Discussion
Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be iid r.v.s with \(U(\theta, \theta + 1)\), then \(E[X_{(n)} - X_{(1)}]\), where \(X_{(n)} = \text{Max } X_{i}\), and \(X_{(1)} = \text{Min } X_{i}\), is given by :
Answer: C
\(E[X_{(n)} - X_{(1)}] =\) \(\dfrac{n - 1}{n + 1}\).
Write \(X_{i} = \theta + U_{i}\) with \(U_{i}\) uniform on (0, 1). The range does not depend on \(\theta\). For uniform (0, 1) samples, \(E[U_{(n)}] = n/(n + 1)\) and \(E[U_{(1)}] = 1/(n + 1)\), so
\[ \begin{aligned} E[X_{(n)} - X_{(1)}] &= \frac{n}{n + 1} - \frac{1}{n + 1} \\ &= \frac{n - 1}{n + 1} \end{aligned} \]
For \(n = 1\) the range is 0, as the formula gives. \(1/(n + 1)\) is the expected minimum alone, and the other two options exceed 1, which no range of points in an interval of length 1 can do.
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