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

CC BY-SA 4.0
2022-01-16T09:06:43.843Z
I mean, I know that if $T$ is big enough it doesn't really matter. I'm asking if there is a default norm.
Michael Gmeiner
CC BY-SA 4.0
2022-01-15T15:57:46.117Z
2022-01-15T15:57:46.117Z
tdm
<p>Suppose we have a panel of data in which we observe <span class="math-container">$N$</span> individuals for <span class="math-container">$T$</span> time periods. We want to estimate:
<span class="math-container">$$y_{it}= \delta y_{it-1}+x_{it}'\beta +\alpha_i +v_{it} $$</span></p>
<p>where <span class="math-container">$\alpha_i$</span> is a time-invariant unobserved error and <span class="math-container">$v_{it}$</span> is an i.i.d. shock independent of all regressors.</p>
<p>The fixed effects transformation would be to subtract the means of each variable. That is, to estimate</p>
<p><span class="math-container">$$y_{it} -\bar{y_i}= \delta y_{it-1}-\bar{y_i}+x_{it}'\beta-\bar{x_i}'\beta +v_{it}-\bar{v_i} $$</span></p>
<p>Due to the lagged dependent variable, we can only use observations at times <span class="math-container">$2$</span> through <span class="math-container">$T$</span>. 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, <span class="math-container">$\bar{y_i}=\frac{1}{T-1}\sum_{t=2}^Ty_{it}$</span> or instead use all data, even the time 1 outcome? That is, <span class="math-container">$\bar{y_i}=\frac{1}{T}\sum_{t=1}^Ty_{it}$</span>.</p>
<p>Is there a standard norm for this?</p>
<p>There is an analogous question regarding the subtraction of the mean for the lagged dependent variable on the right side.</p>
<p>Also, to preemptively eliminate an answer addressing this point, I do in fact know the FE estimator is biased in this setting.</p>

<fixed-effects>
Fixed Effects with lagged dependent variable, how is the average calculated?
Michael Gmeiner
# Fixed Effects with lagged dependent variable, how is the average calculated?

(Asked by: username_0 on 2022-01-14)

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.

**Tags:** fixed-effects


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## 1 answers

### Answer by username_1 on 2022-01-15. Score: 2

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.

> **Comments:**
> **username_0**: I mean, I know that if $T$ is big enough it doesn't really matter. I'm asking if there is a default norm.  

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