Non-Symmetry in Stability Index

Question

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)

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 $\alpha$.