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1

Johnstone and Silverman (2005) claimed that for large x

$\frac{1-\Phi(x)}{\phi(x)} \approx \frac{1}{x}$

where $\Phi(x)$ and $\phi(x)$ are the CDF and PDF for a normal random variable.

I was able to verify the claim numerically. Q: But how would I show this analytically? This seems like it should be easy, but I can't figure it out. Also, Q: Is there a symbolic logic system (e.g., Mathematica) that can generate these sort of approximations?

flag	asked Sep 5 2011 at 0:30 brianjd 53●4

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Hint: integration by parts in the integral for the CDF. This is a very standard trick for finding asymptotic approximations for many kinds of integrals. – Zen Harper Sep 5 2011 at 1:30

@Zen: Ahh, nice trick. Then you generate a power series in $1/x$---the one given by @Robert below. – brianjd Sep 5 2011 at 18:40

Deane Yang · Accepted Answer · 2011-09-05 02:52:53Z UTC

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If you interpret this as the existence of the limit $$ \lim_{x \rightarrow \infty} \frac{x(1-\Phi(x))}{\phi(x)} $$ then it is easy to verify using l'Hopital's rule.

answered Sep 5 2011 at 2:52

Deane Yang
12.6k●1●33●80

@Deane: Ahh yes. Easy. I suspected. :( Second derivatives of numerator and denominator do the trick. – brianjd Sep 5 2011 at 18:19

Robert Israel · Answer 2 · 2011-09-05 01:19:21Z UTC

In Maple:

with(Statistics): Phi:= CDF(Normal(0,1),x): phi:= PDF(Normal(0,1),x): asympt((1-Phi)/phi,x,10);

$\frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} - \frac{15}{x^7} + \frac{105}{x^9} + O\left(\frac{1}{x^{11}}\right)$

See also http://oeis.org/A001147 for the sequence of coefficients

Yuri Bakhtin · Answer 3 · 2011-09-05 20:31:39Z UTC

Reproducing a lemma from the classic Feller book, first we can write

$$ (1-3x^{-4})\phi(x)<\phi(x)<(1+x^{-2})\phi(x). $$

Integrating this from $x$ to $+\infty$, we obtain

$$(x^{-1}-x^{-3})\phi(x)<1-\Phi(x)< x^{-1}\phi(x),$$

so you easily get an approximation rate $x^{-3}\phi(x)$, too.

Brendan McKay · Answer 4 · 2011-09-05 02:17:43Z UTC

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If $Y(x)=(1-\Phi(x))/\phi(x)$, it is easy to check that $Y'(x)=xY(x)-1$ and from this anything you like follows by standard methods.

answered Sep 5 2011 at 2:17

Brendan McKay
9,120●1●13●33

@Brendan. Then by L'hospital's rule on $Y'(x)$ we get that $\lim_{\infty}Y'(x)=0$ so that $\lim_{\infty}xY(x)=1$ or $Y(x) \approx \frac{1}{x}$ as x becomes large. Thx! – brianjd Sep 5 2011 at 19:01

Approximation to the ratio of a Gaussian CDF to PDF

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