I was reading that the Kolmogorov Smirnov 2 sample test is consistent, that is Probability of rejection under $H_1$ is 1 for sample size going to infinity.

Say we have 2 random variables X and Y. K-S Test checks if $F=G$.

The test statistics is: $sup_z|F_n(z)-G_n(z)|$

The test is consistent (for some level $\alpha$) means :

$$Lim_{n\rightarrow \infty}P(sup_z|F_n(z)-G_n(z)|>D_{n,\alpha})=1$$

where $G_n$ is the empirical cdf distribution of Y and $G_n(y)=\sum_{i=1}^n\frac{\mathbb{1}_{Y_i<y}}{m}$ where m is the number of sample of Y.

$F_n$ is the empirical cdf of X, $F_n(X)=\sum_{i=1}^n\frac{\mathbb{1}_{X_i<x}}{n}$ where n is the number of sample of X.

I cannot prove the consistency can anyone help in it ?

Thanks in advance.


1 Answer 1


Assume, under $H_1$, $F\not=G$. This means that for some $z$ and $\epsilon>0$,we have $|F(z)-G(z)|>\epsilon$. Glivenko-Cantelli gives $F_n\to F, G_n\to G$ uniformly in probability. In particular, with probability going to 1, $|F_n(z)-G_n(z)|\to\epsilon>0$. You get the result from $D_{n,\alpha}\to 0$. This follows from the fact that if $F=G$, Glivenko-Cantelli implies that the supremum difference goes to zero in probability, therefore all quantiles of the distribution of that supremum for fixed $\alpha$.

  • $\begingroup$ Why is $D_{n,\alpha}$ go to 0 ? $\endgroup$
    – Andrew741
    May 7, 2023 at 10:20
  • $\begingroup$ @Andrew741 I have added it to the answer. $\endgroup$ May 7, 2023 at 10:56
  • $\begingroup$ So basically to are showing that under $H_0$ , $D_n=D_{n,\alpha}$ goes to 0 , and under $H_1$ , $D_n>\epsilon$ for some $\epsilon$ and thus it follows right ? $\endgroup$
    – Andrew741
    May 7, 2023 at 12:32
  • 1
    $\begingroup$ @Andrew741 Yes (except that I have to guess what $D_n$ is as this had not been defined). $\endgroup$ May 7, 2023 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.