I am reading wikipedia's article on asymptotic expansion, and I would like it if someone can help me clarify some doubts with an example, with the inverse of the cumulative density function of the standard normal variable.
Also in wikipedia it is stated that for small $p$: $$\Phi^{-1}(p) \sim -\sqrt{\ln\frac{1}{p^2}-\ln\ln\frac{1}{p^2}-\ln(2\pi)}:=f(p)$$
A more concrete question is: what does this say about the ratio the limit of $\Phi^{-1}(p)/f(p)$ as $p \to 0^+$? Can we say this limit is 1? I understand the concept of $o$ notations, but it would be good if this claim is true and to understand why it is the case. Also, any references on this assymptotic expansion on the normal quantile function would be appreciated. Thanks!
In terms of the inverse error and complementary error functions $$ \Phi ^{ - 1} (p) = \sqrt 2 \operatorname{erf}^{ - 1} (2p - 1) = - \sqrt 2 \operatorname{erf}^{ - 1} (1 - 2p) = - \sqrt 2 \operatorname{erfc}^{ - 1} (2p). $$ Then using the leading term and estimating the remaining ones of the known asymptotic expansion for $\operatorname{erfc}^{ - 1} $, we deduce $$ \Phi ^{ - 1} (p) = - \sqrt { \ln \frac{1}{{p^2 }} - \ln \ln \frac{1}{{4p^2 }} - \ln (2\pi )} + \mathcal{O}\!\left( {\frac{{\ln \left| {\ln p} \right|}}{{\left| {\ln p} \right|^{3/2} }}} \right) $$ as $p\to 0^+$. Now $$ \ln \ln \frac{1}{{4p^2 }} = \ln \ln \frac{1}{{p^2 }} + \mathcal{O}\!\left( {\frac{1}{{\left| {\ln p} \right|}}} \right) $$ as $p\to 0^+$, which gives \begin{align*} \Phi ^{ - 1} (p) &= - \sqrt {\ln \frac{1}{{p^2 }} - \ln \ln \frac{1}{{p^2 }} - \ln (2\pi )} + \mathcal{O}\!\left( {\frac{1}{{\left| {\ln p} \right|^{3/2} }}} \right) + \mathcal{O}\!\left( {\frac{{\ln \left| {\ln p} \right|}}{{\left| {\ln p} \right|^{3/2} }}} \right) \\ & = - \sqrt {\ln \frac{1}{{p^2 }} - \ln \ln \frac{1}{{p^2 }} - \ln (2\pi )} + \mathcal{O}\!\left( {\frac{{\ln \left| {\ln p} \right|}}{{\left| {\ln p} \right|^{3/2} }}} \right) \end{align*} as $p\to 0^+$, which agrees with the one cited in the other answer.