Let $X \sim \operatorname{Exp}(\theta)$. Show that $Y = \mathrm{e}^{-X/\theta} \sim \operatorname{Uniform}(0,1]$. 
Let $X \sim \operatorname{Exp}(\theta)$. Show that $Y = \mathrm{e}^{-X/\theta} \sim \operatorname{Uniform}(0,1]$.

Hi, I tried to prove this in my tutorial but I got stuck trying to prove it using MGF. Can someone help me with the first method? I was planning to get the $My(t)$ expression to see if I can see any similarity with that of a uniform distribution. Is my train of thoughts correct?
I also tried to prove using the pdf of $Y$. I have written my questions in red, and I was wondering how to prove the part where the uniform distribution is valid from $0$ (exclusive) to $1$ inclusive.
Attached are my workings, I hope someone can enlighten me on this! thank you!
EDIT:
**I have taken down my detailed workings as I have verified that the workings are correct.*
To answer my above questions, The  $My(t)$ expression can be gotten via a direct integration from E($\mathrm{e}^{tY}$) , and then a comparison of that with the  $Mx(t)$ of a uniform distribution.
Also, the Uniform distribution is valid (0,1] because the exp distribution is valid from x more than or equals to 0.
 A: Construing $X\sim\operatorname{Exp}(\theta)$ to mean the distribution of $X$ is
$$
e^{-x/\theta} \, \dfrac{dx} \theta \text{ for } x\ge 0,
$$
I'd start by showing that
$$
\Pr(X>x) = e^{-x/\theta} \text{ for } x\ge 0.
$$
Then
$$
\Pr(e^{-X/\theta} \le y)  = \Pr\left( X \ge -\theta\log y \right) = e^{-(-\theta\log y)/\theta} = y, \text{ for } 0 \le y\le 1. 
$$
A: You have exponential distribution and you are transforming it into another r.v. by running it through its of cdf, well, one minus its own cdf. That the new r.v. is uniform is then a consequence of probability integral transform. The proof in the link uses standard cdf method for deriving distribution of a function of random variable.
A: $$P(e^{-X/\theta}\le y)=P(\frac{-X}\theta\le \log y)=P(-X\le -\theta\log y)$$
$$ = P(X\ge -\theta\log y)=\int_{-\theta \log y}^\infty \theta e^{-\theta x}\, dx,\qquad 0\le y \le 1.$$
$$F_Y(y)=\int_{-\theta \log y}^\infty \theta e^{-\theta\cdot x}dx.$$
If you differentiate with the Leibnitz rule you get
$$f_Y(y)=\theta^2 y^{\theta^2-1} \text{ with } 0\le y \le 1$$
And this pdf is certainly not of a Uniform Distribution.
If you are considering the pdf as $\frac{1}{\theta}e^{\frac{-x}{\theta}}$. Then you will get the pdf of a uniform distribution.
A: Indeed the distinction between $\geq$ and $>$ is not important, because the exponential distribution is absolutely continuous and every point has zero probability mass.   That is: $~\mathsf P(X=-\theta\ln y)~=~0$
So...
$\begin{align}F_Y(y) &= \mathsf P(Y\leqslant y)
\\ &= \mathsf P(e^{-X/\theta}\leqslant y)
\\ &= \mathsf P(-X/\theta \leqslant \ln y)
\\ &= \mathsf P(X\geqslant -\theta\ln y) 
\\ &= \big(1-F_X(-\theta\ln y\;^-)\big)\cdot\mathbf 1_{(-\theta\ln y)~\in[0;\infty)}~+\mathbf 1_{(-\theta\ln y)~\in(-\infty;0)}
\\ &= e^{-(-\theta\ln y)/\theta}\cdot\mathbf 1_{\ln y~\in(-\infty;0]}~+\mathbf 1_{\ln y~\in(0;\infty)}
\\ &= y \cdot\mathbf 1_{y\in(0;1]}~+\mathbf 1_{y~\in(1;\infty)}
\end{align}$
Which is the CDF of a Uniform(0;1] distribution.

PS: The indicator function's interval comes because $F_X((-\theta\ln y)\,^-)$ is supported when $(-\theta\ln y)\in[0;\infty)$.   Then rearrange.
