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.

  • $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ My apologies. I'm a newbie here and didn't know the protocol well. :) $\endgroup$ – Double E Jul 23 '14 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.