Solved – Kurtosis expressed in raw moments

Question

It is possible to express central moments in terms of raw moments.

E.g. for skewness we get:
$$
{\begin{aligned}\gamma _{1}&=\operatorname {E} \left[\left({\frac {X-\mu }{\sigma }}\right)^{3}\right]\\&={\frac {\operatorname {E} [X^{3}]-3\mu \operatorname {E} [X^{2}]+3\mu ^{2}\operatorname {E} [X]-\mu ^{3}}{\sigma ^{3}}}\\&={\frac {\operatorname {E} [X^{3}]-3\mu (\operatorname {E} [X^{2}]-\mu \operatorname {E} [X])-\mu ^{3}}{\sigma ^{3}}}\\&={\frac {\operatorname {E} [X^{3}]-3\mu \sigma ^{2}-\mu ^{3}}{\sigma ^{3}}}.\end{aligned}}
$$

I was wondering how to express kurtosis in terms of raw moments? I came this far:

$$\beta_4= {E} \left[\left({\frac {X-\mu }{\sigma }}\right)^{4}\right]= \frac{E(X^4)-4E(X^3)\mu+6E(X^2)\mu^2-3\mu^4}{(\sigma^2)^2}$$

Hereby I used Binomial expansion to write out the brackets. If this is correct, can the nominator perhaps be more simplified (like in the skewness example where the variance was taken out in step 3)?

Answer 1

There are many alternative methods to convert from central to raw moments ... the text by Stuart and Ord, Kendall's Advanced Theory of Statistics, volume 1, provides a good treatment. There are also automated functions to do this. Here is an automated conversion of the numerator: the $4^{\text{th}}$ central moment $\mu_4$ in terms of raw moments:

_{(source: tri.org.au)}

using the CentralToRaw function from the mathStatica package.