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The following is a well-known chain of inequalities for the tail of the normal distribution when $a = 1:$ $$ \Big(\frac{1}{x} - \frac{a}{x^3}\Big) \phi(x) \leq \Big(\frac{x}{a + x^2}\Big) \phi(x) \leq \Phi(-x) \leq \frac{1}{x}\phi(x), \qquad x > 0. $$ where $\phi(x) = \frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$ and $\Phi(-x) = \int_x^\infty \phi(t) \,\mathrm{d}t$ are the normal pdf and cdf respectively. Moreover, observe that setting $a = 0$ recovers the upper bound for the normal cdf on the right-hand side of the above chain. My question, therefore, is whether a tighter upper bound for the normal tail exists for some $a \in (0,1)$ when we restrict to $x \geq 1$. Specifically, this yields the following two questions

Prove or disprove: There exists $a \in (0, 1)$ such that for all $x \geq 1$ $$ \Phi(-x) \leq \Big(\frac{1}{x} - \frac{a}{x^3}\Big) \phi(x) $$

Prove or disprove: There exists $a \in (0,1)$ such that for all $x \geq 1$, $$ \Phi(-x) \leq \Big(\frac{x}{a + x^2}\Big) \phi(x) $$

So far, I've exhausted most of my "painless" tricks, and thought I'd post a fun problem that looks to me like it might have a positive answer.

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  • $\begingroup$ It might help to look at the proofs of the inequalities in the case $a=1$ and inspect the looseness of each inequality that appears in the proof. (Although I have not done this myself.) $\endgroup$ – angryavian Aug 17 '18 at 3:56
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Let $$ G(x) = \left(\frac {1} {x} - \frac {a} {x^3}\right)\phi(x) - \Phi(-x)$$ for $a \in (0, 1)$. Then $$ \begin{align} G'(x) &= \left(-\frac {1} {x^2} + \frac {3a} {x^4}\right)\phi(x) + \left(\frac {1} {x} - \frac {a} {x^3}\right)(-x)\phi(x) + \phi(-x) \\ &= \left(-\frac {1} {x^2} + \frac {3a} {x^4} - 1 + \frac {a} {x^2} + 1\right)\phi(x) \\ &= \frac {\phi(x)} {x^4}\left[-(1-a)x^2 + 3a\right] \\ &\begin{cases} <0 & \text{when} & \displaystyle x < -\sqrt{\frac {3a} {1-a}} \\ =0 & \text{when} & \displaystyle x = -\sqrt{\frac {3a} {1-a}} \\ >0 & \text{when} & \displaystyle -\sqrt{\frac {3a} {1-a}} < x < \sqrt{\frac {3a} {1-a}}\\ =0 & \text{when} & \displaystyle x = \sqrt{\frac {3a} {1-a}} \\ <0 & \text{when} & \displaystyle x > \sqrt{\frac {3a} {1-a}} \\ \end{cases} \end{align}$$

Since we are interested in $x > 1$ only, we can safely ignore the first two negative cases. Note that

$$ \sqrt{\frac {3a} {1-a}} > 1 \iff a > \frac {1} {4}$$

So we can conclude that when $\displaystyle 0 < a \leq \frac {1} {4}$, $G$ is strictly decreasing on $[1, +\infty)$. And it is trivial to check that $\displaystyle \lim_{x \to +\infty} G(x)= 0 $ which implies $G(x) \geq 0$ for all $x \in (1, +\infty)$. And this proves the first claim - it holds when $a$ is small enough.

We can also check the case when $\displaystyle \frac {1} {4} < a < 1$, $G$ first strictly increasing on $\displaystyle \left[1, \sqrt{\frac {3a} {1-a}}\right)$, attains the maximum at $\displaystyle x = \sqrt{\frac {3a} {1-a}}$, and then strictly decreasing on $\displaystyle \left(\sqrt{\frac {3a} {1-a}}, +\infty\right)$. So we also need to check the value of $G(1)$ as the boundary point. Note that

$$ G(1) = (1 - a)\phi(1) - \Phi(-1) \geq 0 \iff a \leq 1 - \frac {\Phi(-1)} {\phi(1)} \approx 0.3443205$$

So the inequality also holds when $\displaystyle \frac {1} {4} < a \leq 1 - \frac {\Phi(-1)} {\phi(1)} $

The another claim can also be checked similarly I think.

This is also worth a read: https://mathoverflow.net/questions/19404/approximation-of-a-normal-distribution-function

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  • $\begingroup$ Phenomenal! I bungled a sign when I tried this myself the first time. I'm happy to see it works out. $\endgroup$ – bashfuloctopus Aug 18 '18 at 0:18

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