Confusion with relationships between uniform and exponential rvs

Question

If X is an exponential rv with rate 1 and U is a standard uniform rv, then we can write X = - log U.

At the same time, using the probability integral transform method, we know that $U = F(X) = 1-e^{-X}$, where F is the CDF of X.

These two approaches seem to give inconsistent results - can anyone point out my mistake?

Answer 1

First approach: $X \stackrel{d}{=} - log U$, second approach: $U \stackrel{d}{=} 1-e^{-X}$, they don't contradict each other. Indeed, in the first approach $e^{-X} \stackrel{d}{=} U$, in the second approach $e^{-X} \stackrel{d}{=} 1-U$. But $U \stackrel{d}{=} 1 - U$, because $P(U \le t) = P(1-U \le t) = t$ for all $t$.