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mbakare
Calcite | Level 5
The likelihood ratio test of comparing reduced model with full model differs by fixed factor result to chi-square distribution of zero degree of freedom.
/* reduced model */
proc mixed method = ml;
class block gen;
model rtwt = /ddfm = kr;
random block gen;
run;
 
/* full model*/
proc mixed method = ml;
class block gen;
model rtwt = prop_hav/ddfm = kr;
random block gen;
run;
 
There are 3 degree of freedom from reduced model - block variance, genotype variance, and residual variance. The same degree of freedom for full model despite inclusion of  prop_hav as covariate. The difference in their -2 loglikelihood has zero degree of freedom under chi-distribution. Please could anyone guide me on how to compare these model to ascertain if the full model is significantly different from reduced model.
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sld
Rhodochrosite | Level 12 sld
Rhodochrosite | Level 12

You misunderstand the method. Yes, the two models have the same number of parameters associated with random effects (variances or covariances), but they have different numbers of parameters associated with fixed effects.

 

You seek to assess the significance of a continuous-scale fixed effects factor (prop_hav) using a likelihood ratio test. The full model has 5 parameters: intercept, slope for prop_hav, and 3 variances (rep, gen, residual). The reduced model has 4 parameters: intercept, and 3 variances (rep, gen, residual). The difference in the number of parameters between the two models is 5 - 4 = 1.

 

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sld
Rhodochrosite | Level 12 sld
Rhodochrosite | Level 12

Compute the difference between the "-2 Log Like" of the full model and the reduced model. Then compare this statistic to the  distribution with degrees of freedom equal to the difference in the number of parameters for the two models. The difference in number of estimated parameters for your example should be 1.

 

mbakare
Calcite | Level 5

Thanks for the reply. Please have a look at the SAS script I posted. Both reduced and full model have the same number of covariance parameters which is 3.

/* reduced model*/;

proc mixed method=ml;

class rep gen;

model rtwt=/ddfm=kr;

random rep gen;

run;

quit;

 

/* Full model */;

proc mixed method=ml;

class rep gen;

model rtwt= prop_hav /ddfm=kr;

random rep gen;

run;

quit;

 

As you can see, the full model has prop_hav as a covariate. The only difference between the model is the fixed factor which is covariate. The number of parameters in each case is 3: rep variance, gen variance, and residual variance. This means the difference of -2 log likelihood with have chi-square distribution with zero degree of freedom (3-3) and NOT 1 degree of freedom.

 

Thanks

 

 

sld
Rhodochrosite | Level 12 sld
Rhodochrosite | Level 12

You misunderstand the method. Yes, the two models have the same number of parameters associated with random effects (variances or covariances), but they have different numbers of parameters associated with fixed effects.

 

You seek to assess the significance of a continuous-scale fixed effects factor (prop_hav) using a likelihood ratio test. The full model has 5 parameters: intercept, slope for prop_hav, and 3 variances (rep, gen, residual). The reduced model has 4 parameters: intercept, and 3 variances (rep, gen, residual). The difference in the number of parameters between the two models is 5 - 4 = 1.

 

mbakare
Calcite | Level 5
Thanks a lot for your informative response. Your response helps a lot to figure out the solution.
sld
Rhodochrosite | Level 12 sld
Rhodochrosite | Level 12

I'm pleased you found it helpful. 

mbakare
Calcite | Level 5

Hi,

Just posted questions on externally conditional studentized residual from proc mixed. I would appreciate your guide and assistance on this.

Thanks

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