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Compare 2 regression models

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Yi-xiao Liu
Yi-xiao Liu am 16 Dez. 2020
Kommentiert: Ive J am 18 Dez. 2020
I wonder if there is any tool to compare 2 regression models on the same sample pool. In general more terms you put into the model the closer the fit, but you risk over fitting. For example:
>> mdl=fitglm(FitZV,FitDataV,'linear')
mdl =
Generalized linear regression model:
y ~ 1 + x1
Distribution = Normal
Estimated Coefficients:
Estimate SE tStat pValue
________ __________ _______ ______
(Intercept) 81.101 0.0085111 9528.9 0
x1 -0.22506 0.00058189 -386.77 0
9638 observations, 9636 error degrees of freedom
Estimated Dispersion: 0.698
F-statistic vs. constant model: 1.5e+05, p-value = 0
>> mdl2=fitglm(FitZV,FitDataV,'purequadratic')
mdl2 =
Generalized linear regression model:
y ~ 1 + x1 + x1^2
Distribution = Normal
Estimated Coefficients:
Estimate SE tStat pValue
___________ __________ _______ __________
(Intercept) 81.286 0.012269 6625.6 0
x1 -0.22447 0.00057029 -393.6 0
x1^2 -0.00086632 4.2147e-05 -20.555 6.3993e-92
9638 observations, 9635 error degrees of freedom
Estimated Dispersion: 0.668
F-statistic vs. constant model: 7.83e+04, p-value = 0
Both mdl and mdl2 are statically better than a constant model, but does mdl2 explain the dataset significantly better than mdl (or the oppsite)? From what I have found devianceTest only test the model to constant, but couldn't find a function to compare 2 models? If anyone can point me to the right direction that would be appreciated.

Akzeptierte Antwort

Ive J
Ive J am 16 Dez. 2020
Bearbeitet: Ive J am 16 Dez. 2020
One way is to use likelihood ratio test (similar to R anova function). For more information see here.
y = randn(200, 1); % test response
x = randn(200, 1); % test predictor
model1 = fitlm(x, y)
model2 = fitlm([x, x.^2], y);
LR = 2*(model2.LogLikelihood - model1.LogLikelihood); % has a X2 distribution with a df equals to number of constrained parameters, here: 1
pval = 1 - chi2cdf(LR, 1);
  2 Kommentare
Yi-xiao Liu
Yi-xiao Liu am 18 Dez. 2020
How was the degrees of freedom in chi2cdf determined?
In the example above it is 1, is it because model2 have one more coefficient than model1?
Ive J
Ive J am 18 Dez. 2020
Yes

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