# Sampling distribution of ordinary least squares confusion

I was reviewing the derivation for the variance of ordinary least squares estimators and experienced some confusion.

$$\Large Var(\hat{\beta}) = \frac{\sigma^2}{\Sigma^n_{i=1}(X_i-\bar{X})^2}$$

From this article, there was this formula, which said in an OLS model where $$y_i = \alpha + \beta x_i + \epsilon$$, and $$\epsilon \sim \mathcal{N}(0,\,\sigma^{2})$$, this is a formula that gives the variance of the estimator of $$\beta$$.

I'm confused what is the sampling distribution that $$\hat{\beta}$$ comes from? All possible $$(x_i, y_i)$$ points? Just the $$(x_i, y_i)$$ points given in the sample dataset? I'm confused on what this variance estimator really means. Any clarity would be helpful.

The usual regression model conditions on the $$x$$'s. They stay the same.
$$Y_i$$ (not the observed $$y_i$$ but the random variable that $$y_i$$ is a realization of) has a distribution that depends on their specific $$x_i$$. If you and I were both to independently draw a sample with those same $$x$$'s we'd get different observations, and we'd get different $$\hat{\beta}$$ values. We'd then each have one draw from the sampling distribution of $$\hat{\beta}$$, some estimated coefficient $$b$$. We could call your realized coefficient $$b^{[1]}$$, and mine $$b^{[2]}$$ and so forth for any number of them. They're all draws from the conditional distribution (conditional on the x's) of $$\hat{\beta}$$.
The distribution of $$\hat{\beta}$$ follows from the joint conditional distribution of the $$Y_i$$ (the joint distribution conditional on the $$x$$s).
In the usual model, the vector $$\mathbf{Y} = (Y_1,Y_2,...,Y_n)^\top\sim N_n(\mu,\Sigma)$$ where $$\mu=(\alpha+\beta x_1,\alpha+\beta x_2,...,\alpha+\beta x_n)^\top$$ and $$\Sigma = \sigma^2 I_n$$.
The usual least squares estimator of $$(\alpha,\beta)^\top$$ can be written down and hence under the model assumptions, the joint distribution of $$(\hat{\alpha},\hat{\beta})^\top$$ and the marginal distributions of either obtained. Indeed because $$(\hat{\alpha},\hat{\beta})^\top$$ is linear in the Y's, they're bivariate normal and so the marginal distributions are also normal. It's a few lines of algebra to show that $$E(\hat{\beta})=\beta$$ and $$\text{Var}(\hat{\beta})$$ is as you give it.
When we carry out our exercise of drawing samples with the same x's, your estimate $$b^{[1]}$$, and my $$b^{[2]}$$ are just two draws from that normal distribution.