Independence and Order-Statistics

Question

Let $X_1,...,X_n$ be a random sample from a uniform $U(0,\theta)$ distribution, where $\theta>0$.

Are $X_{(n)}$ and $\left(\frac{X_{(n)}}{X_{(n-1)}},\frac{X_{(n-1)}}{X_{(n-2)}},...,\frac{X_{(2)}}{X_{(1)}}\right)$ independent?

Answer 1

The joint density of the order statistic $X_{(1)}, \ldots, X_{(n)}$ is given by

\begin{align} & n! \theta^{-n} I\{0 < x_1 < x_2 < \ldots < x_n < \theta\} \\[10pt] = {} & n! \theta^{-n}I\{0 < x_n < \theta\} \prod \limits_{i = 1}^{n - 1}I\Big\{0 < \frac{x_i}{x_{i + 1}} < 1\Big\} \end{align}

Now consider the map $$V: (x_1, \ldots, x_n) \mapsto \left(\frac{x_2}{x_1}, \frac{x_3}{x_2}, \ldots, \frac{x_n}{x_{n - 1}}, x_n\right),$$

determine its Jacobian and calculate the density of $V(X_{(1)}, \ldots, X_{(n)})$ by the density transformation theorem.

Answer 2

Note that $ Y_i := X_i/\theta \sim U(0,1)$ for each $i$. Since $\theta>0$, we have the $i$th order statistic $Y_{(i)} = X_{(i)}/\theta$.

Then, for a standard uniform distribution we have the result $Y_{(i)} \sim \text{Beta}(i,n+1-i)$. (see here https://en.wikipedia.org/wiki/Order_statistic#Order_statistics_sampled_from_a_uniform_distribution).

Hence, $\dfrac{X_{(i+1)}}{X_{(i)}} = \dfrac{X_{(i+1)}/\theta}{X_{(i)}/\theta} = \dfrac{Y_{(i+1)}}{Y_{(i)}}$ is $\theta$-free; i.e. ancillary for $i=2,\dots,n-1$. Moreover, you can show $X_{(n)}$ is complete & sufficient for $\theta$.

Finally by Basu's theorem, $X_{(n)}$ is independent of any of the $n-1$ ratios of order statistics.

Answer 3

Sketch of the proof without computing the density or invoking specialized results:

$(X_{(1)},\dots,X_{(n)})$ is uniformly distributed on $\{0\le x_1\le \dots\le x_n\le \theta\}$. Therefore, given $X_{(n)} = x$, $(X_{(1)},\dots,X_{(n-1)})$ is distributed as order statistics from $U[0,x]$. Consequently, the vector $(X_{(1)}/X_{(n)},\dots,X_{(n-1)}/X_{(n)})$ is independent of $X_{(n)}$. The vector $(X_{(n)}/X_{(n-1)},X_{(n-1)}/X_{(n-2)},\dots,X_{(2)}/X_{(1)})$ is a transformation of the former, so independent of $X_{(n)}$ as well.

It can be shown similarly that $(X_{(n)}/X_{(n-1)},X_{(n-1)}/X_{(n-2)},\dots,X_{(2)}/X_{(1)})$ is independent of $(X_{(n)},X_{(n+1)})$, therefore, of $X_{(n)}/X_{(n+1)}$. Hence the independence of components can be shown by induction.