Approximation to the ratio of a Gaussian CDF to PDF - MathOverflow

Approximation to the ratio of a Gaussian CDF to PDF

2011-09-05T00:30:07Z

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?

Answer by Robert Israel for Approximation to the ratio of a Gaussian CDF to PDF

2011-09-05T01:19:21Z

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

Answer by Brendan McKay for Approximation to the ratio of a Gaussian CDF to PDF

2011-09-05T02:17:43Z

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.

Answer by Deane Yang for Approximation to the ratio of a Gaussian CDF to PDF

2011-09-05T02:52:53Z

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.

Answer by Yuri Bakhtin for Approximation to the ratio of a Gaussian CDF to PDF

2011-09-05T20:31:39Z

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.