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Consistency of the estimator of the variance of the error

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

  1. Snoopy

    Snoopy Guest

    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?

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