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I am looking for a little clarity on what the policy gradient theorem means. My confusion lies in the fact that the reward $R$ in reinforcement learning is non-differentiable in the policy parameters. As that is the case how does the central objective of policy gradients, finding the gradients of Reward $R$ wrt the parameters of policy function even make sense?

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We want to find the gradient of policy "return" $V$ wrt. parameters of the policy $\theta$. Where the return $V$ could be written as "how good is an action $Q$ $\times$ probability of taking that action $\pi$".

Consider the policy gradient, $\nabla_\theta V = \sum_a Q \nabla_\theta \pi + \pi \nabla_\theta Q$

The first term tells us to adjust the action probability proportionally to how good it is. To me it could read "if an action yields good, take more". That is to move the peak of $\pi$ to match the peak of $Q$. This is a reasonable thing to do. But of course since $Q$ cannot directly guide us toward its peak, it is up to our $\pi$ to luckily stumble upon the high peak of $Q$. This emphasizes the importance of exploratory nature of $\pi$.

The second term is vice versa. That is to move the peak of $Q$ to match the peak of $\pi$. This is much harder a task because $Q$ is a function of both action and policy, $Q_{\pi_\theta}(s, a)$. We clearly don't have this in a differentiable form i.e. we don't have a universal $Q$ function over the space of all possible $\pi$.

We now have a partial gradient from the first term but we have yet to estimate the second term.

Turns out, the second term can be recursively written solely in the form of the first term but with subsequent actions and states.

$$ \nabla_\theta V_0 = \sum Q_0 \nabla_\theta \pi_0 + \sum Q_1 \nabla_\theta \pi_1 + \sum Q_2 \nabla_\theta \pi_2 + \dots $$

That is to get good policy i.e. policy gradient we only need to move the peaks of $\pi$ to match the peaks of $Q$ not only the first (state, action) but also for all subsequent (state, action)'s. This yields the same result as if we differentiate through the $Q$.

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