# Is the exogeneity condition required for GLS with panel data a form of strict exogeneity?

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!

• 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. Jul 8, 2022 at 5:00

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 $$E(v_i|x_i)=0$$.