Median of Cauchy Distribution

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Theorem

Let $X$ be a continuous random variable with a Cauchy distribution:

$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$

for:

$\lambda \in \R_{>0}$
$\gamma \in \R$

The median of $X$ is $\gamma$.


Proof

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

$\map {f_X} x = \dfrac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda }^2} }$

Note that $f_X$ is non-zero, sufficient to ensure a unique median.

By the definition of a median, to prove that $\gamma$ is the median of $X$ we must verify:

$\ds \map \Pr {X < \gamma} = \int_{-\infty}^\gamma \map {f_X} x \rd x = \frac 1 2$

We have:

\(\ds \int_{-\infty}^\gamma \map {f_X} x \rd x\) \(=\) \(\ds \frac 1 {\pi \lambda} \int_{-\infty}^\gamma \dfrac 1 {\paren {1 + \paren {\frac {x - \gamma} \lambda }^2} } \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 \pi \int_{-\infty}^0 \dfrac 1 {1 + t^2} \rd t\) substituting $t = \dfrac {x - \gamma} {\lambda}$
\(\ds \) \(=\) \(\ds \dfrac 1 \pi \bigintlimits {\arctan t} {-\infty} {0}\) Primitive of $\dfrac 1 {1 + x^2}$
\(\ds \) \(=\) \(\ds \dfrac 1 \pi \paren {0 - \paren {-\dfrac \pi 2} }\) Definition of Arctangent and Arctangent of Zero is Zero
\(\ds \) \(=\) \(\ds \frac 1 2\)

$\blacksquare$


Sources

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