research24 min studyLesson 633 in Engineering

Fit comparison probabilities while keeping reward coordinates identifiable

Derive logistic pair likelihoods, explain additive score ambiguity and comparison-graph limits, and separate a fitted ranking scale from universal utility.

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What you’ll understand

  • Write a Bradley–Terry likelihood for an identified response pair.
  • Explain prompt-specific additive invariance and graph connectivity limits.
  • Interpret finite and regularized fits without turning scores into utility truth.

A reward model assigns scores of seven and six to two answers, and another checkpoint assigns one and zero. Both imply the same Bradley–Terry comparison probability when the logistic scale is fixed. The scores matter through their difference for that prompt. Learning such a scale can predict observed preferences, but its coordinate origin is not identified by pair labels. Coverage, separation, annotation noise, and model misspecification all affect what the fitted numbers can legitimately mean.

Use a probabilistic comparison model

For a prompt x and two candidate responses, the Bradley–Terry construction assigns positive strengths proportional to exponentiated scores. The probability of choosing A is exp(r_A)/(exp(r_A)+exp(r_B)), equivalently sigmoid(r_A−r_B). This is a model of the annotation event under the collection protocol, not a proof that every human has one stable scalar utility. Fix the link scale or temperature convention before interpreting score gaps. Multiplying scores changes probabilities at a fixed temperature, while adding the same constant to both leaves the probability unchanged.

Fit binary outcomes with their log likelihood, keeping label direction and display mapping correct. A preferred response contributes negative log sigmoid of its score advantage. Soft labels or tie models require an explicitly chosen extension; forcing ties into arbitrary binary winners changes the data model. For neural reward models, one scalar head may share parameters across many prompts, but that parameter sharing is a modeling assumption about generalization. It does not create direct observed comparisons between unrelated prompts or grant a universal interpretation to their raw score levels.

p(A≻B∣x)=σ(Δr)=11+e−Δr,Δr=r(x,A)−r(x,B)p(A\succ B\mid x)=\sigma(\Delta r)=\frac{1}{1+e^{-\Delta r}},\qquad\Delta r=r(x,A)-r(x,B)
Scores and their difference are finite real values on a fixed dimensionless logistic scale, with ties excluded or separately modeled. A comparison probability describes the specified annotation event; it is not a probability that A is objectively correct.↔ Scroll wide equations horizontally.

Identify the likelihood’s ambiguities

Adding a prompt-specific constant c(x) to all response scores preserves every within-prompt difference. A centering constraint or anchor selects coordinates but does not provide new preference evidence. In a free-score comparison graph, disconnected components can have independent offsets, leaving relative levels between components unsupported. A connected graph supplies links but is not enough by itself to guarantee a finite unregularized maximum-likelihood estimate. If one response wins every observed comparison in a separable direction, the likelihood can keep improving as the corresponding gap grows without a finite optimum.

Regularization, priors, bounded parameterizations, and additional comparisons can make practical fitting better behaved, but their effects must be reported. A finite optimizer output after a fixed number of steps is not proof that the unconstrained statistical optimum is finite. Examine label conflicts, graph coverage, held-out likelihood, and sensitivity to regularization. A scalar-score model also imposes structure on preference patterns; systematic cycles, subgroup differences, or context omissions can show that one shared scale is inadequate. Treat those observations as model-diagnostic evidence rather than automatically declaring disagreeing annotators wrong.

ℓi=−[zilog⁡pi+(1−zi)log⁡(1−pi)]\ell_i=-\bigl[z_i\log p_i+(1-z_i)\log(1-p_i)\bigr]
For binary label z_i∈{0, 1} and model probability 0<p_i<1, ℓ_i is negative log likelihood in nats using natural logs. With finite score gaps the logistic probabilities lie inside this domain; regularization and observation weights are separate objective choices.↔ Scroll wide equations horizontally.

Check what the fitted score predicts

Evaluate both ranking errors and probabilistic losses on appropriately held-out prompt groups. A large score margin can be confidently wrong under a new response generator or rubric, and an accurate ordering does not guarantee calibrated probabilities. Preserve the model checkpoint, tokenizer, prompt formatting, terminal-token selection, and dataset version used to compute scores. A reward score produced from a truncated answer may describe a different observation from the one annotators judged. Inspect slices and examples rather than compare raw checkpoint score means whose origins or scaling may differ.

The network lab demonstrates one logistic link with a fixed bias. For each separate view, Reset, set both weight w=1 and input x, predict the logit wx−0.5, then inspect. At x=0.5 the logit is zero and output 0.5; at x=1.5 it is one and output about 0.7311. These are abstract score-gap examples, not learned response scores. The instrument has no pair dataset, comparison graph, optimization, or utility measurement. Its fixed bias must be included when mapping the visible controls to a desired logistic probability.

Compute a preference probability and its coordinate invariance

For one fictional prompt, let r_A=ln 3 and r_B=0 on a fixed unit-temperature logistic scale. The observed binary label prefers A. Use natural logarithms.

  1. The score gap is ln 3, so A’s modeled preference probability is 3/(3+1)=0.75.

  2. The label’s negative log likelihood is−ln 0.75≈0.28768 nats. This scores the probability prediction for that label, not the answer’s objective correctness.

  3. Adding five to both scores leaves the gap ln 3 and probability 0.75 unchanged. The data cannot distinguish those two origins; a centering rule would select coordinates without changing this comparison.

The result

The model assigns probability 0.75 and loss about 0.28768 nats to the observed A label, invariant to the common score shift.

Explore the model

Interactive lab

Interactive lab · network

Map a score gap to a logistic probability

For each view: Reset, set both w=1 and x, predict z=wx−0.5, then inspect. Compare x=0.5 and x=1.5: outputs are 0.5 and approximately 0.7311. This fixed-bias sigmoid unit does not fit rewards, train a policy, or establish utility truth.

Map a score gap to a logistic probabilityOne inspectable sigmoid unit, σ(wx − 0.5). Playback traces the computation graph. There is no training, attention mechanism or hidden deep network. Current readings: Product · wx: 1.2; Logit · wx − 0.5: 0.7; Sigmoid output: 0.6682. All parameters are adjustable with the labeled controls below.One input. One weight. One sigmoid.x1.20× w1.20+ b0.70σ0.668output0.668w = 1.00b = −0.50σ(z) = 1 / (1 + exp(−z))
Computation graphSignal trace
Product · wx
1.2
Logit · wx − 0.5
0.7
Sigmoid output
0.6682
t = 0 s

One inspectable sigmoid unit, σ(wx − 0.5). Playback traces the computation graph. There is no training, attention mechanism or hidden deep network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAt a fixed link scale, a comparison depends on its score gap. Include the instrument’s fixed bias rather than treating input x itself as the logit.

Check your understanding

Why do adding the same constant to both response scores leave their Bradley–Terry probability unchanged?

Try it yourself

For one prompt, score A is ln 4 and score B is 0. Compute the modeled A preference probability and the loss for an observed A label. What happens if both scores are reduced by two?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Exponentiated strengths are four and one before a shared rescaling.

Compare with the solution

The probability is 4/5=0.8 and the loss is−ln 0.8≈0.22314 nats. Reducing both scores by two preserves the gap and therefore both quantities. These modeled judgments remain conditional on the chosen link and annotation process.

Take it with you

  • Bradley–Terry fits probabilities of specified comparison outcomes.
  • Pair labels leave common score origins unidentified and can yield poorly behaved unregularized fits.
  • Ranking, calibration, generalization, and utility interpretation require distinct evidence.

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