For the chart for a training data set, each point on the chart represents a terminal node from the tree. The terminal node with the highest event probability is the first point on the chart and appears leftmost. The other terminal nodes are in order of decreasing event probability.
Use the following process to find the x- and y-coordinates for the points.
For example, suppose the following table summarizes a tree with 4 terminal nodes:
A: Terminal node | B: Number of events | C: Number of cases | D: Threshold (B/C) |
---|---|---|---|
4 | 18 | 30 | 0.60 |
1 | 25 | 67 | 0.37 |
3 | 12 | 56 | 0.21 |
2 | 4 | 36 | 0.11 |
Totals | 59 | 189 |
Then the following are the corresponding true positive rates to 2 decimal places:
A: Terminal node | B: Number of events | C: True positive rate |
---|---|---|
4 | 18 | 18 / 59 = 0.31 |
1 | 25 | 25 / 59 = 0.42 |
3 | 12 | 12 / 59 = 0.20 |
2 | 4 | 4 / 59 = 0.07 |
Totals | 59 |
For example, if the terminal node with the highest predicted probability contains 0.16 of the data and the terminal node with the second-highest event probability has 0.35 of the population, then the cumulative percentage of the data for the first terminal node is 0.16 and the cumulative percentage of the population for the second terminal node is 0.16 + 0.35 = 0.51.
The following table shows an example of the computations for a small tree. The values are to 2 decimal places.
A: Terminal node | B: Number of events | C: Number of cases | D: Event probability for sorting (B/C) | E: True positive rate | F: Percent in data (C/ sum of C) | G: Cumulative percent in data, x-coordinate | H: Lift (E /F), y-coordinate |
---|---|---|---|---|---|---|---|
4 | 18 | 30 | 0.60 | 0.31 | 0.16 | 0.16 | 1.94 |
1 | 25 | 67 | 0.37 | 0.42 | 0.35 | 0.51 | 1.20 |
3 | 12 | 56 | 0.21 | 0.20 | 0.30 | 0.81 | 0.67 |
2 | 4 | 36 | 0.11 | 0.07 | 0.19 | 1.00 | 0.37 |
Use the same steps as the training data set case but calculate the event probability from the cases for the test data set.
The procedure to define the x- and y-coordinates on the lift chart with k-fold cross-validation has an additional step. This step creates many distinct event probabilities. For example, suppose the tree diagram contains 4 terminal nodes. We have 10-fold cross-validation. Then, for the ith fold, you use 9/10 portion of the data to estimate the event probabilities for cases in fold i. When this process repeats for each fold, the maximum number of distinct event probabilities is 4 *10 = 40. After that, sort all the distinct event probabilities in decreasing order and combine any distinct event probabilities that need more cases to form a proper bin. After this step, steps from 3 to the end for the training data set procedure apply to find the x- and y-coordinates.