6
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Zen Harper
    Sep 5, 2011 at 1:30
  • $\begingroup$ @Zen: Ahh, nice trick. Then you generate a power series in $1/x$---the one given by @Robert below. $\endgroup$
    – lowndrul
    Sep 5, 2011 at 18:40

5 Answers 5

7
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ @Deane: Ahh yes. Easy. I suspected. :( Second derivatives of numerator and denominator do the trick. $\endgroup$
    – lowndrul
    Sep 5, 2011 at 18:19
6
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ @Robert. VERY useful. I must get Maple (or Mathematica). Thx! $\endgroup$
    – lowndrul
    Sep 5, 2011 at 19:03
  • $\begingroup$ That's a surprisingly simple series expansion. $1*3*5*...*(2n-1)$ $\endgroup$ May 14, 2016 at 16:57
4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @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! $\endgroup$
    – lowndrul
    Sep 5, 2011 at 19:01
4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Something is wrong here - for example the first line is a little too trivial. Could you please provide the exact reference? $\endgroup$
    – domotorp
    Feb 12, 2014 at 22:46
  • $\begingroup$ @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. $\endgroup$ Feb 13, 2014 at 4:33
  • $\begingroup$ 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. $\endgroup$ Feb 13, 2014 at 4:48
  • $\begingroup$ 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... $\endgroup$
    – domotorp
    Feb 13, 2014 at 5:47
  • $\begingroup$ @domotorp: W.Feller. An Introduction to Probability Theory and its Applications, Vol I and II $\endgroup$ Feb 13, 2014 at 18:23
3
$\begingroup$

The stated result can be easily obtained by successive applications of integration by parts. We know that $$ 1 - \Phi(x) = \Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} e^{-t^2/2}dt. $$ To apply integration by parts, multiply and divide the integrand by $\frac{1}{t}$ to obtain $$ \Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \frac{1}{t} \left(t e^{-t^2/2}\right)dt $$ The terms shown in brackets can be integrated to obtain $-e^{-t^2/2}$. Hence, the integral simplifies to $$ \Phi^c(x) = \frac{1}{\sqrt{2\pi}} \frac{e^{-x^2/2}}{x} - \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2}. $$ Now we know that

  • $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ and moreover
  • the integral term on the right hand side is always greater than 0, i.e. specifically, $$ \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2} \geq 0. $$

Hence, combining these observations we find that $$ \Phi^c(x) \leq \frac{\phi(x)}{x} $$ or in other words, $$ \frac{1 - \Phi(x)}{\phi(x)} \leq \frac{1}{x} $$

Now, you could continue to apply integration by parts again on $\Phi^c(x)$ to obtain a lower bound. More specifically, we obtain $$ \Phi^c(x) = \phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) + \int_{x}^{\infty} 3t^{-4} e^{-t^2/2}dt. $$ Again, the integral on the right hand side is positive and hence, $$ \phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) \leq \Phi^c(x). $$ Combining both these inequalities we find that $\frac{1}{x} - o(x^{-1}) \leq \frac{\Phi^c(x)}{\phi(x)} \leq \frac{1}{x}$. In other words, $$ \frac{\Phi^c(x)}{\phi(x)} \approx \frac{1}{x} $$ in the asymptotic limit of $x$.

It turns out that more applications of integration by parts leads to an alternating series which is enveloping. More details can be taken from the textbook by Christopher G Small, Expansions and asymptotics for statistics, Monographs on Statistics and Applied Probability 115. Boca Raton, FL: CRC Press, ISBN 978-1-58488-590-0/hbk; 978-1-4200-1102-9/ebook, pp. xiv+343 (2010), MR2681183, Zbl 1196.62002, which should be available online.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.