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.

  • 1
    $\begingroup$ Hi. What did you do to (dis)prove the proposition? Show us your attempts. Thanks. $\endgroup$ Apr 16 at 12:58
  • $\begingroup$ I was only able to find an upper bound by the Chernoff method. $\endgroup$ Apr 16 at 13:04
  • $\begingroup$ stats.stackexchange.com/… $\endgroup$
    – whuber
    Apr 16 at 15:34

1 Answer 1


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 $-1$.


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