-glm- vs -sem-: Point estimates are the same, but SEs are not

[Bruce Weaver]

Hello folks. My apologies if this has been asked before--I did search, but failed to find anything on it.

When I estimate the same model via -glm- and -sem-, I get the same point estimates, but different SEs (see example below). Given that both procedures use MLE, I expected everything (including SEs) to be identical. Was I wrong to expect that? Are there any options with different defaults that I need to tinker with?

Thanks for any insight you can offer.

By the way, I am using v13.1 (for Windows).

Cheers,
Bruce

. sysuse auto, clear
(1978 Automobile Data)

.
. glm mpg weight foreign        // estimate same model via -glm-

Iteration 0:   log likelihood = -194.18306  

Generalized linear models                          No. of obs      =        74
Optimization     : ML                              Residual df     =        71
                                                   Scale parameter =  11.60805
Deviance         =  824.1717613                    (1/df) Deviance =  11.60805
Pearson          =  824.1717613                    (1/df) Pearson  =  11.60805

Variance function: V(u) = 1                        [Gaussian]
Link function    : g(u) = u                        [Identity]

                                                   AIC             =  5.329272
Log likelihood   = -194.1830644                    BIC             =  518.5831

------------------------------------------------------------------------------
             |                 OIM
         mpg |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
      weight |  -.0065879   .0006371   -10.34   0.000    -.0078366   -.0053392
     foreign |  -1.650029   1.075994    -1.53   0.125    -3.758939    .4588806
       _cons |    41.6797   2.165547    19.25   0.000     37.43531     45.9241
------------------------------------------------------------------------------

. sem (mpg <- weight foreign)   // estimate same model via -sem-

Endogenous variables

Observed:  mpg

Exogenous variables

Observed:  weight foreign

Fitting target model:

Iteration 0:   log likelihood =  -822.2459  
Iteration 1:   log likelihood =  -822.2459  

Structural equation model                       Number of obs      =        74
Estimation method  = ml
Log likelihood     =  -822.2459

------------------------------------------------------------------------------
             |                 OIM
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
Structural   |
  mpg <-     |
      weight |  -.0065879   .0006241   -10.56   0.000     -.007811   -.0053647
     foreign |  -1.650029   1.053958    -1.57   0.117    -3.715748    .4156902
       _cons |    41.6797   2.121197    19.65   0.000     37.52223    45.83717
-------------+----------------------------------------------------------------
   var(e.mpg)|   11.13746   1.830987                       8.06955    15.37173
------------------------------------------------------------------------------
LR test of model vs. saturated: chi2(0)   =      0.00, Prob > chi2 =      .

.
. // The -glm- and -sem- commands both use MLE; and the
. // defaults for -glm- are normal error distribution with
. // identity link function.  So I expected -glm- and -sem-
. // to yield identical results.  The point estimates are
. // the same, but the SEs are not.

[Richard Williams]

The sem manual says

"When you estimate a linear regression by using sem, you obtain the same point estimates as you would with regress and the same standard errors up to a degree-of-freedom adjustment applied by regress."

You can probably dig through the manual and get the exact formulas if you really really want them. But my guess is it is something like using N-k-1 vs using N-k.

[Joseph Coveney]

If you look at the header for the glm output, you'll see that the scale parameter is the same as the Deviance (sum squared residuals) divided by the degrees of freedom. If you instead use Deviance divided by N as the scale parameter, then the coefficient standard errors will be the same. (The coefficients are scaled by the square root of the scale parameter.)

Try the following.

quietly glm mpg weight foreign // just to get the residual sum squares in the ereturn scalar
glm mpg weight foreign, scale(`=e(deviance) / e(N)') nolog

The scale parameter will now be the same as the residual variance coefficient fitted by sem (or mixed with an empty random-effects equation), and you'll now get the same regression coefficient standard errors as by sem (or mixed).

mixed mpg weight foreign, nolog

[Richard Williams]

The formula (if I can retype it correctly) is

sem se = sqrt((N-k-1)/N) * regression se

So in this case,

. di sqrt((74-2-1)/74) * .0006371
.00062405

. di sqrt((74-2-1)/74) * 1.075994
1.0539577

. di sqrt((74-2-1)/74) * 2.165547
2.1211966

[Bruce Weaver]

Thanks very much, Joseph & Richard. That certainly explains the difference I'm seeing. However, I'm still not clear on why the scale parameter is not computed using the same denominator for all three procedures (glm, sem, mixed). But that's a different question.

Thanks again.
Cheers,
Bruce