CDF and PDF of non-monotonic transformations

Question

$y = x(1-x)$. Distribution of $x$ is uniform on $[0,1]$ for $x$ in $(0,1)$.

Find $p(y)$ and $P(Y)$

Can you guys give me hints on how to proceed?

I think the general way for this problem is to partition the function into monotone parts. But when I proceed, I got stuck with the function $y = x(1-x)$ as I cannot express $x$ in terms of $y$.

Thank you

Answer 1

With $y=x(1-x)$ you have

• $y$ increasing from $0$ to $\frac14$ as $x$ increases from $0$ to $\frac12$
• $y$ decreasing from $\frac14$ to $0$ as $x$ increases from $\frac12$ to $0$
• $x=\frac12 \pm \frac12\sqrt{1-4y}$
• $\mathbb P(Y \le y) = \mathbb P\left(X \le \frac12 -\frac12\sqrt{1-4y} \right)+\mathbb P\left(X \ge \frac12 +\frac12\sqrt{1-4y} \right) = 1-\sqrt{1-4y}$ when $0 \le y \le \frac14$
• $p_Y^{\,}(y) = \frac{d}{dy}\mathbb P(Y \le y) = \frac{2}{\sqrt{1-4y}}$ when $0 \le y \le \frac14$