In policy gradient method, there's a trick to reduce a variance of policy gradient. We use causality, and remove part of the sum over rewards so that only actions happened after the reward are taken into account (See here http://rail.eecs.berkeley.edu/deeprlcourse/static/slides/lec-5.pdf, slide 18).

Why does it work? I understand the intuitive explanation, but what's the rigorous proof of it? Can you point me to some papers?


An important thing we're going to need is what is called the "Expected Grad-Log-Prob Lemma here" (proof included on that page), which says that (for any $t$):

$$\mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta}(a_t \mid s_t) \right] = 0.$$

Taking the analytical expression of the gradient (from, for example, slide 9) as a starting point:

$$\begin{aligned} \nabla_{\theta} J(\theta) &= \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \left( \sum_{t=1}^T \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \right) \left( \sum_{t=1}^T r(s_t, a_t) \right) \right] \\ % &= \sum_{t=1}^T \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \sum_{t'=1}^T r(s_{t'}, a_{t'}) \right] \\ % &= \sum_{t=1}^T \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \sum_{t'=1}^{t-1} r(s_{t'}, a_{t'}) + \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \sum_{t'=t}^T r(s_{t'}, a_{t'}) \right] \\ % &= \sum_{t=1}^T \left( \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \sum_{t'=1}^{t-1} r(s_{t'}, a_{t'}) \right] \\ + \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \sum_{t'=t}^T r(s_{t'}, a_{t'}) \right] \right) \\ \end{aligned}$$

At the $t^{th}$ "iteration" of the outer sum, the complete sum $\sum_{t'=1}^{t-1} r(s_{t'}, a_{t'})$ is independent of the trajectory $\tau$ due to the Markov property, which means we're allowed to pull that expression out of the expectation:

$$\nabla_{\theta} J(\theta) = \sum_{t=1}^T \left( \sum_{t'=1}^{t-1} r(s_{t'}, a_{t'}) \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \right] \\ + \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \sum_{t'=t}^T r(s_{t'}, a_{t'}) \right] \right)$$

The first expectation can now be replaced by $0$ due to the lemma mentioned at the top of the post:

$$ \begin{aligned} \nabla_{\theta} J(\theta) &= \sum_{t=1}^T \left( \sum_{t'=1}^{t-1} r(s_{t'}, a_{t'}) \times 0 + \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \sum_{t'=t}^T r(s_{t'}, a_{t'}) \right] \right) \\ % &= \sum_{t=1}^T \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \left[ \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \sum_{t'=t}^T r(s_{t'}, a_{t'}) \right] \\ % &= \mathbb{E}_{\tau \sim \pi_{\theta}(\tau)} \sum_{t=1}^T \nabla_{\theta} \log \pi_{\theta} (a_t \mid s_t) \left( \sum_{t'=t}^T r(s_{t'}, a_{t'}) \right). \\ \end{aligned} $$

The expression on slide 18 of the linked slides is an unbiased, sample-based estimator of this gradient:

$$\nabla_{\theta} J(\theta) \approx \frac{1}{N} \sum_{i=1}^N \sum_{t=1}^T \nabla_{\theta} \log \pi_{\theta} (a_{i, t} \mid s_{i, t}) \left( \sum_{t'=t}^T r(s_{i, t'}, a_{i, t'}) \right)$$

For a more formal treatment of the claim that we can pull $\sum_{t'=1}^{t-1} r(s_{t'}, a_{t'})$ out of an expectation due to the Markov property, see this page: https://spinningup.openai.com/en/latest/spinningup/extra_pg_proof1.html

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