For a given random variable $X$, we define $X^+=\max\{0,X\}$ and $X^-=\max\{0,-X\}$.
My question is how to express the probabilities of the two R.V $X^+$ and $X^-$.
If we write: \begin{equation} F_{X^+}(x)=\mathbb{P}(X\leq x)=\mathbb{P}(X \leq x; X\geq 0) =\mathbb{P}(X \leq x).\mathbb{P}(X\geq 0) \end{equation} And deduce: \begin{equation} F_{X^+}(x)= F_X(x).(1-F_X(0)) \end{equation} and the same thing goes with $X^-$: \begin{equation} F_{X^-}(x)= F_X(0).(1-F_X(x)) \end{equation} is it right? Particularly, in this case neither $F_{X^+}$ nor $F_{X^-}$ equals to 1 when $x \rightarrow +\infty$. Please, if you have any suggestions, I am all ears.
The events $\{X \le x\}$ and $\{X \ge 0\}$ are in general not independent, so you can't factor the CDF in this way.
Instead, remember that $\max(a, b) \le c$ is equivalent to $a \le c$ and $b \le c$. This yields: $$P(X^+ \le x) = P(\max(0, X) \le x) = P(X \le x, 0 \le x) = \begin{cases} 0 & x < 0 \\ P(X \le x) & x \ge 0 \end{cases}.$$
Similar calculations using $\min(a, b) > c \iff (a > c) \wedge (b > c)$ yield the CDF of $X^-$.