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Suppose I have a sample $(X_1 \dots X_n)$ and $(Y_1 \dots Y_n)$, all of which are $N(0,1)$ random variables. I am interested in the asymptotic behaviour of

$$\frac{1}{n} \sum_{i=0}^n X_{(i)}Y_{(i)} $$

Intuitively this converges to 1. But how do you prove this? It's easy to prove for when $X$ , $Y$ are uniform, but I'm not sure how to handle this case (or the general case to show that $Cov(X,Y)$ converges to $Var(X)$ when they have the same distribution, if indeed that is true).

kjetil b halvorsen
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user194513
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    Assuming all $2n$ variables are independent (and some such assumption is essential for this question to be answerable), the exact distribution of this expression is derived at https://stats.stackexchange.com/questions/85916. It's then trivial to demonstrate the convergence to $0$ (not to $1$!) by observing its mean is $0$ and its variance shrinks to $0$. This also demonstrates that the covariance generally does *not* converge to the variance. – whuber Feb 07 '18 at 22:34
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    I believe what is intended is $$\frac{1}{n}\sum_{i=1}^n X_{(i)}Y_{(i)},$$ which does seem to converge to 1 (and seems much harder to prove because it is not a straight LLN application). The difficulty seems to stem from the dependence between the terms in the sum, as well as the fact that the distribution of each term changes as n gets big. – Aaron P Feb 08 '18 at 14:08
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    @Aaron That's a perceptive comment and likely is the intended interpretation. One way to prove convergence to $1$ would be to relate it to the correlation coefficient of the paired order statistics and that in turn can be bounded below by a function of the Kolmogorov-Smirnov statistic, which converges to zero (whence the function converges to 1). But upon suitable standardization, this expression must have some nondegenerate limiting distribution. It's not immediately clear to me that it will be Normal. An analysis of this question would be interesting. – whuber Feb 08 '18 at 14:39
  • Upon further thought, it's clear the answer depends on the multivariate distribution, not just the marginals. As an example, let $Z$ be a standard Normal variable and define $X_n=Y_n=Z$ for all $n\ge 0.$ This satisfies all the conditions in the question, but now the mean shown in the question equals $Z^2(n+1)/n$ for any $n$ and that is a multiple of a $\chi^2(1)$ distribution. – whuber Apr 28 '20 at 11:59

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This looks like you may be able to get the answer from some kind of nonparametric rank/permutation-based hypothesis testing result. I feel this way mainly because suitably rescaled, your quantity would converge to some non-degenerate distribution, and then this distribution would be usable for hypothesis testing.

Aaron P
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