1
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ 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
    Commented Jul 8, 2022 at 5:00

1 Answer 1

2
$\begingroup$

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 $E(v_i|x_i)=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.