ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Weighted ELBO term differs from the simple diffusion loss

Example

Assume AC, take the one-dimensional schedule $T = 2$ , $β_{1} = \frac{1}{4}$ , $β_{2} = \frac{1}{2}$ , and set the reverse variance at step two to $σ_{2}^{2} = {\tilde{β}}_{2}$ . For a fixed $x_{0}$ , define the Borel predictor at step two by $ϵ_{θ} (x_{2}, 2) = \frac{x_{2} - \sqrt{3 / 8} x_{0}}{\sqrt{5 / 8}} + 1.$ For the step-one decoder, define $ϵ_{θ} (x_{1}, 1) = \frac{x_{1} - \sqrt{3 / 4} x_{0}}{\sqrt{1 / 4}} .$ Along the forward chain, the step-two noise-prediction error is identically one, while the step-one decoder mean equals $x_{0}$ . Thus the exact step-two ELBO contribution has coefficient $2$ , whereas the step-two contribution to the uniform-time $L_{s i m p l e}$ is $\frac{1}{2}$ .

Example references

Exact Gaussian noise-prediction weight and the unweighted objective: Ho, Jain, and Abbeel, Eqs. (12) and (14).

Pedagogical derivation of weighted and unweighted noise-prediction terms; its finite-horizon ELBO simplification approximates the terminal-prior term by zero: Prince, Chapter 18, “The diffusion ELBO” and “Simplifying it”.

Facts & Assumptions

Given: AC, $T = 2$ , $β_{1} = \frac{1}{4}$ , $β_{2} = \frac{1}{2}$ , $σ_{2}^{2} = {\tilde{β}}_{2}$ , and fixed $x_{0} \in R$ .

[A1]

AC is inherited from the Gaussian model and ELBO interfaces. The Axiom of Choice.

[F1]

The finite schedule has $α_{t} = 1 - β_{t}$ and ${\overset{ˉ}{α}}_{t} = \prod_{s = 1}^{t} α_{s}$ . Finite Gaussian forward chain and Gaussian reverse model.

[F2]

For $t \geq 2$ , the exact negative-ELBO step term is $C_{t} + w_{t} ∥ ϵ - ϵ_{θ} (x_{t}, t) ∥_{2}^{2}$ , with $w_{t} = \frac{β_{t}^{2}}{2 σ_{t}^{2} α_{t} (1 - {\overset{ˉ}{α}}_{t})}$ , while the simple objective is the uniform average over $t = 1, \dots, T$ of the unweighted squared noise-prediction error. DDPM variational objective and weighted noise-prediction terms.

Verification

technique · compute the schedule, weighted coefficient, and two objective contributions exactly

Given: The hypotheses in the Example.

1.1F1givenalgebra

The schedule is $α_{1} = \frac{3}{4}$ , $α_{2} = \frac{1}{2}$ , ${\overset{ˉ}{α}}_{1} = \frac{3}{4}$ , and ${\overset{ˉ}{α}}_{2} = \frac{3}{8}$ . Hence $1 - {\overset{ˉ}{α}}_{2} = \frac{5}{8}$ and ${\tilde{β}}_{2} = \frac{(1 / 2) (1 - 3 / 4)}{1 - 3 / 8} = \frac{1}{5}$ .

1.2A1F1givenalgebra

For the forward sample $x_{2} = \sqrt{3 / 8} x_{0} + \sqrt{5 / 8} ϵ$ , the step-two predictor in the Example returns $ϵ + 1$ , so the squared error is $1$ for every $ϵ$ . For $x_{1} = \sqrt{3 / 4} x_{0} + \sqrt{1 / 4} ϵ$ , the step-one predictor returns $ϵ$ , and its reverse mean is $\frac{1}{\sqrt{3 / 4}} (x_{1} - \frac{1 / 4}{\sqrt{1 / 4}} ϵ) = x_{0}$ . Therefore the reconstruction log-likelihood contribution is the fixed Gaussian density at its mean, independent of the step-two error.

2.1F2step 1.1algebra

Substituting $σ_{2}^{2} = {\tilde{β}}_{2} = \frac{1}{5}$ into [F2] gives $w_{2} = \frac{(1 / 2)^{2}}{2 (1 / 5) (1 / 2) (5 / 8)} = 2$ and $C_{2} = \frac{1}{2} (1 - 1 + \log 1) = 0$ . Thus the exact step-two ELBO contribution is $2$ times the expected squared noise error.

3.1A1F2step 1.2step 2.1algebra∎

The uniform-time simple objective in [F2] has step-two contribution $\frac{1}{2} \cdot 1 = \frac{1}{2}$ (and step-one noise error zero), while the exact weighted negative-ELBO term at step two is $2$ . Thus the two step-two objective contributions differ.

Source notes

Exact weighted DDPM term: DDPM variational objective and weighted noise-prediction terms.

Uniform-time simple-loss definition: DDPM variational objective and weighted noise-prediction terms.

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Sources

Jonathan Ho, Ajay Jain, and Pieter Abbeel, Denoising Diffusion Probabilistic Models (2020) (standard reference, not scraped)
Simon J. D. Prince, Understanding Deep Learning, Chapter 18: Diffusion models (2026), pp. 363–387 (standard reference, not scraped)