Constraining coefficients to equality for all non-base categories

[Giulio Zanella]

When using a categorical dependent variable (like in the multinomial Logit), one can constrain coefficients for x to equality for, say, categories cat2 and cat3 (cat1 is the base category) by defining the following constraint:

constraint 1 [cat2=cat3]: x

Suppose that one has N categories and wants to constrain coefficients for x to equality for all the N-1 non-base categories. Is there a way to do this by defining a single constraint? I mean somthing like

constraint 1 [(all categories)]: x

where (all categories) is what I don't know how to specify.

Many thanks to those who can provide some clue.

[Clyde Schechter]

The easiest way to do this is to not use constraints at all. Create a new variable equal to 0 for the base category of x and 1 for all the non-base categories (and missing for any observations where x is missing.) Then do your regression using this new variable instead of x.

[Giulio Zanella]

Thanks Clyde, but I'm not sure I get your point. Here's a simple example of what I would like to do:

clear all
set seed 666

set obs 1000
gen id = _n

gen x = runiform()
gen eps = runiform()

gen tot = x+eps
sum tot, d

gen y = .
replace y = 0 if tot <= .71
replace y = 1 if tot >  .71 & tot <= 1.01
replace y = 2 if tot > 1.01 & tot <= 1.31
replace y = 3 if tot > 1.31

drop tot

tab y

constraint 1 [1=2]: x
constraint 2 [2=3]: x
mlogit y x,  baseoutcome(0) constraints(1,2)

The way I understand your suggestion is:

gen x1 = y != 0
mlogit y x1,  baseoutcome(0)

which does not work because there's no variability in x1 among the non-base categories

[Leonardo Guizzetti]

I think you just need one more constraint: a set to make the coefficients equal, then one more to set the constrained value. Your don't get sensible results in your example because it's just a toy dataset, but it shows the idea.

constraint 1 [1=2]: x
constraint 2 [2=3]: x
constraint 3 [1]: x=1
mlogit y x,  baseoutcome(0) constraints(1 2 3)

[Clyde Schechter]

Sorry, I misread what you wrote. I thought you were trying to constrain the levels of an explanatory variable, not an outcome. Here's how it works in that situation, in case somebody is interested:

. sysuse auto, clear
(1978 automobile data)

.
. regress price mpg i.rep78

      Source |       SS           df       MS      Number of obs   =        69
-------------+----------------------------------   F(5, 63)        =      4.39
       Model |   149020603         5  29804120.7   Prob > F        =    0.0017
    Residual |   427776355        63  6790100.88   R-squared       =    0.2584
-------------+----------------------------------   Adj R-squared   =    0.1995
       Total |   576796959        68  8482308.22   Root MSE        =    2605.8

------------------------------------------------------------------------------
       price | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
         mpg |  -280.2615   61.57666    -4.55   0.000    -403.3126   -157.2103
             |
       rep78 |
          2  |   877.6347   2063.285     0.43   0.672     -3245.51     5000.78
          3  |   1425.657   1905.438     0.75   0.457    -2382.057    5233.371
          4  |   1693.841   1942.669     0.87   0.387    -2188.274    5575.956
          5  |   3131.982   2041.049     1.53   0.130    -946.7282    7210.693
             |
       _cons |   10449.99   2251.041     4.64   0.000     5951.646    14948.34
------------------------------------------------------------------------------

.
. constraint 1 3.rep78 = 2.rep78

. constraint 2 4.rep78 = 3.rep78

. constraint 3 5.rep78 = 4.rep78

. cnsreg price mpg i.rep78, constraints(1 2 3)

Constrained linear regression                        Number of obs =        69
                                                     F(2, 66)      =      9.18
                                                     Prob > F      =    0.0003
                                                     Root MSE      = 2614.9289

 ( 1)  - 2.rep78 + 3.rep78 = 0
 ( 2)  - 3.rep78 + 4.rep78 = 0
 ( 3)  - 4.rep78 + 5.rep78 = 0
------------------------------------------------------------------------------
       price | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
         mpg |  -226.7809   54.05666    -4.20   0.000    -334.7085   -118.8533
             |
       rep78 |
          2  |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
          3  |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
          4  |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
          5  |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
             |
       _cons |   9326.899   2169.696     4.30   0.000      4994.96    13658.84
------------------------------------------------------------------------------

.
. recode rep78 (3/5 = 2), gen(rep78_red)
(59 differences between rep78 and rep78_red)

. regress price mpg i.rep78_red

      Source |       SS           df       MS      Number of obs   =        69
-------------+----------------------------------   F(2, 66)        =      9.18
       Model |   125498634         2  62749316.9   Prob > F        =    0.0003
    Residual |   451298325        66  6837853.41   R-squared       =    0.2176
-------------+----------------------------------   Adj R-squared   =    0.1939
       Total |   576796959        68  8482308.22   Root MSE        =    2614.9

------------------------------------------------------------------------------
       price | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
         mpg |  -226.7809   54.05666    -4.20   0.000    -334.7085   -118.8533
 2.rep78_red |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
       _cons |   9326.899   2169.696     4.30   0.000      4994.96    13658.84
------------------------------------------------------------------------------

Note that the single coefficient for rep78_red gets exactly the same results as the several constrained coefficients for the original levels of rep78.

You can, in fact, do something similar with an outcome variable, but it doesn't work quite as well because you lose all the information about the constant terms at those levels.

. constraint 1 [2=3]:mpg

. constraint 2 [3=4]:mpg

. constraint 3 [4=5]:mpg

. mlogit rep78 mpg, constraints(1 2 3) baseoutcome(1)

Iteration 0:  Log likelihood = -93.692061  
Iteration 1:  Log likelihood = -93.689471  
Iteration 2:  Log likelihood = -93.689468  

Multinomial logistic regression                         Number of obs =     69
                                                        Wald chi2(1)  =   0.01
Log likelihood = -93.689468                             Prob > chi2   = 0.9431

 ( 1)  [2]mpg - [3]mpg = 0
 ( 2)  [3]mpg - [4]mpg = 0
 ( 3)  [4]mpg - [5]mpg = 0
------------------------------------------------------------------------------
       rep78 | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
1            |  (base outcome)
-------------+----------------------------------------------------------------
2            |
         mpg |   .0090203   .1263047     0.07   0.943    -.2385324    .2565731
       _cons |   1.195531   2.768717     0.43   0.666    -4.231054    6.622117
-------------+----------------------------------------------------------------
3            |
         mpg |   .0090203   .1263047     0.07   0.943    -.2385324    .2565731
       _cons |   2.517287   2.752113     0.91   0.360    -2.876755     7.91133
-------------+----------------------------------------------------------------
4            |
         mpg |   .0090203   .1263047     0.07   0.943    -.2385324    .2565731
       _cons |   2.006462   2.756148     0.73   0.467    -3.395488    7.408412
-------------+----------------------------------------------------------------
5            |
         mpg |   .0090203   .1263047     0.07   0.943    -.2385324    .2565731
       _cons |   1.513985   2.762554     0.55   0.584     -3.90052    6.928491
------------------------------------------------------------------------------

.
. mlogit rep78_red mpg, baseoutcome(1)

Iteration 0:  Log likelihood =  -9.052649  
Iteration 1:  Log likelihood = -9.0500588  
Iteration 2:  Log likelihood = -9.0500558  
Iteration 3:  Log likelihood = -9.0500558  

Multinomial logistic regression                         Number of obs =     69
                                                        LR chi2(1)    =   0.01
                                                        Prob > chi2   = 0.9426
Log likelihood = -9.0500558                             Pseudo R2     = 0.0003

------------------------------------------------------------------------------
   rep78_red | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
1            |  (base outcome)
-------------+----------------------------------------------------------------
2            |
         mpg |   .0090207    .126305     0.07   0.943    -.2385325    .2565739
       _cons |   3.320776    2.74877     1.21   0.227    -2.066714    8.708266
------------------------------------------------------------------------------

As you can see, the coefficient of mpg in the rep78_red regression is equal, except for tiny rounding errors that are common with maximum likelihood estimation, to the constrained mpg coefficients in the constrained model. But, the constrained model still gives you separate constant terms at each level, which are lost in my approach.

[Andrew Musau]

My reading of this question is not that the OP does not know how to specify multiple constraints. He is asking whether there is a convenient way to specify multiple constraints. With tuples from SSC, one can write a program that simplifies this.

ssc install tuples, replace

The following can be generalized to any variable with a finite number of levels (within limits). Only change the highlighted.

macro drop _all
*LOAD DATASET
sysuse auto, clear
*SPECIFY VARIABLE (HIGHLIGHTED)
local var "rep78"
levelsof `var', local(all)
local all= substr("`all'", 2,.)
tuples `all', max(2) min(2) conditionals(1)
local constraints
forval i= 1/`ntuples'{
    constraint `i' `=word("`tuple`i''", 1)'.`var' = `=word("`tuple`i''", 2)'.`var'
    local constraints "`constraints' `i'"
}
cnsreg price mpg i.rep78, constraints(`constraints')

Res.:

 
. cnsreg price mpg i.rep78, constraints(`constraints')

Constrained linear regression                        Number of obs =        69
                                                     F(2, 66)      =      9.18
                                                     Prob > F      =    0.0003
                                                     Root MSE      = 2614.9289

 ( 1)  2.rep78 - 5.rep78 = 0
 ( 2)  2.rep78 - 4.rep78 = 0
 ( 3)  2.rep78 - 3.rep78 = 0
------------------------------------------------------------------------------
       price | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
         mpg |  -226.7809   54.05666    -4.20   0.000    -334.7085   -118.8533
             |
       rep78 |
          2  |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
          3  |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
          4  |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
          5  |    1696.45   1876.498     0.90   0.369      -2050.1    5442.999
             |
       _cons |   9326.899   2169.696     4.30   0.000      4994.96    13658.84
------------------------------------------------------------------------------

.

[Giulio Zanella]

Thank you all, these are very useful tips!