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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.

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1 Answer 1

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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 $\alpha$.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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