Kolmogorov–Smirnov test

Remarks

Under $H_0$ the law of $D_n$ is the same for every continuous $F_0$, because $F_0(X_i)$ is uniform on $[0, 1]$. As $n \to \infty$, $\sqrt{n} D_n$ converges in distribution to the Kolmogorov distribution, the law of $\sup_{0 \le t \le 1} \abs{B(t)}$ for a Brownian bridge $B$: $$\lim_{n \to \infty} P(\sqrt{n} D_n \le t) = 1 - 2 \sum_{k=1}^{\infty} (-1)^{k-1} e^{-2k^2 t^2}, \qquad t > 0.$$ For ordered data $x_{(1)} \le \dots \le x_{(n)}$ the supremum is a finite maximum: $D_n = \max_{i} \max\big(i/n - F_0(x_{(i)}), \, F_0(x_{(i)}) - (i - 1)/n\big)$.

Example: $n = 1$, $X_1 = 0.3$ and $F_0(x) = x$ on $[0, 1]$. Then $F_1$ jumps from $0$ to $1$ at $0.3$, and $D_1 = \max(0.3, 1 - 0.3) = 0.7$.

Non-example: testing normality with the mean and variance estimated from the same data is not this test. Then $F_0$ is not fixed in advance, and the null law of $D_n$ changes; the corrected procedure is the Lilliefors test.