Approximation to the ratio of a Gaussian CDF to PDF

Question

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 1

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 2

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 3

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 4

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.

Answer 5

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.