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.


1 Answer 1


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.


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