Consistency of estimators in simple linear regression Under what conditions are the estimators $\beta_0$ and $\beta_1$ in simple linear regression consistent?
I derived that $S_{xx}$ should go to infinity as n goes to infinity, but I do not get any further. 
I hope you can help me out!
 A: We'll look at $\hat{\beta}_0 = \bar{y} - \hat{\beta}_1 \bar{x}$ first.  The law of large numbers says that $\bar{y}$ converges to $\text{E}(y) = \beta_0 + \beta_1 \text{E}(x)$ and if $\hat{\beta}_1 \to \beta_1$ then $\hat{\beta}_1 \bar{x}$ converges to $\beta_1 \text{E}(x)$.  This means $\hat{\beta}_0$ will be consistent if $\hat{\beta}_1$ is.  Now looking at $\hat{\beta}_1$ and assuming all variances and covariances are finite and well-defined we have
\begin{align}
\hat{\beta}_1 &= \frac{\sum_{i=1}^{n} (y_i - \bar{y})(x_i - \bar{x})}{\sum_{i=1}^{n}(x_i - \bar{x})^2} \\
&\to \frac{\text{Cov}(y, x)}{\text{Var}(x)} \\
&= \frac{\text{Cov}(\beta_0 + \beta_1 x + \epsilon, x)}{\text{Var}(x)} \\
&= \beta_1 + \frac{\text{Cov}(\epsilon, x)}{\text{Var}(x)}
\end{align}
which equals $\beta_1$ so long as $\text{Cov}(\epsilon, x) = 0$.
To prove the stronger claim that the estimators are consistent in mean square we can start with the variance covariance matrix for $(\hat{\beta}_0, \hat{\beta}_1)$ which equals $\sigma^2 (X^T X)^{-1}$.  Here $X$ is the data matrix and for simple linear regression this is just $[1 ; x]$ where $1$ is a vector of ones and $x = (x_1, x_2, \ldots, x_n)$ is the predictor set.  If we go through the linear algebra we get
$$
(X^T X)^{-1} = \begin{bmatrix} n^{-1} \sum_{i=1}^{n} x_i^2 & - \bar{x} \\ - \bar{x} & 1 \end{bmatrix} \frac{1}{\sum_{i=1}^{n} x_i^2 - n \bar{x}^2} 
$$
and the denominator $\sum_{i=1}^{n} x_i^2 - n \bar{x}^2$ is nothing but the sum of squares for $x$.  This means that as long as $\sum_{i=1}^{n} (x_i - \bar{x})^2 \to \infty$ as $n \to \infty$ every element of this matrix goes to zero, including the diagonal elements.
A: Consider the classic situation in which you assume that the true model is of the form:
$y = \beta_0 + \beta_1x + u$, with $E[u] = 0$ . In this case, the OLS estimator will asymptotically converge to $\beta_0$ and $\beta_1$ provided that $E[xu] = 0$
If you mean consistency in the sense of convergence to the parameters of the best linear approximation to $E[y|x]$, then all you need is that the data you get, $(y_i,x_i)$ is iid. 
