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.) $\endgroup$ Commented 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
    Commented 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
    Commented Jan 7 at 16:02
  • $\begingroup$ Now, the concept should be clearer. $\endgroup$
    – Tony
    Commented 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
    Commented 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
    Commented 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
    Commented Jan 6 at 21:42
  • $\begingroup$ @Rebecca: see the edit. $\endgroup$
    – Bertrand
    Commented 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
    Commented Jan 7 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.