# Covariance of order statistics convergence?

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).

• 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 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 7 '18 at 22:34
• 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 8 '18 at 14:08
• @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 8 '18 at 14:39