Suppose that there are three candidate predictors, x1 , x2, and x3, for a final regression model. Suppose further that the intercept term, β0 is always included in all the model equations. How many models must be estimated and examined if one applies all possible regressions approach?




Our models vary based on which predictors are selected to be in a particular model.

How many ways can we make a model with $0$ of the predictors? With $1$? With $2$? With all $3$?

  • $\begingroup$ - 0 predictors give 1 model (only the constant term) - 1 predictors give 1 model (only x1) - 2 predictors give 3 possible models (x1, x2 or x1&x2) - 3 predictors give 7 possible models (x1, x2, x3, x1&x2, x2&x3, x1&x3 or x1&x2&x3) So in the nth case, what is the amount of models that can be created? Is it n!+1? $\endgroup$ – Vidar Trojenborg Jun 4 '19 at 7:16
  • $\begingroup$ Not quite. You have listed out all of the possibilities, but some of them you have listed out more than once. (I see x1 up there three times). When I ask how many models use one predictor, I am referring to the models with x1 only or x2 only or x3 only. Once you have figured out how many models there are for the cases of $0$, $1$, $2$, $3$ then you add up the possibilities to get the total $\endgroup$ – WaveX Jun 4 '19 at 11:14
  • $\begingroup$ Alright, sorry if I'm not following, but in that case, we can; - Use 0 predictor in 1 way - Use 1 predictor 3 ways - Use 2 predictors in 3 ways - Use all 3 predictors in 1 way? Then would that mean that having 3 predictors we can make 8 different models? $\endgroup$ – Vidar Trojenborg Jun 4 '19 at 11:40
  • $\begingroup$ Yes. This is the correct answer! $\endgroup$ – WaveX Jun 4 '19 at 11:53
  • $\begingroup$ Great, thanks! And in the nth case, how would one approach it then? What would the formula be? $\endgroup$ – Vidar Trojenborg Jun 4 '19 at 12:29

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