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.
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!