What does $\lim_{n \to \infty} \frac{X'X}{n}=Q$ imply?

Question

Suppose $X$ is a $n \times m$ matrix with elements $x_{ij}$ and $$\lim_{n \to \infty} \frac{X'X}{n}=Q,$$ where $Q$ is positive definite. Let $x_i$ be the $i$-th row vector of $X$. For fixed $m$, a conjecture is $$\max_{i=1, ..., n} x_i'(X'X)^{-1}x_i =0, \text{ as } n \to \infty.$$

For $m=1$, it can be proved. But how to prove or disprove in a general case? This might be related here

Answer 1

Upon reconsideration of your question, I believe that you are trying to ask the following.

Fix elements $x_{ij}$ for $i = 1,\dots,m$ and $j = 1,2,3,\dots$. Let $X_n$ denote the $n \times n$ matrix with entries $x_{ij}$. If $$\lim_{n \to \infty} \frac{X_n'X_n}{n}=Q,$$ where $Q$ is positive definite, is it true that for every column $x_i$ of $X$, we have $$ \max_{i=1, ..., n} x_i'(X_n'X_n)^{-1}x_i =0, \text{ as } k \to \infty? $$

If this is what you meant, please say so.

For this version of the question, the conjecture is correct. In particular, $$ \lim_{n \to \infty} x_i'(X_n'X_n)^{-1}x_i = \lim_{n \to \infty} \frac 1n x_i'\left(\frac{X_n'X_n}{n}\right)^{-1}x_i= \lim_{n \to \infty} \frac 1n x_i'Qx_i = 0. $$

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I assume that what you're trying to say is something like this:

Suppose that for $k > 0$, $X(k)$ is a $n \times m$ matrix with elements $x_{ij}$ and $$\lim_{k \to \infty} \frac{X'(k)X(k)}{k}=Q,$$ where $Q$ is positive definite. A conjecture is $$\max_{i=1, ..., n} x_i'(X'(k)X(k))^{-1}x_i =0, \text{ as } k \to \infty.$$

The answer to this question is no. For example, take $X(k) = \sqrt{k}I$. Then $\max_{i=1,\dots,n}x_i'(X'(k)X(k))^{-1}x_i = 1$ for all $k$.

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For an more general example (with a non-square $X$), take $X(k) = \pmatrix{\sqrt{k}I_n & 0_{n \times (m-n)}}$. Again, we find that $\max_{i=1,\dots,n}x_i'(X'(k)X(k))^{-1}x_i = 1$ for all $k$.