Assume $X$ and $Y$ are random variables, F and G are their cdf's. If my work is correct,
$$G(y)=\frac{1}{2}[1 + F(y) - F(-y)]$$
is the CDF of
$$ Y = \left\{ \begin{array}{} X, & \text{if}~ \lfloor X \rfloor ~\text{is even} \\ -X, & \text{if}~ \lfloor X \rfloor ~\text{is odd} \end{array} \right.$$
In relation to another question of mine, I want to estimate $$ \begin{array}{} C(u,v) &= \mathbb{P}(X \leq F^{-1}(u), Y \leq G^{-1}(v))\\ &= \frac{1}{2}\left[\mathbb{P}(X \leq F^{-1}(u), X > -G^{-1}(v)) + \mathbb{P}(X \leq F^{-1}(u), X \leq G^{-1}(v)) \right]\\ &=\frac{1}{2}\left[\mathbb{P}(X \leq \min\{F^{-1}(u), G^{-1}(v)\}) + \mathbb{P}(X \in [-G^{-1}(u), F^{-1}(v)]) \right] = \\ &= \frac{1}{2}\left[ v - F(G^{-1}(v)) + F(\min\{F^{-1}(u), G^{-1}(v)\}) \right] \end{array} \tag{1}$$
It seems I might be able to untangle this as a function of $(u,v)$ if I find $F(G^{-1}(v))$. However, here I'm not too sure that the usual approach of finding the inverse works.
$$ y = G(G^{-1}(y)) =\frac{1}{2}[1 + F(G^{-1}(y)) - F(-G^{-1}(y))] $$ $$ 2y -1 = F(G^{-1}(y)) - F(-G^{-1}(y)) $$ $$ F^{-1}(2y -1) = F^{-1}\left( F(G^{-1}(y)) -F(-G^{-1}(y)) \right) \tag{2}$$
But CDF's aren't necessarily subadditive, are they?
My question: the goal is to simplify $(1)$ to some convenient form, my idea is that finding the expression for $G^{-1}(y)$ would help, but I got stuck at $(2)$.
Is there another way to solve this? Or am I just missing something? Can we further work with $(2)$ to extract the $G^{-1}(y)$ expression?