The pathwise estimator is not always the quieter one

For $z\sim\mathcal N(\mu,1)$, $\E[\cos(\omega z)]=\cos(\omega\mu)e^{-\omega^2/2}$. Use $\cos^2t=\tfrac12(1+\cos2t)$.

The truth. $\E_q[\sin(10z)]=\sin(10\mu)e^{-50}$, so the derivative at $\mu=0$ is $10e^{-50}=1.9\times10^{-21}$: zero to any printed precision.

Pathwise. The estimate is $10\cos(10z)$. Its mean is the derivative, and its second moment is $100\,\E[\cos^2(10z)]=50\big(1+e^{-200}\big)$. Its variance is therefore $50$ to every printed digit.

Score function. The estimate is $\sin(10z)\,z$. Its mean is also the derivative. Its second moment is $\E[\sin^2(10z)\,z^2]=\tfrac12\E[z^2]-\tfrac12\E[z^2\cos(20z)]$, and the second average is $-399\,e^{-200}$, so the variance is $0.5$ to every printed digit.

So the pathwise estimate is a hundred times noisier here. It reads the slope of $f$ at the draw, and a function that oscillates ten times per unit has slopes of size ten everywhere, which cancel only on average. The score function reads the value of $f$, which never exceeds one. The claim broken is that reparameterisation always lowers the variance. It does for the smooth, slowly varying log-likelihoods of a VAE decoder; it does not in general.