How to calculate the mean of the Generalized Pareto Distribution? The mean of the GPD distribution is $ \mu +{ \frac {\lambda }{1-\xi }}\,\;$, always for $(\xi <1) $. I tried to solve this integral but it did not seem clear to find the result given. Can someone help me?
$$\mathop{\mathbb{E}}(X) = \mu= \int_{-\infty}^{\infty} x f(x) \, {\rm d}x$$
$$\int_{-\infty}^{\infty} x f(x) \, {\rm d}x = \int_{-\infty}^{\infty} x \frac 1\lambda(1+\frac \xi\lambda(x-m))^{-(1+ \frac 1 \xi)} \, {\rm d}x $$
 A: (I'm replacing $\lambda$ by $\sigma$.) The GDP has the density you use only in case $\xi\neq0$, so I consider only the case $\xi\neq0$ here (the case $\xi=0$ is rather easy).
In case $\xi\neq0$:
$
\mathbb{E}(X) =  \int_{\mu}^{\infty} x \frac{1}{\sigma} (1 + \xi \frac{x-\mu}{\sigma} )^{-(1/\xi +1)}  ~{\rm d}x$
With the Substitution $1+\xi\frac{x-\mu}{\sigma} = u$ you get ${\rm d}x =  \frac{\sigma}{\xi} du$ and thus
\begin{align}
\mathbb{E}(X) &= \int x\frac{1}{\sigma}(1 + \xi\frac{x-\mu}{\sigma} )^{-(1/\xi +1)} ~{\rm d}x\\
&=  \int ((u-1)\frac{\sigma}{\xi} + \mu)\frac{1}{\sigma}u^{-(1/\xi +1)}\frac{\sigma}{\xi} ~{\rm d}u\\
&=  \int ((u-1)\frac{\sigma}{\xi} + \mu)\frac{1}{\xi}u^{-(1/\xi +1)} ~{\rm d}u\\
&=  \frac{1}{\xi}\left(\int (u-1)\frac{\sigma}{\xi}u^{-(1/\xi +1)} ~{\rm d}u +  \int \mu u^{-(1/\xi +1)} ~{\rm d}u \right) \\
&=  \frac{1}{\xi}\left(\frac{\sigma}{\xi}\int u^{-(1/\xi)} ~{\rm d}u - \frac{\sigma}{\xi}\int u^{-(1/\xi +1)}~{\rm d}u +  \mu\int u^{-(1/\xi +1)} ~{\rm d}u \right)
\end{align}
These three integrals evaluate straightforward into
\begin{align}
\mathbb{E}(X) &= \frac{1}{\xi} \left(\frac{\sigma}{\xi} \left[\frac{u^{1-(1/\xi)}}{1-(1/\xi)}\right]_{l}^{h} - \frac{\sigma}{\xi} \left[-\xi u^{-1/\xi} \right]_{l}^{h} + \mu\left[ -\xi u^{-(1/\xi)} \right]_{l}^{h} \right)\\ 
\end{align}
where $l$ and $h$ are the lower and upper bound of the integral. If you now plug in $l=1$ (why?) and $h=0$ (again, why?) in case $\xi<0$ resp. $l=1$ and $h=\infty$ in case $\xi>0$ you will find in the latter case the the expresion is finite only if $\xi<1$. Doing the calculations for the remaing cases ($0<\xi<1$ and $\xi<0$) will lead to the desired result.
