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Part CCSIR NET December 2024the-minimum-of-n-exponentials-has-rate-n

The minimum of n exponentials has rate n

Let X1,X2,…,XnX_{1}, X_{2}, \dots, X_{n} be a random sample from an exponential distribution with probability density function f(x|θ)=e(−(x−θ))\theta) = e^(-(x-\theta)) if x≥θx \ge \theta, and 0 otherwise, where θ∈R\theta \in \mathbb{R} is an unknown parameter. If X(1)=min⁡X_{(1)} = \min{X1,X2,…,XnX_{1}, X_{2}, \dots, X_{n}}, then which of the following statements are true?

  1. A.(X(1)−(2/n)ln⁡5,X(1))(X_{(1)} - (2/n)\ln 5, X_{(1)}) is a 97% confidence interval for θ\theta.
  2. B.(X(1)−(2/n)ln⁡5,X(1))(X_{(1)} - (2/n)\ln 5, X_{(1)}) is a 96% confidence interval for θ\theta.✓ Correct
  3. C.(X(1)−(1/n)ln⁡20,X(1))(X_{(1)} - (1/n)\ln 20, X_{(1)}) is a 95% confidence interval for θ\theta.✓ Correct
  4. D.(X(1)−(1/n)ln⁡20,X(1))(X_{(1)} - (1/n)\ln 20, X_{(1)}) is a 96% confidence interval for θ\theta.

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 standard counterexample.

See pricing

50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: The MLE is unbiased

More on this topic

The chapter behind this: Maximum likelihood and the method of moments — free to read

From Estimation › MLE and method of moments

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