I'm given a PDF of X, such as $f_x(x) = \frac 12 \lvert x \lvert$ for $-2 \le x \le 2$, and $ 0 $ otherwise, and told to find the CDF for Y where $Y = X^2 $. Trying to do this problem, and others like it, there's two places that trip me up every time and I was hoping someone could clear them up.
I get to the point where I have: $F_Y(y) = P(Y \le y) = P(X^2 \le y) = P(-\sqrt y \le X \le \sqrt y) = \int_{-\sqrt y}^\sqrt y f_x(X) dx $
From here, I'm unsure of how to split the integral up. I know the range of Y is $[0, 4]$, but I'm not sure how to split that up. In other problems I've seen, the range is split up into $[0, 1]$ and then $[1, 4]$ with two separate integrals. I know this is a pretty basic concept, but it's been a while and I just can't recall it.
My second question is, how does the absolute value factor in for when I'm splitting up the integral into multiple ones?
EDIT: After doing a bit of further thinking, I think that for this problem specifically, you don't need to split up the integral, and you can simply integrate as is and plug in the $f_x(x)$ function, since all the y values from 0 to 4 can be obtained from the existing -2 to 2 X range. If for example, you had X having a range of -2 to 1, then I think you would need one integral for when X is -1 to 1, i.e. y is 0 to 1, and another integral for when x > 1, because in that case it's outside the bounds for x>1, but still inside the bounds on the negative side. So then I think you'd have an integral with bounds of $- \sqrt y $ to 1. I hope I didn't word that explanation too badly, but am I right in my thinking?
