Consider the random variable X with probability density function $$f(x) = \begin{cases} \frac{24}{x^4}; & \text{ if, } x>2 \\ 0; & \text{ otherwise } \end{cases}$$ Find the probability density function of $Y=\frac{8}{X^3}$.
I tried using the formula $$f_Y(y)=f_X(g^{-1}(y)) \vert\frac{\partial}{\partial y}g^{-1}(y)\vert.$$
where
$Y=g(x)=\frac{8}{x^3}$,
$g^{-1}(Y)=\frac{2}{y^\frac{1}{3}}$
but it ended up as $f_Y(y)=$$\frac{y^\frac{4}{3}}{|y\frac{4}{3}|}$
Can anyone tell me where it's gone wrong?
2nd attempt: $P(Y \leq y) = P(\frac{8}{X^3} \leq y) = P(X \geq \frac{2}{y^\frac{1}{3}})$ = $1$- $P(X \leq \frac{2}{y^\frac{1}{3}})$ = $\frac{8}{y}$ using the integral on $f(x)$ over $2$ to $\frac{8}{y\frac{1}{3}}$
What you have done is correct for $0<y<1$. There is no need for absolute value so the density is $1$. Note that $X>2$ implies that $\frac 8 {X^{3}} \in (0,1)$ so $Y$ is uniformly distributed on $(0,1)$.
In your second method $P(Y \leq y)=P(X \geq 2 y^{-1/3})=\int_{ 2 y^{-1/3}}^{\infty} \frac {24}{x^{4}}dx=y$ if $y$ is between $0$ and $1$.