Squared transformations, distribution functions

Question

$Y = X^2$

$f_{X}(x) = 1$

Find the PDF of $Y$.

$$f_{Y}(y) = \dfrac{1}{2\sqrt{y}}\left[f_{X}\left(\sqrt{y}\right)+f_{X}\left(-\sqrt{y}\right)\right]$$

Right?

But what sense does the term $f_{X}(\sqrt{y})$ make, considering $f_{X}(x) = 1$?

Answer 1

Since $X$ is a continuous $\text{Unif}(0,1)$ RV, we have that

$f_X(x) = \begin{cases} 1,& \text{if } 0 \leq x \leq 1\\ 0, & \text{otherwise} \end{cases} $

As noted in the comments above, the formula you provided works for any continuous random variable. That is, if $Y = X^2$, then for $y \geq 0$,

\begin{align} f_Y(y) &= \frac{1}{2\sqrt{y}}\left[f_X(\sqrt{y}) + f_X(-\sqrt{y})\right],\\ \end{align} As for your question, what does the term $f_X(\sqrt{y})$ actually mean? Since $0 \leq X \leq 1$, we have that $0 \leq Y \leq 1$ as well, and $f_X(\sqrt{y}) = 1$ along the entire support of $Y$. That makes the above equation

\begin{align} f_Y(y) &= \frac{1}{2\sqrt{y}}\left[1 + f_X(-\sqrt{y})\right],\\ \end{align}

I'll let you tackle the derivations for $f_X(-\sqrt{y})$ on your own, but you should get that it's equal to zero along the entire support of $Y$.