Index: The Book of Statistical Proofs ▷ Probability 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:

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

The probability density function of the exponential distribution is:

\[\label{eq:exp-pdf} f_X(x) = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; x < 0 \\ \lambda e^{-\lambda x} \; , & \text{if} \; x \geq 0 \; . \end{array} \right.\]

Since

\[\label{eq:exp-pdf-eq0} f_X(0) = \lambda\]

and

\[\label{eq:exp-pdf-neq0} 0 < e^{-\lambda x} < 1 \quad \text{for any} \quad x > 0 \; ,\]

it follows that

\[\label{eq:exp-mode-qed} \mathrm{mode}(X) = 0 \; .\]
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Metadata: ID: P51 | shortcut: exp-mode | author: kantundpeterpan | date: 2020-02-12, 15:53.