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Active network tomography and identifiability by design — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

A shared-link Gaussian tree with nonidentifiable link means

Example

Assume AC for the library's standard-normal-law convention. 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. On each probe, the three link delays are independent with laws $X_{i} \sim N (μ_{i}, 1)$ , $i = 0, 1, 2$ , and the observed path delays are $Y_{1} = X_{0} + X_{1}$ and $Y_{2} = X_{0} + X_{2}$ . The distinct link-mean vectors $μ = (0, 0, 0)$ and $μ^{'} = (1, - 1, - 1)$ induce the same joint law of $(Y_{1}, Y_{2})$ ; hence the three link means are not identifiable. The same equality holds for any number of independent repeated probes with these fixed link laws.

The diagram labels the shared and branch links by their delay variables; the two rightmost vertices are the receivers.

Example references

Standard normal and normal laws

Axiom of Choice

Identifiability of an estimand

Lawrence, Michailidis, and Nair, Example 3

Facts & Assumptions

Given: The stated rooted tree, independent link delays with fixed means $μ_{i}$ and variance $1$ , and the two observed path sums. Assume AC only for the cited library definition of the normal laws.

[A1]

The Axiom of Choice: AC says every family of nonempty sets has a choice function; it is assumed here only because the cited normal-law definition declares it.

[F1]

Standard normal and normal laws: $N (μ, 1)$ is the pushforward of the standard normal law under $z \mapsto μ + z$ ; the definition assumes AC.

[F2]

Identifiability of an estimand: The identity target is identifiable exactly when equal sampling laws force equal parameter vectors.

Verification

1.1givenA1F1construct

On one probability space carrying independent $Z_{0}, Z_{1}, Z_{2} \sim N (0, 1)$ , set $X_{i}^{μ} = μ_{i} + Z_{i}$ and $X_{i}^{μ^{'}} = μ_{i}^{'} + Z_{i}$ . By [F1], each vector has the required independent link-delay laws.

2.1givenF1step 1.1algebra

For $μ = (0, 0, 0)$ the observation vector is $(Z_{0} + Z_{1}, Z_{0} + Z_{2})$ . For $μ^{'} = (1, - 1, - 1)$ it is $((1 + Z_{0}) + (- 1 + Z_{1}), (1 + Z_{0}) + (- 1 + Z_{2})) = (Z_{0} + Z_{1}, Z_{0} + Z_{2})$ . Thus the two observation vectors are equal on this coupling and have the same joint law; independent repetitions also have the same law.

3.1givenF2step 2.1algebra∎

The mean vectors are distinct, so [F2] implies that the identity link-mean target is not identifiable from the observed path-delay law.

ExampleConstruction: AI-generatedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

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.

Active network tomography and identifiability by design — Examples and Counterexamples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

A shared-link Gaussian tree with nonidentifiable link means

Example

Example references

Facts & Assumptions

Verification

Unicast products lose link identifiability; multicast restores it

Example

Example references

Facts & Assumptions

Verification

Sources