This paper continues a study, initiated in [1], of the stochastic independence of $V = X - Y$ and the mean of order $\gamma$, \begin{equation*}\tag{1}M_\gamma\begin{align*} = (1/\gamma) \log \lbrack(e^{\gamma X} + e^{\gamma Y})/2\rbrack\quad\text{for} \gamma \neq 0 \text{and} \gamma \neq \pm \infty\\ = (X + Y)/2\quad\text{for} \gamma = 0\\ = \max (X, Y)\quad\text{for} \gamma = + \infty\\ = \min (X, Y)\quad \text{for} \gamma = - \infty\end{align*}\end{equation*} for independent $X$ and $Y$. The case $\gamma = 0$ is solved by the well known result, if, for independent $X$ and $Y$, the variables $X + Y$ and $X - Y$ are independent, then both $X$ and $Y$ have normal distributions with a common variance. The first proof of this result may be found in Kac [6]. (Although the statement of this result there is much weaker, Kac's proof applies exactly to the statement above.) A corresponding result for the gamma distribution, proved in restricted form by Hogg [5], and proved without restrictions by Lukacs [7], is as follows: if $Z_1$ and $Z_2$ are independent positive random variables for which $Z_1 + Z_2$ and $Z_1/Z_2$ are independent, then both $Z_1$ and $Z_2$ have gamma distributions with a common scale parameter. This result may be used to complete the study of the independence of $M_\gamma$ and $V$ for independent $X$ and $Y$ when $\gamma$ is finite and non-zero, by letting $Z_1 = \exp (\gamma X)$ and $Z_2 = \exp (\gamma Y)$. It is clear, then, from the result of Hogg and Lukacs that both $\exp (\gamma Y)$ and $\exp (\gamma Y)$ must have gamma distributions with a common scale parameter. The resulting distributions of $X$ and $Y$ have been studied in [1]. Therefore, of this study, there remain yet to be completely solved only the cases $\gamma = +\infty$ and $\gamma = - \infty$. These two cases are essentially the same since if $M_{+\infty}$ and $V$ are independent, then so are $\min (-X, -Y)$ and $V$. Thus, the negatives of the distributions for which $M_{+\infty}$ and $V$ are independent will yield independent $M_{-\infty}$ and $V$. We consider only the case $\gamma = -\infty$. This problem is considered in a previous paper, [2], in which the distributions were restricted to be discrete. The main result of that paper is as follows: if $X$ and $Y$ are independent, non-degenerate, discrete random variables, and if $U = \min (X, Y)$ and $V = X - Y$ are independent, then $X$ and $Y$ both have geometric distributions with common location and scale parameters, but possibly different geometric parameters. If the probability mass function of the geometric distribution is written as \begin{equation*}\tag{2}p(x) = (1 - p)p^{(x-\theta)/c}\quad\text{for} x = \theta, \theta + c, \theta + 2c, \cdots\end{equation*} then, the parameter $\theta$ is a location parameter, $c$ is a scale parameter, and $p$ is called the geometric parameter, $c > 0, 0 < p < 1$. In this paper, the distributions of $X$ and $Y$ are restricted to be absolutely continuous. This will result in a characterization of the so-called exponential distribution whose density is \begin{equation*}\tag{3}f(x) = (1/\sigma) \exp (-(x - \theta)/\sigma)\quad\text{if} x > \theta\end{equation*} and zero otherwise. The parameter $\theta$ is a location parameter, and the parameter $\sigma > 0$ is a scale parameter, which is also the standard deviation of the distribution. In these terms the main result of this paper may be stated as follows: if $X$ and $Y$ are independent random variables with absolutely continuous distributions, and if $U = \min (X, Y)$ and $V = X - Y$ are independent, then both $X$ and $Y$ have exponential distributions with a common location parameter but with possibly different scale parameters. This result then gives a characterization of the exponential distribution, since it is easy to show the converse, that if $X$ and $Y$ are independent and if each has an exponential distribution with a common location parameter but with possibly different scale parameters, then $U$ and $V$ are independent. For related results, the reader is referred to papers by Fisz [3] and Rogers [8].
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