Given a r.v. Uniform in [0,1], What is the distribution of 1-U?

Question

I have an excercise and would like to know if the way I've solved it is correct or not
Given $U \sim \mathcal{Uniform}(0,1)$ What is the distribution of $X = 1-U$
My solution is..
$$ P(X\leq u)=\\ P(1-U \leq u)=\\ P(1-u \leq U)=\\ 1-P(U \leq 1-u)=\\ 1 - \int_{0}^{1-u}1du=\\ 1 - (1 - u)=\\ u $$ So $F_x(u)=u$ and therefore, the pdf is $f_x(u)=1$
It is just another Uniform(0,1)

Is the procedure correct? Am I missing something? Thanks a lot in advance for any commment!

Answer 1

Keep in mind $U=\phi(X)=1-X$ is a decreasing function, so when you want to find $P(\phi(X)<u) = P(X>\phi^{-1}(u))$, i.e. the inequality sign changes.