Let $Q^{-1}$ be the inverse function of a standard normal CDF. For $0 < \epsilon < p,p' < 1 - \epsilon$, how much does the function $Q^{-1}$ change as a function of $|p - p'|$? Any useful upper bounds would be helpful.
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2$\begingroup$ Of course, if one specified $0<\epsilon \leq p,p'\leq 1-\epsilon$ one might get somewhere. $\endgroup$– kodluCommented Jan 29, 2015 at 6:13
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$\begingroup$ Yes, thanks. Let us assume that is the case. $\endgroup$– VedarunCommented Jan 29, 2015 at 12:28
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2 Answers
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If $F$ is the standard normal CDF, $$(Q^{-1})'(p) = \dfrac{1}{F'(t)} = \sqrt{2\pi} \exp(t^2/2)$$ where $p = F(t)$. The maximum for $\epsilon \le p \le 1-\epsilon$ is at the endpoints. So $$|Q^{-1}(p) - Q^{-1}(p')| \le \sqrt{2\pi} \exp(t^2/2) |p - p'|$$ where $t = Q^{-1}(1-\epsilon)$. Asymptotically as $t \to +\infty$, $$F(t) \sim 1 - \dfrac{\exp(-t^2/2)}{\sqrt{2\pi} t}$$ so as $\epsilon \to 0+$ $$t \sim \sqrt{W(1/(2\pi \epsilon^2))}$$ where $W$ is the Lambert W function, and then $$\sqrt{2\pi} \exp(t^2/2) \sim \dfrac{1}{\epsilon \sqrt{W(1/(2\pi \epsilon^2))}}$$
answered Jan 29, 2015 at 19:32
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$\begingroup$ Thanks that's exactly what I had in mind in my comment to the OP. $\endgroup$– kodluCommented Jan 30, 2015 at 1:33
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Let $Q^{-1}$ be the inverse function of a standard normal CDF. For $0 < p,p' < 1$, how much does the function $Q^{-1}$ change as a function of $|p - p'|$? Any useful upper bounds would be helpful.
The only upper bound is $\infty$, which is also true for any unbounded random variable in place of the normal one.
answered Jan 29, 2015 at 6:03
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$\begingroup$ I mean a bound that depends on $|p - p'|$ $\endgroup$– VedarunCommented Jan 29, 2015 at 12:28
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$\begingroup$ @Vedarun Yes, that's what I mean too. $\endgroup$ Commented Jan 29, 2015 at 16:55