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When $Z \sim N(0,1)$, then $P(\vert Z\vert \gt t)\le\sqrt\frac{2}{\pi}\times \frac{exp(-t^2/2)}t$.

My question is: I can get the aboving inequality, but how can I get the inequality such that $P(\vert Z\vert \gt t)\le\sqrt\frac{2}{\pi}\times \frac{t}{1+t^2} \times exp(-t^2/2)$?

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  • $\begingroup$ Does this new $Z$ follow normal(0,1) $\endgroup$ – Qwerty Jun 7 '16 at 13:45
  • $\begingroup$ Yes. Z follows standard normal distribution. $\endgroup$ – Vivian Zhang Jun 7 '16 at 13:47
  • $\begingroup$ @Qwerty Yes, $Z\sim \mathcal N(0,1)$. Or what do you mean by 'new $Z$ ' ? $\endgroup$ – callculus Jun 7 '16 at 13:47
  • $\begingroup$ I dont understand your question. How can the same Z follow the 2nd inequality if it follows the first but the second is not implied by the first? $\endgroup$ – Qwerty Jun 7 '16 at 13:49
  • $\begingroup$ I can get the first inequality. But for my supervisor asking, I need to figure out the second inequality. It's the same Z, not a new one. $\endgroup$ – Vivian Zhang Jun 7 '16 at 13:55
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We have $$1-\text{erf}(x)=\frac{x}{\sqrt{\pi}}\exp(-x^2)\frac{1}{x^2+(1/2)/q}$$ with $q\geq 1$. Thus $$1-\text{erf}(x)\geq\frac{x}{\sqrt{\pi}}\exp(-x^2)\frac{1}{x^2+(1/2)}$$ Now $$P(|Z|>t)=P(Z>t)+P(Z<-t)\\ =2P(Z>t)\\ =2\left(1-\frac{1}{2}(1+\text{erf}(t/\sqrt{2}))\right)\\ =1-\text{erf}(t/\sqrt{2})$$ $$1-\text{erf}(t/\sqrt{2})\geq \sqrt{\frac{2}{\pi}}\exp(-t^2/2)\frac{t}{t^2+1}$$ This looks right.

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  • $\begingroup$ What I need is "less or equal". $\endgroup$ – Vivian Zhang Jun 21 '16 at 15:03
  • $\begingroup$ Your inequality is wrong. You can see this by setting $t=0$: $P(|Z|>0)=1/2$ , $\sqrt{2/\pi}\frac{0}{1+0}\exp(0)=0$. Greater than or equal is correct. $\endgroup$ – Wouter Jun 21 '16 at 15:27
  • $\begingroup$ From the first one inequality, I can derive that $t \neq 0$. Now I want to get the boundary for t such that $t \ge 1$, that's why I need to get the second inequality. $\endgroup$ – Vivian Zhang Jun 22 '16 at 14:31
  • $\begingroup$ So If do not consider $t = 0$, then how can I drive the second inequality by using error function. $\endgroup$ – Vivian Zhang Jun 22 '16 at 14:34
  • $\begingroup$ These inequalities hold for all $t$ (at least all positive $t$). Your original inequality is never true. $\endgroup$ – Wouter Jun 22 '16 at 16:18

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