The mechanism
Replay mixes old training examples into adaptation. It changes the expected objective and consumes part of the same update budget. Comparing a longer replay run with a shorter new-only run confounds mixture with computation. Retention must be evaluated on reserved old-domain examples, not merely on repeated training items. In the lifecycle task the same prompt has opposite desired answers in copy and swap. No deterministic single-answer policy without a domain identifier can satisfy both simultaneously. That impossibility is a property of this constructed task, not a universal claim about continual learning.
- Hold the starting checkpoint, total draws and optimizer policy fixed.
- Select a declared old-domain replay probability.
- Record realized domain draws rather than assuming the expected fraction occurred exactly.
- Compare both reserved-domain losses, and explain whether conflict or insufficient acquisition limits the result.
Science and assumptions
Retention, replay and transfer · worked relation
Reason about a fixed-budget retention tradeoff without treating replay as a guarantee.
Symbols and units
- p
- Probability of the old-domain answer
dimensionless
- q
- Old-domain mixture weight
dimensionless
- L
- Expected binary target loss
nats per target
- p*
- Optimal probability for the specified conflicting prompt
dimensionless
For one conflicting binary prompt let p be the old-answer probability and q the old-example mixture fraction. Expected NLL is −q ln p−(1−q)ln(1−p), minimized at p=q. With q=0.25, old/new losses at that optimum are ln4=1.386294 and −ln0.75=0.287682 nats. Greedy output selects the new answer, so old accuracy is zero even though replay has raised old-answer probability above zero.
Assumptions One identical prompt with contradictory binary labels, unlimited fitting of its probability and no domain tag. Finite decoder optimization need not reach this ideal optimum.
Engineering example
For one conflicting binary prompt let p be the old-answer probability and q the old-example mixture fraction. Expected NLL is −q ln p−(1−q)ln(1−p), minimized at p=q. With q=0.25, old/new losses at that optimum are ln4=1.386294 and −ln0.75=0.287682 nats. Greedy output selects the new answer, so old accuracy is zero even though replay has raised old-answer probability above zero.
Distinctions and tradeoffs
It is a mistake to conclude: Likelihood and accuracy are the same statistic.
Replay allocates scarce updates to retention; it can reduce new-task progress and cannot solve an information-theoretic label conflict by itself.
Adapt an existing checkpoint while measuring retained and acquired behavior. Reason about a fixed-budget retention tradeoff without treating replay as a guarantee.
Specifications and interfaces
- Learning mode
- executable replay; worked ideal optimum
The executable decoder has 423 parameters; worked extensions do not imply an additional simulator.
- Worked contract
- Reason about a fixed-budget retention tradeoff without treating replay as a guarantee.
One identical prompt with contradictory binary labels, unlimited fitting of its probability and no domain tag. Finite decoder optimization need not reach this ideal optimum.
Interfaces
- Inputs and state Input
- Hold the starting checkpoint, total draws and optimizer policy fixed.
- Result and check Output
- Reason about a fixed-budget retention tradeoff without treating replay as a guarantee.
Check your understanding
Why can replay improve old-answer likelihood without restoring old-answer accuracy?
- Probability can rise while remaining below the greedy decision threshold.
- Likelihood and accuracy are the same statistic.
- Replay guarantees both incompatible labels are always correct.
Answer and reasoning
Probability can rise while remaining below the greedy decision threshold.
Probability can rise while remaining below the greedy decision threshold. One identical prompt with contradictory binary labels, unlimited fitting of its probability and no domain tag. Finite decoder optimization need not reach this ideal optimum.
Evidence and scope
Mode: executable replay; worked ideal optimum. One identical prompt with contradictory binary labels, unlimited fitting of its probability and no domain tag. Finite decoder optimization need not reach this ideal optimum. Numerical examples are authored calculations, except explicitly identified automated core fixtures. Primary sources support the scoped mechanism; no human specialist review or learner outcome is implied.
Model lifecycle studio: executable replay; worked ideal optimum. One identical prompt with contradictory binary labels, unlimited fitting of its probability and no domain tag. Finite decoder optimization need not reach this ideal optimum.