Inverse of CDF for truncated normal distributions

Let $a_1 < b_1 < a_2 < b_2$ and assume $z \in (a_2,b_2)$ fixed. Let $F = \Phi^{[a_2,b_2]}_{(\cdot), \sigma^2}(z)$ and $G = \Phi^{\cup_i[a_i,b_i]}_{(\cdot), \sigma^2}(z)$, where $\Phi^S_{\mu, \sigma^2}$ denotes the CDF of a $N(\mu, \sigma^2)$ distribution truncated to the set $S$ and $\sigma^2 > 0$ is known and fixed. That is, \begin{align} F(\mu) &= \Phi^{[a_2, b_2]}_{\mu, \sigma^2}(z) = \frac{\Phi_{\mu, \sigma^2}(z) - \Phi_{\mu, \sigma^2}(a_2)}{\Phi_{\mu, \sigma^2}(b_2) - \Phi_{\mu, \sigma^2}(a_2)}, \\ G(\mu) &= \Phi^{\cup_i [a_i, b_i]}_{\mu, \sigma^2}(z) = \frac{\Phi_{\mu, \sigma^2}(z) - \Phi_{\mu, \sigma^2}(a_2) + \Phi_{\mu, \sigma^2}(b_1) - \Phi_{\mu, \sigma^2}(a_1)}{\Sigma_i [\Phi_{\mu, \sigma^2}(b_i) - \Phi_{\mu, \sigma^2}(a_i)]}. \end{align}

Note that $F$ and $G$ are functions of the mean parameter and hence not CDFs themselves. It can easily be seen that both $F$ and $G$ are differentiable, strictly decreasing and that $F<G$. Let $f = F^{\prime}$ and $g = G^{\prime}$. I now wish to prove that $\vert (F^{-1})^{\prime}\vert > \vert (G^{-1})^{\prime}\vert$ or similarly $f(F^{-1}) > g(G^{-1})$.

Can anyone help?

For intuition I have included plots of $F$ & $G$ (top) and $(F^{-1})^{\prime}$ & $(G^{-1})^{\prime}$ (bottom) for the simple example of $[a_1, b_1] = [-1, 0]$, $[a_2, b_2] = [1, 2]$, $z = 1.3$ and $\sigma = 0.5$.

• How can cdf $F$ be strictly decreasing? – wolfies Jan 17 '17 at 16:59
• $F$ and $G$ are functions of the mean parameter. $F(\mu) = \Phi_{\mu, \sigma^2}^{[a_2, b_2]}(z)$ $\forall \mu \in \mathbb{R}$ for a fixed $z \in (a_2, b_2)$. I will edit the question to make this more clear. – Rune Christiansen Jan 18 '17 at 8:26