9

That can be done. For example, Chapter 13 of the 2nd edition of Sutton and Barto's Reinforcement Learning book (page 332) has pseudocode for "Actor Critic with Eligibility Traces". It's using $G_t^{\lambda}$ returns for the critic (value function estimator), but also for the actor's policy gradients. Note that you do not explicitly see the $G_t^{\lambda}$ ...


7

Recent actor-critic algorithms do use $\lambda$-returns, but they are disguised as something called the Generalized Advantage Estimator defined as $A^{GAE}_t = \sum_{i=0}^{\infty} (\gamma\lambda)^i \delta_{t+i}$ where $\delta_t = r_t + \gamma V(s_{t+1}) - V(s_t)$. This turns out to be identically equal to $[G^\lambda_t - V(s_t)]$, i.e. the $\lambda$-return ...


6

Eligibility traces is a method of weighting between temporal-difference "targets" and Monte-Carlo "returns". In practice, for example, instead of using the one-step TD target, $r_t + \gamma V (s_{t+1})$, as in the temporal difference update $V (s_t) \leftarrow V (s_t) + \alpha (r_t + \gamma V (s_{t+1}) − V (s_t))$, you use the so-called &...


4

$TD(\lambda)$ return has the following form: \begin{equation} G_t^\lambda = (1 - \lambda) \sum_{n=1}^{\infty} \lambda^{n-1} G_{t:t+n} \end{equation} For you MDP $TD(1)$ looks like this: \begin{align} G &= 0.64 (r_0 + r_2 + r_4 + r_5 + r_6) + 0.36(r_1 + r_3 + r_4 + r_5 + r_6)\\ G &\approx 6.164 \end{align} $TD(\lambda)$ looks like this: \begin{...


4

Let us denote the state we are in at time $t$ by $S_t$. Then at iteration $t$ we create a placeholder $V_{old} = V(S_{t+1})$ for the state we will transition into. We then update the value function $V(s) \; \forall s \in \mathcal{S}$ - i.e. we update the value function for all states in our state space. Let us denote this updated value function by $V'(S)$. ...


1

Theoretically, nothing precludes the use of $\lambda$-returns in actor-critic methods. The $\lambda$-return is an unbiased estimator of the Monte Carlo (MC) return, which means they are essentially interchangeable. In fact, as discussed in High-Dimensional Continuous Control Using Generalized Advantage Estimation, using the $\lambda$-return instead of the MC ...


1

TD($\lambda$) can be thought of as a combination of TD and MC learning, so as to avoid to choose one method or the other and to take advantage of both approaches. More precisely, TD($\lambda$) is temporal-difference learning with a $\lambda$-return, which is defined as an average of all $n$-step returns, for all $n$, where an $n$-step return is the target ...


1

The previous answer from Brale is mostly correct but is missing a large detail to get the precise answer. Given this is a question from a GT course homework, I only want to leave pointers so those seeking help can understand the required concept. 𝑇𝐷(𝜆) equation is a summation over infinite K-steps (𝐺0:1 -> 𝐺0:∞) and should be included in our equation ...


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