Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Exponential distribution ▷ Mode

Theorem: Let $X$ be a random variable following an exponential distribution:

$\label{eq:exp} X \sim \mathrm{Exp}(\lambda) \; .$

Then, the mode of $X$ is

$\label{eq:exp-mode} \mathrm{mode}(X) = 0 \; .$

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

$\label{eq:mode} \mathrm{mode}(X) = \operatorname*{arg\,max}_x f_X(x) \; .$ $\label{eq:exp-pdf} f_X(x) = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; x < 0 \\ \lambda \exp[-\lambda x] \; , & \text{if} \; x \geq 0 \; . \end{array} \right.$

Since

$\label{eq:exp-pdf-eq0} \lim_{x \to 0} f_X(x) = \infty$

and

$\label{eq:exp-pdf-neq0} f_X(x) < \infty \quad \text{for any} \quad x \neq 0 \; ,$

it follows that

$\label{eq:exp-mode-qed} \mathrm{mode}(X) = 0 \; .$
Sources:

Metadata: ID: P51 | shortcut: exp-mode | author: JoramSoch | date: 2020-02-12, 15:53.