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I have to prove that for a random variable $X\sim N(0,1)$,

\begin{equation*} \mathbb{P}(X>t) \leq \frac{1}{2}\,e^{-\frac{t^2}2}. \end{equation*}

I tried using the fact that for $h(t)=e^{\alpha t}$,

\begin{equation*} \mathbb{P}(|X|>t)\leq \frac{E(h(X))}{h(t)} \end{equation*}

but didn't manage to advance that much.

Jack D'Aurizio
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Nathan Sikora
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    I think this was answered here: http://math.stackexchange.com/questions/988822/normal-distribution-tail-probability-inequality – b yen May 19 '15 at 19:37
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    And here: http://math.stackexchange.com/questions/658398/bound-on-the-q-function-related-to-chernoff-bound – b yen May 19 '15 at 20:08

1 Answers1

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By using the density of the standard normal variable we have: $$\begin{eqnarray*}\mathbb{P}[X>t]&=&\frac{1}{\sqrt{2\pi}}\int_{t}^{+\infty}e^{-x^2/2}\,dx=\frac{e^{-t^2/2}}{\sqrt{2\pi}}\int_{0}^{+\infty}e^{-xt-x^2/2}\,dx\\&\leq&\frac{e^{-t^2/2}}{\sqrt{2\pi}}\int_{0}^{+\infty}e^{-xt}\,dx=\color{red}{\frac{e^{-t^2/2}}{t\sqrt{2\pi}}},\end{eqnarray*}$$ hence a stronger inequality is true for sure for any $t>\sqrt{\frac{2}{\pi}}$. If $t\leq\sqrt{\frac{2}{\pi}}$, we just need to exploit: $$ \frac{e^{-t^2/2}}{\sqrt{2\pi}}\int_{0}^{+\infty}e^{-xt-x^2/2}\,dx\leq \frac{e^{-t^2/2}}{\sqrt{2\pi}}\int_{0}^{+\infty}e^{-x^2/2}\,dx = \color{red}{\frac{e^{-t^2/2}}{2}} $$ hence we have:

$$ \forall t>0,\qquad \mathbb{P}[X>t]\leq \frac{e^{-t^2/2}}{\max(2,t\sqrt{2\pi})}.$$

That can also be improved up to: $$ \forall t>0,\qquad \mathbb{P}[X>t]\leq \frac{e^{-t^2/2}}{\sqrt{4+2\pi t^2}}.$$

Have a look at the Wikipedia page about the Mills Ratio and its continued fraction representation.

Jack D'Aurizio
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