Find the distribution of U = X/Y

Question

I have the following situation:

The joint distribution of X and Y is defined as

$$ f_{XY}(x,y) = 2 \mathbb{I}_{(0,y)}(x) \mathbb{I}_{(0,1)}(y) $$

I need to find the distribution of U = X/Y.

I tried to find the marginal distribution of X and Y ($X \sim \beta(1,2)$ e $ Y\sim \beta(2,1)$) and create a auxiliary random variable V = X to use jacobian transformation.

I think I'm taking the wrong way. Someone could help me with this problem?

Thanks for all help!

Answer 1

Well...

Using a auxiliary random variable $V=Y$ we have:

\begin{cases} x = uv\\ y = v \end{cases}

The Jacobian is

$$\begin{bmatrix} \dfrac{\partial v}{\partial u} & \dfrac{\partial uv}{\partial u}\\ \dfrac{\partial v}{\partial v} & \dfrac{\partial uv}{\partial v} \end{bmatrix} = \begin{bmatrix} 0 & v \\ 1 & u \end{bmatrix} $$

So $|Det(J)| = v$ and $f_{UV}(u,v) = f_{XY}(h(u,v))|Det(J)|$ and

$$ f_{UV}(u,v) = 2v \ \mathbb{I}_{(0,1)}(u), \mathbb{I}_{(0,1)}(v). $$

Am I correct?

Answer 2

You can compute the cumulative distribution:

$$F_U(u)=P(U\le u) =P(X \le u Y )$$

Now, notice that $(X,Y)$ are uniform over a tringle, then computing the above probability corresponds to computing the area of a (another) triangle. Can you go on from here?