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
  • 1
    $\begingroup$ 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

  • $\begingroup$ 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.


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.