There have recently been a number of Statalist postings concerning the
heteroskedastic probit model. hetprob.ado represents code that James
Hardin and I <jhardin@stata.com> and <wgould@stata.com>) *THINK* estimate
such models correctly. I want to emphasize the word *THINK*. Neither
James nor I have
read the articles we should and, in fact, most of what follows comes from
James and I hearing the words "heteroskedastic probit" and then thinking
to
ourselves, "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
seek
reassurance. In particular, our concerns have to do with assumed
normalizations which might make the coefficients we estimate
multiplicatively
different from those estimated by some other package.
Before providing the code, let us reveal our thinking.
below
are:
The sections
Development of probit model
The heteroskedastic case
Derivation of a heteroskedastic-corrected probit model
The -hetprob- command
Options
An example using -hetprobComment 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
here
is one:
We have a continuous variable y_j, j=1, ..., N, and y_j is given by the
linear
model
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
= 0
if
if
y_j > c
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' )
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
and
then we will generalize beyond it. Pretend we have two groups of data
and
each 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.
This
means that, if we observed y, we could estimate each model separately and
we
would 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
-------------a1' = (a-c)/s1
b1' =
b/s1
group 2
-------------a2' = (a-c)/s2
b2' =
b/s2
The probit coefficients are different in each group, but related!
a2'/a1'
b2'/b1'
=
=
[(a-c)/s2] / [(a-c)/s1]
[
b/s2] / [
b/s1]
=
=
s1/s2
s1/s2
This is a very un linear-regression like result but hardly surprising.
We do
not observe y, we observe whether y>c or y<c, and if the variance of the
process increases, our prediction of Pr(y>c) must move toward .5;
coefficients
must move toward zero.
This issue is *NOT* addressed by the Huber/White/Robust correction to the
standard 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))
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 the
coefficients 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(#)
<maximize-options>
]
-----
-
and, as with all estimation commands, -hetprob- typed without arguments
redisplays 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 where
s^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.
type
If I
. hetprob works educ sex, v(age)
I am assuming s^2 = exp(g1*age). The same variables can even appear
among the
standard explanatory variables and the explanatory variables for
variance, but
realize that you are pushing things.
. hetprob works age, v(age)
amounts to estimating
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 which
the variance is assumed to depend.
nochi suppresses the calculation of the model chi-squared statistic -the
likelihood-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.
should never have to specify them.
You
An example using -hetprob--------------------------. hetprob foreign mpg weight, v(price)
Estimating constant-only model:
Iteration 0: Log Likelihood = -45.03321
Iteration 1: Log Likelihood = -44.836587
<output omitted>
Iteration 8: Log Likelihood = -44.66722
Estimating full model:
Iteration 0: Log Likelihood = -26.844189
(unproductive step attempted)
Iteration 1: Log Likelihood = -26.572242
<output omitted>
Iteration 7: Log Likelihood = -24.833819
Number of obs
=
Model chi2(2)
=
Prob > chi2
=
74
39.67
0.0000
Log 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
Comment on estimated standard errors and likelihood-ratio tests
---------------------------------------------------------------If you look at the output above, the Wald tests (tests based on the
conventionally estimated standard errors) and the likelihood-ratio tests
yield
considerably 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)
( 2)
[foreign]mpg = 0.0
[foreign]weight = 0.0
chi2( 2) =
Prob > chi2 =
3.40
0.1831
(The tests for coefficients in the ln_var equation, at least in this
case, are
in 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 some
caution in interpreting Wald-test results. We would check results
against the
likelihood-ratio test before reporting them.
Calculating predictions from the -hetprob- results
--------------------------------------------------Here is how you obtain predicted probabilities form the -hetprobresults:
.
.
.
.
predict
predict
replace
gen p =
i1
s, eq(ln_var)
s = exp(s/2)
normprob(i1/s)
Request for comments
--------------------We seek comments and, in particular, we seek comparison of results from
this
command with other implementations.
--- Bill
wgould@stata.com
-- James
jhardin@stata.com