Assuming we can interchange the derivative and integral operator, $$\nabla_\phi \int f(z) q_\theta(z) dz = \int f(z) \nabla_\phi q_\theta(z), $$ and then note that by the chain rule, $$\nabla_\phi \ln q_\theta(z) = \frac{\nabla q_\phi(z) }{q_\phi(z)} \Rightarrow \nabla q_\phi(z) = \nabla_\phi \ln q_\theta(z) q_\phi(z), $$ and the result just follows.


While using the harmonic mean of the simulation densities as an estimator of one is "the worst Monte Carlo method ever", I checked its convergence by coding the simulation from $u(x,\cdot)$ on my own and I did not spot any discrepancy: > mean(1/propz(simox())) [1] 0.9945046 > mean(1/propz(simox())) [1] 1.001786 Here is my R code for completion. wrap&...


Observe that \begin{align*} \log \psi(x) &= \log \sum_{k\in\mathbb Z}\phi(x+k) \\ &= \log \sum_k \exp\left[ \log \phi(x+k) \right] \\ &= m + \log \sum_k \exp\left[ \log \phi(x+k) - m \right] \end{align*} The goal will be to perform the sum in log-space, and then exponentiate at the very end to get $\psi(x)$. First, if we take $m := \max_k\{ \...

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