Suppose that $X$ and $Y$ are chosen randomly and independently according to the uniform distribution from $(0,1)$. Define $$ Z=\frac{\max(X,Y)}{\min(X,Y)}.$$
Compute the probability distribution function of $Z$.
Can anyone give me some hints on how to proceed?
I can only note that $\mathbb{P}[Z\geq 1]=1$ and $$F_Z(t)= \mathbb{P}[Z \leq t]=\mathbb{P}[X\leq Y, Y\leq tX]+\mathbb{P}[Y \leq X, x\leq tY]$$
On
We find the cdf $F_Z(x)$ of the random variable $Z$. By symmetry, it is enough to find $\Pr(Z\le z|Y\gt X)$, that is, $$\frac{\Pr((Z\le z) \cap (Y\gt X))}{\Pr(Y\gt X)}.$$ The denominator is $\frac{1}{2}$. so we concentrate on the numerator.
The rest of the argument is purely geometric, and will require a diagram. Draw the square with corners $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$.
Draw the line $x+y=1$ joining $(1,0)$ and $(0,1)$.
For a fixed $z$ (make it $\gt 1$) draw the line $y=zx$.
We will need labels. Let $P$ be the point where the lines $x+y=1$ and $x+y=1$ meet. Let $Q$ be the point where the line $y=zx$ meets the line $y=1$, and let $R=(0,1)$. Let $A=(1,0)$ and $B=(1,1)$. The quadrilateral $ABQP$ is the region where $Y\ge X$ and $Y\le zX$. So the probability that $Y\gt X$ and $Z\le z$ is the area of this region.
It is easier to first find the area of $\triangle PQR$. One can show that $QR=z$ and that the height of $\triangle PQR$, with respect to $P$, is $\frac{1}{1+z}$. Thus $\triangle PQR$ has area $\frac{1}{2}\frac{z}{1+z}$.
It follows that the area of $ABQP$ is $\frac{1}{2}-\frac{1}{2}\frac{1}{1+z}$. Divide by $\frac{1}{2}$ and simplify. We find that $\Pr(Z\le z)=\frac{z}{1+z}$.
Thus $F_Z(z)=\frac{z}{1+z}$ (for $z\gt 1$). If we want the density function, differentiate.
You might find this more helpful easier to write $F_Z(t)= \mathbb{P}[Z \leq t]$ as $1-(\mathbb{P}[X\leq Y/t] +\mathbb{P}[Y \leq X/t])$ for $t\ge 1$.
You can calculate this from the unit square
or by simple integration $$1-\left(\int_{y=0}^{1} \int_{x=0}^{y/t} dx \; dy + \int_{x=0}^{1} \int_{y=0}^{x/t} dy \; dx\right)$$