I want to know how to prove the following inequality that seems to be true numerically.

Let $n(x)$ be the density of the standard normal, and $N(x)$ be the cdf of standard normal. Then, for $x\geq 0$,

$\left(\frac{n(x)}{1-N(x)}-x \right)\left(\frac{2n(x)}{1-N(x)}-x\right)\geq 1$

Thanks, and sorry that I don't know how to write the math symbols in this environment.

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    $\begingroup$ Typeset in LaTeX please, it's unreadable $\endgroup$ – james42 Jul 8 '15 at 22:26
  • $\begingroup$ Thanks Ale42, I just edited the inequality. $\endgroup$ – Mehmet Jul 8 '15 at 23:39
  • $\begingroup$ Excuse me but why am I voted down? I am a new user and please advise. $\endgroup$ – Mehmet Jul 9 '15 at 0:05
  • $\begingroup$ Well I don't know... Maybe somebody doesn't like your question. Anyway I'll give you one upvote :) $\endgroup$ – james42 Jul 9 '15 at 0:09

I get that $\left(\frac{n(x)}{M(x)}-x \right)\left(\frac{2n(x)}{M(x)}-x\right) - 1 \approx \frac{3}{x^2} $ for large $x$.

Since $N(x) =\int_{-\infty}^x n(t)dt $, $1-N(x) =\int_x^{\infty} n(t)dt $. Call this $M(x)$.

The inequality becomes $\left(\frac{n(x)}{M(x)}-x \right)\left(\frac{2n(x)}{M(x)}-x\right)\geq 1 $.

Multiplying by $M^2(x)$, I get $(n(x) - xM(x))(2n(x) - xM(x)) \ge M^2(x) $ or $2n^2(x) -3xn(x)M(x)+x^2M^2(x) \ge M^2(x) $ or $2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x) \ge 0 $.

The purported inequality above thus becomes $P(x) =2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x) \ge 0 $.

Since $M(0) = 1/2$ and $n(0) =1/\sqrt{2\pi} \approx 0.4 $, $P(0) \approx 0.32-1/4 = 0.07 > 0 $.

Let's see what happens for large $x$.

Asymptotically, $M(x) \approx n(x)/x $.

Therefore, for large $x$,

$\begin{array}\\ P(x) &= 2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x)\\ &\approx 2n^2(x) -3xn(x)(n(x)/x)+(x^2-1)(n(x)/x)^2\\ &= 2n^2(x) -3n^2(x)+(1-1/x^2)n^2(x)\\ &= -n^2(x)/x^2\\ \end{array} $

Therefore, modulo errors on my part, $P(x) \approx -n^2(x)/x^2 $, so it is negative and small.

Note: A comment said that this was not correct, so I'll take an additional term in the expansion and see what happens.

Asymptotically, $M(x) \approx \frac{n(x)}{x}(1-\frac1{x^2}) $.

Therefore, for large $x$,

$\begin{array}\\ P(x) &= 2n^2(x) -3xn(x)M(x)+(x^2-1)M^2(x)\\ &\approx 2n^2(x) -3xn(x)(n(x)/x)(1-1/x^2)+(x^2-1)(n(x)/x)^2(1-1/x^2)^2\\ &= 2n^2(x) -3n^2(x)(1-1/x^2)+n^2(x)(1-1/x^2)^3\\ &= 2n^2(x) -3n^2(x)+3n^2(x)/x^2+n^2(x)(1-3/x^2+3/x^4-1/x^6)\\ &= -n^2(x)+3n^2(x)/x^2+n^2(x)-n^2(x)(3/x^2-3/x^4+1/x^6)\\ &= n^2(x)(3/x^4-1/x^6)\\ &\approx 3n^2(x)/x^4\\ \end{array} $

Therefore, modulo errors on my part, $P(x) \approx 3n^2(x)/x^4 $, so it is positive and small.

Looks like the comment was correct.

Since I multiplied by $M^2(x)$ to get this inequality, the difference has to be divided by $M^2(x)$, so it is $\frac{3n^2(x)/x^4}{M^2(x)} \approx \frac{3n^2(x)/x^4}{(n(x)/x)^2} =\frac{3}{x^2} $.

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    $\begingroup$ Thanks Marty but I think your answer is not correct. Your approximation has small errors that are at the same order as $\frac{n(x)^2}{x^2}$. $\endgroup$ – Mehmet Jul 9 '15 at 2:52
  • $\begingroup$ Not surprised . I might try more terms later. $\endgroup$ – marty cohen Jul 9 '15 at 5:43
  • $\begingroup$ You were right. $\endgroup$ – marty cohen Jul 9 '15 at 19:03

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