I need some help on the following problem:
Let $Y_1$ and $Y_2$ be two random variables with the following density function:$$f_1(y_1)= \begin{cases} 6y_1(1-y_1), & \text{if } 0\le y_1\le 1 \\ 0, & \text{otherwise} \\ \end{cases} \\ f_2(y_2)=\begin{cases} 3y_2^2, & \text{if } 0\le y_1\le 1 \\ 0, & \text{otherwise} \\ \end{cases}$$ If $Y_1$ and $Y_2$ are independent find the pdf of random variables $U=\frac{Y_1}{Y_2}$ .
Attempt:
$(i)$
First I get the joint pdf of random variables $f_{Y_1,Y_2}(y_1,y_2)=\begin{cases}
18y_1y_2^2(1-y_1), & \text{if } 0\le y_1\le 1,0\le y_2\le 1 \\
0, & \text{otherwise} \\
\end{cases}$
Then Let $V=Y_2$ and got $(y_1,y_2)=(uv,v)$ .
Let find $$f_{U,V}(u,v) = |\det(J)|f_{Y_1,Y_2}(y_1,y_2) \qquad u\in ?,v \in ?\\
|J|=\begin{pmatrix}\frac{\partial y_1}{\partial u} & \frac{\partial y_1}{\partial v} \\ \frac{\partial y_2}{\partial u} & \frac{\partial y_2}{\partial v}\end{pmatrix} = \begin{pmatrix}v & u \\ 0 & 1 \end{pmatrix}=|v|$$
Then $f_U(u)=\int_{\text{all} \ v} 18uv^4(1-uv)|v| dv$ .
(ii)
Domain part: When I try to find domain for $u,v$ what i get is $$0\le y_1 \le 1\Rightarrow 0\le\frac{y_1}{y_2}\le\frac{1}{y_2}\Rightarrow 0\le u\le\frac{1}{v}\\0\le y_2 \le 1\Rightarrow 0\le v \le 1$$ Am I right $?$ But after doing the integration over the limit of $v$ I didn't get the right answer .Is my domain are wrong $?$
I think I do mistakes when I try to find the domain .Can anyone give me some hints and intuitive way to find it out .In order to get a clear idea I add some extra problem:
(i)If $0\le y_2 \le y_1 \le 1$ and $U=Y_1 -Y_2$ then what about $u \in ?,v\in ?$
(ii)If $0\le y_2 \le 1,0\le y_1 \le 1$ and $U=Y_1Y_2$ then what about $u \in ?,v\in ?$
Sorry if I ask too many question but I think without clearing my concept I can't read it all .And Thanks in advance .
Edit: This answer is wrong, the correct way of doing this problem is here find the region for CDF. Nevertheless, I will not delete this answer.
Let's do it. Let $F_U$ be the CDF of $U$.
$$F_U(t) = P(U \leq t) = P \left(\dfrac{Y_1}{Y_2} \leq t\right) = P(Y_1 \leq t Y_2).$$
Well, $t$ could be any number: $0.9$, $100$ or $2.5$. We have to break this problem into two intervals, $t \in [0,1]$ and $t \in [1,\infty)$. Let's begin with $t \in [0,1]$. If you draw $f_1(y_1)$ and $f_2(y_2)$, you find that $P(Y_1 \leq t Y_2)$ is true when $y_1 \geq t/(t+2)$. Why? When does the inequality happen?
$$6y_1 (1 - y_1) \leq 3 y_2^2 t.$$
If $y_1 = y_2$, find $y_1$ (or $y_2$) in terms of $t$.
To find $P(Y_1 \leq t Y_2)$, we use the PDF of $Y_1$ and $Y_2$:
$$ \begin{align} P(Y_1 \leq t Y_2) &= \int_{0}^{1} \left( \int_{t/(t+2)}^{1} 6y_1(1-y_1)\, dy_1 \right) 3y^2_2 \, dy_2 \\ &=\dfrac{t^2 (t+6)}{(t+2)^3}. \end{align} $$
We can multiply $Y_1$ and $Y_2$ because they are independent. So the PDF is:
$$ \begin{align} f_U(t) &= \dfrac{d}{dt} \left( \dfrac{t^2 (t+6)}{(t+2)^3} \right) \\ &= \dfrac{24 t}{(t+2)^4}, \quad \text{valid for $t \in [0, 1]$.} \end{align} $$
Now, for $t \in [1, \infty)$. We do the same but the limit for $y_1$ changes.
$$ \begin{align} P(Y_1 \leq t Y_2) &= \int_{0}^{1} \left( \int_{0}^{t/(t+2)} 6y_1(1-y_1)\, dy_1 \right) 3y^2_2 \, dy_2 \\ &=\dfrac{t^2(t+6)}{(t+2)^3}. \end{align} $$
So the PDF is:
$$ \begin{align} f_U(t) &= \dfrac{d}{dt} \left(\dfrac{t^2(t+6)}{(t+2)^3} \right) \\ &= \dfrac{24 t}{(t+2)^4}, \quad \text{valid for $t \in [1, \infty)$.} \end{align} $$
As you can see, we got the same function for both intervals. You can check that:
$$\int_0^{\infty} \! \dfrac{24 t}{(t+2)^4} dt = 1.$$