I am giving the following question:
Consider a random variable, $X$, having pdf: $f_x(x) = (x^2)/3, I(-2,1) (x)$
(a) Give the cdf of $Y = X^2$
I believe the support of $X^2$ is $[0, 4)$, is that correct?
When trying to find the cdf, I look for $P(Y <= X^2) = P(-y^{(1/2)} <= X <= y^{(1/2)})$
I then integrate from $-y^{(1/2)}$ to $y^{(1/2)}$ which gives me $\frac{2}{9}y^{(2/3)}$ which gives me $0$ when I plug in $0$ but I don't get $1$ when I plug in the upper bound support of $4$.
Since the cdf must be bounded by $0$ and $1$, I don't understand why I'm not able to get the correct answer. I've also tried integrating from $-2$ to $-y^{1/2}$ and then from $0$ to $y^{1/2}$ and that doesn't get me the correct answer either.
Any help is much appreciated!
The cdf of $X$ is $F_X(x)=\frac{x^3+8}{9}$ for $(-2\le x\le 1)$
$F_Y(y)=P(Y\le y)=P(X^2\le y)=P(-\sqrt{y}\le X\le \sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})$
$=\frac{8+\min(1,y^\frac{3}{2})}{9}-\frac{8-y^\frac{3}{2}}{9}$, since $F_X(x)=1$ for $x\ge 1$.
Summary: $F_Y(y)=\frac{2y^\frac{3}{2}}{9}$ for $0\le y\le 1$ and $F_Y(y)=(\frac{1+y^\frac{3}{2}}{9})$ for $1\lt y\le 4$