What are the minimum conditions needed for the consistency of OLS estimator in the following linear regression model?

Question

Suppose $Y_i=X_i'\beta+\epsilon_i$ with $E(\epsilon_i|X_i)=0$. Consider the usual OLS estimator for $\beta$ using a random sample $\{X_i,Y_i\}_{i=1}^n$: $\widehat{\beta}=(\frac{1}{n}\sum_{i=1}^nX_iX_i')^{-1}\frac{1}{n}\sum_{i=1}^n X_iY_i$.

Substitute $Y_i=X_i'\beta+\epsilon_i$ into the expression gives $\widehat{\beta}=\beta+(\frac{1}{n}\sum_{i=1}^nX_iX_i')^{-1}\frac{1}{n}\sum_{i=1}^n X_i\epsilon_i$. The way to prove consistency is to show that $\frac{1}{n}\sum_{i=1}^nX_iX_i'\overset{p}{\rightarrow} E(X_iX_i')$, and $\frac{1}{n}\sum_{i=1}^n X_i\epsilon_i\overset{p}{\rightarrow} E(X_i\epsilon_i)=0$ by weak law of large numbers and then by continuous mapping theorem. Note that the weak law of large numbers only requires the existence of expected values: $E(X_iX_i')$ and $E(X_i\epsilon_i)$, where $E(X_i\epsilon_i)=E(X_iE(\epsilon_i|X_i))=0$ always hold under our model.

Thus it seems that all I need to assume is that $E(X_iX_i')<\infty$ and $E(X_iX_i')$ being invertible and I only need these two assumptions. Am I right?

Answer 1

$\DeclareMathOperator{\pl}{\operatorname{plim}}$

We have \begin{align}\pl \hat{\boldsymbol\beta}&= \boldsymbol \beta + \pl \left(\frac{\mathbf{X^\top X}}{n}\right)^{-1}\cdot\pl \left(\frac{\mathbf X^\top\boldsymbol \varepsilon}{n}\right)\tag 1, \label 1\end{align}

The bone of contention could be $\pl \left(\frac{\mathbf{X^\top X}}{n}\right)=:\mathbf Q.$ When $\bf X$ is of full column rank, then we can assume $\mathbb E\left[\mathbf x_i\mathbf x_i^\top\right] = \mathbf Q$ and the rest is what you asserted.

The bare minimum or "very weak" assumptions that $\mathbf X$ should follow is Grenander conditions.

Observe that once $\lim_{n\to\infty}\lambda_\text{smallest}(\mathbf{X^\top X}) = \infty, ~\hat{\boldsymbol\beta}$ becomes consistent. (See this relevant post of mine).

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Reference:

$\rm [I]$ Advanced Econometrics, Takeshi Amemiya, Harvard University Press, $1985,$ sec. $3.5,$ p. $95.$