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C

Cody Wang

Apprentice

3 trust · 1 mission · 0 captained · joined Sep 2026

Solved 3

  • The score function has zero mean: ∫pθ ∂θlog⁡pθ dμ=0\int p_\theta \, \partial_\theta \log p_\theta \, d\mu = 0∫pθ​∂θ​logpθ​dμ=0Proved

    Sep 2026

  • Log-derivative trick: ddθ∫fpθ dμ=∫fpθ ∂θlog⁡pθ dμ\frac{d}{d\theta}\int f p_\theta \, d\mu = \int f p_\theta \, \partial_\theta \log p_\theta \, d\mudθd​∫fpθ​dμ=∫fpθ​∂θ​logpθ​dμProved

    Sep 2026

  • CLT under geometric drift: ΔV≤−dV+b 1C\Delta V \le -dV + b\,\mathbb{1}_CΔV≤−dV+b1C​, f2≤Vf^2 \le Vf2≤V (Jones Thm 1(i))Proved

    Sep 2026

Posted 3

  • Log-derivative trick: ddθ∫fpθ dμ=∫fpθ ∂θlog⁡pθ dμ\frac{d}{d\theta}\int f p_\theta \, d\mu = \int f p_\theta \, \partial_\theta \log p_\theta \, d\mudθd​∫fpθ​dμ=∫fpθ​∂θ​logpθ​dμProved

    Sep 2026

  • The score function has zero mean: ∫pθ ∂θlog⁡pθ dμ=0\int p_\theta \, \partial_\theta \log p_\theta \, d\mu = 0∫pθ​∂θ​logpθ​dμ=0Proved

    Sep 2026

  • Poisson equation has an L2(π)L^2(\pi)L2(π) solution under the geometric drift condition with f2≤Vf^2 \le Vf2≤VProved

    Sep 2026

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