I'm given that the random variable $X$ is distributed as $UNIF(0,1)$.
Moreover, I'm given that:
$U = X(1-X)$
Using the CDF method (i.e. finding the CDF, then taking the derivative), how would I find the PDF of $U$? I can't isolate X in the equation, so how would I go about getting the PDF of $U$ as a result? Thank you
On
This is a solution using the density.
Let's assume $X\sim U(0,1)$ and take the transformation $u=X(1-X)$, now you can find $X$ in terms of $u$ by using $-X^2+X-u=0$ this is a quadratic equation so by using the traditional formula you get $X=\frac{-1\pm\sqrt{1-4u}}{-2}$.
Note that you can get the derivative of $X$ by using implicit differentiation (before solving the quadratic equation) $-2X*X'+X'-1=0$ then $X'=\frac{1}{1-2X}$.
By replacing the value of X found in the quadratic equation you get $\arrowvert X'\arrowvert=\frac{1}{\sqrt{1-4u}} $.
Now the transformation you proposed is not one-to-one, (that's why it has two solutions for X) then by using the transformation formula you get $\frac{1}{\sqrt{1-4u}}\left(f_X\left(\frac{-1+\sqrt{1-4u}}{-2}\right)+f_X\left(\frac{-1-\sqrt{1-4u}}{-2}\right)\right)=\frac{2}{\sqrt{1-4u}}$
Furthermore the maximum value for u is $1/4$ so $0<u<1/4$
Let $F(u) = \Pr(U \le u)$. The minimum and maximum values of $U$ are $0$ and $1/4$, respectively. Therefore, $F(u)=0$ when $u \le 0$, and $F(u)=1$ when $u \ge 1/4$.
Below is the calculations for $0 \le u \le 1/4$. Observe that
$$F(u) = \Pr(X(1-X)\le u) = \Pr(X-X^2 \le u) = \Pr(X-X^2-1/4 \le u-1/4).$$ But $-(X-1/2)^2 = -(X^2-X+1/4)=X-X^2-1/4$. Therefore, $$F(u) = \Pr\left(-\left(X-\frac{1}{2}\right)^2 \le u-\frac{1}{4}\right) \\ = \Pr\left(\left(X-\frac{1}{2}\right)^2 \ge \frac{1}{4}-u\right) \\ = \Pr\left(\left|X-\frac{1}{2}\right| \ge \sqrt{\frac{1}{4}-u}\right).$$ Consequently, $$F(u) = \Pr\left(X \le \frac{1}{2} - \sqrt{\frac{1}{4}-u}\right) + \Pr\left(X \ge \frac{1}{2} + \sqrt{\frac{1}{4}-u}\right).$$ But $\Pr(X \le x)=x$ and $\Pr(X \ge x) = 1-x$. So $$F(u) = 1-\sqrt{1-4u}.$$