Suppose that $X$ has a uniform distribution on $\left [ 0, \frac{1}{2} \right ]$. Then find pdf of $X^2$

Question

My solution: Let $Y=X^{2}$. Then $$F_{y}(y)=P(Y \leq y ) = P(X^{2} \leq y) = P(X^{2} \leq \frac{\sqrt{y}}{2}).$$

So $$f_{y}(y) = \frac{1}{4 \sqrt{y}}.$$ Then $$f_{x^{2}}(x) = \frac{1}{4 x}$$

Is my solution correct?

Answer 1

No. $X^{2} \leq y$ means $X \leq \sqrt y$, not $X \leq \sqrt y /2$.

$P(X^{2} \leq y)=P(X \leq \sqrt y)=2\sqrt y$ and the density is $\frac 1 {\sqrt y}$ on $(0,\frac 1 4)$.