In
this thread
the first two moments of the two-parameter GPD are given, where the distribution might be defined as
$G(y)= \begin{cases} 1-\left(1+ \frac{\xi y}{\beta} \right)^{-\frac{1}{\xi}} & \xi \neq 0 \\
1-\exp\left(-\frac{y}{\beta}\right) & \xi=0 \end{cases}$
Now I require the formula for the first four moments and the conditions under which they exist / are finite. Wikipedia
didn't help since only other types of the GPD are discussed here. The same holds for this page.
Best Answer
The wikipedia page for the Generalized Pareto is the same distribution as yours - you have $\mu=0$ and $\sigma=\beta$.
That page gives the mean and variance, and also gives the skewness and excess kurtosis, from which you can back out the quantities you need:
1) $E(Y)= \frac{\beta}{1-\xi}\, \; (\xi < 1)$
2) $\text{Var}(Y)=\frac{\beta^2}{(1-\xi)^2(1-2\xi)}\, \; (\xi < 1/2)$
$\quad\quad\quad\quad\;=E(Y)^2\frac{1}{(1-2\xi)}\, \; (\xi < 1/2)$
3) $\text{skewness}(Y) = \mu_3/\text{Var(Y)}^{3/2} = \frac{2(1+\xi)\sqrt(1-{2\xi})}{(1-3\xi)}\,\;(\xi<1/3)$
Hence $\mu_3 = \frac{\beta^3}{[(1-\xi)^2(1-2\xi)]^{3/2}} \frac{2(1+\xi)\sqrt(1-{2\xi})}{(1-3\xi)}\,\;(\xi<1/3)$
$\quad\quad\quad\quad= \frac{\beta^3}{(1-\xi)^3} \frac{2(1+\xi)}{(1-2\xi)(1-3\xi)}\,\;(\xi<1/3)$
$\quad\quad\quad\quad= E(Y)^3 \frac{2(1+\xi)}{(1-2\xi)(1-3\xi)}\,\;(\xi<1/3)$
4) Excess kurtosis = $\frac{3(1-2\xi)(2\xi^2+\xi+3)}{(1-3\xi)(1-4\xi)}-3\,\;(\xi<1/4)$
Hence $\text{kurtosis}(Y) = \mu_4/\text{Var(Y)}^{2} = \frac{3(1-2\xi)(2\xi^2+\xi+3)}{(1-3\xi)(1-4\xi)}\,\;(\xi<1/4)$
so $\mu_4 = \frac{\beta^4}{[(1-\xi)^2(1-2\xi)]^2}\frac{3(1-2\xi)(2\xi^2+\xi+3)}{(1-3\xi)(1-4\xi)}\,\;(\xi<1/4)$
$\quad\quad\;\, = E(Y)^4\frac{3(2\xi^2+\xi+3)}{(1-2\xi)(1-3\xi)(1-4\xi)}\,\;(\xi<1/4)$
If you want the raw moments rather than the central moments, the raw moments can readily be obtained from them.