On a Locally most Powerful Boundary Randomized Similar Test for the Independence of Two Poisson Variables

Abstract

By definition $(X, Y)$ is a bivariate Poisson vector if $(X, Y) = (X^\ast + U, Y^\ast + U)$ where $X^\ast, Y^\ast$ and $U$ are three independent Poisson variables with, say, respective expectations $a, b$ and $d$. Let $(X_n, Y_n)n = 1, 2, \cdots, N$ be independent observations on a bivariate Poisson vector $(X, Y)$. It is shown that no test for the independence of $X$ and $Y$ can be both boundary randomized similar in $a$ and $b$ and also uniformly most powerful. However, a test of the form \begin{equation*} \varphi_0(Z\mid s, t) = \begin{cases} 1 \\ \gamma 0 \\ 0 \end{cases} \text{according as} Z = \sum^N_{n = 1} X_nY_n \gtreqqless k_0(N, s, t)\end{equation*} given $S = s$ and $T = t$, where, $\sum^N_{n = 1} X_n = S,\quad\sum^N_{n = 1} Y_n = T$, is boundary randomized similar and locally most powerful. Using a lemma on the convergence to a Normal probability distribution function of the conditional probability distribution function of $\sum^N_{n = 1} (X_n - sN^{-1}) (Y_n - tN^{-1})$ given $S = s$ and $T = t$, asymptotic formulae for the values of the $k_0(N, s, t)$ corresponding to a given level of significance are derived. In addition, it is shown that the asymptotic power of the test can be obtained from an approximation to a Normal probability function and that, in case instead of $k_0(N, s, t)$ its value calculated from the asymptotic formulae is used, the modified test is asymptotically locally most powerful in the sense of Definition 2. To extend the domain of application of this test, we replace $U$ in the definition of the bivariate Poisson vector by another random variable $W$ also taking non-negative integral values according to the probability function $P\{W = w \mid \sigma \geqq 0\} = \int^\infty_0 e^{-\sigma t}\frac{{(\sigma t)}^w} {w!} f(t) dt$ where $f(t)$ is any continuous probability function satisfying $\frac{\partial}{\partial\sigma} \int^\infty_0 e^{-\sigma t} \frac{{(\sigma t)}^w}{w!} f(t) dt \bigg |_{\sigma = 0} = \int^\infty_0 \frac{\partial}{\partial\sigma} \big\{e^{-\sigma t} \frac{{(\sigma t)}^w}{w!}\big\}|_{\sigma = 0} f(t) dt$ $\sigma \geqq 0, w = 0, 1, 2, \cdots$ and $\int^\infty_0 tf(t) dt < \infty$. In this way a class of bivariate probability functions is obtained such that for every member of this class also, regarded as a probability function of the random vector $(X, Y)$, the locally most powerful boundary randomized similar test for the independence of the two random variables $X$ and $Y$ is the same as the one given in the Poisson case.