Inverse CDF (Quantile) of Piecewise Function [duplicate]

Question

This question may be insanely simple, but I'm unsure.

Let's say we have the following function: $$ f(x) = \begin{cases} x & 0 \leq x < 1 \\ x-1 & 1 \leq x < 2 \\ 0 & \text{otherwise} \end{cases}$$

The CDF is obviously $$ F(x) = \begin{cases} \frac{1}{2}x^2 & 0 \leq x < 1 \\ \frac{1}{2}x^2-x+1 & 1 \leq x < 2 \\ 0 & \text{otherwise} \end{cases} $$

How would I calculate the Inverse CDF from this?

(Note: The +1 is required to make it add to 1 and not suddenly become negative. I don't know if that matters for the Inverse CDF.)

Answer 1

In practice, the first step would be to generate a random deviate for the first branch of the specified CDF with half of the probability. So, let ½ U (where U is a Uniform random deviate on (0,1] ) = F(x) = ½ ${x^2}$, where x lies between 0 and 1, producing the obvious answer (per the Monte Carlo inversion approach for deriving random deviates) that ${X = SQRT (U)}$.

Step 2, for the 2nd branch similarly generate a random deviate from ½ U’ (where U is a Uniform random deviate on (1,2] ) = F(x’) = ½${(x’-1)^2}$ + ½ where x’ lies between 1 and 2 (so, ${x = x’ - 1}$). The inversion random deviate is correspondingly derived as follows:

${½ U’ = ½(x’-1)^2 + ½ }$

Or: ${U’ = (x’-1)^2 + 1}$

As: ${U = U’ -1}$ and ${x = x’ - 1}$

Or: ${U = x^2}$

And again:

${X = SQRT (U)}$

So, per this approach, there is no inconsistency in the generation process for random deviates of the inverse of the complete two-branch CDF as specified in the question.