TAOCP 3.3.1 Exercise 11
Let F_n(x)=\frac{1}{n}\#\{j:X_j\le x\} be the empirical distribution function of the original sample
Section 3.3.1: General Test Procedures for Studying Random Data
Exercise 11. [**] [10] Solve exercise 10 substituting the KS test for the chi-square test.
Verified: yes
Solve time: 10m50s
Let
$$ F_n(x)=\frac{1}{n}#{j:X_j\le x} $$
be the empirical distribution function of the original sample
$$ X_1,X_2,\dots,X_n, $$
and let
$$ D_n=\sup_x |F_n(x)-F(x)| $$
be the corresponding Kolmogorov-Smirnov statistic, where $F(x)$ is the hypothesized distribution function.
Now repeat the test on the same $n$ observations again. Thus the combined sample of size $2n$ is
$$ Y_1,\dots,Y_{2n} = X_1,\dots,X_n,X_1,\dots,X_n. $$
The empirical distribution function based on these $2n$ observations is
$$ F_{2n}(x) = \frac{1}{2n}#{j:Y_j\le x}. $$
Since every observation $X_j$ appears twice in the combined sample,
$$ #{j:Y_j\le x} = 2#{j:X_j\le x}. $$
Hence
$$ F_{2n}(x) = \frac{2#{j:X_j\le x}}{2n} = \frac{1}{n}#{j:X_j\le x} = F_n(x). $$
Therefore the empirical distribution function is exactly the same after duplicating the data. It follows that the KS statistic is also unchanged:
$$ D_{2n} = \sup_x |F_{2n}(x)-F(x)| = \sup_x |F_n(x)-F(x)| = D_n. $$
Thus, when the same observations are repeated, the Kolmogorov-Smirnov statistic for the combined sample of size $2n$ is identical to the original statistic. ∎