Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Exponential distribution ▷ Quantile function

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

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

Then, the quantile function of $X$ is

\[\label{eq:exp-qf} Q_X(p) = \left\{ \begin{array}{rl} -\infty \; , & \text{if} \; p = 0 \\ -\frac{\ln(1-p)}{\lambda} \; , & \text{if} \; p > 0 \; . \end{array} \right.\]

Proof: The cumulative distribution function of the exponential distribution is:

\[\label{eq:exp-cdf} F_X(x) = \left\{ \begin{array}{rl} 0 \; , & \text{if} \; x < 0 \\ 1 - \exp[-\lambda x] \; , & \text{if} \; x \geq 0 \; . \end{array} \right.\]

The quantile function $Q_X(p)$ is defined as the smallest $x$, such that $F_X(x) = p$:

\[\label{eq:qf} Q_X(p) = \min \left\lbrace x \in \mathbb{R} \, \vert \, F_X(x) = p \right\rbrace \; .\]

Thus, we have $Q_X(p) = -\infty$, if $p = 0$. When $p > 0$, it holds that

\[\label{eq:exp-qf-s1} Q_X(p) = F_X^{-1}(x) \; .\]

This can be derived by rearranging equation \eqref{eq:exp-cdf}:

\[\label{eq:exp-qf-s2} \begin{split} p &= 1 - \exp[-\lambda x] \\ \exp[-\lambda x] &= 1-p \\ -\lambda x &= \ln(1-p) \\ x &= -\frac{\ln(1-p)}{\lambda} \; . \end{split}\]
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Metadata: ID: P50 | shortcut: exp-qf | author: JoramSoch | date: 2020-02-12, 15:48.