Let $X_{1},X_{2},\ldots,X_{n}$ be an independent and equally distributed random sample whose distribution is uniform on the interval $(\theta,\theta+1)$, $-\infty<\theta<+\infty$. Then consider the order statistics $X_{(1)} < X_{(2)} < \ldots < X_{(n)}$. The exercise asks us to prove that $R = X_{(n)} - X_{(1)}$ is an ancillary statistics.
To be quite frank, I have no idea as how to prove it. Could someone help me out?
PS: this is not homework. I am just curious about the result.
An ancillary statistic is a statistic with a distribution that does not depend on the parameters of the model. In this case it is particularly easy to prove this property, without having to derive the distribution. For all values $i=1,...,n$ we can define the random variables $Y_i \equiv X_i-\theta$ so that $Y_1,...,Y_n \sim \text{IID U}(0,1)$. The joint distribution of these latter random variables does not depend on $\theta$, so we have:
$$\begin{equation} \begin{aligned} R_n &\equiv X_{(n)} - X_{(1)} \\[6pt] &= (X_{(n)} - \theta) - (X_{(1)} - \theta) \\[6pt] &= Y_{(n)} - Y_{(1)} \\[6pt] &\sim \text{Distribution that does not depend on } \theta. \\[6pt] \end{aligned} \end{equation}$$
Define Zi = Xi - theta Zi ~ U[0, 1] as Xi are uniformly distributed R == Xn - Xn == Zn - Z1, because of the order statistics the distribution is thus independent of the theta as well
