I am reading one paper https://arxiv.org/abs/1207.7209 In proposition 4.1 the author mentioned a fact $$p \sqrt{k_1 \log (1/p)} \leq \phi \circ \Phi^{-1}(p)$$ where $k_1 = 1/2, p \in (0, 1/2], \phi, \Phi^{-1}$ are pdf and inverse cdf for standard normal distribution respectively.

I'm not sure how this fact comes. Can anyone prove this inequality? Thanks!

Tips: the above inequality may follow from $\phi(x) - x\bar{\Phi}(x) \geq 0$ for $x > 0$, where $\bar{\Phi}(x) = 1 - \Phi(x)$


1 Answer 1


That is not what is claimed in the paper.

The paper says

For $\kappa_1=1/2$, $p\in(0,1/2]$, the fact that $p\sqrt{\kappa_1\log(1/p)}\leq\phi\circ\Phi^{\leftarrow}(p)$ follows from $\phi(x)-x\overline{\Phi}(x)\geq 0$ for $x>0$.

where the notation $\Phi^{\leftarrow}$ is the (generalized) inverse to $\Phi$ (immediately before Definition 2.4) and $\overline{\Phi}$ means $1-\Phi$ (Definition 2.6) has been introduced previously.

  • $\begingroup$ yes, thanks! fixed the mistakes. Any ideas about how to prove? $\endgroup$
    – Alyssa
    Jun 16, 2019 at 6:53

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