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Altman's Z score


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I have two simple questions about Alman's Z score I ws hoping you could answer for me.

Are we supposed to memorize the formula for this or just how to interpret it? It seem s a little silly to have to memorize a bunch of betas and their corresponding variables, but if that's what they want I guess I will do it.

What do you mean by the "Ignorant zone"? I assume this just means that we just have no decision as far as default or no default is concerned. Is this accurate?


David Harper CFA FRM

David Harper CFA FRM
Staff member
Hi Shannon,

No, the test will *never* ask you to know the Altman's Z coefficients; it's use to us is merely that it's the leading example of a linear discriminant (a linear combination of independent variables gives a line used to divide "probably will not default" from "probably will default").

The "zone of ignorance" (it occurs to me this would make a nice question!) functions like the YELLOW ZONE in the Basel backtest: in both cases, there cannot be certainty, there are two errors that can be made. So these zones represent zones where the power/size of the test is insufficient such that, in a sense, the model does not want to make a formal decision.

In Altman's I *think* the null hypothesis is "the firm will default" such that:
  • Z of less than 1.81 (http://en.wikipedia.org/wiki/Altman_Z-score; interesting, looks like it's also called a "Grey zone") signifies "let's accept the null;" i.e., we predict default
  • Z of greater than 2.99 signifies: reject the null and predict NON-default
So (this part i am just musing), maybe they decided that, under this definition of the null, a Type I error (predict non-default when the firm does; i.e., mistakenly reject a true null) is worse than a Type II error (predict default but the firm does not).
... in which case maybe a 2.50 predicts non-default (a firm with this score falls on the "wrong side" of the line), but the probability of a Type I error is just too high. So they carve out a zone, bumping up the score to 2.99 is the way, even after crossing the point where the line would predict "non-default," of simply reducing the probability of costly Type I error.

Thanks, David