4
$\begingroup$

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

$\endgroup$
3
  • 2
    $\begingroup$ hint: $\mu_3$ is the third moment about the mean, not just the third moment. $E[X^k]\neq E[(X-(EX))^k]$ unless $E(X)=0$. $\endgroup$ 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! $\endgroup$ Jan 26, 2014 at 14:35

1 Answer 1

3
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ You need to multiply this answer by the cube of the scale factor, $\beta^3$. $\endgroup$
    – whuber
    Jan 26, 2014 at 14:46
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 26, 2014 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.