Does the variance in the estimates of the coefficients for standard linear regression decrease as $O\bigg(\frac{1}{n}\bigg)$?

Question

I am interested in how the estimate of the regression $\beta_1$ decreases with respect to sample size $n$. Does this look ok or am I missing something? $$ \begin{align} \text{Var}(\hat \beta_1) & = \frac{\sigma^2}{\sum_{i=1}^n (x_i - \bar x)^2} \\ & \le \frac{\sigma^2}{\sum_{i=1}^n \min_i \{(x_i - \bar x)^2\}} \\ & = \frac{\sigma^2}{n \min_i\{(x_i - \bar x)^2\}} \\ & = O\bigg(\frac{1}{n}\bigg). \end{align} $$

Edit: It seems that the first answer in this post seems to agree with my result. However, the first answer in this post seems to say something else as it says that $\sigma^2$ is itself estimated (by the MSE) and that this estimate has a dependence on $n$.

Can someone clarify exactly how the variance decreases with $n$?

Answer 1

The true variance of $\hat\beta_1$ is $\mathcal O(1/n)$, as you correctly proved. The estimated variance of $\beta_1$ is a random variable and its asymptotic behavior is better described using big O probability notation. Since the MSE is a consistent sinais of $\sigma^2$, we have $\hat\sigma^2=\sigma^2+\mathcal o_p(1)$. If we replace $\sigma^2$ by $\hat\sigma^2$ in your proof, we get that the estimated variance of $\beta_1$ is $\mathcal O_p(1/n)$.

TL;DR: the true variance is $\mathcal O(1/n)$, while the estimated variance is $\mathcal O_p(1/n)$ (that is, it is asymptotically bounded by $1/n$ with probability 1).