Do nice properties of MLE still hold in Classical Linear Regression Model?

Question

For simplicity, let's assume that we have the true DGP

$$y_i= \beta_0 + \beta_1x_i + \epsilon_i$$ where $\epsilon_i \sim N(0, \sigma^2)$ $(i = 1,2,...,n)$

Assume that these following usual assumptions meet (for the classical linear regression model)

1. $y_i= \beta_0 + \beta_1x_i + \epsilon_i$ (linearity)
2. $x_i$ is deterministic
3. $\epsilon_i \sim N(0, \sigma^2)$ and $Cov[\epsilon_i, \epsilon_j] = 0$ for $i \neq j$ (normality, homoskedasticity and uncorrelatedness)

If we use Maximum Likelihood Estimation Method (MLE) to estimate $\beta_0$ and $\beta_1$, is it correct to state that the nice asymptotic properties as n approaches infinity (asymptotic efficiency, asymptotic consistency, ...) still hold for the MLE estimators of $\beta_0$ and $\beta_1$?

I believe that the answer is no because of the second assumption and, for example, $MSE[\hat{\beta}_{1MLE/OLS}] = \mathbb{V}[\hat{\beta}_{1MLE/OLS}] = \frac{\sigma^2}{Dev(x)}$ which implies there is no guarantee that $MSE[\hat{\beta}_{1MLE/OLS}]$ will approach 0 as n approaches infinity (sufficient but not necessary condition for convergence in probability), but I want to know whether or not it is correct.

Answer 1

Requested in comments

With simple linear regression, the MLE estimators the same as those from minimising the sum of squares of the residuals.

It depends on what additional $x_i$ you include as $n$ increases. The variance of $\hat \beta_1$ is $\frac{\sigma^2}{\sum(x_i-\bar x)^2}$ which tends towards $0$ so long as the denominator increases without bound and you do not concentrate the additional $x_i$ around the existing $\bar x$.

So for example you are correct to worry if $x_1=1$ and $x_2=-1$ and all the other $x_i=0$: you will never mitigate the impact of $\epsilon_1$ and $\epsilon_2$ on $\hat \beta_1$.

But this is not realistic: the point of expanding the sample for OLS is to see the impact of different values of $x_i$ on $y_i$.