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Typically under the model that looks like this:

$$ y_{it} = \rho y_{it-1} + u_i + e_{it} = \rho y_{it-1} + v_{it} $$

Define the OLS estimator as the estimator for $\rho$ that estimates the following equation.

$$ y_{it} = \rho y_{it-1} + u_{it} $$

Define the FE estimator as the estimator for $\rho$ that includes fixed effects:

$$ y_{it} = \rho y_{it-1} + u_i + e_{it} $$

We know that if $\rho>0$, then the OLS estimator for $\rho$ is biased upwards and the fixed effects estimator for $\rho$ is biased downwards, according to the Nickell bias, and therefore we can bound the coefficient by using OLS and FE estimators.

However, I'm wondering if the case where we have this model:

$$ y_{it} = \rho y_{it-1} + \beta x_{it} + u_i + e_{it} $$

Is the OLS estimator for $\beta$ (not for $\rho$) still biased upwards and the fixed effects estimator biased downwards? What if we use a nonparametric estimator for $\rho$? aka instead of estimating the original equation, we now have:

$$ y_{it} = f(y_{it-1}) + \beta x_{it} + u_i + e_{it} $$

What if we don't know whether the $t-2,...$ variables factor in?

$$ y_{it} = f(y_{it-1}, y_{it-2}, y_{it-3}, ...) + \beta x_{it} + u_i + e_{it} $$

Thanks for your time in advance!

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    $\begingroup$ My first thought is that, even if you know the direction of the expected bias, in any finite sample the estimate could be very different. Thus, there's no guarantee for performance. Having said that, no I don't know, but I'm guessing you can create fake data and simulate this situation, then get an idea. Good luck! $\endgroup$ Commented Jun 11 at 8:36
  • $\begingroup$ ok, fair, but that could be said of any biased estimator. $\endgroup$
    – Daycent
    Commented Jun 13 at 18:57

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