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Let $G$ be a standard normal variable, and let $a>0$. Show that

$$P(G>a) \leq exp(-a^2/2)$$

I have tried writing $P(G >a) = (2 \pi)^{-1} exp(-a^2/2) \int_a^{\infty}exp[-(x^2-a^2)/2]dx$ and then completing the square but it did not work. Also seen this result, but did not help much either.

Any help is appreciated. Thanks.

user1110
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1 Answers1

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$P(G\gt a)=\frac{1}{\sqrt{2\pi}}\int\limits_a^\infty e^{-\frac{x^2}{2}}dx\lt\frac{1}{\sqrt{2\pi}}\int\limits_a^\infty xe^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2\pi}} e^{-\frac{a^2}{2}}$

herb steinberg
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  • Thank you for your answer. I am not 100% sure though why your strict inequality holds. – user1110 Apr 12 '22 at 23:01
  • Minor error. Holds for $a\gt 1$, since $x\gt 1$ for entire integral domain. For $0\le a\le 1$, needs a little work. – herb steinberg Apr 13 '22 at 22:13
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    For $0\le a\le 1$, $P(G\gt a)\lt \frac{1}{\sqrt{2\pi}}\int\limits _a^\infty xe^{-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2\pi}}\int\limits _a^1 (1-x)e^{-\frac{x^2}{2}}dx \lt \frac{2}{\sqrt{2\pi}}e^{-\frac{a^2}{2}}$ – herb steinberg Apr 14 '22 at 19:19
  • Sorry, but why was again the second integral in your comment smaller than $\frac{1}{2\pi}e^{-a^2/2}$? – user1110 Apr 27 '22 at 14:45
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    $\frac{1}{\sqrt{2\pi}}\int_a^1(1-x)e^{-\frac{x^2}{2}}dx\le \frac{1}{\sqrt{2\pi}}\int_a^1(1-a)e^{-\frac{a^2}{2}}dx\le \frac{1}{\sqrt{2\pi}}(1-a)^2e^{-\frac{a^2}{2}}$ – herb steinberg Apr 27 '22 at 18:40