How do you know a fine-tuned AI model is actually better?

You measure it. Hold out a portion of the client's real examples so the model never sees them during training, score the plain base model on those held-out questions first, fine-tune, then score again on the identical questions with identical scoring. The difference is the lift. Because the only thing that changed is the fine-tune, the improvement is attributable to it.

Does fine-tuning actually improve accuracy, or is the base model already good enough?

It depends on the task and, critically, on how much data you have. On a public benchmark of 510 lawyer-annotated commercial contracts, a base Qwen3-4B extracted short contract fields at 62.2% accuracy. Fine-tuned on 396 examples it reached 83.5-85.4%, a lift of 21.3 to 23.2 points across three runs. The same model trained on only 48 examples produced a lift inside the noise floor, so the amount of data was the deciding factor rather than the model's ceiling.

How much of our own data do we need to fine-tune a model?

Fewer examples than most people expect, but more than a handful. In our contract-extraction measurements, 48 training examples produced no measurable improvement while roughly 400 produced a 22-point gain. The honest answer for any specific case is that it depends on how varied and how consistent the examples are, which is what a paid readiness assessment measures on the client's own corpus.

Why can't you just use a public AI benchmark score?

Public benchmarks measure general ability on standardised tests, not your work. A model that scores well in the abstract can still be mediocre at reading your contracts or following your policies. Public benchmarks are also increasingly targeted during training, so a high score can reflect test preparation as much as capability. The only benchmark that answers your question is built from your own tasks.

Research record

Does the Transition Create Features, or Select Them?

By Joshua Bolick, Awareness Software Group ~6 min read August 2026

Ongoing research. This is an experimental result from active work, not a settled conclusion. The numbers are what we measured and the method is described so you can judge it, but the programme is still running and later experiments may revise what it means. More about this programme.

EXPLORATORY. Not a preregistered study. Local CPU, 42 training runs, no GPU, no cost.

Program v2 tier 2, item F3. Decisive computation: analysis/echo_state.py. Output: analysis/echo_state.json. Reproduce with python analysis/echo_state.py in about twenty-five minutes on a throttled laptop CPU.

The question

Freeze the recurrent weights at initialisation and train only the embedding and the output head. That is reservoir computing: the recurrence becomes a fixed random dynamical system and learning happens only at the two ends.

**A model that cannot rewire itself never gets there.** Two ways of training the same model at five sizes. The upper line trains everything. The lower line leaves the model's memory fixed at the random values it started with and trains only the input and output layers, so the model can learn to read its memory but never to change it. Normal training has essentially solved the task by the third size. The frozen version never catches up, even at four times that size and four times the training. It is still learning something real, well above guessing, but it gets there gradually rather than suddenly: roughly nineteen times more slowly through the steepest part of its progress. Whatever makes learning abrupt here needs the model to be able to rewire its memory, not just reinterpret it.

If random recurrent dynamics already support the task, the transition is not about creating features. It is about learning to select and read features that were always present, and the width at which a frozen reservoir fails would measure the deficit training has to close.

F2 gave this a real prior. At step 1, before meaningful training, a linear probe on the hidden state read 0.1761 against the model's 0.0368, about 5.6x chance. Some of the answer is present in a near-untrained recurrent state.

The answer is that it is nowhere near enough.

Part A: the deficit never closes

Three seeds per cell, 900 steps, full training against frozen recurrence:

Width	Full	Frozen	Gap	Trainable share, frozen arm
16	0.6705	0.1832	`+0.4873`	39%
32	0.9592	0.2698	`+0.6894`	25%
48	0.9951	0.3057	`+0.6894`	18%
96	0.9999	0.4516	`+0.5483`	10%
192	0.9997	0.5742	`+0.4255`	5%

A fixed random recurrence never reaches full-model accuracy at any width tested. Full training is essentially solved from width 48; the reservoir at four times that width still sits at 0.574.

The reservoir is not useless, 0.574 against a chance floor of 0.0346 is real learning, and it improves steadily with width. It is just not the same thing.

Part B: and it does not transition

At width 48, five seeds, using the confirmation's own definitions:

Arm	Transitioned	Transition step	Residual-energy rise	Final accuracy
full	5 / 5	89	`+0.2230` `[+0.2048, +0.2412]`	0.9930
frozen	0 / 5	,	,	0.3015

No frozen run gained enough accuracy to have a transition to measure, and with no transition there is no expansion signature either.

Part C: the compute-matched arm, which is where the real answer is

Part B compares at equal step counts, and a frozen step is cheaper because it computes no gradient for the recurrent weights. This program's standing controls require compute-matched baselines, so Part C gives the frozen arm four times the step budget, a generous allowance in the reservoir's favour rather than a precise FLOP match, and measures not just where it lands but how sharply it gets there: the steps taken to cross from a quarter to three quarters of its own accuracy range.

Width	Arm	Steps	Final accuracy	Transitioned	25% → 75% took
48	full	900	0.9951	3 / 3	63 steps
48	frozen	3,600	0.4746	3 / 3	1,175 steps
192	full	900	0.9997	3 / 3	42 steps
192	frozen	3,600	0.7535	3 / 3	775 steps

Given enough steps the reservoir does satisfy the eligibility rule, so "it cannot learn" would be the wrong claim. What it cannot do is learn abruptly. Its improvement is spread over 18.7x more steps at width 48 and 18.5x more at width 192, and it still lands far short.

The transition builds the mechanism, it does not merely select it. A fixed random recurrence supports slow, partial, gradual learning. The sharp transition requires a recurrence that is itself learning.

How this squares with F2

These two results look opposed and are not. F2 established that information about the answer is linearly present in the state earlier than the model can use it: including, weakly, at initialisation. F3 establishes that this early information is not sufficient: a system built entirely on it plateaus around 0.30 to 0.75 rather than 0.99, and gets there gradually.

Both are true. Random recurrent dynamics give a real head start, worth roughly 10x chance at width 48 with a trained readout. The transition is where the recurrence acquires the rest, and the rest is most of it.

What this changes for the program

The expansion is a recurrence phenomenon. It does not appear when the recurrence is frozen, which is consistent with it tracking the recurrence acquiring structure rather than the readout fitting. That is now evidence rather than assumption.
A2 stays undermined, and F3 explains why differently. D6 showed the post-peak fall is a measurement-frame effect. F3 adds that the thing being measured is a property of the learning recurrence, so any signal derived from it is unavailable in a frozen-feature regime.
Sharpness is now a measurable endpoint, not just a description. The 25%-to-75% crossing width separates two runs that reach the same accuracy by different routes, and it belongs in any future comparison that claims a transition.

Limits

One architecture (tiny-gru), one task, one optimizer (AdamW), one learning rate. The frozen arm was not separately tuned, and a reservoir may prefer a different learning rate for its readout; the deficit is large enough that tuning is unlikely to close it, but it was not tested.
Reservoir computing normally uses much wider reservoirs with tuned spectral radius. This tests a frozen GRU at its default initialisation, which is the honest comparison for "what did this model's recurrence have to learn", and not a fair test of reservoir computing as a method.
Part C's four-times budget is an allowance, not a measured FLOP match. It is generous in the reservoir's favour, so the conclusion is conservative in the direction that matters.

The plain-language version. This page is the record: the design, the numbers and the limits. If you would rather have the story, Does the Transition Create Features, or Select Them?, explained without the jargon, covers the same result for a reader who has never seen a training curve.

About this record. This page is the write-up of an experiment we ran ourselves. Every number on it comes from a run we recorded, and the script that produced each one is named in the text above. We publish these results and the data behind them here, including the ones that went against what we expected.

All research from this programme →