Rates of convergence of an OLS estimator

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I have a linear regression model $$ y_t=x_t\beta+e_t,\quad t=1,\ldots,N. $$ Here $x_t$ is non-random and given by $(1,\delta_t t)$ where $\delta_t$ is 1 for odd $t$ and $0$ otherwise. Moreover, $e_t$ is i.i.d. with mean $0$ and variance $\sigma^2$.

I'm supposed to find the rates of convergence (as sample size $\to\infty$) for the OLS estimator $b$ and the estimated variance of $b$.

Attempt: we care about large-sample properties so let's assume for convenience that $N=2k+1$. This allows us to construct the matrices of regressors and error terms $$ x=\begin{pmatrix} 1&1\\ 1&0\\ 1&3\\ \vdots&\vdots\\ 1&0\\ 1&2k+1 \end{pmatrix},\quad e=\begin{pmatrix}e_1\\\vdots\\e_N\end{pmatrix}. $$ Then, $$ b-\beta=(x'x)^{-1}x'e=% \left[\begin{pmatrix} 2k+1&(k+1)^2\\ (k+1)^2&\frac{(k+1)(2k+1)(2k+3)}{3} \end{pmatrix}\right]^{-1} \sum_{t=1}^Nx_t'e_t \\= \left[\frac{1}{N}\begin{pmatrix} 2k+1&(k+1)^2\\ (k+1)^2&\frac{(k+1)(2k+1)(2k+3)}{3} \end{pmatrix}\right]^{-1} \frac{1}{N}\sum_{t=1}^Nx_t'e_t\\ =\left[\frac{1}{2k+1}\begin{pmatrix} 2k+1&(k+1)^2\\ (k+1)^2&\frac{(k+1)(2k+1)(2k+3)}{3} \end{pmatrix}\right]^{-1} \frac{1}{N}\sum_{t=1}^Nx_t'e_t. $$ But how do I now proceed further please?

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$\newcommand{\var}{\operatorname{var}}$ \begin{align} \var(b) = \var(\ \underbrace{(x'x)^{-1}x'}_{\begin{smallmatrix} \text{This is constant,} \\ \text{i.e. not random.} \end{smallmatrix}}\ \ y) & = (x'x)^{-1} x' \Big( \var(y)\Big) x (x'x)^{-1} \\[8pt] & = (x'x)^{-1} x' \Big( \sigma^2 I_{2\times 2}\Big) x (x'x)^{-1} \\[8pt] & = \sigma^2 (x'x)^{-1} x'x (x'x)^{-1} \\[8pt] & = \sigma^2 (x'x)^{-1} \in \mathbb R^{2\times 2}. \end{align}

An estimated variance would then require an estimate of $\sigma^2$. Often one uses $$ \frac{\|y - xb\|^2}{n-2} $$ which is unbiased. If you make certain assumptions about normality, homeskedasticity, and independence, then the MLE is the same thing but with $n$ rather than $n-2$ in the denominator.

By "rates of convergence" I'm assuming you mean as $k\to\infty$. Probably one would need to take into account the special form of your $k\times 2$ matrix.