How to prove something at Uniform distribution... $X\sim U (0,1)$.
The point $X$ divides $[0,1]$ to two parts.
$Y=\frac{\text{The big part}}{\text{The small part}}$. ($Y$ is the ratio... $Y\ge1$).
What is the density function of $Y$?
I'd like to get ideas how to prove it because I'm not sure that I'm do it right...
Thank you!
 A: You want to find the probability density function (pdf) f(Y) but in this sort of problem it is easiest to start from the the cumulative density function (cdf) F(Y), which is defined for all z by saying that F(z) is the probability that Y will be less than or equal to z.  Then f(Y) will be the derivative of F(Y).
For your function, we can find F(Y) easily. First, since Y(X=1-u) = Y(x=u), we can without changing the distribution of Y take x to be uniform on (0,1/2) [I am calling this lower case x to note that it is not really our same variate X); then we don't have to work with a conditional in our definition of Y, since 1-X will always be the bigger part.
So now, we know that F(z) = Prob (Y <  z) = Prob ( (1-x)/x < z ) = Prob (x z > 1-x )
= Prob ( x(z+1) > 1 ) = Prob ( x > 1/(z+1) ).
By the way, the domain of F(Z) is z >= 1, since the bigger part cannot be less than the smaller part.
Notice that when we have a uniform variate (X in this case) we are always trying to get X isolated in a probability like this. {Thank you, Gian-Carlo Rota, who first taught me this in probability class!}
Since x is uniform on (0,1/2), Prob (x > w) = 2( 1/2 -w) = 1 - 2w, for all 0 < w < 1/2. 
So F(z) = Prob (x > 1/(z+1)) = 1 - 2/(z+1).
Finally, given F(Y) we take the derivative to find f(Y):
f(Y) = 2/(Y+1)^2
A: Let
$
X_1=\max(X, 1-X).
$
Then,
$$
Y=\frac{X_1}{1-X_1}.
$$
So what is the distribution of $X_1$?
$$
F_{X_1}(x_1)=P(X_1\leq x_1)=P(\max(X, 1-X)\leq x_1)=P(X\leq x_1, 1-X\leq x_1)\\
=P(1-x_1\leq X\leq x_1)=\int_{1-x_1}^{x_1}dx=x_1-(1-x_1)=2x_1-1, \quad x_1\in[1/2, 1]. 
$$
Since $Y=X_1/(1-X_1)$, you have that
$$
F_Y(y)=P(Y\leq y)=P\left(\frac{X_1}{1-X_1}\leq y\right)=P\left(X_1\leq y(1-X_1)\right)=P\left(X_1\leq\frac{y}{1+y}\right)\\
=\frac{2y}{1+y}-1=\frac{y-1}{y+1}, \quad y\in[1, \infty).
$$
This is the cdf, and the pdf is
$$
f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{2}{(1+y)^2}.
$$
