deriving pdf with two methods, with two different results

Question

I have this problem to determine a density function $f_{Y}(y)$ of $Y$ given $X$ as a random variable with uniform distribution on $[−1, 1]$ and $Y = X^2$. So: $f_X(x) = \frac{1}{2}$ for $x \in [-1, 1]$. I approached this problem with two different ways, 1) through determining first the $F_Y$ as the $CDF$, then derive from it $PDF$ 2) by transformation.

For the first method I got $f_Y(y) = \frac{1}{2\sqrt(y)}$ for $y \in ]0, 1]$ This seems correct. However, for transformation method, I tried the following:

$g(x) = x^2$ and $g'(x) = 2x$

$x_1 = g^{-1}(y) = \sqrt(y)$

so $f_Y(y) = \frac{f_X(x_1)}{|g'(x_1)|} = \frac{1/2}{2\sqrt(y)} = \frac{1}{4\sqrt(y)}$

Also, there is the example of $Y = X^3$, then:

$g(x) = x^3$ and $g'(x) = 3x^2$

$x_1 = g^{-1}(y) = \sqrt[3](y)$

so $f_Y(y) = \frac{f_X(x_1)}{|g'(x_1)|} = \frac{1/2}{3\sqrt[3](y)^2} = \frac{1}{6\sqrt(y)^2}$ but the correct answer is $\frac{1}{3\sqrt(y)^2}$

Answer 1

This idea that you proposed only works if the transform $g(\cdot)$ is one to one.

https://www.math.arizona.edu/~jwatkins/f-transform.pdf

This is why the following reasoning is incorrect, for instance, if $g(x)=x^2$

\begin{align} f_Y(y)&=f_X(g^{-1}(y))\Big| \frac{d}{dy}g^{-1}(y) \Big| \\ &= \frac{1}{2} \Big| \frac{1}{2} y^{-1/2} \Big| \end{align}

The correct would be

\begin{align} F_Y(y) &= P(Y \leq y) = F_X(\sqrt{y})- F_X(-\sqrt{y}) \\ f_Y(y)&= \frac{1}{2\sqrt{y}}(f_X(\sqrt{y})+f_X(-\sqrt{y})) = \frac{1}{2\sqrt{y}} \end{align}