2
  • $X \sim \mathcal{N}(0,1)$
  • $ k \geq 2$
  • We want to prove that $ P(|X| >k) \leq e^{-k}$

My attempt :

$ \begin{align*} P(|X| >k) & =P( e^{ |X| } > e^k ) \\ &\leq e^{-k} E( e^{|X|} ) \\ &= e^{-k} \int_{0}^{ \infty} \dfrac{2}{ \sqrt{2 \pi} }e^{-t} e^{-\frac{t^2}{2} }dt \\ &= e^{-k} \int_{0}^{ \infty} \dfrac{2}{ \sqrt{2 \pi} } e^{ -\frac{(t+1)^2-1 }{2} }dt\\ \end{align*} $

zestiria
  • 877

1 Answers1

1

Let's proceed as follows. With your notations, by symmetry, it suffices to show that $$\mathbb{P}(X > k) \leq \frac{1}{2}\,\mathrm{e}^{-k}$$ holding for each $k>2$. Let's define the function $$\Phi \,\colon y \mapsto \frac{1}{\sqrt{2\,\pi}}\int_y^\infty \mathrm{e}^{-x^2/2} \, \mathrm{d}x,$$ so it suffices to show that $\Phi(k) < \frac{1}{2}\,\mathrm{e}^{-k}.$ In general, we will show that $$\Phi(y) < \tfrac{1}{\sqrt{2\,\pi}\,y}\mathrm{e}^{-y^2/2}$$ for each $y > 0$. Indeed, by the definition of $\Phi$, we have $$\Phi(y) \leq \frac{1}{\sqrt{2\,\pi}} \int_y^\infty \tfrac{x}{y}\mathrm{e}^{-x^2/2} \, \mathrm{d}x = \tfrac{1}{\sqrt{2\,\pi}\,y}\mathrm{e}^{-y^2/2}.$$ Can you conclude from there?

Fei Cao
  • 2,830
  • Remark: A very similar approach yields a lower bound as well, meaning that for each $y > 0$, we will have $$\Phi(y) \geq \tfrac{1}{\sqrt{2,\pi},(y+1)}\left(\mathrm{e}^{-y^2/2} - \mathrm{e}^{-(y+1)^2/2}\right).$$ Combing these yields the asymptotic equivalence $$\Phi(y) \sim \tfrac{1}{\sqrt{2,\pi},y}\mathrm{e}^{-y^2/2}$$ for all large $y$. – Fei Cao Apr 19 '21 at 00:06
  • $ \Phi(y) \geq \dfrac{1}{ \sqrt{ 2 \pi }} \int_{y}^{y+1} x e^{ -\frac{x^2}{2}}$ – zestiria Apr 19 '21 at 09:47
  • Thanks so much. – zestiria Apr 19 '21 at 09:47