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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?

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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 '11 at 1:30
@Zen: Ahh, nice trick. Then you generate a power series in $1/x$---the one given by @Robert below. – lowndrul Sep 5 '11 at 18:40
up vote 4 down vote accepted

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.

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@Deane: Ahh yes. Easy. I suspected. :( Second derivatives of numerator and denominator do the trick. – lowndrul Sep 5 '11 at 18:19

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 for the sequence of coefficients

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@Robert. VERY useful. I must get Maple (or Mathematica). Thx! – lowndrul Sep 5 '11 at 19:03
That's a surprisingly simple series expansion. $1*3*5*...*(2n-1)$ – Thomas Dybdahl Ahle May 14 at 16:57

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.

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@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! – lowndrul Sep 5 '11 at 19:01

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.

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Something is wrong here - for example the first line is a little too trivial. Could you please provide the exact reference? – domotorp Feb 12 '14 at 22:46
@domotorp: I don't happen to have Feller's book anywhere around, but I do not see what seems to be a problem. Yes, the first line of inequalities is trivial, but it is only a differential version of the second one. – Yuri Bakhtin Feb 13 '14 at 4:33
The coefficients on the l.h.s. are chosen so that a certain cancellation occurs when you differentiate the l.h.s. of the second chain of inequalities. – Yuri Bakhtin Feb 13 '14 at 4:48
I see, nice trick! What I meant was - what is "Feller's book"? For someone not knowing much about the topic, this does not reveal much... – domotorp Feb 13 '14 at 5:47
@domotorp: W.Feller. An Introduction to Probability Theory and its Applications, Vol I and II – Yuri Bakhtin Feb 13 '14 at 18:23

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