Why does the discrete chain converge to the Variance-Preserving SDE?

Scaling. Place steps on with and set for a fixed continuous function . The per-step noise variance must be of order : only then does the accumulated noise converge to Brownian motion. Order would blow up, and order would vanish.

Euler–Maruyama form. Since ,

which is one Euler–Maruyama step of .

Convergence. The first two conditional moments per unit time converge, and , and the fourth moment is , which excludes jumps. The drift is Lipschitz, so the limiting martingale problem is well posed. The classical diffusion-limit theorem for Markov chains (Stroock–Varadhan; Ethier–Kurtz) then gives weak convergence of the interpolated chain to the SDE.

Cumulative factor. Taking logarithms turns the product into a sum:

so . This is the formula used in the Noise Schedule.

Why not differentiate the marginal formula? Differentiating in with fixed gives a deterministic ODE. After replacing by its conditional expectation it becomes the Probability Flow ODE, which has the same marginals as the SDE but different paths. Marginals alone do not determine a diffusion; the Markov-chain limit selects the SDE.

References

  1. Y. Song et al. (2021). Score-Based Generative Modeling through Stochastic Differential Equations. ICLR 2021, App. B. arXiv:2011.13456
  2. S. N. Ethier, T. G. Kurtz (1986). Markov Processes: Characterization and Convergence. Wiley. doi:10.1002/9780470316658
  3. D. W. Stroock, S. R. S. Varadhan (1979). Multidimensional Diffusion Processes. Springer (reprinted in Classics in Mathematics). doi:10.1007/3-540-28999-2