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?


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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 $k > 0$, $\mathbb E[\epsilon^{2k+1}] = 0$.


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