Show that $\min(U,1-U)$ and that $\max(U,1-U)$ are uniform Let $U$ be uniform on $(0,\ 1)$. Show that $\min(U,\ 1-U)$ is uniform on $(0,\ 1/2)$ and that $\max(U,\ 1-U)$ is uniform on $(1/2,\ 1)$.
I'm not sure how to approach... the only hint i have is that a Random Variable U is Uniform on (0,1) if for all 0<=u<=1, P[U<=u]=u
the only thing that I have seen is that the function min(U,1-U)=U for all 0<=U<=1/2 and that is is quite like the behave of a uniform variable on (0,1/2) but not sure is that's proof enough
 A: If $X = max(U, 1-U)$ then it's distribution function is 
$F_X = Pr(X \leq x)$ $ = Pr\{ max(U, 1-U)  \leq x \}$
You should note that the maximum between two numbers is less than x iff both numbers are below x, so
$F_X =  Pr\{ max(U, 1-U)  \leq x \} = Pr\{U \leq x , (1 -U ) \leq x\} = Pr\{(1-x) \leq U \leq x \}$
This event has zero probability if x is below $\frac{1}{2}$ (in that case you would be asking for the probability of an impossible event, since $0.5 \leq 1- x$ and $x \leq 0.5$). In terms of distributions, then, you would have:
$F_X  = Pr\{(1-x) \leq U \leq x \} = [F_U(x) - F_U(1-x)].I_{(x \geq 0.5)}$
Taking derivatives (since they exist except in a finite number of points), you can find $X$ density:
$f_X = [f_U(x) +f_U(1-x)].I_{(x \geq 0.5)} = [I_{(0,1)}(x)+I_{(0,1)}(1-x)].I_{(x \geq 0.5)}$
Since $I_{(0,1)}(x)=I_{(0,1)}(1-x)$, you finally get:
$f_X = 2.I_{(0,1)}(x).I_{(x \geq 0.5)}=2.I_{(0.5,1)}(x)$
So the maximum is distributed uniformly between $0.5$ and $1$.
For the minimum, you have that $min(U,1-U) = 1 - max(U,1-U)$ you have just proven that the maximum is uniform between $.5$ and $1$, so this equality means that the minimum is uniform between $0$ and $.5$ (being a linear function of a uniform r.v.)
