2
$\begingroup$

In the RL textbook by Sutton & Barto section 7.4, the author talked about the "True online TD($\lambda$)". The figure (7.10 in the book) below shows the algorithm.

At the end of each step, $V_{old} \leftarrow V(S')$ and also $S \leftarrow S'$. When we jump to next step, $\Delta \leftarrow V(S') - V(S')$, which is 0. It seems that $\Delta$ is always going to be 0 after step 1. If that is true, it does not make any sense to me. Can you please elaborate on how $\Delta$ is updated?

enter image description here

$\endgroup$
4
$\begingroup$

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 $V'(S)$.

At iteration $t+1$ we calculate $\Delta = V'(S_{t+1}) - V_{old} = V'(S_{t+1}) - V(S_{t+1})$, which does not necessarily equal 0 because the placeholder $V_{old}$ was created using the value function before the last update.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Ah, I see! I missed the fact that when TD updates, it updates the whole state space! Thank you so much, David! $\endgroup$ – roy Jun 4 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.