I know that the negative log of a uniform random variable is an exponential random variable. I am trying to prove the reverse, but I seem to have arrived at something that doesn't make sense.
Define $Y \sim \text{Exponential}(\lambda)$, and $X = -\exp(Y)$.
\begin{align} P(X \leq x) = P(-\exp(Y) \leq x) = P(Y \geq -\ln(x)) \\ = \int_{-\ln(x)}^{{\infty}} \lambda \exp (-\lambda y) dy = -\exp(- \lambda y) \big|_{-\ln(x)}^{\infty} \\ = \exp(\lambda \ln(x)) \\ = x^{\lambda} \end{align}
If you take the derivative of this to get the pdf of $x$, you see that it is a function of $x$, which isn't constant, hence $X$ isn't a uniform random variable. So I am confused, is the negative exponential of a exponential random variable not a uniform random variable, or did I make a mistake in my derivation?
A fix in red: what does have a $U(0,\,1)$ distribution is $\color{blue}{1-}\exp(-\color{red}{\lambda}Y)$, where the blue part is unnecessary but makes the transformation order-preserving. In fact, it's $Y$'s CDF, say $F$. This is no coincidence:$$P(F(Y)\le f)=P(Y\le F^{-1}(f))=F(F^{-1}(f))=f\implies F(Y)\sim U(0,\,1).$$