»  Home »  Resources & support »  FAQs »  Define constraints for parameters

## How do I impose the restriction that rho is zero using the heckman command with full ml?

 Title Define constraints for parameters Author Weihua Guan, StataCorp

In this particular model, heckman does not estimate the parameter rho directly, but estimates a transformation:

atanh_rho = 1/2*ln[(1+rho)]/(1−rho)]

It is estimated in a constant-only equation athrho. Thus we need to constrain the constant term of equation athrho to be 0 (rho=0 implies atanh_rho=0).

. constraint define 1 _b[/athrho]=0

. heckman ..., select(...) constraint(1)

Now let’s extend the answer to more general cases: how to define constraints on parameters of a model in Stata. The syntax is generally

constraint define # [ exp=exp | coefficientlist ]

When we want to fix a parameter at a certain value, it becomes

constraint define # [equation_name]coefficient_name = #

The equation_name may not be necessary for a single-equation model such as OLS. It is easy to apply this rule to the coefficient of a covariate.

constraint define # [equation_name]covariate_name = #

One can find the equation_name easily from the output. Often it is just the name of the dependent variable.

But how about other parameters in the model, such as rho in heckman? This needs some understanding on how Stata estimates those parameters. In ML estimation, Stata always defines them in separate equations, i.e., one equation for one parameter. Those equations are constant-only, and the estimated constants will be the estimated parameters. Often, some transformations are needed to fit the parameter spaces. For instance, the standard deviation sigma of a normal distribution should be always greater than 0, so a log-transformation will be used to allow the estimation (ln(sigma)) from −infinity to +infinity. One can check the Methods and Formulas section of the estimation command to find out if any transformation is applied.

Now let’s go back to the question in heckman. As described in the short answer, heckman does use a transformation to estimate rho.

atanh_rho = 1/2*ln[(1+rho)]/(1−rho)]	(p.976 of [R] heckman)

Using the example in the manual

. use http://www.stata-press.com/data/r14/womenwk, clear

. heckman wage educ age, select(married children educ age)

(output omitted)
 Coef. Std. Err. z P>|z| [95% Conf. Interval] wage education .9899537 .0532565 18.59 0.000 .8855729 1.094334 age .2131294 .0206031 10.34 0.000 .1727481 .2535108 _cons .4857752 1.077037 0.45 0.652 -1.625179 2.59673 select married .4451721 .0673954 6.61 0.000 .3130794 .5772647 children .4387068 .0277828 15.79 0.000 .3842534 .4931601 education .0557318 .0107349 5.19 0.000 .0346917 .0767718 age .0365098 .0041533 8.79 0.000 .0283694 .0446502 _cons -2.491015 .1893402 -13.16 0.000 -2.862115 -2.119915 /athrho .8742086 .1014225 8.62 0.000 .6754241 1.072993 /lnsigma 1.792559 .027598 64.95 0.000 1.738468 1.84665 rho .7035061 .0512264 .5885365 .7905862 sigma 6.004797 .1657202 5.68862 6.338548 lambda 4.224412 .3992265 3.441942 5.006881 LR test of indep. eqns. (rho = 0): chi2(1) = 61.20 Prob > chi2 = 0.0000

Here are four equations: wage and select equation for those covariates, athrho for atanh_rho, and lnsigma for ln(sigma). The constant-only equation for a parameter is often displayed as /equation_name in the output table. The last row of the table displays the estimated value for rho, sigma, and lambda, which are transformed back from the estimation results.

Now we can impose the constraint on rho, which is actually on the constant term of equation athrho.

use http://www.stata-press.com/data/r14/womenwk, clear

. local athrho=1/2*ln((1+0)/(1-0))

. constraint define 1 _b[/athrho]=0

. heckman wage educ age, select(married children educ age) constraint(1)

Iteration 0:   log likelihood = -5283.1781
Iteration 1:   log likelihood = -5230.2173
Iteration 2:   log likelihood = -5208.9358
Iteration 3:   log likelihood = -5208.9038
Iteration 4:   log likelihood = -5208.9038

Heckman selection model                         Number of obs      =      2000
(regression model with sample selection)        Censored obs       =       657
Uncensored obs     =      1343

Wald chi2(2)       =    456.00
Log likelihood = -5208.904                      Prob > chi2        =    0.0000

( 1)  [athrho]_cons = 0
 wage Coef. Std. Err. z P>|z| [95% Conf. Interval] wage education .8965829 .0497504 18.02 0.000 .7990738 .994092 age .1465739 .0186926 7.84 0.000 .1099371 .1832106 _cons 6.084875 .8886241 6.85 0.000 4.343204 7.826546 select married .4308575 .074208 5.81 0.000 .2854125 .5763025 children .4473249 .0287417 15.56 0.000 .3909922 .5036576 education .0583645 .0109742 5.32 0.000 .0368555 .0798735 age .0347211 .0042293 8.21 0.000 .0264318 .0430105 _cons -2.467365 .1925635 -12.81 0.000 -2.844782 -2.089948 /athrho 0 (constrained) /lnsigma 1.694868 .0192951 87.84 0.000 1.65705 1.732686 rho 0 (omitted) sigma 5.445927 .1050797 5.243821 5.655824 lambda 0 (omitted)
Wald test of indep. eqns. (rho = 0): chi2(1) = . Prob > chi2 = .

The output shows that the constraint is applied correctly.