# A Normal Distribution Inequality

Let $n(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$, and $N(x) = \int_{-\infty}^x n(t)dt$. I have plotted the curves of the both sides of the following inequality. The graph shows that the following inequality may be true. $$f(x)\equiv (x^2+1)N + xn-(xN+n)^2 > N^2$$ where the dependency of $n$ and $N$ on $x$ are absorbed into the function symbols. However, I have not succeeded in providing a full proof except for $x$ above some positive number, with the help of various Mill's Ratio $\frac{m}{n}$ bounds.

I am asking for help in proving the above inequality or providing an $x$ that violates the above inequality. Judging from the aforementioned plot I am pretty confident the validity of the inequality, though.

The left hand side is actually the variance of a truncated normal distribution. I am trying to give it a lower bound. More explicitly, $$f(x)\equiv\int_0^\infty t^2n(t+x)dt-\Big(\int_0^\infty t\,n(t+x)dt\Big)^2>\Big(\int_0^\infty n(t-x)dt\Big)^2.$$

The form of the inequality is probably more transparent if we set $m=1-N$ and the inequality is equivalent to $$g(x)\equiv m[(x^2+1)(1-m)+2xn]-n(x+n) > 0.$$

Incidentally, I have proved that $N$ is the upper bound of the left hand side of the first inequality, i.e. $$(x^2+1)N + xn-(xN+n)^2 < N$$ or $$h(x)\equiv x^2 m(1-m)-n[x(1-2m)+n]<0$$ as follows.

$h$ is an even function and $h(0)<0$, so we only need to consider $x>0$. From the integration by part of $m(x)$ and dropping a negative term, we have $$xm<n, \forall x>0.$$ The first term of $h(x)$ is then bounded and \begin{eqnarray} h(x)&<&x(1-m)n-n[x(1-2m)+n] \\ &=& n(xm-n) \\ &<& 0, \end{eqnarray} where last inequality is obtained by using $xm<n$ again.

The lower bound of $f(x)$ appears to be more difficult since it requires tighter approximation of $m$ without singularity at $x=0$. I can prove the lower bound for $x$ greater than some positive number. I know I need to stitch the small and large regions of positive $x$ together, but I have not carried the detailed computation out yet. Does anyone have more clever trick to accomplish this task?

Here is the proof for $g(x)>0, \forall x\ge\sqrt{\frac{4}{3}}$. \begin{align} \frac{dg}{dx} &= 2n[xr(1-m)-2(0.5-m)] \\ &= 2n^2[(xr-1)n^{-1}+(2-xr)r] \end{align} where $r:=\frac{m}{n}$. In what follows we will use the first expression. The second expression is an alternative which I keep just for maybe future reference. Since $$r<\frac{1}{x}\Big(1-\frac{1}{x^2+3}\Big), \forall x>0,$$ \begin{align} \frac{dg}{dx} &< \frac{2n^2}{x^2+3}(-n^{-1}+(x^2+4)r) \\ &<\frac{2n^2}{x^2+3}\Big(-n^{-1}+x\Big(1+\frac{4}{x^2}\Big)\Big), \end{align} where on the last line we apply the $r$ bound again. Choose $x\ge x_0:=\sqrt{\frac{4}{3}}$, $$n^{-1}-x\Big(1+\frac{4}{x^2}\Big)>n^{-1}-4x.$$ It can be shown that $n^{-1}-4x$ is positive at $x=x_0$ and its derivative is always positive for $x\ge x_0$. We thus have $$\frac{dg}{dx}<0, \forall x\ge x_0.$$ It is easy to see that $g(x)>0$ for sufficiently large $x$. Therefore, $g(x)>0, \forall x\ge x_0$.

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When you say "prove the following inequality" - do you mean that you already know the inequality is true? If so, what is your source? If not, then what evidence do you have for why the inequality might be true? –  Yemon Choi Jul 3 '13 at 4:10
@WillJagy et al: I have edited my original post to describe the problem with accuracy, provide reason for my speculation of its validity, and give more context. This is not an easy problem. There is a subject called normal approximation with Stein's Method. Besides, browsing through the forum, I have seen several other more trivial looking but legitimate posts. I would like to ask for the reason for deeming this question "off topic" and a review of the classification. –  Hansen Jul 3 '13 at 13:22
I am inclined to believe this may not be easy, as you say, but if you have a write-up of the results you've obtained so far that you can link to, you might have more success in convincing others that the problem is definitely non-trivial. (This is too far from my areas of research for me to weigh in with any authority, so I won't vote to reopen. I suspect however it might be MO-worthy.) –  Todd Trimble Jul 3 '13 at 14:57
@Hans: Note that the argument for the upper bound, as given, isn't quite correct since $x^2 m (1-m) < n x (1-m)$ holds only for nonnegative $x$. You are saved by the fact that $h(x)$ happens to be an even function, so it suffices to consider only nonnegative $x$. Also, $g$ is even. Could you please edit to specify precisely what truncation of a normal you are considering. Perhaps a somewhat more indirect approach might yield something if we know a little more about the problem you are considering. Cheers. –  cardinal Jul 5 '13 at 14:19

Yes, the conjectured lower bound is true and can be proved using fairly simple, if somewhat tedious, analysis of derivatives.

First define $$b := f - N^2 = x(xN + n) - (xN + n)^2 + N(1-N)\>.$$ The plan is to show that $b$ is a decreasing function bounded below by zero.

Let $u := x N + n$, so that $b = (x-u)u + (1-N)N = (x-u)u + (1-u')u'$. Since $u(-x) = -(x-u(x))$ and $N(-x) = 1-N(x)$, $b$ is an even function and so we restrict ourselves to the case $x \geq 0$.

Observe that $u' = N$, $u'' = n$, and $b(0) = (1/4) - (1/2\pi) > 0$.

By using the classical inequalities, valid for $x > 0$, $$\frac{xn}{x^2+1} \leq 1-N \leq \frac{n}{x} \>,$$ on $(x-u)u$, it is straightforward to verify that $\lim_{x\to\infty} b(x) = 0$.

Now, using the fact $u = x u' + u''$, $$b' = 2u(1-u') - 2 u' u'' = 2 u' u''\left(\frac{(1-u')u}{u'u''} - 1\right) \>.$$ So, if we can show that $\frac{(1-u')u}{u'u''} \leq 1$, we will be done. Plugging in the definitions yields $\frac{(1-u')u}{u'u''} = \frac{1-N}{n}(x+n/N)$.

Lemma 1. For $x \geq 0$, $n/N \leq a e^{-a x}$ where $a = \sqrt{2/\pi}$.

Proof. Define $g := a^{-1} e^{ax} n - N$. Then $g(0) = 0$ and $$g' = (1-x/a - e^{-ax})e^{ax} n < 0 \>.$$

In particular, we have, $x+n/N \leq x + a e^{-a x}$ for any $x \geq 0$.

Lemma 2. For $x \geq 0$, $(1-N)/n \leq (x+a e^{-ax})^{-1}$.

Proof. Set $g := (x+ae^{-ax})^{-1} n - (1-N)$. Then, $g(0) = 0$ and $$g' = (a+ae^{-ax} + x - a^{-1} e^{ax}) \frac{a e^{-ax} n}{(x+a e^{-ax})^2}\>.$$ The fraction on the right is positive, so we concentrate on the first term on the right. Let $z := a + a e^{-ax} + x - a^{-1} e^{ax}$. Then $z(0) = 2a - 1/a > 0$ and $\lim_{x\to\infty} z(x) = -\infty$. Furthermore, $$z' = - a^2 e^{-ax} + 1 - e^{ax} < 0 \>.$$ Hence, $g'$ is positive for small $x$ and negative for large $x$. Since $\lim_{x\to\infty} g(x) = 0$, we conclude that $g \geq 0$.

This allows us to complete the proof, since by applying Lemma 1 and then Lemma 2, we have $$\frac{1-N}{n} (x + n/N) \leq \frac{1-N}{n} (x+a e^{-ax}) \leq 1 \>.$$

Hence, $b' < 0$, so $b > 0$ as desired.

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Beautiful proof! I think the introduction of $u:=\int_{-\infty}^x n(t)dt$ is the key. Lemma 1 and Lemma 2 are useful result in their own rights. They connect the behavior of $N$ or $1-N$ near $0$ and $\infty$ smoothly. I will wait a while for others to check the computation, before I will check it as THE accepted answer, even though it is too pretty to be wrong. Meanwhile, could you please describe your motivation in coming up with the function $a e^{ax}$? –  Hansen Jul 8 '13 at 2:36
It is unfortunate that there is only 1 point up vote allow for each person per answer. Otherwise, I would have put in ticked more. :-) –  Hansen Jul 8 '13 at 2:40
Dear @Hans: Regarding motivation: Note that $n/N$ is decreasing and so I first tried the crudest thing possible, i.e, $x + n/N \leq x + n(0)/N(0) = x + a$. However, this doesn't work since it turns out by a similar argument to Lemma 2 that $(1-N)/n \geq (x+a)^{-1}$. So, I needed a function that decreased but stayed above $n/N$, while also decreasing fast enough that I'd still get an upper bound on $(1-N)/n$. Note that, actually, the same basic analysis as Lemma 2 will yield $(1-N)/n \leq (x+a e^{-bx})^{-1}$ where $b = \sqrt{\pi/2}-\sqrt{2/\pi}$, which is a little sharper, but unneeded here. –  cardinal Jul 8 '13 at 3:08
@Hans: (Also, just a minor typo in your first comment: $u := \int_{-\infty}^x N(u)\,\mathrm du$. Cheers.) –  cardinal Jul 8 '13 at 3:09
I see your rationale, but can you describe what makes you think of the particular form of the exponential function $e^{-ax}$? Just a first lucky choice? And thanks for pointing out my typo. –  Hansen Jul 8 '13 at 4:46
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We may see that the inequality is true for every $|x|<0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2$$ is always positive. In other words the inequality is true for this $x$ as soon as the discriminant $\Delta$ of this quadratic is negative (the coefficient of $y^2$ being positive).

The discriminant is $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<1/(2\pi)$, the inequality will be true for every $x$ such that $4N^2(1-N)^2>1/(2\pi)$.

Thus the inequality is true for every $x$ such that $0.275214<N<0.724786$. This corresponds to the condition $|x|<0.597$.

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@ juan : You wrote "Since the derivative of your function is easily bounded". Could you explain that place in detail? The expression for $f'(x)$ (which can be downloaded from rapidshare.com/files/3032281090/derivative.pdf ) is not so simple. –  user64494 Jul 7 '13 at 5:11
@user64494 To apply the maximal slope principle you only need a rough bound of the derivative. For example substitute all exp(-x^2/2) by 1 and all N(x) by 1, all x by 0.597. All in absolute value and this bound will suffice. –  juan Jul 7 '13 at 8:11
@ juan: You don't answer my request concerning the estimate of the derivative. So called slope principle is the next step in your answer. –  user64494 Jul 7 '13 at 8:17
@user64494 My answer proof completely the inequality for $x<-0.597$ or $x>0.597$. For this you have no need of a bound of the derivative. Now to show $f(x)>0$ (my $f$ is different from yours) on the interval $|x|<0.597$ you may apply the maximal slope principle. This need a bound of the derivative on $|x|<0.597$ (a rough bound suffice). This is very easy to get. And you finish without difficulty the proof with a little computation (see the paper cited in my answer). –  juan Jul 7 '13 at 8:18
I am not sure to follow: the good regime $4N^2(1-N)^2\gt1/2\pi$ is when $|x|\lt x_*$ for some $x_*$, not the other way round (as an aside, note that the inequality one is interested in holds at $x=0$ hence also in a neighborhood of $x=0$). –  Did Jul 7 '13 at 11:52

In view of $$f(x):= (x^2+1)N(x)+xn(x)-(x+N(x))^2-N(x)^2=$$ $$\left( {x}^{2}+1 \right) \left( 1/2+1/2\, {{\rm erf}\left(1/2\,\sqrt {2}x\right)} \right) +1/2\,{\frac {{{\rm e} ^{-1/2\,{x}^{2}}}\sqrt {2}x}{\sqrt {\pi }}}-$$ $$\left( x+1/2+1/2\, {{\rm erf}\left(1/2\,\sqrt {2}x\right)} \right) ^{2}- \left( 1/2+1/2\, {{\rm erf}\left(1/2\,\sqrt {2}x\right)} \right) ^{2}$$ and its taylor expansion at $x=0$ $$f(x)=-x+ \left( -1/2\,{\pi }^{-1}+{\frac {1}{2}}- \left( 1+1/2\,{\frac { \sqrt {2}}{\sqrt {\pi }}} \right) ^{2} \right) {x}^{2}+O \left( {x}^{3 } \right)$$ the inequality under consideration seems to fail for small positive values of $x$.

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I think the series at 0 is $$(\pi-2)/(4\pi) +(1/4-1/\pi) x^2+ O(x^3)$$ –  juan Jul 7 '13 at 8:05
See the taylor expansion found with Maple in the worksheet exported as a pdf file rapidshare.com/files/4225333834/taylor.pdf –  user64494 Jul 7 '13 at 8:15
@ juan: Substitute $x=0$ in your expression and in the inequality under consideration. –  user64494 Jul 7 '13 at 8:27
but your function has a term $-(x+N)^2$ instead of $-(x N+n)^2$. –  juan Jul 7 '13 at 8:32
@ juan: Thank you. You are right. I must be more careful. –  user64494 Jul 7 '13 at 8:56

The Maple command $$asympt((x^2+1)*N(x)+x*n(x)-(x*N(x)+n(x))^2-N(x)^2, x, 8)$$ produces $$\left( {\frac {\sqrt {2}}{\sqrt {\pi }{x}^{3}}}-6\,{\frac {\sqrt {2}} {\sqrt {\pi }{x}^{5}}}+O \left( {x}^{-7} \right) \right) {\frac {1}{ \sqrt {{{\rm e}^{{x}^{2}}}}}}.$$ Thus the inequality is true for big positive $x$. I leave the investigation of it on the finite interval on your own. The above asymptotics can be obtained by hand too.

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You did not say which finite interval. –  Did Jul 6 '13 at 8:33
@ Did: A positive result is obtained by me. What can you do? –  user64494 Jul 6 '13 at 8:43
Sorry but I do not understand your comment. You might want to explain (or to delete the comment). –  Did Jul 6 '13 at 8:48
@ Did: Indeed, I did not say it. What can you suggest to this end? –  user64494 Jul 6 '13 at 9:17
@user64494: Did you read the last few sentences of my original post? "I can prove the lower bound for x greater than some positive number. I know I need to stitch the small and large regions of positive x together, but I have not carried the detailed computation out yet. Does anyone have more clever trick to accomplish this task?" The lower bound can be easily verified for very small $x$ too. The difficulty lies in specifying what you call "finite interval" show that the valid finite interval overlaps with the large $x$ interval. –  Hansen Jul 6 '13 at 16:15