where $$ Y_{it}=\alpha_i+\psi_{j(i,t)}+\epsilon_{it} $$ for $i=1,...,N$ (individual index), $t=1,...,T$ (time index), $j=1,...,J$ (firm index), and $j(i,t)$ is the firm employing worker $i$ at time $t$. $\alpha_i$ is called worker $i$'s fixed effect. $\psi_{j(i,t)}$ is called firm $j(i,t)$'s fixed effect.

By stacking observations over individuals, the equation above can be rewritten as $$ \underbrace{Y_i}_{T\times 1}=\underbrace{D_i}_{T\times N}\underbrace{\alpha}_{N\times 1} +\underbrace{F_i}_{T\times J} \underbrace{\psi}_{J\times 1}+\underbrace{\epsilon_i}_{T\times 1} $$

Assume that:

[A1] We have a sample of observations $\{Y_i, F_i\}_{i=1}^N$ i.i.d., for $N$ large and $T$ small.

[A2] $E(\epsilon_i| D_i, F_i)=0$ for each $i=1,...,N$.

As shown here, the identification of $\alpha$ and $\psi$ passes through dividing the sample into connected groups. In turn, under A1-A2 and for each of these groups $g=1,...,G$, $\alpha_g$ and $\psi_g$ are identified if and only if

  • One element of the vector $(\alpha_g, \psi_g)$ is set equal to a known constant (for example, zero).

  • $N_g>1$ and/or $J_g>1$, where $N_g$ (resp. $J_g$) are the numbers of individuals (resp. firms) in group $g$.


(Q1) Is the element of $(\alpha_g, \psi_g)$ to be normalised picked arbitrarily? That is, suppose the vector $(\alpha_g, \psi_g)$ has length $5$. Then, from a purely mathematical perspective, am I totally free to fix its third element equal to $0$ or, equivalently, its fifth element equal to zero?

(Q2) Suppose $G=2$ and that the two groups have the same number of firms and workers. It is often said that the estimates of the fixed effects cannot be compared across groups due to the normalisation. However, if I am totally free to normalise the value of any element of the vector $(\alpha_g, \psi_g)$ - subject to the answer to Q1 -, I could set in both groups the first element of the vector $(\alpha_g, \psi_g)$ equal to zero. Would I then be allowed to make comparisons?


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