Consider the following dataset with features X, Y and Z
[table]
X | Y | Z
Jack | 0 | A Jack | 1 | B Amy | 0 | A Amy | 1 | B Sam | 0 | B Sam | 1 | B
[/table] Let H denote the entropy function. Which of the following is true?

Question

BlackTom AI · Accepted Answer

Question restatement: We are given a dataset with features X, Y, and Z and are asked to determine which statement about entropy H is true for each feature. The dataset rows are: (Jack, 0, A), (Jack, 1, B), (Amy, 0, A), (Amy, 1, B), (Sam, 0, B), (Sam, 1, B).

Now evaluate each option option by option:

Option a: H(X) = 1.213, H(Y) = 1, H(Z) = 0.154
- This is likely incorrect because H(Y) = 1 is plausible since Y is balanced (0 and 1 appear equally often). However, H(X) is computed from the distribution of X values {Jack, Amy, Sam}, each appearing twice, which yields probabilities of 2/6 for each unique X, giving H(X) = -∑ p log2 p = -3*(2/6) log2(2/6) = log2(3) ≈ 1.585 bits, not 1.213. Therefore H(X) value here is wrong.
- H(Z) would depend on the distribution of Z values {A, B}. In the data, A occurs twice and B occurs four times, so p(A)=1/3 and p(B)=2/3, giving H(Z) = -(1/3)log2(1/3) - (2/3)log2(2/3) ≈ 0.918 bits, not 0.154. So this option is inconsistent overall.

Option b: H(X) = 1.785, H(Y) = 1, H(Z) = 0.656
- H(Y) = 1 is fine, but H(X) = 1.785 implies a distribution with more than three equally likely outcomes or uneven probabilities that yield higher entropy; here there are exactly 3 distinct X values (each occurs twice), so H(X) = log2(3) ≈ 1.585, not 1.785. This makes the option incorrect.
- H(Z) = 0.656 is close to the actual 0.918, but since H(X) is already incorrect, this option fails.

Option c: H(X) = 1.585, H(Y) = 0, H(Z) = 0.918
- H(X) = 1.585 is correct as shown above, since X has three equally likely values (Jack, Amy, Sam) each with probability 1/3.
- H(Y) = 0 would imply that Y has no variability (a degenerate distribution where all samples have the same Y value). But in the dataset Y takes both 0 and 1 equally often, giving H(Y) = 1. This makes this option incorrect.
- H(Z) = 0.918 matches the computed value for Z’s distribution (A with probability 1/3 and B with probability 2/3), so this portion is correct.

Option d: H(X) = 1.585, H(Y) = 1, H(Z) = 0.9183
- H(X) = 1.585 is correct, as explained.
- H(Y) = 1 is correct since Y has two equally likely values 0 and 1 (each occurs 3 times in 6 samples).
- H(Z) = 0.9183 is correct (exactly the binary entropy for p(A)=1/3, p(B)=2/3).
- All components align with the data, so this option is fully correct.

Summary: The key ideas are (1) X takes three equally likely values, giving H(X) = log2(3) ≈ 1.585; (2) Y is balanced with two outcomes, giving H(Y) = 1; (3) Z has probabilities 1/3 and 2/3, giving H(Z) ≈ 0.918. Among the options, only option d matches all three entropies exactly.

COMP9417-Machine Learning & Data Mining - T3 2025

Consider the following dataset with features X, Y and Z [table] X | Y | Z Jack | 0 | A Jack | 1 | B Amy | 0 | A Amy | 1 | B Sam | 0 | B Sam | 1 | B [/table] Let H denote the entropy function. Which of the following is true?

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