Let $(X_1,\dots,X_n)$ be independent random variables with common distribution $\mathcal{N}(0,1)$. The order statistics satisfy $X_{(1)} \leq X_{(2)} \leq \dots \leq X_{(n)}$. I am interested in the estimation of $\mathbb{E} X_{(k)}$ where $1\leq k \leq n$. It is easily seen that \begin{align*} \mathbb{E} X_{(k)} = \mathbb{E} \Phi^{-1}(U_{(k)}) \end{align*} where $U_{(k)}$ is the $k$-th order statistic of a uniform $n$-sample over the segment $[0,1]$ and $\Phi$ is the c.d.f. of $\mathcal{N}(0,1)$. A taylor expansion of $\Phi^{-1}$ gives: \begin{align*} \Phi^{-1}(x) = \Phi^{-1}(a) + \frac{1}{\Phi'\circ\Phi^{-1}(a)}(x-a) + R(x) \end{align*} where $R(x)$ is a remainder that satisfies $R(x) = o_{x \rightarrow a}(x-a)$. It is well known that $U_{(k)}$ has the Beta$(k,n-k+1)$ distribution, so letting $a = \mathbb{E} U_{(k)} = k/(n+1)$ we have: \begin{align*} \mathbb{E} X_{(k)} &= \mathbb{E} \Phi^{-1}(U_{(k)}) \\ & = \Phi^{-1}\left( \frac{k}{n+1} \right) + \mathbb{E} R(U_{(k)}). \end{align*} Because $U_{(k)} \approx k/(n+1)$, we should expect $\mathbb{E} R(U_{(k)})$ to be small. I am struggeling to derive an upper bound on:
\begin{align*} \left| \mathbb{E} X_{(k)} - \Phi^{-1}\left( \frac{k}{n+1} \right) \right| \end{align*}
Some people have asked related questions concerning the approximation of expected order statistics, but unfortunately I did not fin any discussion on the theoretical precision of those approximations.
