# Consistency of the estimator of the variance of the error

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 of 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 $$\mathrm{E}[u] = 0$$ and $$\mathrm{Var}[u] = \sigma^2I_n$$ 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 $$\mathrm{Var}[\frac{u'u}{n}] = \frac{2\sigma^4}{n} \rightarrow 0$$. (A somewhat weaker additional condition would be that the errors are not just uncorrelated but i.i.d, un which case a stronger theorem, due to Khintchine, can be invoked.)

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?

The goal is to show that plim$$\left(\frac{u'u}{n}\right)=\sigma^2$$ as $$n\to\infty$$. What the textbook is doing is proving limit in probability by proving something stronger, that $$\left(\frac{u'u}{n}\right)$$ converges in quadratic mean to $$\sigma^2$$. Sufficient conditions for this are that
\begin{align*} \mathbb{E}\left(\frac{u'u}{n}\right)&\to\sigma^2\\ \mathbb{V}\left(\frac{u'u}{n}\right)&\to0,\:\:\:\text{ as }\:\:n\to\infty\\ \end{align*} The first condition is met, but is not sufficient since we need the variance to converge to $$0$$ as well. One way to guarantee this is by assuming normality, as the textbook does.
• I am still not clear on the following. 1. $\mathrm{plim} \: (\frac{u'u}{n}) = \mathrm{plim} \: \frac{1}{n}\sum_{i = 1}^{n} u_i^2 = \mathrm{E}[\frac{u'u}{n}] = \sigma^2$. This shows what $\frac{u'u}{n}$ converges to. So $\mathrm{plim}$ of $\frac{u'u}{n}$ is shown. We are done. Why to deal with the variance in addition? We already showed consistency. 2. I agree that showing that the variance goes to 0 is providing an additional argument that the random term converges to the pop. mean. But we do not have to do this. This is extra. So not clear to me why we are told "This is not sufficient...". – Snoopy Oct 9 '18 at 9:15