R.V. transformation without using the transformation theorem

Question

Let $X$ have the pdf

$f(x) = \frac{1}{2} (1+x), \quad -1 < x < 1$.

Find the pdf of $Y = X^2$.

I want to solve this question without using the transformation theorem, but I always arrive that $P(Y \leq y) = \sqrt{y}$. Any suggestions?

Answer 1

Sorry if I wasted anyone's time.

For those wondering, what I messed up is that I found $F_Y(y) = \sqrt{y}$ when I needed to find $f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{1}{2} y^{-\frac{1}{2}}$.

Answer 2

For $y\in[0;1]$ we have:

$$\begin{align} f_Y(y) & = \frac{\mathrm d\;}{\mathrm d y }\mathsf P(Y\leq y) \\ & = \frac{\mathrm d\;}{\mathrm d y }\mathsf P({-}\surd y \leq X \leq {+}\surd y) \\ & = \frac{\mathrm d\;}{\mathrm d y }\int_{-\surd y}^{+\surd y}\tfrac 12 (1+x)\operatorname d x\end{align}$$

Then we use: $\displaystyle \dfrac{\mathrm d}{\mathrm d y}\int_{h(y)}^{g(y)} f(x)\operatorname d x = f\big(g(y)\big)\cdot\frac{\mathrm g(y)}{\mathrm d y} - f\big(h(y)\big)\cdot\frac{\mathrm dh(y)}{\mathrm d y}$

$$\begin{align}f_Y(y) & = \tfrac 12 \big(\tfrac 1{2\surd y}(1+\surd y)+\tfrac 1{2\surd y}(1-\surd y)\big)\\ & = \frac 1 {2\surd y} \\ \Box\end{align}$$