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I have a technical question concerning the choice of the splitting criteria for the recursive partitioning. Having selected the most significant variable, I would like to know why the optimal splitting criteria is found using only the test statistics as the objective function of the optimization and not with the p-value of the test. I don't know if I'm right but the optimal splitting criteria could not be significant if we computed its p-value instead of just using the value of the test statistic (even if the selected variable is significant).

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  • $\begingroup$ I'm not sure what the question is here. Are you asking why potentially different tests are used for selecting the splitting variable vs. the splitting point? $\endgroup$ Jul 26, 2018 at 7:55
  • $\begingroup$ I wanted to know why the optimal splitting point is selected using the value of the statistic and not the p-value. I had a case where the optimal split maximizing the value of the statistic was associated with a p-value higher than 5%. Therefore, the splitting variable was significant but not the splitting point of this variable. $\endgroup$
    – R.B
    Aug 2, 2018 at 11:17

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By default ctree() uses different test statistics for selecting the split variable and the split point, respectively. The default test for the former is a kind of correlation test but for the latter a maximally-selected two-sample test. See:

In this notation, the transformation $g(\cdot)$ used for selecting the variable is by default $g(x) = x$ for a numeric variable. The transformation for selecting the splitpoint is then formed by considering all transformations of type $g(x) = (1, 1, 1, \dots, 0, 0)^\top$ encoding all possible binary splits (subject to subset size requirements), and then taking the maximum.

The reason for choosing the maximalliy-selected statistic in the second step is straightforward, I guess, but the association test in the first step might be surprising. Of course, the maximally-selected test could also be used there but (a) is much more costly to compute and (b) might have less power against monotonic associations. But it is surely possible to find situations when one or the other test performs better. If you want to deviate from ctree's default you need to modify the xtrafo argument.

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