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I want to establish the following inequality for $x>0$:

$$\phi(x) \left( \frac{1}{x} - \frac{1}{x^3}\right)\leq 1- \Phi(x) \leq \phi(x) \frac{1}{x}$$ with $\phi(x)=\frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2} x^2} $ and $\Phi(x)= \int_{-\infty}^x \frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2} t^2} dt $

Can somebody give me a hint to show this inequality?

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  • $\begingroup$ Thanks for editing:) Sorry for my little mistakes. $\endgroup$ Commented Apr 28, 2019 at 11:18
  • $\begingroup$ Are you sure you typed the inequality correctly? I think $\phi(x) ( \frac{1}{x} - \frac{1}{x^3}) \leq \phi(x) \frac{1}{x} \leq 1- \Phi(x)$ for $1 ≤ x$ and $x$ real is more likely according to my Desmos graph. $\endgroup$ Commented Apr 28, 2019 at 11:23
  • $\begingroup$ I forgot to mention, that $x>0$ The rest is right. $\endgroup$ Commented Apr 28, 2019 at 11:28
  • $\begingroup$ For a hint I should use partial integration, but how? $\endgroup$ Commented Apr 28, 2019 at 11:28
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    $\begingroup$ I have replaced in your definition of $\Phi$, $\int_{0}^x$ by $\int_{-\infty}^x$ (the reason for which @Toby Mak wasn't understanding). $\endgroup$ Commented Apr 29, 2019 at 17:20

1 Answer 1

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This kind of (rather classical) inequalities are usually established using integration by parts. I would like to present here a different proof based on a certain continued fraction.

Let

$$\varphi(x):=e^{x^2/2}\int_x^{+\infty}e^{-t^2/2}dt\tag{1}$$

It is easy to prove that

$$\varphi(x)=\dfrac{1-\Phi(x)}{\phi(x)}\tag{2}$$

Thus, we have to establish that :

$$\text{for all} \ x>0 : \ \ \left( \frac{1}{x} - \frac{1}{x^3}\right) \leq \varphi(x) \leq \dfrac{1}{x}\tag{3}$$

(3) is going to be a rather easy consequence of the following "beautiful" continued fraction decomposition, valid for any $x>0$ : $$\varphi(x)=\cfrac{1}{x +\cfrac{1}{x +\cfrac{2}{x +\cfrac{3}{\ddots x + \cfrac{n}{x+\cdots}}}}}\tag{4}$$

Reference for (4) : this paper about inequalities dealing with "erf" function.

Indeed, due to the positivity of $x$, this decomposition allows to "bracket" $\varphi(x)$ between its two first "convergents" :

$$\text{for all} \ x>0 : \ \ \dfrac{1}{x+\frac{1}{x}} \leq \varphi(x) \leq \dfrac{1}{x}\tag{5}$$

Besides :

$$\text{for any} \ x>0, \ \ \dfrac{1}{x}-\dfrac{1}{x^3}<\dfrac{1}{x+\frac{1}{x}}\tag{6}$$

(due to the fact that $x^2-1<x^4$ for any $x$).

Combining (5) and (6) we get the awaited result (3).

Remarks :

1) One has remarked that (5) is a stronger bracketing than (3).

2) Oddly, https://en.wikipedia.org/wiki/Error_function#Continued_fraction_expansion doesn't give a continued fraction similar to (4).

3) The author of the cited publication, Omran Kouba, has an interesting personal site : https://www.sites.google.com/site/koubamath/home/reaserch-papers

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  • $\begingroup$ Thank you for your proof:) When I do that with partial integration, I add the factor $1 = \frac{t}{t}$. But how do I get to the left side of my inequality with the same trick? $\endgroup$ Commented Apr 28, 2019 at 13:24
  • $\begingroup$ It is hard for me to understand what you call "adding factor $1=\tfrac{t}{t}$", thus to answer your question... $\endgroup$ Commented Apr 28, 2019 at 13:26
  • $\begingroup$ $$ \int_x^{\infty} \frac{1}{t} t \phi(t) dt$$. Then I would go for partial integration. Thus, I get the right side of the inequality. $\endgroup$ Commented Apr 28, 2019 at 13:30

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