Lower bound of Gaussian tail?
By Abigail Rogers •
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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$\begingroup$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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