Eric Rasmusen’s Weblog

Somebody asked me about Wald tests, and I thought I’d write up my answer as a weblog entry, so someone can correct me if I’m wrong. Here’s some STATA output:

						Wald chi2(5)    =     40.86
                                                  Prob > chi2     =    0.0000

------------------------------------------------------------------------------
  LnTax04Tob |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
    attrny05 |  -.0002878   .0000518    -5.56   0.000    -.0003893   -.0001863
      flunks |  -.0252277   .0092373    -2.73   0.006    -.0433325   -.0071228
      utokyo |  -.1934326   .1203548    -1.61   0.108    -.4293237    .0424585
     exprnce |  -.0007321   .0034238    -0.21   0.831    -.0074426    .0059784
         Sex |   .8102301    .349133     2.32   0.020     .1259419    1.494518
       _cons |   8.559199   .3726405    22.97   0.000     7.828837    9.289561
------------------------------------------------------------------------------
Instrumented:  attrny05
Instruments:   flunks utokyo exprnce Sex TotMuseum concerts schinternet
               collgrad
------------------------------------------------------------------------------
Wald test of exogeneity:     chi2(1) =     7.75           Prob > chi2 = 0.0054

There are two Wald tests here, at the top and the bottom.

The first Wald test, with the value of 40, tests whether the explanatory variables are jointly significant. This is usually a boring test– of course they are significant, unless the regression is totally misguided. Note the 5 degrees of freedom , Wald chi2(5), one for each variable.

That Wald Test is analogous to an F-test in a linear regression.

The second Wald test, of exogeneity, is special to the instrumental variables technique. In doing IV, we assume that variable X is endogenous, so we intstrument for it using variable Z.

I think what that Wald Test is doing is testing for whether X is really endogenous or not. Note the 1 degree of freedom, chi2(1). My guess is that a value of 7.75 says that if X is really exogenous, then we should get a value as big as 7.75 only with the tiny probability .0054, so X is probably endogenous.

That Wald Test is new since I took econometrics. What I found on the web was:

This test is mentioned along with the theory behind -ivprobit- in Wooldridge’s “Econometric Analysis of Cross Section and Panel Data” (2002, pp. 472-477).

For the maximum likelihood variant with a single endogenous variable, the test is simply a Wald test that the correlation parameter rho is equal to zero. That is, the test simply asks whether the error terms in the structural equation and the reduced-form equation for the endogenous variable are correlated. If there are multiple endogenous variables, then it is a joint test of the covariances between the k reduced form equations’ errors and the structural equation’s error.

In the two-step estimator, in the second stage we include the residuals from the first-stage OLS regression(s) as regressors. The Wald test is a test of significance on those residuals’ coefficients.

I have a hard time understanding that, but I think it confirms what I thought. The second paragraph says that we do a first-stage regression of X on Z to get fitted values, Xhat, and errors, Ehat that are orthogonal to them. Then we run Y = XHAT + EHAT and see if EHAT comes in significant. If it were that simple, we’d have a Z-test. But, actually, a Chi-squared with 1 degree of freedom is a Normal squared, I think, and so would give the same results as a z-test.

This entry was posted on Wednesday, October 11th, 2006 at 2:08 pm and is filed under Thoughts-New, Math, Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.