Transformation of random variable

Question

I want to prove the following: $$\text{Let F be a distribution function of any random variable $\\$ and G(x) the quantile function (or inverse) of } \frac 1 {1-F(x)}$$ $$\text{Then, for a standard exponentially (exp(1)) distributed random variable E,$\\$ the random variable } G\Bigg(\frac 1 {1-e^{-E}}\Bigg)=\frac 1 {1-F(\frac 1 {1-e^{-E}})} \text{ has distribution function F.}$$ I tried simply transforming the random variables, but failed at getting the desired result. Any ideas?

Answer 1

Let $X=G\left(\dfrac 1 {1-\mathrm e^{-E}}\right)$. If $G$ is the inverse of $\dfrac1 {1-F}$, then the identity $X=\dfrac 1 {1-F\left(\dfrac 1 {1-\mathrm e^{-E}}\right)}$ written in the post is incorrect. Actually, $X=G\left(\dfrac 1 {1-\mathrm e^{-E}}\right)$ translates as $\dfrac1 {1-F(X)}=\dfrac 1 {1-\mathrm e^{-E}}$ hence $F(X)=\mathrm e^{-E}$ and, for every $x$, $$[X\leqslant x]=[F(X)\leqslant F(x)]=[\mathrm e^{-E}\leqslant F(x)]=[E\geqslant-\log F(x)].$$ For every $u\geqslant0$, $P(E\geqslant u)=\mathrm e^{-u}$ hence $$P(X\leqslant x)=P(E\geqslant-\log F(x))=F(x),$$ which proves that the CDF of $X$ is $F$, as desired.