Moments of the two-parameter generalized Pareto distribution (GPD) needed 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.
 A: 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.
