# 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 $$X_{(1)} is the order statistics. I suppose it should approach 1 as $$n\to\infty$$ but do not know how to justify my guess.

• I think the polarization identity $X_{(1)}X_{(n)}+X_{(n)}X_{(1)} = \left(X_{(1)}+X_{(n)}\right)^2 - X_{(1)}^2-X_{(2)}^2$ might be helpful. – Lior Silberman Feb 3 '19 at 3:18

$$\newcommand{\D}{\overset{\text{D}}=}$$ 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 $$$$U_{(i)}=\frac{S_i}{S_n},$$$$ where $$S_i:=Y_1+\dots+Y_i$$ and the $$Y_i$$'s are iid random variables each with the standard exponential distribution. So, $$$$R_n:=\frac{X_{(1)}X_{(n)}+X_{(2)}X_{(n-1)}+\cdots+X_{(n)}X_{(1)}}{n} \D\sum_{i=1}^n\Phi^{-1}\Big(\frac{S_i}{S_n}\Big)\Phi^{-1}\Big(\frac{S_{n+1-i}}{S_n}\Big)\frac1n=:J_n,$$$$ where $$\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$$. So, it is plausible that $$$$J_n\to J:=\int_0^1\Phi^{-1}(u)\Phi^{-1}(1-u)\,du.$$$$ Using the symmetry and the change $$u=\Phi(x)$$ of variables, we have $$$$J=-\int_0^1\Phi^{-1}(u)^2\,du=-\int_{-\infty}^\infty x^2\,d\Phi(x)=-1.$$$$ Thus, once "plausible" is replaced here by "proved", your guess that $$|R_n|\to1$$ (say, in probability) will be confirmed.