Yя= β0+ β1Икся+ β2Zя+ ϵя
я = 1 , … , нИкс*я= ( хя- х¯) / sd ( x )Z*я= ( zя- з¯) / sd ( z)
и вместо этого подгоните модель (используя обычные наименьшие квадраты)
Yя= β*0+ β*1Икс*я+ β*2Z*я+ ϵя
Тогда измененные параметры (бета-версии) изменятся, но они изменятся таким образом, что вы можете вычислить с помощью простой алгебры из примененного преобразования. Так что, если мы называем оценочные бета-версии из модели с использованием преобразованных предикторов дляβ*1 , 2 и обозначают беты из нетрансформированной модели с β^1 , 2, we can calculate one set of betas from the other one, knowing the means and standard deviations of the predictors. The realtionship between the transformed and untransformed parameters is the same as between their estimates, when based on OLS. Some algebra will give the relationship as
β0=β∗0−β∗1x¯sd(x)−β∗2z¯sd(z),β1=β∗1sd(x),β2=β∗2sd(z)
So standardization is not a necessary part of modelling. (It might still be done for other reasons, which we do not cover here). This answer depends also upon us using ordinary least squares. For some other fitting methods, such as ridge or lasso, standardization is important, because we loose this invariance we have with least squares. This is easy to see: both lasso and ridge do regularization based on the size of the betas, so any transformation which change the relative sizes of the betas will change the result!
And this discussion for the case of linear regression tells you what you should look after in other cases: Is there invariance, or is it not? Generally, methods which depends on distance measures among the predictors will not show invariance, so standardization is important. Another example will be clustering.