Tensions in Practice: Compact pair checks in tension with retaining joint structure¶
Three binary indicators in a stipulated probability model
A and B are independent fair bits, taking 0 or 1. C is 1 exactly when A and B differ. Every pair is independent, yet knowing A and B determines C. Keeping only pairwise summaries answers pair questions compactly. Keeping the joint table exposes the forbidden triples, at the cost of a larger representation.
Keep summaries compact
Retain only the pairwise relations needed for pair questions.
Preserve joint constraints
Answer questions about all three indicators together.
Why these aims pull against each other
Pairwise margins omit higher-order constraints; full joint records grow with the number of variables.
Choose an arrangement to see what changes and what remains difficult.
Rows hold the A,B pair and columns hold C. “Not fixed” means not recoverable from the retained pairwise summaries; zero means impossible under the stated joint law. In the joint view, shading distinguishes impossible triples from those assigned nonzero probability.
What this choice protects
What it costs
When it fits
Compare the arrangements
Retain pairwise summaries
Keep each two-variable distribution; all four combinations in every pair have probability 1/4.
| A,B total | C = 0 | C = 1 | |
|---|---|---|---|
| A0 B0 | 1/4 | Not fixed | Not fixed |
| A0 B1 | 1/4 | Not fixed | Not fixed |
| A1 B0 | 1/4 | Not fixed | Not fixed |
| A1 B1 | 1/4 | Not fixed | Not fixed |
- What it protects
- Pair questions need only the retained margins.
- What it costs
- Triple probabilities are not determined by those summaries.
- When it fits
- Fits when downstream questions really involve only pairs and no triple factorization is claimed.
Illustration note: Each A,B row has total 1/4; its split across C remains unknown from pairwise margins alone.
Retain the joint law
Keep all eight triple probabilities, including zeros.
| A,B total | C = 0 | C = 1 | |
|---|---|---|---|
| A0 B0 | 1/4 | 1/4 | 0 |
| A0 B1 | 1/4 | 0 | 1/4 |
| A1 B0 | 1/4 | 0 | 1/4 |
| A1 B1 | 1/4 | 1/4 | 0 |
- What it protects
- The parity constraint and impossible triples remain inspectable.
- What it costs
- Joint storage and validation grow rapidly for more variables.
- When it fits
- Fits when joint outcomes matter and the full law can be specified or estimated reliably.
Illustration note: Here four triples have probability 1/4; multiplying three fair marginals would incorrectly assign 1/8 to every triple.
What this illustration does—and does not—establish
The source supplies the tension. The invented setting, alternatives and any numbers illustrate a limited comparison; each arrangement retains its stated costs and conditions.
- This is an exact invented law, not a finite-sample statistical test.
- Statistical independence is a distributional property, not proof of absent causal connections.
- Only this small joint table is shown; real estimation costs depend on assumptions and data.
Source entries
Statistical Independence
This source passage supplies the contextual tension. The concrete arrangements and schematic examples are editorial illustrations, not measured findings.
Pairwise versus Mutual (scalar)
For three or more variables, pairwise independence does not imply mutual independence; the factorization must hold over the full joint, not just every pair.
The source operation
Two variables are statistically independent when learning the value of one gives no probabilistic information about the other: the conditional distribution equals the marginal.