0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 hope this helps.

$\endgroup$
  • $\begingroup$ 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...". $\endgroup$ – Snoopy Oct 9 '18 at 9:15
  • $\begingroup$ 1. I need to read the entire statement of the proposition you are reading, but because you're mentioning it's the classical regression model, I would agree with you that by the Weak Law of Large Numbers, consistency of sample means is straightforward and, like you said, we would be done. 2. This is where I guessed when I said that the author, maybe, was trying to prove consistency by showing convergence in quadratic mean (I insist, I haven't read the full comment/remark/statement but what you mention here). 3. It's literally that - deterministic convergence, so it follows immediately. $\endgroup$ – MauOlivares Oct 9 '18 at 16:03
  • $\begingroup$ I edited the original post and added a sentence which I should have done before. All what is happening is proving consistency. My confusion is due to how the arguments are developed in the book which got me confused. Thanks Mau. I am not sure if the post is still useful. I could delete it. $\endgroup$ – Snoopy Oct 11 '18 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.