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In the draft for Sutton's latest RL book, page 270, he derives the REINFORCE algorithm from the policy gradient theorem. The first part is the equivalence

$$\sum_{s}d_\pi(s)\sum_{a}{q_\pi(s,a)\nabla\pi (a|s,\theta)} = \mathbf{E}[\gamma^t\sum_{a}{q_\pi(S_t,a)\nabla\pi (a|S_t,\theta)}] $$

where $$d_\pi(s) = \sum_{k=0}^{\infty}{\gamma^kP(S_k = s | S_0, \pi)}$$

This makes intuitive sense (we just sample our trajectory, and we expect that we will average the long term trajectory), but I'm having trouble deriving this analytically. The discount factor $\gamma^t$ stops us from treating $d_\pi$ as simply a probability distribution, so we aren't just taking the expected value over the states. Can anyone point me in the right direction please?

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The expectation is over states we sample when running the policy. Suppose $D$ is the multiset containing all the states we've visited in our dataset, and $D_t$ is the same thing, but only containing those states visited at exactly time $t$: $$\nabla \eta(\theta) = E_{\pi}\left[\gamma^t \sum_a q_{\pi}(S_t, a) \nabla_{\theta} \pi(a|S_t,\theta)\right] $$ Then we can write this as an expectation over the distribution of possible multisets D. (Let the size of D be $n$) $$= E\left[ \frac{1}{n} \sum_{s \in D} \gamma^t \sum_a q_{\pi}(S_t, a) \nabla_{\theta} \pi(a|S_t,\theta) \right] \\ = \sum_t E\left [ \frac{1}{n} \sum_{s \in D_t} \gamma^t \sum_a q_{\pi}(S_t, a) \nabla_{\theta} \pi(a|S_t,\theta) \right] \\ = \sum_t \sum_{s \in S} \frac{1}{n} P(s \in D_t) \gamma^t \sum_a q_{\pi}(S_t, a) \nabla_{\theta} \pi(a|S_t,\theta) \\ = \sum_t \sum_{s \in S} P(S_t = s | S_0, \pi) \gamma^t \sum_a q_{\pi}(S_t, a) \nabla_{\theta} \pi(a|S_t,\theta) $$ Then reorder the sums $$= \sum_{s \in S} \sum_t P(S_t = s | S_0, \pi) \gamma^t \sum_a q_{\pi}(S_t, a) \nabla_{\theta} \pi(a|S_t,\theta) \\ = \sum_{s \in S} d_{\pi}(s) \sum_a q_{\pi}(S_t, a) \nabla_{\theta} \pi(a|S_t,\theta) $$ Which is the result we want.

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As a remark:

There are some different variations of the REINFORCE formula in different sources.

Sutton & Barto end up deriving the formula:

$ \frac{\partial}{\partial \theta}v(S_0;\theta)=\sum_{t=0}^{T-1}\gamma^tG_t \frac{\partial}{\partial \theta} \ln \pi(A_t|S_t;\theta) $

Another variation of the formula are given in http://www.scholarpedia.org/article/Policy_gradient_methods

Here the formula derived are;

$ \frac{\partial}{\partial \theta}v(S_0;\theta)= G_0 \sum_{t=0}^{T-1} \frac{\partial}{\partial \theta} \ln \pi(A_t|S_t;\theta) $

As far as I can tell the reason for the deviation are a fault in Sutton & Barto. One step in their derivation assume that $ \frac{\partial}{\partial \theta}r(s',s,a)=0 $. However $ \theta $ affect the policy, which affect $ a $ and $ s' $. As a result, the reward $ r $ are not invariant under changes in $ \theta $.

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@flush_bingo you can get $\frac{\partial r(s',s,a)}{\partial \theta_1} = 0$ by considering the classical definition of partial derivative i.e. for a given tuple of $(s',s, a)$

$$\frac{\partial r(s',s,a)}{\partial \theta_1} = \frac{r(s',s,a)_{\theta_1+\Delta\theta_1} - r(s',s,a)_{\theta_1}}{\Delta\theta_1}$$

Since the value of reward for taking a action $a$ in state $s$ and getting to state $s'$ is going to be the same no matter what the value of $\theta_1$ is the partial derivative is zero. Different values of $\theta_1$ are only going to affect the frequency with which $r(s',s,a)$ appears.

You can follow the same logic for the derivative of transition model with respect to policy parameters i.e.

$$\frac{\partial P(s'|s,a)}{\partial \theta_1} = 0$$

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