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Variational inference and the evidence lower bound — Examples

1 · Prerequisites

2 · Summary

These examples separate the exact finite-family case from two failure boundaries. An exact discrete family attains the log normalizer; a supported factorized density with failed entropy integrability can produce a coordinate exponential with infinite normalizer; and a candidate that puts mass outside the target support is not admissible. The calculations are local and make no generic convergence claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Exact discrete variational family attains log evidence

Example

Let $Ω = {0, 1}$ with counting measure, $γ (0) = 2$ , and $γ (1) = 1$ . Then $Z = 3$ and $π = (2 / 3, 1 / 3)$ . Taking $q = π$ attains $ELBO (q) = \log 3$ . The distinct admissible density $q = (1 / 2, 1 / 2)$ gives $ELBO (q) = \frac{3}{2} \log 2 < \log 3$ .

Facts & Assumptions

Given: The two-point counting space, target weights $γ = (2, 1)$ , and ELBO convention.

[F1]

Variational family and evidence lower bound: $Z = \int γ d μ$ , $π = γ / Z$ , and $ELBO (q) = \sum_{x} q (x) (\log γ (x) - \log q (x))$ for admissible $q$ .

[F2]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\log$ is strictly increasing, $\log 1 = 0$ , and it satisfies the product and quotient laws on positive reals.

Verification

Given: The data in the Example.

1.1F1algebra

Counting measure gives $Z = 2 + 1 = 3$ and $π = (2 / 3, 1 / 3)$ . For $q = π$ , each positive coordinate has $\log q (x) = \log γ (x) - \log 3$ , so the defining sum is $ELBO (π) = \log 3$ . Both terms are finite and $q$ is supported on $π$ .

2.1F1F2step 1.1algebra∎

For $q = (1 / 2, 1 / 2)$ , the defining sum is $\frac{1}{2} \log 2 + \log 2 = \frac{3}{2} \log 2$ . Since $(2^{3 / 2})^{2} = 8 < 9 = 3^{2}$ and both bases are positive, $2^{3 / 2} < 3$ . The log laws give $\log 3 - \frac{3}{2} \log 2 = \frac{1}{2} \log (9 / 8) > 0$ , because $9 / 8 > 1$ and [F2] makes $\log$ strictly increasing with $\log 1 = 0$ . This density is admissible because both target weights are positive and the space is finite.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

A candidate outside posterior support is not admissible

Example

On $Ω = {0, 1}$ with counting measure, take $γ = (1, 0)$ . Then $Z = 1$ and $π = (1, 0)$ . The probability density $q = (0, 1)$ is supported entirely outside the target support, so it is not admissible and has no finite ELBO.

Facts & Assumptions

Given: The two-point counting space, target density $γ = (1, 0)$ , and candidate $q = (0, 1)$ .

[F1]

Variational family and evidence lower bound: admissibility requires $q = 0$ on ${π = 0}$ and $\int q (∣ \log γ ∣ + ∣ \log q ∣) < \infty$ .

Verification

Given: The data in the Example.

1.1F1algebra

Counting measure gives $Z = 1$ and $π = (1, 0)$ . At the point $1$ , $q (1) = 1$ but $π (1) = 0$ , so the support condition in [F1] fails.

2.1F1step 1.1algebra∎

Also $γ (1) = 0$ and $q (1) = 1$ , so $q (1) ∣ \log γ (1) ∣ = + \infty$ . The absolute-integrability condition in [F1] fails as well; the finite-ELBO identity is not applicable to this candidate.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

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.