Consistency of unbiased estimator of error term variance in Multiple regression Let $Y=X\beta+\epsilon$. We know that $\frac{e'e}{n-k}$ is an unbiased estimator of $Var(\epsilon)$, where $e$ is the vector of residuals, and $\epsilon$ is multivariate normal distributed in this model. 
How can we prove, if it is, that this estimator converges to the right value? Or even if it's consistent?
Any help would be appreciated. 
 A: The residuals can be expressed in terms of the true error terms as
$$e_i = \epsilon_i - \mathbf x_i'(\hat \beta - \beta)$$
Squaring,
$$e_i^2 = [\epsilon_i - \mathbf x_i'(\hat \beta - \beta)]^2 = \epsilon_i^2 -2 (\hat \beta - \beta)'\mathbf x_i\epsilon_i + (\hat \beta - \beta)'\mathbf x_i\mathbf x_i'(\hat \beta - \beta)$$
Then 
$$\frac{\mathbf e'\mathbf e}{n-K}=\frac {1}{n-K}\sum_{i=1}^n e_i^2 = \frac {1}{n-K}\sum_{i=1}^n \epsilon_i^2 -\frac {2}{n-K}\sum_{i=1}^n (\hat \beta - \beta)'\mathbf x_i\epsilon_i   \\ +  \frac {1}{n-K}\sum_{i=1}^n(\hat \beta - \beta)'\mathbf x_i\mathbf x_i'(\hat \beta - \beta)$$
Note that $(\hat \beta - \beta)$ does not depend on the sum index so 
$$\frac {1}{n-K}\sum_{i=1}^n e_i^2 = \frac {1}{n-K}\sum_{i=1}^n \epsilon_i^2 -(\hat \beta - \beta)'\frac {2}{n-K}\sum_{i=1}^n \mathbf x_i\epsilon_i + (\hat \beta - \beta)'\left[\frac {1}{n-K}\sum_{i=1}^n\mathbf x_i\mathbf x_i'\right](\hat \beta - \beta)$$
Considering the limit $n \to \infty$,  


*

*Under the assumptions of ergodicity, that the 2nd moment of the error term exists, and that $E(\epsilon) = 0$ we have by Kinchin's LLN


$$\frac {1}{n-K}\sum_{i=1}^n e_i^2 \to_p E(\epsilon^2) = \text{Var}(\epsilon)$$


*Under the assumption $E(\mathbf x_i\epsilon_i)=\mathbf 0$ and given ergodicity,


$$(\hat \beta - \beta)' \to_p 0, \;\;\;\frac {2}{n-K}\sum_{i=1}^n \mathbf x_i\epsilon_i \to_p 2E(\mathbf x_i\epsilon_i)=0$$
so the second term goes to zero in probability


*Given the Grenander conditions on the regressor matrix, and ergodicity,


$$\left[\frac {1}{n-K}\sum_{i=1}^n\mathbf x_i\mathbf x_i'\right] \to_p \mathbf Q < \infty$$
and since $(\hat \beta - \beta)' \to_p 0$, the the third term also goes to zero in probability.
So we are left only with the first term, and we have consistency
A: You know that $S^2_e = \frac{1}{n-k}\mathbf{e}'\mathbf{e}$ is an unbiased estimator of $\sigma^2$.  So, if you show that $Var(S^2_e) \to 0$ as $n \to \infty$ you've shown that $S^2_e$ is consistent (i.e., converges to the right value).
This isn't difficult: $\mathbf{e}'\mathbf{e} \sim \sigma^2\cdot\chi^2_{n-k}$.  Consequently, the variance of $\mathbf{e}'\mathbf{e}$ is $2(n-k)\sigma^4$.  The variance of $S^2_e$ then is $\frac{1}{(n-k)^2}\cdot Var(\mathbf{e}'\mathbf{e}) = \frac{2\sigma^4}{n-k}$, which obviously tends to $0$ as $n$ grows without bound.
