Skewness of Chi-Squared Distribution
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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:
\(\ds \gamma_1\) | \(=\) | \(\ds \frac {n^3 + 6 n^2 + 8 n - 6 n^2 - n^3} {\paren {\sqrt {2 n} }^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt {\paren {8 n}^2} } {\sqrt {\paren {2 n}^3} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sqrt {64 n^2} } {\sqrt {8 n^3} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac 8 n}\) |
$\blacksquare$