Distribution function with exponential-distribution

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Given the r.v. $X$ with distribution function $F(x) = (1-e^{-x})_+$ (i.e. $X\sim\operatorname{Exp}(1)$), what is the distribution of $e^{-X}$?

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\begin{align} \Pr( e^{-X}\le 0) & = 0 \\[10pt] \Pr(e^{-X} < 1) & = 1 \\[10pt] \Pr( e^{-X} \le x) & = \Pr(-X \le \log x) = \Pr(X\ge -\log x) \\[10pt] & = e^{-(-\log x)} = e^{\log x} = x. \end{align} $$ \text{So } F_{e^{-X}} (x) = \begin{cases} 0 & \text{if } x\le0, \\ x & \text{if } 0<x<1, \\ 1 & \text{if } x\ge 1. \end{cases} $$ There is a standard name for that distribution.