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

<p>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.</p>

CC BY-SA 4.0
2020-03-21T17:49:54.800Z
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?
night_owl89
CC BY-SA 4.0
2020-03-20T17:29:07.210Z
2020-03-20T17:29:07.210Z
chan1142
<p>What would happen when, starting from a panel data model with group fixed effects <em>and</em> time fixed effects I apply the Mundlak approach?</p>

<p>This is the model: </p>

<p><span class="math-container">$ 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} $</span> </p>

<p>where <span class="math-container">$B_t$</span> are time dummies (time fixed effects).</p>

<p>Should I also include the sample averages for each <span class="math-container">$B_t$</span> so that the equation becomes: </p>

<p><span class="math-container">$ 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}$</span></p>

<p>wher <span class="math-container">$\bar{\bullet}$</span> denotes the sample average computed over each panel for variable <span class="math-container">$\bullet$</span>.   </p>

<econometrics>
Mundlak Approach when panel data include time fixed effects
night_owl89
# Mundlak Approach when panel data include time fixed effects

(Asked by: username_0 on 2020-03-19)

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 $\bullet$.

**Tags:** econometrics

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## 1 answers

### Answer by username_1 on 2020-03-20. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_0**: 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?  

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