Why Compare Repeated Fits?
A model fitted to one training sample may produce a precise-looking prediction, but a different plausible sample from the same source population could produce a different fitted model. This matters when the prediction informs a decision: a small change in a predicted probability may cross a referral threshold, or a fitted relationship may change direction. Comparing repeated fits helps assess whether the result depends strongly on the particular training data.
Here, model instability across repeated fits means that predictions or conclusions vary materially when the fitting procedure is applied to plausible alternative training samples. The question is not whether the model performs well on average, nor whether a prediction changes when its input changes. The question is whether refitting on alternative samples changes the result for a fixed prediction target.
A common design is to resample the training observations, refit the model to each resample, and score the same fixed validation cases each time. Sampling with replacement is often called bootstrap resampling. Repeated subsampling or repeated training splits are other possibilities. These procedures answer related but not identical questions: a bootstrap assessment approximates variation under resampling from the observed data, while repeated splits assess sensitivity to the chosen training split.
The resampling design must reflect how observations were collected. If observations are clustered by policyholder, household, or location, resampling individual rows as though they were independent may understate or misrepresent variation. For time-ordered data, resampling that destroys the time structure can also be inappropriate. The assessment is only as credible as the alternative samples it treats as plausible.
A Consistent Comparison Procedure
The aim is to change the training sample while keeping the comparison meaningful. Before refitting, specify which parts of the workflow remain fixed. For example, the model family, predictors, outcome definition, and preprocessing rules can be held constant. If the operational procedure includes tuning or variable selection, decide whether to repeat those steps in every fit. Repeating them measures variation in the whole procedure; freezing them measures variation conditional on those choices.
Choose the predictions or conclusions to assess, such as predicted claim probabilities for a fixed set of cases or the proportion of cases assigned to a referral group.
State how training data will be resampled and what structure must be preserved, such as clusters or time order.
Apply the same declared modeling and preprocessing steps to every resample. Record the results for the same target cases.
Report a suitable spread measure and, when decisions are involved, how often the decision changes. Explain the scope and limitations of the comparison.
For a single target case, useful summaries include the mean and sample standard deviation of its repeated predictions, the minimum and maximum, and the number of fits in which a threshold-based decision changes. These summaries should not be treated as interchangeable. A range is sensitive to extreme fits; a standard deviation describes typical squared deviations around the mean; and a decision-change rate directly addresses a threshold-based action.
Worked Examples
Worked Example: Variation in a Fixed Risk Prediction
A hypothetical insurer repeatedly fits the same specified claim-frequency model to five bootstrap resamples. Each fit predicts the annual probability of a claim for the same policyholder. The recorded probabilities are 0.08, 0.10, 0.09, 0.13, and 0.10. Calculate the mean prediction, sample standard deviation, and range. Interpret the results in percentage points.
Method. Treat the five values as repeated fit results for one fixed target. Use the sample standard deviation to summarize spread across the five fits. The common mistake is to divide the sum of squared deviations by five; the sample standard deviation uses four in the denominator.
Mean. The sum is \(0.08+0.10+0.09+0.13+0.10=0.50\), so the mean is \(0.50/5=0.10\), or 10%.
Sample standard deviation. Deviations from 0.10 are \(-0.02, 0, -0.01, 0.03, 0\). Their squared values sum to \(0.0004+0+0.0001+0.0009+0=0.0014\). Divide by \(5-1=4\): the sample variance is \(0.0014/4=0.00035\). Its square root is approximately \(0.0187\), or 1.87 percentage points.
Range. The maximum is 0.13 and the minimum is 0.08, so the range is \(0.13-0.08=0.05\), or 5 percentage points.
The average is 10%, but individual fits range from 8% to 13%. That spread may be material if a decision is sensitive to a few percentage points. The arithmetic checks: the five probabilities sum to 0.50, and both the range and standard deviation are expressed on the probability scale.
Worked Example: How Often Does a Decision Change?
A hypothetical review team refers a case for manual review when its predicted probability of a costly event is at least 0.12. Across six repeated fits on plausible training samples, one fixed case receives predictions of 0.11, 0.14, 0.13, 0.09, 0.15, and 0.12. Calculate the fraction of fits that trigger review and assess the stability of the decision.
Method. Apply the stated rule separately to each prediction, including equality at the threshold. A common mistake is to count only values strictly greater than 0.12, even though the rule says “at least 0.12.”
The predictions 0.14, 0.13, 0.15, and 0.12 meet the rule. Thus, 4 of the 6 fits trigger review, giving \(4/6=0.6667\), or 66.67% of fits. The other 2 predictions, 0.11 and 0.09, do not trigger review.
The decision is not stable across these fits: it changes between review and no review. This does not establish that either decision is correct, nor does it prove that 66.67% of future decisions will trigger review. It describes the results for this specified set of resamples and this fixed case. The boundary value of 0.12 must be counted because the rule includes equality.
Worked Example: Stable Aggregate, Unstable Case
A hypothetical model is refitted four times and used to score the same four cases. A decision is made at a predicted probability of 0.45 or higher. The predictions are shown below. Compare the average prediction across all 16 scores with the stability of the decision for Case C.
| Case | Fit 1 | Fit 2 | Fit 3 | Fit 4 |
|---|---|---|---|---|
| A | 0.08 | 0.09 | 0.07 | 0.08 |
| B | 0.21 | 0.23 | 0.22 | 0.20 |
| C | 0.44 | 0.42 | 0.46 | 0.43 |
| D | 0.71 | 0.70 | 0.74 | 0.73 |
Overall average. Fit 1 sums to \(0.08+0.21+0.44+0.71=1.44\). Fit 2 sums to \(0.09+0.23+0.42+0.70=1.44\). Fit 3 sums to \(0.07+0.22+0.46+0.74=1.49\). Fit 4 sums to \(0.08+0.20+0.43+0.73=1.44\). The total is \(1.44+1.44+1.49+1.44=5.81\). Dividing by 16 scores gives an overall average of \(5.81/16=0.363125\).
Case C. Its four predictions sum to \(0.44+0.42+0.46+0.43=1.75\), so its mean is \(1.75/4=0.4375\). Its range is \(0.46-0.42=0.04\). Only the Fit 3 prediction reaches 0.45 or higher, so the decision is positive in 1 of 4 fits.
The overall average is not enough to describe the case-level decision: it combines cases with very different risk levels and hides instability near the threshold for Case C. The case-specific calculations check against the table: three of Case C’s four predictions are below 0.45, and one is above it.
Why the Sample-Variance Formula Works
Suppose \(Z_1,\ldots,Z_B\) are results from \(B\) repeated fits for the same target and the same resampling procedure. For this justification, assume the results are independent draws from the same distribution, with finite mean \(\mu_Z\) and variance \(\sigma_Z^2\). The usual sample variance is
To see why the denominator is \(B-1\), first use the identity
Taking expectations, the first sum on the right has expectation \(B\sigma_Z^2\). Because the \(Z_b\) are independent with common variance \(\sigma_Z^2\), the variance of their mean is \(\operatorname{Var}(\bar{Z})=\sigma_Z^2/B\). Therefore the expectation of the second term is \(B(\sigma_Z^2/B)=\sigma_Z^2\). The expected sum of squared deviations from \(\bar{Z}\) is consequently \(B\sigma_Z^2-\sigma_Z^2=(B-1)\sigma_Z^2\). Dividing by \(B-1\) gives \(E[s_Z^2]=\sigma_Z^2\).
This result explains the sample-variance denominator under the stated assumptions; it does not guarantee that a chosen resampling scheme represents every source of uncertainty. In practice, bootstrap resamples drawn from one observed data set are not independent new studies. Their spread is a resampling-based diagnostic whose interpretation depends on the data and resampling design.
Interpreting Instability Carefully
Repeated-fit variation is a warning about dependence on the observed training sample, not automatically evidence of poor predictive performance. A model can have similar average performance across fits while giving inconsistent predictions for a particular case. Conversely, noticeable coefficient variation does not necessarily imply equally large prediction variation for every case. Report the quantity that answers the stated question, and avoid replacing case-level evidence with a single aggregate.
A useful communication states the resampling method, what was held fixed, the number of fits, the target being compared, and the observed spread or decision-change frequency. For example: “Across 50 bootstrap refits using the same preprocessing and model specification, this fixed case’s predicted probability ranged from 0.09 to 0.14; the referral decision changed in 18 fits at the 0.12 threshold.” That statement describes what was measured without claiming the result applies to every future sample.
Common Mistakes
- Changing the target along with the training sample. Comparing predictions for different cases confounds fit variation with differences between cases. Use a common target set.
- Changing the workflow without saying so. If preprocessing, tuning, or variable selection changes across fits, the measured variation includes those steps. State whether they were repeated or held fixed.
- Calling a spread estimate a guarantee. A resampling assessment describes variation under its design; it does not guarantee future predictions or decisions.
- Reporting only an overall average. An aggregate can conceal instability for cases close to a decision threshold. Examine the targets relevant to the decision.
- Using an unsuitable resampling unit. When data are clustered or time-ordered, a row-wise resampling scheme may not preserve the structure that makes alternative samples plausible.
Check Your Understanding
For each question, use the stated resampling setup and decision rule.
- Four refits give a fixed case predictions of 0.20, 0.24, 0.22, and 0.18. Find the mean and range. (Answer: mean \(=0.21\); range \(=0.06\).)
- A rule triggers an action at a predicted probability of at least 0.30. Five fits give 0.28, 0.31, 0.30, 0.27, and 0.33. What fraction trigger the action? (Answer: \(3/5=0.6000\), or 60%.)
- Three repeated-fit predictions are 0.40, 0.50, and 0.60. Calculate the sample variance and sample standard deviation. (Answer: sample variance \(=0.0100\); sample standard deviation \(=0.1000\).)
- A data set contains multiple records per household, and the intended assessment concerns variation across plausible training samples of households. Why might resampling individual records be unsuitable, and what should be resampled instead? (Answer: Row-wise resampling can break within-household dependence; resample whole households so each household’s records remain together.)
- Across repeated fits, the average predicted probability over a portfolio is almost unchanged, but one fixed case crosses the action threshold in half the fits. What conclusion is justified? (Answer: The aggregate is stable, but the action for that case is fit-sensitive; communicate the case-level instability and avoid treating the aggregate as proof of decision stability.)