## Abstract

In order to solve nonparametric statistical problems, it is often found useful to apply those criteria of optimality which are employed in parametric problems. In the present paper, we are concerned with nonparametric two-sample problems of testing the null hypothesis that two populations have the same distribution against certain nonparametric alternative hypotheses, and generalize the parametric optimality conditions of locally most powerfulness. A rank test for a two-sample problem in this paper is called locally most powerful if it is locally most powerful against a one-parameter family of alternatives. A criterion is constructed by which it is possible to solve the problem whether or not a given rank test is locally most powerful for a two-sample problem in which the set of all possible pairs of cumulative distribution functions is convex and closed (in the weak$^{\ast}$ topology). Let $X_1, \cdots, X_{n_1}$ be sample elements from the first population, $X_{n_1^{+1}}, \cdots, X_n (n = n_1 + n_2)$ from the second population, and the statistic $Z_j = 0$ or 1, according to whether the $j$th smallest observation is from the first population or the second, $j = 1, \cdots, n$. It will be shown that any locally most powerful rank test has the following form: \begin{align*}\tag{*} \begin{cases}\text{Reject the null hypothesis if} \sum_j a_jZ_j > c \\ \text{Accept the null hypothesis if} \sum_j a_jZ_j < c\end{cases}\end{align*} where $a_1, \cdots, a_n$ are constant numbers. For the two-sided two-sample problem, any rank test of the form $(^{\ast})$ is locally most powerful. For the one-sided two sample problem, a non-trivial rank test of the form $(^{\ast})$ is locally most powerful if, and only if, $c_j = \big\lbrack 1\big/\binom{n - 2}{j}\rbrack \sum^{n-2}_{s=j} \binom{s + 1}{j + 1} (a_{s+2} - \bar a),\quad j = 0, 1, \cdots, n - 2,$ where $\bar a = \frac{1}{n} (a_1 + \cdots + a_n),$ have all non-negative Hankel determinants (the precise definition of the Hankel determinants is given in Section 8 below). For the symmetric two-sided two-sample problem, a non-trivial rank test of the form $(^{\ast})$ is locally most powerful if, and only if, $\sum^n_{s=j+1} \binom{s - 1}{j} (a_{n+1-s} - a_s) = 0,\quad\text{for} j = 1, \cdots, n.$ Finally, it will be shown that for the two-sample problem, in which the alternative hypothesis is that the expectation of the first cumulative distribution function with respect to the second distribution is not less than $\frac{1}{2}$, a non-trivial rank test of the form $(^{\ast})$ is locally most powerful if, and only if, $\sum^{\lbrack(n+1)/2\rbrack}_{s=1} \big(\frac{n + 1}{2} - s\big) (a_{n-s} - a_s) \geqq 0.$

## Citation

Hirofumi Uzawa. "Locally Most Powerful Rank Tests for Two-Sample Problems." Ann. Math. Statist. 31 (3) 685 - 702, September, 1960. https://doi.org/10.1214/aoms/1177705795

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