# An integral involving an inequality expression

I have not ever seen this type of integral evaluation before. Does it have a specific name. Would you please guide me in understanding how to solve it. I have tried researching it but have not found anything on it as I do not even know the name of it.

The question goes as follows:

$x$ and $y$ are uniformly distributed in the unit interval and we define $Z$ as:

$$Z = \frac{x}{y-x}$$

Now consider $0 < Z < z$. For $Z>0$ we have $Y>x$. For $Z<z$ we have $Y>x ( \frac{1}{z} + 1)$.

Hence integrate the following:

$$\int_{0}^{1} \left\{ \max \left(x,x \left( \frac{1}{z} + 1 \right) \right) <y<1 \right\}dy$$

Apparently the solution is:

$$\frac{z - x-xz}{z}$$

where

$$z > \frac{x}{1-x}$$

Do not quite know where to start even. Your guidance is much appreciated.

• Has it already been given that $z=\frac{x}{1-x}$ in the question? If possible, type the entire question.
– user371838
Dec 18, 2017 at 8:59
• I guess you are missing a $\mu$, standing for the measure of some set. Dec 18, 2017 at 9:17
• @Rohan My apologies, it was $z > \frac{x}{1-x}$. I've typed the entire question as given. Dec 18, 2017 at 9:45