The problem comes from reading this [0] paper but I think I can express it in a self contained question.
Consider the implicit function $H(z)$ defined by the relation:
$$F_z(z+H(z))-F_z(z-H(z))=0.5$$
Where $F_z$ is the CDF of an absolutly continuous, unimodal distribution.
The authors point out that when $F'_z=f_z=\max(1-|z|,0)$ (the triangle distribution with parameters $a=-1,c=0$ and $b=1$),
$$H(z)= \begin{cases} 1-\sqrt{1/2-z^2} & \text{if } |z| \leq 1/2 \\ |z| & \text{if } |z| > 1/2 \end{cases}$$
(p387, bottom of the article I linked to)
My questions are:
- How was the explicit formulation for $H(z)$ derived?
- Can it also be done for other values of $c$, i.e. $c=1$
[0] Ola Hössjer, Peter J. Rousseeuw and Christophe Croux. Asymptotics of an estimator of a robust spread functional. Statistica Sinica 6(1996), 375-388.
Here is a piecewise formula for $f_z$. $$ f_z(z) = \begin{cases} z+1 & -1\leq z\leq 0 \\ 1-z & 0\leq z\leq 1 \\ 0 & \text{else}. \end{cases} $$ Computing a formula for $F_z$ is just a matter of integrating. $$ F_z(z) = \int_{-1}^z f_z(z) \, dz = \begin{cases} 0 & z\leq -1 \\ \frac{z^2}{2}+z+\frac{1}{2} & -1<z\leq 0 \\ -\frac{z^2}{2}+z+\frac{1}{2} & 0<z\leq 1 \\ 1 & z>1. \end{cases} $$ Note that the constants are chosen to ensure continuity. With this, your implicit formula $$F_z(z+H(z))-F_z(z-H(z))=1/2$$ can be analyzed on certain intervals. Suppose, for example, that $0\leq z+H(z) \leq 1$ and that $-1\leq z-H(z) \leq 0$, which imples that $-1/2\leq z \leq 1/2$. Then the implicit formula becomes $$\left(-\frac{1}{2}(z+H(z))^2 + (z+H(z)) + 1/2\right) - \left(\frac{1}{2}(z-H(z))^2 + (z-H(z)) + 1/2)\right) = \frac{1}{2}.$$ Solving this formula for $H(Z)$ yields the formula for $H(z)$ yields the portion of your piecewise expression valid for $-1/2\leq z \leq 1/2$. The other terms follow in similar fashion.
I'm certain that a similar thing can be done with different choices of your constant $c$.