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I am trying to show that $$P(X>t)\leq \frac{1}{2}e^\frac{-t^2}{2}$$ for $t>0$ where $X$ is a standard normal random variable. Perhaps this is simple. I have been starting with

$$ \int_{t}^{\infty} \frac{1}{\sqrt{2\pi}} e^\frac{-x^2}{2}dx \leq \int_{t}^{\infty}\frac{x}{t}\frac{1}{\sqrt{2\pi}}e^\frac{-x^2}{2}dx = \frac{1}{t\sqrt{2\pi}}e^\frac{-t^2}{2}$$

but this is not what I want...I am looking for a much stronger inequality. Any help in the direction of getting the $1/2$?

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  • $\begingroup$ so your approach is valid for $t\geq \sqrt{\pi/2}$. i do not know how to do it, so maybe, you start by assuming otherwise. $\endgroup$ – Lost1 Oct 24 '14 at 8:46
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For every $t\gt0$, $$P(X\gt t)=\int_0^\infty\frac1{\sqrt{2\pi}}\mathrm e^{-(x+t)^2/2}\mathrm dx,$$ and, for every $x\geqslant0$, $$\mathrm e^{-(x+t)^2/2}\leqslant\mathrm e^{-t^2/2}\mathrm e^{-x^2/2},$$ hence $$P(X\gt t)\leqslant\mathrm e^{-t^2/2}\int_0^\infty\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\mathrm e^{-t^2/2}\,P(X\gt0).$$

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    $\begingroup$ nice and easy... $\endgroup$ – Lost1 Oct 24 '14 at 11:53

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