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Setup. Let $\alpha\in(0,1)$ fixed; and $\tau\in[0,1]$ (think of it very close to one).

Suppose $X_1,\dots,X_n$ are i.i.d. standard normal. Let $Y_1,\dots,Y_n$ be another sequence of standard normals with the following properties:

(a) $Y_i$ are i.i.d. $\mathcal{N}(0,1)$, $1\le i\le n$.

(b) $\mathbb{E}[X_iY_i] = \tau$ for $1\le i\le n$.

Consider now their ordered version: $$ X_{i_1}<X_{i_2}<\cdots<X_{i_n}\qquad\text{and}\qquad Y_{j_1}<Y_{j_2}<\cdots<Y_{j_n}; $$ and let us look at the first $M\triangleq \lfloor \alpha n\rfloor$ of them. That is define the sets of indices: $$ S_X = \{i_1,\dots,i_M\}\qquad\text{and}\qquad S_Y = \{j_1,\dots,j_M\}. $$ Question. What can we say about the random variable $|S_X\cap S_Y|$?

Obviously, $|S_X\cap S_Y|=M$ when $\tau=1$, whereas a PMF (albeit somewhat messy) can be found for $\tau=0$.

Essentially, I am interested in showing something like this: if $\tau=1-o_n(1)$, $|S_X\cap S_Y| = M(1-o_n(1))$ with high probability as $n\to\infty$.

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  • $\begingroup$ Do you assume that the pairs $(X_i,Y_i)$ are independent? If so, I think you need $1-\tau=o(1/n^2)$ to get the desired conclusion. $\endgroup$ May 26, 2021 at 21:46
  • $\begingroup$ Yes, the pairs are independent, though the $\tau$ I typically consider are of form $1-o(1/n)$. Still very curious to see your argument though Iosif. $\endgroup$
    – ttteessttt
    May 26, 2021 at 22:17
  • $\begingroup$ I only have certain heuristics. $\endgroup$ May 27, 2021 at 0:37

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