Skewness of Gamma Distribution
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 1
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 2 {\sqrt \alpha}\) |
$\blacksquare$
Proof 2
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 |
$\blacksquare$