Effacer les filtres
Effacer les filtres

Compare 2 regression models

34 vues (au cours des 30 derniers jours)
Yi-xiao Liu
Yi-xiao Liu le 16 Déc 2020
Commenté : Ive J le 18 Déc 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.

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Ive J
Ive J le 16 Déc 2020
Modifié(e) : Ive J le 16 Déc 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 commentaires
Yi-xiao Liu
Yi-xiao Liu le 18 Déc 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 le 18 Déc 2020
Yes

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