Consider the following regression model using panel data: $$y_{it}=x_{it} + u_{it}$$

According to Wooldridge (2010), Chapter 7.4, the first assumption required for Generalized Least Squares (GLS) in a panel data setting can be stated as:

Assumption SGLS.1: $$E(X_i \otimes u_i) = 0 $$ Where $X_i$ is a matrix containing all $T$ observations of the dependent variables for individual $i$ and and $u_i$ is a vector containing all $T$ idiosyncratic error terms for individual $i$.

I am confused as to whether this assumption implies strict exogenetiy or not. As far as I understand, given what the kronecker product means, SGLS.1 only implies uncorrelatedness of x with all elements of u, which does not necesarily imply zero conditional mean on all elements of u, which is needed for strict exogeneity. However, in a later section (Chapter 10.3), Wooldridge introduces the assumptions for Random Effects (RE) estimation by stating the assumption $$E(v_i | x_i)=0 , $$ where $v$ is the composite error $v_i = c_i + u_i$ of a model with unobserved heterogeneity $c_i$: $$y_{it}=x_{it} \beta + v_{it}$$ He refers to this assumption as satisfying "the exogeneity assumption SGLS.1 (see Chapter 7)".

How comes he can make this reference if SGLS.1 does not necessarily imply strict exogeneity? I would appreciate some orientation here. Thanks in advance!

  • $\begingroup$ Welcome. It's just a matter of how your "strict exogeneity" is defined. If it's defined in terms of covariances, $E(X_i\otimes u_i)=0$ is strict exogeneity; if it's defined in terms of conditional means, not. SGLS.1 is written in terms of covariances. Zero conditional mean implies zero covariance, not vice versa. $\endgroup$
    – chan1142
    Jul 8, 2022 at 5:00

1 Answer 1


Wooldridge's wording is possibly confusing, but I believe it is correct.

Wooldridge is saying that if $E(v_i|x_i)=0$ then $E(x_i\otimes v_i)=0$.

Wooldridge is not saying that if $E(x_i\otimes v_i)=0$ then $E(v_i|x_i)=0$.


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