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In the middle of some proof, I have faced an expression $\Phi^{-1}(1-x) =O(\sqrt{\log{x^{-1}}})$, where $\Phi(\cdot)^{-1}$ is a quantile function of the standard normal distribution and $x \in (0,1)$.

Can someone help me how to prove this or give me a quick reference?

Thanks in advance.

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  • $\begingroup$ Can you include the definition of the function $\Phi(x)$? $\endgroup$ – Semiclassical Jul 22 '14 at 19:09
  • $\begingroup$ In the Landau notation $O\left(\sqrt{\log\frac{1}{x}}\right)$ are we assuming that $x\to 0$ or $x\to 1$ ? $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 19:24
  • $\begingroup$ Re: Semiclassical... Sorry about that. $\Phi(\cdot)$ is the cdf of the standard normal distribution. $\endgroup$ – Double E Jul 22 '14 at 19:24
  • $\begingroup$ Re: Jack D'aurizio... Yes, we are interested in the tail rate of those quantile sequences. We are interested in the case of $x\rightarrow 0$. $\endgroup$ – Double E Jul 22 '14 at 19:28
  • $\begingroup$ There exist very tight bounds for such tails due to the fact that the erf integral admits a fast converging representation in terms of a continued fraction, you can find it by googling "Mills ratio" and "continued fraction", for instance. For our purposes, it is enough to isolate the main term of the asymptotics. $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 19:52
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Given that $X\sim N(0,1)$, we have: $$\int_{w}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2\pi}}e^{-\frac{w^2}{2}}\int_{0}^{+\infty}\exp\left(-xw-\frac{x^2}{2}\right)dx\leq\frac{1}{w\sqrt{2\pi}}\exp\left(-\frac{w^2}{2}\right)$$ and, by assuming $w\gg 1$: $$\int_{w}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\geq \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{w^2}{2}\right)\int_{0}^{\sqrt{2}}e^{-xw}\left(1-\frac{x^2}{2}\right)dx\geq\frac{w^2-1}{w^3\sqrt{2\pi}}\exp\left(-\frac{w^2}{2}\right),$$ hence: $$\Phi^{-1}(1-x)=\Theta\left(\sqrt{-\log x}\right)$$ as long as $x\to 0$. $$\Phi^{-1}(1-x)\approx\sqrt{W\left(\frac{1}{2\pi x^2}\right)}$$ is an even more precise approximation in terms of the Lambert W-function.

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    $\begingroup$ Got it! Thanks so much. $\endgroup$ – Double E Jul 22 '14 at 22:59
  • $\begingroup$ @user165795: I politely ask you to accept my answer, if you think it is satisfactory. $\endgroup$ – Jack D'Aurizio Jul 22 '14 at 23:01
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    $\begingroup$ My apologies. I'm a newbie here and didn't know the protocol well. :) $\endgroup$ – Double E Jul 23 '14 at 1:13

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