# Lower bound of Gaussian tail?

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 $x \ge 1$,$$\textbf{P}(N \ge x) \ge K_2 x^{-1} e^{-x^2/2}\text{ ?}$$

• What did you try? (And didn't you already post this recently?)
– Did
Sep 12 '15 at 20:48
• 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? Sep 13 '15 at 20:37
• 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? Sep 15 '15 at 2:51
• 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 $K_1$. ${}\qquad{}$ Sep 16 '15 at 21:15

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 $K_2 = 1/(2\sqrt{2\pi})$.