Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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Marginals and positive-mass conditionals of a finite joint law are probability vectors

Statement

Let U,V be finite nonempty alphabets and let pUV(u,v)≥0 satisfy ∑u,vpUV(u,v)=1. Define pU(u)=∑vpUV(u,v) and pV(v)=∑upUV(u,v). Then both marginals are probability vectors. For each v with pV(v)>0, the conditional vector pU∣v(u):=pUV(u,v)/pV(v) is a probability vector; likewise for conditioning on positive-mass u. Consequently, if pUVW is a finite joint pmf, then for every w with pW(w)>0, pUV∣w and its two marginals are probability vectors.

Facts & Assumptions

Given: Finite nonempty alphabets U,V and a joint pmf pUV.

[F1]

Finite distributions, entropy, divergence, and stochastic matrices: a finite probability vector has nonnegative coordinates whose sum is 1.

[F2]

Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule: for finite sets S,T and a real array as,t, the sum over S×T equals either iterated sum, and finite sums are invariant under bijective reindexing.

Proof

1.1givenF1F2algebra

Every marginal coordinate is nonnegative because it is a finite sum of nonnegative entries of pUV. By [F2], ∑upU(u)=∑u∑vpUV(u,v)=∑(u,v)pUV(u,v)=1, and the same calculation with the order reversed gives ∑vpV(v)=1. Thus both marginals are probability vectors by [F1].

2.1givenF1step 1.1algebra∎

Fix v with pV(v)>0. Each ratio pUV(u,v)/pV(v) is nonnegative, and ∑upU∣v(u)=(∑upUV(u,v))/pV(v)=pV(v)/pV(v)=1. Hence pU∣v is a probability vector by [F1]. The same argument with u fixed gives pV∣u. If pUVW is a joint pmf and pW(w)>0, first apply this argument to the pair of finite alphabets U×V and W; then apply step 1.1 to the resulting conditional pmf on U×V. This gives the last assertion.

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