# 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. $$f(x):= (x^2+1)N + xn-(xN+n)^2 > N^2$$ 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, $$f(x):=\int_0^\infty t^2n(t+x)dt-\Big(\int_0^\infty t\,n(t+x)dt\Big)^2>\Big(\int_0^\infty n(t-x)dt\Big)^2.$$

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

$N$ is the upper bound of $f$, i.e. $$(x^2+1)N + xn-(xN+n)^2 < N$$ or $$h(x)\equiv x^2 m(1-m)-n[x(1-2m)+n]<0$$

Proof: $h$ 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?

$g(x)>0, \forall x\ge\sqrt{\frac{4}{3}}$

Proof: \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, $g(x)>0, \forall x\ge x_0$.

• 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? Jul 3, 2013 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.
– Hans
Jul 3, 2013 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.) Jul 3, 2013 at 14:57
• Posted this meta.mathoverflow.net/questions/223/requests-for-reopen-votes/… on meta Jul 3, 2013 at 16:55
• @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. Jul 5, 2013 at 14:19

Here is a complete solution. The idea is to kill the entries of $$N$$ in two steps, by applying two appropriately constructed first-order differential operators, which will result in a simple elementary expression:

Let $$b:=f-N^2$$. As noted by cardinal, $$b$$ is an even function. So, it is enough to show that $$b>0$$ on $$[0,\infty)$$. Let $$b_0(x):=\frac{b(x)}{x^2+1}$$ and $$b_1(x)=\pi\, \left(x^2+1\right)^2 e^{x^2/2}\, b_0'(x).$$ Then $$b_1'(x)=-e^{-\frac{x^2}{2}} \left(x^2+1\right)<0$$, so that $$b_1$$ is decreasing. Also, $$b_1(0)=0$$. Hence, $$b_1(x)<0$$ for $$x>0$$, and so, $$b_0$$ is decreasing on $$[0,\infty)$$. Moreover, $$b_0(x)\to0$$ as $$x\to\infty$$. So, $$b_0>0$$ and hence $$b>0$$.

• I will verify the algebra later but it looks like an ingenious proof! Would you be so kind as to share your insight and motivation that leads you to the definition of $b_0$ and subsequent $b_1$?
– Hans
Jun 4, 2015 at 19:49
• Thank you for your comment Hans. The "worst" in the expression of $b(x)$ are the "like terms" containing $N(x)^2$, which can be combined, with the total coefficient $x^2+1$ (maybe times some constant). So, the division by $x^2+1$ leaves only a constant coeff. at $N(x)^2$, and the derivative of $N^2$ is $2Nn$, which is "simpler" than $N^2$. So, $b_0'$ is "simpler" than $b$. Similarly proceeding further, till I get something really simple. Jun 4, 2015 at 20:32
• I am revisiting this problem. I see your motivation clearly now after having finally checked out the algebra, which is correct. It is, as you said, to keep reducing the integration and its powers of $n$ while the derivatives of $n$ is itself multiplied by polynomials. Again, ingenious. +1 I do not now what the etiquette is here, but if it is not frowned upon and not seen as diminishing the excellency of cardinal's proof, I would accept your answer.
– Hans
Jan 10, 2018 at 9:18

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.

• 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 $a e^{ax}$?
– Hans
Jul 8, 2013 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. :-)
– Hans
Jul 8, 2013 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. Jul 8, 2013 at 3:08
• @Hans: (Also, just a minor typo in your first comment: $u := \int_{-\infty}^x N(u)\,\mathrm du$. Cheers.) Jul 8, 2013 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.
– Hans
Jul 8, 2013 at 4:46

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 $|x|<0.597$.

• @ 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. Jul 7, 2013 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, 2013 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. Jul 7, 2013 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, 2013 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, 2013 at 11:52

This is just to make explicit the functions in Iosif Pinelis' ingenious answer for posterity.

$$b_0(x):=\frac{b(x)}{x^2+1}=-N^2+\Big(1-\frac{2xn}{x^2+1}\Big)N+\frac{x-n}{x^2+1}n,$$ and $$b_0'(x)=\frac{2n}{(x^2+1)^2}\,(-2N+1+xn).$$ $$b_1(x):=\frac{(x^2+1)^2}{2n}b_0'(x),$$ implying $$b_1'(x)=-(1+x^2)n>0.$$

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.

• You did not say which finite interval.
– Did
Jul 6, 2013 at 8:33
• @ Did: A positive result is obtained by me. What can you do? Jul 6, 2013 at 8:43
• Sorry but I do not understand your comment. You might want to explain (or to delete the comment).
– Did
Jul 6, 2013 at 8:48
• @ Did: Indeed, I did not say it. What can you suggest to this end? Jul 6, 2013 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.
– Hans
Jul 6, 2013 at 16:15

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 $x$.

• I think the series at 0 is $$(\pi-2)/(4\pi) +(1/4-1/\pi) x^2+ O(x^3)$$
– juan
Jul 7, 2013 at 8:05
• See the taylor expansion found with Maple in the worksheet exported as a pdf file rapidshare.com/files/4225333834/taylor.pdf Jul 7, 2013 at 8:15
• @ juan: Substitute $x=0$ in your expression and in the inequality under consideration. Jul 7, 2013 at 8:27
• but your function has a term $-(x+N)^2$ instead of $-(x N+n)^2$.
– juan
Jul 7, 2013 at 8:32
• @ juan: Thank you. You are right. I must be more careful. Jul 7, 2013 at 8:56