Variance of Chi Distribution

Theorem

Let $n$ be a strictly positive integer.

Let $X \sim \chi_n$ where $\chi_n$ is the chi distribution with $n$ degrees of freedom.

Then the variance of $X$ is given by:

$\var X = n - 2 \paren {\dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} } }^2$

where $\Gamma$ is the gamma function.

Proof

By Variance as Expectation of Square minus Square of Expectation, we have:

$\var X = \expect {X^2} - \paren {\expect X}^2$

By Square of Chi Random Variable has Chi-Squared Distribution:

$X^2 \sim \chi^2_n$

So, by Expectation of Chi-Squared Distribution:

$\expect {X^2} = n$

By Expectation of Chi Distribution:

$\expect X = \sqrt 2 \dfrac {\map \Gamma {\paren {n + 1} / 2} } {\map \Gamma {n / 2} }$

Hence the result.

$\blacksquare$