# Proof of upper-tail inequality for standard normal distribution

$X \sim \mathcal{N}(0,1)$, then to show that for $x > 0$, $$\mathbb{P}(X>x) \leq \frac{\exp(-x^2/2)}{x \sqrt{2 \pi}} \>.$$

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See math.stackexchange.com/questions/74151/… for a relted question. –  Mariano Suárez-Alvarez Oct 20 '11 at 23:10

Since for $t \geq x > 0$ we have that $1 \leq \frac{t}{x}$, $$\mathbb{P}(X > x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty 1 \cdot e^{-t^2/2} \,\mathrm{d}t \leq \frac{1}{\sqrt{2\pi}} \int_x^\infty \frac{t}{x} e^{-t^2/2} \,\mathrm{d}t = \frac{e^{-x^2/2}}{x \sqrt{2\pi}} .$$
@Harpreet, read the very beginning of the statement. The integrand to the left of $\leq$ is thus less than the integrand on the right. The inequality of the integrals follows from the monotonicity property of integrals. –  cardinal Mar 24 '11 at 2:44
Integrating by parts, \begin{align*} Q(x) &= \int_x^{\infty} \phi(t)\mathrm dt = \int_x^{\infty} \frac{1}{\sqrt{2\pi}}\exp(-t^2/2) \mathrm dt\\ &= \int_x^{\infty} \frac{1}{t} \frac{1}{\sqrt{2\pi}}t\cdot\exp(-t^2/2) \mathrm dt\\ &= - \frac{1}{t}\frac{1}{\sqrt{2\pi}}\exp(-t^2/2)\biggr\vert_x^\infty - \int_x^{\infty} \left( - \frac{1}{t^2} \right ) \left ( - \frac{1}{\sqrt{2\pi}} \exp(-t^2/2) \right )\mathrm dt\\ &= \frac{\phi(x)}{x} - \int_x^{\infty} \frac{\phi(t)}{t^2} \mathrm dt. \end{align*} The integral on the last line above has a positive integrand and so must have positive value. Therefore we have that $$Q(x) < \frac{\phi(x)}{x} = \frac{\exp(-x^2/2)}{x\sqrt{2\pi}}~~ \text{for}~~ x > 0.$$ This argument is more complicated than @cardinal's elegant proof of the same result. However, note that by repeating the above trick of integrating by parts and the argument about the value of an integral with positive integrand, we get that $$Q(x) > \phi(x) \left (\frac{1}{x} - \frac{1}{x^3}\right ) = \frac{\exp(-x^2/2)}{\sqrt{2\pi}}\left (\frac{1}{x} - \frac{1}{x^3}\right )~~ \text{for}~~ x > 0.$$ In fact, for large values of $x$, a sequence of increasingly tighter upper and lower bounds can be developed via this argument. Unfortunately all the bounds diverge to $\pm \infty$ as $x \to 0$.
(+1) There are fairly general techniques related to this approach to "fix up" the behavior near zero and still obtain a bound valid on all of $[0,\infty)$. –  cardinal Oct 13 '11 at 1:55