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| Title | Two-stage least-squares regression | |
| Author | Vince Wiggins, StataCorp |
Note: This model could also be fit with
sem, using
maximum likelihood instead of a two-step method.
You can find examples for recursive models fit with sem in
the “Structural models: Dependencies between response
variables” section of [SEM] intro 5 — Tour of models.
Someone posed the following question:
I then want to instrument \(W\) with \(Q\). I know the first-stage regression is supposed to be
(for instance, use all the exogenous variables in the first stage). Actually this is automatically done if I use the ivregress command. However, I only want to use \(Q\) to instrument \(W\) without using \(X\) and \(Z\) in the first stage. Is there a way I can do it in Stata? I can regress \(W\) on \(Q\) and get the predicted \(W\), and then use it in the second-stage regression. The standard errors will, however, be incorrect.
ivregress will not let you do this and, moreover, if you believe \(W\) to be endogenous because it is part of a system, then you must include \(X\) and \(Z\) as instruments, or you will get biased estimates for b, c, and d.
Consider the system
Warning: Assume we are estimating structural equation (1); if \(X1\) and \(X2\) are exogenous, then they must be kept as instruments or your estimates will be biased. In a general system, such exogenous variables must be used as instruments for any endogenous variables when the instrumented value for the endogenous variables appears in an equation in which the exogenous variable also appears.
Consider the reduced forms of your two equations:
where \(e\#\) and \(f\#\) are combinations of the \(a\#\) and \(b\#\) coefficients from (1) and (2) and \(u1\) and \(u2\) are linear combinations of \(e1\) and \(e2\).
All exogenous variables appear in each equation for an endogenous variable. This is the nature of simultaneous systems, so efficiency argues that all exogenous variables be included as instruments for each endogenous variable.
Here is the real problem. Take (1): the reduced-form equation for \(Y2\), (2r), clearly shows that \(Y2\) is correlated with \(X2\) (by the coefficient \(f2\)). If we do not include \(X2\) among the instruments for \(Y2\), then we will have failed to account for the correlation of \(Y2\) with \(X2\) in its instrumented values. Since we did not account for this correlation, when we estimate (1) with the instrumented values for \(Y2\), the coefficient \(a3\) will be forced to account for this correlation. This approach will lead to biased estimates of both \(a1\) and \(a3\).
For a brief reference, see Baltagi (2011). See the whole discussion of 2SLS, particularly the paragraph after equation 11.40, on page 265. (I have no idea why this issue is not emphasized in more books.)
Failing to include \(X4\) affects only efficiency and not bias.
However, there is one case where it is not necessary to include \(X1\) and \(X2\) as instruments for \(Y2\). That is when the system is triangular such that \(Y2\) does not depend on \(Y1\), but you believe it is weakly endogenous because the disturbances are correlated between the equations. You are still consistent here to do what ivregress does and retain \(X1\) and \(X2\) as instruments. They are, however, no longer required. Then you could do what you suggested and just regress on the predicted instruments from the first stage.
If you do use this method of indirect least squares, you will have to perform the adjustment to the covariance matrix yourself. Consider the structural equation
where you have an instrument \(z1\) and you do not think that \(y2\) is a function of \(y1\).
The following example uses only \(z1\) as an instrument for \(y2\). Let's begin by creating a dataset (containing made-up data) on \(y1\), \(y2\), \(x1\), and \(z1\):
. sysuse auto (1978 automobile data) . rename price y1 . rename mpg y2 . rename displacement z1 . rename turn x1
Now we perform the first-stage regression and get predictions for the instrumented variable, which we must do for each endogenous right-hand-side variable.
. regress y2 z1
| Source | SS df MS | Number of obs = 74 | |
| F(1, 72) = 71.41 | |||
| Model | 1216.67534 1 1216.67534 | Prob > F = 0.0000 | |
| Residual | 1226.78412 72 17.0386683 | R-squared = 0.4979 | |
| Adj R-squared = 0.4910 | |||
| Total | 2443.45946 73 33.4720474 | Root MSE = 4.1278 |
| y2 | Coefficient Std. err. t P>|t| [95% conf. interval] | |
| z1 | -.0444536 .0052606 -8.45 0.000 -.0549405 -.0339668 | |
| _cons | 30.06788 1.143462 26.30 0.000 27.78843 32.34733 | |
| Source | SS df MS | Number of obs = 74 | |
| F(2, 71) = 12.41 | |||
| Model | 164538571 2 82269285.5 | Prob > F = 0.0000 | |
| Residual | 470526825 71 6627138.38 | R-squared = 0.2591 | |
| Adj R-squared = 0.2382 | |||
| Total | 635065396 73 8699525.97 | Root MSE = 2574.3 |
| y1 | Coefficient Std. err. t P>|t| [95% conf. interval] | |
| y2hat | -463.4688 117.187 -3.95 0.000 -697.1329 -229.8046 | |
| x1 | -126.4979 108.7468 -1.16 0.249 -343.3328 90.33697 | |
| _cons | 21051.36 6451.837 3.26 0.002 8186.762 33915.96 | |
Now we correct the variance–covariance by applying the correct mean squared error:
. rename y2hat y2hold . rename y2 y2hat . predict double res, residual . rename y2hat y2 /* put back real y2 */ . rename y2hold y2hat . replace res = res^2 (74 real changes made) . summarize res
| Variable | Obs Mean Std. dev. Min Max | |
| res | 74 7553657 1.43e+07 117.4375 1.06e+08 |
| y1 | Coefficient Std. err. t P>|t| [95% conf. interval] | |
| y2hat | -463.4688 127.7267 -3.63 0.001 -718.1485 -208.789 | |
| x1 | -126.4979 118.5274 -1.07 0.289 -362.8348 109.8389 | |
| _cons | 21051.36 7032.111 2.99 0.004 7029.73 35072.99 | |
