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 $k=2$

  • 1
    $\begingroup$ 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. $\endgroup$
    – Dinesh
    Oct 16, 2010 at 0:30

3 Answers 3


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).

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    $\begingroup$ 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. $\endgroup$ Oct 16, 2010 at 2:26
  • $\begingroup$ Thanks for pointing out the rearrangement inequality, it seems to make the proof very easy! $\endgroup$ Oct 16, 2010 at 3:32
  • $\begingroup$ Ah!!! This is such a lovely proof. $\endgroup$
    – Dinesh
    Oct 16, 2010 at 3:43
  • $\begingroup$ @Willie: Rearrangement Inequality? I thought that was something completely different. Wikipedia seems to agree: en.wikipedia.org/wiki/Rearrangement_inequality. Can you please provide a reference which uses this name for the problem you state? $\endgroup$
    – Aryabhata
    Oct 16, 2010 at 15:03
  • $\begingroup$ @Moron: Beautifully done! I will delete my answer. $\endgroup$
    – user940
    Oct 16, 2010 at 15:31

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.


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.

  • 1
    $\begingroup$ It's actually not clear to me how to show fact 1. How do you show this? $\endgroup$
    – dmh
    Oct 12, 2022 at 14:00
  • $\begingroup$ @dmh Thanks for comment! I edited it. $\endgroup$
    – River Li
    Oct 12, 2022 at 15:02

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