TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Finite conditional mutual information identities and equality

Statement

For finite $X, Y, Z$ ,

$I (X; Y ∣ Z) = H (X ∣ Z) + H (Y ∣ Z) - H (X, Y ∣ Z) \geq 0.$

Equality holds if and only if, for every $z$ with $p_{Z} (z) > 0$ , the conditional joint pmf factors as $p_{X Y ∣ z} = p_{X ∣ z} p_{Y ∣ z}$ ; this is conditional independence given $Z$ .

Facts & Assumptions

Given: A joint pmf $p_{X Y Z}$ on finite alphabets; all entropies and conditional mutual information use the nats convention.

[F1]

Finite conditional mutual information: $I (X; Y ∣ Z) = \sum_{z : p_{Z} (z) > 0} p_{Z} (z) D (p_{X Y ∣ z} ∥ p_{X ∣ z} \otimes p_{Y ∣ z})$ .

[F2]

Finite conditional entropy: $H (U ∣ Z) = \sum_{z : p_{Z} (z) > 0} p_{Z} (z) H (p_{U ∣ z})$ , with zero-mass conditioning values omitted.

[F3]

Finite distributions, entropy, divergence, and stochastic matrices: $H (r) = - \sum_{r_{i} > 0} r_{i} \ln r_{i}$ , with $0 \ln 0 = 0$ .

[F4]

Finite KL divergence is nonnegative with equality exactly at equal laws: for finite probability vectors $r, s$ , $D (r ∥ s) \geq 0$ , with equality exactly when $r = s$ .

[F5]

Marginals and positive-mass conditionals of a finite joint law are probability vectors: each positive-mass conditional joint vector and its marginals are probability vectors.

Proof

1.1F1F3F5algebra

Fix $z$ with $p_{Z} (z) > 0$ . By [F5], $p_{X Y ∣ z}$ and its marginals are pmfs. If $p_{X Y ∣ z} (x, y) > 0$ , then both marginals are positive at $x, y$ . Expanding [F1] and collecting terms using [F3] gives $D (p_{X Y ∣ z} ∥ p_{X ∣ z} \otimes p_{Y ∣ z}) = H (p_{X ∣ z}) + H (p_{Y ∣ z}) - H (p_{X Y ∣ z})$ ; zero coordinates contribute $0$ , so no logarithm of zero is taken.

2.1F1F2F4step 1.1algebra∎

Multiply the identity in step 1.1 by $p_{Z} (z)$ and sum over positive-mass $z$ . By [F1] and [F2], this gives the stated entropy identity. Each conditional KL is nonnegative by [F4], and every weight in the sum is positive. Hence the total is zero exactly when each conditional KL is zero, which by [F4] is exactly $p_{X Y ∣ z} = p_{X ∣ z} \otimes p_{Y ∣ z}$ for every positive-mass $z$ .

Depends on

Used by

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Sources

Robert M. Gray, Entropy and Information Theory, corrected first edition (standard reference, not scraped)
Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, author prepublication (standard reference, not scraped)