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

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

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

Then, the median of $X$ is

\[\label{eq:exp-med} \mathrm{median}(X) = \frac{\ln 2}{\lambda} \; .\]

Proof: The median is the value at which the cumulative distribution function is $1/2$:

\[\label{eq:median} F_X(\mathrm{median}(X)) = \frac{1}{2} \; .\]

The cumulative distribution function of the exponential distribution is

\[\label{eq:exp-cdf} F_X(x) = 1 - \exp[-\lambda x], \quad x \geq 0 \; .\]

Thus, the inverse CDF is

\[\label{eq:exp-cdf-inv} x = -\frac{\ln(1-p)}{\lambda}\]

and setting $p = 1/2$, we obtain:

\[\label{eq:exp-med-qed} \mathrm{median}(X) = -\frac{\ln(1-\frac{1}{2})}{\lambda} = \frac{\ln 2}{\lambda} \; .\]
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Metadata: ID: P49 | shortcut: exp-med | author: JoramSoch | date: 2020-02-11, 15:03.