This question is about how to show identification of the fixed effects in a static panel linear model.

A1 (model): The model is $$ Y_{it}=\alpha_i+X_{it}^\top \beta+\epsilon_{it} $$ for each $i=1,...,N$ and $t=1,...,T$, where $i$ indexes individuals and $t$ indexes time periods.

A2 (data): We assume to have an i.i.d. sample of $N$ observations $\{Y_{i1}, X_{i1},\dots, Y_{iT}, X_{iT}\}_{i=1}^N$ with $N$ large.

A3 (exogeneity): $E(\epsilon_{it}| X_{i1},..., X_{iT}, \alpha_i)=0$ for each $t=1,...,T$ and $i=1,...,N$.

Question: In the so called "fixed effect model", $\alpha_1,..., \alpha_N$ are treated as parameters (together with $\beta$) and possibly estimated. How can we show that $(\alpha_1,..., \alpha_N, \beta)$ are identified under A1, A2, A3?

Remark: I think that $T$ large is also needed to identify $\alpha_1,..., \alpha_N$. Feel free to add this assumption.

My thoughts and doubts:

I have found several sources discussing how to estimate $(\alpha_1,..., \alpha_N, \beta)$ or how to identify/estimate $\beta$ alone (by differencing out $\alpha_1,..., \alpha_N$), but no papers or books explaining the joint identification of $(\alpha_1,..., \alpha_N, \beta)$.

I'm aware of the incidental parameter problem which prevents consistent estimation of $(\alpha_1,..., \alpha_N)$ when $T$ is fixed and $N\rightarrow \infty$. Hence, I suppose that $(\alpha_1,..., \alpha_N)$ are not identified when $T$ is fixed and $N\rightarrow \infty$. The incidental parameter problem disappears if we also let $T\rightarrow \infty$. Does this imply that identification of $(\alpha_1,..., \alpha_N)$ can be established? How?

In what follows, I report my incomplete attempt.

$Y_i\equiv (Y_{i1},..., Y_{iT})$ and $X_i\equiv (X_{i1},..., X_{iT})$. $K$ is the size of $\beta$. I assume $NT> N+K$.

First, I rewrite the model as $$ Y_{it}=\sum_{\ell=1}^N\alpha_l 1\{i=\ell\}+X_{it}^\top \beta+\epsilon_{it}, $$ where I consider $\alpha_1,..., \alpha_N$ as parameters and the index $i$ as a random variable. Second, I rewrite A3 $$ E(\epsilon_{it}| i, X_{i1},..., X_{iT})=0, $$ for each $i=1,...,N$ and $t=1,...,T$.

By A3, for each $i=1,\dots, N$, there exists a realisation of the $T\times K $ matrix $X_{i}\equiv (X_{i1},..., X_{iT})$ (which I denote by $x_{i}\equiv (x_{i1},..., x_{iT})$) such that $$ \begin{cases} E(\epsilon_{i1}|i=1, X_{i} =x_1)=0 \\ \vdots\\ E(\epsilon_{iT}|i=1, X_{i} =x_1 )=0 \\ \vdots\\ E(\epsilon_{i1}|i=N, X_{i} =x_N )=0 \\ \vdots\\ E(\epsilon_{iT}|i=N, X_{i} =x_N )=0 \\ \end{cases} $$ In turn, by A1, $$ \begin{cases} E(Y_{i1}|i=1, X_{i} =x_1 )=\alpha_1+\beta x_{11} \\ \vdots\\ E(Y_{iT}|i=1, X_{i} =x_1 )=\alpha_1+\beta x_{1T} \\ \vdots\\ E(Y_{i1}|i=N, X_{i} =x_N )=\alpha_N+\beta x_{N1} \\ \vdots\\ E(Y_{iT}|i=N, X_{i} =x_N )=\alpha_N+\beta x_{NT} \\ \end{cases} $$ I can more compactly rewrite this system of equations as $$ \underbrace{Y}_{NT\times 1}=\overbrace{\underbrace{\begin{pmatrix} D & X \end{pmatrix}}_{NT \times (N+K)}}^{\equiv \Gamma} \overbrace{\underbrace{\begin{pmatrix} \alpha \\ \beta \end{pmatrix}}_{(N+K)\times 1}}^{\equiv \phi}. $$ Next, $$ \Gamma^\top Y= \Gamma^\top \Gamma \phi. $$ Thus, $$ \phi=(\Gamma^\top \Gamma)^{-1} \Gamma^\top Y $$ under the assumption that $\Gamma^\top \Gamma$ is invertible.

I would be done with the proof if I could claim that $Y$ is known for $N$ large under assumption A1. I don't think this is the case, though. I suppose that somehow we also need large $T$, but I don't see clearly how.

  • $\begingroup$ There are a number of problem here. To begin with, the term identifiability is normally used in such a way that parameters are identifiable if the models parametrised by them are different. This is the case for your $\alpha_1,\ldots,\alpha_N$. The incidental parameter problem says that you can't estimate them consistently, which is possible even if they are identifiable. $\endgroup$ Aug 3, 2021 at 20:43
  • $\begingroup$ You will need some assumptions about the $X$. Some would have the $X$ so that their first row is constant 1, so the forst component in $\beta$ is an intercept. If this is the case, the $\alpha_1,\ldots,\alpha_N$ are not identifiable but confounded with the intercept, and you'd need an additional condition that they sum up to zero (for example). Generally $X$ could be chosen so that there are identifiability problems and conditions are needed to forbid this. $\endgroup$ Aug 3, 2021 at 20:46
  • $\begingroup$ Generally both identifiability and the existence of consistent estimators deal with the model assuming that potentially infinitely many observations can be had, rather than a finite sample. So "$N$ large" or "$T$ large" won't help proving anything. If something doesn't work for finite $N$ or $T$, asking them to be "large" won't help. However if the model allows $N\to\infty$, say, you don't need to be concerned about finite $N$. $\endgroup$ Aug 3, 2021 at 20:50
  • $\begingroup$ "First, I rewrite the model..." - I think that using $i$ as a random variable will not lead to anything good. If you have $i$ running from 1 to $N$, you're explicitly saying that it isn't random, aren't you? (I suspect you don't need its randomness anyway.) $\endgroup$ Aug 3, 2021 at 20:54
  • $\begingroup$ Can't edit the first comment anymore, but you'll see this was imprecise. $\alpha_1,\ldots,\alpha_N$ are not automatically identifiable here, you'll require conditions on the $X$ for this. Just to be in line with the second comment. $\endgroup$ Aug 3, 2021 at 20:59

1 Answer 1


If your $D$ is what I think it is, and taking into account the comments I made regarding the use of terminology and notation, then the additional assumption about $X$ I mentioned in the comments is just that $\Gamma^T\Gamma$ is invertible, which you figured out yourself. (It is not quite that trivial to find out what this means for your $X$ though. Large $N$ and $T$ may make it easier to fulfill this.)

What may be confusing is that you used $Y$ in the last equations for a vector that actually has the expected values of the $Y_{it}$ rather than the $Y_{it}$ themselves. The latter are random and not known, whereas the former are defined in the model framework. So you're basically done.

Note that for identifiability the $N$ and $T$ can be taken as fixed, however, as identifiability is about the model and not about a fixed sample, implicitly this assumes an infinite number of realisations of any $\epsilon_{it}$, or rather, more precisely and clearly, knowledge of the full distribution of $\epsilon_{it}$ (which carries all the randomness - I'm assuming that you treat the $X$ as fixed, otherwise you'd have to worry about identifying their distribution, too). This may look counterintuitive to you as you may interpret $N$ and $T$ as some kinds of sample sizes that potentially can go to $\infty$. In fact you need $T\to\infty$ (and some further conditions on $X$!) to have a consistent estimator (which as I wrote before is stronger than identifiability). Note that $N\to\infty$ will increase the number of $\alpha$-parameters, so not only will you have more observations at your disposal, you will also have to estimate more parameters, and this is not a standard identifiability problem. (It may however help with estimating $\beta$.)

  • $\begingroup$ Thank you for all your comments. Just a few clarifications (in line with your insights): 1) $X$ does not contain an intercept term. 2) $D$ is an $NT\times N$ matrix stacking for each $i$ a $T\times N$ matrix with $i$-th column of ones and the other elements equal to zero. My understanding from your answer is that $\Gamma^\top \Gamma$ should be invertible for identifiability. $\endgroup$
    – Star
    Aug 4, 2021 at 19:28
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    $\begingroup$ "My understanding from your answer is that $\Gamma^T\Gamma$ should be invertible for identifiability." Yes exactly. (A) Identifiability does not involve what you observe. Identifiability (at least according to its standard definition) means that two different parameters do not parametrise the same model. That the parameters are in this way identifiable does not necessarily imply that you can observe them in practice. $\endgroup$ Aug 4, 2021 at 23:16
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    $\begingroup$ (B) The conditioning in these conditions is technically wrong because nothing on which you "condition" is actually a random variable. Unless $X$ is random (in which case the whole thing becomes more complicated), $\epsilon_{it}$ are the only random variables here, and $E(\epsilon_{it})=0$ is all you need. $\endgroup$ Aug 4, 2021 at 23:19
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    $\begingroup$ I repeat myself but a key thing to understand is that the existence of consistent estimators (which is not possible for finite $T$ due to the incidental parameters problem) is not the same as identifiability, but rather a stronger property of a parameter that requires stronger conditions. $\endgroup$ Aug 4, 2021 at 23:24
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    $\begingroup$ Yes, unless the $\alpha_i+X_{it}^T\beta$ are the same for all $i$. $\endgroup$ Aug 5, 2021 at 13:14

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