Suppose $X \sim \mathcal{Exponential}(1/b)$ with pdf:
$f(x) = \frac{1}{b}\,e^{-x/b}\;,\; x \geqslant 0$ and $0$ o.w
Find the pdf and identify the distribution of $U=1-e^{-X/b}$.
What I have so far is this:
$U= h(X) = 1-e^{-X/b}$ which can be rewritten as: $X= -b \ln(1-U)$, $X \geqslant 0$ and $U\geqslant 0$
Then I took the derivative of $-b\ln(1-U)$ wrt $U$ and get $b/(1-U)$.
I don't really know where to go from there. Especially with the part about the distribution of $U$. I believe I could find the pdf of $U$ eventually, but I'm not sure what is meant by the distribution of $U$.
Thanks in advance!
Note: Since $X$ is exponential (1/b), the CDF is $\mathsf P(X\leq x)=(1-e^{-x/b})$ for $x\geqslant 0$ (and $0$ otherwise). (PS: This may be found by integrating the given pdf.)
So, by substitution, algebraic rearrangement, and use of this CDF:
$$\begin{align}\mathsf P(U \leqslant u)&=\mathsf P(1-e^{-X/b} \leqslant u)\\[1ex]&=\mathsf P(e^{-X/b} \geqslant 1-u)\\[1ex]&=P\mathsf (-\frac X b \geqslant\ln (1-u))\\[1ex]&=\mathsf P(X \leqslant -b\ln (1-u))\\[1ex]&=1-e^{\ln (1-u)}\quad[-b\ln(1-u)\geqslant 0]\\[1ex]&= 1-(1-u)\quad[\ln(1-u)\leqslant 0]\\[1ex]&=u\qquad[0<u\leqslant 1]\end{align}$$
Hence, taking the differential gives us the probability density function of $f_U(u)=1$ for $0<u\leqslant 1$ and $0$ otherwise.
Now, all that is left is to identify what distribution has this density function.