1
$\begingroup$

This question already has an answer here:

By comparison with the integral of $ \frac xae^{-x^{2}}$

Show that $\int_a^\infty e^{-x^{2}}dx≤ \frac 1{2a}e^{-a^{2}} $ Given that $a>0$.

$\endgroup$

marked as duplicate by Dilip Sarwate, Ali Caglayan, E. Joseph, Alex M., tired Dec 20 '16 at 12:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The hint is direct. You should have tried to evaluate $\int_a^\infty\frac{x}{a}e^{-x^2}dx$. In general, you should show any attempted work / thoughts. $\endgroup$ – yurnero Dec 12 '16 at 22:58
2
$\begingroup$

For $x\ge a>0$, one has $$\color{red}{1}\le\color{blue}{\dfrac{x}a},$$ integrating gives $$ \int_a^\infty e^{-x^{2}}\cdot \color{red}{1}\:dx≤ \int_a^\infty e^{-x^{2}}\cdot\color{blue}{\dfrac{x}a}\:dx=\left[-\frac{e^{-x^2}}{2a} \right]_a^\infty=\frac 1{2a}e^{-a^{2}}. $$

$\endgroup$
1
$\begingroup$

Let me indulge in the undeniable pleasure of an overkill.
By studying the behaviour of a sequence of moments it is not difficult to prove that the complementary error function has the following continued fraction:

$$\frac{1}{\sqrt{2\pi}}\int_{z}^{+\infty}e^{-x^2/2}\,dx = \frac{e^{-z^2/2}}{\sqrt{2\pi}}\cdot \frac{1}{z+\frac{1}{z+\frac{2}{z+\frac{3}{\ldots}}}}$$ that makes the present question trivial. The LHS is the probability that a random variable with gaussian $N(0,1)$ distribution takes values in $(z,+\infty)$, i.e. a "tail probability".

$\endgroup$
  • $\begingroup$ Overkill but nice! Do you have a nice proof of it? (+1) $\endgroup$ – Olivier Oloa Dec 12 '16 at 23:01
  • $\begingroup$ @OlivierOloa: when dealing with moments the trick is always the same: to define $$I_n = \int x^n \omega(x)\,dx $$ then study the recurrence relation for the ratios $$\frac{I_{n+2}+I_{n+1}}{I_{n+1}+I_n}.$$ If $\omega(x)=e^{-x}$ or $\omega(x)=e^{-x^2}$ makes actually little difference. $\endgroup$ – Jack D'Aurizio Dec 12 '16 at 23:05
  • $\begingroup$ Ok, thank you @Jack D'Aurizio. $\endgroup$ – Olivier Oloa Dec 12 '16 at 23:26

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