# Asymptotic variance of linear regression with homoskedasticity assumption (Wooldridge Panel book Eq. (4.10))

Jeffrey M. Wooldridge Econometric Analysis of Cross Section and Panel Data

Chapter 4 The Single-Equation Linear Model and OLS Estimation

Section 4.2 Asymptotic Properties of OLS

Subsection 4.2.2 Asymptotic Inference Using OLS

Assumption OLS.3: $$E\left(u^{2}x^{\prime}x\right)=\sigma^{2}E\left(x^{\prime}x\right)\text{ where }\sigma^{2}\equiv E\left(u^{2}\right)$$

Eq. (4.10) states $$\widehat{Avar\left(\hat{\beta}\right)}=\hat{\sigma}^{2}\left(X^{\prime}X\right)^{-1}$$ where $$\hat{\sigma}$$ is a consistent estimator of $$\sigma$$

I think it should be $$\widehat{Avar\left(\hat{\beta}\right)}=\hat{\sigma}^{2}\left(\frac{1}{N}X^{\prime}X\right)^{-1}$$

which one is correct:

$$\widehat{Avar\left(\hat{\beta}\right)}=\hat{\sigma}^{2}\left(\sum_{i=1}^{N}x_{i}^{\prime}x_{i}\right)^{-1}$$

$$\widehat{Avar\left(\hat{\beta}\right)}=\hat{\sigma}^{2}\left(\frac{1}{N}\sum_{i=1}^{N}x_{i}^{\prime}x_{i}\right)^{-1}$$?

A copy of Wooldridge is not in proximity. But the author is correct in stating equation no. $$(4.10).$$

Because it seems to obfuscate the thinking in OP, it warrants a brief casting of light on the matter.

The variance of the least square coefficient vector $$\bf b$$ is

\begin{align} \mathbb{Var} [\mathbf b|\mathbf X] & =\left(\mathbf{ X^\mathsf TX}\right) ^{-1}\mathbf X^\mathsf T\mathbb E \left[\boldsymbol{\varepsilon\varepsilon}^\mathsf T|\mathbf X\right]\left(\left(\mathbf{ X^\mathsf TX}\right) ^{-1}\mathbf X^\mathsf T\right) ^\mathsf T\\ &= \sigma^2\left(\mathbf{ X^\mathsf TX}\right) ^{-1}.\tag 1 \end{align}

For estimation of $$(1),$$ one needs to estimate $$\sigma^2.$$ It is an easy routine to deduce based on sum of squared residuals

$$\mathbb E\left[\mathbf e^\mathsf T\mathbf e|\mathbf X\right] = (n - k) \sigma^2.\tag 2$$

From $$(2) ,$$ one gets an (unbiased) estimator of $$\sigma^2$$ and subsequently

$$\operatorname{Est.Var} [\mathbf b|\mathbf X] = s^2\left(\mathbf{ X^\mathsf TX}\right) ^{-1},\tag 3$$ where $$s^2 := \frac{\mathbf e^\mathsf T\mathbf e}{n-k}.$$

Now, to ensure the data in large samples is "well-behaved", one assumption taken is

\begin{align}\operatorname{plim}_{n\to \infty} \frac{\mathbf X^\mathsf T\mathbf X}n &:= \mathbf Q \\&\equiv \textrm{a positive definite matrix};\tag{A1}\end{align} this is attainable if it is assumed $$\mathbb E\left[\mathbf x_i\mathbf x_i^\mathsf T\right] = \mathbf Q,$$ which means $$\sum \mathbf x_i\mathbf x_i^\mathsf T/n \overset{\mathbb P}{\to}\mathbf Q$$ and hence $$\operatorname{plim}\left( \frac{\mathbf X^\mathsf T\mathbf X}n\right) ^{-1} =\mathbf Q^{-1}.$$

Now, the asymptotic variance, like $$(1)$$ would be

$$\operatorname{Asy.Var}[\mathbf b] = (\sigma^2/n) \mathbf Q^{-1}.\tag 4$$ Again, to get an estimator of $$(4),$$ one needs to concentrate on the term in parenthesis - $$s^2$$ can be used with proper evaluation of its consistency:

\begin{align}s^2 &= \frac{\boldsymbol \varepsilon^\mathsf T\mathbf M\boldsymbol\varepsilon}{n-k}\\&= \frac{\boldsymbol \varepsilon^\mathsf T\boldsymbol\varepsilon - \boldsymbol\varepsilon^\mathsf T \mathbf X\left(\mathbf{ X^\mathsf TX}\right)^{-1}\mathbf X^\mathsf T\boldsymbol \varepsilon}{n-k}\\ &= \frac{n}{n-k}\left[\frac{\boldsymbol \varepsilon^\mathsf T\boldsymbol\varepsilon}n - \left(\frac{\boldsymbol\varepsilon^\mathsf T \mathbf X}n\right)\left(\frac{\mathbf{ X^\mathsf TX}}n\right)^{-1}\left(\frac{\mathbf X^\mathsf T\boldsymbol \varepsilon}n\right)\right]; \tag 5 \end{align} where $$\mathbf M:= \mathbf I - \mathbf X\left(\mathbf{ X^\mathsf TX}\right)^{-1}\mathbf X^\mathsf T$$ is the residual maker. Using $$(\mathrm A1)$$ and the fact that $$\operatorname{plim} \left(\frac{\boldsymbol\varepsilon^\mathsf T \mathbf X}n\right) = \mathbf O$$ (since $$\mathbb E[\mathbf x\varepsilon] =\mathbf O$$), the second term in the bracket in $$(5)$$ converges to $$0.$$ The factor outside the bracket converges to $$1.$$ The only term left for inspection is $$\sum \varepsilon_i^2/n.$$ Now $$\varepsilon_i^2$$ are independent with same finite expectation $$\sigma^2.$$ To ensure the term's convergence (almost surely to $$\sigma^2$$), Markov's law of large numbers could be employed provided $$\mathbb E\left[\left(\varepsilon_i^2\right)^{1+\delta}\right] <\infty$$ for some $$\delta \in (0, 1)$$ - it would suffice then to assume for all $$\varepsilon_i,$$ there exist finite moments greater than $$2.$$ In any case, $$\operatorname{plim} s^2 = \sigma^2. \tag 6$$ This implies

$$\operatorname{plim} s^2 \left( \frac{\mathbf X^\mathsf T\mathbf X}n\right) ^{-1}= \sigma^2\mathbf Q^{-1}. \tag{ 6.I}$$ Therefore, the appropriate estimator of $$(4)$$ would be $$\operatorname{Est.Asy.Var}[\mathbf b] = s^2\left(\mathbf X^\mathsf T\mathbf X\right) ^{-1}.\tag 7$$

## Reference

Econometric Analysis, William H. Greene, Pearson Education, 2018.