Let $X$ is uniformly distributed over the interval $[0,2]$.I need to find CDF of random variable $Y=X(2-X)$.
My solution:
$$\begin{align}F_Y(t)&=P(Y \le t)
\\[1ex]&=P(X(2-X)\le t)
\\[1ex]& = P(1-(X-1)^2 \le t)
\\[1ex]&= 1- P((X-1)^2 \lt 1-t)
\\[1ex]&=1 -P(1-\sqrt{1-t} \lt X \lt 1+ \sqrt{1-t})
\\[1ex]&=1-F_X(1+\sqrt{1-t}) + F_X(1-\sqrt{1-t})\end{align}$$
But, I'm not sure how to interpret it now.
Best Answer
The plot for $Y=X(2-X)$ over $X\in[0;2]$ is a curve that is symmetric about $X=1$. (Specifically, it is a parabola). $Y$ then ranges over $[0;1]$, and each value of $Y$ corresponds to two values of $X$ (the map contains a fold).
So the event $Y\leq t$ equates, as you've shown, to the event $X\leq 1-\surd(1-t)$ or $X\geq 1+\surd(1-t)$ for some $t$ in $[0;1]$.
$$F_Y(t)~=~F_X(1-\sqrt{1-t})~+~1-F_X(1+\sqrt{1-t})$$
Oh, and we know that $F_X(x)=\tfrac x2\mathbf 1_{x\in[0;2]}$ so the above becomes:
$$\begin{align}F_Y(t)~&=~1+\tfrac 12((1-\sqrt{1-t})-(1+\sqrt{1-t}))\mathbf 1_{t\in[0;1]}\\&=~1+\sqrt{1-t~}\mathbf 1_{t\in[0;1]}\end{align}$$