# Thread: if X is uniform, find density function of Y=X^2

1. ## if X is uniform, find density function of Y=X^2

If $\displaystyle X$ has the uniform distribution $\displaystyle U(-1,3)$, find the probability density function of $\displaystyle Y=X^2$.
Since $\displaystyle -1<x<3, 0<y<9$. Also, $\displaystyle X=\sqrt{Y},-\sqrt{Y}$. I've done some work and ended up with $\displaystyle f(x)=\frac{1}{2\sqrt{y}}$. But when I integrate this from 0 to 9, it does not give me 1. Can I get some help?

2. You have to break it up.

$\displaystyle G(y) = \frac{1}{4} \int_{-\sqrt{y}}^{\sqrt{y}} \ dx = \frac{\sqrt{y}}{2}$ if $\displaystyle 0 \le y<1$

and $\displaystyle G(y) = \frac{1}{2} + \frac{1}{4} \int_{1}^{\sqrt{y}} \ dx = \frac{1}{2} + \frac{1}{4} (\sqrt{y} -1) = \frac{1}{4} + \frac{\sqrt{y}}{4}$ if $\displaystyle 1 < y \le 9$

so $\displaystyle g(y) = \frac{1}{4\sqrt{y}}$ if $\displaystyle 0 \le y<1$

and $\displaystyle g(y)= \frac{1}{8 \sqrt{y}}$ if $\displaystyle 1 \le y \le 9$

3. Originally Posted by Random Variable
$\displaystyle G(y) = \frac{1}{2} + \frac{1}{4} \int_{1}^{\sqrt{y}} \ dx = \frac{1}{2} + \frac{1}{4} (\sqrt{y} -1) = \frac{1}{4} + \frac{\sqrt{y}}{4}$ if $\displaystyle 1 < y \le 9$
How did you get $\displaystyle \frac{1}{2}$?

4. Originally Posted by dori1123
How did you get $\displaystyle \frac{1}{2}$?
G(1) = 1/2. Therefore G(y) = what RV said when $\displaystyle 1 < y \leq 9$.