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$X \sim \mathcal{N}(0,1)$, 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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Since for $t \geq x > 0$ we have that $1 \leq \frac{t}{x}$, $$ \mathbb{P}(X > x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty 1 \cdot e^{-t^2/2} \,\mathrm{d}t \leq \frac{1}{\sqrt{2\pi}} \int_x^\infty \frac{t}{x} e^{-t^2/2} \,\mathrm{d}t = \frac{e^{-x^2/2}}{x \sqrt{2\pi}} . $$

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  • $\begingroup$ @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. $\endgroup$ – cardinal Mar 24 '11 at 2:44
  • $\begingroup$ Look at the limits of integration carefully. $\endgroup$ – cardinal Mar 24 '11 at 2:56
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    $\begingroup$ Too much elegance in too short an answer!!! $\endgroup$ – Qwerty Oct 15 '16 at 18:45
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    $\begingroup$ Just to add something to this nifty answer and for people like me, see above that $1 \leq t/x$, as cardinal suggested, get guided by integral limits! $t$ is always equal or greater than $x$. $\endgroup$ – htennek2k Sep 6 '18 at 22:06
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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 $x \to 0$.

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    $\begingroup$ (+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 $[0,\infty)$. $\endgroup$ – cardinal Oct 13 '11 at 1:55
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    $\begingroup$ I have rolled back the edit made by Thor H. Jonsson who claimed that "integration by parts was wrong as the author forgot to differentiate the latter part of the integrand." There is nothing wrong with the integration by parts that I wrote: with $u=\frac 1t$,$v=-\exp(-t^2/2)$, $\mathrm dv=t\exp(-t^2/2)$, and using $$\int u\,\mathrm dv=uv-\int v\,\mathrm du,$$ we have that $$\int \frac 1t\cdot t\exp(-t^2/2)=\frac 1t\exp(-t^2/2)-\int -\frac{1}{t^2}(-\exp(-t^2/2)\,\mathrm dt.$$ What is this "missing differentiation" that Jonsson is complaining about? $\endgroup$ – Dilip Sarwate Apr 9 '15 at 23:16

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