# System GMM Estimator

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

• 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.) Jan 6 at 12:53
• 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 $v_{it}$. Jan 7 at 11:14

You can exploit additional non-redundant moment conditions for the equations in levels by imposing the following initial condition assumption: $$$$E[\Delta y_{i2} \alpha_i] = 0$$$$ This assumption requires a mean stationarity restriction on the initial conditions $$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, $$$$E[y_{it}|\alpha_i] = \rho E[y_{it}|\alpha_i] + \alpha_i$$$$ Eventually, $$$$E[y_{it}|\alpha_i] = \frac{\alpha_i}{1-\rho},$$$$ and this is the steady-state value of $$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 $$$$y_{i1} = \frac{\alpha_i}{1-\rho} + e_{i1}$$$$ with $$e_{i1}$$ being an i.i.d. innovation.

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

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

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

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: $$$$\Delta y_{it} = \rho \Delta y_{i,t-1} + \Delta \nu_{it} = \rho [\alpha \Delta y_{i,t-2} + \Delta \nu_{i,t-1}] + \Delta \nu_{it} = \rho^2 \Delta y_{i,t-2} + \Delta \nu_{it} + \rho \Delta \nu_{i,t-1}$$$$

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

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): $$$$E[(\alpha_i +\nu_{it}) \Delta y_{i,t-1} ] = 0 \quad \text{for } t = 3, \ldots, T$$$$

can be rewritten as: \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}

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

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

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

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.

• What I'm still missing is why mean stationarity is necessary to make the additional Bundell and Bond moment conditions to hold Jan 7 at 16:02
• Now, the concept should be clearer.
– 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 $$E[ \Delta y_{i,t-1} ( \alpha_i+\nu_{i,t} ) ] = 0.$$

• 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 $\rho=1$? Jan 6 at 17:50
• @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 $\Delta y_{i,t-1} = \alpha_i + \nu_{i,t-1}.$ Jan 6 at 19:05
• 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 $|\rho| < 1$? Jan 6 at 21:42
• @Rebecca: see the edit. Jan 7 at 11:16
• 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 Jan 7 at 12:16