If $X$ is exponentially distributed with parameter $1$, prove that $\exp(-X)$ is uniformly distributed on $[0,1]$.

Question

This is what I have so far:

The PDF of $X$ is $$f_X(x)=e^{-x}$$ when $x\geq0$ and $0$ otherwise. The CDF of $X$ is $$P(X\leq x)=F_X(x)=1-e^{-x}$$ when $x\geq 0$ and $0$ otherwise. I know that I want to end up with the pdf of $Y=e^{-X}$ being $$f_Y=1$$ on $[0,1]$ and $0$ otherwise, hence a uniform distribution. So,

\begin{align}F_Y(y)&=P(Y\leq y)\\ &=P(e^{-X}\leq y)\\ &=P(-\ln(y)\leq X) \end{align} I don't know how to proceed from here. Also, I know that $X=-\ln(Y)$, but I am not sure how to use it/ if I need to.

Answer 1

You're almost there. You only need to observe that $$\Pr (-\ln y \le X) =1 - \Pr(X < -\ln y) = 1 - \left( 1 - e^{\ln y}\right) = y.$$

Hence, $F_Y(y) = y$, and $f_Y(y) = 1$.

Answer 2

$$\begin{align*}F_Y(y) &= P(Y\leq y) \\&= P(\exp(-X)\leq y) \\&= P(X\geq -\log(y)) \\&= 1 - P(X\leq-\log(y)) \\&= 1 - (1 - \exp(\log(y))) \\&= y\end{align*}$$ and hence $$f_y = \frac{\mathrm d}{\mathrm d y} F_Y = 1$$