Building Difference-GMM Instruments In Short Panels

Why Instrument Construction Is Equation-Specific

In a dynamic panel, differencing removes the time-invariant unit effect, but it also creates an error that contains two adjacent innovations. Consequently, a lagged outcome level that is valid for one differenced equation may be invalid for an earlier one. In a short panel, the available equations and instruments can be listed explicitly, making the timing logic especially clear.

This tutorial considers a balanced panel with only a lagged dependent variable as a regressor. The goal is to construct the instruments and the associated moment vector—not to choose a GMM weighting matrix or calculate an estimator.

Model and assumptions: For unit \(i\) and periods \(t=1,\ldots,T\), let \(y_{it}=\alpha_i+\rho y_{i,t-1}+u_{it}\) for \(t\geq 2\). Here \(\alpha_i\) is a time-invariant unit effect and \(u_{it}\) is an innovation. Assume the innovations have mean zero, are serially uncorrelated with variance \(\sigma_u^2\), and are orthogonal to the initial outcome and unit effect and to the past. These timing assumptions imply \(E[y_{i,t-s}\Delta u_{it}]=0\) for \(s\geq2\), where \(\Delta u_{it}=u_{it}-u_{i,t-1}\). The lagged outcome levels are the instruments considered here.

First differencing gives \(\Delta y_{it}=\rho\Delta y_{i,t-1}+\Delta u_{it}\), for \(t=3,\ldots,T\), where \(\Delta y_{it}=y_{it}-y_{i,t-1}\). There is no differenced equation at \(t=2\) in this setup because it would require the unobserved presample difference \(\Delta y_{i1}\). The first available equation is therefore at \(t=3\).

The timing restriction \(s\geq2\) says that the instrument must be at least two periods earlier than the outcome date of its differenced equation. For the equation at \(t=3\), this leaves \(y_{i1}\). For the equation at \(t=4\), it leaves \(y_{i1}\) and \(y_{i2}\). These are the uncollapsed, equation-specific instrument lists.

Three Observed Periods

With \(T=3\), there is just one differenced equation, at \(t=3\). Its dependent variable is \(\Delta y_{i3}\), its regressor is \(\Delta y_{i2}\), and its error is \(\Delta u_{i3}\). The only available lagged-level instrument satisfying the timing restriction is \(y_{i1}\).

\Delta y_{i3}=\rho\Delta y_{i2}+\Delta u_{i3}, \qquad Z_i=\begin{bmatrix}y_{i1}\end{bmatrix}, \qquad E[y_{i1}\Delta u_{i3}]=0.

Here \(Z_i\) is a one-row, one-column instrument matrix. Multiplying the instrument by the differenced error gives the unit-level moment \(g_i=Z_i'\Delta u_{i3}=y_{i1}\Delta u_{i3}\). Across \(N\) units, the sample moment is the sum (or, equivalently for setting it equal to zero, the average) of these unit-level moments.

g_N=\sum_{i=1}^{N}Z_i'\Delta u_{i3} =\sum_{i=1}^{N}y_{i1}\Delta u_{i3}, \qquad \bar g_N=\frac{1}{N}g_N.

Worked Example: The Instrument in a Three-Period Panel

Suppose a unit follows the model with \(\rho=0.5\) and \(\alpha_i=1\). Let \(y_{i1}=2\), \(u_{i2}=0.2\), and \(u_{i3}=-0.1\). The level equations imply \(y_{i2}=1+0.5(2)+0.2=2.2\) and \(y_{i3}=1+0.5(2.2)-0.1=2.0\). Thus \(\Delta y_{i2}=0.2\), \(\Delta y_{i3}=-0.2\), and \(\Delta u_{i3}=-0.1-0.2=-0.3\).

The differenced equation checks: \(\rho\Delta y_{i2}+\Delta u_{i3}=0.5(0.2)-0.3=-0.2=\Delta y_{i3}\). The instrument is \(y_{i1}=2\), and its realized moment contribution is \(g_i=2(-0.3)=-0.6\). This single realized contribution need not be zero; the moment restriction is an expectation across the assumed data-generating process, not a claim that every unit's product vanishes.

Four Observed Periods

With \(T=4\), there are two differenced equations: \(t=3\) and \(t=4\). The first equation can use \(y_{i1}\). The second can use \(y_{i1}\) and \(y_{i2}\). To keep the equation-specific instruments distinct, assign separate columns to them, even when the same level \(y_{i1}\) is used in both equations.

\begin{aligned} \Delta y_{i3}&=\rho\Delta y_{i2}+\Delta u_{i3},\\ \Delta y_{i4}&=\rho\Delta y_{i3}+\Delta u_{i4}, \end{aligned} \qquad \Delta u_i= \begin{bmatrix}\Delta u_{i3}\\\Delta u_{i4}\end{bmatrix}, \qquad Z_i= \begin{bmatrix} y_{i1}&0&0\\ 0&y_{i1}&y_{i2} \end{bmatrix}.

The first row corresponds to the \(t=3\) equation, and the second to \(t=4\). The columns correspond, in order, to \(y_{i1}\) used for \(t=3\), \(y_{i1}\) used for \(t=4\), and \(y_{i2}\) used for \(t=4\). The zero entries indicate that a column is not an instrument for that equation. Thus \(Z_i\) has two rows and three columns, and the unit-level moment vector has three components.

g_i=Z_i'\Delta u_i = \begin{bmatrix} y_{i1}\Delta u_{i3}\\ y_{i1}\Delta u_{i4}\\ y_{i2}\Delta u_{i4} \end{bmatrix}, \qquad g_N=\sum_{i=1}^{N}g_i.

The moment vector is not a list of three different equations of motion. It is a list of three instrument-error products: one for the \(t=3\) equation and two for the \(t=4\) equation. For a candidate value \(\rho\), the differenced residual vector is \(\Delta y_i-\rho\Delta y_{i,-1}\), where \(\Delta y_i=(\Delta y_{i3},\Delta y_{i4})'\) and \(\Delta y_{i,-1}=(\Delta y_{i2},\Delta y_{i3})'\). Replacing \(\Delta u_i\) by this residual vector gives the sample moments used in estimation.

Worked Example: Constructing and Checking Four-Period Moments

Consider two hypothetical units under a common \(\rho=0.5\). For unit A, set \(\alpha_A=1\), \(y_{A1}=2\), and \((u_{A2},u_{A3},u_{A4})=(0.2,-0.1,0.3)\). The level equations give \(y_{A2}=2.2\), \(y_{A3}=2.0\), and \(y_{A4}=2.3\). Therefore \((\Delta y_{A2},\Delta y_{A3},\Delta y_{A4})=(0.2,-0.2,0.3)\), while \((\Delta u_{A3},\Delta u_{A4})=(-0.3,0.4)\). Both differenced equations hold: \(0.5(0.2)-0.3=-0.2\) and \(0.5(-0.2)+0.4=0.3\).

For unit B, set \(\alpha_B=0.5\), \(y_{B1}=1\), and \((u_{B2},u_{B3},u_{B4})=(-0.1,0.2,-0.2)\). The level equations give \(y_{B2}=0.9\), \(y_{B3}=1.15\), and \(y_{B4}=0.875\). Hence \((\Delta y_{B2},\Delta y_{B3},\Delta y_{B4})=(-0.1,0.25,-0.275)\) and \((\Delta u_{B3},\Delta u_{B4})=(0.3,-0.4)\). Again, \(0.5(-0.1)+0.3=0.25\) and \(0.5(0.25)-0.4=-0.275\), so the differenced equations are consistent with the same \(\rho\).

For unit A, the three moment contributions are \(2(-0.3)=-0.6\), \(2(0.4)=0.8\), and \(2.2(0.4)=0.88\). For unit B, they are \(1(0.3)=0.3\), \(1(-0.4)=-0.4\), and \(0.9(-0.4)=-0.36\). Summing component by component gives \(g_N=(-0.3,\ 0.4,\ 0.52)'\). These are the realized sample moments for these two units at the model's \(\rho\); they are not required to equal zero in this small hypothetical sample. The key construction is the order and pairing of the three instrument columns with their corresponding differenced errors.

Why the Timing Restriction Matters

The exclusion of \(y_{i2}\) from the \(t=3\) equation is not a matter of convention. Under the model, \(y_{i2}=\alpha_i+\rho y_{i1}+u_{i2}\), while \(\Delta u_{i3}=u_{i3}-u_{i2}\). If \(u_{i2}\) and \(u_{i3}\) are serially uncorrelated and \(y_{i1}\) and \(\alpha_i\) are orthogonal to those innovations, then \(E[y_{i2}\Delta u_{i3}]=-\sigma_u^2\), not zero. The same innovation \(u_{i2}\) enters the level instrument positively and the differenced error negatively.

Worked Example: Auditing a Proposed Instrument Matrix

A proposed three-period instrument matrix uses \(y_{i1}\) and \(y_{i2}\) for the sole equation at \(t=3\). Check whether both columns satisfy the required moment condition. Under the assumptions above, \(y_{i1}\) contains no \(u_{i2}\) or \(u_{i3}\), so \(E[y_{i1}(u_{i3}-u_{i2})]=0\). It is a valid instrument.

For the second proposed instrument, expand the product using the model: \(E[y_{i2}\Delta u_{i3}]=E[(\alpha_i+\rho y_{i1}+u_{i2})(u_{i3}-u_{i2})]\). The terms involving \(\alpha_i\) and \(y_{i1}\) have expectation zero by orthogonality to future innovations. The remaining terms are \(E[u_{i2}u_{i3}]-E[u_{i2}^2]=0-\sigma_u^2=-\sigma_u^2\), which is nonzero when the innovation variance is positive. Thus \(y_{i2}\) is invalid for this equation and must not be added as a moment condition.

From Instrument Lists to a Stacked Moment Vector

The construction can be done systematically: list equations in time order, list eligible levels for each equation, then place each eligible level in its own column. In three periods, the list is just \(y_{i1}\) for \(t=3\). In four periods, it is \(y_{i1}\) for \(t=3\), followed by \(y_{i1},y_{i2}\) for \(t=4\). Multiplication by the stacked differenced-error vector then produces one moment component per instrument column.

Write the available differenced equations.
For observations beginning at \(t=1\), these run from \(t=3\) to \(T\).

Apply the timing rule.
For each equation at date \(t\), retain the observed outcome levels \(y_{i,t-s}\) with \(s\geq2\).

Arrange the matrix and moments.
Put equations in rows, give each equation-instrument pair its own column, and compute \(Z_i'\Delta u_i\).

The number of uncollapsed instrument columns for a balanced panel with \(T\) observed periods is \(1+2+\cdots+(T-2)=(T-1)(T-2)/2\). This count is three when \(T=4\), matching the matrix above. It counts equation-instrument pairs, not distinct outcome dates: \(y_{i1}\), for example, appears in two separate columns at \(T=4\) because it instruments two different equations.

A frequent pitfall is to list every past outcome level for every differenced equation. That ignores the moving-average error created by differencing. In particular, using \(y_{i,t-1}\) is generally invalid because it contains \(u_{i,t-1}\), which also appears with a negative sign in \(\Delta u_{it}\). The valid lag depth begins at two, subject to the stated innovation and initial-condition assumptions.

Key takeaway: In a short Difference-GMM panel, instrument construction is equation-specific. With three periods, the sole equation uses \(y_{i1}\); with four periods, the equations use \(y_{i1}\), then \(y_{i1}\) and \(y_{i2}\). The stacked moment vector records each valid instrument's product with its equation's differenced error.

Check Your Understanding

Use the timing rule and matrix structure developed above.

For a three-period panel, write the differenced equation and identify its valid lagged-level instrument.
For a four-period panel, list the valid instruments separately for \(t=3\) and \(t=4\).
In the four-period case, how many columns does the uncollapsed instrument matrix have, and why are there two separate columns containing \(y_{i1}\)?
Explain why \(y_{i2}\) is generally invalid as an instrument for the \(t=3\) differenced equation.
For \(T=5\), use the stated counting formula to find the number of uncollapsed instrument columns.