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There have recently been a number of Statalist postings concerning the heteroskedastic probit model. hetprob.ado represents code that James Hardin and I and ) *THINK* estimate such models correctly. Iwant to emphasize the word *THINK*. Neither James nor I haveread the articles we should and, in fact, most of what follows comes fromJames and I hearing the words "heteroskedastic probit" and then thinking toourselves, "what could that mean?"

While neither James nor I would ever distribute anything -- even informally--- if we thought there was much chance it was completely misguided, we seekreassurance. In particular, our concerns have to do with assumednormalizations which might make the coefficients we estimate multiplicativelydifferent from those estimated by some other package.

Before providing the code, let us reveal our thinking. The sections beloware:

Development of probit model The heteroskedastic case Derivation of a heteroskedastic-corrected probit model The -hetprob- command Options An example using -hetprob- Comment on estimated standard errors and likelihood-ratio tests

Calculating predictions from the -hetprob- estimates Request for comments (signature) The -hetprob- ado-files

Development of probit model----------------------------

There are lots of way to think about and so derive the probit model, but hereis one:

We have a continuous variable y_j, j=1, ..., N, and y_j is given by the linearmodel

y_j = a + X_j*b + u_j, u_j distributed N(0, s^2)

If we observed y_j and X_j, we could estimate b using linear regression. We,however, do not observe y_j. Instead, we observe

d_j = 1 if y_j > c = 0 if y_j < c

Thus,

Pr(d_j==1) = Pr(y_j > c) = Pr(a + X_j*b + u_j > c)

= Pr(u_j > c - a - X_j*b) = Pr(u_j/s > (c-a)/s - X_j*(b/s)) = Pr(-u_j/s < (a-c)/s + X_j*(b/s)) = F( (a-c)/s + X_j*(b/s) )

Thus, when we estimate a probit model

Pr(outcome) = F( a' + X_j*b' )

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The estimates we obtain are

a' = (a-c)/s b' = b/s

In words this means that we cannot identify the scale of the unobserved y.

The heteroskedastic case-------------------------

Let us now consider heteroskedasticity. Let us start with a simple case andthen we will generalize beyond it. Pretend we have two groups of data andeach group is, itself, homoskedastic.

y_j = a + X_j*b + u_j, u_j distributed N(0, s1^2) for group 1

y_j = a + X_j*b + v_j, v_j distributed N(), s2^2) for group 2

Note that we assume the coefficients (a,b) are the same for both groups. Thismeans that, if we observed y, we could estimate each model separately and wewould expect to estimate similar coefficients. In the probit case, however,something surprising happens.

Estimating on each of the groups separately, we would obtain:

group 1 group 2 -------------- -------------- a1' = (a-c)/s1 a2' = (a-c)/s2 b1' = b/s1 b2' = b/s2

The probit coefficients are different in each group, but related!

a2'/a1' = [(a-c)/s2] / [(a-c)/s1] = s1/s2 b2'/b1' = [ b/s2] / [ b/s1] = s1/s2

This is a very un linear-regression like result but hardly surprising. We donot observe y, we observe whether y>c or yc) must move toward .5; coefficientsmust move toward zero.

This issue is *NOT* addressed by the Huber/White/Robust correction to thestandard errors.

Derivation of a heteroskedastic-corrected probit model--------------------------------------------------------

Let us assume

y_j = a + X_j*b + u_j, u_j distributed N(0, s_j^2)

where s_j^2 is given by some function of Z_j which we will specify later.Then:

Pr(y_j > c) == Pr(a + X_j*b + u_j > c)= Pr(u_j > c - a - X_j*b)

= Pr(u_j/s_j > (c-a)/s_j - X_j*(b/s_j))

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= Pr(-u_j/s_j < (a-c)/s + X_j*(b/s)) = F( (a-c)/s_j + X_j*(b/s_j) )

so a' = (a-c)/s_j b' = b/s_j

Let us now specify s_j^2 = exp(s0 + Z_j*g). Then

b' = b/s_j = b/[exp(s0/2)exp(Z_j*g/2)]

and there is obviously a lack of identification problem. We will identify thecoefficients by (arbitrarily) setting s0=0. Then the model is

Pr(outcome) = F( a'/s_j) + X_j*b'/s_j ) = F( (a' + X_j*b') / s_j )where s_j^2 = exp(Z_j*g)

a', b', and g are to be estimated.

The -hetprob- command----------------------

--hetprob- has syntax

hetprob depvar [indepvars] [if exp] [in range], variance(varlist) ------- -- -- -

[ nochi level(#) ] ----- -

and, as with all estimation commands, -hetprob- typed without argumentsredisplays estimation results.

For instance, if I type

. hetprob outcome bp age, v(group1 group2)

I am estimating the model

Pr(outcome) = F( b0 + b1*bp + b2*age wheres^2 = exp(g1*group1 + g2*group2) )

= F( [b0+b1*bp+b2*age]/exp[(g1*group1 + g2*group2)/2] )

The variance() variables are not required to be discrete variables. If Itype

. hetprob works educ sex, v(age)

I am assuming s^2 = exp(g1*age). The same variables can even appear among thestandard explanatory variables and the explanatory variables for variance, butrealize that you are pushing things.

. hetprob works age, v(age)

amounts to estimating

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Pr(works) = F( (b0 + b1*age) / exp(g1*age/2) )

and obviously coefficients g1 and b1 will be highly correlated.

Options--------

variance(varlist) is not optional; it specifies the variables on whichthe variance is assumed to depend.

nochi suppresses the calculation of the model chi-squared statistic -- thelikelihood-ratio test against the constant-only model. Specifying this

option speeds execution considerably.

level(#) specifies the confidence level, in percent, for the confidence intervals of the coefficients; see help level.

maximize_options control the maximization process; see [R] maximize. You should never have to specify them.

An example using -hetprob----------------------------

. hetprob foreign mpg weight, v(price)

Estimating constant-only model:Iteration 0: Log Likelihood = -45.03321Iteration 1: Log Likelihood = -44.836587Iteration 8: Log Likelihood = -44.66722

Estimating full model:Iteration 0: Log Likelihood = -26.844189(unproductive step attempted)Iteration 1: Log Likelihood = -26.572242

Iteration 7: Log Likelihood = -24.833819

Number of obs = 74 Model chi2(2) = 39.67 Prob > chi2 = 0.0000Log Likelihood = -24.8338194

-------------------------------------------------------------------------------foreign | Coef. Std. Err. z P>|z| [95% Conf. Interval]----------+--------------------------------------------------------------------foreign | mpg | -.1651034 .1257984 -1.312 0.189 -.4116637 .081457

weight | -.0057221 .0031172 -1.836 0.066 -.0118317 .0003876 _cons | 17.44533 9.500341 1.836 0.066 -1.174997 36.06566----------+--------------------------------------------------------------------ln_var | price | .0003066 .0001685 1.819 0.069 -.0000237 .0006369-------------------------------------------------------------------------------note, LR test for ln_var equation: chi2(1) = 4.02, Prob > Chi2 = 0.0449

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Comment on estimated standard errors and likelihood-ratio tests----------------------------------------------------------------

If you look at the output above, the Wald tests (tests based on theconventionally estimated standard errors) and the likelihood-ratio tests yieldconsiderably different results.

At the top of the output, the model chi-squared is reported as chi2(2) = 39.67(p=.0000). The corresponding Wald test is

. test mpg weight

( 1) [foreign]mpg = 0.0 ( 2) [foreign]weight = 0.0

chi2( 2) = 3.40 Prob > chi2 = 0.1831

(The tests for coefficients in the ln_var equation, at least in this case, arein more agreement. The reported z statistic is 1.819 (meaning chi2 = 1.819^2= 3.31) and the corresponding likelihood-ratio test is 4.02.)

James and I took a simple example to explore how different results might be.In our example, we simulated data from the true model

y_j = 2*x1_j + 3*x2_j + u_j

u_j distributed N(0,1) for group 0 and N(0,4) for group 2.

d_j = 1 if y_j>average(y_j in the sample)

Here is what we obtained in five particular runs:

Sample-size L.R. test Wald test for each group chi^2 chi^2 ------------------------------------------------- 50 + 50 27.48 13.11

100 + 100 58.10 33.40 1000 + 1000 556.42 297.35 5000 + 5000 2812.04 1478.40

We do not want to make too big a deal about this but we do recommend somecaution in interpreting Wald-test results. We would check results against thelikelihood-ratio test before reporting them.

Calculating predictions from the -hetprob- results---------------------------------------------------

Here is how you obtain predicted probabilities form the -hetprob- results:

. predict i1 . predict s, eq(ln_var) . replace s = exp(s/2) . gen p = normprob(i1/s)

Request for comments---------------------

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We seek comments and, in particular, we seek comparison of results from thiscommand with other implementations.

--- Bill -- James wgould@stata.com jhardin@stata.com