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Have $N$ denote a $N(0, 1)$ random variable. I have found a $K_1$ such that for all $x >0$,$$\textbf{P}(N \ge x) \le K_1 x^{-1} e^{-x^2/2}.$$My question is, does there exist $K_2 > 0$ such that for all $x \ge 1$,$$\textbf{P}(N \ge x) \ge K_2 x^{-1} e^{-x^2/2}\text{ ?}$$

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    $\begingroup$ What did you try? (And didn't you already post this recently?) $\endgroup$
    – Did
    Sep 12 '15 at 20:48
  • $\begingroup$ Hint: $x^{-1} e^{-x^2/2}$ increases without bound as $x \downarrow 0$. Can you figure out the answer to your question from this? $\endgroup$ Sep 13 '15 at 20:37
  • $\begingroup$ Without more constraints on $N$ we could have $P(N\ge 1)=0$ .Or if you want $P(N\ge x)>0 $ for $ x\ge 1$ ,we could have $0< P(N\ge x)< x^{-2} e^{-x^2/2}$ .In both cases $ K_2$ does not exist. Are you referring to the "Bell Curve"? What is an $N(0,1)$ random variable? $\endgroup$ Sep 15 '15 at 2:51
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    $\begingroup$ Certainly the value of $K_1$ can serve in the role of $K_2$. Perhaps your question is whether there is a value of $K_2$ that is less conservative than $K_1$. ${}\qquad{}$ $\endgroup$ Sep 16 '15 at 21:15
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There is a standard lower estimate for $1-\Phi(x) = P(N(0,1)\ge x)$: for all $x>0$, $$ 1-\Phi(x)> \frac{x}{x^2+1}\varphi(x) = \frac{x}{x^2+1}\frac1{\sqrt{2\pi}}e^{-x^2/2}. $$ You can find the proof e.g. here.

So for $x\ge 1$ $$ 1-\Phi(x)> \frac{1}{x+1/x}\varphi(x)\ge \frac{1}{2x}\varphi(x), $$ consequently, the desired inequality holds with $K_2 = 1/(2\sqrt{2\pi})$.

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    $\begingroup$ for future visitors, it may help to know that this result is often known as "Mills' ratio". In Shorack's Probability for Statisticians, it is discussed in Exercise 6.1 (Mills' ratio), pg. 177, along with a slightly stronger one Shorack refers to as "Ito-McKean". $\endgroup$
    – D.R.
    Feb 23 at 19:29

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