Skewness of Chi-Squared Distribution

Theorem

Let $n$ be a strictly positive integer.

Let $X \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.

Then the skewness $\gamma_1$ of $X$ is given by:

$\gamma_1 = \sqrt{\dfrac 8 n}$

Proof

From Skewness in terms of Non-Central Moments, we have:

$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$

where:

$\mu$ is the expectation of $X$.

$\sigma$ is the standard deviation of $X$.

By Expectation of Chi-Squared Distribution we have:

$\mu = n$

By Variance of Chi-Squared Distribution we have:

$\sigma = \sqrt {2 n}$

We also have:

\(\ds \expect {X^3}\)

\(=\)

\(\ds \prod_{k \mathop = 0}^2 \paren {n + 2 k}\)

Raw Moment of Chi-Squared Distribution

\(\ds \)

\(=\)

\(\ds n \paren {n + 2} \paren {n + 4}\)

\(\ds \)

\(=\)

\(\ds n^3 + 6 n^2 + 8 n\)

So:

\(\ds \gamma_1\)

\(=\)

\(\ds \frac {n^3 + 6 n^2 + 8 n - 6 n^2 - n^3} {\paren {\sqrt {2 n} }^3}\)

\(\ds \)

\(=\)

\(\ds \frac {\sqrt {\paren {8 n}^2} } {\sqrt {\paren {2 n}^3} }\)

\(\ds \)

\(=\)

\(\ds \frac {\sqrt {64 n^2} } {\sqrt {8 n^3} }\)

\(\ds \)

\(=\)

\(\ds \sqrt {\frac 8 n}\)

$\blacksquare$