Bounds on CDF of a Transformation of Random Variable

Question

I am attempting to transform a random variable, but am having problems understanding over which range I need to integrate. The random variables in question are $X\sim\operatorname{Uniform}(0,1)$ and $Y=e^{-X}$ My goal is to be able to get the CDF of $Y$

I have

\begin{align} F_Y(y) & =P(Y\le y) \\ & =P(e^{-X}\le y) \\ & =P(-X\le \ln(y)) \\ & =P\left(X\ge \ln\left(\frac 1 y\right)\right) \end{align}

I know furthermore that $x \in (0,1)$ and that $R_Y \in (\frac{1}{e}, 1)$.

If I understand correctly, I can obtain the CDF of $Y$ by some variation of $\int{F_X}\,dx$ which in this case is $\int1\,dx$ I feel like this should be an easy problem, but Ihave just not been able to understand the concept.

Answer 1

You've just about finished the problem, and then you gave up when you were three feet below the summit.

If $1/e \le y \le 1$ then $$ F_Y(y) = \Pr(e^{-X} \le y) = \Pr(X\ge - \ln y) = 1-(-\ln y). $$

So $$F_Y(y) = \begin{cases} 1 +\ln y &\text{if }& \frac 1 e \le y\le 1, \\[6pt] 1 &\text{if }& 1< y, \\[6pt] 0 &\text{if }& y< \tfrac 1e. \end{cases}$$

Answer 2

You are on the right track. Just note that $$P(X \ge \ln(1/y)) = 1 - \ln(1/y) = 1 + \ln y$$ because $X$ is uniform on $(0,1)$.