Transformation of Extreme Value Distribution Let $X$ be a random variable following distribution function (i.e., generalized Pareto distribution)
$$
F_{\gamma, \sigma}(x) = 1-\left( 1+\frac{\gamma x}{\sigma} \right)^{-\frac{1}{\gamma}},
$$
where $\sigma>0, \gamma\in\mathbb R$. In addition,$x\in[0, \infty)$ if $\gamma\geq 0$ and $x\in\left[0, -\frac{\sigma}{\gamma}\right]$ if $\gamma<0$.
It is claimed that $U:=\left( 1+\frac{\gamma X}{\sigma} \right)^{-\frac{1}{\gamma}}$ is uniformly distributed on $(0,1)$. To check this claim it is sufficient to show that the distribution function for $U$ is $F_U(u)=u$.
$$
F_U(u) = \mathbb P(U\leq u) = \mathbb P \left[\left( 1+\frac{\gamma X}{\sigma} \right)^{-\frac{1}{\gamma}}\leq u\right] = F_{\gamma, \sigma}\left[ \frac{\sigma}{\gamma}\left( u^{-\gamma}-1 \right) \right] = 1-u.
$$
From the above calculation, it seems that the claim does NOT check out. Is there any explanation for the claim, please? Thank you!
 A: The inequalities are not conserved in these manipulations.


*

*If $\gamma>0$, then 
\begin{align*}
F_U(u) = \mathbb P(U\leq u) = \mathbb P \left[\left( 1+\frac{\gamma X}{\sigma} \right)^{-\frac{1}{\gamma}}\leq u\right] 
&=\mathbb P \left[1+\frac{\gamma X}{\sigma}\geq u^{-\gamma}\right] \\
&=1-F_{\gamma, \sigma}\left[\frac{\sigma}{\gamma}\left( u^{-\gamma}-1 \right) \right]\\
&=u.
\end{align*}

*If $\gamma<0$, then
\begin{align*}
F_U(u) = \mathbb P(U\leq u) = \mathbb P \left[\left( 1+\frac{\gamma X}{\sigma} \right)^{-\frac{1}{\gamma}}\leq u\right] 
&=\mathbb P \left[1+\frac{\gamma X}{\sigma}\leq u^{-\gamma}\right] \\
&=\mathbb P \left[\frac{\gamma X}{\sigma}\leq u^{-\gamma}-1\right] \\
&=\mathbb P \left[X\geq\frac\sigma\gamma\left(u^{-\gamma}-1\right)\right] \\
&=1-F_{\gamma, \sigma}\left[\frac{\sigma}{\gamma}\left( u^{-\gamma}-1 \right) \right]\\
&=u.
\end{align*}

*If $\gamma=0$, then $F_{\gamma, \sigma}(x)=1-{\mathrm e}^{-x/\sigma}$, and it is easy to check that $\exp\left(-\frac1\sigma X\right)$ is uniformly distributed.
