Let X be a uniformly distributed random variable on the interval $(−1, 1)$. Find the pdf of $Y = X^2$.
So far I know the following:
$$Y\le y = X^2 \le y = X \le y^{1\over2}$$
I also know that $f_x(x)={1\over2}$
Now I am confused about two things. Number one, what is the interval of $Y$?
When I replace $x$ with $y^{1\over2}$, this is what happens:
$$-1\le y^{1\over2}\le1$$
$$(-1)^2\le y\le(1)^2$$
$$1\le y\le 1$$
How does this make sense, and where do I proceed from here? Furthermore, the real solution says that the cdf is $y^{1\over2}$ for $0<y<1$. Where did the zero come from?
Number two, to get the pdf I did a change of variables:
$${1\over2} {\big| {d\over dy} y^{1\over2} \big|}$$
$${1\over4} { y^{-{1\over2}} }$$
However, the real solution says ${1\over2} { y^{-{1\over2}} }$ for $0<y<1$.
Any help would be greatly appreciated.
Let’s find the cumulative distribution function $F_Y$ of $Y$.
We have
$$F_Y(y)=P[Y \le y]= \begin{cases} 0 &\text{for } y <0\\ \int_{-\sqrt{y}}^{\sqrt{y}}f_X(x)\ dx = \sqrt{y} &\text{for } 0\le y <1\\ 1 & \text{for } y \ge 1 \end{cases}$$
as $P[X^2 \le y]= P[-\sqrt{y} \le X \le \sqrt{y}]$.
Therefore $f_Y(y)= \dfrac{1}{2\sqrt{y}}$ for $0\le y \le 1$ by taking the derivative.