Find joint density function, marginal and expected value Let $X_1$ and $X_2$ have joint density function:
$$f_{X_{1},X_{2}}\left(x_{1},x_{2}\right)=4x_{1}x_{2}, 0\leq x_{1}\leq1,
0\leq x_{2}\leq1$$
If $Y_{1}=X_{1}$ and $Y_{2}=X_{2}/X_{1}$
(a) Find the joint density function of $Y_{1}$ and $Y_{2}$.
(b) Find $P\left(Y_{2}\geq2\right)$
Solutions:
(a) I did this by using Jacobian method.
The inverse of the transformation is,
$$T^{-1}: X_{1}=Y_{1}$$ 
$$\qquad\qquad\quad\qquad\;X_{2}=Y_{2}X_{1}=Y_{2}Y_{1}$$
The jacobian of the transformation is:
$$J_{T^{-1}}:\:det\begin{bmatrix}1 & 0\\
y_{2} & y_{1}
\end{bmatrix}=y_{1}$$
Then, the joint density function of $Y_{1}$ and $Y_{2}$ is:
$$f_{Y_{1},Y_{2}}\left(y_{1},y_{2}\right)=4y_{2}y_{1}^{3}, 0\leq y_{1}\leq1
 , 0\leq y_{2}\leq\frac{1}{y_{1}}$$
(b) To do this, first i have to find the marginal density of $Y_2$.
I got the next result,
$$f_{Y_{2}}\left(y_{2}\right)=\int_{0}^{1}4y_{2}y_{1}^{3}dy_{1}+\int_{0}^{1/y_{2}}4y_{2}y_{1}^{3}dy_{1}=y_{2}I_{\left(0,1\right)\left(y_{2}\right)}+\frac{1}{y_{2}^{3}}I_{\left(1,\infty\right)\left(y_{2}\right)}$$
Next, I calculated $P\left(Y_{2}\geq2\right)$ in two ways. These are,


*

*$P\left(Y_{2}\geq2\right)=\int_{2}^{\infty}\frac{1}{y_{2}^{3}}dy_{2}=-\frac{1}{2}\frac{1}{y_{2}^{2}}\biggr|_{y_{1}=2}^{y_{2}=\infty}=\frac{1}{8}$

*$P\left(Y_{2}\geq2\right)=1-P\left(Y_{2}\leq1\right)=1-\int_{0}^{1}y_{2}dy_{2}=\frac{1}{2}y_{2}^{2}\biggr|_{0}^{1}=\frac{1}{2}$
My teacher said that all the procedures i did to calculate the joint density and the marginal are wrong. As the marginal is wrong, then the calculation of $P\left(Y_{2}\geq2\right)$ is wrong. 
Then, I think something is wrong in my calculations because of the diferent results that i get in numbers 1 and 2.
Please help me to clarify the ideas.
 A: 
My teacher said that all the procedures I did to calculate the joint density and the marginal are wrong. 

Nope.   It looks okay.   Running through it my way seems to match with what you have.
The joint is correct.
$$\begin{align}f_{Y_1, Y_2}(y_1,y_2) & = f_{X_1, X_2}(y_1, y_1y_2)\cdot \lVert\frac{\partial(y_1,y_1y_2)}{\partial(y_1, y_2)}\rVert\\[1ex] & = 4(y_1)(y_1y_2)~\mathbf 1_{0\leq y_1\leq 1~,~ 0\leq y_1y_2\leq 1}\cdot\begin{Vmatrix}1 & 0\\y_2 & y_1\end{Vmatrix} \\ &= 4 ~ y_1^3 ~ y_2~\mathbf 1_{0\leq y_1\leq 1~,~ 0\leq y_2\leq y_1^{-1}} \end{align}$$
The marginal is then indeed a piecewise function, like so:
$$\begin{align}f_{Y_2}(y_2) &= \int_0^{\min(1,1/y_2)} f_{Y_1,Y_2}(t, y_2)\operatorname d t\cdot \mathbf 1_{0\leq y_2 <\infty}
\\ & = 4y_2\int_0^1 t^3 \operatorname d t\cdot \mathbf 1_{0\leq y_2<1}+4y_2\int_0^{1/y_2}4t^3\operatorname d t\cdot\mathbf 1_{1\leq y_2<\infty}
\\ & = y_2\mathbf 1_{0\leq y_2<1}+y_2^{-3}\mathbf 1_{1\leq y_2<\infty}
\end{align}$$
And so we have: $$\begin{align}\mathsf P(Y_2\geq 2) &= \int_2^\infty s^{-3}\operatorname d s \\[1ex]&= \frac 18\end{align}$$
Alternatively, ah, there is your mistake.   The complement of $Y_2\geq 2$ is not $Y_2\leq 1$.   This is the way to go:
 $$\begin{align}\mathsf P(Y_2\geq 2) &= 1-\mathsf P(Y_2<2)\\[1ex] &= 1-\int_0^1 s\operatorname d s-\int_1^2 s^{-3}\operatorname d s\\[1ex] & =1-\tfrac 12-\tfrac 38 \\[1ex] & = \tfrac 18\end{align}$$

Just to verify, let us do it the easy way.
$$\begin{align}\mathbf P(Y_2\geq 2) & = \mathbf P(X_1\leq X_2/2) \\[1ex] & =\int_0^1\int_0^{t/2} 4st\operatorname d s\operatorname d t \\[1ex] & = 4\int_0^1 t \int_0^{t/2} s\operatorname d s\operatorname d t \\[1ex] &= \tfrac 12 \int_0^1 t^3\operatorname d t \\[1ex] & = \tfrac 18\end{align}$$

tl;dr Everything checks out until the very end.   Get up and try again!
