# Consistency of the estimator of the variance of the error

Discussion in 'Education' started by Snoopy, Oct 8, 2018.

1. ### SnoopyGuest

In the classical linear regression model, the estimator of the variance of the regression error is $s^2 = \frac{e'e}{n-k} = \frac{u'Mu}{n-k}$ where u is the error vector, e is the residual vector, and M is the projection matrix of X. It can be shown that $\mathrm{plim} \: s^2 = \sigma^2$ because $\mathrm{plim} \: (\frac{u'u}{n}) = \sigma^2$. The last expression is part of the proof pf the consistency of the $s^2$. An econometrics textbook presents this derivation. But just after this, it says the following:

This last step requires a little attention. If $\mathrm{E} = 0$ and $\mathrm{Var} = \sigma^2I_n$ as we always assume, then $\mathrm{E}[\frac{u'u}{n}] = \sigma^2$. This is not sufficient to prove that $\mathrm{plim} \: (\frac{u'u}{n}) = \sigma^2$. We need an additional assumption, for example that the errors are normal. Under normality we have $\frac{u'u}{\sigma^2} \sim \chi^2(n)$, so that $\mathrm{Var[u'u] = 2n\sigma^4}$ and hence $\mathrm{Var}[\frac{u'u}{n}] = \frac{2\sigma^4}{n} \rightarrow 0$.

Why is showing $\mathrm{plim} \: (\frac{u'u}{n}) = \mathrm{plim} \: \frac{1}{n}\sum_{i = 1}^{n} u_i^2 = \sigma^2$ not sufficient? When showing the consistency of a statistic, is not the probability limit argument enough? Do I also need to show that the variance collapses to $\mathrm{E}[\frac{u'u}{n}]$? I thought that either the probability limit or the variance argument is enough to prove consistency. But the excerpt seems to suggest otherwise. What is the point of this excerpt?