Median of Exponential Distribution

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Theorem

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$.

Then the median of $X$ is equal to $\beta \ln 2$.


Proof

Let $M$ be the median of $X$.

From the definition of the exponential distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$

Note that $f_X$ is non-zero, so the median is unique.

We have by the definition of a median:

$\ds \map \Pr {X < M} = \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x = \frac 1 2$

Evaluating this integral:

\(\ds \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x\) \(=\) \(\ds \frac 1 \beta \sqbrk {-\beta e^{-\frac x \beta} }_0^M\) Primitive of Exponential Function
\(\ds \) \(=\) \(\ds 1 - e^{-\frac M \beta}\)


So:

$1 - e^{-\frac M \beta} = \dfrac 1 2$

Then:

$e^{-\frac M \beta} = \dfrac 1 2$

giving:

$-\dfrac M \beta = \map \ln {\dfrac 1 2}$

So, by Logarithm of Reciprocal:

$M = \beta \ln 2$

$\blacksquare$