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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!

Cauchy_96
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  • Why would you need the marginals to use the Jacobian formula? This is exactly the opposite, you need a 2D-to-2D transform $(X,Y)\to(U,V)$ for some suitable $V$ to compute the joint distribution of $(U,V)$, and then you will marginalize it to deduce the distribution of $U$. Your suggestion $V=X$ works pretty well... so, as they say, just do it. – Did Jan 25 '18 at 18:04
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    Sorry what is $\Bbb I$ distribution? – Mostafa Ayaz Jan 25 '18 at 18:18
  • Ok @Did! But what is the support of random variables U and V? – Cauchy_96 Jan 25 '18 at 18:33
  • If $$(U,V)=(X/Y,X)$$ and the $(X,Y)$-domain is $$0<X<Y<1$$ then $$(X,Y)=(V,V/U)$$ hence the $(U,V)$-domain is $$0<V<V/U<1$$ which is also $$0<V<U<1$$ This is entirely automatic... – Did Jan 25 '18 at 19:42
  • If $$(U,V)=(X/Y,Y)$$ and the $(X,Y)$-domain is $$0<X<Y<1$$ then $$(X,Y)=(UV,V)$$ hence the $(U,V)$-domain is $$0<UV<V<1$$ which is also $$0<U<1\qquad 0<V<1$$ This is entirely automatic... – Did Jan 25 '18 at 19:45

2 Answers2

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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?

Cauchy_96
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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?

leonbloy
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