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This question already has an answer here:

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!!

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marked as duplicate by Dilip Sarwate, Nate Eldredge, Namaste, Davide Giraudo, user63181 May 22 '14 at 22:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Duplicate of math.stackexchange.com/q/28751/15941 $\endgroup$ – Dilip Sarwate May 22 '14 at 18:26
  • $\begingroup$ See this answer to the question mentioned above where it is shown via integration by parts that for $x > 0$, $$\phi(x) \left (\frac{1}{x} - \frac{1}{x^3}\right ) < 1-\Phi(x) < \frac{\phi(x)}{x}.$$ The upper and lower bounds approach each other arbitrarily closely as $x \to \infty$. $\endgroup$ – Dilip Sarwate May 22 '14 at 18:34
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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 $$\frac{\int_b^\infty e^{-x^2/2}\,dx}{\frac{e^{-b^2/2}}{b}}=1+O(1/b).$$

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