# How to derive the gradient of the reparameterized score function estimator?

In the paper Evolution Strategies as a Scalable Alternative to Reinforcement Learning, the authors derive the following gradient of the score function estimator

\begin{align} \nabla_\psi\mathbb E_{\theta\sim p_\psi}[F(\theta)]&=\mathbb E_{\theta\sim p_\psi}[F(\theta)\nabla_\psi\log p_\psi(\theta)]\\ &=\nabla_\theta\mathbb E_{\epsilon\sim\mathcal N(0,I)}F(\theta+\sigma\epsilon)\\ &={1\over\sigma}\mathbb E_{\epsilon\sim\mathcal N(0,I)}\{F(\theta+\sigma\epsilon)\epsilon\} \end{align} where $$p_\psi$$ is a multivariate Gaussian, $$F$$ is the objective function(e.g., return on reinforcement learning problems) and $$\theta+\sigma\epsilon$$ is the result of the reparameterization trick. The meaning of $$\theta$$ changes from samples from $$p_\psi$$ to the mean of the $$p_\psi$$ in the second step, following the same convention used by the paper. I'm wondering how the last step is derived?

In the list of equations you have written above, you are mixing up two different approaches. In particular, line 2 (their approach) does not follow from line 1 (the REINFORCE objective). Even if $$p_\psi$$ is a gaussian, this would only follow if the variance is fixed ($$\psi = \theta$$).
As for the step line 2 to line 3, I think this is an approximation result based on Taylor series of $$F(\cdot)$$ about $$a = \theta$$
\begin{align} \mathbb E_{\epsilon\sim\mathcal N(0,I)} \left[\epsilon F(\theta+\sigma\epsilon)\right] &= \mathbb E_{\epsilon\sim\mathcal N(0,I)}\left[ \epsilon \left( F(\theta) + \sigma\epsilon \nabla F(\theta) + \dotso \right) \right]\\ &\approx \mathbb E_{\epsilon\sim\mathcal N(0,I)}\left[ \epsilon F(\theta) + \sigma\epsilon^2 \nabla F(\theta) \right] \end{align} Note that since $$\epsilon \sim \mathcal{N}(0,I)$$, we have that $$E[\epsilon] = 0$$, $$E[\epsilon^2] = \mathbf{I}$$. So the above expression simplifies to \begin{align} \mathbb E_{\epsilon\sim\mathcal N(0,I)} \left[\epsilon F(\theta+\sigma\epsilon)\right] \approx \sigma \nabla F(\theta)\\ \end{align} and rearranging we get,$$\nabla F(\theta) \approx \frac{1}{\sigma}\mathbb E_{\epsilon\sim\mathcal N(0,I)} \left[\epsilon F(\theta+\sigma\epsilon)\right]$$
I am not sure why they have the relationship as an equalilty. Perhaps the approximation is tight since for $$k > 0$$, $$\mathbb E[\epsilon^{2k+1}] = 0$$.