Finding CDFs of Transformations of Uniform RV

Question

Let Random variable $U \sim \text{Uniform}[0,1]$. How can I find CDF of Random Variable $S = U(1-U)$?

I know the answer is $F_{S}(s) = 1 - \sqrt{1 - 4S}$ but I do not understand how to get there.

Can someone please help me explain how to arrive at the answer? I've tried solving for U in terms of S using the quadratic formula, and end up with:

$U = \frac{1}{2} \pm \frac{\sqrt{1 - 4S}}{2} $

This looks close to the real answer, but I'm not sure how to make the connection. Any help would be most appreciated. Thank you!

Answer 1

First, if $0 \le x \le 1$ then $0 \le x(1-x) \le 1/4$. Consequently, $F_S(s)=0$ for $s \le 0$, and $F_{S}(s)=1$ for $s \ge 1/4$. For $0 \le s \le 1/4$, we have \begin{align} \Pr\{S \ge s\} & =\Pr\{U(1-U) \ge s\} \\ & = \Pr\left\{(U-1/2)^2 \le\frac{1-4s}{4}\right\} \\ & =\Pr\left\{\frac{1-\sqrt{1-4s}}{2} \le U \le \frac{1+\sqrt{1-4s}}{2}\right\} \\ & = \frac{1+\sqrt{1-4s}}{2}-\frac{1-\sqrt{1-4s}}{2} \\ & = \sqrt{1-4s}. \end{align}

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Note that $$\Pr\{a \le U \le b\} = b-a$$ for $a,b \in [0,1]$ and $a \le b$.