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$\left\{(\mathbf{X_n}^T\mathbf{X_n}/n)^{-1}\right\}_{n=1}^\infty$

Let $\mathbf{X}_n$ be the usual data matrix in standard multiple regression where I have used the subscript $n$ to indicate the number of observations: $$ \mathbf{X}_n = \begin{bmatrix} \mathbf{x}_1^T \\ \mathbf{x}_2^T \\ \vdots \\ \mathbf{x}_n^T \\ \end{bmatrix} = \begin{bmatrix} 1 & x_{11} & \dots & x_{1P} \\ 1 & x_{21} & \dots & x_{2P} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_{n1} & \dots & x_{nP} \\ \end{bmatrix} $$

Under the standard assumptions of multiple linear regression, the conditional variance of the estimated least squares regression coefficients $\hat \beta$ is $$ \text{Var}[\hat \beta| \mathbf{X_n}] = \sigma^2(\mathbf{X_n}^T \mathbf{X_n})^{-1}, $$ where $\sigma^2$ is the variance of the error.

One of the assumptions of linear regression for large samples is $$ \text{plim}_{n\to \infty }\bigg(\frac{\mathbf{X_n}^T \mathbf{X_n}}{n}\bigg)^{-1} = Q, $$ a positive definite matrix.

Define the sequence of random matrices $\{Y_n\}_{n=1}^\infty$ where $$ Y_n := \bigg(\frac{\mathbf{X_n}^T \mathbf{X_n}}{n}\bigg)^{-1}. $$

I am interested in the uniform integrability (UI) of the elements of $Y_n$. Namely because if they are uniformly integrable, then we can say the elements of $Y_n$ converge in expectation to the elements of $Q$, since convergence in probability + UI implies convergence in expectation.

Is it possible that $\{Y_n\}_{n=1}^\infty$ is automatically uniformly integrable due to the fact that the observations $\mathbf{x}_i$ used to create the data matrix are i.i.d.?

If not, what conditions should we impose on the (elements of the) data matrix $\mathbf{X_n}$ so that we obtain uniform integrability for the (elements of the) sequence of matrices $\{Y_n\}_{n=1}^\infty$

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I am not sure if it is possible to give a thorough answer here (at least for me) given the breadth of the possible issues (dependence, heterogeneity, existence of moments etc.).

That said, here are some thoughts:

Consider the case $P=1$ and no constant for simplicity. Then, letting $Y_n:=(Z_n)^{-1}$, where $$ Z_n=\frac{\sum_{i=1}^nx_i^2}{n}. $$ We have $$ E(Z_n)=E\left(\frac{\sum_{i=1}^nx_i^2}{n}\right)=\frac{\sum_{i=1}^nE(x_i^2)}{n}. $$ If the $x_i$ are iid, this simplifies to $E(x_i^2)$.

Uniform integrability is usually invoked to shut down counterexamples like this one: if $x_i=i$ w.p. $1/i$ and zero w.p. $1-1/i$, we have $x_i\to_p0$ but $E(x_i)=1$ for any $i$, so expectations do not converge (to the plim). Also, $E(x_i^2)=i$, which diverges. Then, $$ E(Z_n)=\frac{n(n+1)}{2n}\to\infty $$ This, however, is evidently not an i.i.d. setup.

Davidson, Stochastic Limit Theory, Theorem 12.11, for example provides sufficient conditions for uniform integrability, such as uniformly bounded moments. That is $\{X_t\}$ is uniformly integrable if $E|X_t|^{1+\theta}<\infty$ for $\theta>0$.

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  • $\begingroup$ Interesting post. Yes I noticed that the examples for when convergence in probability doesn't match convergence in expectation seem to always make a random variable $X_n$ under consideration heavily dependent on the index $n$ of the sequence (such as the image on the Wikipedia page. But for 'conventional' multiple regression in the real world an observation isn't dependent on the particular index $n$ of the sequence of observations. $\endgroup$ – sonicboom Dec 10 '20 at 17:24
  • $\begingroup$ For example if we regress heights against weights for all people in a particular city on a particular day. Then the rows of the data matrix $\mathbf{X}_n$ will feature observations of individual people's weights, and these will not be dependent on $n$. They are simply i.i.d observations. Its for this type of reason that I wonder if the $\{Y_n\}_{n=1}^\infty$ in my original question could be automatically uniformly integrable due to the i.i.d. assumption. $\endgroup$ – sonicboom Dec 10 '20 at 17:28
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    $\begingroup$ Indeed, as I was trying to express (not very well, apparently), I could not come up with an issue when there is iid-ness, and looking at the definition of uniform integrability suggests that it is relevant basically when there indeed is heterogeneity. $\endgroup$ – Christoph Hanck Dec 10 '20 at 18:17
  • $\begingroup$ Section 2 of this paper seems to imply the OLS estimators are uniformly integrable which I assume would mean $\left\{(\mathbf{X_n}^T\mathbf{X_n}/n)^{-1}\right\}_{n=1}^\infty$ is UI since $(\mathbf{X_n}^T\mathbf{X_n})^{-1}$ is used in the construction of the OLS estimators. $\endgroup$ – sonicboom Dec 13 '20 at 16:29
  • $\begingroup$ Actually I think that paper uses fixed regressors not stochastic regressors so it does not correspond to the case I am asking about. $\endgroup$ – sonicboom Mar 15 at 10:04

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