2
$\begingroup$

Consider the following Panel Data model:

$$ y_{it}=x_{it}\beta+\alpha_{i}+u_{it} $$ where $\alpha_{i}$ denotes the individual specific fixed effect, $x$ and $y$ are both scalars for individual $i$ at time $t$. I wish to estimate this equation using fixed effects. One soluton is to put dummy variables for each individuals. Another way is to demean the data by subtracting within individual means via the following transform: $$ y_{it}-\bar{y_{i}}=\left(\bar{x}_{it}-\bar{x}_{i}\right)\beta+u_{it}-\bar{u}_{i} $$ This estimator uses only within individual variation, i.e. it does not consider across individual variation at all. Is there an immediate way of seeing this just staring at this equation? More specifically, I don't see how just by demeaning the data at the individual level restricts the source of variation to within individaul solely. Any help is much appreciated.

$\endgroup$
1
$\begingroup$

If the observations for each individual are demeaned, then the only variation left is how much each individual deviates from their mean over time.

Think about it this way. In the fixed effects model $y_{it} = x_{it}\beta + \alpha_i + u_{it}$, where $\alpha_i$ represents a dummy for each entity in the data, the dummies are all averages for each entity. These are partialled out, so only each $i$'s deviation from its average is used to estimate $\beta_i$.

The demeaned model does the same thing. Subtracting the $i$'s mean from $i$'s observation centers it around 0. The only variation left is each $i$'s deviation from its own mean. If $i$ has no change over time, $x_i - \bar x = 0$, so does not contribute to the estimate of $\beta$. If $i$ has changes over time, this gets picked up in $\beta$. So only changes within $i$ get estimated.

$\endgroup$
1
$\begingroup$

The variation used to identify beta is the variation of the independent variable around the individual's mean for that variable. Hence none of the cross sectional variation is being used to identify beta, only variation within individuals over time. For that reason it is within variation only.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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