Inner product of sorted Gaussian vector Suppose $X_1,\ldots,X_n$ are i.i.d. standard normal. I'm wondering how to analyze the following quantity:
$$\left|\frac{X_{(1)}X_{(n)}+X_{(2)}X_{(n-1)}+\cdots+X_{(n)}X_{(1)}}{n}\right|$$
where $X_{(1)}<X_{(2)}<\cdots<X_{(n)}$ is the order statistics. I suppose it should approach 1 as $n\to\infty$ but do not know how to justify my guess. 
 A: $\newcommand{\D}{\overset{\text{D}}=}$
Without loss of generality (wlog), $X_i=\Phi^{-1}(U_i)$ and hence $X_{(i)}=\Phi^{-1}(U_{(i)})$ for $i=1,\dots,n$, where $\Phi$ is the standard normal cdf and the $U_i$'s are iid random variables each uniformly distributed on $[0,1]$. In turn, wlog 
\begin{equation}
 U_{(i)}=\frac{S_i}{S_n},
\end{equation}
where $S_i:=Y_1+\dots+Y_i$ and the $Y_i$'s are iid random variables each with the standard exponential distribution. So, 
\begin{equation}
 R_n:=\frac{X_{(1)}X_{(n)}+X_{(2)}X_{(n-1)}+\cdots+X_{(n)}X_{(1)}}{n}
 \D\sum_{i=1}^n\Phi^{-1}\Big(\frac{S_i}{S_n}\Big)\Phi^{-1}\Big(\frac{S_{n+1-i}}{S_n}\Big)\frac1n=:J_n, 
\end{equation}
where $\D$ denotes the equality in distribution. 
By the law of large numbers, $\frac{S_i}{S_n}\sim\frac in$ almost surely (a.s.) if $n\to\infty$ and $\frac in$ is bounded away from $0$. So, it is plausible that 
\begin{equation}
 J_n\to J:=\int_0^1\Phi^{-1}(u)\Phi^{-1}(1-u)\,du. 
\end{equation}
Using the symmetry and the change $u=\Phi(x)$ of variables,
we have 
\begin{equation}
 J=-\int_0^1\Phi^{-1}(u)^2\,du=-\int_{-\infty}^\infty x^2\,d\Phi(x)=-1. 
\end{equation}
Thus, once "plausible" is replaced here by "proved", your guess that $|R_n|\to1$ (say, in probability) will be confirmed. 
