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If I understand it correctly from the following equation

$$U(\theta)=\mathbb{E}_{\tau \sim P(\tau;\theta)}\left [ \sum_{t=0}^{H-1}R(s_t,u_t);\pi_{\theta} \right ]=\sum_{\tau}P(\tau;\theta)R(\tau)$$

from this paper, the utility of a policy parameterized by weights $\theta$, is the total expected reward following (all, one?) trajectory $\tau$. After rearranging, taking the gradient, and running the policy over $m$ trials, the equation becomes

$$\nabla_{\theta}U(\theta)\approx \hat{g} = \frac{1}{m}\sum_{i=1}^{m}\nabla_{\theta}logP(\tau^{(i)};\theta)R(\tau^{(i)}).$$

My question is - how does this work? I can't understand it in terms of the utility/loss surface. At most, doesn't taking multiple trials give you a more accurate guess of the point $(\theta,U(\theta))$ alone? I don't get how you can get the gradient (a sort of local tangent plane of a surface) without evaluating the function at different $\theta$'s, like how do you get the slope at a point (2,3) without evaluating it at 2.01, 1.99, etc. at the very least? Is it by virtue of the well defined geometry of the function $U$ we defined?

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From the equation you wrote:

  • $U(\theta)$ is the total expected reward of a policy $\pi_\theta$ (with parameters/weights $\theta$) over all possible trajectories $\tau$ sampled under the environment dynamics while following the policy itself: the term $P(\tau;\theta)$ should include both the env. dynamics (i.e. how it evolves each step given an action from a state), usually denoted as $P(s'\mid s, a)$ where $s'$ is next state, and the fact that we sample actions from the policy $a\sim \pi_\theta(s)$ from a state $s$.
  • For practical purpose, you can't take the expectation over all possible trajectories so you run the policy over $m$ of them and approximate the expectation (integral) with a summation over finite num. of trajectories. Thus, $\hat g = \frac{1}{m}\sum_{i=1}^m\nabla_\theta \log P(\tau^{(i)};\theta)R(\tau^{(i)})$.

My question is how does this work?

Basically the formula that estimates the gradient $\hat g$ makes use of what is called a score estimator (also called REINFORCE from the notable algorithm). It works by computing the probability of a trajectory to occur, then you score that trajectory by $R(\tau^{(i)})$ the total (discounted) reward, and finally compute the expectation as a simple average over $m$ trajectories.

The gradient of the policy $\nabla_\theta$ is so scored by the log-prob times the reward: the log-prob term has the intuitive explanation of increasing the probability of trajectories that score a high $R(\tau)$, while decreasing the ones for which $R$ is bad.

doesn't taking multiple trials give you a more accurate guess of the point $(\theta,U(\theta))$ alone?

Increasing $m$ provides a better estimate of $\hat g$. Indeed, as $m\to\infty$ you recover the true gradient $g$ by computing the full expectation. But that is not feasible in practice.

I don't get how you can get the gradient without evaluating the function at different $\theta$'s

At each update/optimization step the policy is only evaluated at $\theta$: its weights define both the point at which the gradient is computed and the policy itself, since it's a parametrized function approximator (like a neural network.) So, there is no need to know/evaluate at the surroundings of $\theta$.

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  • $\begingroup$ So is the ground truth of $R(\tau)$ fixed? The only thing that changes when varying $\theta$ is the probability of following that trajectory $\tau$? $\endgroup$
    – User
    Commented Apr 25, 2023 at 16:04
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    $\begingroup$ Exactly. For a given trajectory $\tau$, $R(\tau)$ is always the same since it's just a (discounted) sum over the immediate rewards $r_t$ gathered in $\tau$: it depends on the environment not the agent/policy. Instead, the log-prob changes as the parameters $\theta$ got updated. $\endgroup$ Commented Apr 25, 2023 at 19:47
  • $\begingroup$ Sorry, I have just one more question, isn't the $U(\theta)$ function defined in essentially the way that the value function is (i.e. total expected reward of a policy)? Except it doesn't take in the state as an input, rather the parameters themselves? $\endgroup$
    – User
    Commented Apr 26, 2023 at 3:30
  • $\begingroup$ $U(\theta)$ and $V(s)$ are similar (I assume $V(s)=V^{\pi_\theta}(s)$), but not equal. Although both represent sum of rewards, $V(s)$ computes that only from a state, while $U(\theta)$ considers an initial $(s_0,a_0)$ pair: this looks like a Q-value, $Q(s,a)$, but it's still different because $U(\theta)$ also averages over the initial state distribution $s_0\sim\rho(s_0)$. Intuitively, it's like $U(\theta) = \mathbb{E}_{s_0\sim \rho}[V(s_0);\ \pi_\theta]$ (but I'm not sure if 100% correct) as after all the RL objective is to maximize the total reward expected over all possible trajectories. $\endgroup$ Commented Apr 26, 2023 at 8:01

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