# Why are most standard goodness of fit tests based only on continuous distributions?

I tried to search info regarding this fact but I don't really understand why most of the standard goodness of fit tests (e.g. Kolmogorov-Smirnov, Anderson-Darling, a part of the Chi-square test, perhaps!) work only with continuous distributions. Can someone help me? Thank You.

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What do you mean by a "validation test"? Also, please note that the chi-square test of goodness of fit is specifically for discrete distributions; it can be applied to continuous distributions only by binning them (that is, turning them into discrete distributions). –  whuber Sep 15 '11 at 16:36
Sorry, with validation tests i mean goodness of fit tests! in order to decide if a data sample come from a specific distribution. Ok leave the Chi-square test and take into account the others. –  Maurizio Sep 15 '11 at 17:03
You seem to be creating an artificial problem with your exclusion of the chi-square test from consideration. Since the chi-square test works (well) with discrete distributions, what help do you need exactly? –  whuber Sep 15 '11 at 17:07
In other words, i don't understand why different goodness of fit tests works only for continuous distributions. Ok that the chi-square test works well with discrete distributions, but why the other (for example Kolmogorov-Smirnov, Anderson-Darling, etc) don't work for the discrete ones? –  Maurizio Sep 15 '11 at 17:24

The reason for the KS test is that its generality, e.g. it's usefulness for non-parametric models comes from the definition of the test statistic under the assumption of the CDF being continuous.

Where we define the KS Statistic as

$$D_n(F) = \max\left(D_n^+(F), D_n^-(F)\right)$$

$$D_n^+(F) = \sup_{x \in \mathbb{R}} [F_n(x) - F(x)]$$

(and the reverse for $D_n^-(F)$).

Then under the null $D_n^+(F) = \max_{0 \le i \le n} \left( F_n(x_i) - F(X_{(i)}) \right)$

Recall that under the null $F(X_{(i)})$ is continuous uniform on $(0,1)$ so the distribution of $F$ doesn't matter.

So you can create your own K-S like test for any discrete distribution, but it won't be a generalized test.

Reference/Citation, Mathematical Statistics (Shao 2010)

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I suspect this good reply hasn't been upvoted more because it doesn't sufficiently emphasize the important point that the sampling distribution of the K-S statistic has been computed for all continuous distributions (because they become uniform under the probability transformation) while each discrete distribution leads to its own idiosyncratic distribution of the K-S statistic. –  whuber Oct 16 '11 at 16:52