A very popular index to measure the stability of characteristics of a scorecard is defined by the following formula:

$$SI = \frac{1}{n} \sum \text{(actual in %)-(expected in %)} \cdot \log\left(\frac{\text{(actual in %)}}{\text{(expected in %)}}\right)$$

where actual describes the proportion of the characteristics in the build sample and expected the proportion in the validation sample.

For further information have a look at this links:

german source: http://docplayer.org/2070055-Einige-validierungsaspekte-von-scoring-systemen.html

english source: Credit Risk Scorecards: Developing and Implementing Intelligent Credit Scoring by Naeem Siddiqi

I don't know why this index weights decreasing of a variable more then increasing. If you take a look at my example you will get what's in my mind. In both cases there is a shift of 15% in the data in characteristic A and it's shifted homogenous to each other characteristic on the one hand increasing and on the other decreasing. The index don't penalize both shifts in the same way.

Question: Why is the stability index not symmetric for a similar proportion shift? What is the advantage of penalizing decreasing more?

(Decreasing in proportion is penalized much more 0,04120 instead of 0,02599)

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1 Answer 1


Looking at the source link, it seems the formula is actually

$$ \sum_i \left[ \left( a_i - e_i \right) \ln\left( \frac{a_i}{e_i} \right) \right] $$

(although I don't read German).

For some small $\alpha > 0$, suppose that $a_i = e_i (1 + \alpha)$. Then this adds $$e_i \alpha \ln\left(1 + \alpha\right)$$ to the sum. Conversely, if $a_i = e_i (1 - \alpha)$, then this adds $$-e_i \alpha \ln\left(1 - \alpha\right) = e_i \alpha \ln\left(\frac{1}{1 - \alpha}\right).$$ However, for small $\alpha$, the Taylor expansion for the geometric series shows that $\frac{1}{1 - \alpha} \sim 1 + \alpha$, so $$-e_i \alpha \ln\left(1 - \alpha\right) \sim e_i \alpha \ln\left(1 +\alpha\right).$$

As you note, the approximation is less precise for larger $\alpha$. I can't read the German text, but, presumably, this formula is more useful for small $\alpha$.

  • $\begingroup$ So, you mean that the index is made as symmetric index, but has got an inaccuracy because of the approximation and that's all? There is nothing more behind of this difference in penalizing the shift? $\endgroup$
    – T. Beige
    Nov 16, 2017 at 14:17

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