Unit V · Lecture 35 of 40 · B.Sc. Data Science, 1st Year, Semester II
A set of frequency data is said to be consistent if it is logically possible — that is, if there exists at least one valid 2×2 (or larger) table of non-negative frequencies that satisfies all the given marginal totals and cross-class frequencies simultaneously. If no such table exists, the data is inconsistent.
Think of the frequencies as pieces of a jigsaw puzzle. The marginal totals (A), (α), (B), (β) are the edges, and the cross-class frequencies (AB), (Aβ), (αB), (αβ) are the interior pieces. The data is "consistent" if the pieces actually fit together — if the interior pieces add up to match the edges. If someone claims (A) = 100 and (B) = 80 but (AB) = 90, the pieces don't fit — the data is inconsistent because (AB) cannot exceed either (A) or (B).
Given frequencies (A), (B), (AB), and N, the data is consistent if and only if all four of the following conditions hold:
Condition 1: (AB) ≤ (A) — The joint class cannot exceed the marginal class
Condition 2: (AB) ≤ (B) — Same logic for the other attribute
Condition 3: (AB) ≥ (A) + (B) − N — The overlap cannot be too small
Condition 4: (AB) ≥ 0 — Frequencies cannot be negative
Combining: max(0, (A)+(B)−N) ≤ (AB) ≤ min((A), (B))
(αβ) = N − (A) − (B) + (AB) ≥ 0
Therefore: (AB) ≥ (A) + (B) − N
This says: the number of people with both attributes must be at least as large as the "excess" — the amount by which (A) + (B) exceeds N. If (A) + (B) > N, some people must have both attributes.
Upper bound: (AB) can't be bigger than either (A) or (B) — you can't have more people with both attributes than have either attribute alone.
Lower bound: If (A) + (B) > N, then by the pigeonhole principle, at least (A) + (B) − N people must have both attributes. The classes (A) and (B) must overlap by at least this amount.
Given: N = 1000, (A) = 600, (B) = 500, (AB) = 300. Is this data consistent?
Check Condition 1: (AB) ≤ (A)? → 300 ≤ 600? → Yes ✓
Check Condition 2: (AB) ≤ (B)? → 300 ≤ 500? → Yes ✓
Check Condition 3: (AB) ≥ (A) + (B) − N? → 300 ≥ 600 + 500 − 1000? → 300 ≥ 100? → Yes ✓
Check Condition 4: (AB) ≥ 0? → 300 ≥ 0? → Yes ✓
All conditions satisfied → Data is CONSISTENT ✓
| (B) | (β) | Total | |
|---|---|---|---|
| (A) | 300 | 300 | 600 |
| (α) | 200 | 200 | 400 |
| Total | 500 | 500 | 1000 |
Given: N = 200, (A) = 120, (B) = 150, (AB) = 10. Is this data consistent?
Check Condition 1: (AB) ≤ (A)? → 10 ≤ 120? → Yes ✓
Check Condition 2: (AB) ≤ (B)? → 10 ≤ 150? → Yes ✓
Check Condition 3: (AB) ≥ (A) + (B) − N? → 10 ≥ 120 + 150 − 200? → 10 ≥ 70? → NO ✗
Condition 3 is violated → Data is INCONSISTENT ✗
(A) + (B) = 120 + 150 = 270, but N = 200. So 270 − 200 = 70 people must have both attributes (by the pigeonhole principle). But the data claims only 10 have both — that's impossible!
If we try to build the table:
(αβ) = N − (A) − (B) + (AB) = 200 − 120 − 150 + 10 = −60
A negative frequency is impossible. The data is inconsistent.
Given N = 500, (A) = 300, (B) = 250. What values can (AB) take for the data to be consistent?
Lower bound: max(0, (A)+(B)−N) = max(0, 300+250−500) = max(0, 50) = 50
Upper bound: min((A), (B)) = min(300, 250) = 250
For consistency: 50 ≤ (AB) ≤ 250
Any value of (AB) between 50 and 250 (inclusive) gives a consistent dataset.
(AB) = 50, (Aβ) = 250, (αB) = 200, (αβ) = 0
→ Maximum possible independence from B for A-holders
(AB) = 250, (Aβ) = 50, (αB) = 0, (αβ) = 200
→ All B-holders are also A-holders (B ⊂ A)
For three attributes A, B, C with given (A), (B), (C), (AB), (BC), (AC), (ABC), and N, the data is consistent if:
1. Each pair satisfies the two-attribute conditions:
max(0, (A)+(B)−N) ≤ (AB) ≤ min((A), (B))
max(0, (A)+(C)−N) ≤ (AC) ≤ min((A), (C))
max(0, (B)+(C)−N) ≤ (BC) ≤ min((B), (C))
2. The triple intersection satisfies:
(ABC) ≤ min((AB), (BC), (AC))
(ABC) ≥ max(0, (AB)+(AC)−(A), (AB)+(BC)−(B), (AC)+(BC)−(C))
3. All eight cross-class frequencies must be non-negative.
Q1: Data is consistent when:
Q2: Given N=100, (A)=60, (B)=70. The minimum possible value of (AB) is:
Q3: Given N=100, (A)=60, (B)=70. The maximum possible value of (AB) is:
Q4: If (A)+(B) > N, then:
Q5: If (αβ) = −10, the data is:
1. Check consistency: N=300, (A)=180, (B)=200, (AB)=100.
2. Check consistency: N=500, (A)=350, (B)=300, (AB)=50.
3. Given N=400, (A)=250, (B)=200. Find the range of (AB) for consistency.
4. Given N=1000, (A)=400, (B)=500, (C)=600, (AB)=200, (BC)=300, (AC)=250, (ABC)=100. Check consistency by computing all 8 cross-class frequencies.
5. A researcher reports: N=200, (A)=150, (B)=120, (AB)=100, (Aβ)=30. Verify whether this is consistent. If not, identify the error.
Independence and Association of Attributes — When are attributes related?