This is probably a trivial question, but I can't find it online or in the textbooks I have access to.

For a simple model:

$$y_{it} = x_{it}\beta + c_{i} + \epsilon_{it} $$

After a random-effects estimation, what is the method used to predict $c_{i}$?

I know how to obtain them using several software, but I am interest in the methodology behind.

  • $\begingroup$ If you have Greene, I found the answer in section 11.2 $\endgroup$ – VCG Aug 24 '16 at 12:50
  • $\begingroup$ Sorry his textbook: "Econometric Analysis" - it's a 1st year phd econometrics book. But good question. I could easily answer this if you asked about fixed effects but I still don't quite get what random effects actually does when you estimate it. $\endgroup$ – VCG Aug 24 '16 at 13:11
  • $\begingroup$ Thanks. I understand how FE predicts the time-invariant component (it is also in the documentation of Stata's xtreg command), but I cannot find for RE. Could you write an answer so I can accept it? $\endgroup$ – luchonacho Aug 24 '16 at 13:20
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    $\begingroup$ For random effects you compute the BLUP, Best Linear Unbiased Predictor $\endgroup$ – user83346 Aug 24 '16 at 14:52

So the true model has the unobserved individual level time invariant heterogeneity:

$y_{it}=\beta x_{it}+c_i+e_{it}$

So we estimate: $y_{it}=\alpha + \beta x_{it}+u_{it}$, where $u_{it}=c_i-\alpha+e_{it}$

Use pooled ols to get $\hat u_{it} $ and $\hat a$

let $c_i-a=\mu$

$\hat\mu=(1/n)\sum \hat u_{it}$ and $\hat e_{it}=\hat u_{it}-\hat\mu$

More info here: http://www.utdallas.edu/~d.sul/Econo1/lec_note_part3.pdf

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  • $\begingroup$ The $c_{i} - a = \mu$ step makes no sense. Where has the individual heterogeneity go? $\endgroup$ – luchonacho Aug 25 '16 at 12:09
  • $\begingroup$ So I took the observable $c_i$ and added it to the error term but then added in a constant without changing the equation. I then just call this difference $\mu$. The point is simply that we construct $mu$ to be equal to that so we can back out $c_i$. $\endgroup$ – VCG Aug 25 '16 at 12:16

Ok, I managed to get the answer I wanted, which also explains why the estimator is unbiased and consistent. Here it is:

The model is:

$$y_{it} = x_{it}\beta + c_{i} + \epsilon_{it} $$

From RE we obtain an estimation of $\beta$. Define the estimation error $\hat{u}_{it}$:

$$ \hat{u}_{it} \equiv y_{it} - x_{it}\hat{\beta} $$

Now, define the linear predictor $\bar{u}_{it}$ as the mean of the estimation error:

$$ \bar{u}_{it} \equiv \frac{\sum_{t=1}^{T}\hat{u}_{i}}{T} = \bar{y_{it}} - \bar{x}_{i}\hat{\beta} $$

This is, allegedly, the BLUP estimator of $c_{i}$. To confirm this, let us evaluate the statistical properties of this predictor. To do this, replace the original model into the above expression. After some rearranging, the outcome is:

$$ \bar{u}_{it} = \bar{x}_{i}\beta - \bar{x}_{i}\hat{\beta} + c_{i} + \frac{\sum_{t=1}^{T}\epsilon_{it}}{T} $$

The expectation of this estimator is:

$$ E(\bar{u}_{it}) = \bar{x}_{i}\beta - \bar{x}_{i}E(\hat{\beta}) + E(c_{i}) + \frac{\sum_{t=1}^{T}E(\epsilon_{it})}{T} $$

Assume $\hat{\beta}$ is an unbiased estimator of $\beta$ (requires strict exogeneity, unobserved component orthogonal to regressors, and rank condition). Moreover, $E(\epsilon_{it}) = 0$ (trivial when constant included in $x_{it}$). In consequence, $\bar{u}_{it}$ is an unbiased estimator of $E(c_{i})$.

Regarding consistency, the probability limit of this predictor is:

$$ p \lim\limits_{T \rightarrow \infty} \bar{u}_{it} = p \lim\limits_{T \rightarrow \infty} \left( \bar{x}_{i}\beta\right) - p \lim\limits_{T \rightarrow \infty} \left(\bar{x}_{i}\hat{\beta}\right) + p \lim\limits_{T \rightarrow \infty} c_{i} + p \lim\limits_{T \rightarrow \infty} \left( \frac{\sum_{t=1}^{T}\epsilon_{it}}{T}\right) $$

Again, $\hat{\beta}$ is a consistent estimator of $\beta$. This is, $p \lim\limits_{T \rightarrow \infty} \hat{\beta} = \beta $. Furthermore, $p \lim\limits_{T \rightarrow \infty} \left( \frac{\sum_{t=1}^{T}\epsilon_{it}}{T}\right) = E(\epsilon_{it})$, which is zero. Therefore:

$$ p \lim\limits_{T \rightarrow \infty} \bar{u}_{it} = c_{i} $$

Or, equivalently:

$$ \bar{u}_{it} \xrightarrow{P} c_{i} $$

This proves that the predictor is indeed BLUP.

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