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Replacing eligibility trace

Definition

A replacing eligibility trace is a short-term credit-assignment record that is reset to a fixed active value, usually one, whenever its state or feature is visited again. Between visits, it decays. Unlike an accumulating trace, it does not add another unit for a repeated visit, preventing repeated activations from stacking up in the same way.

What it means

Eligibility traces determine which earlier states or parameters should receive part of a current temporal-difference update. A recently visited state has a stronger trace; an older visit usually has a weaker one. This lets a later reward affect earlier predictions without waiting to reconstruct every complete return.

Replacing traces change how repeat visits are remembered. Visiting a state again refreshes its eligibility rather than adding a new layer to its existing trace. The distinction matters in trajectories with loops or repeated features. It does not mean the state value itself is replaced, nor that the agent forgets the learned value from previous episodes.

How it works

In a tabular state-value implementation, initialize all traces to zero at the start of an episode. At each step, decay the previous traces by gamma times lambda, then set the current state's trace to one. Equivalently, the current state receives one and every other state receives its decayed previous trace.

Calculate the TD error from reward and the next-state value, with zero continuation at genuine termination. Update each state's value by alpha times the TD error times its trace. An alternative code ordering can decay traces after updates, provided its time indexing is consistent.

An accumulating trace instead decays and then adds one to the current state's trace. Repeated visits can therefore take it above one. Replacing traces stay between zero and one in the stated tabular form when gamma and lambda lie between zero and one.

For state-action methods, conventions can include clearing traces of other actions at the current state. That is an additional choice, not implied by every replacing-trace implementation. Binary-feature versions reset active feature traces, but arbitrary real-valued or neural representations require further design rather than a mechanical copy of the tabular rule.

Three practical examples

1. A tabular navigation learner uses replacing traces when a trajectory repeatedly revisits the same state before receiving a delayed reward.

2. A simulator comparison measures how accumulating and replacing traces change update magnitudes in looping trajectories.

3. A binary-feature control experiment documents whether traces for nonselected actions are cleared, making its implementation reproducible.

Think of it this way

Think of a fading timestamp marker on a location. Revisiting it restores the marker to full brightness. An accumulating version instead adds another layer of brightness each time. Both remember recency, but only the accumulating version stacks repeated visits into a stronger-than-new marker. Neither marker is the prediction being learned; it determines how strongly that prediction is updated.

Worked example

Consider an original episode visiting A, B, A and C before termination. Rewards on the first three transitions are zero; the final C-to-terminal reward is 2. Initialize every value and trace to zero. Set gamma to 0.9, lambda to 0.8 and alpha to 0.1. The decay multiplier is 0.72.

Use the decay-then-activate convention. At the first A visit, the replacing trace vector in order A, B, C is (1, 0, 0). At B, it is (0.72, 1, 0). On returning to A, decay first, then reset A to one: the vector is (1, 0.72, 0).

At C, the replacing traces become (0.72, 0.5184, 1). All earlier TD errors were zero because rewards and value estimates were zero. The final TD error is 2, so the value increments are 0.1 times 2 times each trace: 0.144 for A, 0.10368 for B and 0.2 for C.

Now calculate accumulating traces on the same episode. At the second A visit, A's trace is 0.72 squared plus one, or 1.5184. At C, it becomes 0.72 times 1.5184, or 1.093248. B and C still have traces 0.5184 and 1.

The final accumulating-trace update to A is therefore 0.2186496, rather than 0.144. The difference comes from retaining contributions from both A visits. The replacing version refreshes A at its most recent visit instead of stacking those contributions.

These calculations isolate the repeated-visit effect with identical rewards, parameters and initial values.

Illustrative case study

In a fictional project, Willow Learning compares trace methods in a maze simulator containing short loops. Some runs with accumulating traces produce larger updates than expected after many repeated visits.

The team inspects trace magnitudes and tests replacing traces under comparable learning rates. It also checks episode resets and update ordering. Its report distinguishes the bounded trace values from the values being learned, which can still be large.

Common misconceptions and mistakes

Replacing traces do not reset learned values to one. They reset the eligibility used to distribute a later update.

They are not the same as Dutch traces or true-online TD, which use different formulas and equivalence properties. Replacing the trace rule alone does not establish exact equivalence to every forward lambda-return implementation.

Do not carry episodic traces into a new independent episode unintentionally. Also avoid claiming universal superiority over accumulating traces. Representation, repeated visits, step sizes and task dynamics determine the practical effect. An off-policy control algorithm may additionally cut or correct traces according to its own policy logic.

Business use

Replacing traces matter to teams building sequential predictors or controllers where delayed outcomes should update recently relevant states. Their practical value is controlling how repeated activations influence credit assignment. Compare learning speed, stability and sensitivity to step size, and document trace-reset conventions. A useful diagnostic records trace magnitudes separately from TD errors so the source of large updates can be identified.

People also ask

Can a tabular replacing trace exceed one?

Not under the stated initialization and decay rule with gamma and lambda in the usual range.

What happens at lambda zero?

Older traces vanish, leaving only the current state's eligibility.

Does replacement remove delayed credit?

No. Unvisited recent states retain decaying traces and still receive updates.

Check your understanding

Why is A's final replacing trace 0.72?

Answer: Its most recent activation occurred one step before C.

Why is its accumulating trace larger?

Answer: The earlier A visit still contributes after the second visit adds another unit.

Related terms

• Eligibility trace
• TD(lambda)
• SARSA(lambda)
• accumulating trace
• Dutch trace
• temporal credit assignment and binary features

Learn next

Next, repeat the trace calculation with three consecutive visits to A. Then inspect how optional clearing of other action traces changes a state-action version.

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Sources

Primary source: Richard S. Sutton and Andrew G. Barto, Reinforcement Learning: An Introduction, author draft, discussion of replacing traces and feature-based extensions: https://web.stanford.edu/class/psych209/Readings/SuttonBartoIPRLBook2ndEd.pdf

The repeated-state episode and fictional case are original teaching material. The calculation uses tabular state traces with explicit timing; it does not assume an arbitrary nonlinear feature extension.

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