The Calculation of Distributions of Two-Sided Kolmogorov-Smirnov Type Statistics

TLDR: In this article, the precision of a truncated development of a two-sided Kolmogorov-Smirnov type has been determined in powers of λ √ 2 for any λ = 1,2, 10, 100.

Abstract: Let $X_1^n \leqq X_2^n \leqq \cdots \leqq X_n^n$ be the order statistics of a size $n$ sample from any distribution function $F$ not necessarily continuous. Let $\alpha_j, \beta_j, (j = 1,2, \cdots, n)$ be any numbers. Let $P_n = P(\alpha_j < X_j^n \leqq \beta_j, j = 1,2, \cdots, n)$. A recursion is given which calculates $P_n$ for any $F$ and any $\alpha_j, \beta_j$. Suppose now that $F$ is continuous. A two-sided statistic of Kolmogorov-Smirnov type has the distribution function $P_{\mathrm{KS}} = P\lbrack\sup n^{\frac{1}{2}}\psi(F) \cdot |F^n - F| \leqq \lambda\rbrack$, where $F^n$ is the empirical distribution function of the sample and $\psi(x)$ is any nonnegative weight function. As $P_{\mathrm{KS}}$ has the form $P_n$, its calculation as a function of $\lambda$ can be carried out by means of the recursion. This has been done for the case $\psi(x) = \lbrack x(1 - x)\rbrack^{-\frac{1}{2}}$. Curves are given which represent $\lambda$ versus $1 - P_{\mathrm{KS}}$ for $n = 1,2, 10, 100$. From additional computations, the precision of a truncated development of $1 - P_{\mathrm{KS}}$ in powers of $\lambda^{-2}$ has been determined.