Median of Continuous Uniform Distribution

Theorem

Let $X$ be a continuous random variable which is uniformly distributed on a closed real interval $\closedint a b$.

Then the median $M$ of $X$ is given by:

$M = \dfrac {a + b} 2$

Proof

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

$\map {f_X} x = \dfrac 1 {b - a}$

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

We have by the definition of a median:

$\ds \map \Pr {X < M} = \int_a^M \frac 1 {b - a} \rd x = \frac 1 2$

This article, or a section of it, needs explaining. In particular: Reconcile the fact that the definition of the median is actually taken from $-\infty$ not $a$. A bit of fancy footwork needed to get the lower limit in place. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

We have, by Primitive of Constant:

$\dfrac {M - a} {b - a} = \dfrac 1 2$

So:

\(\ds M\)

\(=\)

\(\ds a + \frac {b - a} 2\)

\(\ds \)

\(=\)

\(\ds \frac {b - a + 2 a} 2\)

\(\ds \)

\(=\)

\(\ds \frac {b + a} 2\)

$\blacksquare$