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This is a follow-up to this question. I came across a tighter potential bound and checked it numerically for $0\le x\le 5$. I think it holds for all positive $x$, can anyone see a proof?

$$1-\exp(-4x^2/ \pi) \ge \text{erf}(x)^2$$

Note: using analysis for previous question you can show that $1-\exp(-k x^2)$ is an upper bound on $\text{erf}(x)^2$ for $k=2$ and a lower bound when $k=1$. The factor $k=4/\pi$ comes out when numerically searching for tightest upper bound. Not only does it seem to give an upper bound, but it also tracks $\text{erf}(x)^2$ very closely.

Dashed graph below is $\text{erf}(x)^2$, red is $1-\exp(-k x^2)$ for $k=4/\pi$, other two graphs are for $k=1$ and $k=2$

Glorfindel
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Yaroslav Bulatov
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    The inequality is correct and I think it can be proved by some differentiation and Taylor series manipulations. The Taylor series expansion about $x=0$ also tells us why $k=4/\pi$ is the best possible. But I don't know if there is a simple proof of the inequality. Would love to see one. – Dinesh Oct 16 '10 at 00:30

3 Answers3

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As before we consider

$$\text{erf}(x)^2={4\over \pi}\int_0^x\int_0^x \exp{-(s^2+t^2)}\ ds \ dt\,.$$

Now compare this with the same over the area which is given by the quarter of a circle of radius $\displaystyle \frac{2x}{\sqrt{\pi}}$. The area of this is same as the area of the square of side $x$.

Since $\displaystyle e^{-(s^2 + t^2)}$ decreases as $\displaystyle s^2 + t^2$ increases, we are done!

The integral over the non-common area for the quarter circle is greater than the integral over the non-common area of the square (which lies outside the circle and so $\displaystyle s^2 + t^2$ is higher in that region).

Batominovski
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Aryabhata
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    Yep. This is another illustration of the rearrangement inequality: for an arbitrary radial function $f(x) = f(|x|) \geq 0$, such that $\partial_r f \leq 0$, we can ask: of all the sets $\Omega$ of area 1, when is the integral $\int_\Omega f(x) dx$ the greatest? The answer would be the ball centered at the origin with total area 1. – Willie Wong Oct 16 '10 at 02:26
  • Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy! – Yaroslav Bulatov Oct 16 '10 at 03:32
  • Ah!!! This is such a lovely proof. – Dinesh Oct 16 '10 at 03:43
  • @Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: http://en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state? – Aryabhata Oct 16 '10 at 15:03
  • @Moron: Beautifully done! I will delete my answer. –  Oct 16 '10 at 15:31
  • @Moron: it is the same thing. Let $x_1\ldots x_k$ be 1, and $x_{k+1} \ldots x_n$ be 0. Let $y_1\ldots y_n$ be a bunch of positive numbers. Then one arrangement that gives the maximum $x_1 y_{\sigma(1)} + \cdots + x_n y_{\sigma(n)}$ is if you arrange the $y$ in decreasing order. Now you move to the integral case and consider $\int \chi_{\Omega} f dx$ where $\chi_{\Omega}$ is a characteristic function. The rearrangement inequality then says the the norm is the maximum if you put the support of $\chi_{\Omega}$ where $f$ is big... – Willie Wong Oct 16 '10 at 16:34
  • In general, for arbitrary integrable non-negative function $f$, without radial symmetry, the analogous problem can be solved by first defining $A_\lambda$ as the set ${ x : f(x) > \lamba }$. Let $\lambda_0$ be the inf of all $\lambda$ with $|A_\lambda| < 1$. Then any $\Omega\subset $A_{\lambda_0}$ such that $\Omega\supset A_\lambda$ for all $\lambda > \lambda_0$ is a solution. – Willie Wong Oct 16 '10 at 16:42
  • @Willie: Oh I see. What you meant(and said!) regarding the ball at origin was a consequence of rearrangement inequality. I read that as being same as rearrangement inequality and hence the confusion. Thanks for clarifying. – Aryabhata Oct 16 '10 at 16:46
  • (In other words, think of it as rearrangement where the index set is $\mathbb{R}^2$, instead of a finite set of points.) Yes, sorry about the confusion. I meant the fact that followed the colon mark in my first sentence to be the "illustration". – Willie Wong Oct 16 '10 at 16:47
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This is one of the many results that keep being periodically rediscovered. The bound, together with a less accurate lower bound of the same form, was given in a report by J.T. Chu (1954; On bounds for the normal integral). It was noted in the report that the upper bound had previously been obtained independently by G. Polya and J.D. Williams. Both used the nice geometrical argument that has been here expounded.

Roberto Iacono
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Alternative proof:

It suffices to prove that, for $x > 0$, $$\sqrt{1 - \mathrm{e}^{-4x^2/\pi}} - \frac{2}{\sqrt{\pi}}\int_0^x \mathrm{e}^{-t^2} \mathrm{d} t \ge 0.$$ Denote LHS by $f(x)$. We have $$f'(x) = \frac{4x \mathrm{e}^{-4x^2/\pi}}{\pi\sqrt{1 - \mathrm{e}^{-4x^2/\pi}}} - \frac{2}{\sqrt{\pi}}\mathrm{e}^{-x^2}.$$

Fact 1: $f'(x) = 0$ has exactly one positive real solution, denoted by $x_0$.
(The proof is given at the end.)

By Fact 1 and $f'(1) > 0$ and $f'(2) < 0$, we have $f'(x) > 0$ on $(0, x_0)$ and $f'(x) < 0$ on $(x_0, \infty)$. Thus, $f(x)$ is strictly increasing on $(0, x_0)$ and strictly decreasing on $(x_0, \infty)$. Also, $f(0) = 0, f(\infty) = 0$. Thus, $f(x) > 0$ on $(0, \infty)$. We are done.

Proof of Fact 1:

For $x > 0$, we have \begin{align} f'(x) = 0 \quad &\Longleftrightarrow \quad \left(\frac{4x \mathrm{e}^{-4x^2/\pi}}{\pi\sqrt{1 - \mathrm{e}^{-4x^2/\pi}}}\right)^2 = \left(\frac{2}{\sqrt{\pi}}\mathrm{e}^{-x^2}\right)^2\\[6pt] \quad &\Longleftrightarrow \quad \frac{4x^2}{\pi} = \mathrm{e}^{8x^2/\pi} \mathrm{e}^{-2x^2}(1 - \mathrm{e}^{-4x^2/\pi}) \\[6pt] \quad &\Longleftrightarrow \quad \frac{4x^2}{\pi} + \mathrm{e}^{(1-\pi/2)\cdot \frac{4x^2}{\pi}} = \mathrm{e}^{(2-\pi/2)\cdot \frac{4x^2}{\pi}}\\ &\Longleftrightarrow \quad u + \mathrm{e}^{(1-\pi/2)u} = \mathrm{e}^{(2-\pi/2)u}, \quad u = \frac{4x^2}{\pi}. \end{align}

Let $g(u) = u + \mathrm{e}^{(1-\pi/2)u} - \mathrm{e}^{(2-\pi/2)u}$. We have $$g'(u) = 1 + (1-\pi/2)\mathrm{e}^{(1-\pi/2)u} - (2-\pi/2)\mathrm{e}^{(2-\pi/2)u}$$ and $$g''(u) = (1-\pi/2)^2\mathrm{e}^{(1-\pi/2)u} - (2-\pi/2)^2\mathrm{e}^{(2-\pi/2)u}.$$

Let $u_0 = 2\ln (\pi - 2) - 2\ln (4 - \pi)$. It is easy to prove that $g''(u) > 0$ on $(0, u_0)$, and $g''(u) < 0$ on $(u_0, \infty)$.

Thus, $g'(u)$ is strictly increasing on $(0, u_0)$, and strictly decreasing on $(u_0, \infty)$. Also, $g'(0) = 0$ and $g'(\infty) = -\infty$. Thus, $g'(u) = 0$ has exactly one positive real solution, denoted by $u_1$. Thus, $g'(u) > 0$ on $(0, u_1)$, and $g'(u) < 0$ on $(u_1, \infty)$. Thus, $g(u)$ is strictly increasing on $(0, u_1)$, and strictly decreasing on $(u_1, \infty)$. Also, $g(0) = 0$ and $g(\infty) = -\infty$. Thus, $g(u) = 0$ has exactly one positive real solution. As a result, $f'(x) = 0$ has exactly one positive real solution. We are done.

River Li
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