I have this question.
By considering the probability that 2 independent, standard normal random variables, $x_1$ and $x_2$, lie within the square: $\{(x_1,x_2)||x_1|<x,|x_2|<x\}$, prove the Chernoff bound: $$erfc(x)<e^{-x^2}\text{where }x>0$$
I thought of first writing them in terms of Q-function, then convert to erfc. Here's what I have so far.
\begin{align} P(|x_1|<x,|x_2|<x) &= P(|x_1|<x)P(|x_2|<x)\\ &= [1-P(|x_1|>x)][1-P(|x_2|>x)]\\ &= \Bigg[1-\int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-|x_1|^2}{2}}dx_1\Bigg]\Bigg[1-\int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-|x_2|^2}{2}}dx_2\Bigg]\\ &= 1-\int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-x_2^2}{2}}dx_2-\int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-x_1^2}{2}}dx_1+\int_{x}^{\infty}\int_{x}^{\infty}\frac{1}{2\pi}e^{\frac{-(x_1^2+x_2^2)}{2}}dx_1dx_2\\ &= [1-erfc(x)]^2 \end{align}
I know that $$erfc(x) = \frac{2}{\sqrt{\pi}}\int_{x}^{\infty}e^{v^2}dV$$
But I have no idea where the inequality with the $e^{-x^2}$ comes in. Do I need to consider some other area, like a quadrant of a circle with radius $\sqrt{2}x$ or something?
