I am kind of at my wits end for this proof. Given $Y$ is Gamma with parameters $\alpha$ and $\beta$, the MGF is given by $(1-\beta t)^{-\alpha}$. I need to find mean, variance and skewness. So I managed to derive the mean and variance, but cannot prove the result for skewness.
Given the MGF, I calculate 1st, 2nd and 3rd moment
$$
\begin{align}
M_x(t) &= (1-\beta t)^{-\alpha}&\\
M'_x(t)&=\alpha \beta (1-\beta t)^{-\alpha – 1}& M'_x(0)&=\mu_1= \alpha\beta\\
M''_x(t)&=\alpha (\alpha +1)\beta^2 (1-\beta t)^{-\alpha -2}& M''_x(0)&=\mu_2= \alpha(\alpha +1)\beta^2\\
M'''_x(t)&=\alpha (\alpha +1)(\alpha +2)\beta^3 (1-\beta t)^{-\alpha -3}& M'''_x(0)&=\mu_3= \alpha (\alpha +1)(\alpha +2)\beta^3\\
\end{align}
$$
So the variance is
$$
\begin{align*}
\sigma^2&=\mu_2 – (\mu_1)^2\\ &= \alpha\beta^2(\alpha +1) – \alpha^2\beta^2\\&=\alpha\beta^2
\end{align*}
$$
But I can't seem to prove that skewness is $\frac 2 {\sqrt \alpha}$. My proof so far is:
$$
\begin{align}
\frac{\mu_3}{\sigma^3}&=\frac{\alpha\beta^3(\alpha+1)(\alpha+2)}{\alpha^{\frac 32}\beta^3}\\&=\frac 1{\sqrt \alpha}(\alpha+1)(\alpha+2)
\\&=?
\end{align}
$$
Best Answer
$$ \begin{align} \mathbb E[(X-\mu_1)^3]&= \mathbb E(X^3)-3\mu_1^2\mathbb E(X^2)+3\mu_1\mathbb E(X)-\mu_1^3\\&=\mu_3-3\mu_1\mu_2+3\mu_1^2\mu_1-\mu_1^3 \\&= 2\alpha\beta^3 \end{align} $$