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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: $$ \begin{align} \frac{\mu_3}{\sigma^3}&=\frac{\alpha\beta^3(\alpha+1)(\alpha+2)}{\alpha^{\frac 32}\beta^3}\\&=\frac 1{\sqrt \alpha}(\alpha+1)(\alpha+2) \\&=? \end{align} $$

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    $\begingroup$ hint: $\mu_3$ is the third moment about the mean, not just the third moment. $E[X^k]\neq E[(X-(EX))^k]$ unless $E(X)=0$. $\endgroup$
    – LessFaceMoreBook
    Jan 26, 2014 at 12:14
  • $\begingroup$ An easier way to obtain the first three moments is from the MacLaurin series of a cumulant generating function, here defined as $\log(M(t))$ = $-\alpha\log(1-\beta t)$ = $\alpha\beta t + \alpha\beta^2 t^2/2!+2\alpha\beta^3 t^3/3!+\cdots,$ from which the central moments $\mu_2=\sigma^2=\alpha\beta^2$ and $\mu_3=2\alpha\beta^3$ can be read directly. $\endgroup$
    – whuber ♦
    Jan 26, 2014 at 14:26
  • $\begingroup$ Thank you @LessFaceMoreBook I have solved the missing puzzle piece in this proof! $\endgroup$
    – bryan.blackbee
    Jan 26, 2014 at 14:35

1 Answer 1

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$$ \begin{align} \mathbb E[(X-\mu_1)^3]&= \mathbb E(X^3)-3\mu_1^2\mathbb E(X^2)+3\mu_1\mathbb E(X)-\mu_1^3\\&=\mu_3-3\mu_1\mu_2+3\mu_1^2\mu_1-\mu_1^3 \\&= 2\alpha\beta^3 \end{align} $$

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    $\begingroup$ You need to multiply this answer by the cube of the scale factor, $\beta^3$. $\endgroup$
    – whuber ♦
    Jan 26, 2014 at 14:46
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    $\begingroup$ wow that was an awesome catch @whuber. I managed to get this and solve the proof. Simply put, put this as the numerator, divide by the denominator to complete the proof. $\endgroup$
    – bryan.blackbee
    Jan 26, 2014 at 15:24

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