Why Moment Conditions Matter
In random-design regression, both the regressors and the errors vary across observations. Large-sample arguments therefore depend on moments of their joint distribution. A finite second moment for the regressors makes the entries of the population regressor second-moment matrix finite. A central limit theorem for the regression score requires more: the score must have finite covariance. These are distinct requirements, and assumptions that ensure one do not automatically ensure the other.
Let the number of regressors, including any intercept, be a fixed integer \(k\). For observation \(i\), write the model as \(Y_i=x_i^\top\beta+u_i\), where \(x_i\) is a \(k\)-dimensional random vector and \(u_i\) is the error. Assume that the pairs \((x_i,u_i)\) are independent and identically distributed, and that \(\beta\) is fixed. No independence between \(x_i\) and \(u_i\) is assumed. For the moment, the concern is whether relevant expectations exist, not whether the population matrix is nonsingular.
The matrix \(G\) is a second-moment matrix; it is not necessarily a covariance matrix, since the regressors need not be centered. In particular, when \(x_i\) includes an intercept, its first entry is one. The score \(s_i\) is the random vector whose sample average appears in the OLS estimating equations. Its variability depends on the joint behavior of regressors and errors, not just on their separate distributions.
Conditions for Finite Regressor Moments and Score Variance
Use the Euclidean norm \(\|x\|=(\sum_{j=1}^k x_j^2)^{1/2}\). A convenient condition for \(G\) to be finite is \(E\|x_i\|^2<\infty\). The direct condition for finite score covariance is \(E[\|x_i\|^2u_i^2]<\infty\), along with a finite score mean. The latter mean exists under this second condition: by Cauchy–Schwarz, \(E\|x_i u_i\|\leq (E[\|x_i\|^2u_i^2])^{1/2}\).
Proof. For any coordinates \(j,\ell\), \(|x_{ij}x_{i\ell}|\leq (x_{ij}^2+x_{i\ell}^2)/2\leq\|x_i\|^2\). Thus \(E|x_{ij}x_{i\ell}|<\infty\), so the corresponding entry of \(G\) is finite. For the score, its \(j\)th coordinate is \(x_{ij}u_i\), and
Consequently each score coordinate has a finite second moment, and hence a finite first absolute moment by Cauchy–Schwarz. Products of score coordinates are integrable because \(|x_{ij}x_{i\ell}u_i^2|\leq \|x_i\|^2u_i^2\). The covariance entries, which subtract products of finite means from these finite second moments, are therefore finite. \(\square\)
The score condition is deliberately joint: it concerns the product of regressor size and error size. Requiring \(E\|x_i\|^4<\infty\) and \(E|u_i|^4<\infty\) is a familiar sufficient alternative, but can be stronger than necessary. Indeed,
This inequality follows from Cauchy–Schwarz applied to \(\|x_i\|^2\) and \(u_i^2\); it does not require independence. Another useful sufficient condition is \(E[u_i^2\mid x_i]\leq C\) almost surely for some finite constant \(C\), together with \(E\|x_i\|^2<\infty\). The tower property then gives \(E[\|x_i\|^2u_i^2]=E[\|x_i\|^2E(u_i^2\mid x_i)]\leq C E\|x_i\|^2\).
Worked Example: A Heteroskedastic Model With Finite Score Covariance
Let \(Z\sim N(0,1)\) and let \(\varepsilon\sim N(0,1)\) be independent of \(Z\). Take \(x=(1,Z)^\top\) and \(u=(1+Z^2)^{1/2}\varepsilon\). The conditional error mean is zero, and its conditional variance is \(E[u^2\mid Z]=1+Z^2\), which is not bounded by a constant. We can nevertheless check the relevant moments directly.
Since \(E[Z^2]=1\), \(E\|x\|^2=E[1+Z^2]=2\), so \(G\) is finite. For the score covariance, use \(E[\varepsilon^2]=1\):
Here \(E[Z(1+Z^2)]=0\) by symmetry, and \(E[Z^2(1+Z^2)]=E[Z^2]+E[Z^4]=1+3=4\). Thus the score has finite second moments even though its conditional variance grows with the regressor.
From Moments to Laws of Large Numbers and Central Limit Theorems
Under the iid assumption, the strong law of large numbers applies to each integrable coordinate. Thus \(E\|x_i\|^2<\infty\) ensures that the sample average of \(x_i x_i^\top\) converges almost surely, entry by entry, to \(G\). If \(E[x_i u_i]=0\) and \(E[\|x_i\|^2u_i^2]<\infty\), the score coordinates are integrable, so their sample average converges almost surely to zero. The finite-score-covariance condition also permits the iid multivariate central limit theorem for the centered score.
Justification. For the first convergence, each matrix entry \(x_{ij}x_{i\ell}\) is integrable by the bound in the preceding proof. The iid strong law applies to each of the finitely many entries. For the second, finite score second moments imply finite first absolute moments, and the iid strong law applies to each score coordinate; the stated zero mean gives the limit. Finally, the score vectors are iid, have mean zero, and have finite covariance \(\Omega\), so the iid multivariate central limit theorem gives the displayed distributional convergence. These are applications of the standard iid laws; they do not require normal regressors or normal errors. \(\square\)
The distinction between the two limit arguments is important. A law of large numbers needs integrability of the quantity being averaged. A central limit theorem for the score needs finite score covariance. Finite \(E\|x_i\|^2\) alone says nothing about whether the score variance is finite. Likewise, a finite score variance does not itself imply a zero score mean: that is a separate moment restriction, often ensured by the conditional mean assumption \(E[u_i\mid x_i]=0\), when the score is integrable.
Worked Example: Fourth Moments Are Sufficient, Not Necessary
Let \(W\) have a Student \(t\) distribution with 3 degrees of freedom, and let \(u\) be an independent random variable taking values \(-1\) and \(1\) with equal probability. Set \(x=(1,W)^\top\). A \(t_3\) variable has \(E[W]=0\), \(E[W^2]=3\), and \(E[W^4]=\infty\). Therefore
because \(u^2=1\). Also \(E[xu]=E[x]E[u]=0\), using independence and \(E[u]=0\). Thus the score has finite covariance, despite \(E\|x\|^4=\infty\). Imposing fourth moments on both regressors and errors would exclude this example unnecessarily; the direct score-moment condition captures what is needed here.
Worked Example: Finite Marginal Second Moments Do Not Ensure Finite Score Variance
Let \(A\) have a \(t_3\) distribution, let \(S\) be an independent random sign with probabilities \(P(S=1)=P(S=-1)=1/2\), and set \(x=(1,A)^\top\), \(u=AS\). Both \(A\) and \(u\) have finite second moments: \(E[A^2]=E[u^2]=3\). The score has mean zero, since \(E[AS]=0\) and \(E[A^2S]=E[A^2]E[S]=0\).
However, its second coordinate is \(A^2S\), whose squared expectation is
Thus the score has infinite variance, and the finite-variance iid central limit theorem cannot be applied to it. The regressors and errors each have finite second moments, but their dependence makes \(E[\|x\|^2u^2]\) infinite. Checking only marginal moments misses precisely the joint tail behavior that matters for the score.
Common Mistake and Practical Interpretation
A common mistake is to reason, “The regressors have finite variance and the errors have finite variance, so the OLS score has finite variance.” That conclusion is not valid without additional information about their joint distribution. The correction is to check \(E[\|x_i\|^2u_i^2]\) directly, or to verify a stated sufficient condition such as finite fourth moments for both quantities or a bounded conditional error second moment. Conversely, imposing fourth moments automatically can be needlessly restrictive when the direct product moment is finite.
These moment conditions support basic averages and score limits; they do not establish every other requirement for OLS asymptotics. In particular, finite moments do not imply that \(G\) is nonsingular, nor do they address dependence across observations or changing distributions. Keep the roles separate: integrability supports laws of large numbers, finite score covariance supports the usual iid central limit theorem, and a zero score mean is an additional centering condition.
Check Your Understanding
Use the distinction between regressor moments, score moments, and score means to answer the following.
- Why does \(E\|x_i\|^2<\infty\) ensure that each entry of \(E[x_ix_i^\top]\) is finite?
- State a direct moment condition that ensures the score \(x_i u_i\) has finite covariance.
- Why are finite fourth moments of both \(x_i\) and \(u_i\) a sufficient condition for finite score covariance, even when they are dependent?
- Does \(E[u_i\mid x_i]=0\) by itself ensure finite score covariance? Explain which separate moment issue remains.
- In the random-sign example, which score coordinate has infinite second moment, and why do finite marginal second moments fail to prevent it?