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My lecture slides sometimes mention asymptotic full rank as an assumption:

$Q=p\lim_{n \rightarrow \infty} \frac{1}{n}X^TX$ has full rank.

It seems obvious (?) that this is implied by full rank; i.e. if $X$ has full rank, then $X^TX$ has full rank and hence $Q$ has full rank.

Then the only reason to use asymptotic full rank would be if it is weaker than full rank. Is this the case? It would be nice to have an example where $Q$ has full rank while $X$ does not have full rank.

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  • $\begingroup$ Your $X$ should also depend on $n$. More importantly , the condition claims that the limit exists, while full rank is just a side-cindition. $\endgroup$ – Zhanxiong Jan 8 '18 at 17:19
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    $\begingroup$ It does not say that at every n it has full rank. $\endgroup$ – Michael R. Chernick Jan 8 '18 at 17:23
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    $\begingroup$ The simplest possible example would be where the observations comprising the rows of the $n\times 1$ matrix $X$ are iid Bernoulli$(p)$ variables, $p\gt 0$. With probability $(1-p)^n$, $X^\prime X/n$ has rank zero, whereas the limit in probability of the rank is $1$ for any possible value of $p$. $\endgroup$ – whuber Jan 8 '18 at 17:31
  • $\begingroup$ @Zhanxiong So if X has full rank, Q can still fail to have full rank, if $p\lim_{n \rightarrow \infty} \frac{1}{n}X^TX$ fails to converge? Is this what you are saying? :) $\endgroup$ – tmkadamcz Jan 8 '18 at 17:40
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    $\begingroup$ Extend @whuber's example to $X$ $n \times 2$ with all entries iid Bernoulli$(p)$, $p>0$. With probability $p^{2n}$ all the observations have value 1, in which case $X$ is nonzero but not of full rank. $\endgroup$ – jbowman Jan 8 '18 at 18:35
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Let $X_n$ be a $n \times K$ matrix, $n\geq K$. If $\text{plim}_{n \rightarrow \infty} \frac{1}{n}X_n^T X_n$ exists, let $Q=\text{plim}_{n \rightarrow \infty} \frac{1}{n}X_n^T X_n$

Neither of these statements is true in general:

  1. for all $n$, if $X_n$ has full rank $K$, $Q$ has full rank $K$.
  2. for all $n$, if $Q$ has full rank $K$, $X_n$ has full rank $K$.

(1) fails because $\frac{1}{n}X_n^TX_n$ may not converge in probability as $n \rightarrow \infty$, hence $Q$ would not be defined.

(2) fails, by the following counter-example: $X$ is an $n\times2$ matrix with all entries iid $Bernoulli(p)$, $p>0$. With probability $p^{2n}$ all the observations have value 1, in which case $X$ has rank 1. Yet $Q=\begin{bmatrix} p & 0 \\ 0 & p \end{bmatrix}$ has full rank 2.

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