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Classification Trees

Good morning,

I have a small file (see attached file) to better understand what JMP calculates.

The tree is quite small, I set the minimum size of a node to 1, I performed the divisions one by one, I asked for the probabilities to be displayed.

Now, I'm focused on the node "X1>=14" and if I understand correctly:

1. The G^2 (3.82) characterizes the impurity of this knot and is independent of the "best" division that comes below it
2. Logworth (1.05) is calculated using G^2 displayed right next to it.
3. The Logworth depends on the "best" division, which is displayed just below; it is calculated from an adjusted p-value obtained from all possible cuts.

Points 2 and 3 appear to be incompatible.

JMP's learnbot has, for once, stalled, and sees 2 possible but incompatible interpretations.

So, where does this 1.05 come from and how do we interpret it?

Thank you for your response.

Jean-Luc

Predictive Modeling and Machine Learning Re: Classification Trees

Hello @rugolf ,

I think there is confusion in the definition and use of the two metrics G2 and Logworth.
As mentioned in the JMP help documentation , the decision tree in JMP uses the Logworth value as the decision criterion for determining where to perform the split (with the goal of maximizing the Logworth value), and this corresponds to -log10(p-value). This p-value is calculated with an adjustment, not detailed in the help documentation, which ensures a fair and equitable calculation between factors with many or few levels.

The G2 is displayed for categorical responses only, and a low G2 indicates a better fit. For continuous responses, the standard deviation is displayed for each node, along with the Candidate SS (sum of squares for the best split).

I hope this answer will shed some light on how this platform works.

Re: Classification Trees

THANKS,

If I understand correctly, in the case of a categorical response with 2 categories A and B:

1. The G² measures the impurity of the node (also called deviance): 0 if only A or only B, unbalanced classes but no further division from this node
2. the maximum of G^2 (equal to n * 2ln(2) I think) if the classes are balanced (as many A as
3. G^2 decreases with each division while R^2 increases; this is going in the right direction, which is what we want.
4. We could calculate the improvement (the gain) after a division G^2(parent)-(G^2(left)+G^2(right))
5. The best cut is the one that provides the greatest improvement, the one that most significantly reduces the previous G^2 value.
6. logworth = -log(p-value) with p-value associated with a test and adjusted (I'll skip the calculation, no big deal)
7. The logworth displayed next to G^2 refers to the best division displayed just below it.
8. The test is H_0: on the left and right the proportion of A is the same vs ....is not the same
9. Extreme case: p-value=1, therefore logworth=0 when there is the same proportion of A (and in the child nodes, H_0 retained at the end of the test
10. In extreme cases: p-value=0, therefore logworth=infinity when the proportion of A is very different in the child nodes, H_1 retained at the end of the test
11. p-value <0.01, therefore logworth >2, therefore a significant difference in proportion of A (or , a significant discernment
12. The best cut is therefore also the one with the highest logworthiness.
13. The tree stops growing when G^2 is zero, when the minimum size is reached, or when R^2 no longer increases significantly.

Thank you for your help.

Rugolf or Jean-Luc

Re: Classification Trees

Hello Jean-Luc,

Regarding the different parts:

• Yes, G2 is equivalent to other criteria used such as Gini impurity or entropy. G2 corresponds to twice the natural logarithm of entropy. Therefore, if we find a separation that allows us to obtain "pure" nodes (perfect separation), then G2 is zero (zero entropy).
• With each split, G2 will decrease because we will get better and better at classifying observations using factors.
• The G2 improvement is calculated and visible in the node information ( Candidate G2 )
• Regarding the Logworth test, a p-value is indeed calculated to obtain it, but only to guide the selection of the best split. The test behind this p-value corresponds to the hypotheses you mentioned: H0 - the proportions of A and B are equivalent in both nodes vs. H1 - the proportions of A and B are different in the two nodes.
• As mentioned in my first answer, the optimal split is the one corresponding to the highest Logworth.
• The tree stopping criteria are linked to the minimum size for the split ( minimum size split , adjustable in the options), by choosing a tree depth, or a value/evolution of R2 (possibly with a validation sample to identify the moment when R2 no longer evolves/changes little on the validation sample), etc...
  The evolution of R² as a function of splits can be seen in the "Split History":
  
• It's a minor detail, but on the shared dataset, it would seem that the optimal split is rather at X1 > or < 10 (larger Logworth than at 14):
  

  With a value of 14 for X1, the separation is not optimal, and I do not find the values ​​you mentioned in Logworth or G2:

  
• Furthermore, since factors X1 and X2 are very strongly correlated (correlation > 0.99), a decision tree may not be the most appropriate method, as one of the two variables will inevitably be ignored compared to the other; a method that takes this correlation into account would be more appropriate.
  • Either through a penalized logistic regression model (ridge or even lasso since both factors contain the same information), or even a simple logistic model with one factor (X1 or X2) using the Fit Y by X platform:
    
  • Either by choosing a tree-based method that allows managing the correlation/collinearity between factors, such as Random Forest ( Bootstrap Forest platform in JMP), or other Machine Learning models (Support Vector Machine).
    
    There are JMP datasets available to help you learn and understand how certain platforms work; you can find them in the "Help" and "Sample Index" menus. In the "Exploratory Modeling" submenu, several datasets are available that involve the Partition platform: Car Poll, Cereal, Diamonds Data, Mushroom, Titanic Passengers, etc.
    
    

Hopefully this answer clarifies how the Partition platform works.

Re: Classification Trees

Hello Victor,

Thank you very much for this generous response.

I see that I'm on the "right" side.

Otherwise, the choice of the small file is intentional: I do this to clearly see what the software calculates, for example, for a simple linear regression. I have other, larger, and more realistic files.

Thank you again.

Best wishes.

Jean-Luc