Asymptotics for the expectation of the standard normal order statistics What is an asymptotics of the expectation of the order statistics of the standard normal distribution 
$$e(r:n) \approx \Phi^{-1}\Big(\frac{r-\alpha}{n-2\alpha+1}\Big)$$
as $n\rightarrow\infty$?
By asymptotics, it is usually understood that the sought-after function is a "simpler" and more "elementary" function, usually either algebraic or transcendental functions like exponential, logarithmic, or trigonometric function, than the one in question. 
It is obvious that an asymptotics of this question has to approach $-\infty$ slower than $-\sqrt{\ln m}$ (here this upper bound is the composite of the "elementary" functions of logarithm and square root) where $m:=\frac{n-2\alpha+1}{r-\alpha}$ as $m\rightarrow\infty$.
 A: For convenience sake, we make everything positive. So we look at the $r$'th largest random variable. Instead of $\Phi$ we look at $1-\Phi$
$$1-\Phi(x)=\frac1{\sqrt{2\pi}}\int_x^\infty e^{-\frac{t^2}2}\,dt=e^{-\frac{x^2}2}\frac1x-\int_x^\infty \frac{e^{-\frac{t^2}2}}{t^3}\,d\Big(\frac{t^2}2\Big) \tag1$$
as $x\rightarrow\infty$ by integration by parts.
$$\int_x^\infty \frac{e^{-\frac{t^2}2}}{t^3}\,d\Big(\frac{t^2}2\Big)>\frac{e^{-\frac{x^2}2}}{x^3},$$
so
$$\frac1x\Big(1-\frac1{x^2}\Big)<e^{\frac{x^2}2}(1-\Phi(x))<\frac1x \tag2$$
as $\delta>\frac1{x^2}$ for some small positive $\delta$ and large enough $y$ and thus $x$. We have 
$$\frac1x(1-\delta)<e^{\frac{x^2}2}(1-\Phi(x))<\frac1x \tag3$$
For $\frac1y=1-\Phi(x)$ and large $y$, from Equation (1), we know $x>1$. 
Take logarithm of (3). 
$$\sqrt{2(\ln y-\ln x+\ln(1-\delta))}<x<\sqrt{2(\ln y-\ln x)}<\sqrt{2\ln y}. \tag4$$
Taking logarithm on the inequality to the right of $x$ gives
$$\ln x < \frac12(\ln\ln y+\ln 2)$$
Substitute the above back into the inequality left to $x$ of (4), 
$$\sqrt{2\ln y-\ln\ln y+2\ln\frac{1-\delta}{\sqrt2}}<x<\sqrt{2\ln y}$$
Then substitute the left inequality above back into the right inequality of (4). 
We can continue this alternating substitution of the inequalities ad infinitum.

We can also resort to using the Lambert W function and give a more direct answer.
