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I'm sure this is a trivial problem, but I am having trouble committing to an answer.

If $U$~$(0,1)$, then $Y=-ln U$. I know that $\mathbb P(Y \le y)$

= $\mathbb P(g(X) \le y)$

=$\mathbb P(-lnU \le y)$

=$\mathbb P(U \ge e^{-y})$

So would my CDF be $F_Y (y)=e^{-y},0\le y \le \infty$? Then of course, my PDF would be $-e^{-y}$.

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    $\begingroup$ Please refrain from cross-posting (see math.stackexchange.com/questions/2752717/…). If you do feel you need to post versions of your question on more than one SE site, then please be so kind as to summarize the answers you get on each site where you post. $\endgroup$ – whuber Apr 26 '18 at 20:49
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It is an exponential distribution, but your derivation contains a mistake.

$\mathbb P(U \ge e^{-y}) = e^{-y}$, as $U$ is uniform on $(0,1)$. Then, $F_Y(y) = 1-e^{-y}$ and the PDF is $f_Y(y) = e^{-y}$.

You can use your intuition with simple expressions. The expression you obtained for $F$ is decreasing and the expression for the PDF is negative, and we know that CDFs are increasing, and that PDFs are positive.

You can find this in Wikipedia:

https://en.wikipedia.org/wiki/Exponential_distribution#Generating_exponential_variates

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