Cauchy Distribution is Symmetric about Median

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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:

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

$X$ is (reflectionally) symmetric about the vertical line through the median $\gamma$.


Proof

Recall from Median of Cauchy Distribution‎ that $\gamma$ is indeed the median of $X$.

\(\ds \map {f_X} {2 \gamma - x}\) \(=\) \(\ds \frac 1 {\pi \lambda \paren {1 + \paren {\frac {(2 \gamma - x) - \gamma} \lambda}^2} }\)
\(\ds \) \(=\) \(\ds \frac 1 {\pi \lambda \paren {1 + \paren {\frac {\gamma - x} \lambda}^2} }\)
\(\ds \) \(=\) \(\ds \frac 1 {\pi \lambda \paren {1 + \paren {\frac {x - \gamma} \lambda}^2} }\)
\(\ds \) \(=\) \(\ds \map {f_X} x\)

So $X$ is symmetric about the line $x = \gamma$.

$\blacksquare$


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