I have $ X ~ U[0,1]$ and $Y=1-X$ two continuous random variables. I've shown easily that $Y ~ U[0,1]$ also. Now I'm considering $U:=min(X,Y)$ and $V:=max(X,Y)$.
I'm computing the probability density function of U and its mean.
I have $P(U<u)=1-P(U>u)=1-P(X>u,Y>u)=1-P(u<X<1-u)$ $(*)$. Now when $u\in(\frac{1}{2},\infty)$, we have $P(u<X<1-u)=0$ because then $u>1-u$. When $u \in [0,\frac{1}{2}]$, we have $P(u<X<1-u)=\int_u^{1-u} dx=1-2u$. And finally when $u\in(-\infty,1)$ then $P(u<X<1-u)=\int_0^1 dx=1$.
I put these results in $(*)$ and differentiate to obtain the probability density function $2$ when $u\in[0,\frac{1}{2}]$ and $0$ otherwise.
Finally I have $E(U)=\int_0^\frac{1}{2} 2du=1$. I know this is not true because I need $E(V)/E(U)=3$ but $E(U)=E(X)+E(Y)-E(V)$ which would give $E(V)=1$ and so the ratio also $1$. Ideally I should obtain $E(U)=\frac{1}{2}$ right?
I can't find out where I'm wrong, can you help me please? Thanks a lot, I've spent ages on this...
I would take a rather simpler approach. If $X$ is uniform and $Y=1-X$ then the support of $min(X,Y)$ must be $(0,1/2)$ (since one smaller than $1/2$ implies that the other must be larger than 1/2). Moreover, $U$ will be less than a number $x$ if only one of two things occur (and they exclude each other so you can sum their probabilities): whether $X$ is less than $x$ or $Y$ is less $x$, and $x$ must be between $0$ and $1/2$. Therefore, letting $F_U$ be the distribution function of $U$, we have
$F_U(x)=P(U<x)=P(0<X<x)+P(0<Y<1/2)=P(0<X<x)+P(1-x<X<1)=2x$
And from that you can conclude that $U$ has a density function $f_U=2$ on $(0,1/2)$ and $0$ elsewhere. Thereform it is uniformly distributed on $(0,1/2)$. An equally simple reasoning shows you that $V$ is uniform in $(1/2,1)$.
Once you know they are uniform, since the expectation is the halfpoint of the interval, you get $E(U)=1/4$ and $E(V)=3/4$.
About your answer. You did arrive to the correct density function but you have the wrong definition of expectation. If $f_X$ is the density of a random variable $X$ you have $$ \int_{Supp(X)} x f_X(U) \,dx $$.
Your are missing the $x$ in your formula. So if $f_U=2$ then you have
$$\int_0^{1/2} 2x dx = [x^2]_0^{1/2}=1/4$$