r-moments of GPD (generalized Pareto distribution) I was asked to prove the r moments of the GPD with $\kappa$ e $\psi$ parameters.
$$EX^r=\frac{\psi^r}{(-\kappa)^{r+1}}\frac{\Gamma(-r-1/\kappa)}{\Gamma(1-1/\kappa)}r!$$
with $\kappa>-1/r,\,\,\,r\in\mathbb{N}$
So, I did:
The GPD is given by
$$G_{\kappa,\psi}(x)=1-\bigg(1-\kappa\frac{x}{\psi}\bigg)^{1/\kappa},\,\,\,x\in D(\kappa,\psi)$$
where
$$D(\kappa,\psi)=\begin{cases}[0,\infty) & \mbox{if } \kappa\leq0 \\ [0,\psi/\kappa] & \mbox{if } \kappa > 0 \end{cases}$$
$\kappa = 0$ means $\kappa\to0$
The first two moments are accomplished with standard calculations. But the r moments I couldn't do. I tried the moment-generating function too. But I didn't get anywhere. Could someone help me, please? 
 A: You can find a formula for $E_{\kappa,\psi}[X^r]$ as a function of $E_{\kappa^*,\psi^*}[X^{r-1}]$, for a specific set of other parameters $\kappa^* = \frac{1}{\frac{1}{\kappa}+1}$ and $\psi^* = \frac{\psi}{\kappa(\frac{1}{\kappa}+1)}$, and use this to get an inductive formula for $E_{\kappa,\psi}[X^r]$.
Set up the problem as:
\begin{equation}
\int_0^{\frac{\psi}{\kappa}}x^r(1-\frac{\kappa}{\psi}x)^{\frac{1}{\kappa}-1}dx = \int_0^{\frac{\psi}{\kappa}}x^rg(x)dx
\end{equation}
Using integration by parts, with the derivative of $x^r = rx^{r-1}$ and the integral of $g(x)=G(x)$, we get:
\begin{equation}
\begin{split}
\int_0^{\frac{\psi}{\kappa}}x^rg(x)dx & = x^rG(x)|_0^\frac{\psi}{\kappa} - r\int_0^{\frac{\psi}{\kappa}}x^{r-1}G(x)dx\\
& = x^rG(x)|_0^\frac{\psi}{\kappa} - r \int_0^{\frac{\psi}{\kappa}}x^{r-1}dx + r\int_0^{\frac{\psi}{\kappa}}x^{r-1}(1-\frac{\kappa}{\psi}x)^{\frac{1}{\kappa}}dx\\
& = (\frac{\psi}{\kappa})^r - (\frac{\psi}{\kappa})^r + r\int_0^{\frac{\psi}{\kappa}}x^{r-1}(1-\frac{\kappa}{\psi}x)^{\frac{1}{\kappa}}dx\\
& = \frac{r\psi^*}{\psi^*}\int_0^{\frac{\psi^*}{\kappa^*}}x^{r-1}(1-\frac{\kappa^*}{\psi^*}x)^{\frac{1}{\kappa^*}-1}dx\\
& = r\psi^*E_{\kappa^*,\psi^*}[X^{r-1}]
\end{split}
\end{equation}
Knowing that the stated expression is true for $r=1$, plug the new parameters into $r\psi^*E_{\kappa^*,\psi^*}[X^{r-1}]$ to get $E_{\kappa,\psi}[X^r]$.
With $r\psi^*E_{\kappa^*,\psi^*}[X^{r-1}]=r\psi^*\frac{(\psi^*)^{r-1}}{(-1)^{r}(\kappa^*)^{r}}\frac{\Gamma(-r+1-1/\kappa^*)}{\Gamma(1-1/\kappa^*)}(r-1)!$
Doing some calculations:
\begin{equation}
\begin{split}
r\psi^*E_{\kappa^*,\psi^*}[X^{r-1}] & = r\psi^*\frac{(\psi^*)^{r-1}}{(-1)^{r}(\kappa)^{r}}\frac{\Gamma(-r+1-1/\kappa^*)}{\Gamma(1-1/\kappa^*)}(r-1)!\\
 & = \frac{(\psi^*)^{r}}{(-1)^{r}(\kappa^*)^{r}}\frac{\Gamma(-r+1-1/\kappa^*)}{\Gamma(1-1/\kappa^*)}r!\\
& = \frac{(\psi^*)^{r}}{(-1)^{r}(\kappa^*)^{r}}\frac{\Gamma(-r-(1/\kappa^*-1))}{\Gamma(-\frac{1}{\kappa})}r!\\
& = \frac{(\psi^*)^{r}}{(-1)^{r}(\kappa^*)^{r}}\frac{\Gamma(-r-\frac{1}{\kappa})}{\Gamma(-\frac{1}{\kappa})}r!\\
\end{split}
\end{equation}
Noting that $\frac{\kappa^*}{\psi^*} = \frac{\kappa}{\psi}$. we can change the previous expression to:
\begin{equation}
\begin{split}
\frac{\psi^{r}}{(-1)^{r}\kappa^{r}}\frac{\Gamma(-r-\frac{1}{\kappa})}{\Gamma(-\frac{1}{\kappa})}r!\\
\end{split}
\end{equation}
Using the property that $\Gamma(z+1) = z\Gamma(z)$ we can get $\frac{\Gamma(z+1)}{z}=\Gamma(z)$. This gives us that $\Gamma(-\frac{1}{\kappa}) = \frac{\Gamma(1-\frac{1}{\kappa})}{-\frac{1}{\kappa}} = -\kappa\Gamma(1-\frac{1}{\kappa})$.
So the previous expression is:
\begin{equation}
\begin{split}
\frac{\psi^{r}}{(-1)^{r+1}\kappa^{r+1}}\frac{\Gamma(-r-\frac{1}{\kappa})}{\Gamma(1-\frac{1}{\kappa})}r! & = \frac{\psi^{r}}{(-\kappa)^{r+1}}\frac{\Gamma(-r-\frac{1}{\kappa})}{\Gamma(1-\frac{1}{\kappa})}r!
\end{split}
\end{equation}
So the inductive property is satisfied, and you showed it was true for $r=1$, then it's true for all $r$.
