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 the entropy function H. The data rows are: Jack,0,A; Jack,1,B; Amy,0,A; Amy,1,B; Sam,0,B; Sam,1,B. The task is to determine which statement about H(X), H(Y), and H(Z) is true.

Option A: H(X) = 1.585, H(Y) = 0, H(Z) = 0.9183. 
- For H(X), we look at the distribution of the X values across the six records. X takes values {Jack, Amy, Sam}, each appearing twice. So P(Jack)=P(Amy)=P(Sam)=2/6=1/3. The entropy H(X) = -∑ p log2 p = -3*(1/3*log2(1/3)) = log2(3) ≈ 1.585. So the stated H(X) value of 1.585 is plausible.
- For H(Y), Y is 0 or 1, with each appearing three times (three 0s and three 1s). Thus P(Y=0)=P(Y=1)=1/2, giving H(Y) = -[0.5 log2 0.5 + 0.5 log2 0.5] = 1.0, not 0. Therefore H(Y)=0 in this option is wrong.
- For H(Z), Z takes values A or B. In the data, A appears in two rows (Jack,0,A and Amy,0,A), while B appears in four rows (Jack,1,B; Amy,1,B; Sam,0,B; Sam,1,B). So P(Z=A)=2/6=1/3 and P(Z=B)=4/6=2/3. H(Z) = -[(1/3) log2(1/3) + (2/3) log2(2/3)] ≈ 0.9183. This part is consistent, but since H(Y) is incorrect, the whole option is invalid.

Option B: H(X) = 1.213, H(Y) = 1, H(Z) = 0.1544. 
- H(X) would be 1.585 as shown earlier since X has uniform distribution over three values. 1.213 would correspond to a less uniform distribution, which does not match the data. Hence H(X) is incorrect.
- H(Y) = 1 is correct because Y has two equiprobable outcomes (0 and 1) each with probability 0.5; H(Y) = 1.0. So this part is correct, but H(X) and H(Z) need to be checked.
- H(Z) = 0.1544 would imply a distribution heavily skewed toward one value, which is not the case since P(Z=A)=1/3 and P(Z=B)=2/3 yields H(Z) ≈ 0.9183, not 0.1544. Thus this option is inconsistent.

Option C: H(X) = 1.585, H(Y) = 1, H(Z) = 0.9183. 
- H(X) = 1.585 is correct given X has three equally likely values (Jack, Amy, Sam) each appearing twice. This yields log2(3) ≈ 1.585.
- H(Y) = 1 is correct since Y is binary with equal probabilities for 0 and 1, each 0.5.
- H(Z) = 0.9183 is correct because Z has probabilities P(Z=A)=1/3 and P(Z=B)=2/3, and the entropy for this distribution is H(Z) = -(1/3)log2(1/3) - (2/3)log2(2/3) ≈ 0.9183.
- Therefore all three components in this option align with the observed frequencies.

Option D: H(X) = 1.785, H(Y) = 1, H(Z) = 0.656. 
- H(X) = 1.785 would imply a different distribution for X than the actual data; with three equally likely X values, H(X) should be 1.585, not 1.785.
- H(Y) = 1 is correct as before, but since H(X) is already incorrect, this option is invalid.
- H(Z) = 0.656 would be a different entropy value than the calculated 0.9183 for Z, so this is also incorrect.

Summary of reasoning across options:
- The entropy H(X) is maximized for a uniform distribution over three X values, yielding log2(3) ≈ 1.585; the data indeed shows X = {Jack, Amy, Sam} each twice, so H(X) ≈ 1.585. 
- The entropy H(Y) is for a binary distribution with Y=0 and Y=1 each occurring three times, so P=0.5 for both; thus H(Y) = 1.0.
- The entropy H(Z) uses the distribution Z = {A with probability 1/3, B with probability 2/3}, which gives H(Z) ≈ 0.9183.

Concrete conclusions from the analysis:
- Option A fails because H(Y) is listed as 0, which contradicts the actual 0.5-0.5 split for Y.
- Option B fails because H(X) is stated as 1.213 and H(Z) as 0.1544, both inconsistent with the data.
- Option C is consistent with all three entropies computed from the dataset: H(X) ≈ 1.585, H(Y) = 1, H(Z) ≈ 0.9183.
- Option D fails because H(X) = 1.785 and H(Z) = 0.656 are not aligned with the observed frequencies.

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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