Let $(X,Y)$ be uniform on the unit square. Let $Z=\frac{X}{Y}$. Find the density of $Z$.
I found typically when they said $(X,Y)$ has a uniform distribution on the unit square it is defined as
$f(x,y) = \begin{cases}
1, & \text{ if } 0\leq x \leq 1\\
0, & \text{otherwise}
\end{cases}
$
I tried to solve the main problem like this:
First find cdf and from there I calculate the pfd function.
$F_Z(z) = P(Z\leq z)= p(g(x,y) \leq z) = \int_0^1\int_0^1\frac{X}{Y}dxdy$
Probably I should change this to be a function of z which I don't know how. Since I don't know I just continue integrating with respect to x and y I will have
$= \int_0^1\frac{1}{2y}dy = \frac{1}{2}\ln (1) -\ln(0)$ which is not defined
Apparently something is not right here, Can anyone help please?
Best Answer
Seeing the number of similar questions on the site, let us try to delineate some clear and automatic methods.
First step: Write correctly the joint PDF. Here, the PDF $f$ of $(X,Y)$ is defined on the whole plane $\mathbb R^2$ (as every other joint PDF) by the formula $$f(x,y)=\mathbf 1_{0<x<1}\,\mathbf 1_{0<y<1}$$ Again:
Second step: Deduce the joint PDF of a transformed couple. Here, one can consider $(Z,T)=(X/Y,Y)$ (but other choices for the second coordinate $T$ are possible).
Then $(X,Y)=(ZT,T)$ hence $dxdy=tdzdt$ and the change of variable formula based on the Jacobian of the transformation $(x,y)\to(z,t)=(x/y,y)$ yields the joint PDF $g$ of $(Z,T)$ as $$g(z,t)=|t|\,f(x,y)=|t|\,f(zt,t)=t\,\mathbf 1_{0<zt<1}\,\mathbf 1_{0<t<1}$$
Third step: Marginalize the joint PDF to deduce the desired PDF. Here, the PDF $f_Z$ of $Z$ is simply $$f_Z(z)=\int_\mathbb Rg(z,t)dzdt=\int_0^1t\,\mathbf 1_{0<t<1/z}dt=\mathbf 1_{z>0}\int_0^{\min\{1,1/z\}}tdt=\tfrac12\min\{1,1/z\}^2\mathbf 1_{z>0}$$ An equivalent formulation is $$f_Z(z)=\tfrac12\mathbf 1_{0<z<1}+\tfrac1{2z^2}\mathbf 1_{z>1}$$