1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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} $.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.