One number added to the intercept repairs the level. Solving it from the averages misses, and solving it from the change in bad rates double counts.
To recalibrate a logistic PD model to a new default rate, add one constant to every applicant's log-odds of default and solve for the constant that makes the average predicted PD equal the observed rate. On 2,400 SME applications where the card predicts 7.90 per cent and 12.625 per cent defaulted, the solve gives a shift of 0.5623 in log-odds, or 16.22 scorecard points. Computing it from the averages gives 0.5214 and falls short; computing it from the change in cohort bad rates gives 0.7622 and overshoots.
Worked in full in Machine Learning for Finance by Julian R. Sterling, with every figure reproduced in a free workbook.See the book on Amazon →
The model is the SME scorecard of Machine Learning for Finance, scaled at 20 points to double the odds, a factor of 28.8539. It was developed on 2015 to 2019, where 6.32 per cent of loans went bad. The out-of-time cohort, 2020 to 2021, holds 2,400 applications and 303 bads: 12.625 per cent. Poured into the ten score bands fixed on the development sample, the card says this:
| Band | Applications | Bads | Observed | Predicted | Gap, pts |
|---|---|---|---|---|---|
| 1 | 355 | 113 | 31.83% | 22.22% | 9.61 |
| 2 | 315 | 51 | 16.19% | 10.90% | 5.29 |
| 3 | 308 | 42 | 13.64% | 7.82% | 5.82 |
| 4 | 285 | 33 | 11.58% | 5.96% | 5.62 |
| 5 | 246 | 18 | 7.32% | 4.77% | 2.55 |
| 6 | 203 | 15 | 7.39% | 3.87% | 3.52 |
| 7 | 219 | 13 | 5.94% | 3.11% | 2.83 |
| 8 | 179 | 10 | 5.59% | 2.47% | 3.12 |
| 9 | 162 | 7 | 4.32% | 1.86% | 2.46 |
| 10 | 128 | 1 | 0.78% | 1.20% | −0.42 |
| All | 2,400 | 303 | 12.625% | 7.90% | 4.72 |
The ranking survived: bad rates still fall almost band by band. The level did not. The card expects 189.6 bads and 303 arrived. That is the case for recalibration rather than rebuilding.
A logistic model predicts log-odds of default as an intercept plus weighted characteristics. Adding a constant s to the intercept moves every applicant's log-odds by s, multiplies every applicant's odds of default by es, and moves every score by s × factor points. The order of applicants does not change, so Gini, KS and the area under the curve are untouched. Only the level moves, which is exactly what is wrong.
Scaling the PDs instead, by the ratio 12.625 / 7.90 = 1.598, is tempting and wrong: it is linear in probability, breaks above 100 per cent for any PD over 62.6 per cent, and over-corrects the riskiest band.
The condition is that the average of the shifted PDs equals the target. It has no closed form, because the average of logistic curves is not a logistic curve, but Newton's method converges in three or four steps. Each step adds the remaining gap divided by the average of p(1 − p), which is the slope of the mean PD with respect to the shift.
Find s such that Σ nb × 1 / (1 + e−(logit(pb) + s)) / N = 12.625%
Newton step: sk+1 = sk + (target − mean PD(sk)) / mean[p(1 − p)]
Excel, band PDs in C, counts in B, shift in F1: =SUMPRODUCT(B2:B11,1/(1+EXP(-(LN(C2:C11/(1-C2:C11))+F1))))/SUM(B2:B11), then Goal Seek F1 to 0.12625
| Step | Shift | Mean PD | Slope |
|---|---|---|---|
| 0 | 0.0000 | 7.90% | 0.0684 |
| 1 | 0.6905 | 13.97% | 0.1091 |
| 2 | 0.5671 | 12.67% | 0.1011 |
| 3 | 0.5623 | 12.625% | 0.1008 |
The shift is 0.5623 in log-odds. It multiplies every applicant's odds of default by 1.7546 and is worth 0.5623 × 28.8539 = 16.22 points. The banded solve treats every applicant in a band as having the band's average PD. Solved on all 2,400 individual scores, the same method gives 0.573620, or 16.55 points; the difference is the averaging, not the method.
| Band | Observed | Predicted after | Gap before | Gap after |
|---|---|---|---|---|
| 1 | 31.83% | 33.39% | 9.61 | −1.56 |
| 2 | 16.19% | 17.67% | 5.29 | −1.48 |
| 3 | 13.64% | 12.95% | 5.82 | 0.69 |
| 4 | 11.58% | 10.01% | 5.62 | 1.57 |
| 5 | 7.32% | 8.08% | 2.55 | −0.76 |
| 6 | 7.39% | 6.59% | 3.52 | 0.80 |
| 7 | 5.94% | 5.33% | 2.83 | 0.60 |
| 8 | 5.59% | 4.25% | 3.12 | 1.33 |
| 9 | 4.32% | 3.21% | 2.46 | 1.11 |
| 10 | 0.78% | 2.08% | −0.42 | −1.30 |
The largest under-prediction falls from 9.61 points to 1.57. Six bands are still under-predicted, the two worst bands are now slightly over-predicted, and the best band's over-prediction widens. One parameter fixes the average; it cannot fix the shape. If the shape matters for pricing or provisioning, a two-parameter fix (intercept and slope on the score) is the next step.
| Target | Newton shift | Points | Shift from averages | Mean PD it reaches |
|---|---|---|---|---|
| 8.00% | 0.0144 | 0.42 | 0.0136 | 7.99% |
| 10.00% | 0.2770 | 7.99 | 0.2587 | 9.85% |
| 12.625% | 0.5623 | 16.22 | 0.5214 | 12.22% |
| 15.00% | 0.7824 | 22.57 | 0.7213 | 14.31% |
The shortcut, logit(12.625%) − logit(7.90%) = −1.9345 + 2.4559 = 0.5214, falls short at every target in the table, and by more the further the target moves. Here it leaves 9.8 bads unexplained.
The more damaging error is computing the shift from the two cohorts' observed bad rates: odds of 0.1445 out of time against 0.0674 in development, a ratio of 2.1430 and a shift of 0.7622. That double counts. The out-of-time population already scores lower, which is why the card predicts 7.90 per cent rather than the development 6.32; part of the rise is already in the scores. Applied anyway, the 0.7622 shift predicts 14.77 per cent, 51.5 more bads than arrived, and moves the scale by 21.99 points instead of 16.22. Always solve against the mean predicted PD of the population you are calibrating, never against the old sample's bad rate.
The shift has a policy consequence. An applicant at the old cut-off of 550 had a predicted PD of 10.16 per cent; after recalibration the same score means 16.56 per cent. To keep the same risk appetite the cut-off has to read 566.22. What leaving it alone cost is worked in what an unchanged score cut-off costs.
The solve on all 2,400 individual scores, with each Newton step a live formula, is in the free workbook for this case.
Because probabilities cannot be scaled linearly without breaking at the top. Here the ratio is 1.598, and a band predicted at 22.22 per cent would go to 35.5, while a PD of 70 per cent would exceed 100. A shift in log-odds keeps every PD between 0 and 1 and preserves the ranking exactly: the Gini and KS of the card do not move.
No. Adding the same constant to every log-odds moves every score by the same number of points, here 16.22, so the order of applicants is unchanged and every ranking measure is identical. Recalibration fixes level, not discrimination. If the ranking has also decayed, the model needs re-estimating, not a shift.
Multiply the shift by the factor, PDO divided by ln 2. At 20 points to double the odds the factor is 28.8539, so a shift of 0.5623 in log-odds is 16.22 points. A cut-off of 550 would have to read 566.22 to keep the same predicted PD; on the full 2,400 individual scores the exact shift is 0.573620, or 16.55 points.
This article is one calculation from Machine Learning for Finance. The book takes the same case from first principles to the decision, chapter by chapter, and every figure it prints is a live formula in the free companion workbooks.
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