TITLE: Distribution of $-\log X$ if $X$ is uniform.
QUESTION [18 upvotes]: For $X$ and $Y$ random variables; $X$ follows the uniform distribution.
(1): if $Y=-\log X$
(2): then it can be shown that $-\log X$ is distributed as $\exp(1)$ {i.e. exponential with mean 1}.
Why is this so? Intuitively statement (2) make sense to me. But i'd like a mathematical proof.
-Probably wrong working:
(1) seems to imply $\exp(-Y) = X$ (which is like saying $X$ is exponentially distributed, which is a contradiction, since its actually uniform!); or is it incorrect for me to do this since $X$ and $Y$ are random variables?
Ultimately how do I prove (2)?
Thanks

REPLY [8 votes]: reiterate @S4M's answer but with $Y$ = -log($X$), rather than $Y$ = log($x$), according to OP's post.
$$ P(Y<y) = P(-log(X)<y) = P(X>exp(-y))$$
Given that $X$ is uniformly distributed between [0,1], $$P(Y<y) = P(x>exp(-y)) = F(1 -exp(-y)) $$, which is the CDF of exponential distribution, with $\lambda=1$.