Je dirais que c'est un problème avec la façon dont les résultats sont rapportés. Ne pas «battre le tambour bayésien», mais approcher l'incertitude du modèle d'un point de vue bayésien en tant que problème d'inférence aiderait grandement ici. Et cela ne doit pas non plus être un grand changement. Si le rapport contenait simplement la probabilité que le modèle soit vrai, ce serait très utile. Il s'agit d'une quantité facile à estimer à l'aide du BIC. Appelez le BIC pour le mième modèle . La probabilité que le mème modèle soit le "vrai" modèle, étant donné que M modèles étaient adaptés (et que l'un des modèles est vrai) est donnée par:BICmM
P(model m is true|one of the M models is true)≈wmexp(−12BICm)∑Mj=1wjexp(−12BICj)
=11+∑Mj≠mwjwmexp(−12(BICj−BICm))
wjj= 1, cependant, vous pouvez avoir des modèles "théoriques" dans votre classe que vous vous attendez à améliorer avant de voir des données.
Now if somebody else doesn't report all the BIC's from all the models, then I would attempt to infer the above quantity from what you have been given. Suppose you are given the BIC from the model - note that BIC is calculable from the mean square error of the regression model, so you can always get BIC for the reported model. Now if we take the basic premise that the final model was chosen from the smallest BIC then we have BICfinal<BICj. Now, suppose you were told that "forward" or "forward stepwise" model selection was used, starting from the intercept using p potential variables. If the final model is of dimension d, then the procedure must have tried at least
M≥1+p+(p−1)+⋯+(p−d+1)=1+p(p−1)−(p−d)(p−d−1)2
different models (exact for forward selection), If the backwards selection was used, then we know at least
M≥1+p+(p−1)+⋯+(d+1)=1+p(p−1)−d(d−1)2
Models were tried (the +1 comes from the null model or the full model). Now we could try an be more specific, but these are "minimal" parameters which a standard model selection must satisfy. We could specify a probability model for the number of models tried M and the sizes of the BICj - but simply plugging in some values may be useful here anyway. For example suppose that all the BICs were λ bigger than the one of the model chosen so that BICm=BICj−λ, then the probability becomes:
11+(M−1)exp(−λ2)
So what this means is that unless λ is large or M is small, the probability will be small also. From an "over-fitting" perspective, this would occur when the BIC for the bigger model is not much bigger than the BIC for the smaller model - a non-neglible term appears in the denominator. Plugging in the backward selection formula for M we get:
11+p(p−1)−d(d−1)2exp(−λ2)
Now suppose we invert the problem. say p=50 and the backward selection gave d=20 variables, what would λ have to be to make the probability of the model greater than some value P0? we have
λ>−2log(2(1−P0)P0[p(p−1)−d(d−1)])
Setting P0=0.9 we get λ>18.28 - so BIC of the winning model has to win by a lot for the model to be certain.