# Inequality for standard normal distribution with composite function of pdf and inverse cdf

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

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.

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