ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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Finite evidence and finite expected logs do not replace admissibility

Example

There are a finite-evidence target $γ$ and a supported product probability density $q$ for which $\int q ∣ \log γ ∣ d μ < \infty$ and the expected-log coordinate function is identically zero on a countably infinite counting space, but its exponential has infinite normalizer. This $q$ fails the entropy-integrability part of admissibility; finite evidence and pointwise finite expected log alone do not justify a coordinate update.

Facts & Assumptions

Given: The finite product space and densities constructed in the Verification.

[F1]

Variational family and evidence lower bound: an admissible $q$ is a probability density supported by $π = γ / Z$ with $\int q (∣ \log γ ∣ + ∣ \log q ∣) < \infty$ .

[F2]

Coordinate-ascent variational inference: the coordinate factor is proportional to $e^{h (n)}$ , and a usable update requires $0 < \sum_{n} e^{h (n)} < \infty$ .

[F3]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: nonnegative sums and product integrals may be evaluated by iterated integration.

[F4]

The integral logarithm $L (x) : = \int_{1}^{x} \frac{d t}{t}$ for $x > 0$ and The integral logarithm $L$ is the published natural logarithm: for $x > 0$ , $\log x = \int_{1}^{x} d t / t$ .

[F5]

If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ : pointwise bounds on integrable functions over $[a, b]$ pass to corresponding bounds on their integrals.

Verification

Given: Let $I = {1, 2, \dots}$ be the set of positive integers, let $μ_{X}$ be counting measure on $I$ , and define the atomic measure $μ_{Y} (S) = \sum_{m \in S} w_{m}$ for $S \subseteq I$ , where the positive finite weights $w_{m}$ are specified in step 1.1. Each singleton has finite mass under both measures; since the countable singletons cover $I$ , both reference spaces are sigma-finite (Measures on sigma-algebras, Finite, sigma-finite, and semifinite measures).

1.1F3F4algebra

Put $r_{n} = 2^{- n}$ , $b_{n} = - 2 \log (n + 1)$ , $a_{n} = - (1 - r_{n}) b_{n} / r_{n} = 2 (2^{n} - 1) \log (n + 1) > 0$ , and $w_{m} = e^{- 2 a_{m}}$ . Define $γ (n, m) = e^{a_{n}}$ when $m = n$ and $γ (n, m) = e^{b_{n}} = (n + 1)^{- 2}$ otherwise. By [F3], $Z = \sum_{m} e^{- 2 a_{m}} (e^{a_{m}} + \sum_{n \neq m} (n + 1)^{- 2})$ . For every $m$ , $a_{m} \geq 2 \log (m + 1) > 0$ , so $e^{- a_{m}} \leq (m + 1)^{- 2}$ . Also $\sum_{n} (n + 1)^{- 2} < \infty$ because $(n + 1)^{- 2} \leq [n (n + 1)]^{- 1} = 1 / n - 1 / (n + 1)$ . Hence $0 < Z \leq \sum_{m} e^{- a_{m}} + (\sum_{m} e^{- 2 a_{m}}) (\sum_{n} (n + 1)^{- 2}) < \infty$ .

2.1F1F3step 1.1

Give the $X$ -factor masses $q_{X} (n) = r_{n}$ and the $Y$ -factor density $q_{Y} (m) = r_{m} / w_{m}$ relative to $μ_{Y}$ . Each factor has total mass $\sum_{n} r_{n} = 1$ , so $q (n, m) = r_{n} r_{m} / w_{m}$ is a product probability density by [F3]. It is positive wherever $γ$ is positive, so it satisfies the support condition in [F1].

3.1F3F4F5step 2.1algebra

For fixed $n$ , the expectation of $\log γ (n, Y)$ under $q_{Y}$ is $r_{n} a_{n} + (1 - r_{n}) b_{n} = 0$ . Its absolute expectation is $r_{n} a_{n} + (1 - r_{n}) ∣ b_{n} ∣ = 2 (1 - r_{n}) ∣ b_{n} ∣ \leq 4 \log (n + 1)$ . Therefore $\int q ∣ \log γ ∣ d μ \leq 4 \sum_{n} 2^{- n} \log (n + 1) < \infty$ . For $n \geq 1$ , [F4] and [F5] give $\log (n + 1) = \int_{1}^{n + 1} d t / t \leq n$ , since $0 < 1 / t \leq 1$ on $[1, n + 1]$ . Also, for $n \geq 4$ one has $n \leq 2^{n / 2}$ by induction, so $n 2^{- n} \leq 2^{- n / 2}$ and the tail is bounded by a convergent geometric series.

3.2F1F3F4step 2.1algebra

For $m \geq 1$ , induction gives $2^{m} - 1 \geq m$ , and monotonicity of the logarithm gives $\log (m + 1) \geq \log 2 > 0$ . Hence $a_{m} = 2 (2^{m} - 1) \log (m + 1) \geq 2 m \log 2 \geq (m + 1) \log 2$ . On the row $n = 1$ , $\log q (1, m) = - (m + 1) \log 2 + 2 a_{m} \geq a_{m}$ . Its contribution to the nonnegative entropy integral is therefore at least $r_{1} \sum_{m} r_{m} a_{m}$ . But $r_{m} a_{m} = 2 (1 - 2^{- m}) \log (m + 1) \geq \log 2$ for every $m \geq 1$ , so that series diverges. By [F3], $\int q ∣ \log q ∣ d μ = \infty$ , and [F1] shows that $q$ is not admissible.

4.1F2step 3.1algebra∎

For the $X$ -coordinate update with $q_{Y}$ fixed, step 3.1 gives $h (n) = E_{q_{Y}} \log γ (n, Y) = 0$ for every $n$ . Therefore the update normalizer in [F2] is $\sum_{n = 1}^{\infty} e^{h (n)} = \sum_{n = 1}^{\infty} 1 = \infty$ . Thus finite evidence, support and finite pointwise expected log do not suffice: no normalized coordinate factor exists, and step 3.2 identifies the failed admissibility hypothesis.

Remarks

The earlier label “despite admissibility” was incorrect: the entropy series diverges. The construction and its divergent coordinate normalizer are retained with that failure explicit. The A-page finite-normalizer lemma proves that an admissible product satisfying the stated expected-log conditions cannot have this divergence.

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