Modify from this problem, if I have:
$f(x) = \begin{cases} x^{-2}, & \text{ if } 1 \leq x\\ 0, & \text{otherwise}. \end{cases}$
Find the probability density function of $Y=X^2$.
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Since $ 1 \leq x$, the distribution of $Y = X^2$ is $1 \leq y$.
$\displaystyle F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq\sqrt{y})=F_X(\sqrt{y})=\int_1^\sqrt{y}x^{-2}\,dx=-\frac{1}{x}\Bigg |_1^{\sqrt{y}}$ $=1 - y^{-1/2}$
Taking the derivation to get the PDF of $Y$:
$f_Y(y)=\frac{1}{2}\cdot y^{-\frac{3}{2}}$
Is my calculation correct?
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Michael's beautiful method is explained in this video, which applies only to monotonic cases:
$$f_{Y}(y) = f_{X}(h(y))\left|\frac{dh}{dy}(y)\right|.$$
Yes and using the cumulative distribution function $P(Y \leq y) = P(X^2 \leq y)$ is in my opinion the best way of solving these problems. As we restrict our attention to an area $x \geq 1$ where the function from $X$ to $Y$ is 1-to-1 we can also use the transformation rule:
If we let $x = h(y) = \sqrt{y}$ such that $h'(x) = -\frac{1}{2}y^{-1/2}$. Then $$ f_Y(y) = f_X(\sqrt{y}) \left \lvert \frac{\partial h}{\partial y} \right \rvert = (\sqrt{y})^{-2} \left \lvert -\frac{1}{2}y^{-1/2} \right \rvert = \frac{1}{2} y^{-3/2}. $$