Consider the dynamic linear model given by: \begin{equation} y_{it} = \rho y_{i,t-1} + \alpha_i + \nu_{it} \end{equation} where $\alpha_i$ represents individual fixed effects. The GMM two-step estimator imposes $0.5(T-1)(T-2)$ moment restrictions, specifically: \begin{equation} E[y_{i,t-s} \Delta \nu_{it}] = 0 \end{equation} for $t=3,\ldots,T$ and $s \ge 2$.

The System-GMM estimator (Bundell and Bond, 1998) extends these moment restrictions beyond the first differenced equations to also include the levels equation. This is achieved by assuming a mild stationarity assumption on the series $y_{it}$. Can you provide insight into how imposing a mild stationarity assumption on the series ensures the validity of the following moment restrictions: \begin{equation} E[\Delta y_{i,t-1} (\alpha_i + \nu_{it})] = 0 \end{equation} for $t=3,\ldots,T$

  • $\begingroup$ If you do not get an answer here in, say, a week, you may consider posting the question on Cross Validated Stack Exchange. (Posting on multiple Stack Exchange sites simultaneously is frowned up, so I would wait a week.) $\endgroup$ Jan 6 at 12:53
  • $\begingroup$ A simple intuition is that when $y_{it}$ is mean stationary, $\Delta y_{it}$ (and thus $\Delta y_{i,t-1}$) is uncorrelated with $\alpha_i$. Also $\Delta y_{i,t-1}$ is uncorrelated with $v_{it}$. $\endgroup$
    – chan1142
    Jan 7 at 11:14

2 Answers 2


You can exploit additional non-redundant moment conditions for the equations in levels by imposing the following initial condition assumption: \begin{equation} E[\Delta y_{i2} \alpha_i] = 0 \end{equation} This assumption requires a mean stationarity restriction on the initial conditions $y_{i1}$.

Recall the concept of a mean stationary series: $x_{it}$ is stationary in mean if $E[x_{it}] = E[x_{is}] = \mu < \infty$, where the first moment is time-independent and finite.

Now, take the expected value of $y_{it}$ conditioned on $\alpha_i$. Due to mean stationarity, \begin{equation} E[y_{it}|\alpha_i] = \rho E[y_{it}|\alpha_i] + \alpha_i \end{equation} Eventually, \begin{equation} E[y_{it}|\alpha_i] = \frac{\alpha_i}{1-\rho}, \end{equation} and this is the steady-state value of $y_{it}$, where the series converges for the individual $i$.

Now, exploiting the restriction on the initial conditions process generating $y_{i1}$, I write $y_{i1}$ as \begin{equation} y_{i1} = \frac{\alpha_i}{1-\rho} + e_{i1} \end{equation} with $e_{i1}$ being an i.i.d. innovation.

Considering your DGP for the first period observed (at $t=2$): \begin{equation} y_{i2} = \rho y_{i1} + \alpha_i + \nu_{it} \end{equation}

Subtracting $y_{i1}$ from both sides of this equation: \begin{equation} \Delta y_{i2} = (\rho - 1) y_{i1} + \alpha_i + \nu_{i2} \end{equation} and using the expression for $y_{i1}$: \begin{equation} \Delta y_{i2} = (\rho - 1) \left(\frac{\alpha_i}{1-\rho}+ e_{i1}\right) + \alpha_i + \nu_{i2} \end{equation} And, \begin{equation} \Delta y_{i2} = (\rho - 1) e_{i1} + \nu_{i2} \end{equation}

Thus, you see that this moment condition is equivalent to the first one: \begin{equation} E[\Delta y_{i2} \alpha_i] = E\{ [(\rho - 1) e_{i1} + \nu_{i2}]\alpha_i \} = 0 \end{equation}

Since, in defining the DGP, you assumed that the innovations $\nu_{it}$ are uncorrelated with the individual effects, the moment condition holds if $E[e_{i1} \alpha_i] = 0$.

Indeed, assuming $E[\Delta y_{i2} \alpha_i] = 0$, given the AR(1) structure, we have $E[\Delta y_{is} \alpha_i] = 0$, for $s = 2, \ldots, T$. To see this: \begin{equation} \Delta y_{it} = \rho \Delta y_{i,t-1} + \Delta \nu_{it} = \rho [\alpha \Delta y_{i,t-2} + \Delta \nu_{i,t-1}] + \Delta \nu_{it} = \rho^2 \Delta y_{i,t-2} + \Delta \nu_{it} + \rho \Delta \nu_{i,t-1} \end{equation}

Generalizing, \begin{equation} = \rho^{t-2} \Delta y_{i2} + \sum_{s=0}^{t-3} \rho^s \Delta \nu_{i,t-s} \end{equation} for $t = 3, \ldots, T$. This implies an additional $T - 2$ non-redundant linear moment conditions for the equations in levels, which can be written as \begin{equation} E[ \Delta y_{i,t-1} (\alpha_i +\nu_{it})] = 0 \quad \text{for } t = 3, \ldots, T \end{equation}

Intuition: The mean stationarity assumption suggests that individual entities indexed by $i$ can temporarily deviate from their respective steady states. However, these deviations are not systematic, and over the long run, the series tends to converge back to its steady state $\frac{\alpha_i}{1-\rho}$. For example, a positive fixed effect alone consistently boosts $y$ in each period, akin to how investment bolsters the capital stock. However, under the assumption that $|\rho| < 1$, this incremental effect (under stationarity) is offset by reversion to the mean in the long run.

The additional moment conditions from Blundell and Bond(1998): \begin{equation} E[(\alpha_i +\nu_{it}) \Delta y_{i,t-1} ] = 0 \quad \text{for } t = 3, \ldots, T \end{equation}

can be rewritten as: \begin{equation} \begin{aligned} E\left[ (\alpha_i + \nu_{it})(y_{it-1} - y_{it-2}) \right] &= E\left[ (\alpha_i + \nu_{it})(\rho y_{it-2} + \alpha_i + \nu_{it-1} - y_{it-2}) \right] \\ &= E\left[ (\alpha_i + \nu_{it})((\rho - 1)y_{it-2} + \alpha_i + \nu_{it-1}) \right] \\ &= E\left[ \alpha_i((\rho - 1)y_{it-2} + \alpha_i) \right] \\ &= 0 \end{aligned} \end{equation}

which is equivalent to: \begin{equation} E\left[ \alpha_i((\rho - 1)y_{it} + \alpha_i) \right] \quad \text{for } t \ge 1 \end{equation}

By dividing this condition by $(1-\rho)$:

\begin{equation} E\left[ \alpha_i \left( y_{it} - \frac{\alpha_i}{1-\rho} \right) \right] = 0 \end{equation}

Deviations from long-run means must not be correlated with the fixed effects (due to the stationarity hypothesis). If $E\left[ \alpha_i \left( y_{it} - \frac{\alpha_i}{1-\rho} \right) \right] = 0$ holds in $ t$, then it also holds in all subsequent periods. Effectively, this is a condition on the initial observation.

  • $\begingroup$ What I'm still missing is why mean stationarity is necessary to make the additional Bundell and Bond moment conditions to hold $\endgroup$
    – Rebecca
    Jan 7 at 16:02
  • $\begingroup$ Now, the concept should be clearer. $\endgroup$
    – Tony
    Jan 7 at 17:37

Without stationarity, the moment condition is not valid. We can consider the case where $\rho=1$. Then $\Delta y_{it}=\alpha_i + \nu_{it} $ and $$ E[\Delta y_{i,t-1} (\alpha_i + \nu_{it})] = E[ (\alpha_i + \nu_{i,t-1}) (\alpha_i + \nu_{it})],$$ which is nonnegative if $\nu_{it}$ is not correlated with $(\alpha_i,\nu_{i,t-1})$.

EDIT: With mean stationarity, as defined in Tony's post, it follows that $E[y_{it}|\alpha_i] = \rho E[y_{i,t-1}|\alpha_i] + \alpha_i,$ and so, $$ \Delta E[ y_{i,t-1} | \alpha_i ] = \rho \Delta E[ y_{i,t-2} | \alpha_i )] = 0, $$ hence $ E[ \Delta y_{i,t-1} \alpha_i ] = 0$. If in addition $ E[ \Delta y_{i,t-1} \nu_{i,t} ] = 0,$ then $$ E[ \Delta y_{i,t-1} ( \alpha_i+\nu_{i,t} ) ] = 0. $$

  • $\begingroup$ What do you mean? We know that $y_{i,t-1} = \rho y_{i,t-2} + \alpha_i + \nu_{t,-1}$, why does the term $y_{i,t-2}$ vanish in $E[\cdot]$ for $\rho=1$? $\endgroup$
    – Rebecca
    Jan 6 at 17:50
  • $\begingroup$ @Rebecca: If $\rho=1$, your first equation yields $y_{it} = y_{i,t-1} + \alpha_i + \nu_{it}$ or in other words $\Delta y_{i,t} = y_{it} - y_{i,t-1} = \alpha_i + \nu_{it},$ and $\Delta y_{i,t-1} = \alpha_i + \nu_{i,t-1}.$ $\endgroup$
    – Bertrand
    Jan 6 at 19:05
  • $\begingroup$ Right, it was trivial. However, my confusion lies in understanding why, when $|\rho| < 1$, the moment condition holds. To illustrate this, consider $y_{it} - y_{i,t-1} = (\rho - 1)y_{i,t-1} + \alpha_i + \nu_{it}$. Lagging by one period, the covariance in population with $(\alpha_i + \nu_{i,t})$ reads: $E[((\rho - 1)y_{i,t-2} + \alpha_i + \nu_{i,t-1})(\alpha_i + \nu_{i,t-1})]$. Could you explain why this expectation is expected to be zero when $|\rho| < 1$? $\endgroup$
    – Rebecca
    Jan 6 at 21:42
  • $\begingroup$ @Rebecca: see the edit. $\endgroup$
    – Bertrand
    Jan 7 at 11:16
  • $\begingroup$ But, the expectation of $y_{i,t-1}$ conditioned on $\alpha_i$ is not zero, as pointed out by Tony. I also do not get how you can move the difference operator outsitde the $E[\cdot]$ term $\endgroup$
    – Rebecca
    Jan 7 at 12:16

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