( -\frac{a-1}{x^2} \right) + \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \left( \frac{a-1}{x} - b \right)^2 \\ &= \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \left( \left( \frac{a-1}{x} - b \right)^2 - \frac{a-1}{x^2} \right) \; . \end{split}\]

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

\[\label{eq:gam-mode-s1} \begin{split} f'_X(x) = 0 &= \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \left( \frac{a-1}{x} - b \right) \\ &\Leftrightarrow \\ 0 &= \frac{a-1}{x} - b \\ bx &= a-1 \\ x &= \frac{a-1}{b} \; . \end{split}\]

By plugging this value into the second deriative \eqref{eq:gam-pdf-der2}, we find:

\[\label{eq:gam-mode-s2} \begin{split} f''_X\left( \frac{a-1}{b} \right) &= \frac{b^a}{\Gamma(a)} \left( \frac{a-1}{b} \right)^{a-1} \exp\left[-b \left( \frac{a-1}{b} \right)\right] \left( \left( \frac{a-1}{\left( \frac{a-1}{b} \right)} - b \right)^2 - \frac{a-1}{\left( \frac{a-1}{b} \right)^2} \right) \\ &= \frac{b^a}{\Gamma(a)} \left( \frac{a-1}{b} \right)^{a-1} \exp\left[-(a-1)\right] \left( \left( b - b \right)^2 - \frac{b^2}{a-1} \right) \\ &= \frac{b^a}{\Gamma(a)} \left( \frac{a-1}{b} \right)^{a-1} \exp\left[-(a-1)\right] \left( -\frac{b^2}{a-1} \right) \; . \end{split}\]

Thus, if $a \geq 1$, then $f’'_X\left( (a-1)/b \right)$ is negative, such that $f_X(x)$ reaches its maximum at $x = (a-1)/b$. However, if $a < 1$, this value is negative and not in the support of the random variable $X$. In this case, the mode of $X$ is $0$, since $f_X(x)$ tends towards infinity at $x=0$:

\[\label{eq:gam-mode-s3} \begin{split} \lim_{\substack{x \rightarrow 0 \\ a < 1}} f_X(x) &= \lim_{\substack{x \rightarrow 0 \\ a < 1}} \frac{b^a}{\Gamma(a)} x^{a-1} \exp[-b x] \\ &= \frac{b^a}{\Gamma(a)} \lim_{\substack{x \rightarrow 0 \\ a < 1}} x^{a-1} = \infty \; . \end{split}\]
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Metadata: ID: P487 | shortcut: gam-mode | author: JoramSoch | date: 2025-02-06, 12:35.