Fixed Effects with lagged dependent variable, how is the average calculated?

Question

Suppose we have a panel of data in which we observe $N$ individuals for $T$ time periods. We want to estimate: $$y_{it}= \delta y_{it-1}+x_{it}'\beta +\alpha_i +v_{it} $$

where $\alpha_i$ is a time-invariant unobserved error and $v_{it}$ is an i.i.d. shock independent of all regressors.

The fixed effects transformation would be to subtract the means of each variable. That is, to estimate

$$y_{it} -\bar{y_i}= \delta y_{it-1}-\bar{y_i}+x_{it}'\beta-\bar{x_i}'\beta +v_{it}-\bar{v_i} $$

Due to the lagged dependent variable, we can only use observations at times $2$ through $T$. Thus, when we subtract the mean of the outcome, there is some ambiguity. Should we average the outcome only for the outcomes actually used, that is, $\bar{y_i}=\frac{1}{T-1}\sum_{t=2}^Ty_{it}$ or instead use all data, even the time 1 outcome? That is, $\bar{y_i}=\frac{1}{T}\sum_{t=1}^Ty_{it}$.

Is there a standard norm for this?

There is an analogous question regarding the subtraction of the mean for the lagged dependent variable on the right side.

Also, to preemptively eliminate an answer addressing this point, I do in fact know the FE estimator is biased in this setting.

Answer 1

You have the equation: $$ y_{it} = \delta y_{it-1} + x_{it}'\beta+ \alpha_i + v_{it}. $$ The left hand side runs from $t=2$ to $t = T$ as there is a $y_{it-1}$ on the right hand side. So taking the average over $t = 2$ to $t=T$ of this equation gives: $$ y_{it} - \bar y_{i[2,T]} = \delta y_{it-1} - \bar y_{i[1,T-1]} + x_{it}'\beta - \bar x_{i[2,T]} + v_{it} - \bar v_{i[2,T]} $$ where $$ \begin{align*} &\bar y_{i[2,T]} = \frac{1}{N-1} \sum_{t = 2}^T y_{it},\\ &\bar y_{i[1,T-1]} = \frac{1}{N-1} \sum_{t = 1}^{T-1} y_{it}\\ &\bar x_{i[2,T]} = \frac{1}{N-1} \sum_{t = 2}^T x_{i,t}\\ &\bar v_{i[2,T]} = \frac{1}{N-1} \sum_{t = 2}^T v_{i,t}. \end{align*} $$ So the mean you subtract from the left hand side is different from the mean you subtract from the right hand side. If $T$ is big enough, the difference between $\bar y_{i[2,T]}$ and $y_{i[1,T-1]}$ should be very small, so it won't really matter which average you take.