# Asymptotic equivalence of the survival function of a standard Gaussian [duplicate]

My statistics teacher told us the following asymptotic result: $$X \sim N(0,1)$$ $$P(X > u) \underset{u \rightarrow +\infty}{\sim} \frac{1}{u} \exp\left(-\frac{u^2}{2}\right).$$

Do you know how to demonstrate this (or if there is a mistake?) Thank you.

• Hi. What did you do to (dis)prove the proposition? Show us your attempts. Thanks. Apr 16 at 12:58
• I was only able to find an upper bound by the Chernoff method. Apr 16 at 13:04
• stats.stackexchange.com/…
– whuber
Apr 16 at 15:34

The expression you gave is close (you missed the factor $$\frac{1}{\sqrt{2\pi}}$$) to the correct asymptotic behavior of the survival function of standard normal distribution:
\begin{align*} 1 - \Phi(x) \sim \frac{1}{x}\varphi(x), \tag{1} \end{align*} where $$\Phi$$ and $$\varphi$$ are CDF and PDF of standard normal random variable respectively.
$$(1)$$ is a corollary of the inequality (for fixed $$x > 0$$) \begin{align*} (x^{-1} - x^{-3})\varphi(x) < 1 - \Phi(x) < x^{-1}\varphi(x). \tag{2} \end{align*}
To prove $$(2)$$, notice the obvious inequality: \begin{align*} (1 - 3t^{-4})\varphi(t) < \varphi(t) < (1 + t^{-2})\varphi(t), \; t > x. \tag{3} \end{align*} Integrating three expressions above from $$x$$ to $$\infty$$ yields $$(2$$) -- note that each term in $$(2)$$ is the primitive function of each term in $$(3)$$ times $$-1$$.