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Part CCSIR NET June 2025the-expected-range-is-short-by-n-minus-1-over-n-plus-1-and-must-be-inflated

The expected range is short by n minus 1 over n plus 1 and must be inflated

Let X1,X2,…,Xn(n≥3)X_{1}, X_{2}, \dots, X_{n} (n \ge 3) be a random sample from the uniform distribution on the interval (θ1−θ2,θ1+θ2)(\theta_{1} - \theta_{2}, \theta_{1} + \theta_{2}), where θ1∈R\theta_{1} \in \mathbb{R} and θ2>0\theta_{2} > 0 are unknown parameters. Let X(j)X_{(j)} be the jᵗʰ order statistic, j = 1, 2, …, n, and let Xˉ=(1/n)∑i{\bar{X}} = (1/n)\sum_{i}₌1nXi_{1}^{n} X_{i}. Here, (X(1),X(n))(X_{(1)}, X_{(n)}) is a complete and sufficient statistic for (θ1,θ2)(\theta_{1}, \theta_{2}). Then, which of the following statements are true?

  1. A.X̄ is an unbiased estimator of θ1\theta_{1}✓ Correct
  2. B.(Xˉ−X(1))({\bar{X}} - X_{(1)}) is an unbiased estimator of θ2\theta_{2}
  3. C.(X(1)+X(n))/2(X_{(1)} + X_{(n)})/2 is the uniformly minimum variance unbiased estimator of θ1\theta_{1}✓ Correct
  4. D.(n+1)(X(n)−X(1))/(2(n−1))(n + 1)(X_{(n)} - X_{(1)})/(2(n - 1)) is the uniformly minimum variance unbiased estimator of θ2\theta_{2}✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: A sufficient statistic is complete

More on this topic

The chapter behind this: Sufficiency, completeness and the UMVUE machine — free to read

From Estimation › Sufficiency, completeness, UMVUE, Cramér–Rao

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