Why Work Through Multiple Instruments?
With one endogenous regressor and more than one excluded instrument, 2SLS does not estimate a separate coefficient for each instrument and then automatically average them. It first projects the endogenous regressor onto the space spanned by all the instruments. The final slope uses that fitted component. Working through the projection makes clear both how the instruments jointly contribute and why the answer can differ from each instrument's separate covariance ratio.
The examples below focus on the point estimate. Instrument validity is a substantive assumption, not something established by calculating 2SLS: the instruments must be relevant and satisfy the required exogeneity and exclusion conditions. A numerical example can check whether the sample projection is defined, but cannot establish those population conditions.
Model and Sample 2SLS Formula
For the sample calculation, center \(y\), \(d\), and each instrument by subtracting their sample means. Write the centered vectors as \(\tilde y\), \(\tilde d\), and the \(n\)-by-2 matrix of centered instruments as \(\tilde Z\). This centering is equivalent to including an intercept in both stages. Provided the two instrument columns are linearly independent, their projection matrix is
The projection \(P_{\tilde Z}\tilde d\) is the part of centered \(d\) that lies in the instruments' sample span. Denote this fitted vector by \(\widehat{\tilde d}\). The 2SLS slope and intercept are
The formula requires a nonzero fitted component of \(d\), so that \(\widehat{\tilde d}'\widehat{\tilde d}>0\). This is the sample counterpart of the rank condition for identifying the coefficient. The expression is a scalar here because there is one endogenous regressor. It also shows what “combining the instruments” means: the fitted values come from projection onto their joint span, rather than from choosing one instrument alone.
Worked Example: Computing 2SLS With Two Orthogonal Instruments
Suppose a hypothetical sample has four observations. The variables are centered, and the data are:
| Observation | \(z_1\) | \(z_2\) | \(d\) | \(y\) |
|---|---|---|---|---|
| 1 | -1 | -1 | -2 | -5 |
| 2 | -1 | 1 | -2 | -3 |
| 3 | 1 | -1 | 0 | -1 |
| 4 | 1 | 1 | 4 | 9 |
Check the projection. Both instrument columns have mean zero, are linearly independent, and are orthogonal: \(z_1'z_2=0\). The data satisfy \(d=2z_1+z_2+q\), where \(q=(1,-1,-1,1)'\). Since \(q\) is orthogonal to both instruments, projecting \(d\) onto their span gives
For instance, \(z_1'd=8\), \(z_1'z_1=4\), \(z_2'd=4\), and \(z_2'z_2=4\). Orthogonality means the projection coefficients are \(8/4=2\) and \(4/4=1\), respectively. Likewise, because \(y=4z_1+3z_2+2q\) and \(q\) is orthogonal to the instruments, its projection is \(P_{\tilde Z}\tilde y=4z_1+3z_2\).
Compute the slope. The numerator and denominator in the projection formula are
Therefore, \(\hat\beta_{\mathrm{2SLS}}=44/20=2.200\). Since the variables are centered, \(\bar d=\bar y=0\) and the fitted intercept is zero. The result is a sample slope of 2.200 outcome units per one unit of \(d\); it is a descriptive estimate under the IV model, not proof that the instruments satisfy their assumptions.
Instrument-Specific Ratios and the Orthogonal Case
A single instrument \(z_j\), considered on its own, gives the centered sample IV ratio
provided \(z_j'\tilde d\ne0\). This is the sample covariance ratio for that instrument; the common sample-size factor in the two covariances cancels. It is useful to compare these ratios with 2SLS, but they are not generally the formula for the multiple-instrument estimate.
When the centered instrument columns are mutually orthogonal, the projection matrix can be written as a sum of individual projections:
Substituting this expression into the 2SLS formula gives, for one endogenous regressor,
To see the last equality, replace \(z_j'\tilde y\) by \((z_j'\tilde d)\hat\beta_j\) in each numerator term. The resulting factor multiplying \(\hat\beta_j\) is exactly \(w_j\). Each weight is nonnegative, and it is positive when that instrument has a nonzero sample association with \(d\). Thus, in this orthogonal case, the joint estimate is a weighted average of the instrument-specific ratios with weights determined by relevance and instrument scale.
Worked Example: Comparing Ratios and Their 2SLS Weights
Use the data from the preceding example. The first instrument's ratio is
because \(z_1'\tilde y=16\) and \(z_1'\tilde d=8\). The second instrument's ratio is
To verify how 2SLS combines them, calculate the weights:
The combined estimate is therefore
The first instrument receives the larger weight in this sample because its association with \(d\), measured relative to its squared length, is larger. This calculation agrees with the projection estimate. The numerical comparison does not establish that one instrument is more credible or valid; those are separate questions about the data-generating process.
When the Instruments Are Not Orthogonal
For nonorthogonal instruments, the joint projection still defines 2SLS, but the projection matrix includes cross-products between instrument columns. It cannot generally be decomposed into a sum of separate, independent instrument projections. Consequently, the orthogonal-case positive-weight formula does not generally apply. In particular, it is unsafe to treat the 2SLS estimate as an ordinary average of the separate ratios.
Worked Example: Projection With Nonorthogonal Instruments
Consider a second hypothetical, centered sample with \(z_1=(-1,-1,1,1)'\), \(z_2=(-1,0,0,1)'\), \(d=(-1,-2,0,3)'\), and \(y=(-3,-5,-1,9)'\). The columns are linearly independent, but not orthogonal: \(z_1'z_2=2\). Let \(q=(1,-1,-1,1)'\). Directly, \(z_1'q=z_2'q=0\), and the data can be written as
Because \(q\) is orthogonal to both instruments, the projections are \(\widehat{\tilde d}=z_1+z_2=(-2,-1,1,2)'\) and \(P_{\tilde Z}\tilde y=2z_1+4z_2=(-6,-2,2,6)'\). Hence
The separate ratios are \(\hat\beta_1=(z_1'y)/(z_1'd)=16/6\approx2.667\) and \(\hat\beta_2=(z_2'y)/(z_2'd)=12/4=3.000\). They differ from one another and from the joint estimate. The projection formula remains valid; the shortcut based on separate positive weights is what no longer follows. In this particular sample the joint estimate happens to lie between the two ratios, but nonorthogonality alone does not establish a general weighted-average interpretation.
How to Read the Calculation
The projection view clarifies the roles of the instruments. First, their joint span determines the fitted component of the endogenous regressor. Second, the outcome is related to that fitted component to obtain the 2SLS slope. The resulting estimate depends on the full instrument space, not merely on a list of individual ratios. With orthogonal instruments, this joint calculation simplifies to a positive weighted average; with correlated instruments, the joint projection accounts for their overlap.
Overidentification means there is more instrument information than is needed to identify one coefficient under the model. It does not mean that every instrument yields the same finite-sample ratio, nor does the arithmetic average of those ratios generally equal 2SLS. A discrepancy between instrument-specific estimates can be informative to investigate, but it does not by itself identify which instrument is invalid or establish why the estimates differ.
These examples calculate point estimates only. The ordinary least-squares standard error from a regression on fitted \(d\) is not, in general, a valid 2SLS standard error. Inference must account for the IV estimator and the assumptions used for its variance calculation.
Check Your Understanding
Use the projection formula and the assumptions stated above to answer these questions.
- With one endogenous regressor and three excluded instruments, is the model underidentified, exactly identified, or overidentified? State the comparison that determines your answer.
- Why does centering \(y\), \(d\), and the instruments allow the displayed slope formula to include an intercept in the original model?
- In the orthogonal-instrument example, which instrument-specific ratio receives more weight, and what sample quantities determine that weight?
- For the nonorthogonal example, calculate \(z_1'z_1\), \(z_2'z_2\), and \(z_1'z_2\). Which of these quantities shows that the instruments are not orthogonal?
- Does a computed 2SLS coefficient establish that the instruments satisfy the exclusion restriction? Explain the distinction between calculation and identification assumptions.