1. Supplementary Text 2

2. If the relationship between two random variables X and Y is approximately illustrated in the following figure, then how can we discretize the sample values of them? μ1μ1 μ2μ2 N(μ1, σ1)N(μ1, σ1) N(μ2, σ2)N(μ2, σ2) t 1,1 t 1,2 t 1,3 t 1,4 t 2,1 t 2,2 t 2,3 t 2,4 Fig. A Fig. B

3. To uniformly describe the relation ship between X and Y, we can map them to normal distribution. And the corresponding relationship would be change into that illustrated in the figure below. 0 0 N(0, 1) t1t1 t2t2 t3t3 t4t4 t1t1 t2t2 t3t3 t4t4 Fig. C Fig. D

4. The data transformation between Figure A and C is implemented with the Equation (14) in the paper. And the data transformation between Figure B and D is also implemented with Equation (14) in the paper. Then our goal is to find the best common partitioning points t 1, t 2, …, t k. The Purpose of Equation (15) in the paper is to determine the best common partitioning points for both variable X and Y, while simultaneously making both of them as evenly partitioned as possible. The structures of the variables shown in Figure B and D reveal the essence of the relationship between the two random variables of the same kind. This is of more importance when the number of random variables becomes large. It makes the partition not “casual” just in order to adapt to the data, some of which are just noise.

5. Example 3 The gene expression data is partially listed below. Suppose that Gene #1 and #2 have been divided in the same group G 1, while Gene #3, #4, and #5 are in the same group G 2. Time 1Time 2Time 16 Gene No. Array1 (Low) Array2 (High) Array3 (Low) Array4 (High) ……Array31 (Low) Array32 (High) 1D 1,1,L D 1,1,H D 1,2,L D 1,2,H ……D 1,16,L D 1,16,H 2D 2,1,L D 2,1,H D 2,2,L D 2,2,H ……D 2,16,L D 2,16,H 3D 3,1,L D 3,1,H D 3,2,L D 3,2,H ……D 3,16,L D 3,16,H 4D 4,1,L D 4,1,H D 4,2,L D 4,2,H ……D 4,16,L D 4,16,H 5D 5,1,L D 5,1,H D 5,2,L D 5,2,H ……D 5,16,L D 5,16,H

6. For the data in the table above, we define 15 vectors as follows. Time 1Time 2Time 16 Gene No. Array1 (Low) Array2 (High) Array3 (Low) Array4 (High) ……Array31 (Low) Array32 (High) 1D 1,1,L D 1,1,H D 1,2,L D 1,2,H ……D 1,16,L D 1,16,H 2D 2,1,L D 2,1,H D 2,2,L D 2,2,H ……D 2,16,L D 2,16,H 3D 3,1,L D 3,1,H D 3,2,L D 3,2,H ……D 3,16,L D 3,16,H 4D 4,1,L D 4,1,H D 4,2,L D 4,2,H ……D 4,16,L D 4,16,H 5D 5,1,L D 5,1,H D 5,2,L D 5,2,H ……D 5,16,L D 5,16,H

7. To compute the mutual information between the two groups G 1 and G 2 about low pressure treatment described in Equation (17) in the paper, we can use the following detailed equation in stead of (17):

8. To compute the mutual information between the two groups G 1 and G 2 about high pressure treatment described in Equation (18) in the paper, we can use the following detailed equation in stead of (18):

9. To compute the mutual information between the two groups G 1 and G 2 about both high pressure and the high pressure treatment described in Equation (19) in the paper, we can use the following detailed equation in stead of (19):

10. To compute the mutual information between the treatment low pressure and the treatment high pressure within the group G 1 described in Equation (20) in the paper, we can use the following detailed equation in stead of (20):

11. To compute the mutual information between the treatment low pressure and the treatment high pressure within the group G 2 described in Equation (20) in the paper, we can use the following detailed equation in stead of (20):