Convert a uniform distribution to two uniform distributions

Question

I have a distribution function: $$ F(x) = \begin{cases} 0 & x \leqslant 0, \\ \frac{x}{2} & 0 < x \leqslant 1 \\ \frac{2}{3} & 1 < x \leqslant 2 \\ \frac{2}{3} + \frac{x-2}{3} & 2 < x \leqslant 3, & \\ 1 & x > 3 \end{cases} $$ I want to find such $ f $ so that $ \xi \in U _ {[0; 1]} $, $ U $ is uniform, so that $ f (\xi) $ is distributed like $ F $, I know that in the strictly monotone case, take $ F ^ {- 1} $ as $ f $, but here I don't know what to do, although I see that $ F $ is a union of two uniform ones, I don't know what to do.

Answer 1

If I understood well, the problem is correct but it is bad explained. Moreover, usually the CDF is right continuous, yours is left continuous....it is not forbidden but rare to see...

The question, IMHO, is the following:

"Is it possible, starting from a uniform distribution $Y\sim U[0;1]$ , with a transformation derive $F_X(x)$?"

The answer is Yes, using $F_X^{-1}(y)$ that is the following transformation function

$$F_X^{-1}(y) = \begin{cases} 2y, & \text{if $0\leq y<\frac{1}{2}$} \\ 2, & \text{if $\frac{1}{2}\leq y<\frac{2}{3}$} \\ 3y, & \text{if $\frac{2}{3}\leq y<1$} \\ 3, & \text{if $y \geq 1$} \end{cases}$$