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