How are unobserved components predicted in random effect models? 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.
 A: 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
A: 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.
