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What would happen when, starting from a panel data model with group fixed effects and time fixed effects I apply the Mundlak approach?

This is the model:

$ y_{i,t} = c + \beta x_{i,t} + \alpha_i + \delta_2 B2_t + \delta_3 B3_t +....\delta_T BT_t +\epsilon_{i,t} $

where $B_t$ are time dummies (time fixed effects).

Should I also include the sample averages for each $B_t$ so that the equation becomes:

$ y_{i,t} = c + \beta x_{i,t} + \gamma \bar{x}_{i} + \delta_2 B2_t + \delta_3 B3_t +....\delta_T BT_t + \tilde{\delta}_2 \bar{B2} + \tilde{\delta}_3 \bar{B3} +....+\tilde{\delta}_T \bar{BT} +\nu_{i,t}$

wher $\bar{\bullet}$ denotes the sample average computed over each panel for variable $\bullet$.

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If the panel data set is balanced, the averages will cause perfect collinearity and will all be dropped. So you lose nothing.

If it is unbalanced, the averages of the time dummies will explain some aspects of the structure of the data set. For example, let T=3 so we have two dummies. B2bar(i) is 1/T(i) if observed at t=2, and 0 otherwise. Similarly, B3bar(i) depends on whether t=3 is observed and on T(i). So, I think the individual averages of the time dummies control for heterogeneity due to the period effects and data structure heterogeneity. I personally believe that it is safe to include those Btbar(i) terms. But there may be more fundamental issues involved in this unbalanced case as it relates to modeling selectivity.

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  • $\begingroup$ as a follow up: 2 groups and T=4; all y_(i,t) are available for both groups. group i=2 is missing the first 2 x_(i,t). In this case is it correct to say B2bar(1) = 1/T(1) = 1/4 and B2bar(2) = 0? But even in this case there is perfect collinearity, or am I missing something? $\endgroup$ – night_owl89 Mar 21 at 17:49

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