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Why is there still a block structure in the covariance-variance matrix, and consequently a need for clustering of standard errors in fixed-effects models? Shouldn't demeaning solve the serial correlation issue? Derivations and intuitive explanations are both appreciated!

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  • $\begingroup$ If serial correlation in the error term ($u_i + v_{it}$) solely comes from the time invariant part ($u_i$), you are right. But serial correlation may exist in $v_{it}$ (the time-varying part), which is not dealt with by the within-group transformation. $\endgroup$
    – chan1142
    Oct 8 '21 at 23:40
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Consider the following specification: $$ Y_{i,g} = X_{i,g}\beta + u_{i,g} $$ Where the residuals have different mean across groups and have within group correlation: $$ \begin{align*} &\mathbb{E}(u_{i,g}) = \alpha_g,\\ &cov(u_{i,g} u_{j,g}) = \rho_{i,j},\\ &cov(u_{i,g}, u_{j,g'}) = 0 \text{ for } g \ne g' \end{align*} $$ Taking means gives: $$ \mathbb{E}(Y_{i,g}) = \mathbb{E}(X_{i,g}) \beta + \alpha_g $$ Now use notation $\widehat{Z} = Z - \mathbb{E}(Z)$ for the de-meaned variable. Then: $$ \widehat{Y_{i,g}} = \widehat{X_{i,g}} \beta + \widehat{u_{i,g}} $$ Now the mean of $\widehat{u_{i,g}}$ has become zero but there is still correlation within each group: $$ \begin{align*} &cov(u_{i,g}, u_{j,g}) = \mathbb{E}(\widehat{u_{i,g}},\,\ \widehat{u_{j,g}}) = \rho_{i,j},\\ &cov(u_{i,g}, u_{j,g'}) = \mathbb{E}(\widehat{u_{i,g}},\,\ \widehat{u_{j,g'}}) = 0 \text{ for } g \ne g' \end{align*} $$ So de-meaning does not get rid of the within-group correlation.

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Usually, the conditional mean and the conditional variance of a random variable are independent (of course there are exceptions).
While the fixed effects introduce flexibility in the specification of the conditional mean, clustering introduce flexibility in the specification of the conditional variance.

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