limiting behavior of standard normal survivor function

Question

How do you show that $\lim_{x\to \infty} 1-\Phi(x) \sim \phi(x)/x$? In the previous, I'm using $\Phi$ to refer to the standard normal CDF and $\phi$ to refer to the standard normal pdf. Thanks!!

Answer 1

We omit the $\frac{1}{\sqrt{2\pi}}$ of the density function, since it is hard to type and makes no difference to the ratio. For large $b$, we want to compare $\int_b^\infty e^{-x^2/2}\,dx$ with $\frac{e^{-b^2/2}}{b}$.

Integrate by parts, using $u=\frac{1}{x}$ and $dv=xe^{-x^2/2}\,dx$. Then $du=-\frac{1}{x^2}\,dx$ and we can take $v=-e^{-x^2/2}$. Thus $$\int_b^\infty e^{-x^2/2}\,dx=\frac{e^{-b^2/2}}{b} -\int_b^\infty \frac{e^{-x^2/2}}{x^2}\,dx .$$

We crudely bound the remaining integral. For positive $b$,
$$\int_b^\infty \frac{e^{-x^2/2}}{x^2}\,dx \le \frac{e^{-b^2/2}}{b^2}\int_b^\infty e^{-(x^2-b^2)/2}\,dx.$$ But $\int_b^\infty e^{-(x^2-b^2)/2}\,dx$ is bounded: let $x=b+t$. It follows that $$\frac{\int_b^\infty e^{-x^2/2}\,dx}{\frac{e^{-b^2/2}}{b}}=1+O(1/b).$$