# 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 $m\rightarrow\infty$.

• Since the empirical distribution function converges uniformly with probability 1 to the underlying distribution, it follows that the expectations of the order statistics converge to the true quantiles of the distribution. en.wikipedia.org/wiki/Empirical_distribution_function helps out here. – jbowman Mar 22 '18 at 20:06
• @jbowman: I do not understand how what you are saying relates to my question. I am searching for an asymptotic formula for $e(r:n)$ as a function of $n$. – Hans Mar 22 '18 at 20:17
• Did you try looking at this question stats.stackexchange.com/questions/9001/…, which seems essentially identical? ... also, you should explicitly ask for a formula that's a function of $n$, just to clarify, otherwise people like me might interpret "asymptotics" as meaning "what does it converge to and how (in probability...)". – jbowman Mar 22 '18 at 20:29
• @jbowman: That link you cited is in my question in the first place to explain the motivation of the question. I do not see how the asymptotics could be confused when there is no probability mentioned. Probabilistic convergence is but one special subset of convergence. – Hans Mar 22 '18 at 20:36
• @whuber: By "simple" and "elementary", I mean some algebraic or "simple" transcendental functions like exponential, logrithmic and trigonometric functions. I can add these descriptions in the question. – Hans Mar 22 '18 at 22:47

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.