Suppose $X$ is a random variable with pdf $f(x)=\frac{x^2}{3}$, $-1<x<2$ and zero otherwise. Consider the tranformation $Y=X^2$, which is not one-to-one obviously. Find the CDF of Y.
My book says that
$$F(y)=P[X^2<y]=P[-\sqrt{y}<X<-\sqrt{y}]=\int_{-\sqrt{y}}^{\sqrt{y}} \frac{x^2}{3} dx = \frac{2y^{3/2}}{9}.$$
When $x=2$, $F(x)=1$. So I would also expected $F(y)=1$ when $y=4$, since $y=x^2$. But this is clearly not the case. $F(y)=16/9 > 1$, for $y=4$.
Could anyone explain me what's going on ?
Thanks!
The answer in the book is not complete and is only valid under the condition that $0\leq y\leq1$.
For $y<0$ we have $F(y)=0$.
For $y\geq4$ we have $F(y)=1$.
For $1<y<4$ we have: $$F(y)=F(1)+P(1<X^2\leq y)=\frac29+P(1<X\leq\sqrt y)=\frac29+\left[\frac{x^3}9\right]_1^{\sqrt y}=\frac19+\frac{y^{\frac32}}{9}$$