# Independent Uniformly Distributed Random Variables

I have this problem in my textbook and I just can't understand the answer provided.

It says "Let $X$ and $Y$ be independent random variables that are uniformly distributed on the interval $[0,1]$. What is the $PDF%$ of the random variable $Z=Y/X$?"

It then goes on to first find the $CDF$ of $Z$ and then differentiate to get the $PDF$. I understand that step, but I can't figure out where they're getting these answers for the $CDF$.

It says the $CDF$ of $Z$ from $[0,1]$ is $z/2$. I flat out do not understand how that is correct and would really appreciate it if someone could explain it.

Then it says the $CDF$ of $Z$ if $z>1$ is $1-1/(2z)$. I don't understand this one either...

I'm pretty sure this is supposed to be a relatively simple example, so I'm a little worried that I don't understand it. Some explanation here would be incredibly appreciated.

## 3 Answers

We have $Z\le z$ if $Y/X\le z$, i.e. $Y\le zX$. For $z\le1$, the probability for this, given $X=x$, is $zx$, and then integrating over $x$ yields

$$\int_0^1zx\,\mathrm dx=\frac z2\;.$$

To get the result for $z\gt1$, consider $1/Z=X/Y$ and use the result for $z\le1$.

The first CDF is the area of the green triangle, while the second one is 1 minus the area of the red triangle:

First of all, if $z<0$, then $P(Z\leq z)=P(Y\leq zX)=0$ because $zX\leq 0$ a.s. If $z\in [0,1]$, then $zX\in [0,1]$ almost surely. Now, $$P(Z\leq z)=P(Y\leq zX)=\int_0^1\int_0^{zx} 1\,\mathrm dy\,\mathrm dx=\int_0^1 zx\,\mathrm dx=\frac{1}{2}z.$$ If $z>1$, then use the symmetry argument given by joriki.