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Suppose $x$ is a non-negative random variable with CDF denoted $F$ and let $v = x+e^{-(x+1-\alpha)}$ for some $\alpha\geq 0$. Finally, let $G(v)=1-e^{-(v-\alpha)}$.

Is there a $F$ and $\alpha$ such that the CDF of $v$ is equal to $G(v)=1-e^{(v-\alpha)}$ for $v \in [\alpha,\infty)$?

More generally, if a know $v\sim G $ and $v=T(x)$ for some random variable $x \sim F$, is there a method to find $F$?

Bastiani
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  • Certainly not a general method. Sometimes from $v=T(x)$ you can find an expression for $x$ in $v$ and go on from there, but for that $T$ must be at least be injective. – drhab Aug 10 '17 at 16:43

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