Exercise 12: Eligibility traces

Exercise 12: Eligibility traces#

Note

  • This page contains background information which may be useful in future exercises or projects. You can download this weeks exercise instructions from here:

  • Slides: [1x] ([6x]). Reading: Chapter 10.2; 12-12.7, [SB18].

  • You are encouraged to prepare the homework problems 1 (indicated by a hand in the PDF file) at home and present your solution during the exercise session.

  • To get the newest version of the course material, please see Making sure your files are up to date

The main exercise today will be the tabular version of the \(\textrm{TD}(\lambda)\) algorithm described at http://incompleteideas.net/book/first/ebook/node77.html. The algorithm will be described in Todays lectures before the version which uses function approximators.

Solutions to selected exercises#

Solution to the conceptual problem 12.1

First, the \(n\)-step returns are defined in (Sutton and Barto [SB18], Equation 12.1) as:

\[G_{t:t+n} = R_{t+1} + \gamma R_{t+2} + \dots + \gamma^{n-1} R_{t+n} + v(S_{t+n}, w_{t+n-1} )\]

From this we can see that

\[\begin{split}G_{t+1:t+1} & = v(S_{t+1}, w_t) \\ G_{t:t+n} & = R_{t+1} + \gamma G_{t+1:t+n}\end{split}\]

And we also need the power series: \((1-\lambda)\sum_{k=1}^\infty \lambda^{k-1} = 1\). Given this, the result is a fairly straight-forward but tedious calculation:

\[\begin{split}G_t^\lambda & = (1-\lambda)\sum_{n=1}^\infty \lambda^{n-1} G_{t:t+n} \\ & = (1-\lambda) \sum_{n=1}^\infty \lambda^{n-1} (R_{t+1} + \gamma G_{t+1:t+1+(n-1) } ) \\ & = R_{t+1} + (1-\lambda) \sum_{n=1}^\infty \lambda^{n-1} G_{t+1:t+1+(n-1) } \\ & = R_{t+1} + (1-\lambda) \gamma ( G_{t+1:t+1} + \lambda \sum_{n=2}^\infty \lambda^{n-2} G_{t+1:t+1+(n-1) } ) \\ & = R_{t+1} + (1-\lambda) \gamma (v(S_{t+1}, w_t) + \lambda \sum_{m=1}^\infty \lambda^{m-1} G_{t+1:t+1+m } ) \\ & = R_{t+1} + (1-\lambda) \gamma v(S_{t+1}, w_t) + \lambda \gamma (1-\lambda) \sum_{m=1}^\infty \lambda^{m-1} G_{t+1:t+1+m } \\ & = R_{t+1} + (1-\lambda) \gamma v(S_{t+1}, w_t) + \lambda \gamma G_{t+1}^\lambda \\ & = R_{t+1} + \gamma v(S_{t+1}, w_t) + \lambda \gamma (G_{t+1}^\lambda - v(S_{t+1}, w_t) )\end{split}\]

Intuitively, the first two terms in this decomposition is the TD(0) error, whereas the expression in the last term will have mean 0 since both \(G_{t+1}^\lambda\) and \(v(S_{t+1}, w_t)\) are estimates of the value function at the next state.

Problem 12.1: Tabular Sarsa(Lambda)

Problem 12.2: Tabular Sarsa(Lambda) and the open gridworld environment

Problem 12.3: Semi-gradient Sarsa(Lambda)

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