http://www.stata.com/support/faqs/stat/chow2.html

How can I do a Chow test with the robust variance estimates, that is, after estimating with regress, vce(robust)?

Title		Chow and Wald tests
Author		William Gould, StataCorp
Date		July 1999; minor revision August 2007

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First, see the FAQ How can I compute a Chow test statistic?. The point of that FAQ is that you can do Chow tests using Stata’s test command and, in fact, Chow tests are what the test command reports.

Well, that’s not exactly right. test uses the estimated variance–covariance matrix of the estimators, and test performs Wald tests,

 W = (Rb-r)'(RVR')-1 (Rb-r) 

where V is the estimated variance–covariance matrix of the estimators.

For linear regression with the conventionally estimated V, the Wald test is the Chow test and vice versa.

You might say that you are performing a Chow test, but I say that you are performing a Wald test. That distinction is important, because the Wald test generalizes to different variance estimates of V, whereas the Chow test does not. After regress, vce(robust), for instance, test uses the V matrix estimated by the robust method because that is what regress, vce(robust) left behind.

Thus the short answer is that you estimate your model using regress, vce(robust) and then use Stata’s test command. You then call the result a Wald test.

If you are bothered that a Wald test produces F rather than chi-squared statistics, also see the FAQ Why does test sometimes produce chi-squared and other times F statistics?

http://www.stata.com/support/faqs/stat/chow3.html

Can you explain Chow tests?

Title		Chow tests
Author		William Gould, StataCorp
Date		January 2002; updated August 2005

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Privately I was asked yet another question on Chow tests. The question started out “Is a Chow test the correct test to determine whether data can be pooled together?” and went on from there. Like many on the list, I am tired of seeing and answering questions on Chow tests. I do not blame the questioner for asking; I blame their teachers for confusing them with what is, these days, unnecessary jargon.

In the past, I have always given in and cast my answer in Chow-test terms. In this reply, I try a different approach and, I think, the result is more useful.

This reply concerns linear regression (though the technique is really more general than that), and I gloss over the detail of pooling the residuals and whether the residual variances are really the same. For the last, I think I can be forgiven.

Here is what I wrote:

Is a Chow test the correct test to determine whether data can be pooled together?

A Chow test is simply a test of whether the coefficients estimated over one group of the data are equal to the coefficients estimated over another, and you would be better off to forget the word Chow and remember that definition.

History:   In the days when statistical packages were not as sophisticated as they are now, testing whether coefficients were equal was not so easy. You had to write your own program, typically in FORTRAN. Chow showed a way you could perform a Wald test based on statistics that were commonly reported, and that would produce the same result as if you performed the Wald test.

What does it mean “whether data can be pooled together”? Do you often meet nonprofessionals who say to you, “I was wondering whether the data could be pooled?” Forget that phrase, too: it is another piece of jargon for testing whether the behavior is the same, as measured by whether the coefficients are the same.

Let’s pretend that you have some model and two or more groups of data. Your model predicts something about the behavior within the group based on certain characteristics that vary within the group. Under the assumption that each group's behavior is unique, you have

 y_1 = X_1*b_1 + u_1 (equation for group 1) y_2 = X_2*b_2 + u_2 (equation for group 2) 

and so on. Now, you want to test whether the behavior for one group is the same as for another, which means you want to test

 b_1 = b_2 = ... 

How do you do that? Testing coefficients across separately estimated models is difficult to impossible, depending on things we need not go into right now. A trick is to “pool” the data to convert the multiple equations into one giant equation:

 y = d1*(X_1*b1 + u1) + d2*(X_2*b2 + u2) + ... 

where y is the set of all outcomes (y_1, y_2, ...), and d1 is a variable that is 1 when the data are for group 1 and 0 otherwise, d2 is 1 when the data are for group 2 and 0 otherwise, ....

Notice that from the above I can retrieve the original equations. Setting d1=1 and d2=d3=...=0, I get the equation for group 1; setting d1=0 and d2=1 and d3=...=0, I get the equation for group 2; and so on.

Now, let’s start with

 y = d1*(X_1*b1 + u1) + d2*(X_2*b2 + u2) + ... 

and rewrite it by a little algebraic manipulation:

 y = d1*(X_1*b1 + u1) + d2*(X_2*b2 + u2) + ... = d1*X_1*b1 + d1*u2 + d2*X_2*b2 + d2*u2 + ... = d1*X_1*b1 + d2*X_2*b2 + ... + d1*u1 + d2*u2 + ... = X_1*d1*b1 + X_2*d2*b2 + ... + d1*u1 + d2*u2 + ... = (X_1*d1)*b1 + (X_2*d2)*b2 + ... + d1*u1 + d2*u2 + ... 

By stacking the data, I can get back estimates of b1, b2, ...

I include not X_1 in my model, but X_1*d1 (a set of variables equal to X_1 when group is 1 and 0 otherwise); I include not X_2 in my model, but X_2*d2 (a set of variables equal to X_2 when group is 2 and 0 otherwise); and so on.

Let’s use the auto dataset and pretend that I have two groups.

 . sysuse auto,clear . generate group1=rep78==3 . generate group2=group1==0 

I could fit the models separately:

 . regress price mpg weight if group1==1 Source | SS df MS Number of obs = 30 -------------+------------------------------ F( 2, 27) = 16.20 Model | 196545318 2 98272658.8 Prob > F = 0.0000 Residual | 163826398 27 6067644.36 R-squared = 0.5454 -------------+------------------------------ Adj R-squared = 0.5117 Total | 360371715 29 12426610.9 Root MSE = 2463.3 ------------------------------------------------------------------------------ price | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- mpg | 13.14912 184.5661 0.07 0.944 -365.5492 391.8474 weight | 3.517687 1.015855 3.46 0.002 1.433324 5.60205 _cons | -5431.147 6599.898 -0.82 0.418 -18973.02 8110.725 ------------------------------------------------------------------------------ . regress price mpg weight if group2==1 Source | SS df MS Number of obs = 44 -------------+------------------------------ F( 2, 41) = 5.16 Model | 54562909.6 2 27281454.8 Prob > F = 0.0100 Residual | 216614915 41 5283290.61 R-squared = 0.2012 -------------+------------------------------ Adj R-squared = 0.1622 Total | 271177825 43 6306461.04 Root MSE = 2298.5 ------------------------------------------------------------------------------ price | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- mpg | -170.5474 93.3656 -1.83 0.075 -359.103 18.0083 weight | .0527381 .8064713 0.07 0.948 -1.575964 1.68144 _cons | 9685.028 4190.693 2.31 0.026 1221.752 18148.3 ------------------------------------------------------------------------------ 

I could fit the combined model:

 . generate mpg1=mpg*group1 . generate weight1=weight*group1 . generate mpg2=mpg*group2 . generate weight2=weight*group2 . regress price group1 mpg1 weight1 group2 mpg2 weight2, noconstant  Source | SS df MS Number of obs = 74 -------------+------------------------------ F( 6, 68) = 91.38 Model | 3.0674e+09 6 511232168 Prob > F = 0.0000 Residual | 380441313 68 5594725.19 R-squared = 0.8897 -------------+------------------------------ Adj R-squared = 0.8799 Total | 3.4478e+09 74 46592355.7 Root MSE = 2365.3 ------------------------------------------------------------------------------ price | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- group1 | -5431.147 6337.479 -0.86 0.394 -18077.39 7215.096 mpg1 | 13.14912 177.2275 0.07 0.941 -340.5029 366.8012 weight1 | 3.517687 .9754638 3.61 0.001 1.571179 5.464194 group2 | 9685.028 4312.439 2.25 0.028 1079.69 18290.37 mpg2 | -170.5474 96.07802 -1.78 0.080 -362.2681 21.17334 weight2 | .0527381 .8299005 0.06 0.950 -1.603303 1.708779 ------------------------------------------------------------------------------ 

What is this noconstant option? We must remember that when we fit the separate models, each has its own intercept. There was an intercept in X_1, X_2, and so on. What I have done above is literally translate

 y = (X_1*d1)*b1 + (X_2*d2)*b2 + d1*u1 + d2*u2 

and so included the variables group1 and group2 (variables equal to 1 for their respective groups) and told Stata to omit the overall intercept.

I do not recommend you fit the model the way I have just illustrated because of numerical concerns—we'll get to that later. Fit the models separately or jointly, and you will get the same estimates for b_1 and b_2.

Now we can test whether the coefficients are the same for the two groups:

 . test _b[mpg1]=_b[mpg2], notest ( 1) mpg1 - mpg2 = 0 . test _b[weight1]=_b[weight2], accum ( 1) mpg1 - mpg2 = 0 ( 2) weight1 - weight2 = 0 F( 2, 68) = 5.61 Prob > F = 0.0056 

That is the Chow test. Something was omitted: the intercept. If we really wanted to test whether the two groups were the same, we would would test

 . test _b[mpg1]=_b[mpg2] ( 1) mpg1 - mpg2 = 0 . test _b[weight1]=_b[weight2], accum ( 1) mpg1 - mpg2 = 0 ( 2) weight1 - weight2 = 0 . test _b[group1]=_b[group2], accum ( 1) mpg1 - mpg2 = 0 ( 2) weight1 - weight2 = 0 ( 3) group1 - group2 = 0 F( 3, 68) = 4.07 Prob > F = 0.0102 

Using this approach, however, we are not tied down by what the "Chow test" can test. We can formulate any hypothesis we want. We might think that weight works the same way in both groups but that mpg works differently, and each group has its own intercept. Then, we could test

 . test _b[mpg1]=_b[mpg2] ( 1) mpg1 - mpg2 = 0 F( 1, 68) = 0.83 Prob > F = 0.3654 

by itself. If we had more variables, we could test any subset of variables.

Is “pooling the data” justified? Of course it is: we just established that pooling the data is just another way of fitting separate models and that fitting separate models is certainly justified—we got the same coefficients. That’s why I told you to forget the phrase about whether pooling the data is justified. People who ask that don’t really mean to ask what they are saying: they mean to ask whether the coefficients are the same. In that case, they should say that. Pooling is always justified, and it corresponds to nothing more than the mathematical trick of writing separate equations,

 y_1 = X_1*b_1 + u_1 (equation for group 1) y_2 = X_2*b_2 + u_2 (equation for group 2) 

as one equation

 y = (X_1*d1)*b1 + (X_2*d2)*b2 + d1*u1 + d2*u2 

There are many ways I can write the above equation, and I want to write it a little differently because of numerical concerns. Starting with

 y = (X_1*d1)*b1 + (X_2*d2)*b2 + d1*u1 + d2*u2 

let’s do some algebra to obtain

 y = X*b1 + d2*X_2*(b2-b1) + d1*y1 + d2*u2 

where X = (X_1, X_2). In this formulation, I measure not b1 and b2, but b1 and (b2−b1). This is numerically more stable, and I can still test that b2==b1 by testing whether (b2−b1)=0. Let’s fit this model

 . regress price mpg weight mpg2 weight2 group2 Source | SS df MS Number of obs = 74 -------------+------------------------------ F( 5, 68) = 9.10 Model | 254624083 5 50924816.7 Prob > F = 0.0000 Residual | 380441313 68 5594725.19 R-squared = 0.4009 -------------+------------------------------ Adj R-squared = 0.3569 Total | 635065396 73 8699525.97 Root MSE = 2365.3 ------------------------------------------------------------------------------ price | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- mpg | 13.14912 177.2275 0.07 0.941 -340.5029 366.8012 weight | 3.517687 .9754638 3.61 0.001 1.571179 5.464194 mpg2 | -183.6965 201.5951 -0.91 0.365 -585.9733 218.5803 weight2 | -3.464949 1.280728 -2.71 0.009 -6.020602 -.9092956 group2 | 15116.17 7665.557 1.97 0.053 -180.2075 30412.56 _cons | -5431.147 6337.479 -0.86 0.394 -18077.39 7215.096 ------------------------------------------------------------------------------ 

and, if I want to test whether the coefficients are the same, I can do

 . test _b[mpg2]=0 ( 1) mpg2 = 0 . test _b[weight2]=0, accum ( 1) mpg2 = 0 ( 2) weight2 = 0 F( 2, 68) = 5.61 Prob > F = 0.0056 

and that gives the same answer yet again. If I want to test whether *ALL* the coefficients are the same (including the intercept) I can use

 . test _b[mpg2]=0, notest ( 1) mpg2 = 0 . test _b[weight2]=0, accum notest ( 1) mpg2 = 0 ( 2) weight2 = 0 . test _b[group2]=0, accum ( 1) mpg2 = 0 ( 2) weight2 = 0 ( 3) group2 = 0 F( 3, 68) = 4.07 Prob > F = 0.0102 

Just as before, I can test any subset.

Using this difference formulation, if I had three groups, starting with

 y = (X_1*d1)*b1 + (X_2*d2)*b2 + (X_3*d3)*b3 + d1*u1 + d2*u2 + d3*u3 

as

 y = X*b1 + (X_2*d2)*(b2-b1) + (X_3*d3)*(b3-b1) + d1*u1 + d2*u2 + d3*u3 

Let’s create the group variables and fit this model:

 . sysuse auto,clear . generate group1=rep78==3 . generate group2=rep78==4 . generate group3=(group1+group2)==0 . generate mpg1=mpg*group1 . generate weight1=weight*group1 . generate mpg2=mpg*group2 . generate weight2=weight*group2 . generate mpg3=mpg*group3 . generate weight3=weight*group3 . regress price mpg weight mpg2 weight2 group2 /// > mpg3 weight3 group3 Source | SS df MS Number of obs = 74 -------------+------------------------------ F( 8, 65) = 5.80 Model | 264415585 8 33051948.1 Prob > F = 0.0000 Residual | 370649811 65 5702304.78 R-squared = 0.4164 -------------+------------------------------ Adj R-squared = 0.3445 Total | 635065396 73 8699525.97 Root MSE = 2387.9 ------------------------------------------------------------------------------ price | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- mpg | 13.14912 178.9234 0.07 0.942 -344.1855 370.4837 weight | 3.517687 .9847976 3.57 0.001 1.55091 5.484463 mpg2 | 130.5261 336.6547 0.39 0.699 -541.8198 802.872 weight2 | -2.18337 1.837314 -1.19 0.239 -5.85274 1.486 group2 | 4560.193 12222.22 0.37 0.710 -19849.27 28969.66 mpg3 | -194.1974 216.3459 -0.90 0.373 -626.27 237.8752 weight3 | -3.160952 1.73308 -1.82 0.073 -6.622152 .3002481 group3 | 14556.66 9167.998 1.59 0.117 -3753.101 32866.41 _cons | -5431.147 6398.12 -0.85 0.399 -18209.07 7346.781 ------------------------------------------------------------------------------ 

If I want to test whether the three groups were the same in the Wald-test sense, I can use

 . test (_b[mpg2]=0) (_b[weight2]=0) (_b[group2]=0) /* > */ (_b[mpg3]=0) (_b[weight3]=0) (_b[group3]=0) ( 1) mpg2 = 0 ( 2) weight2 = 0 ( 3) group2 = 0 ( 4) mpg3 = 0 ( 5) weight3 = 0 ( 6) group3 = 0 F( 6, 65) = 2.28 Prob > F = 0.0463 

which I could more easily type as

 . testparm mpg2 weight2 group2 mpg3 weight3 group3 ( 1) mpg2 = 0 ( 2) weight2 = 0 ( 3) group2 = 0 ( 4) mpg3 = 0 ( 5) weight3 = 0 ( 6) group3 = 0 F( 6, 65) = 2.28 Prob > F = 0.0463 

http://www.stata.com/support/faqs/stat/awreg.html

How can I pool data (and perform Chow tests) in linear regression without constraining the residual variances to be equal?

Title		Pooling data and performing Chow tests in linear regression
Author		William Gould, StataCorp
Date		December 1999; updated August 2005

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1. Pooling data and constraining residual variance
   
2. Illustration
   
3. Pooling data without constraining residual variance
   
4. Illustration
   
5. The (lack of) importance of not constraining the variance
   
6. Another way to fit the variance-unconstrained model
   
7. Appendix: do-file and log providing results reported above
   

   7.1   do-file
   7.2   log
   

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1. Pooling data and constraining residual variance

Consider the linear regression model

y = β₀ + β₁x₁ + β₂x₂ + u, u ~ N(0, σ² )

and let us pretend that we have two groups of data, group=1 and group=2. We could have more groups; everything said below generalizes to more than two groups.

We could estimate the models separately by typing

 . regress y x1 x2 if group==1 and . regress y x1 x2 if group==2 

or we could pool the data and estimate a single model, one way being

 . gen g2 = (group==2) . gen g2x1 = g2*x1 . gen g2x2 = g2*x2 . regress y x1 x2 g2 g2x1 g2x2 

The difference between these two approaches is that we are constraining the variance of the residual to be the same in the two groups when we pool the data. When we estimated separately, we estimated

group 1: y = β₀₁ + β₁₁x₁ + β₂₁x₂ + u₁, u₁ ~ N(0, σ₁²)

and

group 2: y = β₀₂ + β₁₂x₁ + β₂₂x₂ + u₂, u₂ ~ N(0, σ₂²)

When we pooled the data, we estimated

y = β₀₁ + β₁₁x₁ + β₂₁x₂ + (β₀₂-β₀₁)g₂ + (β₁₂-β₁₁)g₂x₁ + (β₂₂-β₂₁)g₂x₂ + u, u ~ N(0, σ²)

If we evaluate this equation for the groups separately, we obtain

y = β₀₁ + β₁₁x₁ + β₂₁x₂ + u, u ~ N(0,σ²) for group=1

and

y = β₀₂ + β₁₂x₁ + β₂₂x₂ + u, u ~ N(0,σ²) for group=2

The difference is that we have now constrained the variance of u for group=1 to be the same as the variance of u for group=2.

If you perform this experiment with real data, you will observe the following:

1. You will obtain the same values for the coefficients either way.
   
   
2. You will obtain different standard errors and therefore different test statistics and confidence intervals.

If u is known to have the same variance in the two groups, the standard errors obtained from the pooled regression are better—they are more efficient. If the variances really are different, however, then the standard errors obtained from the pooled regression are wrong.

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2. Illustration (See the do-file and the log with the results in section 7)

I have created a dataset (containing made-up data) on y, x1, and x2. The dataset has 74 observations for group=1 and another 71 observations for group=2. Using these data, I can run the regressions separately by typing

 [1] . regress y x1 x2 if group==1 [2] . regress y x1 x2 if group==2 

or I can run the pooled model by typing

 . gen g2 = (group==2) . gen g2x1 = g2*x1 . gen g2x2 = g2*x2 [3] . regress y x1 x2 g2 g2x1 g2x2 

I did that in Stata and let me summarize the results. When I typed command [1], I obtained the following results (standard errors in parentheses):

 y = -8.650993 + 1.21329*x1 + -.8809939*x2 + u, Var(u) = 15.8912 (22.73703) (.54459) (.405401) 

and when I ran command [2], I obtained

 y = 4.646994 + .9307004*x1 + .8812369*x2 + u, Var(u) = 7.56852 (11.1593) (.236696) (.1997562) 

When I ran command [3], I obtained

 y = -8.650993 + 1.21329*x1 + -.8809939*x2 + (17.92853) (.42942) + (.3196656) 13.29779*g2 + -.2825893*g2x1 + 1.762231*g2x2 + u, Var(u) = 12.5312 (25.74446) (.6123452) (.459958) 

The intercept and coefficients on x1 and x2 in [3] are the same as in [1], but the standard errors are different. Also, if I sum the appropriate coefficients in [3], I obtain the same results as [2]:

 Intercept: 13.29779 + -8.650993 = 4.646797 ([2] has 4.646994) x1: -.2825893 + 1.21329 = .9307004 ([2] has .9307004) x2: 1.762231 + -.8809939 = .8812371 ([2] has .8812369) 

The coefficients are the same, estimated either way. (The fact that the coefficients in [3] are a little off from those in [2] is just because I did not write down enough digits.)

The standard errors for the coefficients are different.

I also wrote down the estimated Var(u), what is reported as RMSE in Stata’s regression output. In standard deviation terms, u has s.d. 15.891 in group=1, 7.5685 in group=2, and if we constrain these two very different numbers to be the same, the pooled s.d. is 12.531.

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3. Pooling data without constraining residual variance

We can pool the data and estimate an equation without constraining the residual variances of the groups to be the same. Previously we typed

 . gen g2 = (group==2) . gen g2x1 = g2*x1 . gen g2x2 = g2*x2 . regress y x1 x2 g2 g2x1 g2x2 

and we start exactly the same way. To that, we add

 . predict r, resid . sum r if group==1 . gen w = r(Var)*(r(N)-1)/(r(N)-3) if group==1 . sum r if group==2 . replace w = r(Var)*(r(N)-1)/(r(N)-3) if group==2 [4] . regress y x1 x2 g2 g2x1 g2x2 [aw=1/w] 

In the above, the constant 3 that appears twice is 3 because there were three coefficients being estimated in each group (an intercept, a coefficient for x1, and a coefficient for x2). If there were a different number of coefficients being estimated, that number would change.

In any case, this will reproduce exactly the standard errors reported by estimating the two models separately. The advantage is that we can now test equality of coefficients between the two equations. For instance, we can now read right off the pooled regression results whether the effect of x1 is the same in groups 1 and 2 (answer: is _b[g2x1]==0?, because _b[x1] is the effect in group 1 and _b[x1]+_b[g2x1] is the effect in group 2, so the difference is _b[g2x1]). And, using test, we can test other constraints as well.

For instance, if you wanted to prove to yourself that the results of [4] are the same as typing regress y x1 x2 if group==2, you could type

 . test x1 + g2x1 == 0 (reproduces test of x1 for group==2) and . test x2 + g2x2 == 0 (reproduces test of x2 for group==2) 

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4. Illustration

Using the made-up data, I did exactly that. To recap, first I estimated separate regressions:

 [1] . regress y x1 x2 if group==1 [2] . regress y x1 x2 if group==2 

and then I ran the variance-constrained regression,

 . gen g2 = (group==2) . gen g2x1 = g2*x1 . gen g2x2 = g2*x2 [3] . regress y x1 x2 g2 g2x1 g2x2 

and then I ran the variance-unconstrained regression,

 . predict r, resid . sum r if group==1 . gen w = r(Var)*(r(N)-1)/(r(N)-3) if group==1 . sum r if group==2 . replace w = r(Var)*(r(N)-1)/(r(N)-3) if group==2 [4] . regress y x1 x2 g2 g2x1 g2x2 [aw=1/w] 

Just to remind you, here is what commands [1] and [2] reported:

 y = -8.650993 + 1.21329*x1 + -.8809939*x2 + u, Var(u) = 15.8912 (22.73703) (.54459) (.405401) y = 4.646994 + .9307004*x1 + .8812369*x2 + u, Var(u) = 7.56852 (11.1593) (.236696) (.1997562) 

Here is what command [4] reported:

 y = -8.650993 + 1.21329*x1 + -.8809939*x2 + (22.73703) (.54459) (.405401) 13.29779*g2 + -.2825893*g2x1 + 1.762231*g2x2 + u (25.3269) (.6050657) (.451943) 

Those results are the same as [1] and [2]. (Pay no attention to the RMSE reported by regress at this last step; the reported RMSE is the standard deviation of neither of the two groups but is instead a weighted average; see the FAQ on this if you care. If you want to know the standard errors of the respective residuals, look back at the output from the summarize statements typed when producing the weighting variable.)

Technical Note:   Note that in creating the weights, we typed . sum r if group==1 . gen w = r(Var)*(r(N)-1)/(r(N)-3) if group==1 and similarly for group 2. The 3 that appears in the finite-sample normalization factor (r(N)-1)/(r(N)-3) appears because there are three coefficients per group being estimated. If our model had fewer or more coefficients, that number would change. In fact, the finite-sample normalization factor changes results very little. In real work, I would have ignored it and just typed . sum r if group==1 . gen w = r(Var) if group==1 unless the number of observations in one of the groups was very small. The normalization factor was included here so that [4] would produce the same results as [1] and [2].

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5. The (lack of) importance of not constraining the variance

Does it matter whether we constrain the variance? Here, it does not matter much. For instance, if after

 [4] . regress y x1 x2 g2 g2x1 g2x2 [aw=1/w] 

we test whether group 2 is the same as group 1, we obtain

 . test g2x1 g2x2 g2 ( 1) g2x1 = 0.0 ( 2) g2x2 = 0.0 ( 3) g2 = 0.0 F( 3, 139) = 307.50 Prob > F = 0.0000 

If instead we had constrained the variances to be the same, estimating the model using

 [3] . regress y x1 x2 g2 g2x1 g2x2 

and then repeated the test, the reported F-statistic would be 300.81.

If there were more groups, and the variance differences were great among the groups, this could become more important.

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6. Another way to fit the variance-unconstrained model

Stata’s xtgls, panels(het) command (see help xtgls) fits exactly the model we have been describing, the only difference being that it does not make all the finite-sample adjustments and so its standard errors are just a little different from those produced by the method just described. (To be clear, xtgls, panels(het) does not make the adjustment described in the technical note above and it does not make the finite-sample adjustments regress itself makes, which is to say, variances are invariable normalized by N, the number of observations, rather than N-k, observations minus number of estimated coefficients.)

Anyway, to estimate xtgls, panels(het), you pool the data just as always,

 . gen g2 = (group==2) . gen g2x1 = g2*x1 . gen g2x2 = g2*x2 

and then type

 [5] . xtgls y x1 x2 g2 g2x1 g2x2, panels(het) i(group) 

to estimate the model. The result of doing that with my fictional data is

 y = -8.650993 + 1.21329*x1 + -.8809939*x2 + (22.27137) (.53344) (.397099) 13.29779*g2 + -.2825893*g2x1 + 1.762231*g2x2 + u (24.80488) (.5925734) (.442610) 

These are the same coefficients we have always seen.

The standard errors produced by xtgls, panels(het) here are about 2% smaller than those produced by [4] and in general will be a little smaller because xtgls, panels(het) is an asymptotically based estimator. The two estimators are asymptotically equivalent, however, and in fact quickly become identical. The only caution I would advise is not to use xtgls, panels(het) if the number of degrees of freedom (observations minus number of coefficients) is below 25 in any of the groups. Then, the weighted OLS approach [4] is better (and you should make the finite-sample adjustment described in the above technical note).

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7. Appendix: do-file and log providing results reported above

7.1 do-file

The following do-file, named uncv.do, was used. Up until the line reading “BEGINNING OF DEMONSTRATION’, the do-file is concerned with constructing the artificial dataset for the demonstration: uncv.do

7.2 log

The do-file shown in 7.1 produced the following output: uncv.log

版主真是辛苦了……很好的东西……

这几篇东西真好 前阵子我妹问题邹氏检验的问题 我也没用过 就查到了上面几篇说明 看了还是挺容易明白的