find the CDF of Y based on X I am given the following:
$Y=X(X+2)$, $X \le 0$
$Y=0, 0<X \le 2$
and
$Y=X(X-2), X>2$.
And I also know the PDF of $X$, which is $f(x)=\frac {a}{\pi (a^2 + x^2)}$
I am supposed to calculate the CDF of $Y$. What confuses me here is how exactly am I supposed to use what I am given for $Y$. I mean how do I know the limits for $Y$ based on the limits of $X$. Thanks
 A: Look at the graph of the function $Y(X)$. 

If $y\geq 0$ then $\mathbb P(Y\leq y) = \mathbb P(x_1\leq X \leq x_2)$, where $x_1\leq -2$ is the solution of equation
$$x(x+2)=y, $$
and $x_2\geq 2$ is the solution of equation
$$x(x-2)=y
$$
Solve this equations. 
$$x^2+2x=y$$ 
$$x^2+2x+1=y+1$$
$$(x+1)^2=y+1$$
$$x_1=-\sqrt{y+1}-1$$
And the second equation:
$$x^2-2x=y$$ 
$$x^2-2x+1=y+1$$
$$(x-1)^2=y+1$$
$$x_2=\sqrt{y+1}+1$$
We obtained that for $y\geq 0$
$$\mathbb P(Y\leq y) = \mathbb P(x_1\leq X \leq x_2)=\mathbb P\left(-\sqrt{y+1}-1\leq X \leq \sqrt{y+1}+1\right).$$
Calculate this probability using pdf $f_X(x)$ given.
If $-1<y<0$ then 
$$\mathbb P(Y\leq y) = \mathbb P(x_1\leq X \leq x_2)=\mathbb P\left(-\sqrt{y+1}-1\leq X \leq \sqrt{y+1}-1\right).$$
Here $x_1<x_2<0$ are the solutions of equation $x(x+2)=y$. Calculate this probability using pdf $f_X(x)$ given. 
The CDF that will be obtained after calculation of this probabilities should be discontinuous at zero. Jump height of this function at zero equals to $\mathbb P(0<X\leq 2)$. You can separately calculate this probability and then check your CDF comparing the jump height with this value. 
