\begin{equation}
y_{it} = \beta_0 + \beta_1y_{i(t-1)} + \beta_2y_{i(t-2)} + \alpha_i + \varepsilon_{it}
\end{equation}
Your instruments in this case will only use two lags which in the example I gave you before gives:
\begin{eqnarray} t=6 & \quad y_{t-4}, y_{t-3} \\ t=5 & \quad y_{t-3}, y_{t-2}\\ t=4 & \quad y_{t-2}, y_{t-1} \\ t=3 & \quad y_{t-1} \end{eqnarray} Ultimately what you did was to reduce the instrument set by including at most 2 lags.
y_{it} = \beta_0 + \beta_1y_{i(t-1)} + \beta_2y_{i(t-2)} + \alpha_i + \varepsilon_{it}
\end{equation}
Your instruments in this case will only use two lags which in the example I gave you before gives:
\begin{eqnarray} t=6 & \quad y_{t-4}, y_{t-3} \\ t=5 & \quad y_{t-3}, y_{t-2}\\ t=4 & \quad y_{t-2}, y_{t-1} \\ t=3 & \quad y_{t-1} \end{eqnarray} Ultimately what you did was to reduce the instrument set by including at most 2 lags.
