Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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Data processing for a finite Markov chain, with equality condition

Statement

Let X,Y,Z have a joint pmf on finite alphabets and suppose that, for every y with pY(y)>0,

pXZ∣y(x,z)=pX∣y(x)pZ∣y(z).

This is the finite Markov condition X→Y→Z. Then

I(X;Z)≤I(X;Y),

with equality if and only if I(X;Y∣Z)=0, equivalently X and Y are conditionally independent given every positive-mass value of Z.

Facts & Assumptions

Given: A finite joint pmf pXYZ satisfying the stated Markov factorization.

[F1]

Finite entropy and mutual-information chain rules: I(X;Y,Z)=I(X;Y)+I(X;Z∣Y)=I(X;Z)+I(X;Y∣Z).

[F2]

Finite conditional mutual information identities and equality: finite CMI is nonnegative and vanishes exactly at conditional independence on each positive-mass conditioning value.

Proof

1.1givenF2algebra

For each y with pY(y)>0, the hypothesis makes pXZ∣y=pX∣y⊗pZ∣y. The corresponding conditional KL term in the definition of I(X;Z∣Y) is therefore D(pXZ∣y∥pX∣y⊗pZ∣y)=0, so the finite weighted sum gives I(X;Z∣Y)=0.

2.1F1F2step 1.1algebra∎

Equating the two expansions in [F1] and using step 1.1 gives I(X;Y)−I(X;Z)=I(X;Y∣Z). This is nonnegative by [F2], proving data processing. The difference vanishes exactly when I(X;Y∣Z)=0, and [F2] identifies that equality case with conditional independence of X,Y given each positive-mass z.

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