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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.

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  • $\begingroup$ 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. $\endgroup$ – Lior Silberman Feb 3 '19 at 3:18
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$\newcommand{\D}{\overset{\text{D}}=}$ 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.

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  • $\begingroup$ The "plausible" part seems to be crucial. How can we prove that? $\endgroup$ – neverevernever Feb 4 '19 at 1:41
  • $\begingroup$ I think I can do it, will try to find time for this. Can you let me know how this problem arises? $\endgroup$ – Iosif Pinelis Feb 4 '19 at 4:32
  • $\begingroup$ Sure. I'm considering how permuting coordinates of a random vector changes the direction of the original vector. So I start from the worst case, i.e. the permutation that matches the smallest to the largest, etc. $\endgroup$ – neverevernever Feb 4 '19 at 18:43

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