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Let \(X \sim \mathcal{N}(0, 1)\) and \(x > 0\).

\(
\mathbb{P}(X > x)
\)
\(
= \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} \,\mathrm{d}t
\)
\(
\leq \frac{1}{\sqrt{2\pi}} \int_x^\infty \frac{t}{x} e^{-t^2/2} \,\mathrm{d}t
\)
\(
= \frac{e^{-x^2/2}}{x \sqrt{2\pi}}.
\)

Quite elegant!

Source: math.stackexchange.com/questio

Another elegant bound that's stronger for small x is

P(|X| > x) ≤ exp(-x² / 2)

for X ~ N(0, 1) and x >0.

(but I'm not aware of a proof quite as short as the previous one)

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