I came across quite interesting question related to the computation of very particular CDF:
Question
Let $\Omega:=[0,3] \subset \mathbb{R}$, $\textbf{P}$ - normalized Lebesgue measure on $\Omega$. Determine CDF of the following random variable defined as:
$X(\omega):= \begin{cases} 2\omega+1 &: \omega \in [0,1] \\ -\omega^{2}+2&: \omega \in (1,2) \\ 3 & :\omega \in [2,3] \end{cases} $
attempts
1.)I know that by definition $$\forall t \in \mathbb{R} :F_{X}(t) = P(X < t) = \lambda(\{\omega \in \Omega | X(\omega)\in (-\infty,t)\})$$, where $\lambda$ - Lebesgue measure
2.) I know that Lebesgue measure in this case will have the form $$\textbf{P}(d\omega)=\frac{1}{3}\times I_{[0,3]}(\omega) d\omega$$
, where $I_{[0,3]}(\omega):= \begin{cases} 1 &; \omega \in [0,3] \\ 0 &: \omega \in \mathbb{R} \setminus [0,3] \end{cases}$
While computation of standard problems related to CDFs are straightforward, this one appears to be quite tricky for me.
I would be thankful for help!
First $X$ rises from $1$ at $\omega=0$ to $1$ at $\omega=3$, then discontinuously falls to $1^-$ for $\omega=1^+$ and continues falling to $(-2)^+$ for $\omega=2^-$, and finally discontinuously rises to $3$. Thus the range of $X$ has $\left(-2,\,3 \right]$. The cdf $F_X$ has trivial behaviour at its ends, viz. $$x\le -2\implies F_X(x)=0,\,x\ge 3\implies F_X(x)=1.$$For $x\in\left(-2,\,1\right)$, $X=x$ iff $\omega=\sqrt{2-x}\in\left(1,\,2\right)$ and $$F_X(x)=P(X\le x)=\frac{2-\sqrt{2-x}}{3}.$$Finally, $$x\in\left[1,\,3\right]\implies F_X(x)=\frac{x}{3}.$$Thus $F$ gradually rises from $0$ at $x=-2$ to $\frac{1}{3}$ at $x=1$ and $1$ at $x=3$.