Skewness of Gamma Distribution/Proof 1

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 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 Gamma Distribution, we have:

$\mu = \dfrac \alpha \beta$

By Variance of Gamma Distribution, we have:

$\sigma = \dfrac {\sqrt \alpha} \beta$

To calculate $\gamma_1$, we must calculate $\expect {X^3}$.

From Moment in terms of Moment Generating Function:

$\expect {X^n} = \map { {M_X}^{\paren n} } 0$

where $M_X$ is the moment generating function of $X$.

From Moment Generating Function of Gamma Distribution: Third Moment:

$\map { {M_X}} t = \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\paren {\beta - t}^{\alpha + 3} }$

Setting $t = 0$:

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

\(=\)

\(\ds \dfrac {\beta^\alpha \alpha \paren {\alpha + 1} \paren {\alpha + 2} } {\paren {\beta - 0}^{\alpha + 3} }\)

\(\ds \)

\(=\)

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

So:

\(\ds \gamma_1\)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \frac {\alpha^3 + 3 \alpha^2 + 2 \alpha - 3 \alpha^2 - \alpha^3} {\paren {\alpha \sqrt \alpha } }\)

$\beta^3$ cancels

\(\ds \)

\(=\)

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

$\blacksquare$