<p>You can exploit additional non-redundant moment conditions for the equations in levels by imposing the following initial condition assumption:
<span class="math-container">\begin{equation}
    E[\Delta y_{i2} \alpha_i] = 0
\end{equation}</span>
This assumption requires a mean stationarity restriction on the initial conditions <span class="math-container">$y_{i1}$</span>.</p>
<p>Recall the concept of a mean stationary series: <span class="math-container">$x_{it}$</span> is stationary in mean if <span class="math-container">$E[x_{it}] = E[x_{is}] = \mu &lt; \infty$</span>, where the first moment is time-independent and finite.</p>
<p>Now, take the expected value of <span class="math-container">$y_{it}$</span> conditioned on <span class="math-container">$\alpha_i$</span>. Due to mean stationarity,
<span class="math-container">\begin{equation}
    E[y_{it}|\alpha_i] = \rho E[y_{it}|\alpha_i] + \alpha_i
\end{equation}</span>
Eventually,
<span class="math-container">\begin{equation}
    E[y_{it}|\alpha_i] = \frac{\alpha_i}{1-\rho},
\end{equation}</span>
and this is the steady-state value of <span class="math-container">$y_{it}$</span>, where the series converges for the individual <span class="math-container">$i$</span>.</p>
<p>Now, exploiting the restriction on the initial conditions process generating <span class="math-container">$y_{i1}$</span>, I write <span class="math-container">$y_{i1}$</span> as
<span class="math-container">\begin{equation}
    y_{i1} = \frac{\alpha_i}{1-\rho} + e_{i1}
\end{equation}</span>
with <span class="math-container">$e_{i1}$</span> being an i.i.d. innovation.</p>
<p>Considering your DGP for the first period observed (at <span class="math-container">$t=2$</span>):
<span class="math-container">\begin{equation}
    y_{i2} = \rho y_{i1} + \alpha_i + \nu_{it}
\end{equation}</span></p>
<p>Subtracting <span class="math-container">$y_{i1}$</span> from both sides of this equation:
<span class="math-container">\begin{equation}
    \Delta y_{i2} = (\rho - 1) y_{i1} + \alpha_i + \nu_{i2}
\end{equation}</span>
and using the expression for <span class="math-container">$y_{i1}$</span>:
<span class="math-container">\begin{equation}
    \Delta y_{i2} = (\rho - 1) \left(\frac{\alpha_i}{1-\rho}+ e_{i1}\right) + \alpha_i + \nu_{i2}
\end{equation}</span>
And,
<span class="math-container">\begin{equation}
    \Delta y_{i2} = (\rho - 1) e_{i1} + \nu_{i2}
\end{equation}</span></p>
<p>Thus, you see that this moment condition is equivalent to the first one:
<span class="math-container">\begin{equation}
    E[\Delta y_{i2} \alpha_i] = E\{ [(\rho - 1) e_{i1} + \nu_{i2}]\alpha_i \} = 0
\end{equation}</span></p>
<p>Since, in defining the DGP, you assumed that the innovations <span class="math-container">$\nu_{it}$</span> are uncorrelated with the individual effects, the moment condition holds if <span class="math-container">$E[e_{i1} \alpha_i] = 0$</span>.</p>
<p>Indeed, assuming <span class="math-container">$E[\Delta y_{i2} \alpha_i] = 0$</span>, given the AR(1) structure, we have <span class="math-container">$E[\Delta y_{is} \alpha_i] = 0$</span>, for <span class="math-container">$s = 2, \ldots, T$</span>. To see this:
<span class="math-container">\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}</span></p>
<p>Generalizing,
<span class="math-container">\begin{equation}
    = \rho^{t-2} \Delta y_{i2} + \sum_{s=0}^{t-3} \rho^s \Delta \nu_{i,t-s}
\end{equation}</span>
for <span class="math-container">$t = 3, \ldots, T$</span>. This implies an additional <span class="math-container">$T - 2$</span> non-redundant linear moment conditions for the equations in levels, which can be written as
<span class="math-container">\begin{equation}
    E[ \Delta y_{i,t-1} (\alpha_i +\nu_{it})] = 0 \quad \text{for } t = 3, \ldots, T 
\end{equation}</span></p>
<p><strong>Intuition:</strong>
The mean stationarity assumption suggests that individual entities indexed by <span class="math-container">$i$</span> 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 <span class="math-container">$\frac{\alpha_i}{1-\rho}$</span>. For example, a positive fixed effect alone consistently boosts <span class="math-container">$y$</span> in each period, akin to how investment bolsters the capital stock. However, under the assumption that <span class="math-container">$|\rho| &lt; 1$</span>, this incremental effect (under stationarity) is offset by reversion to the mean in the long run.</p>
<p>The additional moment conditions from Blundell and Bond(1998):
<span class="math-container">\begin{equation}
    E[(\alpha_i +\nu_{it}) \Delta y_{i,t-1} ] = 0 \quad \text{for } t = 3, \ldots, T 
\end{equation}</span></p>
<p>can be rewritten as:
<span class="math-container">\begin{equation}
\begin{aligned}
    E\left[ (\alpha_i + \nu_{it})(y_{it-1} - y_{it-2}) \right] &amp;= E\left[ (\alpha_i + \nu_{it})(\rho y_{it-2} + \alpha_i + \nu_{it-1} - y_{it-2}) \right] \\
    &amp;= E\left[ (\alpha_i + \nu_{it})((\rho - 1)y_{it-2} + \alpha_i + \nu_{it-1}) \right] \\
    &amp;= E\left[ \alpha_i((\rho - 1)y_{it-2} + \alpha_i) \right] \\
    &amp;= 0
\end{aligned}
\end{equation}</span></p>
<p>which is equivalent to:
<span class="math-container">\begin{equation}
E\left[ \alpha_i((\rho - 1)y_{it} + \alpha_i) \right]  \quad \text{for } t \ge 1
\end{equation}</span></p>
<p>By dividing this condition by <span class="math-container">$(1-\rho)$</span>:</p>
<p><span class="math-container">\begin{equation}
E\left[ \alpha_i \left( y_{it} -  \frac{\alpha_i}{1-\rho} \right) \right] = 0
\end{equation}</span></p>
<p>Deviations from long-run means must not be correlated with the fixed effects (due to the stationarity hypothesis). If <span class="math-container">$E\left[ \alpha_i \left( y_{it} -  \frac{\alpha_i}{1-\rho} \right) \right] = 0$</span> holds in <span class="math-container">$ t$</span>, then it also holds in all subsequent periods. Effectively, this is a condition on the initial observation.</p>

CC BY-SA 4.0
2024-01-07T16:02:38.567Z
What I'm still missing is why mean stationarity is necessary to make the additional Bundell and Bond moment conditions to hold
Rebecca 
CC BY-SA 4.0
2024-01-07T17:37:52.463Z
Now, the concept should be clearer.
Tony
CC BY-SA 4.0
2024-01-06T18:30:55.190Z
2024-01-07T21:11:11.807Z
2024-01-07T21:11:11.807Z
Tony
<p>Without stationarity, the moment condition is not valid. We can consider the case where <span class="math-container">$\rho=1$</span>. Then <span class="math-container">$\Delta y_{it}=\alpha_i + \nu_{it} $</span> and
<span class="math-container">$$ E[\Delta y_{i,t-1} (\alpha_i + \nu_{it})] = E[ (\alpha_i + \nu_{i,t-1}) (\alpha_i + \nu_{it})],$$</span>
which is nonnegative if <span class="math-container">$\nu_{it}$</span> is not correlated with <span class="math-container">$(\alpha_i,\nu_{i,t-1})$</span>.</p>
<p><strong>EDIT</strong>: With mean stationarity, as defined in Tony's post, it follows that <span class="math-container">$E[y_{it}|\alpha_i] = \rho E[y_{i,t-1}|\alpha_i] + \alpha_i,$</span> and so,
<span class="math-container">$$ \Delta E[ y_{i,t-1} | \alpha_i ] = \rho \Delta E[ y_{i,t-2} | \alpha_i )] = 0, $$</span>
hence <span class="math-container">$ E[ \Delta y_{i,t-1} \alpha_i ] = 0$</span>. If in addition <span class="math-container">$ E[ \Delta y_{i,t-1} \nu_{i,t} ] = 0,$</span> then <span class="math-container">$$ E[ \Delta y_{i,t-1} ( \alpha_i+\nu_{i,t} ) ] = 0. $$</span></p>

CC BY-SA 4.0
2024-01-06T17:50:24.797Z
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 $\rho=1$?
Rebecca 
CC BY-SA 4.0
2024-01-06T19:05:12.733Z
@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 
$\Delta y_{i,t-1} = \alpha_i + \nu_{i,t-1}.$
Bertrand
CC BY-SA 4.0
2024-01-06T21:42:36.463Z
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 $|\rho| < 1$?
Rebecca 
CC BY-SA 4.0
2024-01-07T11:16:33.943Z
@Rebecca: see the edit.
Bertrand
CC BY-SA 4.0
2024-01-07T12:16:16.130Z
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
Rebecca 
CC BY-SA 4.0
2024-01-07T16:55:19.923Z
@Rebecca: Because the expectation is linear, $ \Delta E[ y_{i,t-1} | \alpha_i ] = E[ y_{i,t-1} | \alpha_i ] - E[ y_{i,t-2} | \alpha_i ]= E[ y_{i,t-1} - y_{i,t-2} | \alpha_i ] = E[ \Delta y_{i,t-1} | \alpha_i ].$ This term is null under mean stationarity.
Bertrand
CC BY-SA 4.0
2024-01-06T14:22:42.913Z
2024-01-07T13:30:28.327Z
2024-01-07T13:30:28.327Z
Bertrand
<p>Consider the dynamic linear model given by:
<span class="math-container">\begin{equation}
y_{it} = \rho y_{i,t-1} + \alpha_i + \nu_{it}
\end{equation}</span>
where <span class="math-container">$\alpha_i$</span> represents individual fixed effects. The GMM two-step estimator imposes <span class="math-container">$0.5(T-1)(T-2)$</span> moment restrictions, specifically:
<span class="math-container">\begin{equation}
E[y_{i,t-s} \Delta \nu_{it}] = 0
\end{equation}</span>
for <span class="math-container">$t=3,\ldots,T$</span> and <span class="math-container">$s \ge 2$</span>.</p>
<p>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 <span class="math-container">$y_{it}$</span>. Can you provide insight into how imposing a mild stationarity assumption on the series ensures the validity of the following moment restrictions:
<span class="math-container">\begin{equation}
E[\Delta y_{i,t-1} (\alpha_i + \nu_{it})] = 0
\end{equation}</span>
for <span class="math-container">$t=3,\ldots,T$</span></p>

CC BY-SA 4.0
2024-01-06T12:53:30.527Z
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.)
Richard Hardy
CC BY-SA 4.0
2024-01-07T11:14:48.147Z
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 $v_{it}$.
chan1142
<econometrics><paneldata>
System GMM Estimator
Rebecca 
# System GMM Estimator

(Asked by: username_0 on 2024-01-06)

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 $t=3,\ldots,T$

**Tags:** econometrics, paneldata
> **Comments:**
> **username_1**: 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.)  
> **username_2**: 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 $v_{it}$.  

---

## 2 answers

### Answer by username_3 on 2024-01-06. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: What I'm still missing is why mean stationarity is necessary to make the additional Bundell and Bond moment conditions to hold  
> **username_3**: Now, the concept should be clearer.  

---

### Answer by username_4 on 2024-01-06. Score: 1

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 $$ E[ \Delta y\_{i,t-1} ( \alpha\_i+\nu\_{i,t} ) ] = 0. $$

> **Comments:**
> **username_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 $\rho=1$?  
> **username_4**: @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  $\Delta y_{i,t-1} = \alpha_i + \nu_{i,t-1}.$  
> **username_0**: 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 $|\rho| < 1$?  
> **username_4**: @Rebecca: see the edit.  
> **username_0**: 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  
> **username_4**: @Rebecca: Because the expectation is linear, $ \Delta E[ y_{i,t-1} | \alpha_i ] = E[ y_{i,t-1} | \alpha_i ] - E[ y_{i,t-2} | \alpha_i ]= E[ y_{i,t-1} - y_{i,t-2} | \alpha_i ] = E[ \Delta y_{i,t-1} | \alpha_i ].$ This term is null under mean stationarity.  

---

