ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Unicast products lose link identifiability; multicast restores it

Example

Let a rooted tree have root $o$ , one shared link $e_{0} = (o, v)$ , and two branch links $e_{1} = (v, r_{1})$ and $e_{2} = (v, r_{2})$ to leaf receivers $r_{1}, r_{2}$ . Their success probabilities are $p_{0}, p_{1}, p_{2} \in (0, 1)$ . Compare separate unicast designs to $r_{1}$ and $r_{2}$ with a multicast design that sends each probe to both receivers and records both outcomes. For any fixed positive probe counts, the distinct parameter vectors $(1 / 2, 1 / 2, 1 / 4)$ and $(1 / 3, 3 / 4, 3 / 8)$ give the same separate-unicast laws but different multicast laws. In fact, the multicast law identifies all three link probabilities for every interior parameter vector.

The diagram labels each edge by its link-success probability; the two rightmost vertices are the receivers.

Example references

Finite-tree multicast loss experiments

Lawrence, Michailidis, and Nair, Examples 1–2

Xi, Michailidis, and Nair, §3.2

Facts & Assumptions

Given: The three-edge tree and the two interior parameter vectors in the example. Link successes are independent across links and probes; unicast outcomes are observed separately, whereas a multicast probe records both receiver indicators.

[F1]

Finite-tree multicast loss experiments: For receiver $r$ , its success indicator is the product of the link-success variables on its root path.

[F2]

Finite-tree multicast loss experiments: A singleton receiver set is a unicast scheme; a receiver set of size at least two is a multicast scheme.

[F3]

Finite-tree multicast loss experiments: The complete experiment observes $N_{d}$ independent outcomes with law $P_{p, d}$ for each design, independently across designs.

Verification

1.1givenF1F2F3algebra

Under unicast to $r_{i}$ , success has probability $p_{0} p_{i}$ . Both parameter vectors give $p_{0} p_{1} = 1 / 4$ and $p_{0} p_{2} = 1 / 8$ . Independence across probes makes each unicast outcome sequence have the same Bernoulli-product law under both vectors, and the separate schemes therefore have the same joint law.

1.2givenF1F2F3algebra

Under multicast, the receiver marginals are $m_{1} = p_{0} p_{1} = 1 / 4$ and $m_{2} = p_{0} p_{2} = 1 / 8$ , while the joint success probability is $m_{12} = p_{0} p_{1} p_{2}$ . It is $1 / 16$ for the first vector and $3 / 32$ for the second, so their multicast laws differ.

2.1givenF1algebra∎

For any interior vector, the multicast law supplies $m_{1}, m_{2}, m_{12}$ , with $m_{12} > 0$ . The equations give $p_{0} = m_{1} m_{2} / m_{12}$ , $p_{1} = m_{1} / p_{0}$ , and $p_{2} = m_{2} / p_{0}$ . Thus the multicast law determines every link probability.

Depends on

Finite-tree multicast loss experiments

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Lawrence, G. Michailidis, and V. N. Nair, Statistical Inverse Problems in Active Network Tomography, IMS Lecture Notes–Monograph Series 54 (2007), 24–44 (standard reference, not scraped)
B. Xi, G. Michailidis, and V. N. Nair, Estimating Network Loss Rates Using Active Tomography, Journal of the American Statistical Association 101 (2006), 1430–1448 (standard reference, not scraped)