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I'm making a credit scoring model and I get that one variable has Information Value (IV) more than 1, is that possible?

Formulas are pretty simple for weight of evidence(WoE) and information value(IV)

$$WoE_i = \log \left( \dfrac{\dfrac{g_i}{g}}{\dfrac{b_i}{b}} \right)$$ where $g_i$ represents the number of goods (no default) in category $i$ of variable $x_i$, $b_i$ represents the number of bads (default) in category $i$ of variable $x_i$, $g$ represents the number of goods (no default) in the entire dataset, $b$ represents the number of bads (default) in the entire dataset, $N(x)$ is the number of levels in the variable $x$, that is number of categories $$IV = \sum_{i=1}^{N(x)}\left( \dfrac{g_i}{g} - \dfrac{b_i}{b} \right) \cdot WoE_i$$

Also, what is a perfect fit in the model?

By the perfect fit I understand that there is just two categories in x: the first includes all goods and the second includes all bads. In that case when computing $WoE_1$ I get 0 in $log$ denominator, because $b_1 = 0$. When computing $WoE_2$ I get 0 in $log$ numerator, because $g_2 = 0$. Does that make sense?

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IV of greater than 1 is possible, and very common. Assume you are using the natural logarithm? You can deduce the limit by using two categories. Say the model/feature can separate the good and bad perfectly, and let $G_1, B_1, G_2, B_2$ be the proportion of goods and bads falling in the two categories. Then let $G_1 \to 1, B_1 \to 0, G_2 \to 0, B_2 \to 1$. So we are assuming a large pool and the Model is perfectly separating the goods/bads. Then easy to check that that the IV equals:

$\ln \left( \frac{G_1}{B_1}\right) -\ln \left( \frac{G_2}{B_2}\right) $

And minus sign in front of the second means you can invert the second ratio, and hence it is 2 times the log of a quantity that goes to infinity, so you can get infinity. But in practice you should be getting IV of less than 10 most of the time.

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