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I've seen the following lower bound for the complementary error function (erfc) but I haven't been able to prove it. Does anyone know how to establish the following?

$$erfc(x) > \frac{ x \exp(-x^2) }{ \pi(1 + 2x^2) }$$

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3 Answers 3

up vote 6 down vote accepted

Durrett, Probability: Theory and Examples, 3rd edition, p. 6 gives

$$(x^{-1} - x^{-3}) e^{-x^2/2} \le \int_x^\infty e^{-y^2/2} \: dy $$

The proof Durrett gives is from the observation that

$$ \int_x^\infty (1-3y^{-4}) e^{-y^2/2} \: dy = \left( x^{-1} + x^{-3} \right) e^{-x^2/2} $$

which I suspect can be found by integration by parts, although I haven't written it out; in any case, differentiate it to check.

After this, some changes of variables give

$$ \left( {1 \over z} - {1 \over 2z^3} \right) e^{-z^2}/\sqrt{\pi} \le erfc(z). $$

Finally, $z/(1+2z^2) < 1/z-1/(2z^3)$ for $z > 2^{-1/4}$, giving your bound for $z > 2^{-1/4}$ if $\pi$ is replaced with $\sqrt{\pi}$.

Obviously this is a hack trying to get your proposed bound in the form of the bound I already knew, but hopefully it helps.

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Paradoxically, it is quite easier to prove stronger bounds. Let $X$ be a random variable with gaussian distribution and density $$ f(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2).$$ Now let, for any $k\in\mathbb{R}^+$, $$A_k = \sqrt{2\pi}\;\exp(k^2/2)\;\mathbb{P}[X>k] = \sqrt{\frac{\pi}{2}}\;\exp(k^2/2)\;\operatorname{Erfc}\left(\frac{k}{\sqrt{2}}\right).$$ Since $\mathbb{E}\left[\left(X-\mathbb{E}[X]\right)^2\right]\geq 0$, $\mathbb{E}[X^2]\geq\mathbb{E}[X]^2$, and the same holds for the conditional expected values, under the hypothesis $X>k$. That gives, in terms of $A_k$: $$ (1-k\; A_k)^2 \leq A_k\;\left(-k+(1+k^2)\; A_k\right), $$ that can be restated as: $$(\heartsuit)\quad A_k^2+k\;A_k-1\geq 0.$$ From: $$ A_k \geq \frac{2}{k+\sqrt{k^2+4}}, $$ we immediately have a lower bound for the $\operatorname{erfc}$ function: $$ e^{k^2}\;\operatorname{erfc}(k) \geq \frac{2}{\sqrt{\pi}}\left(\frac{1}{k+\sqrt{k^2+2}}\right).$$ An interesting fact is that the "reverse inequality" $$(\spadesuit)\quad (1-k\; A_k)^2 \geq \frac{2}{\pi}\;A_k\;\left(-k+(1+k^2)\; A_k\right), $$ equivalent to: $$(\spadesuit_2)\quad\pi\left(\int_{k}^{+\infty}(x-k)\;e^{-x^2/2}\;dx\right)^2\geq 2\int_{k}^{+\infty}e^{-x^2/2}\;dx\int_{k}^{+\infty}(x-k)^2\;e^{-x^2/2}\;dx,$$ holds, too. Since: $$A_k =\int_{0}^{+\infty}\exp\left(-x^2/2-kx\right)\,dx, $$ by using the Fubini-Tonelli theorem and switching to polar coordinates one can see that $(\spadesuit_2)$ is equivalent to: $$\int_{0}^{+\infty}\rho^3 e^{-\rho^2/2}\int_{0}^{\pi/4}(\pi\cos(2\theta)-2)\; e^{-k\rho\sqrt{2}\cos\theta}\;d\theta \; d\rho\geq 0, $$ also equivalent to (by integrating by parts): $$\int_{0}^{+\infty}\rho^3 e^{-\rho^2/2}\int_{0}^{\pi/4}(\pi\sin\theta\cos\theta-2\theta)\;k\rho\sqrt{2}\sin\theta\; e^{-k\rho\sqrt{2}\cos\theta}\;d\theta \; d\rho\geq 0, $$ that is trivial since over $[0,\pi/2]$, by the concavity of the sine function, we have $\sin\phi\geq\frac{2\phi}{\pi}$. Following the line of the previous proof, the $(\spadesuit)$-inequality can be used to have tight upper bound for the $e^{k^2}\;\operatorname{erfc}(k)$ - function. Moreover, from the $(\spadesuit_2)$-inequality the convexity of $\frac{1}{A_k}$ follows.

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Here's an approach that will establish the inequality, but it doesn't provide any insight into where the inequality came from. Let f(x) be the left side minus the right side, i.e.

f(x) = erfc(x) - \frac{ x \exp(-x^2) }{ \pi(1 + 2x^2) }

Clearly f(x) > 0 and \lim_{x\to\infty} f(x) = 0. A calculation shows that f'(x) < 0 for all x > 0, and so f(x) must be positive for all x > 0. See these notes for details. The notes also state improved bounds but without proof.

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