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# CSE510 Deep Reinforcement Learning (Lecture 15)
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## Motivation
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For policy gradient methods over stochastic policies
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$$
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\pi_\theta(a|s) = P[a|s,\theta]
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$$
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Advantages
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- Potentially learning optimal solutions for multi-agent settings
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- Dealing with partial observable settings
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- Sufficient exploration
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Disadvantages
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- Can not learning a deterministic policy
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- Extension to continuous action space is not straightforward
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### On-Policy vs. Off-Policy Policy Gradients
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On-Policy Policy Gradients:
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- Training samples are collected according to the current policy.
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Off-Policy Algorithms:
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- Enable the reuse of past experience.
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- Samples can be collected by an exploratory behavior policy.
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How to design off-policy policy gradient?
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- Using importance sampling
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## Off-Policy Actor-Critic (OffPAC)
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Stochastic Behavior Policy for exploration.
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- For collecting data. Labelled as $\beta(a|s)$
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The objective function is:
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$$
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\begin{aligned}
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J(\theta)=\mathbb{E}_{s\sim d^\beta}[V^{\pi}(s)]
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&= \sum_{s\in S} d^\beta(s) \sum_{a\in A} \pi_\theta(a|s) Q^{\pi}(s,a)\\
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\end{aligned}
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$$
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$d^\beta(s)$ is the stationary distribution under the behavior policy $\beta(a|s)$.
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### Solving the Off-Policy Policy Gradient
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$$
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\begin{aligned}
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\nabla_\theta J(\theta) &= \nabla_\theta \mathbb{E}_{s\sim d^\beta}\left[\sum_{a\in A} \pi_\theta(a|s) Q^{\pi}(s,a)\right]\\
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&= \mathbb{E}_{s\sim d^\beta}\left[\sum_{a\in A} \nabla_\theta \pi_\theta(a|s) Q^{\pi}(s,a)+\pi_\theta(a|s) \nabla_\theta Q^{\pi}(s,a)\right]\\
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&= \mathbb{E}_{s\sim d^\beta}\left[\sum_{a\in A} \nabla_\theta \pi_\theta(a|s) Q^{\pi}(s,a)\right]\\
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&= \mathbb{E}_{s\sim d^\beta}\left[\sum_{a\in A} \beta(a|s) \frac{1}{\beta(a|s)} \nabla_\theta \pi_\theta(a|s) Q^{\pi}(s,a)\right]\\
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&= \mathbb{E}_{s\sim d^\beta}\left[\sum_{a\in A} \beta(a|s) \nabla_\theta \log \beta(a|s) Q^{\pi}(s,a)\right]\\
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&= \mathbb{E}_{\beta}\left[\frac{1}{\beta(a|s)} \nabla_\theta \pi_\theta(a|s) Q^{\pi}(s,a)\right]\\
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&= \mathbb{E}_{\beta}\left[\frac{\pi_\theta(a|s)}{\beta(a|s)} Q^{\pi}(s,a)\nabla_\theta \log \pi_\theta(a|s)\right]\\
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\end{aligned}
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$$
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To compute the off-policy policy gradient, $Q^{\pi}(s,a)$ is estimated given data collected by $\beta$.
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Common solution:
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- Importance sampling
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- Tree backup
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- Gradient temporal-difference learning
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- Retrace [Munos et al., 2016] [IMPALA](https://arxiv.org/abs/1802.01561)
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### Importance Sampling
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Assume that samples come in the form of episodes.
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$M$ is the number of episodes containing $(s,a), t_m$ be the first time when $(s,a)$ appears in episode $m$.
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The first-visit importance sampling estimator of $Q^{\pi}(s,a)$ is:
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$$
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Q^{IS}(s,a)\coloneqq \frac{1}{M}\sum_{m=1}^M R_m w_m
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$$
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$R_m$ is the return following $(s,a)$ in episode $m$.
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$$
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R_m\coloneqq r_{t_m +1}+\gamma r_{t_m +2}+\cdots+\gamma^{T_m-t_m -1} r_{T_m}
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$$
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$w_m$ is the importance sampling weight:
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$$
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w_m\coloneqq \frac{\pi(a_{t_m}|s_{t_m})}{\beta(a_{t_m}|s_{t_m})}\frac{\pi(a_{t_m+1}|s_{t_m+1})}{\beta(a_{t_m+1}|s_{t_m+1})}\cdots\frac{\pi(a_{T_m}|s_{T_m})}{\beta(a_{T_m}|s_{T_m})}
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$$
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### Per-decision algorithm
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Consider the parts we used in importance sampling:
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$$
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R_m w_m=\sum_{i=t_m+1}^{T_m}\gamma^{i-t_m-1} r_i \frac{\pi(a_{t_m}|s_{t_m})}{\beta(a_{t_m}|s_{t_m})}\cdots \frac{\pi(a_{t_{i-1}}|s_{t_{i-1}})}{\beta(a_{t_{i-1}}|s_{t_{i-1}})}\frac{\pi(a_{t_i}|s_{t_i})}{\beta(a_{t_i}|s_{t_i})}\cdots \frac{\pi(a_{T_m-1}|s_{T_m-1})}{\beta(a_{T_m-1}|s_{T_m-1})}
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$$
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Intuitively, $r_i$ should not depend on the actions taken after $t_i$.
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This gives the per-decision importance sampling estimator:
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$$
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Q^{PD}(s,a)\coloneqq \frac{1}{M}\sum_{m=1}^M \sum_{k=1}^{T_m-t_m} \gamma^{k-1} r_{t_m+k}\prod_{i=t_m}^{t_m+k-1} \frac{\pi(a_{t_i}|s_{t_i})}{\beta(a_{t_i}|s_{t_i})}
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$$
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The per-decision importance sampling estimator is consistence and unbiased estimator of $Q^{\pi}(s,a)$.
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Proof as exercise.
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<details>
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<summary>Hints</summary>
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- Show the expectation of $Q^{PD}(s,a)$ is the same as $Q^{IS}(s,a)$.
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- $Q^{IS}(s,a)$ is a consistence and unbiased estimator of $Q^{\pi}(s,a)$.
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</details>
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## Deterministic Policy Gradient (DPG)
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## Deep Deterministic Policy Gradient (DDPG)
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§ Extensions of DDPG
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The objective function is:
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$$
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J(\theta)=\int_{s\in S} \pho^{\mu}(s) r(s,\mu_\theta(s)) ds
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$$
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where $\pho^{\mu}(s)$ is the stationary distribution under the behavior policy $\mu_\theta(s)$.
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Proof along the same lines of the standard policy gradient theorem.
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$$
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\nabla_\theta J(\theta) = \mathbb{E}_{\mu_\theta}[\nabla_\theta Q^{\mu_\theta}(s,a)]=\mathbb{E}_{s\sim \pho^{\mu}}[\nabla_\theta \mu_\theta(s) \nabla_a Q^{\mu_\theta}(s,a)\vert_{a=\mu_\theta(s)}]
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$$
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### Issues for DPG
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The formulations up to now can only use on-policy data.
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## Deep Deterministic Policy Gradient (DDPG)
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Math4501: {
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Math416: {
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