For error function $\text{Erf}(x)$ I mean $$\operatorname{Erf}(x) = \int_{-\infty}^x\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}u^2\right)\mathrm{d} u.$$

My statistics professor said that

$$1-\operatorname{Erf}(x) \leq \frac{1}{x}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2} x^2\right)$$ and that $$\lim_{x\to+\infty}\frac{1-\operatorname{Erf}(x)}{\frac{1}{x}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}x^2\right)} = 1.$$

Proving the first fact is super easy, what about the second one?

EDIT: there were some mistakes, now it is all fixed.

  • 3
    $\begingroup$ Usually $\operatorname{Erf}(x)$ would mean the integral of the function you've got, i.e. $$\operatorname{Erf} (x) = \int_{-\infty}^x \frac 1 {\sqrt{2\pi}} e^{-u^2/2} \,du,$$ or else a rescaled version of that. $\qquad$ $\endgroup$ Feb 4 '17 at 18:21
  • $\begingroup$ You're using both $x$ and $z$ to represent something. Are they both supposed to be the same thing? $\endgroup$ Feb 4 '17 at 18:22
  • $\begingroup$ Personal attempts? Explanations about why the "super easy" proof does not work to prove the second fact? Anything? $\endgroup$
    – Did
    Feb 4 '17 at 18:24
  • $\begingroup$ @MichaelHardy fixed. $\endgroup$
    – Nisba
    Feb 4 '17 at 18:26
  • 2
    $\begingroup$ Well, using the change of variable $$u=x+\frac{v}x$$ one gets that, for every positive $x$, $\sqrt{2\pi}(1-\mathrm{erf}(x))$ equals $$\int_x^\infty e^{-u^2/2}du=\int_0^\infty e^{-x^2/2-v-v^2/(2x^2)}\frac{dv}x=\frac{e^{-x^2/2}}x\int_0^\infty e^{-v-v^2/(2x^2)}dv$$ and, when $x\to\infty$, the last integral increases to $$\int_0^\infty e^{-v}dv=1$$ as desired. (No L'Hopital, no fancy continued fraction, just a good ol' linear change of variable...) $\endgroup$
    – Did
    Feb 4 '17 at 21:13

The (complementary) error function has a well known continued fraction expansion:

$$ \frac{1}{\sqrt{2\pi}}\int_{z}^{+\infty}e^{-x^2/2}\,dx = \frac{e^{-z^2/2}}{\sqrt{2\pi}}\cdot\frac{1}{z+\frac{1}{z+\frac{2}{z+\frac{3}{z+\ldots}}}}\tag{1}$$ that is not difficult to prove by studying the recurrence relation fulfilled by the moments $$ M_n = \frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty} x^n e^{-x^2/2}\,dx.\tag{2}$$ Anyway, to prove the second limit it is enough to apply de l'Hospital rule.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.