Why is the third derivative of cumulant generating function = skewness?

Question

From what I know, for random variable $X$, skewness is defined as

$$\mathbb{E}\left(\frac{X-\mathbb{E}(X)}{\sigma}\right)^3$$

or

$$\frac{\mathbb{\mathbb{E}}(X^3)-3\mathbb{E}(X)\sigma^2-\mathbb{E}(X)^3}{\sigma^3}$$

where $\sigma$ is the standard deviation of $X$.

and the third derivative of cumulant generating function is

$$\mathbb{E}(X^3)-3\mathbb{E}(X)\mathbb{E}(X^2)+2[\mathbb{E}(X)]^3$$

These $2$ formulae look totally different, but why the third derivative of cumulant generating function is defined as the skewness of $X$?

Answer 1

I am not sure if your second formula is correct. Please check: \begin{align} \mathbb{E}\textstyle\left[\left(\frac{X-\mathbb{E}(X)}{\sigma}\right)^3\right]&=\frac{\mathbb E\big[X^3-\mathbb E[X]^3-3\,X^2\,\mathbb E[X]+3\,X\,\mathbb E[X]^2\big]}{\sigma^3}\\ &=\frac{\mathbb E[X^3]-\mathbb E[X]^3-3\,\mathbb E[X^2]\mathbb E[X]+3\,\mathbb E[X]^3}{\sigma^3}\tag{*}\\[3mm] &=\frac{\mathbb E[X^3]-\mathbb E[X]^3-3\,\mathbb E[X]\,(\mathbb E[X^2]-\mathbb E[X]^2)}{\sigma^3}\\[3mm] &=\frac{\mathbb E[X^3]-\,\mathbb E[X]^3-3\,\mathbb E[X]\,\sigma^2}{\sigma^3}\,. \end{align} The third derivative of the cumulant generating function you have correctly as \begin{align} \mathbb E[X^3]-3\,\mathbb E[X]\,\mathbb E[X^2]+2\mathbb E[X]^3\,. \end{align} Up to a division by $\sigma^3$ this agrees with (*). Therefore, the derivative of the cumulant generating function is $$ \mathbb E\big[(X-\mathbb E[X])^3\big]\,. $$ This is the form in Wikipedia Cumulant and it is related to $$ \mathbb{E}\textstyle\left[\left(\frac{X-\mathbb{E}(X)}{\sigma}\right)^3\right] $$ by a simple factor of $1/\sigma^3\,.$

Answer 2

It suffices to check the case $\Bbb EX=0,\,\Bbb EX^2=1$, for which the cgf is of the form $-\frac12t^2+\frac16ct^3+o(t^3)$, of third derivative $c$. Exponentiating, the mgf is then $1+(-\tfrac12t^2+\frac16ct^3)+o(t^3)$, implying $c=\Bbb EX^3$. Note, however, the fourth cumulant & moment don't coincide, because$$\exp\left(-\frac12t^2+\frac16ct^3+\frac{1}{24}dt^4+o(t^3)\right)=1-\frac12t^2+\frac16ct^3+\frac{d+6}{24}t^4+o(t^4).$$