Skewness of Gamma Distribution/Proof 2

Theorem

Let $X \sim \map \Gamma {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Gamma$ is the Gamma distribution.

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

$\gamma_1 = \dfrac 2 {\sqrt \alpha}$

Proof

From Expectation of Power of Gamma Distribution‎, we have:

$\expect {X^n} = \dfrac {\alpha^{\overline n} } {\beta^n}$

where $\alpha^{\overline n}$ denotes the $n$th rising factorial of $\alpha$.

Hence:

\(\ds \gamma_1\)

\(=\)

\(\ds \expect {\paren {\dfrac {X - \mu} \sigma}^3}\)

Definition of Skewness

\(\ds \)

\(=\)

\(\ds \frac {\expect {X^3 - 3 X^2 \mu + 3 X \mu^2 - \mu^3} } {\sigma^3}\)

Cube of Difference

\(\ds \)

\(=\)

\(\ds \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect X - \mu^3} {\sigma^3}\)

Expectation is Linear

\(\ds \)

\(=\)

\(\ds \frac {\expect {X^3} - 3 \expect X \expect {X^2} + 3 \paren {\expect X}^3 - \paren {\expect X}^3} {\paren {\sqrt {\var X} }^3}\)

Definition of $\mu$ and $\sigma$

\(\ds \)

\(=\)

\(\ds \paren {\expect {X^3} - 3 \expect X \expect {X^2} + 2 \paren {\expect X}^3} \paren {\sqrt {\dfrac {\beta^2} \alpha} }^3\)

simplification, Variance of Gamma Distribution

\(\ds \)

\(=\)

\(\ds \dfrac {\alpha^{\overline 3} } {\beta^3} - 3 \dfrac {\alpha^{\overline 1} } {\beta^1} \dfrac {\alpha^{\overline 2} } {\beta^2} + 2 \paren {\dfrac {\alpha^{\overline 1} } {\beta^1} }^3 \dfrac {\beta^3} {\alpha \sqrt \alpha}\)

Expectation of Power of Gamma Distribution‎

\(\ds \)

\(=\)

\(\ds \paren {\dfrac {\alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\beta^3} - 3 \dfrac {\alpha} \beta \dfrac {\alpha \paren {\alpha + 1} } {\beta^2} + 2 \paren {\dfrac \alpha \beta}^3} \dfrac {\beta^3} {\alpha \sqrt \alpha}\)

Definition of Rising Factorial

\(\ds \)

\(=\)

\(\ds \dfrac {\alpha^3 + 3 \alpha^2 + 2 \alpha - 3 \alpha^3 - 3 \alpha^2 + 2 \alpha^3} {\beta^3} \dfrac {\beta^3} {\alpha \sqrt \alpha}\)

multiplying out

\(\ds \)

\(=\)

\(\ds \dfrac 2 {\sqrt \alpha}\)

simplification

$\blacksquare$