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SSEP empirical-field martingale and bracket

Statement

Assume the Axiom of Choice. Let $d \geq 1$ , $N \geq 3$ , $T \geq 0$ , and let $(η_{t})_{0 \leq t \leq T}$ be the SSEP from the diffusive-model definition with any initial probability vector. For a real smooth periodic $G : T^{d} \to R$ , set

$M_{t}^{N, G} : = ⟨ π_{t}^{N}, G ⟩ - ⟨ π_{0}^{N}, G ⟩ - \int_{0}^{t} ⟨ π_{s}^{N}, Δ_{N} G ⟩ d s .$

Then $M^{N, G}$ is a square-integrable martingale. Its continuous predictable bracket is

$A_{t}^{N, G} = N^{2 - 2 d} \int_{0}^{t} \sum_{i = 1}^{d} \sum_{x \in Λ_{N}} (η_{x + e_{i}} (s) - η_{x} (s))^{2} (G ((x + e_{i}) / N) - G (x / N))^{2} d s,$

in the continuous predictable compensator convention of the finite-chain martingale supplier. In particular,

$A_{T}^{N, G} \leq T N^{- d} \sum_{i = 1}^{d} ∥ \partial_{i} G ∥_{\infty}^{2} .$

Statement references

Symmetric exclusion process under diffusive scaling

Martingale and predictable bracket for a finite-state chain

Facts & Assumptions

Given: The Axiom of Choice, a finite SSEP rate matrix, its uniformized realization, a smooth real periodic $G$ , and a finite horizon $T$ .

[A1]

Assume the Axiom of Choice, meaning every family of nonempty sets has a choice function; the finite-chain martingale supplier requires it (The Axiom of Choice).

[F1]

Symmetric exclusion process under diffusive scaling: $Ω_{N}$ is finite; each occupation-changing unordered nearest-neighbor swap has rate $N^{2}$ . Equal endpoint occupations give no off-diagonal transition and zero observable increment, so generator and squared-jump sums may be extended over all bonds with coefficient $N^{2}$ . Also $π_{t}^{N} = N^{- d} \sum_{x} η_{x} (t) δ_{x / N}$ and $⟨ π, G ⟩ = N^{- d} \sum_{x} η_{x} G (x / N)$ .

[F2]

Martingale and predictable bracket for a finite-state chain: for a finite rate matrix $L$ , bounded $f$ , and its uniformized realization, $f (X_{t}) - f (X_{0}) - \int_{0}^{t} L f (X_{s}) d s$ is a square-integrable martingale with continuous predictable bracket $\int_{0}^{t} \sum_{j \neq i} L_{i j} (f (j) - f (i))^{2} d s$ along the chain.

Proof

1.1A1F1F2algebra

Put $f_{G} (η) = N^{- d} \sum_{x} η_{x} G (x / N)$ . For the bond ${x, x + e_{i}}$ , $f_{G} (η^{x, x + e_{i}}) - f_{G} (η) = N^{- d} (η_{x + e_{i}} - η_{x}) (G (x / N) - G ((x + e_{i}) / N))$ . Summing its rate- $N^{2}$ contribution over all $x, i$ and reindexing the terms containing $η_{x + e_{i}}$ gives $L_{N} f_{G} (η) = N^{- d} \sum_{x} η_{x} N^{2} \sum_{i} [G ((x + e_{i}) / N) + G ((x - e_{i}) / N) - 2 G (x / N)] = ⟨ π^{N} (η), Δ_{N} G ⟩$ . Thus the finite-chain martingale in [F2], applied to $f_{G}$ , is exactly $M_{t}^{N, G}$ and has bracket given by the sum of rates times squared jumps.

2.1F1F2step 1.1algebra∎

A bond with equal endpoint occupations has zero jump; otherwise its squared jump is $N^{- 2 d} (G ((x + e_{i}) / N) - G (x / N))^{2}$ . Substituting these jumps and rate $N^{2}$ into [F2] gives the displayed bracket. By the fundamental theorem of calculus along the coordinate segment, $∣ G ((x + e_{i}) / N) - G (x / N) ∣ \leq N^{- 1} ∥ \partial_{i} G ∥_{\infty}$ , while $(η_{x + e_{i}} - η_{x})^{2} \leq 1$ . There are $N^{d}$ bonds in each coordinate direction, so the bracket is increasing, continuous, predictable, and bounded by $T N^{- d} \sum_{i} ∥ \partial_{i} G ∥_{\infty}^{2}$ . The finite state space makes $f_{G}$ bounded, hence [F2] also gives square integrability.

SSEP empirical-field martingale and bracket

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources