# Upper bound of difference of squares of quantile standard normal

Let $\Phi$ denotes the cummulative standard normal distribution and $\Phi^{-1}$ denotes its inverse. Given $u,v\in[0,1)$. I'am going to find an upper bound of $$\left|\left\{\Phi^{-1}(v)\right\}^2-\left\{\Phi^{-1}(u)\right\}^2\right|$$ in term of "easy" functions like polynomial, trigonometric, logaritmic, etc.

So for, I have obtained as follows:

One can prove that $\phi(\Phi^{-1}(x))\ge \sqrt{\frac2\pi}(x\wedge(1-x)),\forall x\in[0,1),$ where $\phi=\Phi^\prime$. Hence, \begin{align} \left|\Phi^{-1}(v)-\Phi^{-1}(u)\right|&=\left|\int_u^v\frac1{\phi(\Phi^{-1}(w))}dw\right|\\ &\le \sqrt{\frac2\pi}\left|\int_u^v\frac1{w\wedge(1-w)}dw\right|\\ &\le \sqrt{\frac2\pi}\left|\int_u^v\left(\frac1{w}+\frac1{(1-w)}\right)dw\right|\\ &= \sqrt{\frac2\pi}\left|\log\left(\frac{v}{1-v}\right)-\log\left(\frac{u}{1-u}\right)\right|.\\ \end{align}

But then, I haven't been able to handle $$\left|\Phi^{-1}(v)+\Phi^{-1}(u)\right|.$$ Could anyone help me? Or is there a better approach than mine?

Well one answer is right in front of you! By the inequality you are claiming, you have, in particular, for $u = 1/2$, that $$\left| \Phi^{-1}(v) \right| \leq \sqrt{\frac{2}{\pi}} \left| \log{\frac{v}{1-v}}\right|.$$ Then apply the triangle inequality to get $$\left| \Phi^{-1}(v) + \Phi^{-1}(u) \right| \leq \sqrt{\frac{2}{\pi}} \left(\left| \log{\frac{v}{1-v}}\right| + \left| \log{\frac{u}{1-u}}\right| \right).$$
• Ah... That is what I wanted @John M. I forgot to consider $u=1/2$. Thanks for your help. I'll give the bounty for you. – Jlamprong May 30 '14 at 19:48